Class 10 Mathematics - BeTOPPERS IIT Foundation Series [2022 ed.]

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IIT FOUNDATION Class X

MATHEMATICS

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PREFACE Our sincere endeavour in preparing this Book is to enable students effectively grasp & understand the Concepts of Mathematics and help them build a strong foundation in this Subject. From among hundreds of questions being made available in this Book, the Student would be able to extensively practice in each concept exclusively, throughout that Chapter. At the end of each Chapter, two or three Worksheets are provided with questions which shall cover the entire Chapter, helping each Student consolidate his / her learning. This Book help students prepare for their respective Examinations including but not limited to i.e. CBSE, ICSE, various State Boards and Competitive Examinations like IIT, NEET, NTSE, Science Olympiads etc. It is compiled by our inhouse team of experts who have a collective experience of more than 40 years in their respective subject matter / academic backgrounds. This books help students understand concepts and their retention through constant practice. It enables them solve question which are ‘fundamental / foundational’ as well questions which needs ‘higher order thinking’. Students gain the ability to concentrate, to be self-reliant, and hopefully become confident in the subject matter as they traverse through this Book. The important features of this books are: 1.

Lucidly presented Concepts: For ease of understanding, the ‘Concepts’ are briefly presented in simple, easy and comprehensible language.

2.

Learning Outcomes: Each chapter starts with ‘Learning Outcomes’ grid conveying what the student is going to learn / gain from this chapter.

3.

Bold-faced Key Terms: The key words, concepts, definitions, formulae, statements, etc., are presented in ‘bold face’, indicating their importance.

4.

Tables and Charts: Numerous strategically placed tables & charts, list out etc. summarizes the important information, making it readily accessible for effective study.

5.

Box Items: Are ‘highlighted special topics’ that helps students explore / investigate the subject matter thoroughly.

6.

Photographs, Illustrations: A wide array of visually appealing and informative photographs are used to help the students understand various phenomena and inculcate interest, enhance learning in the subject matter.

7.

Flow Diagrams: To help students understand the steps in problem-solving, flow diagrams have been included as needed for various important concepts. These diagrams allow the students visualize the workflow to solve such problems.

8.

Summary Charts: At the end of few important concepts or the chapter, a summary / blueprint is presented which includes a complete overview of that concept / chapter. It shall help students review the learning in a snapshot.

9.

Formative Worksheets: After every concept / few concepts, a ‘Formative Worksheet’ / ‘Classroom Worksheet’ with appropriate questions are provided from such concept/s. The solutions for these problems shall ideally be discussed by the Teacher in the classroom.

10. Conceptive Worksheets: These questions are in addition the above questions and are from that respective concept/s. They are advised to be solved beyond classroom as a ‘Homework’. This rigor, shall help students consolidate their learning as they are exposed to new type of questions related to those concept/s.

11. Summative Worksheets: At the end of each chapter, this worksheet is presented and shall contain questions based on all the concepts of that chapter. Unlike Formative Worksheet and Conceptive Worksheet questions, the questions in this worksheet encourage the students to apply their learnings acquired from that entire chapter and solve the problems analytically. 12. HOTS Worksheets: Most of the times, Summative Worksheet is followed by an HOTS (Higher Order Thinking Skills) worksheet containing advanced type of questions. The concepts can be from the same chapter or as many chapters from the Book. By solving these problems, the students are prepared to face challenging questions that appear in actual competitive entrance examinations. However, strengthening the foundation of students in academics is the main objective of this worksheet. 13. IIT JEE Worksheets: Finally, every chapters end with a IIT JEE worksheet. This worksheet contains the questions which have appeared in various competitive examinations like IIT JEE, AIEEE, EAMCET, KCET, TCET, BHU, CBSE, ICSE, State Boards, CET etc. related to this chapter. This gives real-time experience to students and helps them face various competitive examinations. 14. Different Types of Questions: These type of questions do appear in various competitive examinations. They include:

• Objective Type with Single Answer Correct

• Non-Objective Type

• Objective Type with > one Answer Correct

• True or False Type

• Statement Type - I (Two Statements)

• Statement Type - II (Two Statements)

• MatchingType - I (Two Columns)

• MatchingType - II (Three Columns)

• Assertion and Reasoning Type

• Statement and Explanation Type

• Roadmap Type

• FigurativeType

• Comprehension Type

• And many more...

We would like to thank all members of different departments at BeTOPPERS who played a key role in bringing out this student-friendly Book. We sincerely hope that this Book will prove useful to the students who wish to build a strong Foundation in Mathematics and aim to achieve success in various boards / competitive examinations. Further, we believe that as there is always scope for improvement, we value constructive criticism of the subject matter, as well as suggestions for improving this Book. All suggestions hopefully, shall be duly incorporated in the next edition. Wish you all the best!!!

Team BeTOPPERS .

CONTENTS 1.

Real Numbers

..........

01 – 16

2.

Polynomials

..........

17 – 28

3.

Pair of Linear Equations in Two Variables

.........

29 – 48

4.

Quadratic Equations

..........

49 – 60

5.

Progressions

..........

61 – 74

6.

Functions

..........

75 – 90

7.

Coordinate Geometry

..........

91 – 100

8.

Some Applications of Trigonometry

..........

101 – 112

9.

Binomial Theorem

..........

113 – 122

10.

Triangles

..........

123 – 136

11.

Circles

..........

137 – 142

12.

Surface Areas and Volumes

..........

143 – 162

13.

Statistics

..........

163 – 178

14.

Key and Answers

..........

179 – 498

Real Numbers Chapter - 1

Learning Outcomes By the end of this chapter, you will understand  Introduction  Euclid’s Division Lemma  The Fundamental Theorem of Arithmetic  Revisiting Irrational Numbers  Revisiting Rational Numbers and Their Decimal Expansions

1.

Introduction We begin with two very important properties of positive integers , namely the Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. Euclid’s division algorithm, as the name suggests, has to do with divisibility of integers. Stated simply, it says any positive integer a can be divided by another positive integer b in such a way that it leaves a remainder r that is smaller than b. Many of you probably recognise this as the usual long division process. Although this result is quite easy to state and understand, it has many applications related to the divisibility properties of integers. We touch upon a few of them, and use it mainly to compute the HCF of two positive integers. The Fundamental Theorem of Arithmetic, on the other hand, has to do something with multiplication of positive integers. You already know that every composite number can be expressed as a product of primes in a unique way this important fact is the Fundamental Theorem of Arithmetic. Again, while it is a result that is easy to state and understand, it has some very deep and significant applications in the field of mathematics.

We use the Fundamental Theorem of Arithmetic for two main applications. First, we use it to prove the irrationality of many of the numbers , such as

2, 3 and 5 second,

we apply this theorem to explore when exactly the decimal expansion of a rational number, say

p  q  0  is terminating and when it is q nonterminating repeating. We do so by looking at the prime factorisation of the denominator q of

p q

You will see that the prime factorisation of q will completely reveal the nature of the decimal expansion of

2

p . q

Euclid’s Division Lemma Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0  r < b. This result was perhaps known for a long time, but was first recorded in Book VII of Euclid’s Elements. Euclid’s division algorithm is based on this lemma.

10th Class Mathematics

2 Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. Recall that the HCF of two positive integers a and b is the largest positive integer d that divides both a and b. Let us see how the algorithm works, through an example first. Suppose we need to find the HCF of the integers 455 and 42. We start with the larger integer, that is, 455. Then we use Euclid’s lemma to get 455 = 42 × 10 + 35 Now consider the divisor 42 and the remainder 35, and apply the division lemma to get 42 = 35 × 1 + 7 Now consider the divisor 35 and the remainder 7, and apply the division lemma to get 35 = 7 × 5 + 0 Notice that the remainder has become zero, and we cannot proceed any further.

We claim that the HCF of 455 and 42 is the divisor at this stage, i.e., 7. You can easily verify this by listing all the factors of 455 and 42. Why does this method work? It works because of the following result. So, let us state Euclid’s division algorithm clearly. To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below: Step 1 :Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0  r < d. Step 2 : If r = 0, d is the HCF of c and d. If r  0, apply >the division lemma to d and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes t h e HCF of c and d, etc.

SOLVED EXAMPLES Example 1: Use Euclid’s algorithm to find the HCF of 4052 and 12576. Solution : Step 1 : Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get 12576 = 4052 × 3 + 420 Step 2 : Since the remainder 420  0, we apply the division lemma to 4052 and 420, to get 4052 = 420 × 9 + 272 Step 3 : We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get 420 = 272 × 1 + 148 www.betoppers.com

Real Numbers We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get 272 = 148 × 1 + 124 We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get 148 = 124 × 1 + 24 We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get 124 = 24 × 5 + 4 We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get 24 = 4 × 6 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4, the HCF of 12576 and 4052 is 4. Notice that 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052). Euclid’s division algorithm is not only useful for calculating the HCF of very large numbers, but also because it is one of the earliest examples of an algorithm that a computer had been programmed to carry out. Remarks : 1. Euclid’s division lemma and algorithm are so closely interlinked that people often call former as the division algorithm also. 2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be extended for all integers except zero, i.e., b < 0. However, we shall not discuss this aspect here. Euclid’s division lemma/algorithm has several applications related to finding properties of numbers. We give some examples of these applications below:

3 Example 3: Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer. Solution : Let us start with taking a, where a is a positive odd integer. We apply the division algorithm with a and b = 4. Since 0  r < 4, the possible remainders are 0, 1, 2 and 3. That is, a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient. However, since a is odd, a cannot be 4q or 4q + 2 (since they are both divisible by 2). Therefore, any odd integer is of the form 4q + 1 or 4q + 3. Example 4: A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose? Solution : This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algorithm to find their HCF. We have : 420 = 130 × 3 + 30 130 = 30 × 4 + 10 30 = 10 × 3 + 0 So, the HCF of 420 and 130 is 10. Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.

Formative Worksheet 1.

Example 2: Show that every positive even integer is of the form 2q, and that every positive odd integer is of the form 2q + 1, where q is some integer. Solution : Let a be any positive integer and b = 2. Then, by Euclid’s algorithm, a = 2q + r, for some integer q   0, and r = 0 or r = 1, because 0  r < 2. So, a = 2q or 2q + 1. If a is of the form 2q, then a is an even integer. Also, a positive integer can be either even or odd. Therefore, any positive odd integer is of the form 2q + 1.

2.

3.

Use Euclid’s division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255 Show that any positive odd integer is of the form 6q + 1 or 6q + 3, or 6q + 5, where q is some integer. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

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10th Class Mathematics

4 4.

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m +1.]

5.

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

6.

Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

7.

Prove that the square of any positive integer of the form 5q + 1 is of the same form.

8.

If the H.C.F. of 657 and 963 is expressible in the form 657x + 963 × (–15), find x.

9.

What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.

10.

144 cartons of coke cans and 90 cartons of Pepsi cans are to be stacked is a canteen. If each stack is of same height and is to contains cartons of the same drink, what would be the greatest number of cartons each stack would have ?

Conceptive Worksheet 1.

Using Euclid Division Lemma, find H.C.F. of 180670 and 2937.

2.

Using Euclid Division Lemma, find H.C.F. of 274170 and 17017.

3.

Using Euclid Division Lemma, find H.C.F. of 92690, 7378 and 7161.

4.

Show that every positive even integer is in the form 2n, and that every positive odd interger is of the form 2n+1, where n is some integer.

5.

Show that any positive odd integer is in the form 4n + 1 or 4n + 3, where n is integer.

6.

Find the H.C.F. of 65 and 117 and express it in the form 65m + 117n.

7.

Using Euclid Division Lemma, find H.C.F. of 36575, 2223 and 1330.

8.

Using Euclid Division Lemma, find H.C.F. of 26565, 25806 and 20930.

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3.

The Fundamental Theorem of Arithmetic Every composite number can be expressed ( factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes. Actually it says more. It says that given any composite number it can be factorised as a product of prime numbers in a ‘unique’ way, except for the order in which the primes occur. That is, given any composite number there is one and only one way to write it as a product of primes, as long as we are not particular about the order in which the primes occur. So, for example, we regard 2 × 3 × 5 × 7 as the same as 3 × 5 × 7 × 2, or any other possible order in which these primes are written. This fact is also stated in the following form: The prime factorisation of a natural number is unique, except for the order of its factors. In general, given a composite number x, we factorise it as x = p1p2 ... pn, where p1, p2,..., pn are primes and written in ascending order, i.e., p1  p2  . . .  pn. If we combine the same primes, we will get powers of primes. For example, 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13 Once we have decided that the order will be ascending, then the way the number is factorised, is unique.The Fundamental Theorem of Arithmetic has many applications, both within mathematics and in other fields. Let us look at some examples.

SOLVED EXAMPLES Example 5: Consider the numbers 4n, where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero. Solution : If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 4n would contain the prime 5. This is not possible because 4n = (2)2n; so the only prime in the factorisation of 4n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 4n. So, there is no natural number n for which 4n ends with the digit zero.

Real Numbers

5

You have already learnt how to find the HCF and LCM of two positive integers using the Fundamental Theorem of Arithmetic in earlier classes, without realising it. This method is also called the prime factorisation method. Let us recall this method through an example. Example 6: Find the LCM and HCF of 6 and 20 by the prime factorisation method. Solution: We have : 6  2  3 and

 21  3 20  2  2  5

1

So, HCF (6, 72, 120) = 2 × 3 = 2 × 3 = 6 23, 32 , 51 are the greatest powers of the prime factors 2, 3 and 5 respectively involved in the three numbers. So, LCM (6, 72, 120) = 23 × 32 × 51 = 360 Remark : Notice, 6 × 72 × 120,  HCF (6, 72, 120) × LCM (6, 72, 120). So, the product of three numb er s is not equal to the product of their HCF and LCM.

Formative Worksheet 11.

 2 2  51 You can find HCF(6, 20) = 2 and LCM(6, 20) = 2 × 2 × 3 × 5 = 60, as done in your earlier classes. Note that HCF(6, 20) = 21 = Product of the smallest power of each common prime factor in the numbers. LCM (6, 20) = 22 × 31 × 51 = Product of the greatest power of each prime factor, involved in the numbers. From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20) = 6 × 20. In fact, we can verify that for any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers, if we have already found the HCF of the two positive integers. Example 7: Find the HCF of 96 and 404 by the prime factorisation method. Hence, find their LCM. Solution : The prime factorisation of 96 and 404 gives 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2 5 × 3 404 = 2 × 2 × 101 = 2 2 × 101 Therefore, the HCF of these two integers is 22 = 4. Also, LCM (96, 404) =

96  404 96  404   9696 HCF  96, 404  4 Example 8: Find the HCF and LCM of 6, 72 and 120, using the prime factorisation method. Solution : We have : 6 = 2 × 3, 72 = 23 × 32, 120 = 23 ×3×5 Here, 21 and 31 are the smallest powers of the common factors 2 and 3 respectively.

1

12.

Express each number as product of its prime factors: (i) 140

(ii) 156

(iv) 5005

(v) 7429

(iii) 3825

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92

13.

(iii) 336 and 54

Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 and 25

(ii) 17, 23 and 29

(iii) 8, 9

14.

Given that HCF (306, 657) = 9, find LCM (306, 657).

15.

Check whether 6 n   can  end  with  the  digit  0  for any natural number n.

16.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

17.

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

18.

Determine the prime factors of 45470971.

19.

Find the LCM and HCF of 84, 90 and 120 by applying the prime factorisation method.

20.

In a morning walk three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps?

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10th Class Mathematics

6

Conceptive Worksheet 9.

Find the missing numbers in the following prime factorisation.

4.

Revisiting irrational numbers You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that 2 , 3, 5 and, in general,

p is

irrational, where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic. Recall, a number ‘s’ is called irrational if it cannot be written in the form

10. 11. 12. 13.

14. 15. 16. 17. 18.

Find the H.C.F. and L.C.M. of 29029 and 1740 by using the fundamental theorem of Arithmetic. Find the H.C.F. and L.C.M. of 25152 and 12156 by using the fundamental theorem of Arithmetic. Find the prime factors of the following numbers: (i) 176 (ii) 256 (iii) 4825 (iv) 12673 Find the missing numbers in the following factor trees.

Check whether (15)n can end with the digit 0 for any n  N . Check whether (18)n can end with the digit 0 for any n  N . Check whether (26)n can end with the digit 5 for any n  N . Check whether 2n can end with the digit 6 for any n  N . Check whether 3n can end with the digit 1 n  1, n  N .

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p where p and q are q

integers and q  0 . Some examples of irrational numbers, with which you are already familiar, are : 2, 3, 15, , 

2 , 010110111011110..., etc. 3

Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic. Theorem :Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer. Proof : Let the prime factorisation of a be as follows : a = p1p2 . . . pn, where p1,p2, . . ., pn are primes, not necessarily distinct. Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p1, p2, . . ., pn. So p is one of p1, p2, . . ., pn. Now, since a = p1 p2 . . . pn, p divides a. Therefore, a 2 = (p 1 p 2 ,.....p n ) (p 1 p 2 ....p n ) = p12 p22 .... pn2 . We are now ready to give a proof that

2 is

irrational. The proof is based on a technique called ‘proof by contradiction’.

Real Numbers Theorem :

7

2 is irrational.

Proof : Let us assume, to the contrary, that

2

is rational. So, we can find integers r and s (  0) such that

2

r Suppose r and s have a common s

factor other than 1. Then, we divide by the common factor to get

2

a , where a and b are coprime. b

So, we can write a = 3c for some integer c. Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2. This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3). Therefore, a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that

3 is rational.

So, b 2  a So, we conclude that 3 is irrational. Squaring on both sides and rearranging, we get Note: 2b2 = a2. Therefore, 2 divides a2. Now, by Theorem 1.3, it follows that 2 divides a. • the sum or difference of a rational and an So, we can write a = 2c for some integer c. irrational number is irrational Substituting for a, we get 2b2 = 4c2, that is, b2 = • the product and quotient of a non-zero 2c2 . rational and irrational number is irrational. This means that 2 divides b2, and so 2 divides b We prove some particular cases here. (again using Theorem 1.3 with p = 2). Therefore, a and b have at least 2 as a common Example 10:Show that 5  3 is irrational. factor. Solution: Let us assume, to the contrary, that 5  3 But this contradicts the fact that a and b have no is rational. common factors other than 1. That is, we can find coprime a and b This contradiction has arisen because of our incorrect assumption that So, we conclude that

2 is rational.

b  0

2 is irrational.

SOLVED EXAMPLES Example 9 : Prove that

Therefore, 5 

3 is irrational.

Solution : Let us assume, to the contrary, that

3 is

rational. That is, we can find integers a and b (  0) such that

such that 5  3 

3

a b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. So, b 3  a Squaring on both sides, and rearranging, we get 3b2 = a2. Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.

a b

a  3 b

Rearranging this equation, we get

3  5

a 5b  a  . b b

Since a and b are integers, we get 5  rational, and so

a is b

3 is rational.

But this contradicts the fact that

3 is

irrational. This contradiction has arisen because of our incorrect assumption that 5  3 is rational. So, we conclude that 5  3 is rational.

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10th Class Mathematics factors between the numerator and denominator and see what we get :

8 Example 11:Show that 3 2 is irrational. Solution:

Let us assume, to the contrary, that 3 2 is rational. That is, we can find coprime a and b

(i) 0.375 

375 3  53 3  3 3 3 3 10 2 5 2

a  b  0  such that 3 2  b .

(ii) 0.104 

104 13  23 13   103 23  53 53

Rearranging, we get

2

a . 3b

(iii) 0.0875 

(iv) 23.3408 

and so

Do you see any pattern? It appears that, we have converted a real number whose decimal expansion

2 is rational. 2 is

rational.

terminates into a rational number of the form

Revisiting Rational Numbers and Their Decimal Expansions We know that rational numbers have either a terminating decimal expansion or a nonterminating repeating decimal expansion. In this section, we are going to consider a rational number, say

233408 22  7  521  104 54

a Since 3, a and b are integers, is rational, 3b But this contradicts the fact that

5.

875 7  4 4 10 2 5

p  q  0  , and explore exactly when the q

decimal expansion of

p is terminating and when q

it is non-terminating repeating (or recurring). We do so by considering several examples. Let us consider the following rational numbers: (i) 0.375 (ii) 0.104 (iii) 0.0875 (iv) 23.3408. Now, (i) 0.375 

375 375  1000 103

(ii) 0.104 

104 104  1000 103

(iii) 0.0875 

875 875  4 10000 10

(iv) 23.3408 

233408 233408  10000 104

As one would expect, they can all be expressed as rational numbers whose denominators are powers of 10. Let us try and cancel the common www.betoppers.com

p , q

q  0 where p and q are coprime, and the prime factorisation of the denominator (that is, q) has only powers of 2, or powers of 5, or both. We should expect the denominator to look like this, since powers of 10 can only have powers of 2 and 5 as factors. Even though, we have worked only with a few examples, you can see that any real number which has a decimal expansion that terminates can be expressed as a rational number whose denominator is a power of 10. Also the only prime fators of 10 are 2 and 5. So, cancelling out the common factors between the numerator and the denominator, we find that this real number is a rational number of the form

p , q  0 where the prime factorisation q

of q is of the form 2 n5 m, and n, m are some nonnegative integers. Theorem: Let x be a rational number whose decimal expansion terminates. Then x can be expressed in the form

p , where p and q are q

coprime, and the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. You are probably wondering what happens the other way round in Theorem 1.5. That is, if we have a rational number of the form prime factorisation of q is of the form

p , and the q

Real Numbers

9

2n5m, where n, m are non negative integers, then does

Theorem : Let x 

p have a terminating decimal q

such that the prime factorisation of q is not of the form 2n 5m, where n, m are non-negative integers. Then x, has a decimal expansion which is nonterminating repeating (recurring). From the discussion above, we can conclude that the decimal expansion of every rational number is either terminating or non-terminating repeating.

expansion? Let us see if is some obvious reason why this is true. You will surely agree that any rational number of the form

a , where b is a power of 10. b

Let us go back to our examples above and work backwards.

i 

3 3 3  53 375     0.375 8 23 23  53 103

 ii 

13 13 13  23 104     0.104 125 53 23  53 103

6. 1.

2.

 iii 

7 7 7  53 875  4  4 4  4  0.0875 80 2  5 2  5 10

 iv 

14588 22  7  521  625 54



26  7  521 233408   23.3408 24  54 104

So, these examples show us how we can convert

3.

p a rational number of the form , where q is of q the form 2n 5m, to an equivalent rational number of the form

a 4. , where b is a power of 10. Therefore, b

the decimal expansion of such a rational number 5. terminates . 6. Let us write down our result formally. Theorem :Let x 

p be a rational number,, q

p be a rational number, such q

that the prime factorisation of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates. We are now ready to move on to the rational 7. numbers whose decimal expansions are nonterminating and recurring.

Summary In this chapter, you have studied the following points: Euclid’s division lemma : Given positive integers a and b, there exist whole numbers q and r satisfying a = bq + r, 0  r < b. Euclid’s division algorithm : This is based on Euclid’s division lemma. According to this, the HCF of any two positive integers a and b, with a > b, is obtained as follows: Step 1 : Apply the division lemma to find q and r where a = bq + r, 0  r < b. Step 2 : If r = 0, the HCF is b. If r  0, apply Euclid’s lemma to b and r. Step 3 : Continue the process till the remainder is zero. The divisor at this stage will be HCF (a, b). Also, HCF(a, b) = HCF(b, r). The Fundamental Theorem of Arithmetic : Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur. If p is a prime and p divides a2, then p divides q, where a is a positive integer. To prove that

2, 3 are irrationals.

Let x be a rational number whose decimal expansion terminates. Then we can express x in the form

p , where p and q are coprime, and the q

prime factorisation of q is of the form 2n 5m, where n, m are non-negative integers. Let x 

p be a rational number, such that the q

prime factorisation of q is not of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which is non terminating repeating (recurring). www.betoppers.com

10th Class Mathematics

10

29.

Formative Worksheet

whether has terminating

21.

Prove that

22.

Prove that 3 + 2 5 is irrational Prove that the following are irrationals:

23.

5 is irrational.

1 2  ii  6 + 2

i 

 ii  7

30.

5

2 is an irrational number..

25. 26.

Prove that 3  5 is an irrational number.. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a nonterminating repeating decimal expansion.n:

 ix  27.

28.

17 8 15  iv  1600 129  vii  2 7 5 257 77 x 210

 ii 

35 50

29 343 6  viii  15

v

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p/q , what can you say about the prime factor of q? (i) 43.123456789 (ii) 0.120120012000120000… (iii)

43.123456789 www.betoppers.com

decimal

(ii) 0.123456789

Conceptive Worksheet

Prove that

13 3125 64  iii  455 23  vi  3 2 25

13 3125

expansion or not. What can you say about the prime factorisations of the denominators of the following rationals : (i) 43.123456789

24.

i

Without actually performing the long division, state

19.

Show that 3 2 is irrational.

20.

Show that

21.

7 is anirrational number. By contradiction method. Prove that n is not a rational number, if n is not a perfect square.

22.

Prove that

2  3 is an irraitonal number..

23. 24. 25.

Show that 3 6 is not a rational number.. Is a  rational number? Classify the following numbers as rational or irrationla numbers: (i) 10



(iii) 2  3 26.



(iv)

9 3 8

2 In the following equations find which variables, x, y, z represent rational or irrational numbers.

15 (ii) x2 = 0.9 2 (iii) z3 = 64 (iv) z2 = 0.09 Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansionor a nonterminating repeating decimal expansion: 2 (i) x 

27.

(ii)

Real Numbers

11

11 12 (ii) 3 3 4 5 .2  7 75 Which of the following decimal expansions are

32.

p rational.If they are rational q , then write down

333. If a = 3 9, b = 4 11, c = 6 17 , then

(i) 28.

(A) a  b  c (C) c  b  a

the prime factors of q. (i) 125.085

   

Two irrational numbers of same order can be easily compared by just comparing their radicands otherwise to convert them to the same order and then compare. Irrational numbers having the same irrational factor are called similar or like irrational numbers. Irrational numbers having no common irrational factor are known as unlike irrational numbers. n

a  n b  n ab If the irrational numbers to be multiplied are not of the same order, we first reduce then to the same order and then multiply n

  







n

a

n

b An irrational number which contains a single term is called a monomial irrational number An irrational number which consists of two monomial irrational number or a monomial irrational number and a rational number is called a binomial irrational. Rationalisation of irrational number : The process of converting an irrational number to a rational number by multiplying it with a suitable rationalising factor, is called the rationalisation of the irrational number. When the product of two irrational numbers is a rational number, then each one of the two irrational numbers is a rationalising factor of the other. If any expression is formed by the addition or subtraction of two or more irrational numbers, then each expression is simplified seperately.

If x =

5  122

(A) x  y (C) x  y

3

and y =

3  48 2

(C)

35.

36.

 x  1  x  1  x  1  x  1

  - 3x +  x -1

x 3 - 3x + x 2 -1

x2 - 4 - 2

x3

x2 - 4 + 2

2

x2 x2

(B)

x2 x2

(D)

 x  1  x  1  x  1  x  1

is

x2 x2 x2 x2

The value of x x + x 2 + x 3 + ............ is (A)

x 3  x 4  x 5  ............

(B)

x 3  x 6  x11  ............

(C)

x 3  x 6  x 9  ............

(D)

x 3  x 6  x12  ............

The value of

x2 + 1 + x2 -1

+

x 2 + 1 - x 2 -1

x 2 +1 - x 2 -1 x 2 + 1 + x 2 -1

is

37.

15

864

(B)

15

(A)

3

(C)

3

9

72 The conjugate of (A)

, then

7 4

(C)  7  4 40.

5

24 (C) 250 (D) 648 The rationalising factor of 5 3 3 is 3

38.

(D) x2

(A) 2x 2 (B) 2x (C) x 3 5 The value of 2 × 3 is (A)

39.

(B) x  y (D) x  y

The value of

(A)

a b

Formative Worksheet 31.

(B) b  c  a (D) a  c  b

(ii) 11.21478 34.



If α = 7 - 5, β = 13 - 11 , then (A)    (B)    (C)    (D) none

1 3 (D) All the above

(B)

3

7 + 4 is (B)  7  4 (D) All the above

The rationalising factor of  3 + 5 is (A)  3  5 (C) 2 3  10

(B)

3 5 (D) All the above www.betoppers.com

10th Class Mathematics

12

Conceptive Worksheet 29.

30.

31.

32.

The ascending order of x = 3 - 2, y = 6 - 5, z = 8 - 7 is (A) x, y, z (B) z, y, x (C) y, z, x (D) None 1 1 4 The descending order of x = 3 , y = 2,z = 6 2 3 is (A) x, y, z (B) y, x, z (C) z, x, y (D) y, z, x If a = 7 + 15, b = 7 , then (A) a  b (B) a  b (C) a  b (D) a  b If

A a

=

B

=

C

D

=

(A) 0

(A) 0

5 + 15

3

34.

The value of

÷

8 3 +6 5 3 3 +3 5

(C)

3

389017 4

29 (A) 3 23 (C) 3 The value of

81

5

(D) 2

37.

5 3+

1.

Find a rational numbers between –2 and 6.

2.

Find a rational number between

3.

Insert 10 rational numbers between

4 3 -2 7

4.

Insert 100 rational numbers between

(D) 1 5.

Express

is

343 × 5 16807 × 6 117649

6.

3 and 13

7 in the decimal form by long division 8

Convert

35 in to decimal form by long division 16

method. is

16 . 45

7.

Find the decimal representation of

8.

Show that 1.272727 = 1.27 can be expressed in the form

9.

p , where p and q are integers and q  0 . q

Express 0.99999 in the form

= a + b 21 , then (a, b) is

p , where p and q q

are integers and q  0 .

 37 7  ,  (A)   10 10 

 37 7  ,  (B)   10 10 

 37 7  (C)  ,   10 10 

 37 7  (D)  ,   10 10 

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3 and 11

method.

49

7

2 1 and . 3 4

9 . 13

(A) 3 (B) 3 5 (C) 49 (D) None The number of rationalising factors of an irrational number is (A) One (B) Two (C) Three (D) Infinite If

Summative Worksheet

is

5

36.

1 (C) xy (D) 1

8 . 11

13 (B) 3 73 (D) 3 3

35.

is x +a + x -a 1 1 (A) y (B) x

=

(C) –1

2 15 + 8

(B)

2xy , then the value of y 2 +1

x +a - x -a

A+ B+C+ D

(B) 1

The value of

If x > 0, y  1 and a =

, then

b c d Aa + Bb + Cc + Dd

a+b+c+d

33.

38.

10. 11. 12.

Identify

45 as rational number or irrational

number. Write three numbers whose decimal expansions are non-terminating recurring. Find the six rational numbers between 3 and 4.

Real Numbers 13.

13

Find three rational number lying between

1 6

23.

3 2 3 2 and , find the x2 + y2. 3 2 3 2

14.

If x 

15.

Express the following in decimal form

16.

(a) 15.712 (b) 4.32 Find two irrational numbers between 0.12 and 0.13.

17.

Rationalize :

18.

The value of 0.423 is

423 1000

24.

1 . 2 3 5

(B)

25.

423 100

(C)

423 990

7 8

20.

21.

(B) 16

1 8

(C) 14

7 8

1 8

The value of x, when 2x+4. 3x+1 = 288 (A) 1 (B) - 1 (C) 0 (D) 2 Value of

26.

1 1 1  + 5 6 6 7 + 4 5

27.

22.

If a = 2 +

(C) 4

(A)

5 2

(B)

2 5

(B) -14 (C) 8 3

(D) 8 3

(C)

5 3

8

28.

29.

equal to (A) 14

(B) x 1/3 y 5/3

30.

(D)

3 5

13

 2  2     What is the value of  5    5  1 2

(B) 1 (C)

(D) 5

3 and b = 2 -

xy 2  x 2 y

(C) x 5/3 y1/3 (D) x1/3 y5/3 What is the value of

(A)

1 1 3 then a 2  b 2 is

3

3  53  3 31 5  6 3  556

1 1  7 8 8 9 (B) 3

Which of the following is equal to (A) x 5/3 y 1/3

1 1 1   1 2 2 3 3 4 +

(A) 2

22 7

(C) All rational numbers are terminating or recurring. (D) Product of two irrational number can be rational or irrational

 1    + (64)-1/2 - (-32)4/5 is equal to  64 

(D) 17

5.3 is a real number

(B)  is equal to

0

(A) -15

(A) 3.535 (B) 4.535 (C) 5.53 (D) 2.535 Which of the following statement is not true? (A) The sum of a rational numbers and an irrational number is an irrational number (B) The product of a non-zero rational number and an irrational number is an irrational number. (C) Negative of an irrational number is an irrational number. (D) The sum of two irrational numbers is always irrational number. Which of the following statements is not true? (A)

419 (D) 990 19.

2  1.414, write the value of

5 2 2

1 and . 5

(A)

By taking

22  5 256  1    3 64 2

3 2

(D) 4

2

(A) –2 (B) 2(C) 1 (D) 1/2 3n × 9n + 1  (3n – 1 × 9n – 1) (A) 343 (B) –343 (C) –243 (D) 243

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10th Class Mathematics

14 31.

32.

The number of composite numbers between 101 11. and 120 are ? (A) 11 (B) 12 (C) 13 (D)14 A set of whole numbers satisfy property/properties 12. under multiplication.? (A) Closed (B) Commutative (C) Associative (D) All of these 13.

HOTS Worksheet 1.

2.

3.

4.

5. 6.

The product of the HCF and LCM of two numbers is 39483. If one of the numbers is 123, then the other number is (A) 321 (B) 311 (C) 231 (D) 113 What are the HCF and LCM of 124 and 224? (A) 14 and 1984 (B) 4 and 6944 (C) 8 and 3472 (D) 2 and 13888 What is the prime factorisation of 1224? (A) 23 × 32 × 13 (B) 22 × 33 × 172 (C) 23 × 32 × 17 (D) 22 × 32 × 133 If k is any whole number, then any perfect square can be written in which of the following forms? (A) 4k, 4k + 1 (B) 4k + 1, 4k + 3 (C) 4k + 2, 4k + 3 (D) 4k, 4k + 2 What is the HCF of 24, 32, and 60? (A) 4 (B) 6 (C) 8 (D) 3 Which of the following rational numbers has a nonterminating decimal expansion? (A)

7.

9. 10.

(B)

18 225

(C)

21 700

(D)

15.

16.

17.

45 360

Which of the following rational numbers has a terminating decimal expansion? (A)

8.

16 250

14.

13 507

(B)

27 276

(C)

126 270

(D)

420 240

18.

If n is  any  natural  number,  then  which  of  the following expressions ends in 0? (A) (3 × 2)n (B) (4 × 3)n n (C) (2 × 5) (D) (6 × 2)n What is the LCM of 36, 96, and 72? 19. (A) 144 (B) 254 (C) 288 (D) 336 In a factory, two machines are used for the production of an item. Machine A produces an item in every 15 minutes, while machine B produces an item in every 25 minutes. If the two machines simultaneously produced an item at 12:00 p.m., then at what time will they next produce an item simultaneously? (A) 1 p.m. (B) 1:15 p.m. (C) 1:30 p.m. (D) 2 p.m.

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The HCF of two numbers a and b is 30, while their LCM is 45. What is the value of (a × b)? (A) 1250 (B) 1255 (C) 1350 (D) 1355 1051 bottles of soft drink have to be kept in crates. Each crate can hold a maximum of 28 bottles. How many empty spaces will be left in the last crate? (A) 13 (B) 14 (C) 15 (D) 16 The HCF of 20000 and 25600 can be expressed as (ab × ba). If b is greater than a, then what are the respective values of a and b? (A) 5 and 4 (B) 2 and 5 (C) 3 and 5 (D) 5 and 7 The number 58500 can be prime-factorised as (A) 22 × 53 × 72 × 13 (B) 23 × 32 × 52 × 11 (C) 22 × 32 × 53 × 13 (D) 23 × 53 × 72 × 11 The number 75600 can be factorised as a1  × b2 × c3 × d4 . What is the value of (a + c)? (A) 5 (B) 7 (C) 8 (D) 10 The least common multiple and the greatest common divisor of two numbers are 336 and 8 respectively. If one of the numbers is 48, then the second number is (A) 54 (B) 56 (C) 62 (D) 64 A librarian wants to arrange 154 English books and 84 Mathematics books in shelves such that each shelf contains the same number of books. Also, each shelf should contain either only English books or only Mathematics books. She wants to arrange them in such a way that the least number of shelves is used. What will be the number of books in each shelf? (A) 7 (B) 9 (C) 12 (D) 14 There are 126 bags in a row. It is observed that starting from the first bag, every third bag has a blue ball, every sixth bag has a green ball, and every seventh bag has a red ball. How many bags contain all the three balls? (A) 3 (B) 4 (C) 6 (D) 7 Every ten minutes, a bus departs from Mumbai for Pune. Also, every 15 minutes, a bus departs from Mumbai for Thane. At 9 a.m., two buses simultaneously depart from Mumbai for Pune and Thane. What will be the next instance of time when two buses will simultaneously depart from Mumbai for Pune and Thane? (A) 9:30 a.m. (B) 10:15 a.m. (C) 11:15 a.m. (D) 11:20 a.m.

Real Numbers 20.

21.

22.

A rectangular piece of cloth of dimensions 18 m × 8 m is to be cut in the least number of square pieces such that no cloth is wasted. What is the number of such square pieces that can be cut from this rectangular piece of cloth? (A) 48 (B) 36 (C) 32 (D) 24 Which of the following sets of numbers can be expressed in the form 4q + 1 or 4q + 3, where q is a whole number? (A) Cubes of integers (B) Squares of even integers (C) Positive odd integers (D) Positive even integers Which of the following rational numbers has a terminating decimal expansion? (A)

23.

12 576

(B)

33 528

(C)

11 484

(D)

14 524

Which of the following rational numbers has a nonterminating decimal expansion? (A)

24.

15

132 704

(B)

46 1704

(C)

273 156

(D)

135 216

Which of the following rational numbers has a nonterminating decimal expansion? (A)

11 3 (B) 4356 96

(C)

6 384

(D)

21 168

25.

Which of the following rational numbers has a terminating decimal expansion? (A)

26.

27.

28.

29.

30.

14 196

(B)

21 254

(C)

24 328

(D)

63 672

If the L.(C)M. of the numbers 216 and 576 is of the f orm a 2b  × (a +  1) b , then the values of a and b are  respectively (A) 6 and 2 (B) 3 and 2 (C) 2 and 3 (D) 6 and 3 If the H.(C)F. of the numbers 30375 and 141750 is of the form a4 × ba, then the respective values of a and b are (A) 2 and 5 (B) 3 and 5 (C) 4 and 3 (D) 5 and 4 If 68600 = 2a × (a + 2)2 × (a + 4)a, then what is the value of a? (A) 7 (B) 6 (C) 3 (D) 2 If the number 88200 can be prime factorised as 23 × 32 × 52 × k2, then the value of k is (A) 7 (B) 11 (C) 13 (D) 17 The square of every positive integer can be expressed in terms of m (where m is an integer) (A) 3m (B) 3m or 3m + 2 (C) 3m or 3m + 1 (D) 3m + 2

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16

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10th Class Mathematics

Learning Outcomes By the end of this chapter, you will understand  Recap of Basic terms  Special products & Identities  Polynomial  Synthetic Division (Horner’s method)  Remainder theorem & Factor theorem  Homogeneous, Symmetric & Cyclic Expressions  Division Algorithm  Factorization of Homogenous Cyclic Expressions  Relation between Roots and Coefficients

1. Recap of Basic terms Variable: A symbol which takes various numarical values is known as a variable or literal. Any letter can be used to denote a variable. Examples: x, y, z, etc. Constant: A symbol having a fixed numerical value is called a constant. Examples: 3,10,15,2009, etc. Coefficient: In the product of a variable and a constant, each is called the coefficient of the other. Example: 3x. Here 3 is the numarical coefficient of x and x is variable coefficient of 3. Terms: Several parts of an algebraic expression

separated by + or – signs are called the terms of the expression. Examples: 3x, 4, y, 5y, etc. Algebraic Expression: A combination of constants and variables by the four basic mathematical operations (+, –, ×,  ) is called an Expression or an Algebraic Expression. 1

Examples: 3x2 + 4x – 3, 4 x 3  3 x  5 , 4x–2 + 3x + 5 Note: The powers of variables of an algebraic expression can take any real number.

2. Polynomial An “Expression” of the form P  x   a0  a1 x  a2 x 2  . . .  an x n where

powers of variable are non–negative integer

 n (W) and coefficients are any real numbers  a 0 , a 1 , a 2 , ..., a n  R ) is called a Polynomial in x of nth degree over real numbers. Examples: 2009x3 + 2008 x2 + 2007x + 2006

1 10 x + 2x 5 + 0.25x + 2009 2

Chapter - 2

Polynomials

Note: All polynomials are algebraic expressions but all algebraic expressions are not polynomials. Example: 100x–2 + 1000x3 + 2x-3 + 2009 It is an expression but not a polynomial, as the powers of the variable are negative integers.

Degree of a polynomial in one variable Definition: In case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial. Examples:

7 is a polynomial in x of degree 3. 8 3x5 – 7 is a polynomial in x of degree 5. 4x3 - 7x +

Degree of a polynomial in two or more variables Definition: In the case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial. Examples: 3x2 – 5x3y2 + 9y is a polynomial of degree 5 in x and y. 7x2yz3 + 4xy2z4 – 2x2y is a polynomial of degree 7 in x, y and z.

General Forms of Polynomials First Degree Polynomial or Linear Polynomial: ax +b Second degree Polynomial or Quadratic Polynomial: ax2 + bx + c Third degree Polynomial or Cubic Polynomial: ax3 + bx2 + cx + d nth degree Polynomial:

an x n  an 1 x n 1  ............  a1 x  a0

10th Class Mathematics

18

Types of Polynomials A polynomial with one term is called a Monomial. Examples: 2, 2xy, 3x, etc A polynomial with two terms is called a Binomial. Example: 2 + x2,2009 x2010 + 2010 x2009 , etc. A polynomial with three terms is called a Trinomial. Example: x + 2x2 + 3x3, 10x5 + 100x4 + 1000x3, etc. A polynomial containing more than three terms is called a Multinomial. Example: 2 + 3x + 7x2 + 4x3+ 10x5, 1 + x10 + x100 + x1000, etc.

Dividend = Divisor × Quotient + Remainder f(x) = (x – a) × q(x) + R Put x = a, then f(a) = (a – a) × q(a) + R f(a) = 0 + R = R f(a) = R The above concept is known as Remainder Theorem Example: f(x) = x3 – 6x2 + 12x – 8 is divided by x + 2  Remainder = f(a) = f(–2) = ( – 2)3– 6(– 2)2 +12(– 2) – 8 = – 8 – 24 – 24 – 8 = – 64  Remainder = – 64

Zero of a Polynomial

Factor Theorem

Definition: A real number a is a zero of a polynomial P(x) if P(a) = 0. Example: Let P(x) = x2 – 5x + 6 Put x = 2  P(2) = 22 – 5 (2) + 6 = 4 – 10 + 6 = 0 Put x = 3  P(3) = 32 – 5 (3) + 6 = 9 – 15 + 6 = 0. 2, 3 are the zeroes of the polynomial, P(x) = x2 – 5x + 6 Note: 1) The number of zeros of a polynomial is equal to the degree of the polynomial 2) Every linear polynomial in one variable has a unique zero 3) A non-zero constant polynomial has no zero.

Statement: Let p(x) be a polynomial of degree n > 0. If p(a) = 0 for a real number a, then (x – a) is a factor of p(x). Conversely, if (x – a) is a factor of p(x), then p(a) = 0. Proof: Let f(x) be polynomial of degree n > 0. If f(a) = 0 for a real number a, then (x – a) is a factor of f(x). Conversely, if (x – a) is a factor of f(x), then f(a) = 0 According to division algorithm Dividend = divisor × quotient + remainder f (x) = q (x) × (x – a) + R We know f(a) = R According to remainder theorem, If f(a) = 0 = R i.e., remainder is zero then f (x) = q(x) (x – a) Hence (x – a) is the factor. This above theorem is known as Factor Theorem Example: f (x) = 3x2 – 7x + 4 & (x – 1) divides f (x)  a = 1  Remainder = f (a) = f (1) = 3(1)2 – 7(1) + 4 = 3 – 7 + 4 = 0  (x – 1) is a factor of f(x) = 3x2 – 7x + 4

Division Algorithm Let p(x) be divided by g(x) and f(x) be the quotient and r(x) be the remainder. According to division algorithm, Dividend = Divisor × Quotient + Remainder f(x) = g(x) ×   x  + R

3. Remainder theorem & Factor theorem Remainder theorem Statement: If p(x) is any polynomial of degree greater than or equal to 1 and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a). Proof: Let q(x) be the quotient and R be the remainder obtained when f(x) is divided by (x – a). We know that

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1)

2)

3)

Important Points: If (x – a) divides f(x),then remainder = f(a) (value of x is found by equating the divisor to zero x – a = 0). If for any polynomial f(x) , if f(0) = 0 then x is a factor of f(x) & if f(a) = 0 then (x – a) is a factor of f(x). Degree of any polynomial must always be equal to the degree of the product of its factors.

Polynomials 1) 2) 3) 4) 5)

Useful factors: x2 – y2 = (x –y) (x + y) x3 – y3 = (x – y) (x2 + xy + y2) x4 – y4 = (x – y) (x3 + x2 y + xy2 + y3) x5 – y5 = (x – y) (x4 + x3 y + x2 yi + xy3 + y4) Simmilarly (x – y) (xn–1 + xn–2 y + xn–3 y2 + xn–4 y3 + . . . + xyn–2 + yn–1 )

19 3. 4.

5. 6.

Formative Worksheet Find the remainder when 4x2 – 8x + 3 is divided by 2x – 1. 2. Find the value of b and c if the division of x2 + bx +c by (i) (x – 1) leaves remainder zero and (ii) (x + 2) leaves remainder 12. 3. To find the values of l and m of the expression x4– x 3 + lx 2 + mx + 4 so that it is divisible by ( x2 – x – 2) 4. The expression (ax2 + bx + c) is exactly divisible by (2x – 1) and (x + 2). It leaves a remainder 12 when divided by (x – 2). Find the values of a, b and c. 5. i) Find whether xn + yn is divisible by x + y . ii) Find whether xn + yn is divisible by x – y . iii) Find whether xn – yn is divisible by x + y . iv) Find whether xn – yn is divisible by x – y . 6. i) Show that, if n is odd, ( x + 1) is a factor of ( xn + 1) ii) Show that x2009 + 1 is divisible by x + 1 7. Show that ab is a factor of (a + b)n – (an + bn) for a & b are coprime using factor theorem. 8. If in a polynomial, the coefficients are alternately positive & negative, prove that it cannot have negative roots. 9. 1, a1, a2 … an–1 are called ‘n’ roots of unity, if they are the roots of xn – 1 = 0. Then find the value of (1 – a1)(1 – a2) … (1 – an–1) 10. Prove that there is no polynomial f(x) with integral coefficient such that f(1) = 2001 and f(5) = 2006 11. Which number divides the expression 2903n – 803 n – 464 n + 261 n for any natural number n.

– p), (x – q) and  x  r  respectively when divided

1.

Conceptive Worksheet 1. 2.

Find the value of m, in order that x2 – mx – 2 is the quotient when x3 + 3x2 – 4 is divided by x + 2, If f(x) is divided by (x – 1), (x – 2) gives the remainders 5 and 7 respectively, then find the remainder when same f(x) is divided by (x – 1) (x – 2)

If n is any even positive integer, then what number divides 26n – 62n For any integer m, then what term divides (x3 + 3ax2 + 3a2x + a3)2m +1 + (x3 – 3ax2 + 3a2x – a3)2m + 1 For any integer n, then what number divides 3n + 1 + 3 n + 3 n – 1 Show that the remainder is independent of the coefficients and the degree of the polynomial p(x) in x which leaves remainders p,q,r on division by (x by x3 – Kx3 + Lx + m for K = (p + q + r); L = (pq + qr + rp) and M = –pqr

7.

1  2 for all positive real numbers x then x find the minimum value of If x 

6

1  6 1   x   x  6  2 x  x  f  x   3 1  3 1   x  x  3  x  x  

8. 9.

10.

11.

12. 13.

14. 15.

Show that xn+1 – xn – x + 1 is exactly divisible by (x – 1)2, when n is any positive integer. Show that (b – c)2n+1 + (c – a)2n+1 + (a – b)2n+1 is divisible by (b – c)(c – a)(a – b), n being any positive integer. If f (x) is a polynomial with integral coefficients and suppose f(1) and f(2) both are odd, prove that there exists no integers ‘n’ for which f(x) = 0. Without actual division, prove that (2x4 – 6x3 + 3x2 + 3x – 2) is exactly divisible by (x2 – 3x + 2). What must be subtracted from 4x3 – 3x + 9. So that (2x – 3) is the factor? Resolve into factors using factor theorem i) 2x3 – 11x2 + 17x – 6 ii) 2x3 – x2 – 15x + 18 Find the values of a and b if (x2 – 4) is a factor of (ax4 + 2x3 – 3x2 + bx – 4). Obtain a polynomial of lowest degree with integral coefficients, whose one of the zeros is

3  2  . 16. Find the roots of the equation x4 + x3 – 19x2 – 49x– 30 = 0. Given that the roots are all rational numbers. 17. P(x) is a polynomial of degree 4. Given that P(2) = P(–2) = P(–3) = –1 and P(1) = P(–1) = 1. What is P(9) www.betoppers.com

10th Class Mathematics

20 18. When P(x) = x81 + L x57 + G x41 + H x19 + 2x + 1 is divid by (x – 1), the remainder is 5 and P(x) is divided by (x – 2), the remainder is –4. However, x81 + L x57 + G x41 + H x19 + K x+ R is exactly divisible by (x – 1)(x – 2). If L, G, H, K and R are real, compute the ordered pair (K, R).

Limitations of Remainder theorem Let us find the remainder when x2009 is divided by x2– 1. Here, Dividend = x2009 = f(x) (say) Divisor = x2– 1 In order to find the remainder we have to find the zero of the division and then apply the remainder theorem. Zero of the divisor, x2 – 1 is given by, x2 – 1 = 0  x2 = 1  x  1 Now, f(1) = (1)2005 = 1 and f (–1) = (–1)2005 = –1 From above we observe that we have obtained two remainders i.e., 1 & –1 Which is a CONTRADICTION as we get only one remainder. Thus, the remainder theorem fails when the degree of the divisor is greater than 1. How do you find the remainder when the remainder theorem fails ?

4. Division Algorithm According to division algorithm, Dividend = Divisor × Quotient + Remainder If f(x) is the dividend, g(x) is the divisor,   x  be the quotient and R is the remainder then f(x) = g(x) ×   x  + R. Note: Remainder is atleast one degree less than the degree of the divisor (degree of the divisor is atleast 1). General forms of remainder based on the degree of the divisor

Degree of divisor 2 3 4

General from of Remainder ax + b ax2 + bx + c ax3 + bx2 + cx + d

Method to find the remainder (i)

When divisor is of second degree: The remainder is always one degree less than the degree of the divisor.

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If the divisor is a linear expression of the form (ax + b). the remainder is a numerical number i.e. a constant. But , if the divisor is of second degree like (x – a) (x – b) then the remainder will be linear like (cx + d). In the above cases, the division algorithm will be f(x) = (x – a)(x – b)  (x) + (cx + d) We can find the value of c and d by substituting x = a and x = b in the above, and then solving the equations involving c and d. The values of c and d will enable us to get the required remainder. (ii) When divisor is of third degree: If the divisor is of third degree like (x – a) (x – b) (x – c) then the remainder will be (px2 + qx + r). In the above cases, the division algorithm will be f(x) = (x – a)(x – b) (x – c)  (x) + (px2 + qx + r) We can find the value of p, q, and r by substituting x = a, x = b and x = c in the above and then solving the equations involving p, q and r. The values of p, q, r will enable us to get the required remainder.

Formative Worksheet 12. Find the remainder obtained when x2009 is divided by x2 – 1. 13. The remainder R, when x 100 is divided by x2 – 3x + 2, is a polynomial, then find R. 14. The remainder when x 5 + kx 2 is divided by (x–1)(x–2)(x–3) contains no terms in x2. find k, without performing division. 15. What is the remainder when x11 is divided by 1 + x + x2 + x3 + ........ + x10. 16. Assume that a0x4 + a1x3 + a2x2 + a3x + a4 is the remainder when (x19 + 2x14 + 3x9 + 4x4 + 5) is 4

5

4

3

2

divided by (x – x + x – x + x –1). Find

a

i

.

i=0

17. If x2 + x + 1 divides x4 – 5x2 – bx – 5 exactly. Find the other factor.

Conceptive Worksheet 19. A polynomial f(x) leaves a remainder – 4, –1 and 4 when divided (x – 1), (x – 2) and (x – 3) respectively. Determine the remainder when f(x) is divided by (x3 – 6x2 + 11x – 6). 20. When the polynomial f(x) is divided by (x – 2)2, the remainder is (3x – 3). What is the remainder when (x – 1) f(x) is divided by (x – 1)(x – 2)2.

Polynomials

21

21. When a positive integer x is divided by a positive integer y, the quotient is u and the remainder is v, u and v are integers. Find the remainder when x + 2uy is divided by y.

20. If x1, x2, - - - x1997 are the roots of the polynomial X1997 + a1x1996 + - - - - - - -+ a1996 x + 1 and y1 = x1x2 - - - - x10, y2 = x2x3 - - - - x11, y1997 + a1x1996 x11 - - - x9. 19 98

5. Relation between Roots and Coefficients Let us observe the following. 1) (x – a)(x – b) = x2 – x(a + b) + ab 2) (x – a)(x – b)(x – c) = x(x2 – x(a + b) + ab)– x(x2 – x(a + b) + ab) = x3 – x2b – x2a + xab – x2c + xbc + xac – abc = x3– x2(a + b + c)+ x(ab + bc + ca) – abc = x3– x2(  a) + x(  ab) – abc 3) (x – a)(x – b)(x – c)(x – d) = x4 – x3(  a) + x2(  ab) – x(  abc) + abcd What can be concluded from the above? If x1, x2, … , x are the roots of a polynomial P(x) = anxn + an–1xn–1 + . . . + a1x + a0, then  (x – x1)(x – x2). . (x – xn)= P(x) divided by an  (x – x1)(x – x2)..............(x – xn)

an 1 n 1 an  2 n 2 a0  x n  a x  a x  ...  a n n n

Conceptive Worksheet 22. If ,  and  are the solutions of 1 1 1      23. If (x – a) (x – b) (x – c) (x – d)(x – e) = x5 – 2x4 – 10x3 + 20x2 + 9x + 18, then find i) a + b + c + d + e and ii) the product abcde 24. Let x1, x2, x3 be the roots of x3 + 3x2 – 7x + 1 .

x3 – 2x2 – 11x +12 = 0. Find

2

2

2 25. If ,  are the zeroes of ax  bx  c , then find

an 1 n 1 . . .  a0 x  (–1) (x1x2..xn) = x + an an Equating the coefficients of like terms, we have n

1) Sum of roots  x1 = x1 + x2 + . . . + xn =  an 1 an 2) Sum of roots by taking two at a time

 x1x2 = x1x2 + x2x3 + . . . + xnx1 =

is independent

2 Find x1 + x2 + x3 .

 xn – xn–1(  x1) + xn–2(  x1x2) +...+ n

Then, show that  y1 y 2 ....y1997  of all ai 21. Find x if x+y+z+t=1 x + 3y + 9z + 27t = 81 x + 4y + 16z + 64t = 256 x + 167y + 1672z + 1673t = 1674.

n

   2 free from ,  2  

2 26. If ,  are the zeroes of ax  2bx  c , then find

the value of 33   23  32 free from ,  2 27. If ,  are the zeroes of ax  bx  c , then find the

 1 1  value of  2  2    

an  2 an

3) Product of roots  x1 = x1x2. . . xn =  1 

the value of

a0 an

Formative Worksheet 18. If the product of two of the roots of x4 –11x3 + kx2 + 269x – 2001 is – 69, find the value of k

2

free from , 

2 28. If ,  are the zeroes of ax  bx  c ,then find the

 1  1  free from ,   1  1 29. If the square root of x4 + ax3 + bx2 + cx + d is of the form (x2 + mx + n), then show that

value of

2n  b 

a 2 2c  . 4 a

19. If x3 + px + q =  x    x    x    , then rewrite the polynomial

 x      x      x      , free of , ,  . www.betoppers.com

10th Class Mathematics

22

6. Special products & Identities Useful Expansions 0

1) (a + b) =1 2) (a + b)1 =1a + 1b 3) (a + b)2 = 1a2 + 2ab + 1b2 4) (a + b)3 = 1a3+3a2b+3ab2+1b3 5) (a + b)4 = 1a4 + 4a3b+ 6a2b2 + 4ab3 + b4 6) (a + b)5 =1a5+ 5a4b+10a3b2+10a2b3+ 5ab4+ 1b5 7) (a + b)6 =1a6+ 6a5b+ 15a4b2+ 20a3b3+ 15a2b4+ 6ab5+ b6

Useful Identities 1) (a + b)2 = a2 + 2ab + b2 2) (a – b)2 = a2 – 2ab + b2 3) a2 – b2 = (a + b)(a – b) 4) (a + b)3 = a3 + b3 + 3ab (a + b) 5) (a – b)3 = a3 – b3 – 3ab (a – b) 6) a3 + b3 = (a + b) (a2 – ab + b2) 7) a3 – b3 = (a – b) (a2 + ab + b2) 8) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 9) a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) 10) If a + b + c = 0, then a3 + b3 + c3 = 3abc this is also true when a = b = c

Formative Worksheet 22. Consider the following expressions in a,b,c E1= a2 + b2 + c2 + 2ab + 2bc + 2ca; E2= a2 + b2 + c2 + 2ab – 2bc – 2ca E3= a2 + b2 + c2 – 2ab – 2bc – 2ca; E4= a2 + b2 + c2 + 2ab + 2bc – 2ca Which among the above expressions are not perfect squares 23. Find the value of x3 + y3 + z3 – 2xyz, if x  7  5, y  5  3, z  3  7

24. If a + b + c = 0, consider the expansion of (a + b)5, show that 5ac3 + 10a2c2 + 10a3c + 5a4 = 5bc3 + 10b2c2 + 10b3c + 5b4 25. If x + y + z = 0, then prove that (x2 + xy + y2)3 + (y2 + yz + z2)3+(z2 + zx + x2)3 = 3 (x2 + xy + y2) (y2 + yz + z2) (z2 + zx +x2) 26. Manipulate the equality a2 b2(bc – a2 ) + b2 c 2 (ca – b2) + c2 a2(ab – c2) = a2 b2(b2 – ac) + b2 c2 (c2 – ab) + c2 a2 (a2 – bc) until the equality (a2 + b2) (b2 + c2)(c2 + a2) = abc (a + b) (b + c) (c + a) is obtained.

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Conceptive Worksheet 30. Resolve into factors : a) a2 – ac + ab – bc b) y3 – y2 + y – 1 c) ax – bx + by + cy – cx – ay d) a12 – 1 e) x6 – 2x3 + 1 3 3 f) (a – b) + (b – c) + (c – a)3 g) x2 + 21x + 108 h) p4 – p2q2 – 56q4 i) 119 – 10c – c2 j) 6x3 – 5x4 + x5 4 2 2 4 k) x + x y + y 31. Find (x + 2a)(x + 4a)(x + 6a)(x + 8a) + 7a4 32. Resolve in to factors a 2 x 3 

8a 2 8  x3  3 3 y y

33. If s  a  b  c , then find the value of 4a2b2 – (a2 + b2 – c2)2 34. If x = b - c+ a, y = c - a + b, z = a - b + c, then find the value of (b – a) x + (c – b) y + (a – c) z 35. Simplify (a2 + b2 + c2)2 – (b2 + c2 – a2)2 –(a2 – b2 + c2) + (a2 + b2 – c2)2 36. If a + b + c = 0, then find the value of a4 + b4 + c4 (ac – bd)2 + (ad + bc)2 2

1

37. If x  2  2 3  2 3 then find the value of x3 – 6x2 + 6x + 1. 38. Let the sequence {un} is defined by u1 = 5 and the relation un+1 – un = 3 + 4(n – 1) n  N . Express un as a polynomial in n.

7. Synthetic Division (Horner ’s method) As this is short and simple it is called synthetic division. This is also called Horner’s method. Let us understand the method by considering the the following example. Divide x3 + 6x2 + 11x + 10 by x + 1 –1

1

6

11

10

Row–1

Step 1: Arrange the coefficients of the dividend in descending order of powers of variables (Row–1). 1 6 11 10 Step 2: The required divisor is obtained by taking the zero of the original divisor. i.e., x + 1 = 0  x = – 1

Polynomials –1

23

1

6

11

10

Row–1

0

–1

–5

–6

Row–2

5

6

4

Row–3

1

Step 3: The first element of Row–2 is always a zero (0). Step 4: The first element of Row–3 is obtained by adding first elements of Row–1 and Row–2. 1 + 0 =1 Note: i) nth element of Row–3 = nth element of Row–1 + nth element of Row–2. ii) nth element of Row–2 = divisor × (n – 1)th element of Row–3 [n > 1]. –1

1

6

11

10

Row–1

0

–1

–5

–6

Row–2

5

6

4

Row–3

1

Step 5: The last element of Row–3 is the required remainder. Remainder = 4 Step 6: The number in Row–3 (except the last) represents the coefficients of the quotient, which is one degree less than the dividend. Quotient = x2 + 5x + 6 and Remainder = 4

Formative Worksheet 3

27. Find the quotient and the remainder when 4x – 3x + 9 is divided by 2x – 3. 28. If 2x–1 and 3x+1 are the factors of 6x4 + 29x3 + 30x2 – 11x – 6 then find the other factors by using synthetic division. 29. Solve in positive integer the cubic equation x3 – (x + 1)2 = 2001.

Conceptive Worksheet 3 2 39. Find the factors of x  2 x  5 x  6 40. If x – 2 is a factor of 3x3 – 2x2 + x – 18, then the other factor 41. If (a + 1) is a factor of a4 – 5a3 – 12a2 – 5a + 1, then find the other factors

8. Homogeneous, Symmetric & Cyclic Expressions Homogenous Expression If the sum of the powers of the variables in all the terms of an expression is the same, the expression is called “Homogenous Expression”. Example: 3x4 + x2y2 + 2y4 Power of

Term 1st term  3x4y° nd

2 2

rd

4

2 term  x y

3 term  2y x°

x

y

Sum of powers of x and y

4

0

4+0 =4

2 2

2 4

2+2 =4 0+4 =4

Since the sum of the powers of the variables in all the terms of 3x4 + x2y2 + 2y4 is the same, it is a homogenous expression of the 4th degree. Example: 5x5 + 4x3y2 + xy4 Power Sum of powers Term Of of x and y x y 1st term  5x5 y°

5

0

5+0=5

2 term  4x y

3

2

3+2=5

3rd term  xy 4

1

4

1+4=5

nd

3 2

Since the sum of the powers of the variables in all the terms of 5x5 + 4x3y2 + xy4 is the same, it is a Homogenous expression of the 5th degree. Standard form of Homogenous Expressions: Variables Degree Standard form x, y

ax + by

1 2

x, y

2

ax + bxy + cy 2

x, y

3

ax3 + bx2y + cxy 2 + dy 3

Symmetric Expression An expression f(x, y) is said to be symmetric if the expression does not change when the variables are interchanged. i.e., f(x, y) = f(y, x) Examples: p(a, b) = 10a5 + 100a4b + 1000a3b2 + 1000a2b3 + 100ab4 + 10b5 p(a, b) = 5a3 + 3a2b + 3ab2 + 5b3 Note: A symmetric expression exists only when the expression contains two variables.

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10th Class Mathematics

24

iii) Fifth degree expression.

Cyclic Expression

Apart from (a – b)(b – c)(c – a), or (a + b)(b + c)(c + a) the other factor will be of second degree and will be of the form.

Consider the expression, 2

2

2

f(x, y, z) = x ( y  z )  y ( z  x)  z ( x  y ) Let us replace x by y, y by z , and z by x then, f(y, z, x) =

y 2 ( z  x)  z 2 ( x  y)  x 2 ( y  z ) = f(x,y,z) You will observe that cyclic changes in x,y,z do not result in the change of the expression. Such an expression is called a cyclic expression. Definition: An expression f(x, y, z) is said to be symmetric if the expression does not change when the variables are interchanged. i.e., f(x, y, z) = f(y, z, x) = f(z, x, y) Example: f(a, b, c) = a2 + b2 + c2 + 2ab + 2bc + 2ca

9. Factorization of Homogenous Cyclic Expressions i)

Third degree expression: Let f(a,b,c) be a given expression homogenous, cyclic and of third degree. First treat the expression as an expression in variable a. Find f(b). If f(b) = 0 then (a – b) is a factor . Since the given expression is cyclic in a, b, c, we get (b – c) and (c – a) as other factors . Given expression is a third degree expression and the degree of (a – b)(b – c) (c – a) is also three. Therefore the given expression may have another factor which is a constant Let it be k Then f(a,b,c) = k(a – b)(b – c)(c – a) …….(1) How to get the value of k ? Eq.(1) is an identity in a,b,c, we can find the value of k by giving convenient values for a,b,c, so that right side of (1) is not zero. Instead of f(b) = 0, If f(– b) = 0, then (a + b) is a factor and f (a,b,c) = k(a + b) (b + c)(c + a) and k is to be found out as shown above.

ii) Fourth degree expression: Method followed for third degree expression is followed again and we may get the factors as (a – b)(b–c)(c – a) or (a + b)(b + c)(c + a) according as f(b) = 0 or f(– b) = 0. Since the above products are of third degree and f(a,b,c) being fourth degree expression the fourth factor is a linear cyclic factor of degree one ie (a + b + c)  f(a, b, c) = k(a + b + c) (a – b)(b – c)(c – a) or = k(a + b + c)(a + b)(b + c)(c + a) www.betoppers.com

 k  a 2  b 2  c 2   m  ab  bc  ca     Which is a second degree cyclic expression. The values of k, and m can be obtained by giving suitable values for a,b,c.

Formative Worksheet 30. Use the factor theorem and Prove (x + y + z)3 – (y + z – x)3 – ( z + x – y)3 – (x + y – z)3 = 24xyz. 31. Factorise the cyclic expression (b + c)(b – c)3 + (c + a)(c – a)3 + (a + b)(a – b)3 32. Factorize a(b4 – c4) + b(c4 – a4) + c(a4 – b4).

Conceptive Worksheet 42. Find the continued product of (x + a)(x2 + a2)(x4 + a4)(x8 + a8) 43. Factorise a 2 (b  c)  b2 (c  a )  c 2 (a  b)  2abc 44. Factorise a(b – c)3 + b(c – a)3 + c(a – b)3. 45. Factorise a(b4 – c4) + b(c4 – a4) + c(a4 – b4). 46. Factorize 2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – a4 and show that its value is equal to 4200, when b+c–a=7 c + a – b = 10 a+b–c=3

Summative Worksheet 1.

2.

3.

Degree of [1 + 2x + 3x2 + 4x3 + ...+ (999 terms)] is 1) 1000 2) 999 3) 998 4) zero If the degree of a polynomial is 33, then the degree of its additive inverse is 1) 0 2) 33 3) 1 4) can’t be determined x 4  8 x 2  16 Degree of the polynomial x  2 x  2 is   

1) 0

2) 2

3) 1

4) None

2

4.

If b  4ac  0 and a  0 , then the number of

5.

zeroes of the polynomial ax  bx  c is 1) zero 2) two 3) one 4) four The zeroes of the polynomial x3 – 6x2 + 11x – 6 are 1) 1, 0 and 3 2) 1, 0 and 0 3) 1, 2 and 3 4) 0, 2 and 3

2

Polynomials 6.

7.

8.

9.

10.

11.

12.

13.

25

14. If f(x) be a quadratic polynomial such that The zeroes of the polynomial 2 x + x (c – b) + (c – a) (a – b) are f  2   3 and f  2   21 , then the coefficient 1) (a – c), (a – b) 2) (a – c), (b – a) of x in f(x) is 3) (c – a), (b – a) 4) (c – a), (a + b) 1) 6 2) – 6 3) 4 4) –4 The product of thes zeroes of the quadratic 15. If (x + a) be a common zero of x2 + px + q and polynomial (a + c) (ax2 + 2bx + c) – 2(ac – b2) (x2 x2 + lx + m, then the value of a is + 1) is mq mq mq mq 2 2 2 2 1) 2) 3) 4) c  ac  2b c  ac  2b lp l p l p lp 1) 2 2) 2 4 4 4 a  ac  2b2 c  ac  2b 2 16. If (3 – x) + (2 – x) – (k – 2x) is exactly divisible by(x – 2), then the value of k is 2 2 c  ac  2c 1) 5 2) zero 3) 2 4) 4 3) 2 4) none a  ac  2b2 2 4 17. If x – 3x + 2 is one of the zero of x – px2 + q then If one zero of x2 – (p – 1) x + 10 is 5, then the value the values of p and q are of p is 1) 4 and 6 2) 3 and 5 3) 5 and 6 4) 5 and 4 1) 8 2) 6 3) 4 4) 5 18. The roots x1 and x2 of the equation x2 +px+12 = 0 The values of a and k, if the zeroes of possess the property x1 – x2 = 1, then the value of p is. 3 x 2  2kx  k  2 are the reciprocals of the zeroes 1)  7 2)  6 3)  5 4)  4 of 2ax2 + (k + a) x + 3 are n n+1 19. The remainder when 1 – x – x + x is divided by 1) k = a = 2 2) k = a = 3 1 – 2x + x2 is 3) k = 2, a = 3 4) None of these 1) 1 2) 2 3) 5 4) zero If the ratio of the zeroes of the expression 20. The factors of x4 – 5x3 + x2 + 13x + 6 are x2 + px + q is equal to the ratio of the zeroes of 1) (x + 1) (x – 3) (x2 – 3x – 2) x2 + lx + m, then 2) (x + 3) (x + 3) (x2 + 3x + 2) 2 2 2 2 1) p m = q l 2) pm = q l 3) (x + 1) (x + 3) (x2 – 3x – 2) 3) p2l = q2m 4) p2m = l 2q 4) None of the these If ,  are the zeroes of ax2 + bx + c, then the 21. The factors of 2x3 + 5x2 – 4x – 3 are 1) (x + 1) (2x – 1) (x – 3) 5 8 8 5 value of      is 2) (x + 1) (2x + 1) (x + 3) 3) (x – 1) (2x + 1) (x – 3) 5 3 3 2 c  3ab  b  c  3abc  b  4) (x – 1) (2x + 1) (x + 3) 1) 2) a8 a8 22. 199 + 299 + 399 + 499 + 599 is divisible by 1) 40 2) 30 3) 15 4) 10 c 4  3abc  b 3  c5  3abc  b3  3 5 95 23. 7 + 7 + 7 + ....... + 7 is divisible by 3) 4) a7 a8 1) 48 2) 47 3) 43 4) None 2 n 2 24. If n is any integer, then (x + 7x + 6) – (2 + x)2n If ,  are the zeroes of px + qx + r, then the is divisible by  1  1 1) x 2) 5x2 3) 9x3 + 9 4) 3x + 2 value of  is 2 2   25. The H.C.F. of ax + 2a x + a3, 2ax2 – 4a2x – 6a3, 3(ax + a2)2 is 2r  pq 2r  q 2r  pq 2r  q 1) a(x – a) 2) 2a (x – 3a) 2 2 1) 2) 3) 4) 2 2 r r r r 3) 3a (x + a) 4) a (x + a) 3 2 2 If ,  are the zeroes of px2 – qx + r, then the 26. The H.C.F. of a x – a bx – 6ab x, 2 2 2 2 3 2 a bx – 4ab x + 3b x is value of  4   4 is 1) x (a – 3b) 2) (a + 2b)x 3) x(a – b) 4) None of these 2  q  2 pr   2 p 2 r 2  q2  2 pr   2 p 2 r 2 27. The L.C.M. of 60x4 + 5x3 – 5x2, 1) 2) p4 60x2y + 32xy + 4y, 40x3y – 2x2y – 2xy is p4 1) 20x2y (4x + 1) (3x + 1) (5x + 1) 2 2 2 2 2) 20x2y (4x – 1) (3x + 1) (5x + 1)  q  2 pr   2 p r  q  2 pr   2 pr 3) 4) 3) 20x2y (4x – 1) (3x – 1) (5x – 1) p4 p4 4) None of these www.betoppers.com

10th Class Mathematics

26 28. If a + b + c = 1, a2 + b2 + c2 = 9 and a3 + b3 + c3 = 1, then the value of 1) 0

2) 1

1 1 1 + + is a b c

1 3)  4

4) 3abc

29. Observe the following :x3  y 3  x 2  xy  y 2 ; x y 5

5

7

7

x y  x 4  x 3 y  x 2 y 2  xy 3  y 4 ; x y

6.

Solve

a x 1

1

a 2  ( a  x) 2

Now write the quotient when divided by(x + y) 30. Find H.C.F and L.C.M of i) 15p3 q 4 , 20m2 p 2 q 3 , 30mp3

x

2005

 y 2005  is

1

ax 1

1

 a2.

a 2  (a  x) 2

a  c( a  x ) a  x a   . a  c (a  x) x a  2cx

7:

Solve

8.

Solve the cubic equation 9x3 – 27x2 + 26x – 8 = 0, given that one of the roots of this equation is double to other.

9.

Find the sum of the coefficients of the polynomial obtained after parentheses have been removed and like terms have been collected in the product 2 743

x y  x 6  x5 y  x 4 y 2  x3 y 3  x 2 y 4  xy 5  y 6 . x y



2 744

1  3x  3x  1  3x  3x 

.

10. f(x)is a polynomial of degree atleast two with integer coefficients. Show that when it is divided by (x – a) (x – b). Where a  b , the remainder is  f  a   f  b   a  f b   b  f  a   x  a b ab  

ii) 72k 2 m3 n 4 , 108k 3 m 2 n5 31. If a + b + c = 0, Prove that a2 + ab + b2 = b2 + bc 11. Let f(x) = (x2005 – 1). If the 2005 roots of f(x) are + c2 = c2 + ca + a2 a1, a2, a3, a4, ....... a2005 with one root ai = 1, for 32. Suppose that  is a root of x4 + x2 – 1 what is 1 < i < 2005 then the value of 6 the value of  + 2  4 ? (1 – a1) (1 – a2) (1 – a3) ........ (1 – a2005). 33. If 2s = a + b + c, Prove that (s – a)2 + (s – b)2 + Which is the product of 2004 brackets. (s – c)2 + s2 = a2 + b2 + c2 3 3 34. Resolve m – n +1+ 3mn into factors. 12. Let f(x) = a2005 x2005 + a2003 x2003 +....a1x1 – 2005 3 35. Find the condition that x + (p + q)x + a is divisible Where the terms in the polynomial are ai xi for by (x + p + q) i = 2005, 2003, 2001, .......(2n – 1).....5, 3, 1 (the sequence of odd numbers). OTS orksheet

H

W

1.

The value of f(+2005) for f(–2005) = 1003. Let P(x) = x4 + ax3 + bx2 + cx + d, where a,b,c and d are constants. If P(1) = 10, P(2) = 20 and 13. Let (1 + x + x2) = a + a x + a x2 + .....a x2n be 0 1 2 2n an identity in x. P 12  +P  8  P(3) = 30. Compute If we let s = a0 + a2 + a4 + ......a2n, then s equals 10

2.

Prove that (a + b + c)333 – a333 – b333 – c333 is 3n  1 3n n n+1 3 3 3 3 1) 2 2) 2 3) 4) divisible by (a + b + c) – a – b – c . 2 2 [Each of a,b,c are multiples of 3] 14. If (x + 1)2005 divides p(x) = x2006 + px2 + qx + r Factorise the expression (a10 + a5 + 1) exactly then remainder is zero when (x + k) Prove that the polynomial (x9999 + x8888 + x7777 + x6666 divides p(x). Find k. + x5555 + x4444 + x3333 + x2222 + x1111 + 1) is divisible 15. Find the sum of all coefficients in the expansion of by (x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1) (2009x – 2007y)3. Simplify 16. Given that a, b and c are the roots of the equation (2 x  3)3 ( x  1) ( x  2)  as x3 – 6x2 + 5x – 1 = 0, determine the value of (2 x  3) 4  4 4( x  1) 2  1 4( x  1)2  1 a5 + b5 + c5.

3. 4.

5.

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Polynomials

27 3

2

17. A cubic polynomial f(x) = ax + bx + cx + d has a graph which is tangent to the x-axis at 2, has another X-intercept at (–1), and has y-intercept at (–2) as shown. Find a + b + c + d.

8.

If  is a zero of the polynomial x10 – x9 + x8 – x7 – .....–x +1, then find the value of  44  33   22  11  1 .

4 3 2 1

9.

If x 

–3

–2

–1 –1

1

2

3

4

–3 –4

18. Simplify ax  b 2 x 2  2b 2 y 2  a 2 y 2   by  b 2 x 2  2a 2 x 2  a 2 y 2  ax  by

19. The polynomials f(x) = x3 + (k – 1)x2 – kx +1 and g(x) = x2 + kx +1 have a common root. Find the value of k 20. Factorise polynomial in x: (a2 +1)x3 – (a2 – 1) x2 – (a2 +1)x + (a2 –1)

IIT JEE Worksheet Prove that a2 + b2 + c2 – ab – bc – ca

3.

6.

7.

12. Suppose that f(x) is a polynomial of degree 5 and with leading coefficient 2007. Suppose further that f(1) = 1, f(2) = 3, f(3) = 5, f(4) = 7 and f(5) = 9. What is the value of f(6). 13. Factorise: 2(a – b)2 + (b – c) (c – a) 14. Prove that (a + b + c)3 + (a + b – c)3 + (a – b + c)3 + (a – b – c)3 = 4a (a2 + 3b2 + 3c2). 15. P(x) is a polynomial of degree 100 and Q(x) = P(x +1) – P(x). Find the degree of Q(x).

1 ( a + b + c ) { ( b – c ) 2+ ( c – a ) 2+ ( a – b ) 2) } 2

a0(5x2 + 4)n + 2a1(5x2 + 4)n–1 + 22 a2 (5x2 + 4)n–1

3

5.

11. If P(x) be a polynomial with integer coefficient and a, b, c are three integers satisfying P(a) = 1, P(b) = 2 and P(c) = 3 then prove that the equation P(x) = 5 has at most one integer solution.

16. The roots of the polynomial a0xn + a1xn–1 + a2xn–2 + ..............+a1x + an are all greater than or equal

= 4.

1 . x5

1 [(b – c)2 + (c – a)2 + (a – b)2] 2 If s = a + b + c, prove that (as + bc)(bs + ca) (cs + ab) = (a + b)2 (b + c)2 (c + a)2 Show that (a+b + c)(a2 + b2 + c2 – bc – ca – ab) =

2.

5 + a5, which is equal to the value of x 

10. The polynomial x4 – 50x2 + k = 0 has four distinct real roots and they are in an arithmetic progression. Determine the value of k.

–2

1.

1  k then find the sum of the coefficients x of the polynomial a0k5 + a1k4 + a2k3 + a3k2 + a4k

3

3

= a + b + c – 3abc Show that (a+b+c)3 = a3 + b3 + c3 + 3(b + c)(c + a)(a + b). a, b, c are distinct and p(x) is a polynomial in x which leaves remainders a, b, c on division by (x – a), (x – b), (x – c) respectively find the remainder obtained on division of p(x) by (x – a)(x – b)(x – c). Let p(x) be a polnomial of degree four such that p(2) = p( –2) = p(–3) = –1 and p(1) = p(–1) =1 what is p(0)? Find the largest integer k for which (x – k) (x – 2007) + 1 can be expressed in the form (x + a) (x + b) for a, b  Z , set of integers.

to 2 and are denoted by i , for i = 1, 2, 3, ..., n. Then find the roots of the polynomial +....+2nan in terms of i 17. Given that the equation x4 – 14x3 + 73x2 – 168x + 144 = 0 has two pairs of equal roots, find them. 18. Suppose P(x) is a polynomial with integer coefficients. The remainder when P(x) is divided by (x –1) is 1 and the remainder when P(x) is divided by (x – 4) is 10. If r(x) is the remainder when P(x) is divided by (x –1)(x – 4), find r(314). 19. P(x) is a polynomial in x with real coefficients. Given that the polynomial p2(x) + (9x – 2007)2 has a real root  , determine  and also the multiplicity of  . 20. Factorize the polynomial in x:(t2 + 1)x3 – (t2 – 1)x2 – (t2 + 1)x + (t2 – 1) 21. Factorize a5(b – c) + b5(c – a) + c5(a – b) www.betoppers.com

28

10th Class Mathematics

22. P(x) is a polynomial with ;integral coefficients. P(x) = 4010 for 5 different integral values of x. Prove that there is no integer x such that P(x) = 2005. 23. Prove that there is no polynomial f(x) with integral coefficients such that f(1) = 2001 and f(3) = 2004. 24. f(x) is a polynomial of degree 2003. g(x) = f(x + 1) – f(x). Then g(x) is a polynomial of degree m, where A) m = 2003 B) m = 2002 C) m cannot be uniquely dtermined, but 0  m  2002 D) m may be underfined in some cases E) none of these 25. Factorize a(b2 + c3 – a2) + b(c2 + a2 – b2) + c(a2 + b2 – c2) – 2abc.

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By the end of this chapter, you will understand      

1.

Graphical solution of a pair of linear equations in two variables Expressing given situations mathematically Substitution method of solving pairs of linear equations Elimination method to solve a pair of linear equations Cross-multiplication method of solving pairs of linear equations Equations reducible to a pair of linear equations in two variables

Introduction Linear equations in two variables are equations where we have two variables. We require two equations to find the solution of linear equations in two variables. Let’s understand the basic concept of linear equations in two variables. Point to Remember To solve a linear equation in one variable, only one equation is required. To solve linear equations in two variables, two linear equations in the same two variables are required. Likewise, we can say that to solve linear equations with n number of variables, n numbers of linear equations in the same n number of variables are required. General form of pair of Linear equations The general form of a pair of linear equations in two variables is written as

When their graphs are drawn, there are three possibilities. Let us go through the video to understand the various possibilities.

Solved Examples Example 1. Find whether the given pairs of linear equations are consistent or inconsistent? (a) 5x + 2y = 2 20x + 8y = 1 (b) 4x + y = 8 7x – 2y = 1 Solution: (b) 4x + y = 8 7x – 2y = 1 Here, a1 = 4 b1 = 1 c1 = –8 a2 = 7 b2 = –2 c2 = –1



a1x  b1y  c1  0 , where a1, b1 are real numbers and a12  b12  0 (i.e., a1  b1  0 )

a 2 x  b 2 y  c 2  0 , where a 2 , b 2 ar e real numbers and a 22  b 22  0 (i.e., a 2  b 2  0 ) We know that the solution of a linear equation must satisfy the equation. Conversely, we can say that the value of the variable in the equation, which satisfies the equation, is the solution of the equation. In case of a pair of linear equations, the values of x and y, which satisfy both the equations, are the solutions of the equation. Geometric representation of a Linear Equations Geometrically, a linear equation represents a straight line. Every solution of an equation is a point on the line represented by the equation. Likewise, a pair of linear equations represents two straight lines.

Chapter - 3

Learning Outcomes

Pair of Linear Equations In Two Variables

a1 4 b1 1 1 c 8 = ; = = ; 1 = =8 a 2 7 b2  2 2 c 2 1

We see that

4 1 a b  i.e., 1  1 7 2 a2 b2

Therefore, according to the above rule, this system of equations has a unique solution. It is represented by intersecting lines. (c) 5x + 6y = 9 10x + 12y = 18 Here, a1 = 5 b1 = 6 c1 = –9 a2 = 10 b2 = 12 c2 = –18



a1 5 1 b1 6 1 c1 -9 1 = = ; = = ; = = a 2 10 2 b 2 12 2 c 2 -18 2

We see that

5 6 -9 a b c = = i.e., 1 = 1 = 1 6 12 -18 a 2 b2 c2 Therefore, according to the above rule, this system of equations has infinite solutions. It is represented by coincident lines.

10th Class Mathematics

30

a)

Example 2. There are two lines drawn in each of the following graphs. Each line represents a linear equation in two variables. State whether the pair of linear equations represented in each graph is consistent, inconsistent, or dependent. Solution: In the given graph, the lines representing the two equations are parallel to each other and never intersect. Thus, the lines in this graph represent an inconsistent pair of linear equations in two variables as they do not have any solution.

Formative Worksheet 1.

2.

3.

Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Conceptive Worksheet (b)

In the given graph, the lines representing the two equations intersect at a point. Thus, the lines in this graph represent a consistent pair of linear equations in two variables as they have a unique solution.

1.

2.

(c)

In the given graph, the lines are coincident i.e., the same line represents the two equations. Thus, the lines in this graph represent a dependent pair of linear equations in two variables as they have infinite number of solutions.

3.

Let us take the example that Ms. Lovely goes to International fair, Pragati Maidan with Rs. 105. She wants to have rides on the Musical Chairs and Fair Lady Wheel. The number of times she rides Fair Lady Wheel is half the number of rides she had on the Musical Chairs. If each ride costs Rs.10, and the Fair Lady Wheel games cost Rs.15, she spent Rs.105. Then find the number of times she played Fair Lady Wheel. Represent this situation algebraically and graphically (geometrically). Kashiya went to a stationery shop and purchased 3 pencils and 5 erasers for Rs.21. Her friend Sudha saw the new variety of pencils and erasers with Kashiya, and she also bought 6 pencils and 10 erasers of the same kind for Rs.42. Represent this situation algebraically and graphically. The path of a train A is given by the equation 2x + 3y = 10 and the path of another train B is given by the equation 4x + 6y = 12. Represent this situation geometrically.

2. Graphical solution of a pair of linear equations in two variables Suppose you go to a stationary shop to buy some registers and pens. If you buy two registers and two pens, then you have to pay Rs 30. However, if you buy two registers and four pens, then you have to pay Rs 40. www.betoppers.com

Pair of Linear Equations in Two Variables

(a)

31

Now, how can we represent this information algebraically, i.e., in the form of equations? Let us denote the cost of each register by Rs x and that of each pen by Rs y. Now, it is given to us that two registers and two pens cost Rs 30. 2x + 2y = 30  x + y – 15 = 0 _______________ (1)  We also know that two registers and four pens cost Rs 40. 2x + 4y = 40  x + 2y – 20 = 0 ______________ (2)  Thus, we now have two linear equations in two variables from the given situation. x + y – 15 = 0 ______________ (1) x + 2y – 20 = 0 ______________ (2) These two equations are known as a pair of linear equations in two variables. If we try to individually solve each equation, we will be unable to do so. We can arrive at an answer only if we try and solve both these equations together. The general form of a pair of linear equations in two variables is written as a1x + b1y + c1 = 0

The solution of a pair of linear equations is the point of intersection of their graphs. Now, according to this, a pair of dependent equations is represented by coincident lines.

where a1, b1, c1 are real numbers, and a12 + b12  0 (i.e., a1, b1  0) The lines represented by a pair of linear equations in two variables, a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, have a unique solution (or are consistent) and

Similarly, a table can be drawn for the corresponding values of x and y for equation (ii) as 2x x y= 3 2×3 0 2 3 2 ×6 10 4 3 The two equations can be plotted on a graph as

a

b

1 1 represent intersecting lines if a  b 2 2

(b)

have infinite solutions (or are dependent) and represent coincident lines if

(c)

Example 3. Anshuman and Vikram have 10 marbles with them. Twice the number of marbles with Anshuman is equal to three times the number of marbles with Vikram. Solve this question graphically. Solution: Let us suppose that Anshuman has x marbles and Vikram has y marbles. From the given information, we can form two equations as x + y = 10  x + y – 10 = 0 ––––––––––– (i) 2x = 3y  2x – 3y = 0 –––––––– (ii) A table can be drawn for the corresponding values of x and y for equation (i) as

x

y = 10 – x

0

10 – 0 = 10

10

10 – 10 = 0

have no solution (or are inconsistent) and represent parallel lines if

(i) (ii) (iii)

a1 b1 c1 = = a 2 b2 c2

Solved Examples

a1 b1 c1   a2 b2 c 2

A pair of linear equations represents two straight lines. When their graphs are drawn, there are three possibilities. The two lines may intersect at one point. The two lines may be parallel. They may represent the same line, i.e. they may be coincident. www.betoppers.com

10th Class Mathematics

32 As seen in the graph, the two lines intersect at the point (6, 4). This implies that the solution for the pair of linear equations is x = 6 and y = 4. Thus, Anshuman has 6 marbles while Vikram has 4 marbles. Example 4 Find graphically whether the following pairs of linear equations are consistent or inconsistent? If the equations are consistent, then find their solutions as well. (a) 2x + 3y = 2 4x + 3y = 4 (b) x + y = 5 2x + 2y = 10 Solution: (a) 2x + 3y = 2 4x + 3y = 4 A table can be drawn for the corresponding values of x and y for the equation 2x + 3y = 2 as

x 1 4

2 - 2x 3 2 - 2 ×1 2- 2 0 = = =0 3 3 3 2 - 2× 4 2 - 8 -6 = = = -2 3 3 3 y=

These points can now be plotted and joined to obtain the graphs of the equations as As seen in the graph, the two lines intersect at the point (1, 0).

Thus, the given pair of linear equations is consistent with its solution as x = 1 and y = 0. (b) x + y = 5 2x + 2y = 10 A table can be drawn for the corresponding values of x and y for the equation x + y = 5 as

x

y=5–x

0

5 - 0=5

4

5 - 4=1

Similarly, a table can be drawn for the corresponding values of x and y for the equation 2x + 2y = 10 as

10 - 2x 2 10 - 2 × 3 10 - 6 4 3 = = =2 2 2 2 10 - 2 × 0 10 - 0 10 0 = = =5 2 2 2 These points can now be plotted and joined to obtain the graphs of the lines as

x

Similarly, a table can be drawn for the corresponding values of x and y for the equation 4x + 3y = 4 as 4 - 4x y= 2 x 3 4 - 4 ×1 4 - 4 0 1 = = =0 3 3 3 4- 4× -2 4 +8 12 –2 = = =4 3 3 3 www.betoppers.com

y=

Pair of Linear Equations in Two Variables

33 7.

8.

9.

As seen in the graph, the two equations are represented by the same line. This means that the given pair of linear equations is dependent. Any point lying on this line is a solution to the pair of equations.

Formative Worksheet 4. (i)

(ii)

5.

Form the pair of linear equations in the following problems, and find their solutions graphically. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen. On comparing the ratios

Conceptive Worksheet 4.

5.

6.

a 1 b1 c1 , and a 2 b2 c2 find

out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident: i) 5x – 4y + 8 = 0 ii) 9x + 3y + 12 = 0 iii) 6x – 3y + 10 = 0 7x + 6y – 9 = 0 18x + 6y + 24 = 0 2x – y + 9 = 0 6.

10.

a 1 b1 c1 , On comparing the ratios and a 2 b2 c2 find out whether the following pair of linear equations are consistent, or inconsistent. i) 3x + 2y = 5; 2x – 3y = 7 ii) 2x – 3y = 8; 4x – 6y = 9 iii) 9x – 10y = 14 iv) 5x – 3y = 11; –10x + 6y = – 22 v) 2x + 3y = 12

Which of the following pairs of linear equations are consistent/ inconsistent? If consistent, obtain the solution graphically: i) x + y = 5, 2x + 2y = 10 ii) x – y = 8, 3x – 3y = 16 iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0 iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden. Given the linear equation 2x + 3y – 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is: (i) intersecting lines (ii) parallel lines (iii) coincident lines Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

7.

8.

9.

Solve graphically the following systems of linear equation. 2x – 3y +13 = 0 3x – 2y +12 = 0 Solve graphically the following systems of equation. Also, shade the region bounded by these lines and x-axis. 2x + 3y = – 5 3x – 2y = 12 Solve the following systems of equation graphically and shade the region bounded by these lines and y-axis. 3x + y – 11 = 0 x–y–1=0 Solve the following systems of equation graphically. Also, find out the points, where these lines meet the x-axis. x – 2y = 1 2x + y = 7 Solve the following systems of equation graphically. Also, find out the points, where these lines meet the y-axis. x + 2y – 7 = 0 2x – y + 1 = 0 Use a single graph paper and draw the graph of the following equation. Obtain the vertices of the triangles so obtained. 2y – x = 8 5y – x = 14 y – 2x = 1 www.betoppers.com

10th Class Mathematics

34 10.

Solve the following system of linear equations graphically: 4x – 5y -20 = 0, 3x + 5y – 15 = 0 Determine the vertices of the triangle formed by the lines, representing the above equations, and the y-axis.

3.

Expressing given situations mathematically We come across many situations in real life when it is easy to find the solution, if we express them mathematically. Let us see such a situation. The coach of the school cricket team buys 5 bats and 20 leather balls for Rs 3500. After some time, some more boys joined the team for practice, so he buys another 4 bats and 15 balls for Rs 2750. Suppose that the price of bat and ball does not change in the time period. Can we express this situation mathematically to find out the individual prices of a ball and a bat? Let the price of a bat be Rs x and that of a ball be Rs y. It is given that 5 bats and 20 balls cost Rs 3500. Cost of 5 bats = 5x  And cost of 20 balls = 20y  Cost of 5 bats and 20 balls = 5x + 20y  5x + 20y = 3500 Similarly, it is also given that 4 bats and 15 balls cost Rs 2750.  4x + 15y = 2750 Therefore, the algebraic representation of the given situation is 5x + 20y = 3500 _______________ (1) 4x + 15y = 2750 _______________ (2) After solving these equations, we can find out the individual prices of the ball and the bat.

Solved Examples Example 5. Aman and Yash each have certain number of chocolates. Aman says to Yash, if you give me 10 of your chocolates, I will have twice the number of chocolates left with you. Yash replies, if you give me 10 of your chocolates, I will have the same number of chocolates as left with you. Write this situation mathematically? Solution: Suppose Aman has x number of chocolates and Yash has y number of chocolates. According to the first condition, Yash gives 10 www.betoppers.com

chocolates to Aman so that Aman has twice the number of chocolates than what Yash has.  x + 10 = 2(y – 10) x +10 = 2y – 20 x – 2y + 30 = 0 _______________ (1) According to the second condition, Aman gives 10 chocolates to Yash such that both have equal number of chocolates.  y +10 = x – 10 Thus, the algebraic representation of the given situation is x – y – 20 = 0 __________ (2) Example 6.Cadets in the military academy are made to stand in rows. If one cadet is extra in each row, then there would be 2 rows less. If one cadet is less in each row, then there would be 3 more rows. Express the given situation mathematically? Solution: Let the number of cadets in each row be x and the number of rows be y. Total number of cadets = number of rows number of cadets in each row It is given to us that when one cadet is extra in each row, there are 2 rows less. xy = (y – 2) (x + 1)  xy = xy – 2x + y – 2 2x – y = – 2 _________________ (1) It is also given to us that if one cadet is less in each row, then there are 3 more rows. xy = (y + 3) (x – 1)  xy = xy + 3x – y – 3 3x – y = 3 _______________ (2) Thus, the algebraic representation of the given situation is 2x – y = – 2 _______________ (1) 3x – y = 3 _______________ (2)

4.

Substitution method of solving pairs of linear equations In a class, the number of boys is 7 more than twice the number of girls. Can you find the number of boys and girls in the class? Let the number of boys be x and the number of girls be y. Now, according to the given condition, x – 2y = 7 We cannot find the unique values of x and y by solving this equation because there are multiple values of x and y for which this equation holds true.

Pair of Linear Equations in Two Variables

35

We can write the above equation as x = 2y + 7. Thus, by taking different values of y, we will obtain different values of x. However, if we are given one more condition, then the values of x and y can be evaluated. To solve a linear equation in two variables, two linear equations in the same two variables are required. Let us consider one more condition. 11. Suppose the total number of students in the class is 52. Now, can you find the number of boys and girls in the class? According to this condition, we can write x + y = 52 Therefore, we obtain a pair of linear equations in two variables. x – 2y = 7 x + y = 52 Let us go through the following video and solve this pair of linear equations by Substitution method. Therefore, number of boys = 37 and number of girls = 15 In this method, we have substituted the value of one variable by expressing it in terms of the other variable to solve the pair of linear equations. That is why this method is known as the substitution method. Now, let us solve some examples to understand 12. this concept better. 13.

   

15x + 56x + 96 - 25 =0 5 15x + 56x + 96 – 25 = 0 71x + 71 = 0 71x = – 71

Formative Worksheet

Solved Examples

Example 7. Solve the following pair of linear equations. 7x – 5y + 12 = 0 3x + 8y – 5 = 0 Solution: Given, 7x – 5y + 12 = 0 ––––––––– (i) 3x + 8y – 5 = 0 ––––––––– (ii) From equation (i), we obtain 7x – 5y + 12 = 0

(i) (ii) (iii)

(iv) 7x +12 y= ––––––––– (iii) 5 On substituting the value of y in equation (ii), we obtain  7x +12  3x + 8  -5 = 0  5 



56x + 96 3x + -5 = 0 5

(v)

Solve the following pair of linear equations by the substitution method. (i) x + y = 14 (ii) s – t = 3 x–y=4

(iii) (iv)

s t  6 3 2 3x – y = 13 0.2x + 0.3y = 1.3 9x – 3y = 9 0.4x + 0.5y = 2.3

(v)

2x  3y  0

(vi)

3x 5y   2 2 3

3x  8y  0 x y 13   3 2 6 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3. Form the pair of linear equations for the following problems and find their solution by substitution method. The difference between two numbers is 26 and one number is three times the other. Find them. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them. The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

A fraction becomes

9 , if 2 is added to both the 11

numerator and the denominator. If, 3 is added to www.betoppers.com

10th Class Mathematics Let’s look at the following video to learn how to mathematically represent the information and then solve them using the elimination method.

36 both the numerator and the denominator it becomes (vi)

5 . Find the fraction. 6

Solved Examples

Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Example 8.Solve the following equations using elimination method. 3x + 4y = 0 4x – 3y = 50 Solution: Given, 3x + 4y = 0 ______________ (i) 4x – 3y = 50 ______________ (ii) On multiplying equation (i) by 3 and equation (ii) by 4, we obtain 9x + 12y = 0 ______________ (iii) 16x – 12y = 200 _________ (iv) On adding equation (iii) and (iv), we obtain 25x = 200

Conceptive Worksheet Solve the following pair of linear equations by the substitution method. 11. 7x – 15y = 2, x + 2y = 3 12. 2x + 3y = 9, 4x + 6y = 18 13. 3x – 4y = 10, 4x + 3y = 5 14.

3x 

y7  2  10 11

3x + 3y = 65, find the value of 15.

16.

y x 2

5.

x . y

x  y  0.8 , 2

 10

www.betoppers.com

y=

 Thus,

-24 = -6 4

x = 8 and y = – 6.

Formative Worksheet 14.

Elimination method to solve a pair of linear equations There are many real life situations which can be represented in the form of linear equations. Let us begin with such a situation. Suppose you go to the market with your friend to buy clothes. You buy 3 shirts and 6 pairs of jeans from a shop and the total cost of your purchase comes out to be Rs 7000. Your friend, on the other hand buys 2 shirts and 3 pairs of jeans and the total cost of his purchase comes out to be Rs 3855. Now, we can represent this information mathematically as pair of linear equations in two variables. We can then solve the equations and get the cost of each shirt and each pair of jeans.

200 =8 25

Putting the value of x in equation (i), we obtain 3 × 8 + 4y = 0 4y = – 24 

x+y=a–b ax – by = a2 + b2

7

x=



Solve the following pair of linear equations by the elimination method and the substitution method: i) x + y = 5 and 2x – 3y = 4 ii) 3x + 4y = 10 and 2x – 2y = 2 iii) 3x – 5y – 4 = 0 and 9x – 2y + 7 iv)

15.

(i)

x 2y y   1 and x   3 2 3 3

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method: If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes

(ii)

(iii)

1 2

if we only add 1 to the

denominator.What is the fraction? Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Pair of Linear Equations in Two Variables (iv)

(v)

Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Conceptive Worksheet Solve the following equations by the method of elimination by equating the coefficients: 17. 23x + 17y = 6; 39x – 19y = 58 18.

0.4x + 3y = 1.2, 7x – 2y =

19.

(a + 2b)x + (2a – b) y = 2, (a – 2b)x + (2a + b) y = 3 x – y =0.9

20.

17 6

11  2; x   y xy 21.

x y   2, a b ax – by = a2 – b2; a  0 , b  0

6.

Cross-multiplication method of solving pairs of linear equations We can represent many situations in real life as linear equation in two variables. Let us consider such a situation. Suppose Samay is older than Sumit by 30 years. After 5 years, Samay will be thrice as old as Sumit. Can we find the present ages of Samay and Sumit? Yes, we can find the present ages. However, before that, we have to represent this situation in the form of linear equations in two variables. Let the present age of Samay be x years and the present age of Sumit be y years. It is given that Samay is older than Sumit by 30 years. Thus, according to this condition, we have x – y = 30

37 After 5 years, the age of Samay will be(x + 5) years and the age of Sumit will be (y + 5) years. However, it is given that after 5 years, Samay will be thrice as old as Sumit. Thus, we have x + 5 = 3 (y + 5)  x + 5 = 3y + 15  x – 3y = 10 Now, we obtain the pair of linear equations as follows: x – y = 30 ____________ (1) x – 3y = 10 ____________ (2) We can find the present ages of Samay and Sumit by solving the above equations for variables x and y. We know three methods to solve a pair of linear equations. (i) Substitution method (ii) Elimination method (iii) Graphical method Now, we will follow another method to solve linear equations in two variables. This method is known as Cross-multiplication method. Firstly, let us discuss this method and after that we will solve the above equations for x and y. So, go through the following video to understand the cross-multiplication method and how it is applied to solve the above given equations. In this way, we can solve a pair of linear equations in two variables by cross-multiplication method. We could solve these linear equations in two variables by other methods also. But why should we use cross-multiplication method to solve linear equations? The cross-multiplication method is easier as compared to the other methods if the coefficients of variables in the equations are large numbers or numbers that look complex. To understand this concept better, let us see the following examples.

Solved Examples Example 9.Solve the following pair of linear equations in two variables by cross-multiplication method. ax + by = a2 + b2 and x + y = 2a Solution: The above two equations can be written as ax + by – (a2 + b2) = 0 ________________ (1) x + y – 2a = 0 ________________ (2) From the above two equations, we obtain a1 = a, b1 = b, c1 = – (a2 + b2) and www.betoppers.com

10th Class Mathematics

38 (ii)

a2 = 1, b2 = 1, c2 = – 2a Now,

x y 1 = = b1c 2 - b 2 c1 c1a 2 - c 2a1 a1b 2 - a 2 b1

x  b ×  -2a  -1×  -  a 2 + b 2    

19.

y 1 = -  a + b   ×1-  -2a  ×a a ×1-1× b   

=

2

18.

2

(i)

x y 1 = 2 2 = 2 2 2 -2ab + a + b -a - b + 2a a -b

x

 a  b

2



y 1  2 a b ab 2

x Therefore,

a  b

2

1  a  b and

(ii)

For which value of k will the following pair of linear equations have no solution? 3x + y = 1 (2k – 1) x + (k – 1) y = 2k + 1 Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9 3x + 2y = 4 Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method: A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day. A fraction becomes

1 when 1 is subtracted from 3

1 when 8 is added 4 to its denominator. Find the fraction. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test? Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars? The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

the numerator and it becomes

y 1  2 a b ab 2

x

a  b ab

(iii)

2

2

and y 

 x = a – b and y =

a b ab

2

 a  b  a  b  ab

 x = a – b and y = a + b

(iv)

Formative Worksheet 16.

Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method. (i) x – 3y – 3 = 0 (ii) 2x + y = 5 3x – 9y – 2 = 0 3x + 2y = 8 (iii) 3x – 5y = 20 (iv) x – 3y – 7 = 0 6x – 10y = 40 3x – 3y – 15 = 0 17. (i) For which values of a and b will the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 (a – b) x + (a + b) y = 3a + b – 2 www.betoppers.com

(v)

Conceptive Worksheet Solve the following systems of equaiton by crossmultiplication method: 22. 2x + 3y = 20 2x + y = 4 23. 3x – 4y =10 4x + 3y = 5 24. x + 4y + 9 = 0 3y = 5x – 1

Pair of Linear Equations in Two Variables 25.

26. 27.

7.

2x 3y   17 3 5 3x 2y   19 4 3 4x – 0.5 y = 12.5 ax – ay = 2; (a – 1)x + (a + 1) y = 2 (a2 + 1)

Equations reducible to a pair of linearquations in two variables Consider the following equations.

5 x = y; x = 21y + 15; m + 20n = –13; etc. 7 We can see that the above equations are linear equations in two variables. However, there are many equations, which do not appear to be linear equations at first glance such as

2 3 1 1   20;   5;etc x y x y

Solved Examples Example10.Solve the following pair of linear equations.

5 1  7 x y

2 3   4 x y

Solution:

5 1   7 ––––––––––– (1) x y

2 3   4 ––––––––––– (2) x y From equation (1), we can write,

1 1 5     7 ––––––––––– (3) x y From equation (2), we can write,

1 1 2    3    4 ––––––––––– (4) x y Let

1 1  a and  b x y

Now, equation (3) becomes 5a + b = 7 ___________ (5) and equation (4) becomes 2a – 3b = – 4 ___________ (6)

39 On multiplying (5) by 3, we obtain 15a + 3b = 21 ___________ (7) On adding equations (6) and (7), we obtain 17a = 17  a=1 On putting the value of a in equation (5), we obtain 5×1+b=7 5+b=7  b=7–5=2  Since

1 =a = 1 x



x=1

and

1 =b=2 y



y

Thus,

x = 1 and y 

1 2 1 2

Example11.Solve the following pair of linear equations.

2 5 3 10  7  9 xy xy xy xy Solution: The given equations can be written as

 1   1  2   5   7 ––––––––––– (1) xy xy and

 1   1  3   10    9 ––––––––––– (2)  xy  xy Let

1 1  a and b xy xy

Thus, the equations become 2a + 5b = 7 _________ (3) and, 3a + 10b = 9 _________ (4) Now, on multiplying equation (3) by 2, we obtain 4a + 10b = 14 _____________ (5) On subtracting equation (4) from equation (5), we obtain a = 14 – 9  a=5 Putting the value of a in equation (3), we obtain 10 + 5b = 7  5b = 7 – 10



b=

3 5

www.betoppers.com

10th Class Mathematics

40

Now,

8x  7y  15 xy 2x  4y  5xy 10 2  4  vii  xy xy 1 1 3    viii  3x  y 3x  y 4 15 5   2 xy xy 1 1 1   2  3x  y  2  3x  y  8

1 1  a and b xy xy

 xy

1 5 and x  y   5 3

On adding these two equations, we obtain

2x 

22 15

11 1 and, y  x  15 5 11 1  y  15 5 11  3 14  y  y 15 15 11 14 and y   Thus, x   15 15 x



Formative Worksheet 20.

21. (i)

(ii)

Solve the following pairs of equations by reducing them to a pair of linear equations:

1 1  2 2x 3y

i

2

 ii 

x



3 y

2

4  3y  14 x

 iii 

1 1 13   3x 2y 6 4 x

 iv 



9 y

 1

3  4y  23 x 5 1  2 x 1 y  2

 v

7x  2y 5 xy

 vi 

6x  3y  6xy

6 1  1 x 1 y  2

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(iii)

Formulate the following problems as a pair of equations, and hence find their solutions: Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Conceptive Worksheet Solve the following pairs of equations by reducing them to a pair of linear equations: 28. 29. 30.

31. 32.

4 3  3y  14 ,  4y  23 x x 5 1 6 3   2,  1 x 1 y  2 x 1 y  2 7x  2y 5 xy 8x  7y  15 xy 6x + 3y = 6xy 2x + 3y = 5xy

10 2  4 xy xy 15 5   2 xy xy

Pair of Linear Equations in Two Variables

Summative Worksheet 1.

41 2.

Which graph shows the point of intersection of the equations 4x + 6y = 12 and –2x + 3y = 6?

Which graph shows the point of intersection of equations –3x – 5y = 10 and –2x + 5y = 5?

(A) (A)

(B) (B)

(C)

(C)

(D)

(D)

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10th Class Mathematics

42 3.

Which graph shows the point of intersection of 4. the equations 3x – 2y = –11 and 2x + y = 2?

Which graph shows the point of intersection of the equations 2x – 5y = 16 and x + 2y = –1?

(A)

(A)

(B) (B)

(C)

(C)

(D)

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(D)

Pair of Linear Equations in Two Variables 5.

6.

If the polynomial x3 + ax2 + bx –  8  is  exactly divisible by x 2  + x –  2  and  the  polynomial  x3 + 4x2 – cx + b is divisible by x + 1, then what is the value of the expression a + b + c? (A) 2 (B) 4 (C) 11 (D) 13 If the system of linear equations, ax – 4y –10 = 0 and  2x 

7.

8.

9.

10.

11.

12.

43 13.

14.

b y  5  0 , represents a pair of 3

parallel lines, then what is the value of ab? (A) 6 (B) 8 (C) 12 (D) 24 A motor boat goes 112 km upstream and 96 km downstream in seven hours while in three hours, the boat goes 28 km upstream and 64 km downstream. If x and y represent the speed of the motor boat and the speed of stream respectively, then which pair of linear equations represents the given situation? (A) x + y = 28 and x – y = 32 (B) x + y = 32 and x – y = 28 (C) x + y = 16 and x – y = 36 (D) x + y = 36 and x – y = 16 If the lines representing the equations kx + ay – (k +  2)  =  0  and  3x + by –  5  =  0  coincide,  then what is the value of k? (A) –5 (B) –2 (C) 3 (D) 6 Which of the following systems of equations represents a pair of parallel lines? (A) x – 2y = 6 and 3x – 6y = 2 (B) 2x + 3y + 5 = 0 and 3x – 2y – 12 = 0 (C) x – 2y + 11 = 0 and 3x – 6y + 33 = 0 (D) 4x – 3y + 5 = 0 and 4x + 3y – 10 = 0 The ratio of a two-digit number to the sum of its digits is 7:1 and the digit at tens place is 4 more than the digit at ones place. What is the successor of the number? (A) 43 (B) 62 (C) 85 (D) 96 What is the solution of the pair of equations 35x – 10y = 515 and 11x + 21y = 355? (A) (8, 5) (B) (9, 1) (C) (17, 8) (D) (20, 12) If the system of linear equations –2x + (k2 + 2k – 1)y = 6 and kx + y = 2 has unique solution, then which of the following numbers can be the value of k? (A) –1 (B) –2 (C) 1 (D) 2

15.

16.

17.

18.

19.

For what value of k does the system of equations x – 2y = 6 and kx = 6y have no solution? (A) –2 (B) –3 (C) 2 (D) 3 Which of the following systems of linear equations represents intersecting lines? (A) x + y = 15 and x – y = 8 (B) x – y = 7 and y – x = –7 (C) x + 2y = 12 and 4y +2x = – 3 (D) 2x – 3y = 4 and 4x – 6y = 9 If the system of linear equations 3x – 5y – 2 = 0 and – 3lx + {3 (l + 1) + m}y + 8 = 0 has infinitely many solutions, then what is the value of (l + m)? (A) –6 (B) –9 (C) 6 (D) 9 If the system of linear equations 4(a + b) x – 2b y – 1 = 0 and 2bx + (a – b) y + 8 = 0 has no solution, then what is the value of a? (A) 0 (B) 2 (C) 4 (D) 6 If the system of linear equations and 4x – 3y –38 = 0 is inconsistent, then what is the value of k? (A) 1 (B) 3 (C) 5 (D) 7 Which of the following systems of linear equations has a unique solution? (A) x – y = 5 and 2x – 2y = 10 (B) 3x + 4y = 11 and 6x – 8y = 7 (C) 2x + 6y = 5 and x + 3y = 3 (D) 3x + y = 15 and 12x + 4y = –1 If

xy 3x  4y  7 and  25 , then what is the xy xy

value of xy?

20.

(A)

1 7

(B)

1 9

(C)

1 12

(D)

1 18

If 1 is subtracted from the numerator of a fraction, then the fraction becomes

1 and if 6 is subtracted 4

from the denominator of the fraction, then the fraction becomes

1  . 2

What is the sum of the numerator and denominator of the fraction? (A) 11 (B) 15 (C) 21 (D) 23 www.betoppers.com

10th Class Mathematics

44 21.

22.

23.

24.

If Sonu can row 35 km downstream in 5 hours 29. and 21 km upstream in 7 hours, then what is the speed of rowing in still water? (A) 2 km/hr (B) 3 km/hr (C) 4 km/hr (D) 5 km/hr Which of the following equations is perpendicular to x-axis? (A) x = –4 (B) y = 5 30. (C) x + y = 5 (D) x – y = 2 If the system of equations 3x – ky + 11 = 0 and 15x + 10y + 17 = 0 represents a pair of parallel lines, then what is the value of k? (A) –2 (B) –5 (C) 2 (D) 6 The given figure shows a rectangle ABCD whose perimeter is 40 cm. This rectangle is partitioned into three congruent rectangles as shown in the 31. figure.

32.

25.

26.

27.

28.

What is the perimeter of each congr uent rectangle? (A) 20 cm (B) 24 cm (C) 28 cm (D) 20 cm If the equations x – 3y = 7, 3x – 3y = 15, and x + ky = 3 have a common solution, then what is the value of k? (A) –7 (B) –4 (C) 1 (D) 3 What are the respective coordinates of points where the lines x – y = 5 and x + y = 11 meet xaxis? (A) (–1, 0) and (1, 0) (B) (–5, 0) and (11, 0) (C) (4, 0) and (–3, 0) (D) (5, 0) and (11, 0) If 6x + 3y = 6xy and 2x + 4y = 5xy, then what is the value of x + y? (A) 1 (B) 2 (C) 3 (D) 4 What is the x-coordinate of the point of intersection of the lines 2x + y = 5 and 3x + 2y = 8? (A) 1 (B) 2 (C) –3 (D) –4

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33.

34.

35. 36.

If the interior angles of a triangle are 15x°, 15(x + y)°,  and  25y°  and  the  relation between x and y is given by 5x – 4y + 2 = 0, then what is the measure of the smallest angle of the triangle? (A) 30° (B) 35° (C) 40° (D) 45° For which value of k is the system of linear equations x + ky =  7  and  –3x +  8y =  10 inconsistent? (A) –1 (C) 

8 3

(B) 

3 8

(D) 

1 8

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differs by 30 years. Find the ages of Ani and Biju. One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II) [Hint: x + 100 = 2 (y – 100), y + 10 = 6(x – 10)] A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class. In a ABC, C  3B  3  A  B  . Find the three angles. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

Pair of Linear Equations in Two Variables

HOTS Worksheet 1.

2.

3.

5.

6.

7.

8.

9.

Ten passengers were sitting along with the boatman in a boat. The average weight of the 11 people was 61 kg. If the boatman is replaced by another boatman weighing 52 kg, the average 10. weight of the 11 people would dip by 2 kg. What is the weight of the former boatman? (A) 59 kg (B) 61 kg (C) 68 kg (D) 74 kg In an exam, Gaurav scored 50% of the total marks 11. but was failed by 10 marks. On the other hand, Samik scored 75% of the total marks and passed the exam since he had scored 12.5% more than the passing marks. What is the difference between Gaurav’s and 12. Samik’s scores? (A) 10 (B) 15 (C) 25 (D) 30 If    and    are  the  zeroes  of  the  quadratic 2

4.

45

polynomial x  + 6x + k such that      8 , then what is the value of k? (A) –7 (B) –6 13. (C) 2 (D) 8 A shopkeeper, on selling a shirt at 6% loss and a trouser at 10% profit, can earn a gain of Rs 92. On selling a shirt and a trouser at the profit of 8% each, he can earn a gain of Rs 144. What is the cost price of a shirt? (A) Rs 550 (B) Rs 800 (C) Rs 1, 035 (D) Rs 1,250 What is the point of intersection of the lines 5x + 4y = 16 and 7x + 4y = 8? (A) (–8, 7) (B) (–4, 9) (C) (4, –5) (D) (8, –6) What is the solution of the system of equations x + 3y = 4 and 4x + 7y = 1? (A) x = –5 and y = 3 (B) x = –2 and y = 7 (C) x = 1 and y = 1 (D) x = 5 and y = 4 Three years ago, Anju’s age was five times Jyoti’s 14. age. However, six years later, Anju will be twice as old as Jyoti. What is the present age of Anju? (A) 28 years (B) 23 years (C) 18 years (D) 15 years If the sum and the difference of two numbers are respectively 306 and 36, then what is the larger number? (A) (–8, 7) (B) (–4, 9) (C) (4, –5) (D) (8, –6)

If the system of equations 5x + ay + 7 = 0 and by – 12y + 21 = 0 has infinitely many solutions, then what are the respective values of a and b? (A) –4 and 15 (C) 3 and –5

(B) –3 and 5 (D) 4 and –15

If 5x – 7y = 14 and 5x + 7y = 56, then what is the value of x + y? (A) 10 (C) 20

(B) 15 (D) 26

If 3x – 4y = –2 and x + y = –3, then what is the value of x2y2? (A) 1 (C) 8

(B) 4 (D) 9

A shopkeeper gains Rs 115 on selling a shirt at 10% loss and a trouser at 20% profit whereas on selling a shirt at 20% profit and a trouser at 10% loss, he gains Rs 10. What is the total cost of a shirt and a trouser? (A) Rs 1, 125 (C) Rs 1, 250

(B) Rs 1, 210 (D) Rs 1, 365

A garden is in the shape of a rectangle with square shaped areas attached to two of its adjacent sides. The area of the rectangle is 2 000 m2 and the total area of the garden is 6 100 m2.

The lengths of the sides of the rectangle are (A) 10 m and 200 m (C) 25 m and 80 m

(B) 20 m and 100 m (D) 40 m and 50 m

If the pair of linear equations x – ay + 6 = 0 and bx + y – 6 = 0 has infinite solutions, then what is the relation between a and b? (A) a + b = 0 (C) ab = –1

(B) a – b = 0 (D) ab = 1

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10th Class Mathematics

46 15.

Which graph represents the solution of the pair of equations 3x – 4y + 6 = 0 and 4x – 3y + 1 = 0?

16.

17.

For what value of a does the pair of equations 3x + ay + a + 1 = 0 and a2x – 9y + a – 3 = 0 has infinite solutions? (A) 12 (B) 4 (C) –3 (D) –10 For what condition does the pair of equations – 3x + py + 10 = 0 and 4x + 4y + 17 = 0 represents a pair of intersecting lines? (A) p  3 (B) p  3

(A)

18.

(B)

19.

(C) 20.

(C) p 

Consider the following pair of linear equations in two variables. 5x + 8y = 59         2x + 3y = 23 A student tries to solve these equations by the elimination method and follows the given steps. Step I: [ 5x + 8y = 59 ] × a Step II: [ 2x + 3y = 23 ] × b The student applies this method to eliminate variable x. The equations he got in steps I and II are (i) and (ii) respectively. Consider the following statements. (i) The values of x and y are 7 and 3 respectively. (ii) The values of a and b are 5 and 2 respectively. (iii) Equation (i) is 2x + 15y = 25 (iv) Equation (ii) is 10x + 15y = 115 With respect to the given equations, which two statements are correct? (A) (ii) and (iii) (B) (i) and (iii) (C) (i) and (iv) (D) (ii) and (iv) Which of the following pairs of equations represents a pair of parallel lines? (A) 7x + 2y – 17 = 0 (B) 4x + 3y + 7 = 0 (C) 3x + 5y + 17 = 0 (D) 6x + 15y + 21 = 0 21x + 6y – 34 = 0 8x + 6y + 14 = 0 x + 2y + 7 = 0 4x + 10y + 14 = 0 Which of the following pairs of equations has a unique solution? (A) 3x + 10y – 27 = 0 (B) 2x + 3y – 14 = 0 (C) 5x + 10y – 24 = 0 (D) 4x + 5y – 12 = 0 6x + 20y – 81 = 0 7x + 4y – 40 = 0

6x 

2x + 4y – 27 = 0 21.

(D)

15y  18  0 2

Which of the following relations is correct with respect to two parallel lines a 1x + b1y + c1 =  0 and a2x + b2– y + c2 = 0? (A)

a1 b1 c1   a 2 b 2 c2

(C) a1b 2  a 2 b1 www.betoppers.com

3 3 (D) p   4 4

(B) a1 b1 c1 = a2 b2 c2

(D)

a1 b1 c1   a 2 b2 c2

Pair of Linear Equations in Two Variables 22.

23.

47

Anu purchased 3 bananas and 5 oranges for Rs 21, while Nishi purchased 4 bananas and 2 oranges for Rs 14. How much do one banana and one orange together cost? (A) Rs 2 (B) Rs 4 (C) Rs 5 (D) Rs 7 What is the solution of the pair of equations5x – 29. 7y = 7 and x + 2y = 3?

25.

26.

27.

36 s 7

(B)

(C)

72 s 7

(D)

54 s 7 90 s 7

If f (z) = 3z3 + mz2 – 4z + n and 2 and 3 are the roots of the equation f(z) = 0, then the value of m is

35 8 and y  17 17

(A) 

7 5

(B)

(B) x 

8 35 and y  17 17

(C) 

53 5

(D) 

(C) x 

15 13 and y  17 17

(D) x 

13 15 and y  17 17

27 5 83 5

If the polynomial x4 + ax3 + 3x – b has x2 – 1 as a factor, then (A) a = 3, b = 1 (B) a = –3, b = –1 (C) a = 3, b = –1 (D) a = –3, b = 1

On interchanging the digits of a two digit number, the new number is (8/3) of the original number. What is the original number? (A) 18 (B) 27 (C) 36 (D) 45 A boat, when sailing up the stream, takes 25 minutes to cover a distance of 2 km and takes 20 minutes to return. The speed of the stream is (A) 1.2 km/h (B) 1.1 km/h (C) 0.8 km/h (D) 0.8 km/h The present ages of John and Jamie are in the ratio 3:4. The ratio of their ages will become 4:5 after five years. Jamie’s present age is (A) 10 years (B) 15 years (C) 20 years (D) 25 years The temperature of the liquids when the temperature of both the liquids is the same is

 180   C  7 

(B) 

 108   C  7 

(D) 

(A) 

(C)  28.

(A)

(A) x 

30.

24.

be equal to the temperature of the water?

 144   C  7 

 72   C  7 

A graph has been plotted between time elapsed and the change in the temperature of water and milk on heating. The graph corresponding to water is represented by 3x – 4y + 72 = 0 and that corresponding to milk is represented by 4x – 3y + 36 = 0, where x is in minutes and y is in °C. After how long, the temperature of the milk will www.betoppers.com

48

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10th Class Mathematics

Learning Outcomes By the end of this chapter, you will understand  Similar polygons  Area of similar triangles  Similar triangles

 Pythagoras theorem

 Result on similar triangles

 Converse of Pythagoras theorem

Chapter - 4

Quadratic Equations

 Criteria for similarity of two triangles

1. Introduction The equation is one of the most prominent ideas in mathematics, and is the center of foundation of mathematics itself. Beginning in the “Before Christ” era, the Babylonians were the first to have been recorded demonstrating the equation, circa 400 BC. Throughout the years, the history of mathematics has taken its fair share of changes. It has stretched across the world from the Far East, migrating into the Western Hemisphere. One of the most fundamental and key principles of mathematics has been the quadratic formula. Having been used in several different cultures, the formula has been part of the base of mathematics theory. Pacioli never actually devised his own method of solving the quadratic equation, but instead published a works dealing with the history of certain aspects of mathematics, including idea of all the great mathematicians. Overall, the quadratic formula has actually shaped civilization into the way it is today. The Babylonians used it to irrigate their land, divide out funds to pay workers, and even mobilize armies. The formula to them was not a way of learning or teaching, but a way of life. Before we know further, let us know some pre-requisites of the chapter.

Degree of polynomial Polynomial of one variable: In the case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial. Example: i) 2009z2009 + 2008z2008 + 2007z2007 + ….. + z2 +z+ 1 is a polynomial in z of degree 2009. ii) 10x5 + 2x4 + 3x3 + 4x2 + 5x + 6 The term with the highest power of x is10x5. The degree of the given polynomial is ‘5’.

Polynomial of several variables : In the case of polynomials in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial. Example: 257a2b5c7 – 342 a3 b4 c2 + 532 a5b3c2 – 198 ab9c8 the highest sum of the variables is 18.  The degree of a polynomial is 18 in a, b and c. 2x2yz + 5x2y2z2 – 7x3y2z2 – x3y3z3 The sum of the powers of the variables in the terms of the above polynomial are 4, 6, 7 and 9 respectively. The highest power is 9. Thus, the degree of the given polynomial is 9.

Polynomial

Quadratic Polynomial

It is an expression in which powers of variables are non-negative integers. The general form of a polynomial is anxn + an–1xn–1 + an–2xn–2 + - - - - +a1x+ a0 where an, an–1, an–2 - - - - a0 are real numbers and n is a positive integer.

A polynomial with degree two is called a quadratic polynomial. The general form of a quadratic polynomial is ax2 + bx + c, where a (compulsory condition) ≠0 and a,b,c  R

Examples: 2009x3 + 2008 x2 + 2007x + 2006 1 10 x  2x 5  0.25x  2009 2

Example: x2 + 2, 2x2 + 3x – 4, x 2  2 x ,

3 x2  x  7 etc. We denote quadratic polynomial by p(x), q(x), f(x) etc.

10th Class Mathematics

50

Zeros of a quadratic polynomial Let us now consider a quadratic polynomial in x, say P(x) = x2 + x – 2; we randomly substitute some real values of x and find the corresponding values of P(x) x P(x) –2 0 –1 –2 0 –2 1 0 2 4 We observe that P(x) is zero only when x is –2 and 1 and these values are called zeros, roots or solutions of the polynomial. Definition: The values of a variable, for which a polynomial p(x) becomes zero, are called zeros of a polynomial. If a is zero of a polynomial p(x), then p(a) = 0.

Points to remember: i) It is a fact that every quadratic polynomial can have at most two zeros. Forexample, x2 + 5x + 6 = 0 there are two zeros and they are –2 , –3 ii) Some quadratic polynomials do not have any zero. That is to say, there is no real value of x which makes the value of the polynomial zero. For example, consider the quadratic polynomial x2 + 1. There is no real value of x which makes x2 + 1 zero, since for every real value of x, x2  0 which implies that x2 + 1 > 0. iii) Similarly, some quadratic polynomials have only one zero, i.e., there is only one real value of x which makes the value of the polynomial zero. For example,consider the quadratic polynomial. x2 – 2x + 1.There is only one value of x(x = 1), for which x2 – 2x + 1 is zero.

2. Quadratic Equation A second degree equation is called a Quadratic Equation. or An equation whose highest power of a variable is two is known as a Quadratic Equation. The general form of a quadratic equation is ax 2 + bx + c = 0, where a,b,c  R a  0 (compulsory condition). Since x  0, quadratic equations, in general, are of the following types: i) b = 0, c  0, i.e., of the type ax2 + c = 0

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Example: 5x2 = 125 ii) b  0, c = 0, i.e., of the type ax2 + bx = 0 Example: 2x2 + 3x = 0 iii) b = 0, c = 0, i.e., of the type ax2 = 0 Example: 6x2 = 0 iv) b  0, c  0, i.e., of the type ax2 + bx + c = 0 Example: x2 + 5x + 4 = 0

Solving a quadratic equation We already know how to factorise a quadratic polynomial. We shall now learn how to apply the method of factorisation to solve a quadratic equation. Basically, there are two methods of finding the roots of a Quardratic Equation. 1) Factorization method 2) Formula method (Direct method) Factorization method: Let us consider a quadratic equaiton, x2 + 6x + 5 = 0 Step 1: Compare the given equation with the general form and identify the terms a, b, and c. By comparing x2 + 6x + 5 = 0 with ax2 + bx + c = 0, we get, a = 1, b = 6 and c = 5 Step 2: Multiply the constant with coefficient of x2 i.e., a × c. Here a × c = 5 × 1 = 5. Step 3: Split the product of a and c into as many factors as possible. 5 = 5 × 1 and 5 = –5 × – 1 Step 4: Choose those set of factors whose sum is equal to coefficiant of x (b). Here, b = 6 and 5 + 1 = 6.  The chosen factors are 5 and 1. Step5: Split the middle term into two terms whose coefficients are same as that of the chosen factors. Here, x2 + 6x + 5 = x2 + 5x + x + 5 = 0 Step 6: Factorise the quadratic polynomial by equating it to zero and find the values of variable(x). x2 + 6x + 5 = 0  x2 + 5x + x + 5 = 0  x (x + 5) + 1(x + 5) = 0  (x + 1) (x + 5) = 0 Either x + 1 = 0 or x + 5 = 0

 if a  b  0 then,either a  0or b  0  i.e., x = –1 or x = –5  The roots of the quadratic equation are, x = –1 or x = – 5

Quadratic Equations

51

Formula method: It is not always convenient to factorise the quadratic equation. Let us derive the quadratic formula which is useful for the solving of all quadratic equations.

Formative Worksheet 1.

Statement:

3.

x  3 2x 1   0. 2x  7 x  3 ii) 12x2 + 36a2 = 43ax. Solve the following quadratic equations by formula method. i) 4x2 – 7x – 15 = 0. ii) 9a2b2x2 – 16abcdx – 25c2d2 = 0. Solve for x (3x + 1)(x – 1) = (5x – 3) (2x – 3).

4.

If

i) Solve

If ax2 + bx + c = 0 is a quadratic equation, then the -b ± b 2 - 4ac , 2a Proof:Consider the quadratic equation, ax2 + bx + c = 0, where a,b,c  R a  0 ax2 + bx + c = 0 Dividing throughout by a, we get,

roots are x =

b c x  x 0 a a 2

2.

5.

b b2 b2 c x 2  2  0 a a 4a 4a We can rewrite the equation as

6.

x2 

2

2

2

2

b   b  c   x       0 2a   2a  a  2

b  b2 c  b 2  4ac   x     2a  4a 2 a  4a 2   2

 x

 x=

b b  4ac  2a 2a



+ 2 3



x 2  2 x 1

=

2 , 2 3

7.

x2 + x  6  x + 2 = then find the value of x. I f

x 2  7 x +10 , x ∈ R ,

1  2 1   If 6  x + 2   25  x   +12 = 0 , then find the x x    value of x . Find the number of solutions of the quadratic equation

x +1  x  1 = 4 x  1 .

3x x 1  =2 x 1 3x 1 9. Solve: 5x–1 + 5–x = 1 . 5 x 2  8 x  80 10. Prove that the value of must not lie 2x  8 between –8 and 8, if x be real. 8.

Solve:

11. If x = 2 + 2 + 2 + 2 + ... , then find the value of x. 2

b  b2  4ac b  b2  4ac and are 2a 2a the roots of the quadratic equation . Generally, roots of quadratic equations are denoted  x=

by  and  .

12. If x 3 = 2 , then find the value of x. 13. The length of a hall exceeds its breadth by 8m and its area is 180m2. Find the dimensions of the hall. 14. A man walks a distance of 48 km in a given time. If he walks 2km an hour faster, he will perform the journey 4hours before. Find his normal rate of walking.

Conceptive Worksheet 1.

2

-b + b - 4ac -b - b - 4ac ,β = 2a 2a Here b2 – 4ac is called discriminant and it is denoted by D or  . Here, α =

x 2  2 x +1

 2  log x 1

b  b2  4ac 2a

2

2 + 3 

then find the value of x .

b 2 , we get By adding and subtracting 4a 2

c  b   b   b  x2  2   x         0 a  2a   2a   2a  We can recognise the first three terms as the formula, a2 + 2ab + b2 = (a + b)2

Solve the following quadratic equations by factorisation method.

Solve the following quadratic equations by factorisation method. i) abx2 – (a + b)x + 1 = 0 ii) 3x2 – 11x + 10 = 0 www.betoppers.com

10th Class Mathematics

52 2.

Solve the following quadratic equations by formula method. i) 3a2x2 + 8abx + 4b2 = 0 ii) 9a2 – 9 (a + b)x + [2a2 + 5ab + 2b2] = 0

3.

 1 2 2  22 2 + x   x  9 Solve: 3   + 4 2  x2 + 3   x +9

4.

If

Writing the Quadratic equation Using its roots Let  and  are roots of a quadratic equation then the quadratic equation formed (Constructed) by

   = 7. 

using its roots  and  is x 2       x    0 x2 – (sum of roots) x + (Product of Roots) = 0 Example 1: Let us find the quadratic equation whose roots are 3 and – 2. The required quadratic equation is

2 x + 9  x  4 = 3 , then the find the value of

x. 5. 6. 7. 8.

3x  1

x 2   3   2   x   3  2  = 0

3x  1 3x  1 2 Solve for x : 9x4 – 325x2 + 36 = 0 Solve for x : 3x + 2 + 3–x = 1018.

x2 – x – 6 = 0. Example 2: Find the quadratic equation whose

3 x 2 7  22 . Solve for x : 5 x 3 3

We have sum of roots      = 4, and

Solve:

= 1+

roots are 2  3 and 2  3 .

Product of roots  = 1;  The required quadratic

3. Roots of Quadratic equations

x2 – 4x  1 = 0 . Points to remember: i)

b  b 2  4ac b  b 2  4ac , β 2a 2a

b  b 2  4ax b  b2  4ac , β , 2a 2a we have by addition ,



x f  0 k iv) the equation whose roots are equal but opposite in sign is f(–x) = 0 i.e., the equation whose roots are ,   is f(–x) = 0

  

 

Nature of the roots of quadratic equations The roots  and  of the quadratic equation ax2 + bx + c = 0 are given by 

2

( b  b  4ac )( b  b  4ac ) 4a 2

( b) 2  (b2  4ac) 4ac c  2  . a 4a 4a 2 c  product of roots   a

b  b2  4ac b  b 2  4ac ,  , 2a 2a

Here,   b2  4ac and is called Discriminant of quadratic equation. Then we have the following results:



www.betoppers.com

1 1 , if  

ii) the equation whose roots are   k ,   k is f(x – k) = 0 iii) the equation whose roots are k and k is

If α and β are the two roots of a quadratic equation ax 2 + bx + c = 0, then the two roots are

2

the equation whose roots are

1 f    0 , if c  0 i.e.,   0  x

Sum and Product of roots

b  b 2  4ac  b  b 2  4ac 2a 2b b   2a a b  sum of roots        a and by multiplication, we have

is

x       x    0

For what values of the variable (x) the quadratic equation gets satisfied are called roots of a quadratic equation. Let ax2 + bx + c = 0 is a quadratic equation, then the roots are genarally denoted by α and β , where



equation

2

i)

If   b 2  4ac  0 , then the roots are real and equal.

Quadratic Equations ii)

If   b 2  4ac  0 , then the roots are real and unequal.

iii)

If   b 2  4ac  0 , then the roots are imaginary.. Given that a, b, c are real.   0 if  and  are non-real complex numbers, conjugate to each other. Given that a, b, c are rational numbers,  is positive but not a perfect square if  and  are

iv) v)

conjugate surds like c  d and c  d . By applying these tests the nature of the roots of any quadratic equation can be determined without actually finding the roots.

Important Properties of Roots i)

ii)

iii)

If the roots are equal in magnitude but opposite in sign, then coefficient of x = 0 i.e., b = 0. Example: x2 – 4 = 0  x  2 clearly, b = 0 If unity is a root of ax2 + bx + c = 0 then the other root is c/a. Example: x2 – 4x + 3 = 0  x = 1, x = 3  Other root c/a = 3/1 = 3 Irrational roots of a quadratic equation with rational coefficients occur in conjugate pairs. I f p  q is a root of ax2 + bx + c = 0, then p  q

is also a root.

iv)

Exmaple: 2  3, 2  3 are the roots of x2 – 4x + 1 = 0 Imaginary or complex roots of a quadratic equation with real coefficients occur in conjugate pairs. If p + iq is a root of ax 2 + bx + c, then p – iq is also a root of the equation. Where i  1 Example: 3 + 5i, 3 – 5i are the roots of x2 – 6x + 34 = 0

Common roots of quadratic equations : i)

If both roots are common to two equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0, [where a1, a2  0 and (a1b2 – a2b1)  0]

a1 b1 c1 then, a = b = c 2 2 2 ii)

If only one root is common to two equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0, [where a1, a2  0 and (a1b2 – a2b1)  0] then (c1a2 – c2a1)2 = (b1c2 – b2c1)(a1b2 – a2b1 )

c1a2  c2 a1 and the common root = a b  a b 1 2 2 1

53

Formative Worksheet 15. Solve for x : ( x – 2) ( x – 4) ( x + 3) ( x + 5) = 120 16. Find the sum and product of roots of the following. i) x2 – 15x + 56 = 0 ii) 2x2 + 11x + 5 = 0 17. If  and  are the roots of x2 – px + q = 0, then find the value of i)  2  2 18. Find the

ii) 3  3 number of real

roots

of

3

1  1   x+    x+   0 x  x  19. Find the condition that the roots of the quadratic equation ax2 + bx + c = 0 should be i) reciprocals ii) one of the roots is double the other.

1 1 1 + = are equal in x+a x+b c magnitude but opposite in sign, then find the product of the roots. 21. If the roots of the equation ax2 + bx + c = 0 are of 20. If the roots of

the form

k +1 k+2 and , then find the value of b2 k k +1

– 4ac 22. If α, β are the roots of the quadratic equation 3x2 – 6x + 4 = 0, find the value of

α β 1 1  +  + 2  +  + 3αβ . β α α β 23. If a, b are roots of x2 + px + q = 0 and c, d are roots of x2 + rx + s = 0, then find (a – c) (a – d) (b – c) (b – d) interms of p,q,r,s 24. If the sum of the roots of ax2 + bx + c = 0 is equal to the sum of the squares of the roots, then find the relation between a, b and c. 25. If  ,  are the roots of the quadratic equation px2 + qx + r = 0, p  0 , then form the quadratic equation whose roots are i) 2, 2 iii)

1 1 ,  

ii)  2 , 2 iv)  2, 2

26. If α and β are the roots of the equation ax2 + bx + c = 0, then find the equation whose

1 1 1 roots are α + β , α + β . www.betoppers.com

10th Class Mathematics

54 27. Find the value of p for which the quadratic equation x2 + p(4x + p – 1) + 2 = 0 has equal roots. 28. If the equations x2 – px + q = 0 and x 2 – ax + b = 0 have a common root and the other root of the second equation is the reciprocal of the root of the first, then find the value of (q – b)2. 29

If  and  are the roots of the equation x2 + px + q = 0, form the equation whose roots are:

2 ;  30. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/h from its usual speed. Then find the usual speed of the plane. 31. If  and  be the values of x obtained from the equation m2 (x2 – x) + 2mx + 3 = 0 and if m1 and m2 be the two values of m for which  and  are   4 connected by the relation   then find the   3  2  2 and

m12 m22  . m2 m1 32. If ,  be the roots of the equation x2 + px + q = 0, then show that the quadratic   equation formed by and is qx2 – (p2 – 2q) x   + q = 0. value of

Conceptive Worksheet 9.

Find the value of m so that the equation 3x2 – 2mx – 4 = 0 and x(x – 4m) + 2 = 0 may have a common root 10. Find the sum and product of the following quadratic equations. i) 5x2 + 6x + 7 = 0 ii) –10x2 – 100x – 1000 = 0 11. Frame equation whose roots are 2  i 3 . 12. Find the nature of roots of (x – a)(x – b)+(x – b)(x – c)+(x – c)(x – a)=0. 13. Show that if p, q, r and s be real numbers, and pr = 2(q + s), then atleast one of the equations, x2 + px +q = 0 and x2 + rx + s = 0 has real roots. 14. Find k, if the roots of 5x2 + 7kx + 3 = 0 be the reciprocals of the roots of 3x2 + (8 – k)x + 5 = 0. www.betoppers.com

15. If ax 2 + bx + c = a'x 2 + b'x + c', when x = 183, 281 and 397 respectively, prove that a = a', b = b' and c = c'. 16. If α and β are the roots of x2 + px + 1 = 0 and

γ, δ are the roots of x2 + qx + 1 = 0, then find the value of  α - γ  β - γ  α + δ  β + δ  . 17. Two candidates attempt to solve a quadratic of the form x2 + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. The other starts with a wrong value of q and finds the roots to be 2, –9, then find the roots of the original equation. 18. If  ,  are the roots of the quadratic equation

x2 - 5 x + k = 0, then find the value of k such that α -β =1. 19. If  and  are the roots of the equation x2 + 7x + 12 = 0, then find the equation whose roots 2

2

are  α + β  and  α - β  . 20. If the roots of the equation (b – c)x2 + (c – a) x + (a – b) = 0 be equal, then find the value of b. 21. If the equations x2 + abx + c = 0 and x 2 + acx + b = 0 have a common root, then find the equation which satisfy the other root. 22. If the equation x2 + ax + b = 0 and x2 + bx + a =0 have a common root, then find the value of a + b. 23. If the ratio of the roots of x2 + bx + c = 0 and x2 + qx + r = 0 be the same, then find the relation between b, c, q and r. 24. If sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of the squares of their reciprocals, then show that bc2, ca2, ab2 are in A.P. 25. Find the number of real roots of the equation 2x4 + 5x2 + 3 = 0 26. If a, b, c  R and equation ax2 + bx + c = 0 and x2 + 2x + 9 = 0 have a common root, then find a : b : c. 27. If the equations k(6x2 + 3) + r x + 2x2 – 1 = 0 and 6k (2x2 + 1) + px + 4x2 – 2 = 0 have both the roots common, then find the value of 2r – p.

Quadratic Equations

55

4. Maximum & Minimum values of a Quadratic Expression

Note: a) When a, b, c  R, the quadratic expression ax 2 + bx + c has no maximum value when a > 0 and no minimum value when a < 0. b) When a, b, c  R, and a  0, the graph of the function f(x) = ax2 + bx + c is symmetric with

Value of the Expression: The numerical value obtained after substituting a particular value for the variable in an expression is called value of the algebraic expression. Example: 2x2 + 3x – 5 is an expression . b respect to the line x   . When x = 2 ; 2x 2 + 3x – 5 2a 2 = 2(2) + 3(2) – 5 = 8 + 6 – 5 = 9. Likewise, we want to find the Maximum and ormative orksheet Minimum values of the given Quadratic expression. 33. Find the maximum value of the following quadratic Here, the extreme values of a quadratic expression equations with real coefficients depend on the sign of the x  x R. i) 4x – x2 – 10. ii) 2 coefficient of x2. x 5 x + 9 i) If a  0 , then the expression ax2 + bx + c has 34. Find the minimum value of the following quadratic equations. b i) 3x2 + 4x + 1 ii) x2 – 2x + 10. a maximum value at x   and maximum value 2a 35. If (2x + 3) (x – 2) (x – 1) > 0, then find the real values of x. 2 4ac  b 36. x2 + 6x – 27 > 0, –x2 + 3x + 4 > 0 are both is . simulataneously true, then find the real values of x. 4a 37. Find the minimum value of ii) If a  0 , then the expression ax2 +bx + c has x x 3.7.2 log 7  32 log9  2log 2 3 b a minimum value at x   and minimum value 2a onceptive orksheet

F

W

C

is iii) iv)

4ac  b 2 . 4a 2

The minimum value of k   x  a   a  0  The maximum value of 2

k   x  a   a  0 

W

28. Find the maximum of the following quadratic equations. i) 7 + 10x – 5x2 ii) 12x – x2 – 32 29. Find the minimum of the following quadratic equations. i) 3x2 + 2x + 11 ii) x2 – 8x + 17,  x  R

Example: Find the maximum or minimum of the expressions. x2  1 i) 3x2 + 4x + 1 ii) 4x – x2 – 10 30. If f(x) = 2 for every real number x, then find x +1 i) Here, a = 3, b = 4 and c = 1, Since a = 3 > 0, the minimum value of f(x). the expression 3x2 + 4x + 1 has absolute minimum x +1 1 > . 31. Find the number of integral solutions of 2 2 x +2 4 at x =  and the minimum value is 3 2 32. For all x  R, x  2ax  10  3a  0, then find the 2 2 interval in which a lies. 4ac  b  4(3)(1)  (4)   1 . 33. Find the number of positive integral solutions of 4(3) 3 4a 3 4 x 2  3x  4   x  2  ii) Here a = -1, b = 4 and c = -10, Since 0. 5 6 2 x  5 2 x  7     a = –1 < 0, the expression –x + 4x –10 has absolute maximum at x = 2 and the maximum value is 34. Find the number of negative integral solutions of 2 4ac  b 2  4(1)( 10)  (4)  6. 4(1) 4a

x3  2

x 3  4

x 2 .2 x 1  2  x 2 .2  2 x1 . 35. Find the number of solutions of the equation

x  3  x  1. www.betoppers.com

10th Class Mathematics If the sign of expression is same as and the

56

5. Quadratic InEquation

ii)

discriminant     0 , then range of x is any real

In equation: The combination of two expressions with an inequality signs(,  and  ) is known as an inequation. Examples: 2x – 3 > 5 2x2 – 10x – 1000 < x2 – 1. 2009 x3 – 5x2 + 2x + 3  2010. 10x5 – 100x3 + 1000x  2011.

iii)

the discriminant     0 (equal roots) then the range of x is any real number except x   (where  is the equal root ) i.e., x  R   . The above conclusions can be used to solve quadratic inequalities like x2 – 10x + 21 > 0.

Formative Worksheet

Quadratic inequation: A second degree inequation is known as quadratic inequation. The general form of a quadratic inequation is ax2 + bx + c > 0 or ax2 + bx + c < 0

8 x 2 +16 x - 51 38. Solve: 2 x - 3 x + 4 > 3    39. Find the range of values of x for which

Examples:

x2  2x + 5 1 > 2 3x  2 x  5 2

x2 – 5x + 6 < 0, 5x2 + 125x + 625 > 0

Conceptive Worksheet

Range of Quadratic Inequation An inequation will have infinitely many solutions for which the inequation gets satisfied and the set of all these values is known as Range. Range depends on the sign of the expression and the sign of a in the quadratic inequation.

Range of Quadratic Inequation The following table enables us to find the range of x in the quadratic inequation. Discriminate ( ) & Nature of Roots

Sign of the expression ax2 + bx + c & sign of a

(i) (a)  > 0, real

Same

(b)  > 0, real

i)

Opposite

(ii)

 < 0, imaginary

Same

(iii)

 = 0, real equal

Same

Range of x x does not lie between ,  x >  and x >  x lies between the roots ,  i.e.  < x <  Any value of x i.e. x  R Any value of x except equal root (). i.e. x  R – 

36. Find the integral values of x (5x – 1) < (x + 1)2 < (7x – 3). 37. If b < 0, then the roots x1 and x2 of the equation 2x2 + 6x + b = 0, for what value of k the inequation

 x1   x2         k gets satisfied.  x2   x1  38. Solve: x2 – 6x + 8 > 0. 39. Solve: x2 – 6x + 5 < 0. 40. Find the real values of p for which the equation x2 – (p + 1) x + p2 + p – 8 = 0 has one root smaller than 2 and the other root greater than 2.

Summative Worksheet 2 5x + 9 2 3x + = 2 , then find the value x 1 x + 2 x +x2 of x.

1.

If

2.

5+2 6 + 52 6 = 10 , then find the value/s of x. If 2 x + 9 + x = 13, then find the value of x.

3. 4.

a) If the sign of the expression is same as that a and   0 , then x does not lie between the real

5.

roots  and  . b) If the sign of the expression is opposite to a and   0 , then x lies between the real roots  and  .

6.

     i.e.,   x   . www.betoppers.com

number (i.e., x  R ). If the sign of expression is same as that of a and





x 2 3





x 2 3

If x + 5 + x + 21 = 6 x + 40 , then find the value of x. 1   1  If 2  x 2 + 2  - 9  x +  +14 = 0 , then find the x x    value of x. If

x 1  x 13 + = , then find the value of x. 1 x x 6

Quadratic Equations

57

If α and β are the roots of the equation ax2 + bx + c = 0, then find the equation whose roots 1 1 2 2 are α + β , 2 + 2 . α β a b + = 1 has roots equal in 8. If the equation xa xb magnitude but opposite in sign, then find the value of (a + b) . 9. If two equations x2 + a2 = 1 – 2ax and x 2 + b2 = 1 – 2bx have only one common root, then find the relation between a and b. 10. Find the value of ‘a’ for which the equations x 3 + ax + 1 = 0 and x4 + ax2 + 1 = 0 have a common root. 11. The roots x1 and x2 of the quadratic equation x2 + px + 12 = 0, possess the following property : x1 – x2 = 1. Find the coefficient p. 12. If a and b are positive real numbers and each of the equations x2 + ax + 2b = 0, x2 + 2bx + a = 0 has real roots, then find the smallest possible value of (a + b). 7.

2 13. If 35  2 x  5 x  7  9 5x 2  7

14. Solve: 15.

16. 17. 18.

19.

17  3

2 x 3 , find x.

3x2 – 4x + 3x 2  4 x  6  18. If  and  are the roots of x2 + 2x – 3 = 0 and ,  are roots of x2 + 2x + 3 = 0, then find the vlaue of                   . Find the equation whose roots are 3  2 2 and 3  2 2 . If the equations x2 + 4ax + 3 = 0, 2x 2 + 3ax – 9 = 0 have a common root then find the value of a. The equations x 2 – cx + d = 0, d = 0, x2 – ax + b = 0 have one common root and the second has equal roots, then show that 2( a + d)=ac. If  and  be the roots of the equation

 2 2 2  x + px + q = 0, then find the value of in   terms of p and q. 20. Find the number of real roots of (6 – x)4 + (8 – x)4 = 16. 21. If x  R, then find the minimum value of x2 – 8x + 17. 22. Let  be the roots of x2 + bx + 1 = 0. Then find the equation whose roots are  1 1       and      .    

23. Find the condition that the equation 1 1 1 1    has real roots that are x xb m mb equal in magnitude but opposite in sign. 24. The difference of two numbers is 5 and their product is 84. Then find the numbers. 25. If twice the square of the numbers exceeds 4 times the number by 30, then find the numbers. 26. The perimeter of a rectangular room is 34 m and the length of the diagonal is 13m. Then find the dimensions of the room. 27. In a number of two digits, the digits in the ones place is the square of the digit in the tens place and the number formed by reversing the digits exceeds twice the number by 15. Then find the number. 28. For all x  R , x2 + 2ax + 10 – 3a > 0, then find the interval in which ‘a’ lies. 29. Find the real value of x for which 15x2 + 16x – 15 < 0. 30. Find the real values of x for which the expression –2x2 – x + 6 is positive or negative.

HOTS Worksheet 1.

2.

3.

4.

5. 6. 7.

8. 9.

The length of a hypotenuse of a right triangle exceeds the length of its base by 2 cm and exceeds twice the length of the altitude by 1 cm, then find the length of each side of the triangle (in cm) A two digit number is such that the product of its digits is 12. When 9 is added to the number, the digits interchange their places, then find the number. A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more for the same amount, each book would have cost Re. 1 less. How many books did he buy ? If

x 2  9 x + 20  x 2  12 x  32

=

2 x 2  25 x + 68 , then find the value of x.

5 + 2 6 

x 2 -3



+ 5-2 6



x 2 -3

= 10 , then find the

value/s of x. Solve: x  1  x  3 x  1 x  2  x  2 . Given that a and b are non zero real numbers such a that ab   a  b . Find all possible values of a + b b? Find the sum of all values of x that satisfy x  1  3 x  2  5 x  4  20 Consider the equations x 2 + px + qr = 0, x2 + qx + rp = 0 and x2 + rx + pq = 0. If every pair of given equations has a common root, then find the sum of the three common root. www.betoppers.com

10th Class Mathematics

58

10. If the roots of the equation ax2 + bx + c = 0 are of 21. Find the quadratic equation whose roots α and β are connected by the relation   1 the form and , then find the value of (a α + β = 2, αβ = p and  1  + b + c)2.  4λ 2 +15  1 α 1 β 11. Find the values of x of the quadratic equation + = 2 2  1+β 1 +α  4λ  1  2 3 2 3 2 3 3 3 abx + 4 ab - a +12 b x = 0 22. If one of the roots of the equation ax2 + bx + c = 0 is equal to the nth power of the 2 12. If α and β are the roots of ax + bx + c = 0, then





find the values of

 aα + b 

13.

14.

15.

16.

17.

-2

 aα + b 

-3

-3

+  aβ + b 

and

 

other, then show that, ac n

-2

+  aβ + b  If α an d β are the roots of x2 – p(x + 1) – c = 0, then find the value of α 2 + 2α +1 β 2 + 2β +1 + . α 2 + 2α + c β 2 + 2β + c The ratio of the roots of the equation ax2 + bx + c = 0 is same as the ratio of the roots of the equation px2 + qx + r = 0. If D1 and D2 are the discriminants of ax 2 + bx + c = 0 and px 2 + qx + r = 0 respectively then show that D1 p2  2. D2 q If the equations x2 – px + q = 0 and x 2 – ax + b = 0 have a common root and the other root of second equation are equal, then show that ap = 2(b + q). If α, β are roots of the equation λ (x 2 – x) + x + 5 = 0. If λ1 and λ 2 are the two values of λ for which the roots α, β are connected by the λ λ α β relation + = 4 . Find the value of 1 + 2 . λ 2 λ1 β α If α, β are the roots of the equation x 2 – px + q = 0 and α > 0, β > 0 , then the value of





k

23. If a  b  24. Solve:

2a 1  x 2 x  1  x2 ax

1 a2

 (a 

1 2 x)

1 n 1

 

 an c

1 n 1

b  0.

, find x. 

1

ax 1 a2

 (a 

1 2 x)

 a2.

25. Solve:

1 1  ( x  b)( x  c ) (a  c )(a  b) 

1 1  . ( a  c)(x  c) ( a  b)(x  b)

1 1 1 26. If the roots of x  p + x  q = are equal in r magnitude but opposite in sign, show that, p + q = 2r and the product of roots is equal to 1 2 p  q2 2 27. Form the quadratic equation whose roots are  a + b + a 2  b 2 and . a  b  a 2  b2





28. If ,  are the roots of a1 x 2  b1x  c1  0 and

  ,    are the roots of a2 x 2  b2 x  c2  0 2

1 1 1 4 b12  4a1c1  a1  p + 2 q . Find k. α 4 + β 4 is p + 6 q + 4q   . prove that 2 b  4 a c  a2  2 2 2 18. A two digit number is such that the product of its 29. If the ratio of the roots of ax2 + bx + c = 0 be digits is 12. When 9 is added to the number, the digits interchange their places, then find the number m n b   m : n prove that n m ac 19. If α and β are the roots of ax2 + bx + c = 0, then 30. A stream flows from A to B a distance of 30 km at 1 1 a 2km an hour and a man can row up and down in find the values of aα + b + aβ + b and 8 hours.Find the rate of the man in still water. 31. If  and  be the roots of equation   ax 2 + bx + c = 0,then find + . aα + b aβ + b 2 2      a    b   . 20. If α, β are the roots of x2 + px + q = 0 and also of       α β 2n n n n n x + p x + q = 0 and if , are the roots of x 32. If one roots of equation (l – m)x2 + lx +1 = 0 be β α double of the other and if l be real, then + 1 + (x + 1)n = 0, then show that n is an even 8 8 9 9 1) m  2) m  3) m  4) m  integer 9 9 8 8

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Quadratic Equations

59

33. Find the minimum value of 2log 7

x

3.7.

3

2log9 x

2

9.

log 23

7 times the square 2 root of the number are playing on the shore of a tank. The two remaining ones are playing. with amorous fight, in the water. What is the total number of swans? 35. A piece of cloth costs Rs. 200. If the piece were 5 m longer remained unchanged. How long is the piece and what is its original rate per metre? 34. O Girl! out of a group of swans,

IIT JEE Worksheet 1. 2.

x 2 x 1 is x 2  x 1 1) 1/3 2) 3 3) 1/2 4)1 If x – 2 is a common factor of the expressions

If x is real, then the minimum value of

bd  ca 1) –2 2) –1 3) 1 4) 2 The minimum value of 2x2 + x – 1 is 1) 1/4 2) 3/2 3) – 9/8 4) 9/4 2 x 2  ax  b and x  cx  d , then

3. 4.

If the equations x 2  ax  b  0 and x 2  bx  a  0

a+b= 1) –1 5.

6.

a  b

2) 2

have a common root then

3) 3

4) 4

If ‘3’ is a root of x 2  kx  24  0 it is also root of 1) x 2  5 x  k  0 2) x 2  kx  24  0 3) x 2  kx  6  0 4) x 2  5 x  k  0 If ,  are the roots of x 2  bx  c  0 and 2   h,   h are the roots of x  qx  r  0 then h= 1) b + q 2) b – q

3) ½ 7.

8.

b  q 2



If 2032 x  40 5 1) 

13 2

3) 

4 5

4) ½



b  q 

3 x2 2

then x = 2) 

12 13

4) 

5 4

If ,  are the roots of 9 x 2  6 x  1  0 then the equation with the roots 1/ , 1/  is 1) 2 x 2  3x  18  0 3) x 2  6 x  9  0

2 2) x  6 x  9  0 4) x 2  6 x  9  0

The equation formed by decreasing each root of ax 2  bx  c  0 by 1 is 2 x 2  8 x  2  0 then 1) a = – b 2) b = – c 3) c = – a 4) b = a + c

10. If (3 + i ) is a root of the equation x 2  ax  b  0 then a = 1) 3 2) –3 3) 6 4) –6 11. The maximum value of c + 2bx – x2 is 1) b2c 2)b2 – c 3) c + b2 4) c – b2 12. The minimum value of x2 – 8x + 17,  x  IR is 1) 17 2) –1 3) 1 4) 2 13. If ,  are the roots of the equation ax2 + bx +c = 0, then the quadratic equation whose roots are   :  is 1) a2 x2 + a(b – c) x + bc = 0 2) a2x2 + a(b – c) x – bc = 0 3) ax2 + (b + c)x + bc = 0 4) ax2 – (b + c)x – bc = 0 14. The minimum value of the quadratic expression x2 + 2bx + c is 1) cb2 2) c2b 3) c + b2 4) c – b2 15. If one root of the euation ax2 + bx + c = 0 is 3 – 4i, then a + b + c = 1) 40a 2) 36a 3) -20a 4) 20a 16. If x2 + 6x – 27 > 0; –x2 + 3x + 4 > 0, then x lies in the interval 1) (3,4) 2) {3,4} 3)  ,3  (4, )

4) (-9,4)

17. If  ,  are the roots of ax2 + bx + c = 0, then the equation whose roots are 2 +  , 2 +  is 1) ax2 + x(4a – b) + 4a – 2b + c = 0 2) ax2 + x(4a – b) + 4a + 2b + c = 0 3) ax2 + x(b – 4a) + 4a + 2b + c = 0 4) ax2 + x(b – 4a) + 4a – 2b + c = 0 18. If the ratio of the roots of x2 + bx + c = 0 and x2 + qx + r = 0 is same, then 1) r2c = qb2 2) r2b = qc2 3) rb2 = cq2 4) rc2 = bq2 19. If ,  are the roots of x2 –(a – 2)x –(a +1) = 0 where ‘a’ is a variable, then the least value of

 2   2 is 1) 2 2) 3 3) 5 4) 7 20. Two students while solving a quadratic equation in x, one copied the constant term incorrectly and got the roots as 3 and 2. The other copied the constant term and co-efficient of x2 as-6 and 1 respectively. The correct roots are 1) 3,–2 2) –3,2 3) –6,–1 4) 6,–1 www.betoppers.com

60

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10th Class Mathematics

Progressions Chapter - 5

Learning Outcomes By the end of this chapter, you will understand  Arithmetic Progression  Properties of geometric progression

1.

 Properties of arithmetic progressions

 Geometric mean

 Arithmetic mean

 Harmonic progression

 Geometrical progression

 Harmonic mean

Introduction Federation Internationale de Football Association (FIFA) is organizing the Football World Cup from ........ ... 1904, 1908, 1912,………,1994, 1998,2002, 2006, _________. Can you predict the year of the next FIFA ? It is2010. How is this obtained? It can be observed that each successive year is obtained by adding 4 to the previous year i.e., successive year = previous year + 4. Such a set of numbers is called a sequence.

Sequence An arrangement of numbers in a definite order according to some rule is called a sequence . It is also called a progression or pattern. Every element of a sequence is called a term. For example, in a cricket match number of balls bowled in an innings increase by 6 balls every over. 6, 12, 18, . . . .. balls.

2.

Arithmetic Progression Let us observe the following sequence. Friday will come after every 7 days. 7, 14, 21, 28 ……….. days. What rule does the above sequence follow? In the above sequence, numbers are obtained by adding a fixed number 7 to the preceding (Except first). Such kind of sequences are called Arithmetic sequences or Arithmetic progressions. A sequence in which every term is obtained by adding a fixed constant (positive or negative) to its preceding term except the first term is called an Arithmetic progression(A.P).

or

A sequence in which the difference between any two successive terms is constant is called an Arithmetic Progression(A.P).

Examples: Leap year comes after every four years (if we start from 2000). 2000, 2004, 2008, 2012 ……..

Terms related to A.P. Let us consider an A.P : 4, 7, 10, 13 ……… The terms of an A.P. are generally denoted by t1, t2, t3 ……… Here, the first term (t1) is generally denoted a First term (a) = 4 and common difference (d) = t2 – t1 = 7 – 4 = 3 t1 = 4 ; t 2 = 7 = 4 + 3 = t 1 + 3  t2 – t1 = 3

  

t2 – t1 = 3 t3 – t2 = 3 t4 = 13 =10 + 3 = t3 + 3

 t4 – t3 = 3

tn-1 = …………= tn – 2 + 3  tn-1 – tn – 2 = 3 tn = …………= tn – 1 + 3 t2 = 7 = 4 + 3 = t 1 + 3 t3 = 10 = 7 + 3 = t2 + 3 t4 = 13 =10 + 3 = t3 + 3 tn = …………= tn – 1 + 3

 tn – tn – 1 = 3  t2 – t1 = 3  t3 – t2 = 3  t4 – t3 = 3  tn – tn – 1 = 3

What do you observe ? Here, the difference between successive and preceeding is always constant i.e.. 3 What the above constant is called ? The constant is called common difference and it is denoted by d  In the sequence 4, 7, 10, 13,……. First term = t1 = a = 4 and Common difference = d = 3

10th Class Mathematics

62

General representation of AP nth term of A.P is tn = a + ( n–1)d The first term is ‘a’ and common difference is ‘d’. Note: Sum of 'n' terms of A.P Sn= and Sn 

n  a  l 2

Formative Worksheet 1. 2.

n [ 2a + ( n– 1) d] 2 3.

General Term of an A.P :

4. 5.

6.

7. 8. 9.

10. 11.

12. 13.

14. 15.

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1 5 9 13 , , , ,... ii) 0.6, 1.7, 2.8, 3.9, ….. 3 3 3 3 In which of the following situations, the sequence formed will form an A.P. ? i) Number of students left in the school auditorium from the total strength of 1000 students when they leave the auditorium in batches of 25. ii) The amount of money in the account every year when Rs 100 are deposited annually to accumulate at compound interest at 4% per annum. i)

where l is the last term.

Theorem : Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then, its nth term or general term is given by an = a + (n – 1)d Proof : Let a1, a2, a3, ..., an, ... be the given A.P. Then, a1 = a  a1 = a + (1 – 1)d --------- (1) Since each term of an A.P. is obtained by adding common difference to the preceding term. Therefore, a2 = a + d  a2 = a + (2 – 1)d------- (2) Similarly, we have, a3 = a2 + d  a3 = (a + d) + d  a3 = a + 2d  a3 = a + (3 – 1) d ---- (3) and a4 = a3 + d  a4 = (a + 2d) + d  a4 = a + 3d  a4 = a + (4 – 1)d -------- (4) Observing the pattern in equations (1), (2), (3) and (4), we find that an = a + (n – 1)d Remark : It is evident from the above theorem that General term of an A.P. = First term + (Term number – 1) × (Common difference) From the above, we can conclude that, a sequence in which each term differe from its preceding term by a constant is called an Arithmetic progression, written as A. P. This constant term is called common difference of the A.P. The general form of an A.P is a, a + d, a + 2d, a+3d, ....a + [n-1]d. Here ‘a’ is called first term ‘d’ is the common difference and a + [n-1]d is nth term or general term of A.

Write an arithmetic progression having 4 as the first term and –3 as the common difference. For the following arithmetic progressions write the first term and common difference.

If

2 , , k, 3

5k 8

are in A.P., find the value of

k. The first term of an A.P. is –7 and the common difference 5. Find its 18th term and the general term. i) Find the 20th term from the end of the A.P. –25, –20, –15, ….. 100. ii) Find the 6th term from the end of the A.P. 17, 14, 11, …..– 40. Which term of the sequence – 1, 3, 7, 11, …. is 95 ? How many terms are there in the sequence 3, 6, 9, 12, ….111 ? For what value of n, the nth terms of the following two A.P.s equal ? 23, 25, are 27, 29, ……. and – 17, –10, –3, 4, …. Which term of the A.P. 3, 15, 27, 39, ………. will be greater by 132 than its 54th term? Which term of the sequence

1 1 3 20, 19 , 18 , 17 ,... is the first negative term? 4 2 4 th The 10 term of an A.P. is 31, and 20th term is 71. Find the 30th term. The sixth term of an A.P. is 5 times the 1st one and the eleventh term exceeds twice the fifth term by 3. Find the 8th term. Find the sum of all natural numbers between 100 and 1000 which are multiples of 7. If m times the mth term of an A.P. is equal to n times its nth term, show that the (m + n)th term of the A.P. is zero.

Progressions 16.

17.

18.

63

If (m + 1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m + n + 1)th term. If pth , qth and r th terms of an A.P. are a, b, c respectively, show that (q – r) a + (r – p)b + (p – q) c = 0 In a garden bed there are 23 rose plants in the first row, twenty one in the second row, nineteen in the third row and so on. There are five plants in the last row. How many rows are there of rose plants ?

3.

Properties of progressions

1.

If a constant is added or subtracted from each term of an A.P., then the resulting sequence is also an A.P. with the same common difference. For example, if ‘k’ be added to each term of an A.P. a, a + d, a + 2d, .., then the resulting sequence (a + k), (a + k + d), (a + k + 2d), ...., is also an A.P. in which the first term is (a + k), which is different from the first term of the given sequence, but in both the sequences, the common difference is the same. Similarly, if ‘k’ be subtracted from each term of an A.P. a, a + d, a + 2d, .., then the new sequence (a – k), (a – k + d), (a – k + 2d), ..., is an A.P., whose first term is (a – k), which is different from the first term of the given sequence but the common difference is the same for both the sequences. If each term of a given A.P. (with common difference ‘d’) is multiplied or divided by a nonzero constant ‘k’, then the resulting sequence is also an A.P. with common difference kd or dk. For example, if each term of an A.P. a, a + d, a + 2d, ..., be multiplied by k ( k  0 ), then the new sequence ak, (a + d)k, (a + 2d)k, ..., is also an A.P., whose first term = ak and the common difference = kd. Again, if each term of the A.P. i.e., a, a + d, a + 2d, ..., is divided by k( k  0 ), then the new

2.

a a  d a  2d , ... is also an A.P., , , k k k whose first term = a/k and the common difference = d/k. If the corresponding terms of two A.P.’s be added or subtracted, the resulting sequence is also an A.P. sequence

3.

arithmetic

4.

For example, Let the two A.P’s be a1, a1 + d1, a1 + 2d1, ... and a2, a2 + d2, a2 + 2d2, ... Adding the corresponding terms of the above two A.P.’s, the new sequence a1 + a2, a1 + a2 + d1 + d2, a1 + a2 + 2d1 + 2d2, ... is also an A.P., whose first term = (a1 + a2) and the common difference = (d1 + d2). Similarly, subtracting the terms of the second A.P. from the corresponding terms of the first, we get the new sequence a1 – a2, a1 – a2 + d1 – d2, a1 – a2 + 2d1 – 2d2, ... which is also an A.P. having first term a1 – a2 and common difference d1 – d2. Note : The sequence, formed by multiplying or dividing the terms of one A.P. by the corresponding terms of another A.P., is not necessarily an A.P. For example, the sequence a1a2, (a1 + d1) (a2 + d2), (a1 + 2d1) (a2 + 2d2), ..., the terms of which are formed by the product of the corresponding terms of two A.P’s, is not an A.P. Similarly, the sequence, the terms of which are the quotients of the corresponding terms of two A.P’s, is also not necessarily an A.P. in general. If a1, a2, ..., an are in A.P., then a) a1 + an = a2 + an – 1 = a3 + an – 2 = ...

a r k  a r k , 0knr 2 If nth term of a sequence is a linear expression in n, then the sequence is an A.P. Three numbers a1, a2, a3 are in A.P. if and only if 2a2 = a1 + a3. b) a r 

5. 6.

Sum OF n terms of an AP : Let a be the first term, d, the common difference and l, the nth term of the given A.P. If Sn be the sum of n terms, then Sn = a + (a + d) + (a + 2d) + . + (l – 2d) + (l – d) + l ---------- (1) Rewriting Sn in the reverse order, we have Sn = l + (l – d) + (l – 2d) + .. + (a + 2d) + (a + d) +a ---------- (2) Adding (1) and (2) column wise, we get 2Sn =  a  l    a  l    a  l   . ..   a  l  (n times)  n  a  l 

 Sn =

 Sn =

n n  a  l   [a + a + (n – 1)d] 2 2 ( l = a + (n – 1)d) n [2a + (n – 1)d] 2 www.betoppers.com

10th Class Mathematics

64 Note 1. In the above formula, four quantities Sn, a, n, d are involved. If any three of these are known, the fourth can be found. If two of the quantities are known, then the other two can be related by an equation. 2. If Sn be the sum of n terms of an A.P., whose first term is a and the last term is l, then Sn =

30.

31.

n (a + l) 2 In any sequences Sn – Sn – 1 = (t1 + t2 + t3 + ... + tn –1 + tn) – (t1 + t2 + t3 + ... + tn - 1) = tn

3.

Formative Worksheet 19. 20.

21. 22.

23.

24. 25.

26.

27.

Find the four numbers in A.P. whose sum is 20 and the sum of whose squares is 180. Divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7 : 15. Find the sum of the series 5 + 7 + 9 + 10 + 13 + 13+ 17 + 16 + ……. to 40 terms. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Natural numbers have been grouped in the following way 1 ; (2, 3); (4, 5, 6); (7, 8, 9, 10); ……. The sum of the numbers in the nth group is Find the number of terms of the A.P. 63, 60.57, … so that their sum is 693. For any real numbers a and b, the pattern of numbers with nth term a + nb is always an A.P. What is the common difference ? What is the sum of the first 20 terms ? If the sum of p terms of an A.P. is the same as the sum of q terms, show that the sum of its (p + q) terms is zero. Sum of first p, q and r terms of an A.P. are a, b, c respectively. Prove that a b c  q  r    r  p    p  q  0 p q r

28.

29.

In an A.P. the first term is 2 and the sum of the first five terms is one-fourth of the sum of the next five terms. Show that the 20th term is –112. The sum of n terms of two arithmetic progressions are in the ratio 5n + 4 : 9n + 6. Find the ratio of their eighteenth term.

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A contract on construction job specifies a penalty, for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc; the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much does a delay of 30 days cost the contractor ? If the sum of the first ‘n’ natural numbers is S1 and that of their squares S2 and cubes S3, show that 9S22 = S3 (1 + 8S1).

Conceptive Worksheet 1. 2.

3. 4.

5. 6. 7.

8. 9. 10. 11. 12. 13. 14. 15. 16.

17.

Write an A.P. whose first term and common difference are –1.25 and – 0.25 respectively. Find the common difference of the following A.P. i) 8, 11, 14, 17, 20, ........ ii) 2, 0.5, –1, –2.5, –4, ...... Find a, b such that 18, a, b, –3 are in A.P. The first term of an A.P. is –7 and the common difference 5. Find its 18th term and the general term. Find the 6th term from the end of the A.P. 17, 14, 11, .... – 40. Which term of the sequence 4, 9, 14, 19,..... is 124 ? For what value of n is the nth term of the following two A.P’s the same? i) 1, 7, 13, 19, ...... ii) 69, 68, 67, ..... If the 8th term of an A.P. is 31 and the 15th term is 16 more than the 11th term. Find the A.P. If the 10th term of an A.P. is 52 and 17th term is 20 more than the 13th term, find the A.P. The 10th term of an A.P. is 52 and 16th term is 82. Find the 32nd term and the general term. How many number of two digits are divisible by 5 ? If five times the fifth term of an A.P. is equal to 8 times its eighth term, show that its 13th term is zero. If p times the pth term of an A.P. is equal to q times the qth term of the same A.P., then show that its (p + q)th term is zero. If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n). Split 69 into three parts such that they are in A.P. and the product of two smaller parts is 483. Four different integers form an increasing A.P. One of these numbers is equal to the sum of the squares of the other three numbers, then find the numbers. Find the sum of 25 terms of the A.P.

1 2 3 , , , ... 9 9 9

Progressions 18. 19. 20. 21. 22. 23.

4.

65 Step - 3: Assume the first term (t1) as ‘a’, the last term as tn and substitute the value of ‘a’ in tn to obtain the value of d. t1 = a = 3 ; t6 = 23.  a + (6 – 1)d = 23 [ tn = a+(n – 1)d] 3 + 5d = 23. 5d = 23 – 3. d = 4.  Step -4: Substitute the value of a and d in general form of A.P till we get the last term. Except the first and last, the middle numbers are the required A.M’s between the given numbers. a=3 a+d=3+4=7 a + 2d = 3 + 8 = 11 a + 3d = 3 + 12 = 15 a + 4d = 3 + 16 = 19 a + 5d = 3 + 20 = 23.  7, 11, 15, 19 are the A.M’s between 3 and 23. Example Insert 5 A.M’s between 9 and 51.Let a1, a2, a3, a4, a5 be A.M’s between 9 and 51. 9, a1, a2, a3, a4, a5, 51 which is in A.P t7 = 51 a + 6d = 51  a + 6d = 51  9 + 6d = 51  6d = 51 – 9  d = 7 a=9 a + d = 9 + 7 = 16 a + 2d = 9 + 14 = 23 a + 3d = 9 + 21 = 30 a + 4d = 9 + 28 = 37 a + 5d = 9 + 35 = 44 a + 6d = 9 + 42 = 51  16, 23, 30 37, 44 and 5 A.M’s between 9 and 51.

The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. Find the sum of all the three-digit numbers, which leave the remainder 2, when divided by 5. Find the number of terms of the A.P. 54, 51, 48,.... so that their sum is 513. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms ? If the sum of m terms of an A.P. be n and the sum of n terms be m, show that the sum of (m + n) terms is – (m + n). The sum of n terms of three arithmetical progressions are S1, S2 and S3. The first term of each is unity and the common differences are 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2.

Arithmetic mean If a, b, c are in A.P., the middle one is said to be the arithmetic mean of the other two. It is denoted by A.M. a, b, c are in A.P ; b – a = c – b 2b = a + c

or b 

ac 2

ac 2 If a, b, c are in A.P., then 2b = a + c  Examples: Arithmetic mean of 10, 20, 30 is 20. Arithmetic mean of 2000, 2010, 2020 is 2010.

 AM of a, b, c is b =

Method of Inserting the Arithmetic Mean Given two quantities, it is always possible to insert any number of terms such that the whole series so formed is an A.P. Such inserted terms are known as arithmetic means between the given quantities. Let us understand the method of insertion by considering the following example. Insert 4 Arithmetic means between 3 and 23. Step - 1: Assume the required number of A.M’s between the two numbers. a 1, a 2 , a3 , a4 are the A.M’s between 3 and 23. Step - 2: Insert these A.M’s between the numbers then the sequence so formed will be in A.P. and then find the total number of terms. Insert a1, a2, a3, a4 in between 3 and 23 3, a1, a2, a3, a4, 23 is in A.P.. It contains 6 terms.

Formative Worksheet 32. 33.

34. 35.

Insert 8 arithmetic means between –5 and 13. If a, b, c, be in A.P. and ‘m’ is the A.M between ‘a’ and ‘b’ and ‘n’ is the A.M between ‘b’ and ‘c’, then show that ‘b’ is the A.M between ‘m’ and ‘n’.

 1 1  1 1  1 1 If a    , b    , c    are in A.P., b c a c a b prove that a, b, c, are in A.P. If a, b, c, are inA.P., 1 , b c in A.P.

1 , c a

1 prove that is also a b

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10th Class Mathematics

66

General term of a GP :

Conceptive Worksheet 24.

30.

Insert six arithmetic means between 2 and 16. Also prove that their sum is 6 times the A.M. between 2 and 16. xa xa The A.M. between and is ............. x x There are n A.M.’s between 3 and 54 such that 8th mean : (n – 2)th mean = 3 : 5. Find n. If a, b, c are in A.P., show that (b + c)2 – a2 , (c + a)2 – b2, (a + b)2 – c2 are in A.P. If a, b, c are in A.P., prove that a b + c  b  c + a  c a + b  , , are in A.P.. bc ca ab Prove that a, b, c are in A.P. if and only if 1 1 1 , , are in A.P.. bc ca ab 5, 0.5, 0.05...... a = ?, r = ?, t7, = ? tn =

5.

Geometrical progression

25. 26. 27. 28.

29.



i)

ii)

A sequence (finite or infinite) of non-zero numbers in which every term, except the first one, bears a constant ratio with its preceding term, is called a geometric progression, abbreviated as G.P. Illustration : The sequences given below : i) 2, 4, 8, 16, 32, ... ii) 3, –6, 12, –24, 48, ... 1 1 1 1 1 , ,... iii) , , , 4 12 36 108 324 1 1 1 1 1 , , , , ,... iv) 5 30 180 1080 6480 v) x, x2, x3, x4, x5, ... (where x is any fixed real number),are all geometric progressions. The ratio of any term (i) to the preceding is 2. The corresponding ratios in (ii), (iii), (iv) and (v) are

1 1 2, , and x, respectively. The ratio of any 3 6 term of a G.P. to the preceding term is called the common ratio of the G.P. Thus, in the above examples, the common ratios are 2, –2,

Let a be the first term and r   0  be the common

1 1 , and 3 6

x, respectively. Note : In a G.P. any term may be obtained by multiplying the preceding term by the common ratio of the G.P. Therefore, if any one term and the common ratio of a G.P. be known, any term can be written out, i.e., the G.P. is then completely known. In particular, if the first term and the common ratio are known, the G.P. is completely known. The first term and the common ratio of a G.P. are generally denoted by a and r respectively. www.betoppers.com

iii)

iv)

ratio of a G.P. Let t1, t2, t3, ..., tn denote 1st, 2nd, 3rd, ..., nth terms, respectively. Then, we have t2 = t1r, t3 = t2r, t4 = t3r, ..., tn = tn -1r. On multiplying these, we get t2t3t4 ... tn = t1t2t3 ... tn – 1rn – 1  tn = t1rn –1, but t1 = a. General term = tn = arn –1. Thus, if a is the first term and r the common ratio of a G.P. then the G.P. is a, ar, ar2, ... arn –1 or a, ar, ar2, ... according as it is finite or infinite. If the last term of a G.P. consisting of n terms is denoted by l, then l = arn -1. Note : If a is the first term and r the common ratio of a finite G.P. consisting of m terms, then the nth term from the end is given by arm –n. The nth term from the end of a G.P. with the last term l and common ratio r is l/rn – 1. Three numbers in G.P. can be taken as a/r, a, ar, four numbers in G.P. can be taken as a/r3, a/r, ar, ar3, five numbers in G.P. can be taken as a/ r2, a/r, a, ar, ar2, etc... Three numbers a, b, c are in G.P. if and only if b/a = c/b, i.e., if and only if b2 = ac.

Sum of first n terms of a GP: Let a be the first term and r(  1) be the common ratio of the given G.P. If Sn denotes the sum of n terms, then Sn = a + ar + ar2 + ... + arn –1 ------ (1) Multiplying both sides of (1) by r, we get rSn = ar + ar2 + ar3 + ... + arn ------- (2) Subtracting (2) from (1), we get Sn – rSn = a – arn  (1 – r)Sn = a(1 – rn) 

1 rn Sn = a --------- (3) 1 r This result can also be written as rn 1 --- (4) r 1 If r = 1, then Sn = a + a + a + ... to n terms = na.

Sn = a

Note : If r < 1 numerically, use Sn = a and if r > 1 numerically, use Sn = a

1 rn 1 r

rn 1 . r 1

Progressions

67

Sum of an Infinite G.P. :

4.

Consider the G.P. a, ar, ar2, ... with |r| < 1. We know 1 rn a ar n   1 r 1 r 1 r Since |r| < 1, rn goes on decreasing (numerically) as n goes on increasing. Ultimately, when

that Sn = a

n  , r n  0 . Therefore, when n   , a 1 r In other words, the sum of an infinite number of terms of a G.P. with first term ‘a’ and common ratio ‘r’(|r| G b) the quadratic equation having a, b as its roots is x2 – 2Ax + G2 = 0 c) the two positive numbers are A ± A 2  G 2 .

Formative Worksheet 36.

128, – 96, 72…….. find a, r.

37.

The second term of a G.P. is is

38.

39. 40. 41.

25 and the 8th term 4

16 . Find the G.P.. 625

2 7 Find the value of x so that , x, are three 7 2 consecutive terms of a G.P. 1 1 Which term of the series   1  .... is 256. 4 2 The nth term of two G.M’s ‘5 + 10 + 20 + …..’ and ‘1280 + 640 + 320 …..’ are equal. Then find the value of n. The 4th term of a G.P. is the square of its 2nd term and 1st term is – 3. Determine its 7th term.

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45.

46.

47.

term. Which

3 . Find the G.P. and the 20th 16 term

49.

the

G.P. 2, 6, 3 2, 3 6 ..........is 243 2 ? Find the sum of these terms. If a is the A.M. of two numbers b and c and G1, G2 are two geometric means between b and c then the value of G13 + G23 equals The product of three numbers in G.P. is 1000. If we add 6 to its second number and 7 to its third number the resulting three numbers form an A.P. Find the numbers in G.P. In the G.P. 0.15, 0.015, 0.0015,….; find the sum of 8 terms and 20 terms. How many terms of the G.P. 3, needed to give a sum

48.

of

3 3 , ,.... are 2 4

3069 ? 512

Find the sum of the following sequence to n terms 9, 99, 999, 9999, …..99…9(n9s) Find the sum of n terms of the series 0.5 + 0.55 + 0.555 + . . . n terms

20 80 , ... 7 49

50.

Find the sum to infinity of the G.P.  5,

51.

The first term of a G.P. is 2 and the sum of infinity is 6. Find the common ratio. Value of y = (0.64)

52.

1 1 1  log0.25   2  3  .......upto   3 3 3  53.

54.

55.

56.

If the harmonic mean of two numbers is to their geometric mean as 12 : 13, prove that the numbers are in the ratio 9 : 4. The A.M. between two positive numbers exceeds the G.M. by 3/2 and the G.M. exceeds the H.M. by 6/5. The numbers are___________. If a, b, c are in G.P., then prove that a) a + b, b + c, c + d are in G.P. b) a(b – c)3 = d(a – b)3 A.M. between two numbers, whose sum is 100, is to the G.M. is 5 : 4. Find the numbers.

Progressions

69

Conceptive Worksheet 31.

th

The 6 and 13 terms of a G.P are respectively equal to 24 and

32.

34.

35. 36. 37. 38.

8.

3 . Find the G.P and the 20th term. 16

The second term of a G.P. is is

33.

th

25 and the 8th term 4

16 , find the G.P.. 625

512 Which term of the sequence 18, –12, 8, ... is ? 729 The product of three numbers in G.P. is 1000. If we add 6 to its second number and 7 to its third number the resulting three numbers form an A.P. Then the numbers in G.P. are Find the sum of 50 terms of the sequence 7, 7.7, 7.77, 7.777, ... Find the sum of an infinite G.P. whose first term is 28 and second term is 4 Verify that 10, –9, 8.1 ... is a geometric progression. Find the sum to infinity of the G.P. The sum of first two terms of a G.P is 5/3 and the sum to infinity of the series is 3. Then, find the first term

Harmonic progression A progression in which the reciprocals of the terms of a sequence are in A.P. is called a harmonic progression. It is denoted by H.P.

Example :

9.

Harmonic mean The harmonic mean between two quantities a and b will be x if a,x,b be in harmonical progression.

1 1 1  , , will be in A.P.. a x b 2

1 1 1 a b    x a b ab 2ab  H.M. ab

 x

n Harmonic means between a and b Let x1, x2,................xn, be ‘n’ harmonic means between a and b.  a, x1, x 2.........x n , b are in H.P..

1 1 1 1 1  , , ,........ , are in A.P.. a x1 x 2 xn b

1 1  Tn 2 of A.P.    n  1 d b a 1 1 ab   n  1 d    b a ab ab 1  d . ab n  1 1 1 1 1   T2   d,  T3   2d x1 a x2 a 

1 1  Tn 1   nd xn a On putting for d, we get 1 1 ab 1 nb  a   .  x1 a ab n  1  n  1 ab

1 1 1 1 1 1 , , ......and , , ...... are H.P.'s as 3 5 7 a a  d a  2d their reciprocals 3,5,7........ and a, a + d, a + 2d......... are in A.P.

———— (1) nb  a Interchanging a and b, we get

nth term of HP:

xn 

1 1 1 , , ..... then a a  d a  2d corresponding A.P. is a, a + d, a + 2d,......

Relations between A,G and H :

Tn of A.P. is a   n  1 d

ab 2ab ,G  ab,H  2 ab These three means possess the following properties.

If the H.P. be as

1 Tn of H.P. is a   n  1 d

 x1 

 n  1 ab

 n  1 ab na  b

———— (2)

Let A,G and H be arithmetic, geometric and harmonic means of two numbers a and b. Then,

A

In order to solve the questions on H.P., we should form the corresponding A.P. www.betoppers.com

10th Class Mathematics

70 Property (1) : A  G  H Proof : We have,

Conceptive Worksheet 39.

ab 2ab A ,G  ab and H  2 ab

 AG 

 AG

ab  ab  2



a b 2

40.



2

0

 a  b  2 ab  2ab  ab   a b ab  

2 ab a  b 0 ab  G  H ——— (2) From (1) and (2), we get A  G  H Note that the equality holds only when a = b.





The 7th term of an H.P. is

42.

1 . Find the 20th term and the nth term. 25 If a, b, c are in A.P, p, q, r are in H.P, ap, bq, cr are p r c a in G.P. Prove that r + p = a + c



Property (2) : A, G, H form a G.P., i.e. G2 = AH Proof : We have, 2 a  b 2ab AH    ab  ab  G  2 ab Hence, G2 = AH.



43. 44.

45.



46.

58. 59.

60.

1 7

An insect starts from a point and travels in a straight path 1 mm in the first second and half of the distance covered in the previous second in the succeeding second. In how much time would it reach a point 3mm away from its starting point. 1 1 1   3  2  3  ....upto   3 

log0.25 3

Find the value of y  0.64

48.

If A = 1 + ra + r2a + r3a + .....  and  A 1 

  a  A  B = 1+ r + r2 + r3 + ......  , then b  log  B1   B 

b

Formative Worksheet In a H.P. if the 3rd term is

g12 g 22   2A g 2 g1

47.

ab    A  2 and G  ab   

57.

If a, b, c are in A.P., b, c, d are in G.P., c, d, e are in H.P., prove that a, c, e are in G.P If S1, S2 and S are the sums of n terms, 2n terms and to infinity respectively of a G.P. then show that S1(S1 – S) = S(S1 – S2) If two geometric means between g1 and g2 and one A.M.A be inserted between two numbers, then show that

Property (3) : The equation having a and b as its roots is x 2  2Ax  G 2  0 Proof : The equation having a and b as its roots is x 2   a  b  x  ab  0  x 2  2Ax  G 2  0

1 and 12th term is 10

41.

———— (1)

G  H  ab 

If the 12th term of an H.P. be 1/5 and the 19th term be 3/22, then find the 4th term. If the first two terms of a H.P are 2/5 and 12 / 13 respectively. Then find the largest term.

49.

and the 7th term is

50. 1 then show that the 15th term is 1. 5 If x, y, z are in H.P., prove that log(x + 2) + log(x + z – 2y) = 2 log (x – z) If x, y, z are in H.P., then prove that x y z , , y  z  x x  z  y x  y  z are also in H.P 1.

b

b

a n 1  b n 1 may be a n  bn the harmonic mean between a and b. The A.M. between two positive numbers exceeds the G.M. by 3/2 and the G.M. exceeds the H.M. by 6/5. Then find the numbers.

Find the value of n such that

Summative Worksheet

Insert 6 harmonic means between

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2 2 . and 3 31

2.

Write an A.P. whose first term is 10 and common difference is 3. Determine k so that k + 2, 4k – 6 and 3k – 2 are the three consecutive terms of an A.P.

Progressions 3.

4.

5.

nd

71 st

The 2 , 31 and the last term of an A.P. are

3 1 1 7 , and - 6 , respectively. Find the first term 4 2 2 and the number of terms. Which term of the arithmetic progression 5, 15, 25,....... will be 130 more than its 31st term ?

18. 19. 20.

3 The 2nd, 29th and last terms of an A.P. are 7 , 1 4 3 respectively. Find the first term and the 4 number of terms: Determine the general term of an A.P. whose 7th term is – 1 and 16th term 17. Find the number of integers between 50 and 500 which are divisible by 7. and  6

6. 7.

8.

9.

10. 11. 12. 13.

14.

15.

16.

17.

If the mth term of an A.P. is

1 and the nth term is n

1 , find the (mn)th term. m If pth , qth and r th terms of an A.P. are a, b, c respectively, then show that (a – b) r + (b – c) p + (c – a) q = 0 Find four number in A.P. whose sum is 20 and the sum of whose squares is 120. If the numbers a, b, c, d, e form an A.P., then find the value of a – 4b + 6c – 4d + e. Find the sum of the series 4 + 12 + 20 + 28 + ...+ 100. Find the sum of first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively. In an A.P., if the 5th and 12th terms are 30 and 65 respectively, what is the sum of first 20 terms ? In an A.P., if the 12th term is – 13 and the sum of first four terms is 24. What is the sum of the first 10 terms ? Vijay saves Rs. 1600 during the first year; Rs. 2100 in the second year, Rs. 2600 in the third year. If he continues his savings in this pattern, in how many years will he save Rs. 38500 ? The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find the value of x.

21.

22.

23.

24.

25. 26. 27. 28.

29.

30.

31. 32. 33.

34.

1 2 3 , , , ... 9 9 9 Insert ‘4’ arithmetic means between 8 and 33. Given that n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b, where a, b  R. Suppose further that mth mean between these two sets of numbers is same, then the ratio a : b equals. A) n – m + 1 : m B) n – m + 1 : n C) m : n – m + 1 D) n : n – m – 1 1 1 1 , , If are in A.P. show that a2, b2, bc ca ab c2 are also in A.P. x 1 y , , ...... y x x3 a = ? r = ? t 7 tn = ? 1 The third term of a G.P. is 6 and the 7th term 4 the reciprocal of the third. Which term of this G.P. is unity? Find the four numbers forming a G.P in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18. 2 1 1 81 Find the sum of the series + + + ... + . 9 3 2 32 There are n A.M.’s between 3 and 54 such that 8th mean ; (n – 2)th mean = 3 : 5. Find n. 45 , then find the value of 3 + 3α + 3α 2 + ........ = 8 . Find the 100th term of the sequence 1 1 1 1, , , ,.... 3 5 7 Find three numbers are in H.P., sum of whose 1 reciporcals is 12 and their product is . 48 Find the sum of 25 terms of the A.P.

If a,b,c are in H.P., show that a b c , , are also in H.P.. b+c c+a a +b If a, b, c are in G.P. and ax = by = cz, show that x, y, z are in H.P. If a, b, c be in A.P. and b, c, d be in H.P., prove that ad = bc. The seventh term of a G.P. is 8 times the fourth term. Find the G.P. when its 5th term is 48. Hence, the required G.P. is 3, 6, 12, 24, ... Let a and b be two distinct positive numbers such that a > b. If the G.M between them is m (>1) times the H.M between them, then prove that a : b  m  m 2  1 : m  2 m2  1

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10th Class Mathematics

72 35.

If a1, a2, a3, ...., an be in A.P. and ai, > 0 for all i, then show that

13.

1 1 1 1 n 1    .....   a1a 2 a 2 a 3 a 3 a 4 a n 1a n a1a n

HOTS Worksheet If nth term of the pattern of numbers is a linear expression in n. Then, prove that the pattern of numbers, so formed, is an A.P. and the common difference of this A.P. is equal to the coefficient of n. In an A.P., prove that the sum of the terms equidistant from the beginning and the end is always the same and equal to the sum of the 1st term and last term. Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000 th terms ? Show that the product of the 2nd and 3rd terms of an A.P. exceeds the product of the 1st and 4th by twice the square of the difference between the 1st and the 2nd. Two A.P’s have the same common difference. The difference between their 100th terms is 111 222 333. What is the difference between their Millionth terms ? If the pth term of an A.P. is q and the qth term is p, prove that its nth term is (p + q – n). If a1, a2, a3, ..... an be an A.P. of nonzero terms,

1.

2.

3.

4.

5.

6. 7.

1 1 1 n 1 prove that a a  a a  ........ a a  a a 1 2 2 3 n 1 2 1 n 8.

14.

15.

16.

S =n A Given that n arithmetic means are inserted between two sets of numbers a, 2b and 2a, b, where a, b  R. Suppose further that mth mean between these two sets of numbers is same, then show that

17. 18.

a m  b n  m 1 Find the sum of the series (a + b)2 + (a2 + b2) + (a – b)2 + ... to n terms. Given that ,  are roots of the equation Ax2 –

19.

4x + 1 = 0, and ,  , are the roots of the equation Bx2 – 6x + 1 = 0, then the values of A and B such that , ,  and  are in A.P., are A) A = 3, B = 8 B) A = – 3, B = 8 C) A = 3, B=–8 D) none of these The sum to n terms of the sequence

between a and b, then show that

Find the number of integers between 100 and 1000 that are i) divisible by 7 ii) not divisible by 7 Obtain the sum of all the numbers in the first 1000 integers which are neither divisible by 5 nor by 2.

9.

10. 11. 12.

 x 8 If 1, log81 (3x + 48), log9  3   are in A.P. then 3  find the value of x If S1, S2, S3 be the sum of n, 2n and 3n terms respectively of an A.P., prove that S3 = 3 (S2–S1). 1 If the pth term of an A.P. be and qth term be q 1 . prove that the sum of the first pq terms must p be

1  pq  1 . p

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A thief runs away from a police station with a uniform speed of 100m/minute. After a minute a policeman runs behind the thief to catch him. He goes at a speed of 100m/minute in first minute and increases his speed 10 m each succeeding minute. After how many minutes, the policeman will catch the thief ? In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are n potatoes in the line (see figure). Each competitor starts from the bucket , picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in the bucket, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run ? Given two numbers a and b, let A denote the single A.M. and S denote the sum of n A.M.s

2

20. 21.

22.

2

2

1  2 1   3 1    x   ,  x  2  ,  x  3  ,..... x  x   x   If Sn denotes the sum of n terms of a G.P., then show that (S10 – S20)2 = S10(S30 – S20) The sum of the infinitely decreasing G.P. whose third term, the three times the product of the first and fourth terms and the second term form, an A.P. with the common difference equal to 1/8 is If a, b, c, d are in G.P., then prove that a) a + b, b + c, c + d are in G.P. b) a(b – c)3 = d(a – b)3

Progressions 23.

24. 25. 26.

27.

28.

29.

30.

31.

32.

33.

34.

35.

The ratio of A.M. and G.M. of two numbers is 5 : 4 and the sum of those two numbers is 100, then find the numbers. If the pth term of a H.P. be 'qr' and the qth term be 'rp'. Prove that its nth term is pq. If a,b,c,d are in H.P., then prove that ab + bc + cd = 3ad. If a,b,c be in H.P. and a > c, show that

1 1 4 + > b-c a -b a -c If x, y, z are in A.P., ax, by, cz are in G.P. and a, x z a c b, c are in H.P., then show that + = + . z x c a The number of terms of the series 3  3  3 3  ..... will make the sum

39  13 3 is Show that if the positive numbers a, b, c are in A.P., so are the numbers 1 1 1 , , . b+ c c+ a a+ b Karthik writes letters to four of his friends. He asks each of them to copy the letter and mail to your different persons with the request that they continue the chain similarly. Assuming that the chain is not broken and that it costs 25 paise to mail one latter. If the chain is not broken till 8th person then find the total money spent on latters. The lengths of three unequal edges of a rectangular solid block are in G.P. The volume of the block is 216 cm3 and the total surface area is 252 cm2 . Then find the length of the longest edge If three distinct positive reals x, y, z are in H.P. and x > z, then show that log (x + z) + log (x – 2y + z) = 2 log (x – z) If the roots of the equation a(b – c)x2 + b (c – a)x + c(a – b) = 0 are equal , then show tha a, b, c are in H.P. The fourth power of the common difference of an arithmetic progression with integral terms is added to the product of any four consecutive terms of it. Then show that the resulting sum is square of an integer If H1, H2, ...., Hn are n harmonic means between H  a Hn  b  a and b   a  then value of 1 is H1  a H n  b equal to ______.

73

IIT JEE Worksheet 1.

If between two numbers p and q, there are inserted two arithmetic means A1, A2; two geometric means G1, G2 and two harmonic means H1, H2 then show

G1G 2 A1  A 2 that H H  H  H 1 2 1 2 2.

Let a1, a2, .... a10 be in A.P. and h1, h2, ....., h10 be in H.P. If a1 = h1 = 2 and a10 = h10 = 3, then find a4h7

3.

Let Tr be the rth term of an A.P., for r =1, 2, 3....If for some positive integers m, n then we have

Tm 

1 1 and Tn  ,then show that Tmn= 1 n m

7 If log32, log3(2x – 5) and log3  2x   are in A.P., 2  then find the value of x 5. The interior angles of a polygon are in arithmetic progression. The smallest angle is 120°, and the common difference is 5°. Find the number of sides of the polygon. 6. The harmonic mean of two numbers is 4. Their arithmetic mean A and The geometric mean G satisfy the relation 2A + G2= 27. Find the two numbers 7. If x, y and z are the pth, qth and rth terms respectively of an A.P and also of a G.P., then show that xy – z z – x x– y y z =1 8. If the sum of the first 2n terms of the A.P. 2, 5, 8, ....., is equal to the sum of first n terms of the A.P. 57, 59, 61, .... then find n. 9. Let positive number a, b, c, d be in A.P., then show that abc, abd, acd, bcd are in H.P. 10. The first term of an arithmetic progression is loga and the second term is logb. Express the sum to ‘n’ terms as a logarithm. 11. Prove that in any AP a1, a2, ...... we have, 4.

S



1 1   ........ a1  a 2 a 2  a3 1 n 1  a n 1  a n a1  a n

12. The angles of a plane quadrilateral are in AP and the difference of the greatest and the smallest is a right angle. Find the angles of a quadrelateral. www.betoppers.com

10th Class Mathematics

74 13. If (m + 1)th, (n + 1)th and (r + 1)th terms of an A.P. are in G.P. and m, n, r are in H.P., then show that the ratio of the common difference to the first term

2 . n 14. If x = 1+ y + y2 + ....... to  , then show that in the A.P. is 

x 1 x 15. If the A.M. and G.M. between two distinct positive reals are respectively A and G, then show that the y

G2 A The sum of first ten terms of an A.P. is equal to 155, and the sum of first two terms of a G.P. is 9. Find these two terms of a G.P. is 9. Find these two progression, if the first term of A.P. is equal to the common ratio of the G.P. and the first term of G.P. is equal to the common difference of A.P. pth term of an A.P. is q and qth term is p, then show that its rth term is p + q – r Show that the sum of first n odd natural numbers is n2 The fifth term of G.P. is 2, then find the product of its first 9 terms If the third term of an A.P. is 12 and the seventh term is 24, then find its 10th term The fourth terms of an A.P. is 4. Then find the sum of the first 7 terms. If p, q, r are in A.P. and x, y, z are in G.P., then show that xq – r yr – p zp – q = 1

harmonic mean between the numbers is 16.

17. 18. 19. 20. 21. 22.

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23. If the A.M. between the roots of a quadratic equation is 8 and the GM is 5, then the eqation is 24. If a, b, c, d, e, f are in A.P., then show that e–c=1 25. If 7th and 13th terms of an A.P. be 34 and 64 respectively, then find its 18th term 26. If A.M. between two numbers is 5 and their G.M. is 4, then find their H.M. is 27. The first term and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then find the number of terms 28. If the second term of a G.P. is 2 and the sum of its infinie term is 8, then find its first term 29. In a harmonic progression pth term is q and qth term is p, then show that the (pq)th term is 1 30. The third term of a G.P. is 3. Then find the product of its first five terms 31. If 4th term of an H.P. is 5 and 5th term is 4, then find its 20th term 32. If H is the harmonic mean between a and b, then

Ha Hb  2 Ha Hb 33. Find the two geometric means between 1 and 64 34. If the progressions 3, 10, 17, ..... and 63, 65, 67, ...... are such that their nth terms are equal, then find the value of n 35. If c is the harmonic mean between a and b, then show that

show that

c c  2 a b

Functions Chapter - 6

Learning Outcomes By the end of this chapter, you will understand      

Basic Concepts Relations Introduction to Function Image , Pre image (or) Inverse image Domain, Co-Domain and Range Main Types of functions

 Other types of functions  Graph of a function  Key Points

1. Basic Concepts

A 1

Ordered pair Let A and B are any two non-empty sets. If a  A and b  B then (a, b) is called and ordered pair.. a is called the first component (first coordinate) and b is called the second component (second coordinate) of the ordered pair (a, b). Remarks: i) a is called the first member (First coordinate – x coordinate) ii) b is called the second member (Second coordinate – y coordinate) iii) Two ordered pairs are equal, i.e. (a,b) = (c,d) only if a = c and b = d. Example: The perfect example for ordered pairs is a ‘Point’ in the coordinate plane. Let (2,3) represent a point. Now, (3,2) represent a point. Since the point (2,3)  (3,2), are two different points. (2, 3) (3, 2)

2

B x y

3 A × B = {(1, x), (1, y), (2, x), (2, y), (3, x), (3, y)}. Note: A × B ≠ B × A

Significance of Cartesian product

i) Every element of a cartesian product can be represented on the coordinate plane. ii) It integrates algebra and geometry. iii) Coordinate geometry completely deals with the ordered pairs which are elements of cartesian product only.

Cardinal number of a set The number of elements present in a set A is called cardinal number of a set A. It is denoted by n(A). Example: i) A = {7, 8, 9,10}, Then, n(A) = 4 ii) B = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} n(B) = 7 Note: If n  A = m, n( B) = n then n  A × B  = mn = n( B × A)

2. Relations Cartesian product of sets If A and B are any two non-empty sets, then the cartesian product of A and B is denoted by A × B and it is read as A cross B and it is defined as the set of all ordered pairs (a,b) from A to B where a  A and b  B . Example: If A = {1, 2, 3}, B = {x, y} then,

If A and B are any two non-empty sets, then any subset of A × B is called a relation from A to B. It Is denoted by R. i.e., R  A × B Note: i) We write ( x, y )  R as x R y and say that ‘x’ is in relation with ‘y’ ii) There exist n number of relations between any two sets

10th Class Mathematics

76 iii) Generally we write the relation in set builder form as R = {(x, y) : x is related with y} or R = {(x, y) : x R y} Example: If A = {2, 3, 4, 5}, B = {3, 4, 5}, then, A × B = {(2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5),} i) R1 = {(x, y) :y = x + 1} is a relation from A to B. Then, R1 = {(2, 3), (3, 4), (4, 5)} ii) R2 = {(x, y) : x = y} is a relation from A to B. Then, R2 = {(3, 3), (4, 4), (5, 5)}

Number of relations Let A and B be two non-empty finite sets. Then, the number of relations from A to B = 2n(A×B). Example: If A = {1, 2}, n(A) = 2 B = {4, 5,6} n(B) = 3 n(A × B) = n(A) × n(B) = 2×3 = 6  The number of relations from A to B = 2n(A×B) = 26 = 64

Domain and Co-domain of a Relation Let us consider two sets A = {1, 2, 3, 4}and B = {p, q, r, s, t},

Domain

A

B

1

p

2

q

3

r s t

4

Co-domain

Domain : If R is a relation from A to B then the set of the first coordinates of the ordered pairs of R is called the Domain of Relation. It can be written as Dom R. Thus, Dom R = {x / (x,y)  R} Dom R = {1, 2, 3, 4} Co - Domain :The set of all elements present in set B is called Co - domain of R. Co - dom R = {p, q, r, s, t} Range : The set of all mapped elements in Co – domain is called range of relation R. Thus, Range of R = {y / (x,y)  R} . Range of R = {p, q, r, s, t} Note: Since, Relation R  A × B, we can clearly say,, dom R  A, co-dom R  B and Ran R  B. www.betoppers.com

Inverse of a Relation A

B

B e

b

e f

f

b

c

g

g

c

d

h

h

d

a

R

R–1

A a

Let R be a relation from A to B. It is defined as, R = {(x, y) / (x, y)  R}. Then, R–1 is the relation from B to A. It is denoted by R–1 and is called inverse of R. It is defined as R–1 = {(y, x) / (x, y)  R} . Note: i) Thus: R  A × B  R–1  B × A. ii) Also, dom R–1  B and ran R–1  A. iii) Further dom R–1 = ran R and ran R–1 = dom R. iv) Always (R – 1) – 1 = R

3. Introduction to Function Let A and B be two non-empty sets and f be a relation from A to B. Then, if i) for each element a  A there exists a unique element b  B such that (a, b)  f. ii) all elements of set A are mapped (associated) to a unique element in set B. Then the relation R is called function from A to B. In other words, a function ‘f’ from set A to a set B associates (maps) each element of set A to a unique element of set B. Remarks: i) Terms such as map or mapping and correspondence are used as synonyms for function. ii) If ‘f’ is a function from a set A to a set B then f we write f : A  B or A   B which is read as ‘f’ is a function from A to B or ‘f’ maps A to B. If A = {1, 2, 3} B = {4, 5, 6} A

f

B

1

4

2

5

3

6

Here, f : A  B is a function. f can be written as, f = {(1, 4), (2, 6), (3, 5)} If A = {a, b, c} B = {1, 10, 100}

Functions

77 A

4. Image , Pre image (or) Inverse image

B

f

a

1

b

10

c

100

Here, f: A  B is not a function. f = {(a, 10), (a, 100), (c, 1)} Because b  A has no corresponding element in B, and ‘a’ has two corresponding elements 10 and 100. Example:

If f is a function from A to B and an element a  A is associated to b  B under f, such that f(a) = b, then ‘b’ is called the image of ‘a’. we write this as f(a) = b. Also f(a) is called the image of ‘a’ under f. Example: If A = {a, b, c}; B = {p, q, r} A

B f

a

a

b

b

c

c

Let A  1,2,3,4 and B  a, b, c, d , e be two sets and let f1, f2, f3 and f4 be rules associating elements ( A to elements of B) as shown in the following figures: A

B

A

1

a

1

2

b c d e

2

f1

3 4

f3

1 2 3 4

B a b c d e

3 4

Fig 1 A

f2

Fig 2 B

A

a b c d e

1

B a b c d e

2 3 4

Fig 3

Observations i. f1 is not a function ii. f2 is not a function

f4

Fig 4

Reasons ‘3’ is not mapped two ordered pairs have same first element

iii. f3 is a function iv. f4 is a function

Number of Functions Let A and B be two non-empty finite sets. Then, the total number of functions from A to B = n(B)n(A). Example: Let A = {1, 2} and B = {5, 6,7} n (A) = 2, n (B) = 3 The number of functions from set A to set B = n(B)n(A) = 32 = 9

Here f : A  B is a function

f   a, e  ,  b, d  ,  c, c  Here f(a) = e, f(b) = d and f(c) = c So e, d and c are images of a, b and c under f respectively.

Pre-image (or) Inverse image If f is a function from A to B and an element a  A is associated to b  B under f, such that f(a) = b, then ‘b’ is called the image of ‘a’. we write this as f –1(b) = a. Also f –1(b) is called pre-image (or) inverse image of ‘b’ under ‘f ’. Example : A = {1, 2, 3} ; B = {7, 8, 9} A

B f

1

8

2

7

3

9

Here f : A  B is a function and

f  1, 9  ,  2, 7  ,  3, 8  f –1(9) = 1 ; f –1(7) = 2 ; f –1(8) = 3  1, 2 and 3 are called the inverse images of 9, 7 and 8 respectively.

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10th Class Mathematics

78

5. Domain, Co-Domain and Range Let A = {–2, –1, 0, 1, 2} and B = {0, 1, 2, 3, 4, 5, 6} are any two sets. Then, the function f: A  B is defined as, A –2

f

–1 Domain

0 1 2

B 0 1 2 3 4 5 6

Co - domain

Domain : Let f: A  B be a function. Then the set of all elements of set A is called the domain of f. Domain of function f = {–2, –1, 0, 1, 2} Co - domain : Let f: A  B be a function. Then the set of all elements of set B is called the co-domain of f. Co – Domain f = {0, 1, 2, 3, 4, 5, 6} Range: The set of all images of elements of A is known as the range of f. It is denoted by f(A). or The set of all mapped elements in B is called the range of f. Thus f(A) = {f(x); x  A} = Range of f. Range f = {0, 1, 2, 3, 5} Example: A

f1

1 2 3 4

B

A

a b c d e

1

Fig 1

f2

B a b c

2 3

d e

4 Fig 2

In f1 Co-domain = {a, b, c, d, e} Range = {a, b, d} In f2 Co-domain = {a, b, c, d, e} Range = {b, c, d, e} Let f : A  B be a function; then set A is called the Domain of f, and set B is called the Co-Domain of f. Remarks: i) Co-domain and Range need not be the same sets. ii) Range  co-domain.

Method to find the domain of a function Let f(x) be any function defined on real numbers. Then, to find the domain of the function, following are the steps to be followed. Let us now understand these steps better by taking an example. www.betoppers.com

Let us find the domain of f(x) =

1 6x  x 2  5

Step - 1 : Find the condition for which the function does not exist.

1 the function does not exist 6x  x 2  5 when 6x – x2 – 5x = 0  6x – x2 – 5x  0 Step - 2 : Simplify and obtain the value of a variable from the above condition. (x – 1) (5 – x)  0  x  1, 5 Step - 3 : To obtain the domain of f, extract the obtained values from the real number set.  Domain of f = R – {1, 5} f(x) 

Method to find the range of a function Let f(x) be any function defined on real numbers. Then, to find the range of the function, following are the steps to be followed. Let us now understand these steps better by taking an example. Let us find the range of f(x) =

2 x 2x

Step - 1 : Consider that given function as y i.e., f(x) = y

2 x 2x y 2 x 2x Step - 2 : Simplify and make x as a subject from the above equation and check for what value of y the equation does not exist. f(x) 

2  y  1 2x x y 1 2x Clearly, x is not defined for y + 1 = 0 Step - 3 : To obtain the range of f, extract the obtained values of y from the real number set.  Range of f = R – {– 1}

y

Open and Closed intervals An open interval is an interval that does not include its endpoints. The open interval {x : a < x < b} is denoted by (a, b). Example: (10, 100) is an open interval; it can take all the values in between 10 and 100, excluding 10 and 100.

Functions

79

Open interval {x : 2009 < x < 2000} is denoted by (2009, 2000).

Closed interval : A closed interval is an interval that includes all of its end points (limit points). If the end points are finite numbers like a and b then the interval {x: a  x  b} is denoted by [a, b]. Example: [25, 75] is a closed interval; it can take all the values in between 25 and 75, including 25 and 75. Closed interval {x : 1000 < x < 100000} is denoted by [1000, 100000].

Conceptive Worksheet 1.

If n(A) = 5, n(B) = 2, then find the number of relations which are not functions

2.

Find the domain of f  x  

3.

Find domain of f  x  

4.

Find domain of f  x   x  3

5.

Find the range of f  x  

Formative Worksheet 1. 2. 3.

If n(A) = 5, n(B) = 2, then find the number of relations from A to B If n(A) = 5, n(B) = 3, then find the number of functions from A to B If A = {2, 3, 4, 5}then which of the following relation is a function from A to itself?

 x, y  : y  x  1 ii) f =  x, y  : x  y  6 iii) f =  x, y  : y  x iv) f i) f1 = 2

3

4.

4

=

 x, y  : x  y  7

If b2 – 4ac = 0 and a > 0, then find the domain 3 of the function y = log ax   b  c  x  c

5.

Find the domain of the real function f  x  1  2  3  x

6.

What is the range of the function f  x  

5x 1 2x  7

defined on its maximum domain ? 7. 8. 9.

3 5 x  5 x (7 – x) i) Find the domain of f(x) = Px – 3 ii) Find the Range of f(x) = (7 – x)Px – 3 Which of the two functions are equal ?

Find the domain of f  x  

2

i) f  x    x  x  1 and g  x   x2  1  1  x  x  1  x

ii) h(x) = x2 + x and u  x  

x3  x x

2x  3 x  6x  5 2

1 x2  2 x

x 1  x2

6. Main Types of functions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Following are some main types of functions: One – One function Onto function One - One onto function Many - One function Into function Identity function Constant Function Equal functions Composite function Composition of three functions Inverse function

1. One – One function: A function f : A  B is said to be a one - one function if different elements of set A have different images in set B. This is also called injection. Example 1: Let f : A  B is represented by the following diagram. A a

B f

e

b

f

c

g

d

h

Clearly f : A  B is a one - one function since distinct elements in A has distinct images in set B.

iii) v  x   x 2  1  x and w x   x2  1  1  x  x  1  x

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10th Class Mathematics

80 A

Example 2: Let g : X  Y is represented by

B

g

B

k

o

x

l

p

2

d

m

q

3

e

n

r

4

f

A g

1

From the above diagram, we observe that g : X  Y is not one - one because 2 and 3 in set A have same image under function g. Example 3: f : R  R given by f(x) = x3 + 2 for all x  R . Let x, y be two arbitrary elements of R (Domain of f) such that f(x) = f(y) 3 3 3 3 x +2= y +2 x = y x= y Hence f is a one - one function from R to itself.

Number of one-one functions Let A and B are infinite sets and f : A  B is oneone, n(A) = m and n(B) = n, then the number of possible one-one functions from A to B is: a) nPm, (m  n) b) 0, if m > n. Example 1: If n(A) = 3 and n(B) = 5, then the number of oneone functions from A to B = nPm = 5P3 = 10 Example 2: If n(A) = 2010 and n(B) = 2009, then the number of one-one functions from A to B = 0 when n(A) > n(B).  Number of one-one functions are zero.

2. Onto function: A function f : A  B is said to be an onto function if every element of B is the image of some element of A or The range of f is equal to the co-domain of f. This is also called surjection Example 1: Let f : A  B and g : X  Y be two functions equal to zero represented by the following diagrams. A

f

B

a

e

b

f

c

g

d

h

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Clearly, e is an element in B which does not have pre-image in A. So f : A  B is not an onto function. Under function g every element in Y has its preimage in X. So g : X  Y is an onto function. Example 2: f : R  R given by f(x) = x3 + 2 for all x  R . Let y be an arbitrary element of R. (Co-domain) 1

Then f(x) = y  x3 + 2 = y  x =  y  2  3 Clearly, for all y  R,

1

 y  23

is a real number..

Thus, for all y  R (co-domain) there exists 1

x   y  2  3 in R(domain).

Hence f : R  R is an onto function.

Number of onto functions Case 1: If n(A) = m and n(B) = n, then number of possible onto functions (surjective mappings) from A to B is n C1 (n – 1)m – nC2 (n – 2)m + nC3 (n – 3)m – ………. If m > n. n C1 (n – 1)m – nC2 (n – 2)m + nC3 (n – 3)m – ………. = 0 If m < n. Example: If A = {1, 2, 3, 4}, B= {p, q, r}, then number of onto functions from A to B is n C1 (n – 1)m – nC2 (n – 2)m + nC3 (n – 3)m ………. where m = 4, n = 3. 3C1 (3 – 1)4 – 3C2 (3 – 2)4 + 0 = 3 × 24 – 3(1)4 = 48 – 3 = 45 Case 2: If n(A) = m and n(B) = 2, then number of possible onto functions (surjective mappings) from A to B is 2m – 2.

Functions

81

Example: If n(A) = 3 and n(B) = 2, then the number of onto functions from A to B = 2n(A) – 2 = 23 – 2 = 8 – 2 = 6 Case 3: If n(A) = m and n(B) = m, then number of possible onto functions (surjective mappings) from A to B is m!. Example: If n(A) = 5 and n(B) = 5, then the number of onto functions from A to B = k ! = 5 ! = 5 × 4 × 3 × 2 × 1 = 120

3. Bijective function: A function f : A  B is a bijection if it is one - one(injective) as well as onto(surjective). Example: Let f : A  B be a function represented by the following diagram. A

f

B

a1

b1

a2

b2

a3

b3

a4

b4

Clearly, f is a bijection since it is both injective as well as surjective. Note: If A and B are finite sets and f : A  B is a bijection, then A and B have the same number of 5. elements.

Number bijective functions Let A and B be finite sets and f: A  B a bijection. i) If n(A) = n(B) = k, then the number of bijections from A to B is k !. ii) If n(A)  n(B), then the number of bijections from A to B is 0. Example: If n(A) = 3 and n(B) = 3, find the number of bijections from A to B. Here, n(A) = n(B) = 3 = k. The number of bijections from A to B is k ! = 3 ! = 6

A

f

x1 x2 x3 x4 x5

y1 y2 y3 y4 y5

Clearly, x2, x4 in A have the same f image in B i.e., y1and also x1, x5 in A 0have the same f image in B i.e. y3. Example 2: Let A = {–1, 1, –2, 2} and B = {1, 4, 9, 16} . Consider f : A  B given by f(x) = x2, then f(–1) = 1; f(1) = 1; f(–2) = 4; f(2) = 4. Clearly, 1 and – 1 have the same image. Similarly, 2 and – 2 also have the same image. So f is a many-one function.

Number of many-one functions Let A and B be any two non-empty sets. Then, the total number of many-one functions from A to B = Number of functions – Number of one-one functions = nm – nPm Example: If A = {1, 2, 3}, B = {a, b, c}, then the number of many-one functions = Number of functions – Number of one-one functions = 33 – 3P3 = 27 – 6 = 21

Into function: A function f : A  B is an Into function if theree exists an element in B having no pre image in A. In other words, f : A  B is an into function if it is not an onto function. Example : Let f : A  B and g : X  Y be two functions represented by the following diagrams. A

f

a1 a2 a3 a4

B b1 b2 b3 b4 b5

4. Many - One function: A function f: A  B is said to be many-one if, i) for every element in A there is not more than one corresponding element in B ii) there is at least one element in B for which there are more than one corresponding elements in A. Example 1: Let f : A  B be represented by the following diagram

B

Clearly, b1 and b5 are two elements in B which do into function. X x1 x2 x3 x4 x5

g

Y y1 y2 y3 y4

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10th Class Mathematics

82 Under function g every element in Y has its preimage in X, so g : X  Y is not an into function. Note: If f is an into function, then there exists at least a single element in B having no pre-image in A.

6. Identity function: Let ‘A’ be a non empty set. A function f : A  A is said to be an identity function on set A if f maps every element of set A to the element itself. Thus f : A  A is an identity function if f(x) = x for all x  A . The identity function on set A is generally denoted by IA. Example : Let A = {a, b, c} f : A  A is represented by A

f

A

a

a

b

b

c

c

Here f : A  B is also a constant function because f(4) = 8 ; f(5) = 8 and f(6) = 8 Note : i) A constant function is a function whose range is a singleton set. ii) A constant function is a one - one function if its domain contains singleton element. iii) A constant function is an onto function if its codomain is a singleton set. iv) A constant function is a bijection, if both of its domain and co-domain are singleton sets.

8. Equal functions Two functions f and g are said to be equal if and only if (i) the domain of f = the domain of g (ii) f(x) = g(x) for every their common domain f. Let f : R – {1}  R be defined by f(x) = x + 1 and x2  1 . x 1 We observe that f and g have the same domain R – {1}

g : R – {1}  R be defined by g(x) =

x2  1 . = x + 1, as x  1 x 1  f(x) = g(x)  x  R – {1} f=g Example : Let A = {1, 2} B = {3, 6} f : A  B given by f(x) = x2 + 2 and g : A  B given by g(x) = 3x Then we observe that f(1) = 3 = g(1) and f(2) = 6 = g(2) Hence f and g have the same domain and co-domain. Hence, f = g.

f(x) = x + 1, while g(x) = Thus f(x) = x  x A Hence f is an identity function.

7. Constant Function A function f : A  B is said to be constant function if every element of A has same image under function f i.e., f  x   c x  A and c is a fixed element of B. Then f is said to be a constant function. Example 1: If A = {1, 2, 4}; B = {5} A

f

B

9. Composite function

1 2

Let f : A  B , g : B  C be two functions. Then

5

the function gof : A  C defined by gof (x) = 3

g  f  x   for all x  A is called the composition

Here f : A  B is a constant function. Example 2: If A = {4, 5, 6}; B = {7, 8, 9} A

f

B

1

7

2

8

3

9

of f and g. f A

g B

f: A  B

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C

g: B  C

gof: A  C

Functions Note : It should be noted that ‘gof’ exists if the range of f is equal to the domain of g. similarly fog exists if the range of g is equal to the domain of f.

83 A a

10. Composition of three functions A

f

B

g

C

D

h

a

x

1

2

b

y

2

3

c

z

3

4

f

B p

b

q

c

r

d

s

Observe that f is not bijective. f–1(q) = {a, c}, q is assigned to more than one element in A. So f –1 is not a function. Hence the inverse function of f does not exist. Example 2: Let f : R  R is defined by f(x) = 6x + 5 find f–1(x).

gof: A  C ho (gof) : A  D

Let f : A  B , g : B  C and h : C  D First we can form the composite function gof : A  C and then h : C  D . Then we get the new function ho(gof) : A  D . Note : i) ho(gof) = (hog)of ii) Both have the same Domain A and Co-domain D



iii)  x  A , both give the same image h g  f  x  



Note: i) In the product function gof a) The co-domain of f is the domain of g b) The domain of gof is the domain of f c) The co-domain of gof and g is the same set. d) In general, gof  fog ii) If f : A  B , g : B  C and h : C  D then (hog)of = ho(gof) iii) In the functions (hog)of and ho(gof)z a) both have the same domain A b) both have the same co-domain D. c) for every x  A , both gives the same image h(g(f(x)) d) (hog)of = ho(gof)

Let f(x) = y

 6x + 5 = y  x =

y5 6

y5 for all y  R (co-domain) 6 Also observe that f–1 is one-one and onto. Note : i) Domain of f = Range of f–1 and Range of f = Domain of f –1. ii) An Identity function is its own inverse.  f 1  y  

Formative Worksheet 10. Find whether the function f  x   one-one or not?

x 2  4 x  30 is x 2  8 x  18

11. Find the domain and range of f  x  

x2 , 1  x2

x  R . Is the function one-one ? 12. If A={x1, x2, x3, x4} and B={ y1, y2, y3, y4, y5, y6}, then the number of possible one-one functions defined from A to B 13. The number of many-one functions from {x, y , z} to {1, 2, 3, 4} 14. The number of onto functions from A = {1, 2, 3, 4, 5} to B = {a, b} 15. If A = {a, b, c, d}, B = {p, q, r} then find number 11. Inverse of a function of bijections from A to B. –1 f is a function from A to B, in general f (b), 16. Identify the type of function f : N  Z defined may correspond with one element, more than n 1 n one element or may not with any element. by f  n   when n is odd and f  n   Since it is clear from the definition, 2 2 when n is even 1 f : B  A is a function if f : A  B is a one 17. If A = {1, 2, 3, 4, 5}, then find number of identity - one and onto function. functions from A to A Example 1: Let the function f : A  B be defined 18. If A = {1, 2, 3, 6, 8, 9} and B = {a, b, c, d, e} then find the number of constant functions from A to B by the following diagram. 19. Let f(x) = (x +1) 2 – 2 for x  1 and

g ( x)  2  x  1 for x  2 . Find g(f(x)) for x  1 . www.betoppers.com

10th Class Mathematics

84 20. Let f(x) = x 2 – 2x + 3, g(x) = x + k and f(g(x)) = x2 + 10x + c. Find the value of c. 21. Given f(x) = 2x +1 and h  x   x x , find

7. Other types of functions 1. Polynomial Function: A function ‘f’ defined as f(x) = a0 + a1x + a2x2 + ..... + an xn , where x is a real number, n is any non-negative integer and a0, a1, a2, ......, an are real constants, is called a polynomial function.

–1

(ho f )(2).

 1  22. If f 1    3 x  2 , then find f(x).  x 1 23. If f : R  R  given by f(x) = e2x +3 then find f–1(1)

2. Rational Function:

24. If f : 1,    1,   is given by f(x) = 2x(x – 1), then find f–1(x)

Where g(x) and h(x) are polynomial functions

10 x  10 x 25. Find the inverse of f  x   x 10  10 x

Conceptive Worksheet 6.

type of function Let Z denote the set of all integers, x   x is even  Define f : Z  Z by f  x    2 0  x is odd  

8.

Identify the type of function If f : R  R and g : R  R are defined by 4. f(x) = 4x – 1 and g(x) = x 3 + 2, then find

 fog   5 . 9.

( h  x   0 ) is called a rational function. Examples: i) f (x) 

x 2  3x  2 x2  1

ii) f (x) 

2009x 3  2008x 2  2007x  2006 2010x 2  2009

On the set Z of all integers define f : Z  Z as  x if x iseven follows : f  x    2 then Identify the 0 if x is odd 3.

7.

g ( x) A function ‘f’ defined as f  x   h( x) .

If f : R  R and g : R  R are defined by f(x) = 4x – 1 and g(x) = x 3 + 2, then find 5.

a 1   4  10. If f(x) is an inverse function, and g(x) = 3f(x) + 1, then find g–1(x). 11. Let f(x) = 1 + x2. Find a function g(x) such that (fog)(x) = 1+ x2 – 2x3 + x4. 12. If f(x) = 1   x ,   0 is the inverse of itself, find the value of  .

 gof  

13. If f  x  

2x  5 , then find f–1(x) 3x  4

14. If f : R  R is defined by f(x) = x2 + 5, then find f–1 (21)

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Algebraic functions: An algebraic function is any function y = f(x) which satisfies an equation of the form. P0(x)yn + P1(x)yn–1 + ............+Pn(x) = 0 where P 0 (x), P 1 (x), ..........Pn (x) are certain polynomials in x. Note: Every rational and irrational function is an algebraic function.

Transcendental Function: A function which is not an algebraic function is called a transcendental function. Examples are trigonometric, logarithmic and exponential functions.

Even and Odd Functions: A function ‘f’ is said to be an even function of x if f(–x) = f(x). Odd Function: A function ‘f’ is said to be an odd function of x if f(–x) = – f(x). Example 1: Let f(x) = 5x6 + 6x2 + 7 Replacing x by – x f(–x) = 5(–x)6 + 6(–x)2 + 7 = 5x6 + 6x2 + 7  f(–x) = f(x) Hence f(x) is called an even function. Example 2: Let f(x) = 7x5 + 6x3 + 8x Replacing x by –x, we get f(–x) = 7(–x)5 + 6(–x)3 + 8(–x) = –7x5 – 6x3 – 8x = –[7x5 + 6x3 + + 8x] = – f(x)  f(–x) = –f(x) Hence f(x) is called an odd function.

Functions

85 3

Example 3:Let f(x) = x + x – 2 Replacing x by ‘(–x)’ we get f(–x) = (–x) 3 + (–x) – 2 = –x3 – x – 2 = –(x3 + x + 2)  f(–x)  f(x) and f(–x)  –f(x) Hence f(x) is neither even nor odd. Note : i) A constant function is always an even function Example : f(x) = 2; f(x) = 7 etc. ii) If an integral polynomial function be even with regard to any variable, it can only contain even powers of that variable, if odd it will contain only odd powers of the variable.

6. Explicit and Implicit Functions: If the dependent variable, say y, is Expressed Explicitly in term of the independent variable, the function y = f(x) is called an Explicit Function. Otherwise is said to be an Implicit Function. Example: y = cos3x + x3 is an explicit function where as x3 + y3 + 3yx = 0 is an implicit functions.

7. Absolute Value Functions:   x, x  0 Definition: y | x |   x, x  0 Domain:x  R Range:y  R+ + {0} = [0,  ) Nature: a) many-to-one function b) onto function for co-domain R+ + {0} c) Even function

8. Signum Functions:  1 x  0 |x|  y 0 x0 Definition: x 1 x0 

Domain:x  R Range:y  {–1, 0, 1} Nature: a) Many-to-one b) Into function for co-domain as the set of real numbers c) Odd function.

9. Greatest Integer function : y = f(x) = [x] = greatest integer (x) by definition, [x] also named as integral part function is equal to greatest integer less than or equal to x. It is also known as floor function. Domain: x  R

Range: y  I Nature: i) Many to one ii) into function for co-domain as the set of real numbers iii) If [x] = x, then x must belong to the set of integers Examples: i) [1.72] = 1 ii) [– 3.41] = – 4 iii) [0.22] = 0 iii) [– 0.71] = –1

10. Least integer function: (x) or [x] denotes the least integer function which is greater than or equal to x. It is also known as ceiling of x. Example: i) (3.578) = 4, ii) (0.87) = 1, iii) (4) = 4, iv) [– 8.239] = – 8, v) [– 0.7] = 0.

8. Graph of a function If f: A  B is a real valued function, then f = { (x, f(x)) : x  A } is the set of ordered pairs whose coordinates are real numbers. Therefore we can plot these real numbers on a graph. It consists of all points in the Cartesian plane whose coordinates are (x, f(x)) or (x, y), where x is any point in the domain of the function f(x). Y - ax is y = f(x) f(x) P (x, f(x)) 0

O rigin

x

X - axis

In the point (x, f(x)), x is the distance of the point from y-axis and f(x) is the distance of the point from x-axis as shown in figure.

Identifying the graph Given a graph, how to identify whether it is a function or not? If we know the ordered pairs of the graph, then we have to see the set of ordered pairs of the given relation. If no two ordered pairs have same first coordinate, then it is a function; otherwise, it is not a function. If the ordered pairs of the graph are not known, www.betoppers.com

10th Class Mathematics

86 then draw a line parallel to Y-axis through the graph. If a vertical line cuts the graph at only one point, then it is a function; otherwise, it is not a function. A graph will represent a function if every vertical line cuts the graph at not more than one point. Examples: Let us identify which of the following graphs is a function. i)

Y – axis

P O Origin

X – axis

7. 8.

If f(x) is a function such that f(x + y) = f(x) + f(y), then f(n) = n f(1). If f(x) is a function such that

 1  x f  x  f  y    f    f  xy    0 , 2  y  then f(x) = cos (log x). 9. If f(x) is a function such that f(n + 1) = a f(n) + b(1 – a), then f(n) = an + b. 10. If f(x) = ax + b (linear factor), then f : R  R is a bijection (one-one onto). 11. If f(x) is strictly monotonic function, then f : R  R is a bijection.

12. If y  f  x   The line drawn parallel to Y-axis is touching the graph at only one point at P. Hence, the graph is a function. ii)

O Q

The line drawn parallel to Y-axis is touching the graph at two points, P and Q. Hence, the graph is not a function.

9. Key points The number of one-one onto functions that can be defined from a finite set A onto a finite set B is k, where k = [n(A)]! if n(A) = n(B) and k = 0 if n  A  n  B  .

3.

4. 5. 6.

The number of linear functions from [a, b] into [c, d] is 2. If f(x) is a polynomial such that

1 1 f  x  f    f  x   f   , then f(x) = xn + 1 or  x  x n f(x) = – x + 1 If f(x) is a function such that f(x + y) = f(x) f(y), then f(x) = ax. If f(x) is a function such that f(xy) = f(x) + f(y), then f(x) = log ax. If f(x) is a function such that f(x + y) + f(x – y) = 2 f(x).f(y), then f  x  

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1 13. If f(x) = ax + b, then f  x  

a x  a x 2

x b . a

 b  dx  ax  b 1 , then f  x     .  a  cx  cx  d 15. If A and B are two non empty finite sets, then the number of functions can be defined from A to B in 14. If f  x  

X – axis

Origin

2.

f(y) = x.

Y – axis P

1.

ax  b then (fof) (x) = x or cx  a

n A

.  n  B   16. The number of one - one functions that can be defined from the finite set A to the finite set B is n(B)

Pn(A) if n  B   n  A  and 0 if n(B) < n(A).

17. If A and B are non empty sets and n(B) = 2 then the number of onto functions that can be defined from A to B is 2n  A  2 . 18. If A and B are two sets such that n(A) = n(B), then possible number of bijections from A to B is n(A)!.

Formative Worksheet 26. Find the Range of sinx + cosx 27. Find the Range of a tan x + b cot x + c 28. If f : R  R is defined by f  x   2 x  x , then find f(2x) + f(–x) – f(x) 29. Which of the following functions from A =  x : 1  x  1 to itself is bijection ?

 x  x i) f  x   sin   ii) f  x   2 2   iii) f(x) = x2

iv) f  x   x

Functions

87

ax 1 f x    30. If the real valued function x n  a n  1 is

such that adjacent triangles in the path share a common edge and the path never travels from a lower row to a higher row or revisit a triangle.

even, then find the natural values of n. 31. Which of the following functions is an odd function ? i) f  x   1  x  x 2  1  x  x 2

 ax 1  ii) f  x   x  a x  1   

An example of one such path is illustrated in the given diagram for n = 5. Determine the value of one such path which is illustrated in the given diagram for n = 5. Also determine the value of f(2025).

 1  x2  f x  log    iii) 2  1 x  iv) f(x) = k, k is a constant 32. Find the domain of

1

42. If f  x    3  x n  n , x > 0 then minimum value of

a2  x2

33. If f(mn) = f(m + n) for all m, n  R and f(2) = 2006 then find f(2006 2006 ) 34. Show that if f(x) is a polynomial function

1 satisfying f  x  f    f  x    x n f(x) =  x +1

1 f  x

1 2 43. If 2f(xy) = [f(x)] y + [f(y)] x  x, y  R and 1) 3

2) 4

3) 2

4)

50

f(1) = 2, then the value of

 f r   r 1

1) 2 – 250 2) 251 – 2

x  10 for all x  R , and fn (x) = f1 (fn– 3 (x)) for n  2, Then Find f2006 (x). 1 36. For any function which satisfies the relation f(x) 35. Let f1  x  

2006

+ f(1 – x) = 1, find the value of

  1  f  f  x    f  f    is   x 



i



 f  2007  .

3) 250 4) None 2001  r  4x 44. If f  x   x , then  f   4 2  2002  r 1 1) 100.5 2) 1000.5 3) 1100.5 4) 1110.5

Conceptive Worksheet

i 1

15. Find the range of

an 37. If an 1  1  a for all n  1 and a1 = 2, n find a1729. 38. Let f : N  N be a function defined recursively f(1) = 1 and f(n + 1) = 2 f(n) + 1 for all n  1 . Find the value of f(1) + f(2) + f(3) + ..........f(100). 39. Let f : N  N be a function defined recursively f(1) = 1, f(2) = 1 and f(n + 1) = f(n) + f(n – 1) for 

all n  2 . Find

 n 1

f  n 2n

Given that f(n) = f(n + 1) – 5n – 1 and f(–25) = 1600, find f(0). 41. Consider an equilateral triangle of side n units, which is divided into unit triangles, as shown. Let f(n) be the number of paths from the triangle in the top row to the middle triangle in the bottom row 40.

2  x2 .

1 . 2  cos3x 17. Find the range of f(x) = 9 tan x + 4 cot x + 25. 16. Find the range of

18. If f  x   x  1  x  2  x  3 , then 2 < x < 3 is 1) onto function 3) into function

2) one – one function 4) an identity function 1 19. Domain of f  x   is log x 1) R – {–1, 1} 2) R – {0, 1} 3) R – {–1, 0, 1} 4) None 20. If f(x) = 100 – 8  13x , then find the range of f . 21. Which of the following is an even function ?

 1 x  1) f  x   log    1 x 





2 2) f  x   log x  1  x www.betoppers.com

10th Class Mathematics

88

x x  1 x e 1 2 2x 4) f(x) = e + sin x 22. If f : R  R is defined by f(x) = 2x – 3 if x > 3 _________(1) = x2 if 1  x  3 _________(2) = 3x + 2 if x  1 _________(3) then f(–3) = 1) 7 2) 8 3) –7 4) – 8 3) f  x  

9  x2 .

23. Find the range of 24. If f  x   x 

1 , provethat x

1 3  f  x    f  x   3 f   . x 3

1 25. If 3 f  x   f    log e x 4 for x > 0, find f(ex). x 26. Let f(x) be a real valued function satisfying y

x

2 f  xy    f  x    f  x  for all x, y  R

34. If f(x + y) = f(x) f(y), f(3) = 27, then find f(4). 35. A polynomial function f(x) satisfies the

1 condition f  x  . f    f  x   x 1729, then find f(10).

Summative Worksheet If f  x  

2.

Find the domain of f  x  

3.

1 . 5x  3 Find the domain of f(x) = log (3x – 2).

4.

2 Find the domain of f  x   log x  3 x  2 .

5.

Find the domain of f  x  

6. 7. 8.

n r 1

27. F is a function such that (i) F (0) = 1, (ii) F (1) = 5 and (iii) F(x + y) F(x – y) = F (x) F(y) for all integers x and y. Find F(5) 28. Let f : Z  Z and g : Z  Z . f(x + y) = f(x) + g(y) + 7 for all integers x and y. Let f(x) = x for all negative integers x, and g(3) = 21. Find f(0). 29. Let f : R  R and f(x1 + x2 + x3 + .......+ x101) = f(x1) + f(x2) + f(x3) + ........+ f(x101) – 25 for all x  R . Compute f(0). 30. Let f be a linear function with the properties f 1  f  2  , f  3  f  4  and f(5) = 5. Which of

f (a ) x , then find f ( a  1) . x 1

1.

and f(1) = a, where a is not equal to 1, prove that (a –1)  f(r) = an+1 –a

1 f   if f(12) =  x

9.

1

5 x . log e  Find the domain of f(x) = sec 3x. Find the domain of f(x) = cos (3x + 5).

If f : A  A and A   x /  2  x  2 and f(x) = 2x + 3, then f is... 1) onto 2) bijective 3) not a function 4) one - one x 1 If f  x   where x  1 , then find x 1 (fofofof)(x).

 1 x  3x  x 3 10. If f  x   log  , then  and g  x    1 x  1  3x 2 find fog(x). 11. If f : R   R such that f  x   log3 x then, find f–1 (x). 12. If f : R  R defined by f(x) = x3 + x2 – x –1, then find f–1 (0). 13. The domain of the function f(x) = log 4 (log5 (log3(18 – x2 – 77))) is... 1) x   0, 10 

2) x  8, 10 

the following statements is true ? 3) x   4, 5  4) x  1, 2  (i) f(0) < 0 (ii) f(0) = 0 14. Suppose f :  2,2   R is defined by (iii) f(1) < f(0) < f(–1) (iv) f(0) = 5 (v) f(0) > 5  1 for  2  x  0 f  x   31. Let f(x) = g(x) – 6, where g(x) is a polynomial  x  1 for 0  x  2 containing odd powers of x only and g(0) = 0. If f(–1) = 12, then find f(1). then, x   2, 2 : x  0 and f  x   x  ... 32. If f(x) is an odd periodic function with period 2, then 1   3   1  find f(100).       1) {0} 2) 3) 4) 33. If a function F is such that F(0) = 2, F(1) = 3, 4 2 2 F(n + 2) = 2F(n) – F(n + 1) then find F(5).



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

Functions

89

15. Let g(x) is a polynomial function satisfying g(x)g(y) = g(x) + g(y) + g(xy) – 2 for all x, y  R If g(2] = 5, find g(3). 4x f : R  R f x  16. If is given by   for all 4x  2 x  R , prove that f(x) + f(1 – x) = 1 17. Consider the function f(x) such that f(a b) = f(a + b) for all real numbers a and b. If f(123) = 321, find f(2012).

1 18. Let a function f be such that f    5 f  r   3r , r for every r  0 . Find f(3).     19. Let f : R  R . If [f(xy)]2 = x [f(y)]2 for all x and y f(9) = 21, find f(100). 20. Let f(n) = n2 and define the function g(n) by the formula g(n) = f(n + 1) – f(n). Find the average of the 1000 numbers g(0), g(1), .....,g(999) ? 21. If f is a function such that f(3) = 2, f(4) = 2 and f(n + 4) = f(n + 3) × f(n + 2) for all the integers n  0 , what is the value of f(6) ? la 22. If f  a   log for 0 < a < 1, then find 1 a  2a  f . 2   1 a  

23. If e

f  x

10  x  , x   10, 10  and f(x) = kf 10  x

1 f  x f    f  x   x find f(10).

1 f   and f(2) = 17, then x

HOTS Worksheet 2.

3.

If the function f(x) is defined by f(x) = a+bx and fr = ffff....(replaced r times) then find fr (x). The range of f(x) = 16 – xC2x – 1 + 20 – 3xC4x – 5 is.... 1) [0, 728] 2) [728, 1617] 3) [728, 1744] 4) None The range of the function f(x) = 9x – 3x + 1 is...

3  1)  ,   2)  0,   3)  , 0  4 

4)  ,  

If f  x   x  x and g(x) = g  x  

4x  1 then 4

for (gofogofogofogof)(3) = a  b 3 find (a+b). 1

5. 6. 7.

8.



 200 x   2  , then find k.  100  x  24. Let y = max{(x + 3), (7 – 2x)}. What is the minimum value of y for 2  x  1 ? 1) 4.00 2) 4.50 3) 4.67 4) 4.33 25. If f(x) is a polynomial function such that

1.

4.

9.

It f  x    a  x n  n where a > 0, and n is a positive integer, then show that f(f(x)) = x Let f(x) = 1 + x2 . Find a function g(x) so that f(g(x)) = 1 + x2 – 2x3 + x4 If f : R  R, g : R  R are defined by f(x) = 3x – 4 and g(x) = 2 + 3x, then find (g–1 of –1)(5). 3x  2 If f  x   defined on 5x  3 3 3 R     R    , then f–1(x) =... 5 5 1) x 2) f(x) 3) 2f(x) 4) 3f(x) Find the range of sin2x + cos4x.

10. If f[g(x)] = sin x and g  f  x    sin  x , then f(x), g(x) respectively is... 1)

x , sin3 x

2) sinx, x

4) None x , sin2 x 11. Find a natural number a for 3) n

 f  a  k   16  2

n

which

 1 . where the function

k 1

f satisfies the relation f  x  y   f  x  f  y  for all natural numbers x,y, and further f(1) = 2.

1 1 12. If for non-zero x, af  x   bf     5 , x x where a  b , find f(x). 13. It f : Z  Z be a function defined by f(x + y) = f(x) + f(y) for all x, y  R show that f(x) = xf(1) for all x  R . 14. It f : Q  Q be a function defined f(x + y) = f(x) + f(y) for all x, y  Q . show that f(x) = xf(1) for all x  R . 

15. Given x  1 and

  1 n0

n  n 1 2

xn 

P  x Q  x  , where

P(x) and Q(x) are polynomial functions. Also the greatest common divisor of P(x) and Q(x) is 1. Find

P  x Q  x .

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10th Class Mathematics

90

w x   x2  1  1  x  x  1  x

IIT JEE Worksheet 1.

2.

3. 4.

A function f satisfies f(x) = f(x – 1) if x > 0 and f(x) = |x| if x  0 then find the value of f(2.4) – f(2). If f(m + 1) = m(–1) m + 1 – 2f (m) for integers m  1 and f(1) = f(2001), then compute f(1) + f(2) + .............. + f(2000). f : R  R such that f (x + f(x)) = 4f(X) and f(1) = 4. What is the value of f(5) ? i. Can the following graph be a function. Why or why not? y

4) l  x   log

1 sinx  & 2 8 cos 2 x 1

1

m  x   sinx 

8cos 2 x 10. If k is given a real constant, f(x) is a real quadratic in x such that f(x + k) = f(–x). The coefficient of x2 is 1, then find f(x). 11.

f :  x, y , z  a, b, c is a bijection. It is given that

exactly one of the following statements a, b, c, is true and the remaining two are false. 2) f  y   a

1) f(x) = a x

3) If f  z   b then find f–1(a). 12. If f , g , h : R  R are defined by f(x) = x – 1,

5.

6. 7. 8.

9.

ii. Can it be injective? If f is defined for all real x f(a + b) = f(ab),  a, b  R , f(–1/2) = –1/2, then find f(–311/515) Find an explicit formula (in terms of n  N ) for f(n) = f(n – 1) + 2n – 1. If A = {1, 2, 3, 4, 5, 6} B = {1, 2, 3}, then find the number of surjections from A to B. This question has one or more than one answer is correct. Write all possible answers. The relation R = {(1, 1) (2, 3) (3, 4), (4, 2)} on the set X = {1, 2, 3, 4} 1) is not a function for X  X 2) is a one-to-one function from X  X 3) an onto function of X  X 4) R is a transitive relation on X x

x

1

1

2

2

3

3

4

4

 x2  1 if x  1  g   x 1  2 if x  1 

 x3  2x 2  x  2 if x  1, x  2  2  x  3x  2 h x   2 if x  1  3 if x  2  

f(x) + g(x) – 2h(x) is 1) not a polynomial 2) a polynomial whose degree is undefined 3) a polynomial of degree zero 4) a polynomial of degree one 5) None of these 13. If f(x) is a polynomial function such that

1 1 f  x  f    f  x   f   and f(3) = – 242, then  x  x f(x) = 1) x5 – 1 2) –x5 – 1 3) – x5 +1 4) 1 + x5 14. For any function which satisfies the relation f(x)

Which of the two functions are equal?

g  x  x 1 1  x  x  1 x

2

3) v  x   x  1  x & www.betoppers.com



i



 f  2007  i 1

2

2 2) h  x   x  x & u  x  

2006

+ f(1 – x) = 1, find the value of

1) f  x    x  x 2  1&

then

x3  x x

15. Let f : N  N be a function defined recursively f(1) = 1, f(2) = 1 and f(n + 1) = f(n) + f(n – 1) for all 

n  2 . Find

 n 1

f n 2n

.

10th Class Mathematics

91

Coordinate Geometry Chapter - 7

Learning Outcomes By the end of this chapter, you will understand  Introduction  Distance formula  Section formula  Area of a triangle

1. Introduction Co-ordinate geometry helps us to locate the position of a point on a plane wherein we require a pair of co-ordinate axes. The distance of a point from the y-axis is called its X-co-ordinate, or abscissa. The distance of a point from the X-axis is called its y-co-ordinate, or ordinate. We have seen that the co-ordinates of a point on the x-axis are of the form (x, 0), and that of a point on yaxis are of the form (0, y). A linear equation in two variables, when represented graphically, produces a straight line. Using Co-ordinate geometry we find the distance between two points whose co-ordinates are given, we can also find the co-ordinates of the point which divides the line segment joining two given points in a given ratio.





PQ 2  PT 2  QT 2 [By Pythagoras Theorem] 

2

  y 2  y1 

Y-axis

1.

2.

3. T

4. 5. X

O Y

R

S

X-axis

2

Formative Worksheet

Q (x2, y2)

P (x1, y1)

2

 x 2  x1    y2  y1 

2

(0,0) is given by OP  x 2  y 2 Because the value of x1 – x2 will always be negative and its square will be positive. Similarly, the value of y1 – y2 will always be negative and its square will be positive. Hence, there will not be any change in the value of both the formulae.

2

i.e., We have to find the distance between two points P(x1, y1) and Q(x2, y2). PR and QS perpendiculars have been drawn to the x-axis. A perpendicular from the point P on QS is also drawn which meets at T as shown in figure.

2

=  x 2  x1    y 2  y1 

It is called distance formula. The distance is always non-negative, so we take only the positive square root. The distance of a point P(x, y) from the origin O

The distance between any two points in the plane is the length of the line segment joining them. The distance between two points p(x1, y1) and Q(x2, y2) is given by

 x 2  x1 

PQ2

Hence, PQ =

2. Distance Formula

PQ 

OR = x1, OS = x2 RS = x2 – x1 PT = x2 – x1 [ RS = PT]  Again, PR = y1, QS = y2 QT = y2 – y1 Now, in right  PTQ , we have

Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (– 5, 7), (– 1, 3) (iii) (a, b), (– a, – b) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. In a classroom, 4 friends are seated at the points A, B, C and D as shown in below figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?”

10th Class Mathematics

92 Chameli disagrees. Using distance formula, find which of them is correct.

4.

If two vertices of an equilateral triangle is (0, 0),

 3, 3  , find the third vertex. 5.

6.

Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (i) (– 1, – 2), (1, 0), (– 1, 2), (– 3, 0) (ii) (–3, 5), (3, 1), (0, 3), (–1, – 4) (iii) (4, 5), (7, 6), (4, 3), (1, 2) 7. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9). 8. Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units. 9. If Q (0, 1) is equidistant from P(5, –3) and R(x, 6), find the values of x. Also find the distances QR and PR. 10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (– 3, 4). 11. If the point (x, y) be equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx + ay. 12. Show that the points A(a, a), B(–a, –a) and

C(a 3, a 3) from an equilateral triangle. 13. If A(3, –4); B(4, 2); C(5, –4) and D(4, –10) are the vertices of a quadrilateral ; prove that ABCD is a rhombus.

Conceptive Worksheet 1. 2. 3.

Find a point on X-axis which is equidistant from A(2, –5) and B(–2, 9). Show that the points (1, –1), (5, 2) and (9, 5) are collinear. Show that four points (0, –1), (6, 7) , (–2, 3) and (8, 3) are the vertices of a rectangle. Also find its area.

If P(2, –1), Q(3, 4), R(–2, 3) and S(–3, –2) are four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus. 7. Do the points (3, 2), (–2, –3) and (2, 3) form a triangle? If so, name the type of triangle formed. 8. Show that the points (1, 7), (4, 2), (–1, –1) and (– 4, 4) are the vertices of a square. 9. Find a relation between x and y such that the point (x , y) is equidistant from the points (7, 1) and (3, 5). 10. Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

3. Section formula A point P divides the straight line joining two points A(x1, y1) and B(x2, y2) internally in the ratio m1 : m2, to find the co-ordinates of P. Let the co-ordinate of P be (x, y). AR, BT and PS perpendiculars have been drawn on the X-axis from A, B and P respectively. Now, AQ and PC perpendiculars have been drawn from A and C on PS and BT respectively. In figure. B (x2 , y2) m2 P (x, y) m1

C

A (x 1, y1)

X

R

S

T

X

AQ = RS = OS – OR = x – x1, PC = ST = OT – OS = x2 – x, PQ = SP – SQ = SP – RA = y – y1, and BC = TB – TC = TB – SP = y2 – y. Now, from similar triangles APQ and PCB, we have, AQ AP  PC PB x  x1 m1   x 2  x m2 

m2  x  x1   m1  x 2  x 

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Coordinate Geometry  

m 2 x  m 2 x1  m1x 2  m1 x m1x  m 2 x  m1 x 2  m 2 x 1



x  m1  m2  = m1 x 2  m 2 x1

m1 x 2  m 2 x1 m1  m 2 Again, from similar triangles APQ and PBC, we have PQ AP  BC PB y  y1 m1   y2  y m2

 =



m2  y  y1   m1  y 2  y 

 

m 2 y  m 2 y1  m1 y 2  m1 y m1 y  m 2 y  m1 y 2  m 2 y1



y  m1  m 2   m1y 2  m 2 y1

m1 y 2  m 2 y1 m1  m 2 Hence, the co-ordinates of point P which divide the straight line joining two points (x1, y1) and (x2, y2) internally in the ratio m1 : m2 are,

 y=

 m1 x 2  m 2 x1 m1 y 2  m 2 y1  ,   m1  m 2   m1  m2 COROLLARY Co-ordinates of the mid-point: If P is the mid-point of AB, then it will divide AB in the ratio of 1 : 1, then co-ordinates of P are : 1 1   2  x1  x 2  , 2  y1  y 2    

Formative Worksheet 14. Show that the mid-point of the line segment joining the points (5, 7) and (3, 9) is also the mid-point of the line-segment joining the points (8, 6) and (–0, 10). 15. The co-ordinates of A and B are (1, 2) and (2, 3). AR 4 Find the co-ordinates of R so that  . RB 3 16. Find the co-ordinates of a point R which divides the line-segment joining the points P(–2, 3) and 4 Q(4, 7) internally in the ratio . 7 17. If A and B are (1, 4) and (5, 2) respectively, find AP 3 the co-ordinates of P when  . PB 4

93 18. Find the co-ordinates of a point which divides internally the line-segment joining the points (–3, –4) and (–8, 7) in the ratio 7 : 5. 19. Prove that the points (–2, –1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram. Is it a rectangle? 20. The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex. 21. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. 22. If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b. 23. If the coordinates of the mid-points of the sides of a triangle are (1, 2) (0, –1) and (2, –1). Find the coordinates of its vertices. 24. Show that the mid-point of the line-segment joining the points (5, 7) and (3, 9) is also the mid-point of the line-segment joining the points (8, 6) and (–0, 10). 25. The vertices of a triangle are (a, b – c), (b, c – a) (c, a – b) ; prove that its centroid lies on the xaxis. 26. If k1a – k2b + k3c = 0, family of straight lines ax + by + c = 0 are always concurrent at a point whose coordinate is k k   k k  (A)  1 , 2  (B)   1 , 2   k3 k3   k3 k3 

27.

28.

29.

30.

31. 32.

k  k k  k  (C)  1 ,  2  (D)   1 ,  2   k3 k3   k 3 k3  The diagonals AC and BD of a rhombus intersect at (5, 6). If A  (3, 2) then equation of diagonal BD is (A) y – x = 1 (B) 2y – x = 17 (C) y – 2x + 4 = 0 (D) 2y + x = 17 The coordinates of one end point of a diameter of a circle are (4, –1) and the coordinates of the centre of the circle are (1, –3). Find the coordinates of the other end of the diameter. Find the lengths of the medians of a ABC whose vertices are A(7, –3), B(5, 3) and C(3, –1). If A(5, –1), B(–3, –2) and C(–1, 8) are the vertices of triangle ABC, find the length of median through A and the coordinates of the centroid. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).

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94 33. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as 1 shown in given figure. Niharika runs th the 4 distance AD on the 2nd line and posts a green 1 flag. Preet runs th the distance AD on the 5 eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

34. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6). 35. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. 36. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y. 37. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4). 38. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that 3 AP = AB and P lies on the line segment AB. 7 39. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts. 40. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in 1 order. [Hint : Area of a rhombus = (product of 2 its diagonals)]

10th Class Mathematics

Conceptive Worksheet 11. If the point C(–1, 2) divides internally the line segment joining A(2, 5) and B in ratio 3 : 4, find the co-ordinates of B. 12. Find the ratio in which the point (–3, p) divides the line segment joining the points (–5, –4) and (–2, 3). Hence, find the value of p. 13. The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the co-ordinates of the fourth vertex. 14. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p. 15. If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b. 16. If the co-ordinates of the mid-points of the sides of a triangle are (1, 2), (0, –1) and (2, –1), find the co-ordinates of the vertices. 17. Find the lengths of the medians of a  ABC whose vertices are A(7, –3), B(5, 3) and C(3, –1). 18. If A(5, –1), B(–3, –2) and C(–1, 8) are the vertices of  ABC , find the length of median through A and the co-ordinates of the centroid. 19. Find the coordinates of the point which divides the line segment joining the points (4, – 3) and (8, 5) in the ratio 3 : 1 internally. 20. In what ratio does the point (– 4, 6) divide the line segment joining the points A(– 6, 10) and B(3, – 8)? 21. Find the coordinates of the points of trisection (i.e., points dividing in three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4). 22. Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection. 23. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

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Coordinate Geometry

95

4. Area of a triangle 1   x1  y2  y3   x 2  y3  y1   x 3  y1  y2   2 Hence, the area of  ABC 1   x1  y2  y3   x 2  y3  y1   x 3  y1  y2   2

We have already studied in previous classes about how to calculate the area of a triangle when its base and corresponding height altitude are given. We have used the formula:

Formative Worksheet A(x1,y1) C(x3,y3) B(x2,y2)

X

M

L

N

X

1 Area of triangle   base  height 2 Besides this, we have also studied in previous class about how to find the area of a triangle by using “Heron’s Formula”, When the length of the sides are given. i.e., Area of a triangle =

s.  s  a  s  b  s  c 

abc where s  . 2 Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of  ABC . AL , BM and CN perpendiculars have been drawn from A, B, C to the X-axis. Area of  ABC = Area of trapezium ABML + Area of trapezium ALNC – Area of trapezium BMNC We know that : The area of a trapezium = 1  sumof parallelsides  ×  distanceb/w them  2  Area of  ABC = 1 1  BM  AL   ML   AL  CN   LN 2 2 1   BM  CN   MN 2 1 1   y 2  y1  x1  x 2    y1  y3  x 3  x1  2 2 1   y 2  y3    x 3  x 2  2

41. Find the area of the triangle whose vertices are: (i) (2, 3), (–1, 0), (2, – 4) (ii) (–5, –1), (3, –5), (5, 2) 42. In each of the following find the value of ‘k’, for which the points are collinear. (i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5) 43. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. 44. Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3). 45. You have studied in Class IX, (Chapter 9, and Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for Δ ABC whose vertices are A (4, – 6), B(3, –2) and C(5, 2).

Conceptive Worksheet 24. Find the area of the triangle formed by the points A(5, 2), B(4, 7) and C(7, –4). 25. The vertices of  ABC area A(4, 6), B(1, 5) and

26. 27. 28. 29. 30.

C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively. Such that AD AE 1   . Calculate the area of ADE and AB AC 4 compare it with area of ABC. Find the area of a triangle whose vertices are (1, –1), (– 4, 6) and (–3, –5). Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C (7, – 4). Find the area of the triangle formed by the points P(–1.5, 3), Q(6, –2) and R(–3, 4). Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear. If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

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10th Class Mathematics

96

Summative Worksheet 1.

2.

3.

4.

5.

6.

7.

8.

ABC is a triangle formed by the vertices (0, 5), (6, 13), and (6, −3). If AD⊥BC, then what is the length of AD? (A) 6 units (B) 8 units (C) 6 3 units (D) 8 3 units If the area of triangle formed by the points 9 (−3, 1), (0, h), and (−1, 0) is square units, then 2 what is the value of h? (A) −5 (B) −4 (C) 3 (D) 4 The vertices of a quadrilateral ABCD are A (–1, –2), B (–3, –4), C (–7, –3), and D (8, 4). What is the area of the quadrilateral ABCD? 75 63 (A) square units (B) square units 2 2 49 37 (C) square units (D) square units 2 2 What is the area of a quadrilateral whose vertices in order are (3, 4), (1, 6), (– 3, 3), and (2, – 1)? (A) 19 square unit (B) 21.5 square unit (C) 23 square unit (D) 25.5 square unit If the points (4, 7), (6, ), and (2, 13) lie on a straight line, then the value of  is (A) –2 (B) –1 (C) 1 (D) 2 If the area of a ΔABC is 18 square units, the coordinates of whose vertices are A (7, 3), B (x, 5), and C (4, 9), then the value of x is (A) 8 (B) 9 (C) 12 (D) 13 The points A (5, 2), B (x, y), and C (4, 3) form a triangle ABC such that AB = BC and the area of 7 ΔABC is square units. What are the 2 coordinates of point B? (A) (8, 6) (B) (4, 3) (C) (3, 4) (D) (6, 8) If the points (3, 2), (a, 7), and (–4, 3) are collinear, then the value of a is (A) –32 (B) –28 (C) –24 (D) –18

9.

10. If point (x, y) lies on the line joining the points (3, 5) and (−2, −3), then the relation between xand y is given by the algebraic relation (A) 5x − 8y + 1 = 0 (B) 8x − 5y + 1 = 0 (C) 3x − 6y − 5 = 0 (D) 6x − 3y − 5 = 0 11. The given figure shows triangle ABC. D and E are points on side AB and AC respectively such AD AE 1 that   . AB AC 3

12.

13.

14.

15.

16.

What is the area of quadrilateral BDEC? (A) 60 unit2 (B) 90 unit2 (C) 160 unit2 (D) 180 unit2 What are the coordinates of the point which divides the line segment with end points (–2, 8) and (10, 4) in the ratio 1:3? (A) (1, 7) (B) (7, 1) (C) (1, 5) (D) (5, 1) In what ratio does the point (5, 4) divide the line segment joining the points (3, 2) and (8, 7)? (A) 1: 3 (B) 2: 3 (C) 3: 2 (D) 3: 1 In what ratio does the x-axis divide the line segment whose end points are (–6, 3) and (5, –7)? (A) 3:7 (B) 7:3 (C) 6:5 (D) 5:6 In which ratio does the line 2x + y – 8 = 0 internally divides the line segment joining the points (2, 3) and (5, 7)? (A) 1:7 (B) 2:7 (C) 1:9 (D) 2:9 If a triangle whose vertices are (6, 12), (6, 18),





and 6  3 3,15 is inscribed in a circle, then what is the measure of the radius of the circle? (A)

What is the length of PS? 18 54 (A) units (B) units 5 109 72 27 (C) units (D) units 7 10

3 3 units 2

(B) 2 units

(C) 2 3 units (D) 3 units 17. In which ratio is the line segment joining the points (1, −7) and (−19, 23) divided by the line, 2x + 7y − 4 = 0? (A) 1: 4 (B) 3: 7 (C) 2: 3 (D) 7: 8

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Coordinate Geometry

97

18. A point P lies on the line segment joining the PA 3 points A(−5, 6) and B(9, 6) such that  PB 4 What are the coordinates of point P? (A) (7, 6) (B) (1, 1) (C) (1, 6) (D) (7, 1) 19. If the coordinates of the end-point of the diameter of a circle with centre (−1, 2) are (5, 9), then what are the coordinates of the other end-point of the diameter? (A) (−7, −5) (B) (−3, −3.5) (C) (2, 5.5) (D) (6, 7) 20. ABC is a triangle whose vertices are A(−2, 4), B(3, 6), and C(1, 1). Which type of triangle is ABC? (A) Scalene triangle (B) Equilateral triangle (C) Isosceles triangle (D) Scalene right-angled triangle 21. An insect moves in the x − y plane in such a way that at some point of time t, its position is at the point P(5, −3).

26. If a rectangle ABCD has vertices A (2, 6), B (6, 2), C (3, −1), and D (−1, 3), taken in order, then what is the perimeter of the rectangle? (A) 16 3 units (B) 14 2 units (C) 12 2 units (D) 11 2 units 27. If the points P (1, 3) and Q (−2, 6) lie on the boundary of a circle with the centre at O (5, y), then what is the value of y? (A) 8 (B) 9 (C) 10 (D) 11 28. If (x, y) is equidistant from points (4, 2) and (−2, 3), then which of the following equations gives the relation between x and y? (A) 12x − 2y − 7 = 0 (B) 12x + 10y − 7 = 0 (C) 7x − 12y + 10 = 0 (D) 10x − 7y − 12 = 0

IIT JEE Worksheet I.

1.

2. The coordinates of the points on the x axis which are at a distance of 5 units from the position of the insect are (A) (1, 0) (9, 0) (B) (3, 0) (6, 0) (C) (4, 0) (8, 0) (D) (0, −3) (0, 2) 22. The quadrilateral ABCD whose vertices are A (3, 0), B (–1, 3), C (–4, –1), and D (0, –4) is a (A) rectangle (B) square (C) rhombus (D) kite 23. If (x, y) is equidistant from the points (–2, 8) and (7, 12), then which of the following equations gives the relation between ‘x’ and ‘y’? (A) 9x – 7y = 12 (B) 18x + 8y = 125 (C) 12x – 5y = 20 (D) 11x + 13y = 160 24. The perimeter of a triangle whose vertices are (–1, 6), (11, 2), and (3, –6) is (A) 8

 10  2  units

(B) 16 5 units

(C) 4



(D) 12 10units



5  2 units



3,3  3 is a/an

(A) scalene triangle (B) equilateral triangle (C) isosceles triangle (D) right triangle

(B)

  2 a 2  b 2 cos     4 

(C)

  2 a 2  b2 sin     4 

  a 2  b 2 cos     4  The ratio in which the line segment joining (2, –3) and (5, 6) is divided by (i) x-axis (ii) yaxis is (A) 1 : 2 (internal), 2 : 5 (external) (B) 1 : 2 (internal), 2 : 5 (internal) (C) 1 : 2 (external), 2 : 5 (internal) (D) 1 : 2 (external), 2 : 5 (external) The points (–2, –5), (2, –2), (8, a) are collinear, then the value of a is 5 5 3 1 (A)  (B) (C) (D) 2 2 2 2 The area of the triangle with vertices at the points (a, b + c), (b, c + a), (c, a + b) is (A) 0 (B) a + b + c 1 1 1 (C) ab + bc + ca (D)   a b c (D)

3.

4.

25. The triangle whose vertices are (5, 4), (3, 2), and

4 

Straight Objective Type This section contains multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. The join of (–3, 2) and (4, 6) is cut by x-axis in the ratio (A) 2 : 3 internally (B) 1 : 2 externally (C) 1 : 3 externally (D) 3 : 2 internally Find the distance between the pair of points, (a sin a, –b cos a) and (–a cos a, b sin a).   (A) 2 a 2  b 2 cos     4 

5.

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10th Class Mathematics

98 6.

7.

8.

9.

10.

11.

12.

13.

14.

If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are vertices of a parallelogram PQRS, then (A) a = 2, b = 4 (B) a = 3, b = 4 (C) a = 2, b = 3 (D) a = 3, b = 5 The medians of a triangle meet at (0, –3) and two vertices are at (–1, 4) and (5, 2). Then the third vertex is at (A) (4, 15) (B) (–4, –15) (C) (–4, 15) (D) (4, –15) The area of the triangle with vertices at (–4, 1), (1, 2), (4, –3) is (A) 17 (B) 16 (C) 15 (D) 14 The ratio in which the x-axis divides internally the line segment joining (2, –3) and (3, 5) is (A) 1 : 2 (B) 2 : 3 (C) 3 : 5 (D) 5 : 7 The value of k if the point P(–1, 2) is equidistant from the points A(2, k) and B(k, –1) is 1 1 1 1 (A) (B) (C) (D) 3 2 4 5 If (2, 2p + 2) is the mid-point of (3p, 4) and (–2, 2q), the value of p and q are (A) 2, 4 (B) 3, 6 (C) 7, 9 (D) 8, 10 The co-ordinates of mid points of the sides of a triangle are (4, 2), (3, 3) and (2, 2). Then the coordinate of the centroids of the triangle are  7 (A)  3,  (B) (3, 3)  3 (C) (4, 3) (D) (1, 2) The ratio in which the line x + 2y – 4 = 0 divides the join of (–1, 3) and (3, –1) is (A) 1 : 2 (B) 1 : 4 (C) 1 : 3 (D) 1 : 5 The centroid of the triangle whose vertices are given in this question (4, –8), (–9, 7), (8, 13) is (A) (0, 4) (B) (1, 4) (C) (2, 4) (D) (3, 4)

15. Which point on y-axis is equidistant from (2, 3) and (–4, 1) ? (A) (0, 4) (B) (0, –1) (C) (0, 0) (D) (0, 5) II. Multiple Correct Type This section contains multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 16. The ratio in which the point (8, 4) does not divides internally the line segment joining the points (9, 6) and (5, –2) is (A) 1 : 3 (B) 2 : 3 (C) 3 : 4 (D) 5 : 3 17. A triangle with vertices (4, 0), (–1, –1), (3, 5) is not form a type of figure (A) isosceles and right angled (B) isosceles but not right angled (C) right angled but not isosceles (D) neither right angled nor isosceles 18. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus would not be (A) square (B) circle (C) straight line (D) two intersecting lines 19. The co-ordinates of the points which divide the line-segment joining the points (–4, 0) and (0, 6) in four equal parts are. Some of its points are (A) (–3, 1.5) (B) (–1, 5) (C) (–2, 1.5) (D) (–4, 5) 20. The ratio in which the line joining (1, 3) an (2, 7) is not divided by 3x + y = 9 is (A) 3 : 4 (B) 2 : 4 (C) 1 : 2 (D) 3 : 1

III. Matrix Match Type Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. 21. Column II give distance between pair of points given in column I match them correctly Column I Column II (A) (-5, 7) (-1, 3)

(p)

7

(B) (5, 6), (1, 3)

(q)

8

(C) ( 3  1,1)(0, 3)

(r)

6

(D) (0,0)( 3, 3)

(s) 4 2

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Coordinate Geometry

99

22. Column II gives the coordinates of the point P that divides the line segment joining the points given in Column I, mach them correctly Column I Column II (A) A(-1, 3) and B(-5, 6) internally in the ratio 1 : 2 (p) (7, 3) (B) A(-2, 1) and B(1, 4) internally in the ratio 2 : 1 (q) (0, 3) (C) A(-1, 7) and B(4, -3) internally in the ratio 2 : 3 (r) (1, 3) (D) A(4, -3) and B(8, 5) internally in the ratio 3 : 1 (s) (1, 0) 23. Column II give the triangles whose vertices are given in column I, match them correctly. Column I Column II (A) (2, 3), (-1, 0), (2, -4) (p) 40 (B) (-5, -1), (3, -5), (5, 2) (q) 24 (C) (1, -), (-4, 6), (-3, -5) (r) 32 (D) (0, 0), (8, 0), (0, 10) (s) 10.5 24. Match the column Column I Column II Vertices of  Area of  (in sq. units) (A) A(2, 3), B(-2, 1), C(3, -2) (p) 10 (B) A(0, 0), B(–2, 6), C(3, 1) (q) 0 (C) A(8, 10), B(5, 7), C(-1, 1) (r) 11 (D) A(–5, 1), B(1, –1), C(4, –2) (s) 14 IV. Integers Type This section contains questions where the answer to each of the question is a single digit integer, ranging from 0 to 9. 25. The slope of x-axis is 26. Find the square of distance of the point (1, 2) from the mid-point of the line-segment joining the points (6, 8) and (2, 4) is 27. Area of the triangle formed by the points (2, 4), (2, 6) and (3, 5) is equal to 28. what is the abscissa of circum centre of the triangle formed by the points (4, 0), (0, 6) and (0, 0) ? 29. For what value of p are the points (2, 1), (p, –1) and (–1, 3) collinear? V.

Comprehension Type The following question is based on the paragraph given below 30. ABCD is a quadrilateral formed by the points A(–2, –2), B(5, 1), C(2, 4) and D(–1, 5) (i) What is the area of ACD (ii) What is the area of ABC (iii) What is the ratio of area of ABC and quadrilateral ABCD

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100

10th Class Mathematics

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Some Applications of Trigonometry

Learning Outcomes

By the end of this chapter, you will understand

Chapter - 8

10th Class Mathematics

101

 Introduction  Heights and Distance

1. Introduction Trigonometry is one of the most ancient subjects studied by scholars all over the world. The astronomers used trigonometry to calculate distance from the Earth to the planets and stars. Trigonometry is also used in geography to construct maps, determine the position of an island in relation to the longitudes and latitudes, etc.

2. Heights and distances Let A be the top of a tower and C be the eye of a person from where he is observing the top of a tower, then AC is called the line of sight. The angle BCA , so formed by the line of sight with the horizontal level is called the angle of elevation of the top of tower from the eye of a person. Horizontal level

A



Angle of depression LI NE OF SIG HT

D

level is called the angle of depression. The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal level when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed. Now, in right ABC, we have BC || AD and AC is a transversal.  ACB  CAD  Alternate angles      Hence, Angle of elevation = Angle of depression.

Solved Examples Example 1 Determine the height of a mountain if the elevation of its top at an unknown distance from the base is 30o and at a distance 10 km further off from the mountain, along the same line, the angle of elevation is 15o. [Use tan15o = 0.27] A

h km

Angle of elevation  B C Hence, the line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer, i.e., the angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level. Let C be an object and A be the eye of a person from where he is observing the object C, then AC is called the line of sight. The angle CAD , so formed by the line of sight with the horizontal level is called the angle of depression of the object fro the eye of a person. Hence, the line of sight is the line drawn from the eye of angle observer to the point in the object viewed by the observer, i.e., the angle so formed by the line of sight with the horizontal

30° B

x km

15° C

D

10 km

Solution Let AB be the mountain of height h km. Let C be a point at a distance of x km from the base of the mountain such that the angle of elevation of the top at C is 30o. Let D be a point at a distance of 10 km from C such that the angle of elevation at D is 15o in figure. AB Now, in right ABC , we have tan 30º = BC 1 h   3 x  x  3h

… (i)

10th Class Mathematics

102 In right ABD,

AB we have = tan15o  BD h  0.27  x  10  h = (0.27) (x + 10) … (ii) Substituting the value of x from equation (i) in equation (ii), we get h = (0.27) (x + 10)

 h   0.27 



3 h  10



200 y 200  y 3 

3

… (ii)

Adding (i) and (ii), we get x + y = 200 + 200  xy





3 1 3

 1   x  y  200  1  m 3   x + y = 315.4 m Hence, the width of the river BC = (x + y) m = 315.4 m

 h  0.27  3 h  0.27  10  h  0.27  3 h  27  h 1  0.27  1.732   2.7  h 1  0.46   27  h  0.54  2.7

2.7 0.54 h=5 Hence, the height of the mountain (AB) = h km = 5 km. h

Example 3 Two pillars of equal height are on either side of a road, which is 100 m wide. The angles of elevation of the top of the pillar are 60o and 30o at a point on the road between the pillars. Find the position of the point between the pillars and the height of each pillar. A

Example 2 An aeroplane at an altitude of 200 m observes the angles of depression of opposite points on the two banks of a river to be 45o and 60o. Find the width of the river. P

A

Q 60°

45°

200 m

B

45°

60°

D xm

200 3

C

ym

Solution Let A be an aeroplane and AD be an altitude of 200 m. The observer observes the angles of depression of opposite points B and C on the two banks of a river as 45o and 60o respectively. Let BD = x m and Cd = y m, then we have to find the width of the river, i.e., BC = (x + y) m in figure. AD Now, In right ADB , we have tan 45º = BD 200  1 x  x = 200 m … (i) AD Again, in right ADC, we have tan 60º = DC

C E

h B

60°

30° E

xm

h D

(100-x)m

Solution Let AB and CD be two pillars of equal height h m. Let E be a point on the road such that BE = x m. BD (width of the road) = 100 m. Then ED = (100 – x) m. AEB  60 o and CED  30 o are given in figure. Now, In right ABE , we have AB tan 60o  BE h  3 x h … (i)  x 3 Again, in right CDE, we have CD 1 h tan 30o    ED 3 100  x … (ii)  100  x  3 h Substituting the value of x from equation (i) in equation (ii), we get 100  x  3 h

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Some Applications of Trigonometry

103

100 3  h h  3h  3h  3 3  100 3  h  3h  100 3  4h



… (iii)  h  25 3m  h  25  1.732m  h = 43.3 m Again substituting the value of h in equation (i) from (iii), we get 25 3 h  x m  25m x 3 3 Hence, the position of the point E from the first pillar is 25 m and 75 m from the second pillar. The height of the pillars = hm = 43.3 m

 t

 100 

Example 4: A man on the top of a vertical tower observes a car having at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to charge from 30o to 45o, how soon after this, will the car reach the tower? Give your answer to the nearest second. A E 30° 45°

3  vt  vt  12v

 3 t  t  12

 3 t  t  12

12 



12 3 1



 6

3 1

t 12 3 1



3 1  t  691.732  1





3  1  12

3 1



3 1



3 1

 t  6  2.732 

 t = 16.39 minutes  t = 16 minutes 23 seconds. [ 0.39  60  23 ] Hence, the car will reach the tower from D in 16 minutes and 23 seconds.

KEY POINTS Angle of Elevation  XOM i.e., the angle in which the line joining the object and the eye makes with the horizontal through the eye is called the angle of elevation of M as seen from O. M

h

45° B

vt

30° D

C

12v

Solution Let AB be the tower of height h m. Let C be the initial position of the car and after 12 minutes the car be at D. The angles of depression at C and D are 30o and 45o respectively. Let the speed of the car be v m/s in figure.  Distance traveled by the car in 12 minutes = 12 v m [ distance = speed × time] Let the car take t minutes to reach the tower AB from D. Then DB = vt m. Now, in right ABD , we have AB h tan 45o  1   h = vt … (i) BD vt Again, in right ABC, we have AB 1 h tan 30o    BC vt  12v 3 … (ii)  3 h  vt  12v Substituting the value of h from equation (i) in equation (ii), we get 3 h  vt  12v

X O Angle of Depression  XOP i.e., the angle in which the line joining the object and the eye makes with the horizontal through the eye is called the angle of depression of P as seen from O. O X

Q

or The angle between the horizontal line drawn through the observer’s eye and the line joining the eye to any object is called, the angle of elevation of than object when it is at joining the eye to any object is called, the angle of elevation of than object when it is at higher level than the eye. The angle of depression of the object when it is at a lower level than the eye.

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10th Class Mathematics

104 Formulae 1. The angle of elevation measured from the points A, B on a horizontal line from the floor of the tower are ‘’ and ‘’ if AB = d, then the height of the tower ‘h’ is

4.

D

The angle of elevation cloud from a height‘d’ metres above the level of water in a lake is  and the angle of depression of its image in the lake is . The height of the cloud from the water level in metres is T

h

A

B

2.

d tan .tan  dsin .sin   or  h  tan   tan  sin     

h

d cot   cot  The angle of elevation measured from the points A, B on a horizontal line either side to the floor of the tower are ‘’ and ‘’, If AB = d. Then the height of the tower ‘h’ is

 or 

h

5.

C

3.

d sin      sin     

 or 

h

d  tan   tan   tan   tan 

A B

 or 

h

d h cot   cot 

d tan .tan  tan   tan  The angle of elevation of the top of a tower from the bottom and top of a building of height ‘d’ metres are  and  respectively. The height of the tower is P

8

D a

h

P

B

6.

A balloon is observed simultaneously from the three points P, Q, R on a straight road directly beneath it. the angular elevation at Q is twise that at P and the angular elevation at ‘R’ is thrice that at P. If PQ = a and QR = b then the height of the balloon ‘h’ in terms of a and b is, a h  3b  a  a  b  2b

h

T

T

S d

d 20

Q d sin .cos  h sin     

R

D

d sin .sin   h sin     

 or 

P

h

 cot   cot   or h  d    cot   cot   The angle of elevation of a hill from a point C is . After walking to some point D at a distance ‘a’ metres from C on a slope of inclination , the angle of elevation was found to be , then a sin .sin      h sin     

h A

S

C

d

 h

Q

 or 

R d cot  h cot   cot 

h

S

R

P

Q b

a

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Some Applications of Trigonometry 7.

105

A round balloon of radius r subtends an angle  at the eye of an observer, while the angle of elevation of its centre is . Then the height of the centre of the balloon is  h  r sin .cosec   2 A

90°

2.

3.

O 4.

B h 8.

P A flag staff stands on the top of a tower of height h metres. If the tower and flag staff subtended equal angles at a distance ‘d’ metres from the foot of the tower, then the height of the flag staff in metres is  d2  h 2  h 2 2  d  h  R

h

T

Q

d

Formative Worksheet 1.

If the angle of elevation of a cloud from a point h m above a lake is  and the angle of depression of its reflection in the lake is , prove that the h  tan   tan   height of the chord is . tan   tan  C x 

D 

h

E h

A

B

(x + h)

5.

A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point h metres above the first, the depression of the foot of the tower is 60°. The horizontal distance of the tower from the point is The angle of elevation of the top of a T.V. Tower from three points A, B, C in a st. line thro’ the foot of the tower are , 2, 3 respectively. If AB = a, the height of the tower is A flagstaff on the top of a house subtends the same angel  at two points distance a and b from the house and on the same side of it, then the length of flagstaff is ABC is a triangular park with AB = AC = 100 metres. A clock tower is situated at the mid-point of BC. The angles of elevation of the top of the tower at A and B are cot-13.2 and cosec-1 2.6

respectively. The height of the tower is A man is standing on the deck of a ship, which is 8m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill. 7. A vertical to stands on a horizontal plane and is surrounded by a vertical flag staff of height 6 meters. At point on the plane, the angle of elevation of the bottom and the top of the flag staff are respectively 30° and 60°. Find the height of tower. 8. The angles of depressions of the top and bottom of 8 m tall building from the top of a multistoried building are 30° and 45° respectively. Find the height of multistoried building and the distance between the two buildings. 9. The angle of elevation of an aeroplane from a point on the ground is 45°. After a flight of 15 sec, the elevation changes to 30°. If the aeroplane is flying at a height of 3000 metres, find the speed of the aeroplane. 10. An aeroplane flying horizontally at a height of 1.5 km above the ground is observed at a certain point on earth to subtend an angle of 60º. After 15 seconds, its angle of elevation is observed to be 30º. Calculate the speed of the aeroplane in 6.

km/h  3  1.732 . F ___________________________________________________________________________________________________________________________________________

10th Class Mathematics

106

Conceptive Worksheet 1.

From the top of a light house 60 m high with its base at the sea level the angle of depression of a boat is 15°. The distance of the boat from the foot of light house is 2. At a distance 2h meter from the foot of a tower of height h metre the top of the tower and a pole at the top of the tower subtend equal angles. Height of the pole should be: 3. A house subtends a right angle at the window of an opposite house and the angle of elevation of the window from the bottom of the first house is 60°. If the distance between the two houses be 6 m, then the height of the first house is: 4. The base of a cliff is circular. From the extremities of a diameter of the base of angles of elevation of the top of the cliff are 30° and 60°. If the height of the cliff be 500 m, then the diameter of the base of the cliff is: 5. A flat staff 20 m long standing on a wall 10 m high subtends an angle whose tangent is 0.5 at a point on the ground. If  is the angle subtended by the wall at this point, then: 6. From the top of a cliff of height a, the angle of depression of the foot of a certain tower is found to be double the angle of elevation of the top of the tower of height h. If  be the angle of elevation, then its value is: 7. A boat is being rowed away from a cliff 150 m high. At the top of the cliff the angle of depression of boat changes from 60° to 45° in 2 min. Then find the speed of the boat (in m/h) 8. A tower stands at the top of a hill whose height is 3 times the height of the tower. The tower is found to subtend at a point 3 km away on the horizontal through the foot of the hill, the angle  1 where tan   . The height of the tower is: 9 9. An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60° and after 10s the elevation is observed to be 30°. The uniform speed of the aeroplane(in km/h) is 10. An observer on the top of tree, finds the angle of depression of a car moving towards the tree to be 30°. After 3 minutes this angle becomes 60°. After how much more time, the car will reach the tree:

11. A ladder rests against a wall making an angle  with the horizontal. The foot of the ladder is pulled away from the wall through a distance x, so that it slides a distance y down the wall making an angle  with the horizontal. The correct relation is: 12. From of 60 m high tower angle of depression of the top and bottom of a house are  and  respectively. If the height of the house is 60 sin      is equal, then x: x 13. A spherical balloon of radius r subtends an angle  at the eye of an observer. If the angle of elevation of the centre of the balloon be , the height of the centre of the balloon is: 14. A balloon is observed simultaneously from three points A, B and C on a straight road directly under it. The angular elevation at B is twice and at C is thrice that of A. If the distance between A and B is 200m and the distance betwen B and C is 100 m the height of balloon is given by: 15. The angle of elevation of a cloud from a point 60 m above a lake is 30o and the angle of depression of the reflection of cloud in the lake is 60o. Find the height of the cloud. C hm A 60m

 

F 60m

B

D

(60 + h)m

E

Summative Worksheet 1.

Two ships are observed from the top of a lighthouse at a certain point of time. The ships are sailing on its either side such that the two ships and the lighthouse fall in a straight line. The top of the lighthouse is 60 m above sea level. The angles of depression of the two ships made from the top are 30° and 60° at the given instant.  use 3  1.732 (A) 125.64 m (B) 129.35 m (C) 138.56 m (D) 142.45 m

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Some Applications of Trigonometry 2.

3.

4.

5.

6.

Rohan and Sohan are sitting in a double-decker bus. Rohan is sitting on the upper deck of the bus at a height of 4.5 m from the ground. Sohan is sitting on the lower deck of the bus such that Rohan is vertically above Sohan. Both of them notice a boy on the road. The angles of depression made by them at that instant are 60° and 45° respectively. At what height from the ground is Sohan sitting in the bus? (A) 4.5 3 (B) 3.8 3 4.5 3.8 (C) m (D) m 3 3 A shooter is standing on the ground and is looking at a bird up in the sky, which is at a vertical distance of 11 m. The shooter points the gun towards the bird at an angle of elevation of 30° with the horizontal. If the shooter is holding the gun at a height of 1 m from the ground, then what is the horizontal distance between the bird and the shooter? (A) 10 3 (B) 10 m 10 (C) (D) 3m 3 Two poles of the same height stand opposite to each other on either sides of a road of width 20 m. From a point on the road, the angles of elevation of the top of the poles are 45° and 60°respectively. 3  1.732, 2  1.414 What is the height of each pole? (A) 11.24 m (B) 12.68 m (C) 13.52 m (D) 14.36 m An aeroplane was flying at a height of 7000 3 m above the ground. From two observation points, the angles of elevation of the aeroplane were recorded as 30° and 60° at the same time. Assume that at that particular instance, the aeroplane was directly above the line joining the two observation points. What is the distance between the observation points? (A) 30 km (B) 28 km (C) 20 km (D) 14 km Aman and Suman are standing on a road facing the east and west sides of a building at a distance of x1 and x2 respectively. The building is 30 m long. The angles of elevation made by Aman and Suman to the top of the building are 45° and 30° respectively.  use 3  1.732 What are the respective values of x1 and x2? (A) 10 and 2.73 (B) 20 and 11.6 (C) 20 and 21.3 (D) 30 and 51.9

107 7.

Vicky is standing at a traffic signal and is observing the light. The angle of elevation made by his eyes while observing the traffic signal is 45°. If the height of the traffic signal is 7.8 m, then what is its horizontal distance from Vicky? (A) 7.8 m (B) 8.8 m 7.8 (C) 7.8 3m (D) m 3

IIT JEE Worksheet I.

1.

Straight Objective Type This section contains multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. As observed from the top of a light house, 100 m above area level, the angle of depression of a ship, sailing directly towards it, changes from 30o to 45o. Determine the distance traveled by the ship during the period of observation. A

E 30° 45°

100m

45° B

2.

30° D

xm

C

ym

(A) 10 m (B) 7.32 m (C) 40 m (D) 21 m From the top of a building 60 m high the angles of depression of the top and the bottom of a tower are observed to be 30o and 60o. Find the height of the tower. A (60-h)m 30°

C

xm

hm

hm

D

E

60°

B xm

(A) 20 M (C) 60

(B) 25 (D) 40 M

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10th Class Mathematics

108 3.

A man is standing on the deck of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60o and the angle of depression of the base of the hill as 30o. Calculate the distance of the hill from the ship and the height of the hill.

8.

The angle of elevation of a cloud from a point situated at a metre height from a lake is a and the angle of depression of its reflection is b. Height of the cloud is (A)

C

(C) 9.

hm

60° E

xm

D

30° xm

10 m

4

5.

A 10 m B

(A) 10 M (B) 40 M (C) 80 (D) 35 M The angle of elevation of the top of a tower from the top and bottom of a building of height ‘a’ are 300 and 450 respectively. If the tower and the building stand at the same level, the height of the tower is (A) a 3 (B) a( 3  1) (3  3) (C) a (D) a( 3  1) 2 The angles of elevation of the top of a tower at two points which are at distances a and b from the foot in the same horizontal line and on the same sides of the tower, are complementary. The height of the tower is(A) ab (B) ab

(C) a / b (D) b / a *6. From the top of h metres high cliff, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower is2h h (A) (B) 3 3 2h (C) (D) h 3 3 7. From vertically situated aeroplane to the straight horizontal road, the angle of depression of two consecutive km stones are a and b. If an aeroplane is in vertical plane in between two stones, then the height of the areoplane from the road ( in kilometres) will be (A)

tan  tan  tan   tan 

(B)

(C)

tan   tan  tan   tan 

(D)

tan  tan  tan   tan  tan   tan  tan   tan 

asin      sin      a sin      sin     

m (B)

asin      sin     

m

m (D) a sin(  )

The angle of elevation of the top of a tower at a distance of 500 metres from its foot is 300. Then

height of the tower is 500 3 500( 3  1) (A) (B) m m 3 3 500( 3  1) (C) (D) 500 m m 3 10. Each side of an equilateral triangle subtends an angle of 60° at the top of a tower h m high located at the centre of the triangle. If a is the length of each side of the triangle, then P

B A O

C

(A) 3a2 = 2h2 (B) 2a2 = 3h2 2 2 (C) a = 3h (D) 3a2 = h2 11. A tower subtends an angle of 30° at a point on the same level as its foot, and at a second point h m above the first, the depression of the foot of tower is 60°. The height of the tower isB

h

P 30°

X 60°

O

30°

A

(A) h m

(B) 3h m h (C) 3 hm (D) m. 3 12. A tower subtends an angle of 30° at a point on the same level as the foot of the tower. At a second point, h metre above first, the depression of the foot of the tower is 60°, the horizontal distance of the tower from the point is(A) h cos 60° (B) (h/3) cot 30° (C) (h/3) cot 60° (D) h cot 30°

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Some Applications of Trigonometry 13. The top of a hill observed from the top and bottom of a building of height h is at angles of elevation p and q respectively. The height of the hill ishcot q cot q  cot p h tanp (C) tanp  tan q

(A)

(B)

hcot p cot p  cot q

(D) h cot p

14. The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle 30° with the horizontal, then the length of the wire is (A) 12 m (B) 10 m (C) 8 m (D) 14 m 15. If two towers of heights h1 and h2 subtend angles 60° and 30° respectively at the midpoint of the line joining their feet, then h1 : h2 = (A) 1 : 2 (B) 1 : 3 (C) 2 : 1 (D) 3 : 1 II. Multiple Correct Type This section contains multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 16. A pole 15m long rests against a vertical wall at an angle of 60° with the ground. How high up the wall will the pole reach? 15 3 (B) 8 3 m m 2 (C) 12.99 m (D) 14.38 m 17. A man in a boat rowing away from light house 100 m. He takes 2 minute to change the angle of elevation of the top of light house from 60° to 45°. What is the speed of the boat?

(A)

109

5 18 3  3 m/s (B) 3 1 m / s 18 5 (C) 2.364 m/hr. (D) 1.268 km/h 18. From a point on the ground the angle of elevation of the bottom and top of a transmission elevation of the bottom and top of a transmission tower fixed at the top of a 20m. High building are 45° and 60° respectively. What is the height of the tower? (A)





(A) 20 3 m



(B) 20







3 1 m

(C) 14.64 m (D) 14.94 m 19. The angles of elevation of the top of a rock from the top and foot of a 100m. High tower are respectively 30° and 45°. What is the height of the rock? (A)

3 1

m

(B)

100 3

m 100 3 3 1 (C) 246.5 m (D) 236.5 m 20. A hedgehog wishes to cross a road without being runover. He observer the angle of elevation of a lamp post on the other side of the road to be 45° from the edge of the road and 30° from a point 10m back from the road. how wide is the road? 10 (A) 13.90 m (B) m 3 10 (C) (D) 13.66 m m 3 1

III. Matrix Match Type Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. 21. From the top of a cliff 50 m high, the angles of depression of the top and the bottom of a tower are observed to be 30º and 45º respectively, match them correctly Column I Column II  50  (A) Horizontal distance of point of observation from top of tower (p)  50  m 3  (B) Height of tower (q) 50 m (C) Horizontal distance of bottom of cliff from bottom of tower (r) 50 2m (D) Distance of top of cliff from bottom of tower (s) 50 / 2m

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10th Class Mathematics

110

22. From a window, h metres high above the ground, of a house in a street, the angles of elevation and depression of the top and bottom of another house on the opposite side of the street are and , respectively, then match them correctly Column I Column II (A) DB (p) h(1  tan  cot ) (B) CE (q) h sin  (C) CE (r) h tan cot  (D) AD (s) h cot  23. A ladder (length = l) rests against a wall at an angle  to the horizontal. Its foot is pulled away from the wall through a distance d, so that it slides a distance h down the wall, making an angle  with the horizontal, then match the column Column I Column II (A) d (p) l (sin   sin ) (B) h (q) l (cos   cos ) (C) d/h

(r)

(D) dh/l2 (s) sin(  )  cos(  ) 24. Column I

cos   cos  sin   sin 

Column II

A D

60 ° 45° C B 50m (A) Measure of AD





(p) 6 5  2 2 m

A

C

45° 12m

B 5m D

(B) Perimeter of

(q) 10 m

A

60m B

(C)

30°

45° C D Distance between BC

(D) If the shadow of a tower is 30m. When the sun’s elevation is

 (s) 60  (r) 50

 3  1 m

3 1 m

30°, Find the length of the shadow when sun’s elevation is 60° 26. A statue 1.46 m tall, stands on top of a pedestal. IV. Integers Type This section contains questions where the answer From a point on the ground the angle of to each of the question is a single digit integer, elevation of the top of the statue is 60° and from ranging from 0 to 9. the same point the angle of elevation of the top 25. A bridge across the river makes an angle of 30° of the pedestal is 45°. What is the height of the with the river bank. The length of the bridge pedestal? across the river is 60m. If width of the river is k meter, then what is the value of k/15m. ___________________________________________________________________________________________________________________________________________

Some Applications of Trigonometry 27. The angle of depression of point C, when observed from point A is 45°. If BC = 1m AB determine . 1m 28. The tops of two poles of height 16m and 10m are connected by a wire. If the wire makes an angle of 30° with the horizontal and x is the length of x wire then find 4m 29. In the given figure QR = 20cm, PS = 13 cm, SR PQR  45o , then value of 7cm

111

P 13cm S

45° R Q 20cm V. Comprehension Type The following question is based on the paragraph given below 30. A guard observes an enemy boat, from an observation tower at a height of 200m above sea level to be an angle of depression of 30°. (i) What is the distance of the boat from the foot of the observation tower. (ii) If the boat is 200m from the observation tower, then what is the angle of depression.

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112

10th Class Mathematics

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Binomial Theorem By the end of this chapter, you will understand      

History Pre requisites Pascal’s Triangle or Meru Prastra. Binomial Theorem for Positive Integral Index General term of Binomial Expansion Middle term (or) terms of Binomial Expansion

 Independent term of Binomial Expansion  Coefficient of xth term of Binomial Expansion

Example: (a + b)2, (x + a)5, (x2 + y2)10......... are all binomial expansions. Let us observe the following algebraic identities (x + a)° = 1 (x + a)1 = x + a (x + a)2 = x2 + 2xa +a2 (x + a)3 = x3 + 3x2a +3xa2 + a3 (x + a)4 = x4 + 4x3a +6x2a2 + 4xa3 + a4 Observations:

1. History We have to thank Plague for the binomial theorem In 1665, plague was raging in England, and Isaac Newton, a fresh (and undistinguished) graduate of the University of Cambridge, was forced to spend most of the next two years in the relative safety of his family’s country manor in Woolstrope. It turned out that solitude and free time were just the right stimulus Newton’s creative brain needed. In that 18-month period of retreat, he came up with his proof and extension of the ‘Binomial theorem’, invented Calculus (which he called his method of fluxions), discovered the Law of universal gravitation, and proved that white light is composed of all colours. All of this before the age of 25. Ancient Indian and Chinese mathematicians also knew the binomial theorem. And in Europe, already a century before Newton’s birth, Blaise Pascal’s treatise on the Arithmetical Triangle provided a handy way to generate binomial coefficients. The details of the above shall be witnessed in the upcoming pages. Before we step into the details let us recap some of the prerequisites to understand better.

2. Pre requisites Binomial Expression: Algebraic expressions which have two terms, are called Binomial Expressions. Example: x + a, x + y, 2y + z, z + b, x2 + y2. Binomial Index: It is the power to which a binomial is raised, to give a Binomial Expansion. Example: In a binomial expansion (x + a)7, the binomial index is 7. Binomial Expansions: Binomials when raised to different powers, are called Binomial Expansions.

Chapter - 9

Learning Outcomes

From the above expansions we notice that: i)

The number of terms in an expansions is one more than the power (exponent) of the binomial.

ii)

The power of the first term is same as the exponent of the binomial. The exponent of y in the first term is zero.

iii)

The power of x begins from n and starts decreasing by 1, till it becomes 0, the power of y begins from 0 and starts increasing by 1, till it becomes n.

iv)

The sum of exponents of x and y each term is equal to the exponent of the binomial .

v)

The exponent of x in the last term is zero and that of y is equal the exponent of the binomial . Apart from these , Do you see a pattern among the various coefficients of each of these expansions ? We can find coefficients of different expansions. It’s possible through Pascal’s Triangle.

10th Class Mathematics

114 Index of Power

Coefficients of Various Terms

0

1

1

1

2

1

3

1

4

1

5

1

6 7

3. Pascal’s Prastra

i) ii)

iii)

i) ii)

1 1

Triangle

Meru

1

6

15

We know the binomial expansion upto 3rd index. If the index is greater then we make use of Pascal’s Triangle to get the coeficients and expansions. Observations: Here, Each row starts with 1 and ends in 1. Any other coefficient (except the first and the last) in any row is sum of two coefficients in the preceding row, one on the immediate left and other on the immediate right. For example in sixth row 5 is obtained by adding 1 and 4. We can continue the above process for higher indices. The above symmetrical pattern followed by the coefficients is called Pascal’s Triangle or Meru Prastra. Using Pascal’s Triangle, we can write down the expansion of a power of a binomial when its index (or exponent) is not very large. Examples: (x + a)5 = x5 + 5x4 a +10x3 a2 +10x2 a3+ 5xa4 + a5 (x + a)6 = x6 + 6x5 a +15x4 a2 +20x3 a3+ 15x2 a4 + 6xa5 + a6. However, It is time consuming to construct t h e triangle when the exponent is very large. Then, how to expand the binomial when it’s power very large? It is with the help of Binomial theorem. Before we step into Binomial theorem let’s lea r n the related terms of Binomial theorem.

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3 4

10

21

or

1

3

5

7

2

4

6

1

10 20

1 5

1

15

35

35

6 21

1 7

1

Factorial Representation: The factorial of n is denoted by n! and is defined as, n! = n(n – 1) (n – 2) ... 3.2.1. Also, 0! = 1 n! can also be expressed as, n! = n(n – 1)! = n(n – 1) (n – 2)! ... and so on.

Note: The factorial of n is also denoted by n Examples: 4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

Combinations: Each of the groups or selections which can be made by taking some or all of a number of different things is called a combination. Number of combinations of n dissimilar things taken r at a time is denoted by nCr and it is defined as n Cr 

n! . ( n  r )!r !

Examples: i) Number of combinations of 6 things taken 3at a time is nC3 6

ii)

C3 

6! 456   20 (6  3)!3! 3  2  1

Number of combinations of 10 things taken 7 at a time is 10C7 10

C7 

10! 10  9  8  7!   120 (10  7)!7! 3! 7!

Binomial Theorem

115

4. Binomial Theorem for Positive Integral Index However, when the index of the power is large, it is not desirable to write the expansion using Pascal’s Triangle. We make use of Binomial Theorem for positive integral index. If n is a natural number, x and a are any numbers, then (x + a)n = nC0xn + nC1 xn–1a + n C2xn–2 a2 + .... + n Crxn–r ar + ... + nCnan.

and n c r =

Formative Worksheet 1.

2. 3. 4.

iii) iv) v)

Number of terms in the binomial expansion Number of non-zero terms in the expansion of 1. (x + y)n + (x – y)n is

n +1 when n is odd. 2

2. (x + y)n + (x – y)n is

n+2 when n is even. 2

3. (x + y)n – (x – y)n is 4. (x + y)n – (x – y)n is

n +1 when n is odd. 2 n when n is even. 2





5

 

2 1 

5

2 1

5.

Expand the following:

6.

 2x 3    .   3 2x  Simplify (x + y)6 + (x – y)6 and hence, evaluate

Salient Features of Binomial Expansion:

ii)

9

 

ii) (2x + 3y – 4z)n Expand (x + y2)6 By using binomial theorem, expand (1 – x + x2)4 Using binomial theorem, expand {(x + y)5 + (x – y)5} and hence find the value of



The number of terms in the expansion is one more than the index or the power of the binomial i.e., if the index is n, then there will be (n + 1) terms in its expansion. When the second term of the binomial expression is negative, then the terms in the expansion are th alternatively positive and negative and the n term is positive or negative according as n is even or odd. The power of a begins from 0 and starts increasing by 1, till it becomes n. The power of x begins from n and starts decreasing by 1, till it becomes zero. In successive terms the power of first term of binomial goes on decreasing and that of second term goes on increasing, however, the sum of indices of the powers of the two terms is constant = n.

9



i) 1  5 2 x  1  5 2 x

n!  n  r ! r!

where, n! = 1 × 2 × 3 × . . . . . .× n

i)

Find the number of terms in the expansions of the follwing:

4

 7. 8.

6

 

3 1 

6



3 1 .

Prove that, nCr = nCn–r Find the factorials:

10 ! 2 ! i) 6 !  2 ! ii) 6 !  5 !

iii) 3 ! 2 ! 4 ! 3 !

Conceptive Worksheet 1.

Expand (2x – 3y)4 by binomial theorem. 6

2. 3. 4. 5.

 2 3 Expand  2x   . x  5. Expand (1-2y) If nC5 = nC10, then nC15 = ––––––––––– Prove that 7C5 + 6C5 + 5C5 = 8C6

5. General term Expansion

of

Binomial

We have seen the various terms in the expansion of (x + a)n like T1 = nC0xn a0, T2 = nC1xn–1 a1, T3 = nC2xn–2 a2, Tn–1 = nCn–2x2 an–2 , n n–1 Tn = Cn–1xa and Tn+1 = nCnx0 an. Any of these terms can be written in the from of the general term. The general term in the expansion of (x + a)n is denoted by Tr + 1 and is given as follows: Tr + 1 = nCrxn–r ar by giving different values to r. It is also called r +1th term.

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10th Class Mathematics

116 Note-1: (r +1)th term from end in the expansion of (x + a)n is (r + 1) th term from beginning in the expansion of (a + x)n. The general term of an expansion (x – a) n is Tr + 1 = (–1)r . nCrxn–r ar

8. Coefficient of xth term of Binomial Expansion To find the coefficient of xkth term in any binomial expansion, the following steps are to be followed. Step 1: Write the general term of the given expansion. i.e., Tr + 1. Step 2: Equate the power of x in Tr + 1 with k and find the value of k. Step 3: Substitute the value of k in Tr + 1 and siplify to get the term with xk . Genral form: The coefficient of xk in the expansion

6. Middle term (or) terms of Binomial Expanssion If there are odd number of terms in an expansion, there is only one middle term. If an expansion has an even number of terms, then there are two middle terms. A binomial expansion of (x + a)n has (n + 1) terms and the middle term depends on the nature of n. Case 1: When n is even. If the index n of the Binomial expansion (x + a)n is even, then the number of terms in the expansion is odd, so there is only one middle term

n

np  k  p b  of  ax  q  is nCr an–r br,where r  p  q x  

Formative Worksheet 7

9.

th

n  T and it is   1 term, i.e., n 1 . 2 2   Case 2: When n is odd. If the index n of the Binomial expansion (x + a)n is odd, then the number of terms in the expansion is even, so there are two middle terms  n 1 and these are    2 

i.e.,

Tn 1 2

and

Tn  3 2

th

10

 2 3  2x  x  .   12. Find the middle term in the expansion of

th

20

2 2 3   x   . 2x  3 9

.

The term independent of x, in an expansion is that term which does not have x in it. In such terms, the power of x is considered as ‘0’. Step 1: Write the general term of the given expansion. i.e., Tr + 1. Step 2: Equate the power of x in Tr + 1 with 0 and find the value of k. Step 3: Substitute the value of k in Tr + 1 and siplify to get the constant term . Genral form: The term independent of x in the n

b expansion of  ax p  q  is n Cr an–r br ,where x  

np pq

12

 x 3a  10. Find the 9th term in the expansion of   2  . a x  11. Find the middle term in the expansion of

 n 3 and   terms,  2 

7. Independent term of Binomial Expansion

r

th 3  Find the 4 term of   y  . x 

 x2  2 x   . 13. Find the middle terms of  4   14. Find the term independent of x in the following expansions 11

10 2 x 1  1    i)  2x   ii)  x 2x x    5 15. Find the coefficients of x32 and x–17 in the expansion

15

16.

17. 18.

19.

 4 1 of  x  3  . x   In the expansions of (1 + a)m+n using Binomial Theorem, prove that coefficients of am and an are equal. Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. Determine the value of x in the expansion (x + xlog 10x)5, if the third term in the expansion is 1,000,000. Prove that the middle term in the expansion of 2n

(1+x) is www.betoppers.com

1.3.5...  2n  1 .2n. x n n

.

Binomial Theorem

117

20. Find the number of terms in the expansion of



4

968



500

which are integers.

21. Find the coefficient of x4 in the expansion of (1 + x + x2 x3)11 . n

 2 1  22. In the expansion of  2  x  it is observed that 4   the third term is seven times the second term and the sum of their coefficient is 36. Find the value of x. 23. The sum of the coefficients of terms of even powers of x in (1 – x + x2 – x3) is 1) 213 2) 210 3) 28 4) 214 n

 3 1  24. If in the expansion of  x  2  , n  N , sum of x   5 10 the coefficients of x and x is 0, then the value of n is 1) 10 2) 15 3) 20 4) 25 25. If the sum of the coefficients in the expansion of n (p + q) is 1024, then the greatest coefficient in the expansion is 10 10 10 10 1) C 2) C 3) C 4) C 3

4

5

7

Conceptive Worksheet 2 14

6. 7.

Find the fifth term in the expansion of (2 + 3x ) . Find the 11th term from the end in the expansion of

8.

1    2x  2  . x   Find the general term and the 13th term in the

9.

 1  expansion of  9 x   . 3 x  Find the middle term in the expansion of

25

18

1  2 x – 2 2x  10. Find the

10

3 4 x 13. Find the coefficient of x in  – 2  . 2 x  7 8 14. If the coefficient of x and x in the expansion of n

x   2   are equal, then find the value of n. 3  15. Find the sum of coefficients of the polynomial, x y z u 1001 (p + 2 q – 5 r + 10 s ) . 16. If the last term in the binomial expansion of n  1 1  3 1   2   is  5 2    33

7

x 

10

 2x2 3   2 . 11. Find the middle term in  2x   3

of

10

x2  1

6

  x 



6

x2  1 .



3

445

120



.

20. find sum of the coefficients in the expansion of 2 3 (1 + x + x + x ) . 21. Find the sum of the coefficients of odd powers of x 2 3 in the expansion of (1 – x + x – x ). 22. Find the sum of the coefficients of even powers of 2 3 7 x in the expansion of (1 – x + x – x ) . 10 10 10 10 10 23. Find: C + C + C + C + .... + C 0 1 2 3 10 24. Find the middle term in the expansion 2 3 50 of (1 + 3x + 3x + x ) . 1 log x  25. Find the third term in the expansion of   x 10  x 

5

Summative Worksheet 1.

3.

Number of terms in the expansion of 99 (2x + 3y) is 1) 198 2) 99 3) 98 4) 100 Number of terms in the expansion of 3 2 100 (x + 3x + 3x + 1) is 1) 101 2) 201 3) 301 4) 401 Number of terms in the expansion of

4.

 2 1   x  2  2  is x   1) 23 2) 24 3) 11 4) 21 Find the middle term(s) in the expansion of

10

12. Find the term independent of x in the expan 1    2 3 2 x  3 x  . sion of   

th

, then find the 5 term

19. Find the number of irrational terms in the expansion

2.

 x3  3 x    . 6  

log3 8

from the beginning. 17. Find the number of terms in the expansion of 15 (a + b + c + d) , if a  b  c  d . 18. Find the number of terms in the expansion of

6

  .  middle terms in the expansion of

   

15

2  x   3x  2  ii)   9 y  x   3 

8

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10th Class Mathematics

118 5.

6. 7.

8.

Find the factorial:

7! 3! 8! i) ii) 4! 5! iii) 5! 3! 3! 4! Find the following combination i) 5C3 ii) 10C6 iii) 8C4 Find the i) Independent term and 3 2 21 ii) x coefficient in the expansion of (2x + 1/3x ) Find the value of 4



a) x  2 b)



c)



  x  2 6

 

2 1 



HOTS Worksheet 1.

Show that there exists an independent term in

n np 1   p ax  only when p  q is a positive  q  bx   integer. 2. If O is sum of odd terms and E that of even terms in

n

the expansion of  x  a  , prove that

4

2 2 2 2 i) O  E   x  a 

n

2n

6

ii) 4OE   x  a    x  a 

2 1

2n

2n

9.

2

5

2

x a  x

 

2

2 2 iii) 2  O  E    x  a    x  a 

5

2

x a x



Find the middle term in the expansion of 9 (3x + 4y) .

Prove that 101  9950  10050 .

4.

Find the i) last digit ii) last two digits and

11. Coefficient of x 16

6

1) C 5 16 10 11 3) C 5 5

iii) last three digits of 17 256 . 5. 6.

16

–18

 2 1 in  5x  3  is x   16 10 2) C 5 6 4) none

7. 8. 10

 x 3  2  12. The term independent of x in   3 x 

3 1) 16

5 2) 3

3 3) 5 2 3

9.

n

9

 2 1  15. Which term is independent of x in  3x   is 3x   1) 6th 2) 7th 3) 8th 4) 9th

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Find the coefficient of x3 in the expansion of 9

n

If 1  x   C0  C1 x  C2 x 2  ......  Cn x n show that

 1 13. If the 5 term of  a   does not contain a then a  n= 1) 19 2) 18 3) 10 4) 9 19 14. The coefficient of x in (x + 1) (x + 4) (x + 9) (x + 16) ..... (x + 400) is 1) 2990 2) 1260 3) 2940 4) 1640 th

32

If 7 divides 3232 , then find the remainder . Find the total number of terms in the expansion of (x + y + z + w)n, n  N. If n is any positive integer, show that 23n+3 – 7n – 8 is divisible by 49? 1 1  x  2 x2   2 x 2  3x 

is

9 4) 19

50

3.

10

 2 1  10. The co-efficient of third term of  x  2  is x   1) 30 2) 50 3) 45 4) 60

2n

1  x 

n

 C0  2C1  3C2  ......  (n  1)Cn .

  n  2  22 n 1

10. If the sum of coefficients of the first, second, and third terms of the expansion of is 46, then coefficient n

 2 1 of  x   the term independent of x is x  1) 48 2) 82 3) 84 4) 92 11. If P be the sum of odd terms and Q that of even terms in the expansion of (x + a)n, then the value of [(x + a)2n – (x – a)2n] equals 1) 3 PQ 2) 4 PQ3) 5 PQ4) 6 PQ 12. If (1 + x – 2x2)10 = 1 + a1 x + a2x2 + .......+a20x20, then the value of a2 + a4 + a6 +.........+ a20 is 1) 511 2) 512 3) 513 4) 514

Binomial Theorem

119

n

13. If

 r 0

r

n

2

Crx = C0 + C1 + C1x – C2x + ....+ Cn x ,

3.

99 19 Prove that 80 divides 19  99

4.

5.

If a + b + c = 0, consider the expansion of (a + b)5, then show that 5ac3 + 10a2c2 + 10a3c + 5a4 = 5bc3 + 10b2c2 + 10b3c + 5b4 (SAT-1992) Find the coefficient of xn+1 in (a0+a1x+a2x2 + anxn)2

6.

1  x 

then calculate the value of n

C02  C12  C22  C32  ......  1 Cn2 if n = 15 is equals

2n! 1) 2) 0 16!16!

2n! 3) 12!13!

the coefficients of xn in 1  x  15. If

the

6

th

term

in

2 n 1

the

2n

ar = r 1 a r 1 1) 110 2) 115 3) 120 4) 135  r

is twice

. expansion

7. of

8

 1  2  8 / 3  x log10 x  is 5600, find x. x  16. Find n, if the ratio of the fifth term from the 8 beginning to the fifth from the end in the expansion

1  4  2  4  is 6 :1 . 3  rd 17. The 3 , 4th and 5th terms in the expansion of (x + a)n are respectively 84, 280 and 560, find the values 9. of x, a and n. 18. The term independent of x in

 x 1 x 1   2 1 1  3  x  x3  1 x  x 2

10

   

is

108

IIT JEE Worksheet 1.

2.

1 , then the coefficient of x r in the 2 1  2x expansion of 2 is 1  2x  If x 

1) r.2r 2)  2r  1 2r 3) r.22r 1 4)  2r  1 2r The binomial coeff. which are in decreasing order are 1) 15 c5 ,15 c6 ,15 c7 , 2) 15 c10 ,15 c9 ,15 c8 3)

15

c6 ,15 c7 ,15 c8

4)

15

c7 ,15 c6 ,15 c5

x4 can be expanded in a s c e n d i n g x  5x  6 power of x then the coefficient of x3 is 73 73 71 71 1)  2) 3) 4) 648 648 648 648 If

2

5

10.

1) 140 2) 210 3) 411 4) None 19. Number of terms in the expansion of 2 2  1  1    x     x    is x  x     1) 1 2) 109 3) 54 4) Cannot be determined 20. Last two digits of 3400 are 1) 03 2) 02 3) 01 4) 04

 a 0  a1 x  ..... a15 x15

15

4) None

14. Prove that the coefficients of xn in 1  x 

15

11.

12.

13.

 2 k If the coefficient of x in the expansion of  x   x  is 270, then k = 1) 1 2) 2 3) 3 4) 4 The sum of the coefficients in the expansion of (1 + x + x2)n is 1) 2 2) 2n 3) 3n 4) 4n n In the expansion of (1 + x) the coefficients of pth and (p + 1)th term are respectively p and q then p+q= 1) n 2) n + 1 3) n + 2 4) n + 3 n 1 2 If (1 + x) = c0 + c1.x + c2.x + ...+ cn.xn, Then

C0  2.C1  3.C2 .....  (n  1).Cn  If a0, a1, a2, .....a50 are respectively, the coefficients 1) (n+2)2n-1 2) 2n 1  n.2n of x0, x1, x2, .....x50 in (1 + x + x2)25, then find a1 + n n 3) 2   n  1 2 4) 2n 1  (n  1)2n a2 + .......... + a50 14. The coefficient of x4 in the expansion of 1) Odd 2) Even 3) Zero 4) None ‘p’ is natural number such that no term in the (1  3x ) 2 is p 1  2x x y expansion of    is independent of ‘x’. Then 1) 1 2) 2 3) 3 4) 4 y x

‘p’ is of the form. www.betoppers.com

10th Class Mathematics

120 15. If the coefficient of rth term and (r + 1)th term in the expansion of (1 + x)3n are in the ratio 1:2, then r=

6 1)  n  1 5 3)

4)

1  3n  2  3

(1  x)40 is 1.3.5....39 20 .2 20!

2)

1.3.5....39 20!

40!

3)

 20!

4) 40! 220

2

18. The coeff. of 8th term in the expansion of (1  x)10 is 1) 120 2) 10C3 3) 110 4) 210 19. The term independent of x in the expansion of

(1+x)n, then C1  2.C2  3.C3  ...  n.Cn  2) n.2n

3) n.2n 1 4) n.2n 1 15

 2 3 21. The term independent of x in  2x  3  is x   15 . 5 7 15 . 10 1) C9 2 .3 2) C9 2 . 35 15 . 15 3) C9 2 4) 15C9 . 36.29 22. The sum of the series

23

1   x   , then x  1) T12 = T13 2) x2 – T13 = T12 2 3) T12 = x T13 4) T12 + T13 = 25 27. If C0 ,C 2 C4 ,.... are binomial coefficients in the expansion of (1+x)9 , then C0 + C2 + C4 + C6 + C8 =

1) 27 2) 256 3) 29 4) 258 28. If nC4, nC5, nC6 are in A.P. then n = 1) 12 2) 11 3) 7 4) 8 2n

29. If (1 + 2x + x2)n =

k

2 3  2 3 1)   2)   3) 4) 3 2  3 2 23. The term independent of x in the expansion of (3 + 2x)44, is 1) 4th term 2) 5th term 3) 1st term 4) 7th term

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r

 a x , then a r

r 0

1) 3)



n

Cr 

2n

2

r

=

2) n Cr .n Cr 1 4)

Cr

2n

Cr 1 6

 3  30. The coefficient of x in  x 5   is x3   1) 0 2) 120 3) 420 4) 540 n 31. If (1+x) = C0 + C1x1 + ---- + Cn xn, then C0 – C2 + C4 – C6 + ---- = 3

1) 2 n – 1 3) 2n/2 cos

k k ( k  1) k ( k  1)( k  2) 1    ... is 3 3.6 3.6.9 4

2n.a2 n  ...

(1  x 2 ) 5 (1  x) 4 is 1) 20 2) 30 3) 60 4) 55 26. If T r denotes the rth term in the expansion of

6

 2 1  x   is x  1) –12 2) 15 3) 24 4) –15 20. If Cn is the coefficient of xn in the expansion of

1) 2n

2 n 1

1) 0 2) 1 3) n 4) –n 25. The coefficient of x 5 in the expansion of

 1 16. The coefficient of x  n in (1  x )n .1   is  x n 1) 0 2) 1 3) 2 4) 2nCn 17. The coeff. of the middle term in the expansion of

1)

r 0

a1  2a2  3a3  ....   1

1 2)  3n  1 3

1  n  2 4

2n

24. If (1  x  x 2 ) n   ar x r , then

2) 2n/2 sin

n 4

4) zero n

32.

C0 C1 C2  1 .Cn     .....  1 2 3 n 1 1) 0 3)

2) 2n n 1

4)

1 n 1 2n  1 n 1

n 4

Binomial Theorem

121 n

33. If n is a positive integer,



n

2

Cr  

r 1

1) zero

2)

n

Cn / 2

3)

 2n 

!

4)

n !



9

 

 2n  ! 2 –1  n ! 9



34. The number of non zero terms in the expansion of 1  3 2 x  1  3 2 x is 1) 9 2) zero 3) 5 4) 10 35. If C0 , C1, ----, Cn are binomial coefficients in the expansion of (1 + x)n, then x x2 xn C0  C1 .  C2 .      Cn .  2 3 n 1 n 1 n 1 n 1 n 1 1  x   1 1  x  1  x   1 1  x   1 1) 2) 4)  n  1  n  1 x 3)  n  1 x  n  1

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122

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10th Class Mathematics

10th Class Mathematics

123

Learning Outcomes

Chapter - 10

Triangles

By the end of this chapter, you will understand  Similar polygons  Area of similar triangles  Similar triangles

 Pythagoras theorem

 Result on similar triangles

 Converse of Pythagoras theorem

 Criteria for similarity of two triangles

2. Similar Triangles

Two polygons of the same number of series are said to be similar if (i) their corresponding angles are equal (i.e. they are equiangular) and (ii) their corresponding sides are in the same ratio (or proportion). Example 1 In the polygons ABCD and ABCDE A  A, B  B, C  C, D  D, E  E

3c m

E

A

A A 2.4 cm

AB BC CD DE AE     AB BC CD DA AE

and

Since triangles are also polygons, therefore the same conditions of similarity are applicable to them Two triangles are said to be similar (i) if their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). Example 2 In the triangles ABC and ABC A  A, B  B, C  C and AB AC BC   A B AC BC

2 cm

B

D

D

2.2 cm 6 cm

6 .6

cm C

5 cm

cm

7.5 cm

B

C

Therefore, ABC and ABC satisfy both the condition necessary for them to be similar. Hence, ABC ~  ABC

C

3. Result on Similar Triangles

3cm B

mc 6. 3

E

4 cm A m c 5 . 4

3.3 cm

mc 4. 2

B

4 .4

3.6 cm

1. Similar Polygons

C

Therefore, the two polygons ABCD and ABCDE satisfy the conditions necessary for them to be similar. Hence, polygons ABCDE and ABCDE are similarly. We write ABCDE ~ ABCDE where the symbol ~ stands for ‘is similar to’.

Basic Proportionality Theorem or Thales theorem: If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.  ABC in which DE || BC and DE intersects AB AD AE at D and AC at E. then:  DB EC A

D

B

E

C

10th Class Mathematics

124 Converse of Basic Proportionality Theorem If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. There is a ABC and a line  intersecting AB at D and AC at E, such that AD AE  then, DE || BC DB EC A

D

E

B



C

(ii)

5.

E and F are points on the sides PQ and PR respectively of a PQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm 6. In the following figure, if LM || CB and LN || AM AN CD, prove that = AB AD

Formative Worksheet 1.

Fill in the blanks using correct word given in the brackets: (i) All circles are __________. (congruent, similar) (ii) All squares are __________. (similar, congruent) (iii) All __________ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional) 2. Give two different examples of pair of (i) Similar figures (ii) Non-similar figures 3. State whether the following quadrilaterals are similar or not:

4. (i)

In the given figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

7.

In the following figure, DE || AC and DF || AE. BF BE Prove that = FE EC

8.

In the following figure, DE || OQ and DF || OR, show that EF || QR.

9.

In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

___________________________________________________________________________________________________________________________________________

Triangles

125

10. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. 11. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. 12. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. AO CO Show that = BO DO 13. The diagonals of a quadrilateral ABCD intersect AO CO each other at the point O such that = . BO DO Show that ABCD is a trapezium.

4. Criteria for Similarity of Two Triangles Two triangles are said to be similar if (i) Their correspond angles are equal and (ii) Their corresponds sides are in the same ratio (or proportional). Thus, two triangles ABC and ABC are similar if (i) A  A, B  B, C  C AB BC CA (ii)       A B B C CA Characteristic Property 1 (AAA similarity) If in two triangles, the corresponding angles are equal, then the triangles, are similar A

Conceptive Worksheet 1.

2.

3.

In ABC, D and E are points on the sides AB and AC respectively such that DE||BC. AD = 4x  3,AE  8x  7,BD  3x  1 and CE = 5x – 3, find the value of x. In the given fig. AEF  AFE and E is the BD BF midpoint of CA. prove that  . CD CE In the given figure PQ = 24 cm, QR = 26 cm, PAR = 90°, PA = 6 cm and AR = 8 cm. Find QPR. Q

P A

90°

R

4.

PQ QM In PQR,  , Q  75 and R  45 . PR MR Determine the measure of QPM . P

D

B

E

C

F

In triangles ABC and DEF A  D, B  E and C  F. then ABC ~ DEF Remark: (i) Equiangular triangles are similar triangles. (ii) Similar triangles are equiangular Note: If two angles of a triangle are respectively equal to two angles of another triangle then by the angle sum property of a triangle their third angle will also be equal. There AAA similarity criterion can also be stated as follows. If two angels of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is referred to as the AA similarity criterion for two triangles. Characteristic Property 2 (SSS similarity) If the corresponding sides of two triangles are proportional then they are similar. D

Q

5.

M

A

R

In the adjoining figure exterior. EAB = 110°, CAD = 35°, AB = 5cm, AC = 7 cm and BC = 3 cm. Find CD. E 110°

B

° 35

B

D

C

E

F

There are two triangles ABC and DEF such that AB BC AC   DE EF DF Then ABC ~ DEF

A

C

___________________________________________________________________________________________________________________________________________

10th Class Mathematics

126 Characteristic Property 3 (SAS similarity) If one angle of a triangle is equal to angle of the other and the sides including these angle are proportional then the two triangles are similar. D A

B

C

E

F

In triangle ABC and DEF AB AC such that A  D and  DE DF Then ABC ~ DEF

Formative Worksheet 14. The area of two similar triangles are 81 cm2 and 49 cm2 respectively. If the altitude of the bigger triangle is 4.5 cm, find the corresponding altitude of the smaller triangle. 15. Shown in the given figure XY || AC and XY divides triangular region ABC into two parts AX equal in area. Determine . AB A

X

B

C Y 16. AD is a median of ABC. DE bisects ADB and DF bisects ADC. DE and DF meet AB and AC at E and F respectively. If AE = 3cm, BE = 4 cm and AF = 15 cm, find FC.

18. In the following figure, ODC ∼ OBA, BOC = 125° and CDO = 70°. Find DOC, DCO and OAB

A

E

B

F

D

C

17. State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

19. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, AO OB show that = OC OD

___________________________________________________________________________________________________________________________________________

Triangles 20. In the following figure,

127 QR QT = and 1 = 2 QS PR

Show that PQS ~ TQR

21. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ∼ ΔRTS.

22. In the following figure, if ABE ≅ ACD, show that ADE ∼ ABC.

23. In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

24. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ABE ∼ CFB

25. In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:

26. CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively. If ABC ∼ FEG, Show that: CD AC = i GH FG (ii) DCB ∼ HGE (iii) DCA ∼ HGF 27. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF

28. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ABC ∼ PQR. 29. D is a point on the side BC of a triangle ABC such that ADC = BAC. Show that CA2 = CB.CD. 30. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ABC ~ PQR. 31. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower. 32. If AD and PM are medians of triangles ABC and PQR, respectively where AB AD ABC ~ PQR prove that = PQ PM

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10th Class Mathematics

128

P

Conceptive Worksheet

A

6.

In the given fig. AD = 2.5 cm, DB = 9.5 cm, AE = 4 cm, CE = 3.5 cm, DE = 3.5 cm and BC = x cm. Prove that ADE is similar to ACB. Find the value of x. 7. D is a point on the side BC of a ABC, such CA CB that ADC  BAC . Prove that  . CD CA 8. In the given figure, ABC and DBC are two triangles on the same base BC. Prove that ar  ABC  AO  . ar  DBC  DO 9. Two sides and a median bisecting one of these sides of a triangle are respectively proportional to the two sides and the corresponding median of the other triangle. Prove that the triangle is similarly. 10. If the angels of one triangle are respectively equal to the angels of another triangle, prove that the ratio of their correspond sides is the same as the ratio of their corresponding sides is the same as the ratio of their corresponding (i) medians (ii) angle bisectors (iii) altitudes. 11. ABCD is trapezium in which AB || CD. Its diagonals AC and BD intersect at O. using a OA OB similarity criterion, prove that  . OC OD A

B

C

B

R

Q

In ABC and PQR , if ABC ~ PQR .

Then

ar  ABC  ar  PQR 

AB2 BC2 CA 2   PQ 2 QR 2 RP 2



Formative Worksheet 33. In the trapezium ABCD, AB || CD and AB = 2CD. If the area of OB = 84 cm2, find the area of COD. 34. D and E are points on the sides AB and AC respectively of a ABC such that DE || BC and divides ABC into two parts, equal in area, Find BD . AB 35. Two isosceles triangles have equal vertical angles and their areas are in the ratio 16:25. Find the ratio of their corresponding heights. A

B

D

C

L

E

M

F

36. Shown in figure DE || BC and AD : DB = 5:4. Area  DEF  Find . Area  CFB 

O D

C

A

12. In the given fig. ABCD is a trapezium with AB || DC. If BEC. Prove that AD = BC. A

B D

E B

D

C

5. Area of Similar Triangles The ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

1 4 5 F 2

E

C

37. In ABC, AD is the bisector of A. Determine ar( ABD) the ratio in terms of AB and AC. ar( ACD) 38. Let ABC ~ DEF and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC. 39. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

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Triangles

129

7. Converse of Pythagoras Theorem In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the angel opposite the first side is a right angle. In triangle ABC if AB2  BC 2  AC 2 then B  90 .

Formative Worksheet 47. In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB2 = 4AD2.

48. In the given figure AB = 8 cm, BC = 12 cm, AE = 6 cm. Determine the area of rectangle BCDE. B

12cm

C

8c m

40. In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects area  ΔABC  AO BC at O, show that = . area  ΔDBC  DO 41. If the areas of two similar triangles are equal, prove that they are congruent. 42. D, E and F are respectively the mid-points of sides AB, BC and CA of ABC. Find the ratio of the area of DEF and ABC. 43. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians. 44. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals. 45. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 46. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81

Conceptive Worksheet

6. Pythagoras Theorem [Baudhayan theorem] In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In right triangle ABC, right angled at B, AC 2  AB 2  BC 2 B

A

C

A 6c m

E

D

49. Two poles of height 8 m and 13 m are standing 12 m apart. What is the distance between their tops. D

13m

B 8m

13. Prove that the areas of two similar triangles are in the ratio of the squares of the corresponding altitudes. 14. Prove that the ratio of areas of two similar triangles is the same as the ratio of the squares of their corresponding medians. 15. Prove that the areas of two similar triangles are in the ratio of the squares of the corresponding angle bisector segments. 16. If the areas of two similar triangles are equal, prove that they are congruent.

A

12 m

C

50. In ABC, AB = 4 cm, BC = 8 cm, AC = 4 3 cm . Find the measure of A. 51. If the three sides of a triangle are a, 3 a and 2 a , then what is the measure of the angle opposite the longest side? 52. A semicircle is drawn on AC. Two chords AB and BC of length 8 cm and 6 cm respectively are drawn in the semicircle. What is the measure of the diameter of the circle? 53. Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse. (i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm

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10th Class Mathematics

130 54. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR. 55. In the given figure, ABD is a triangle right angled at A

56. 57. 58. 59.

60.

and AC ⊥ BD. Show that (i) AB2 = BC . BD (ii) AC2 = BC . DC (iii) AD2 = BD . CD ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2. ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle. ABC is an equilateral triangle of side 2a. Find each of its altitudes. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. In the given figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes 1 after 1 hour? 2 64. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops. 65. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2. 66. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3 CD (see Fig). Prove that 2 AB2 = 2 AC2 + BC2.

67. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3 BC. Prove that 9 AD2 = 7 AB2. 68. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. 69. Tick the correct answer and justify: In Δ ABC, AB = 6 3 cm, AC = 12 cm and BC = 6 cm. The angle B is: (A) 120° (B) 60° (C) 90° (D) 45°

Conceptive Worksheet

(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2, (ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2. 61. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall. 62. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? 63. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same

17. A right triangle has hypotenuse of length p cm and one side of length q cm. If p – q = 1, find the length of the third side of the triangle. 18. In PQR, QM  PR and PR 2  PQ 2  QR 2 . Prove that QM2 = PM × MR. 19. Prove that three times the sum of the squares of the sides of a triangle is equal to four times the sum of squares of the medians of the triangle.

A E

F

B

D

C

___________________________________________________________________________________________________________________________________________

Triangles

20.

131

In PQR, Q is obtuse, PM is perpendicular 4. to RQ produced and RN is perpendicular to the PQ produced. Prove that PR 2  PQ.PN  RQ.RM

5.

P

6. M

R

Q

If ABC ~ LMN such that AB = 3LM and AC = 18 cm, then what is the measure of LN? (A) 6 cm (B) 9 cm (C) 21 cm (D) 54 cm If ABC ~ YZX such that A  53 and Z  39 , then what is the measure of C ? (A) 39° (B) 53° (C) 88° (D) 92° Which of the following expressions is equal to the expression AB2 + CE2? A

N

21.

22.

Prove that the internal bisector of an angle of a triangle divides the opposite the side internally in the ratio the sides containing the sides containing the angle. In the given fig. AD is the bisector of BAC, if AB = 10 cm, AC = 6 cm and BC = 12 cm, find 7. BD and DC.

B

E

2

2

(A) AC  + AE (B) (AC2 + 2AE2) (C) AC2 + BE2 (D) (AC2 + 2BE2) In the given figure, ABC is right-angled at B. A

Summative Worksheet 1.

C

L

The given figure shows a right-angled triangle PQR. S and T are the points on PQ and QR

7cm B

1 respectively such that QS = PS and QT = QR. 3 P

8.

24 cm

A

S

l Q

2.

T

m

3.

D

a 2

9.

E

p

A

C

49 13 cm (B) cm (C) 13 cm (D) 14 cm 13 2

D

If AC = 4 cm, then what is the value of EC? (A) 7 cm (B) 8 cm (C) 11 cm (D) 12 cm What is the measure of AC in the given figure?

What is the value of AB2:BD2? (A) 7:9 (B) 9:7 (C) 31:36 (D) 36:31 The areas of two similar triangles are in the ratio 49:169. The length of the shortest side of the larger triangle is 26 cm. What is the length of the shortest side of the smaller triangle? (A)

C

n

A

a

B

R

If 18(PT2 + RS2) = a PR2, then what is the value of a? (A) 16 (B) 18 (C) 20 (D) 30 In the given figure, ABC is right-angled at C and the point D trisects AC.

B

C

What is the measure of LC? (A) 16.76 cm (B) 18.35 cm (C) 21.05 cm (D) 23.04 cm In the given figure, AD = 3 AB and l || m.

E

5cm

B

(A)

D

5cm

C

15 25 cm (B) cm (C) 8 cm (D) 9 cm 4 3 www.betoppers.com

10th Class Mathematics

132

10.

11.

12.

If the isosceles triangles ABC and DEF have the same angles and the ratio of their corresponding sides is 3:4, then what is the ratio of their corresponding heights? (A) 3:4 (B) 4:3 (C) 9:16 (D) 16:9 If the base and hypotenuse of a right-angled triangle are (3a – 1) and (3a + 1) respectively, then what is the height of the triangle? (A) 3 2a (B) 3 2a (C) 2 3a (D) 2 3a The given figure shows a triangle ABC where ABC  ADB . A

15.

What are the respective values of x and y in the given figure? A

2x units

E y units D (x – 1)units C

(A) 39 and 18 (C) 81 and 9

1.

In the given figure, XYZ is equilateral and XL is its altitude. X

B

C

Which of the following ratios is equal to AB: BC? (A) AD: DC (B) AD: BD (C) BD: AC (D) BD: AB The given figure shows a trapezium ABCD, where AD||BC. The diagonals AC and BD intersect at O. Also, BC = 3AD and ar ( AOB ) = 3 ar ( AOD ) A

Y

2.

D

L

Z

Which of the following relations is correct? (A) 3XY2 = 5XL2 (B) 3XL2 = 4XY2 (C) 3XY2  4XY2 In the given figure, ABC and AXY are equilateral such that the ratio of the areas of AXY and ABC is 1:4. X

A

O

B

14.

(B) 40 and 18 (D) 80 and 8

HOTS Worksheet

D

13.

B

C

What is the ratio of the areas of ABD and BCD ? (A) 1:2 (B) 1:3 (C) 1:4 (D) 1:9 In the given figure, the points P and Q trisect AB, the points R and S trisect BC, and the points T and U trisect AC. A P

U

Y B

3.

C

If the perimeter of AXY is 24 cm, then what is the perimeter of ABC ? (A) 36 cm (B) 42 cm (C) 48 cm (D) 52 cm In the given figure, ABC is a right-angled triangle at B and BL  AC . A

Q B

T R

S

C

If the area of ABD is 1296 cm2, then what is the area of the hexagon PQRSTU? (A) 432 cm2 (B) 648 cm2 (C) 864 cm2 (D) 972 cm2

www.betoppers.com

L

B

C

What is the length of CL? (A) 3 cm (B) 4 cm (C) 5 cm (D) 6 cm

Triangles

4.

133

The given figure shows PQR and XYZ . PQ:PR = YZ:XY P

What is the area of the rectangle CDGH?

  (C) 3 1  2 2  cm

X

70

5.

10.

70

45 R

Q

Y

Z

What is the measure of X ? (A) 45° (B) 50° (C) 65° (D) 70° Triangle XYZ is right-angled at Y and XZ2 = 2XY2. The length of XZ is 6 2cm . What is the area of XYZ ?

2 (B) 9 1  2 cm

2

2

The given figure shows an isosceles triangle ABC where AB = AC. The perimeter of ABC is 16 cm and AD = 4 cm.What is the area of ABC ? A

X

B

11. Y

6.

Z

(A) 12 cm2 (B) 18 cm2 (C) 24 cm2 D)28 cm2 In the given figure, ABC is a right-angled triangle at B and AM is the median.

12.

A 10cm

6cm

  (D) 3 1  2  cm

2 (A) 9 1  2 2 cm

D

C

(A) 10 cm2 (B) 12 cm2(C) 18 cm2 (D) 20 cm2 Among the quadrilaterals, parallelogram, rhombus, rectangle, and squares, which quadrilateral is always similar to each other? (A) Parallelogram (B) Rhombus (C) Rectangle (D) Square The given figure shows parallelogram MNOP with MN = 10 cm and NO = 9 cm. XY is a straight line that intersects diagonal NP at point Z. M 2 cm X

B

M

What is the length of the median AM? (A) 7 2cm (B) 8 2cm 7.

(C) 2 13cm (D) 2 10cm The given figure shows a rhombus ABCD. D

13.

O A

8.

9.

4cm

B

A

14.

15.

What is the length of YZ? (A) 2 cm (B) 3 cm (C) 4 cm (D) 6 cm The altitude drawn to the base of an isosceles triangle is of length 4 cm. The perimeter of the triangle is 16 cm. What is the area of the isosceles triangle? (A) 11 cm2 (B) 12 cm2 (C) 24 cm2 (D) 33 cm2 If ABC ~ PQR such that AB = 3 cm, BC = 4 cm, CA = 5 cm, PQ = 1.5 cm, and QR = 2 cm, then what is the length of PR? (A) 5 cm (B) 3 cm (C) 2.5 cm (D) 1.4 cm The given figure shows a rectangle ABCD. G is a point inside the rectangle such that AG = 3 cm and GC = 5 cm. A

B

H G

D E

B G

C

F

Y 4cm O

P

C

What is the value of AC2 + BD2? (A) 16 (B) 32 (C) 64 (D) 68 If the areas of two similar triangles are in the ratio 25:49, then what is the ratio of the corresponding sides of the triangle? (A) 49:25 (B) 25:74 (C) 5:7 (D) 7:5 The given figure shows a regular octagon ABCDEFGH of side 3 cm.

N 4cm Z

C

D

C

What is the sum of the squares of the lengths DG and GB? (A) 24 cm2 (B) 34 cm2 (C) 36 cm2(D)64 cm2 www.betoppers.com

10th Class Mathematics

134 8.

IIT JEE Worksheet I.

1.

2.

3.

4.

5.

6.

7.

Straight Objective Type This section contains multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. If in an isosceles triangle 'a' is the length of the base and 'b' the length of one of the equal side, then its area is a b (A) 4b 2  a 2 (B) 4b2  a 2 4 4 ab 2 ab 2 (C) a  b 2 (D) b  a2 4 4 If an equilateral triangle of area X and a square of area Y have the same perimeter, then (A) X >Y (B) X < Y (C) X = Y (D) X  Y PSR is a triangle right angled at S. D is the midpoint of SR. If the bisector of  PSR and perpendicular bisector of SR meet at O, then triangle OSD is (A) scalene (B) equilateral (C) isosceles right angled (D) acute-angled If any two sides of a triangle are produced beyond its base and the exterior angles thus obtained are bisected, then these bisectors will include an angle equal to (A) half the sum of the base angles (B) sum of the base angles (C) half the difference of the base angles (D) difference of the base angles If x is the length of the median of an equilateral triangle, then its area is 3 2 3 2 1 (A) x2 (B) x (C) x (D) x 2 2 3 2 The area of a right angled triangle is 40 sq. cm. and perimeter is 40 cm. The length of its hypotenuse is (A) 16 cm. (B) 18cm. (C) 17 cm. (D) Data sufficient If each side of triangle ABC is of length 4 and if AD is 1 a ED  AB. What is area of region BCED

(A) 8 3

(B) 4 3

(C) 4.5 3

(D) 3.5 3

Using the given figure, determine x

a x b

9.

c

ac ac (A) (B) bc bc bc 2ac (C) (D) ac bc In the adjacent figure, P and Q are points on the sides AB and AC respectively of a triangle ABC. PQ is parallel to BC and divides the triangle ABC into 2 parts, equal in area. What is the ratio of AP : AB

(A) 1 : 1

(B) ( 2  1) : 2

(C) 1 : 2 (D) ( 2  1) :1 10. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDC is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 11. In triangle ABC, D, E, F are points of trisection of BC, AC and AB respectively. Which of the following statements is not true?

(A) Area EDC = 2/9 area ABC (B) Area FBD = 2/7 area AFDC (C) Area DEF = 2/9 area ABC (D) Area (EDC + DBF +AFE)=2 area DEF 12. In the adjoining figure, if ST || QR. What is the length of PS

(A) 2 cm (B) 4.5 cm

(C) 4 cm (D) 3 cm

Triangles

135

13. If DEF if DE = 6 3 cm, DF = 12 cm and EF = 6 cm, then the angle E is (A) 1200 (B) 900 (C) 600 (D) 450 BD 14. In the adjoining figure, is equal to DA C

17.

D

ABC is a right  in which C  90o and CD  AB, if BC = a, CA = b, AB = c and CD = p, then which of the following is correct? (A) Cp = ab (B) Ca = pb 1 1 1 1 1 1 (C) 2  2  2 (D) 2  2  2 a p b p a b

18. B C 2 2 AB AB  AB   AB  (A)  (C)   (B)  (D) AC AD  AC   AD  15. If ABC is an isosceles triangles and D is a point on BC such that AD  BC, then (A) AB2 – AD2 = BD.DC (B) AB2 – AD2 = BD2 – DC2 (C) AB2 + AD2 = BD.DC (B) AB2 + AD2 = BD2 – DC2

II. Multiple Correct Type 16. In the given figure, B 90o and segment AD  BC, then which of the following is correct? (A) b2 = h2 + a2 – x2 – 2ax (B) b2 = h2 – a2 + x2 + 2ax (C) b2 = h2 + a2 + x2 – 2ax (D) b2 = a2 + c2 – 2ax

If two triangle are similar then (A) Their corresponding angles are equal (B) Their corresponding sides are in same (C) Their corresponding altitude equal are in same (D) None of these 19. If two triangle are congruent then (A) Their corresponding angles are equal (B) Their corresponding sides are same (C) Their corresponding altitude are same (D) None of these 20. ABD is a triangle in which  DAB  90o and

AC  BD , then (A) AB2  BC  BD

(B) AC 2  BC  DC

(C) AD 2  BD  CD

(D)

AB2 BC  AD 2 CD

III. Matrix Match Type This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. 21. If in a ABC, DE||BC and intersects AB in D and AC in E, then match the column Column I Column II (A) AD (p) AC DB AE (B) AB (q) AE AD EC (C) DB (r) AE AB AC (D) AD (s) EC AB AC 22. D is the mid-point of side BC of AC. AD is bisected at point E and BE produced cuts AC at point X, then match the column Column I Column II (A) BE : EX (p) 1:2 (B) XB : MD (q) 2:1 (C) EX : MD (r) 3:1 (D) AX : XM (s) 1:1 ___________________________________________________________________________________________________________________________________________

10th Class Mathematics

136

23. In figure the line segment XY is parallel to the side AC of ABC and it divides the triangle into two parts of equal areas, then match the column Column I Column II (A) AB : XB (p) 2 :1 (B) ar(ABC) : ar(XBY) (q) 2:1 (C) AX : AB (r) ( 2  1)2 : 2 (D) X : A 24. Match the column Column I (A) An ABC, ADBC and AD2 = BD × DC. Then the measure of ABC (B) In ABC, AB = 4cm, BC = 8cm AC  4 3 cm , the measure of CA (C) In the figure PQ = 24cm, QR = 26cm, PAR  90o , PA = 6cm and AR = 8cm, then QPR  Q P

(s)

1:1

Column II (p) 30° (q)

45°

(r)

60°

(s)

90°

90°

(D)

R PQ QM In PQR,  , Q  75o and PR MR R  45o , then QPM 

IV. Integers Type This section contains questions where the answer to each of the question is a single digit integer, ranging from 0 to 9. 25. In a triangle ABC, points P, Q and R are the midpoints the sides AB, BC and CA respectively. If the area of triangle ABC is 20 sq. units, then find the area of the' angle PQR26. The area of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5 cm, find the length of QR 27. A ladder 17m long reaches a window of a building 15m above the ground. Find the distance of the foot of the ladder from building 28. What is the length of the hypotenuse of an isosceles right triangle whose are side is

4 2 cm. 29. Find the length of the second diagonal of a rhombus whose side is 5 cm and one of the diagonals is 8 cm.

V.

Comprehension Type The following questions are based on the paragraph given below 30. In ABC, AD is the bisector of  A A

B

D

C

(i) AB = 8cm, BD = 5cm, DC = 4cm, find AC (A) 4.4 cm (B) 5.4 cm (C) 6.4 cm (D) 7.4 cm (ii) AB = 6cm, AC = 5cm, BD = 3cm, find DC (A) 5.5 cm (B) 4.5 cm (C) 3.5 cm (D) 2.5 cm (iii) AB = 9cm, AC = 4cm, BC = 10.4 cm, find CD and BD (A) CD = 4.2cm, BD = 6.2 cm (B) CD = 3.2cm, BD = 7.2 cm (C) CD = 3.2cm, BD = 5.2 cm (D) CD = 4.2cm, BD = 7.2 cm

___________________________________________________________________________________________________________________________________________

Circles Learning Outcomes By the end of this chapter, you will understand  Introduction

Chapter - 11

10th Class Mathematics

137

 Number of Tangents to a Circle  Some results on tangent to a circle.

1. Introduction

2. Number of Tangents to a Circle

Consider a circle with centre O and line AB in the same plane, then the following figure illustrate the possible relative position of the circle and the line. A

A

The number of tangents that can be drawn to a circle from a point depends on the position of the point relative to the circle, as shown in the following table. Position of the point relative to the circle Inside the circle

Figure

O

O

Number of tangents Zero

P

B

B (i)

(ii)

On the circle

One P

Outside the circle

Two A

Line AB does not touch or intersect the circle.[fig. (i)] Line AB intersects the circle in two distinct points and is called a secant (fig. (ii)) Line AB intersects the circle in two coincident points, i.e. in exactly one point (fig(iii). It is then said to touch the circle and is known as a tangent to the circle at the point. The word tangent comes from the Latin word ‘tangere’, which means to touch. This notice of touching is used in a geometrical meaning of tangent. Hence, a tangent to a circle is a line that intersects the circle in exactly one point. The point of intersection is called the point of contact or the point of tangency. In the given figure. (iii), line AB is tangent and point C is the point of contact. A ray or line segment is a tangent if it is a part of the tangent line and contains the point of tangency (contact). Rays CA and CB are tangent rays and CA and CB are tangent segments

P B

3. Some Results on Tangent to a Circle 

The tangent at any point of a circle is perpendicular to the radius through the point of contact. A circle with centre O and tangent AB at a point P. OP  AB O

A



P

R Q

B

A line drawn through the end of a radius and perpendicular to it is a tangent to the circle. In a circle with centre O in which

10th Class Mathematics

138 (i) OP is radius (ii) AB is line through P such that OP  AB. then AB is a tangent to the circle at point P. O P

A



5.

6. R Q

From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm In given figure, if TP and TQ are the two tangents

B

The length of tangents drawn from an external point to circle are equal. A circle with centre O.A is an external point. Two tangents AP and AQ drawn from a point A to the circle then AP = AQ P O

A Q



If two tangents are drawn to a circle from a external point, then (i) they subtend equal angles at the centre (ii) they are equally inclined to the line segment joining the centre to that point. A circle with centre O and a point A outside it. AP and AQ are tangents drawn to the circle from point A. AOP  AOQ and OAP  OAQ P O Q

Formative Worksheet How many tangents can a circle have? Fill in the blanks: A tangent to a circle intersects it in point (s)_____. (ii) A line intersecting a circle in two points is called a ______. (iii) A circle can have ______parallel tangents at the most. (iv) The common point of a tangent to a circle and the circle is called____. 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is: (A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119cm . 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

to a circle with centre O so that POQ = 110°, then PTQ is equal to (A) 60° (B) 70° (C) 80° (D) 90° 7. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then  POA is equal to (A) 50° (B) 60° (C) 70° (D) 80° 8. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. 9. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre. 10. The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle. 11. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. 12. A quadrilateral ABCD is drawn to circumscribe a circle (see Fig).

1. 2. (i)

In Q.5 to 7, choose the correct option and give justification.

Prove that AB + CD = AD + BC 13. In the given figure, XY and X′Y′ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X′Y′ at B. Prove that  AOB = 90°.

14. Prove that the angle between the two tangents drawn from an external point to a circle

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Circles

139

is supplementary to the angle subtended by the line-segment joining the points of contact at the centre. 15. Prove that the parallelogram circumscribing a circle is a rhombus. 16. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig). Find the sides AB and AC.

A

O

P

B

4.

A quadrilateral ABCD is drawn to circumscribing a circle. Prove that AB + CD = AD + BC. A

B

P

Q

S

5.

C

R

D

In the given figure, a circle is inscribed in a quadrilateral ABCD in which B = 90°. If AD = 23 cm, AB = 29 cm and DS = 5 cm, find the radius (r) of the circle. A

17. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

R

Conceptive Worksheet 1.

If all the sides of a parallelogram touch a circle, show that parallelogram is a rhombus. or Prove that the parallelogram circumscribing a circle is a rhombus. D

R

S

O B

r

S

P C

6.

A circle is inscribed in ABC having sides 8 cm, 10 cm and 12 cm as shown in the figure. Find AD, BE and CF.

7.

Quadrilateral PQRS circumscribes a circle as shown in the figure. Which side of the quadrilateral is equal to PD + QB. Justify your answer.

C Q

A

2.

Q

r D

P

B

Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR. T R Q P

3.

Prove that the angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

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10th Class Mathematics

140 8.

In the given figure, a circle touches all four sides of a quadrilateral PQRS, whose sides are PQ = 6.5 cm, QR = 7.3 cm, and PS = 4.2 cm, find RS.

5.

6.

A circle touches the side BC of a ABC at P and touches AB and AC produced at Q and R respectively, as shown in fig. Show that 1 AQ  (Perimeter of ABC). 2 A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

IIT JEE Worksheet I, 1. 9.

Find the perimeter of quadrilateral ABCD.

2.

10. In the given figure AQ = 4cm, QR = 7 cm, DS = 3 cm. Find x.

3.

4.

Straight Objective Type: Two chords AB and CD of a circle intersect at E such that AE = 2.4 cm, BE = 32. cm and CE = 1.6 cm. The length of DE is (A) 1.6 cm (B) 3.2 cm (C) 4.8 cm (D) 6.4 cm The locus of the middle points of equal chords of a circle with centre at O is (A) a straight line (B) a circle with centre different from O (C) a circle with centre at O (D) a circle intersecting the given circle at end of the chord If a regular hexagon is inscribed in a circle of radius r, then its perpendicular is (A) 3r (B) 6r (C) 9r (D) 12r In the figure if QPR  67 and SPR  72 and RP is a diameter of the circle, then QRS is equal to

Summative Worksheet 1.

In two concentric circles, a chord of length 24 cm of larger circle becomes a tangent to the smaller circle whose radius is 5 cm. Find the radius of the larger circle 5. O

AB = 24 cm, then the distance O1O 2 is equal to

5cm A

2.

3 4.

P

(A) 18° (B) 23° (C) 41° (D) 67° Two circles of radii 20 cm and 37 cm intersect in A and B. If O2 and O2 are their centres and

B

Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle. Prove that the tangents drawn at the ends of a diameter of a circle are parallel. AB is a diameter of a circle APB. AH and BK are perpendiculars from A and B respectively to the tangent at P. Prove that AH + BK = AB.

6.

(A) 44 cm (B) 51 cm (C) 40.5 cm (D) 45 cm AB and CD are two chords of a circle intersecting at the point P outside the circle. If PA = 12 cm, CDE = 7 cm and PC = 15 cm, then AB is equal to (A) 15.5 cm (B) 4 cm (C) 8 cm (D) 10 cm

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Circles 7.

8.

9.

141

If tangents QR, PR, PQ and drawn respectively at A, B, C to the circle circumscribing an acuteangle ABC so as the form another PQR , then the RPQ is equal to (A) BAC (B) 180  BAC 1 (C) (180  BAC) (D) 180  2BAC 2 Three wires of length l1 , l2 , l3 form a triangle surmounted by another circular wire, if l3 is the diameter and l3  2l1 then the angle between l1 and l3 will be (A) 30° (B) 60° (C) 45° (D) 90° The length of a chord of a circle of equal to the radius of the circle. The angle which this chord subtends on the longer segment of the circle is equal to

14. If figure, AB is a chord of circle, and PQ is a tangent at point B of the circle. If ABQ is

(A) 45°

(B) 70°

(C) 55°(D) 35°

II. Multiple Correct Answer 15. In figure, if PA = 12 cm, PC = 15 cm and CD = 7 cm, then AB is less then

(A) 2 cm (B) 10 cm (C) 3 cm (D) 4 cm 16. If figure, if AB = 6 cm, PB = 2 cm an PD = 2.5 cm, then CD is greater then (A) 30 (B) 45 (C) 60 (D) 90 10. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is (A) 7 cm (B) 12 cm (C) 15 cm (D) 24.5 cm 11. In figure, if TP and TQ are the two tangents to a circle with centre O so that POQ  110 , then PTQ is equal to

(A) 60 (B) 70 (C) 80 (D) 90 12. If tangents PQ and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to (A) 50 (B) 60 (C) 70° (D) 80° 13. In figure, PQ is a tangent at point P of a circle and QRB  30 then PRA is

(A) 30°

(B) 90° (C) 120°

(A) 2.9 cm (B) 3.9 cm (C) 3.2 cm (D) 6.4 cm 17. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. The length of PQ is less then are equal to (A) 12 cm (B) 13 cm (C) 8.5 cm (D) 119 cm 18. If AB is a chord of length 6 cm. of a circle of radius 5 cm, the tangents at A and B intersect at a point X (figure), then match the column Column I Column II (A) AY (P) 4 cm (B) OY (Q) 3.75 cm (C) XA (R) 5 cm (D) OA (S) 3 cm 19. For a circle is inscribed in a BC having sides 8 cm, 10 cm and 12 cm (figure), then match the column Column I Column II (A) AD (P) 1 (B) BE (Q) 7 (C) CF (R) 5 (D) AD/AF (S) 3

(D) 60°

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10th Class Mathematics

142 IV. Integer Type Questions 20. If two circles are non intersecting then the number of common tangents are 21. Maximum number of tangents that can be drawn from an exterior point to circle is 22. The length of a tangent from a point A at a distance of 5 cm from the centre of the circle is 4 cm. What is the radius of the circle 23. In the diagram two equal circles of radius 4 cm intersect each other such that each passes through the centre of the other. Find the length of the common chord 3 24. In figure, two chords AB and CD intersects at O. If OA = 8 cm, OC = 4cm and OD = 6 cm, then OB 2

V.

Comprehension Type Questions The number of tangents that can be drawn to a circle from a point depends on the position of the point relative to the circle 25. The number tangents drawn to a circle from a point which is inside the circle is (A) 1 (B) 0 (C) 2 (D) None of these 26. The number tangents drawn to a circle from a point which is outside the circle is (A) 1 (B) 0 (C) 2 (D) None of these 27. The number tangents drawn to a circle from a point which is on the circle is (A) 1 (B) 0 (C) 2 (D) None of these

Learning Outcomes

Surface Areas and Volumes

By the end of this chapter, you will understand  Basic Concepts  Surface Area of a Combination of Solids  Volume of a Combination of Solids  Conversion of Solid from One Shape to Another  Frustum of a Cone

1. Basic Concepts SOLIDS The objects having definite shape and size are called solids. A solid occupies a definite space. CUBOID Solids like matchbox, chalkbox, a tile, a book, an almirah, a room, etc., are in the shape of a cuboid. FORMULAE For a cuboid length = l, length = b and height = h, we have Volume = (l  b  h) cubic units Total surface area = 2(lb + bh + lh) sq units Lateral surface area = [2(l + b)  h] sq units CUBE Solids like ice cubes, sugar cubes, dice, etc., are in the shape of a cube. FORMULAE For cube having each edge = a units, we have Volume a3 cubic units Total surface area = 6a2 sq units Lateral surface area = 4a2 sq units CYLINDER Solids like measuring jars, circular pillars, circular pencils, circular pipes, road rollers, gas cylinders etc., are said to have cylindrical shape. r

Chapter - 12

10th Class Mathematics

143

FORMULAE For a cy0linder of base radius = r and height (or length) = h, we have Volume = (r2h) cubic units Curved surface area = 2rh sq units Total surface area = (2rh + 2r2) sq units = 2r(h + r) sq units HOLLOW CYLINDERS Solids like iron pipes, rubber tubes are in the shape of hollow cylinders. FORMULAE Consider a hollow cylinder having External radius = R, internal radius = r and height = h. Then, we have Volume of material = (external volume) – (internal volume)  ( R 2 h  r 2 h) cubic units

 h(R 2  r 2 ) cubic units Curved surface area of hollow cylinder = (external surface area) – (internal surface area)   2Rh  2rh  sq units  2h(R  r) sq units Total surface area of hollow cylinder =(curved surface area) + (area of the base ring)  (2Rh  2rh)  2( R 2  r 2 ) sq units

 2h  R  r   2  R 2  r 2  sq units CONE Solids like ice-cream cones, conical tents, funnels, etc, are having the shape of a cone

h

l

h r

10th Class Mathematics

144

FORMULAE Consider a cone in which base radius = r, height = h and slant height,

l  h2  r2 . Then, we have 1 Volume of the cone  r 2 h cubic units 3

Curved surface area of the cone  rl  r r 2  h 2 sq units Total surface area of the cone = (curved surface area) + (area of the base)  ( rl  r 2 )  r(l  r) sq units SPHERE Objects like a football, a cricket ball, etc., are said to have the shape of a sphere.

FORMULAE For a sphere of radius r, we have 4  Volume of sphere   r 3  cubic units 3  Surface area of the sphere = 4r 2 sq units





HEMISPHERE A plane through the centre of a sphere cuts it into two equal parts. Each part is called a hemisphere. FORMULAE For a hemisphere of radius r, we have 2 Volume of the hemisphere  r 3 cubic units 3 Curved surface area of the hemisphere = (2r2) sq units Total surface area of the hemisphere = (3r2) sq units

Name of the solid

Figure

Cube a

Lateral/ Curved Surface Area

Total Surface Area

Volume

4a2

6a2

a3

2(l + b)h

2(lb+bh+hl)

lbh

a

a

h

Cuboid b

l

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Surface Areas and Volumes Name of the solid

Right circular cylinder

145 Figure

Lateral/ Curved Surface Area

Total Surface Area

Volume

2rh

2r(r + h)

2rh

2rh

r(l + r)

1 2 r h 3

h r

Right cone

circular

l h r

r

Sphere

4r

2

4r

2

4 3 r 3

r Hemisphere

r

2. Surface Area of a Combination of Solids Let us consider a container (see fig). How do we find the surface area of such a solid?

Now, whenever we come across a new problem, we first try to see, if we can break it down into smaller problems, we have earlier solved. We can see that this solid is made up of a cylinder with two hemispheres stuck at either end. It would look like what we have in following figure, after we put the pieces all together.

2r2

3r2

 3 r 3

If we consider the surface of the newly formed object, we would be able to see only the curved surfaces of the two hemispheres and the curved surface of the cylinder. So, the total surface area of the new solid is the sum of the curved surface areas of each of the individual parts. This gives, TSA of new solid = CSA of one hemisphere + CSA of cylinder + CSA of other hemisphere where TSA, CSA stand for ‘Total Surface Area’ and ‘Curved Surface Area’ respectively. Let us now consider another situation. Suppose we are making a toy by putting together a hemisphere and a cone. Let us see the steps that we would be going through. First, we would take a cone and a hemisphere and bring their flat faces together. Here, of course, we would take the base radius of the cone equal to the radius of

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10th Class Mathematics

146 the hemisphere, for the toy is to have a smooth surface. So, the steps would be as shown in below figure.

9.

10.

At the end of our trial, we have got ourselves a nice round-bottomed toy. Now if we want to find how much paint we would require to colour the surface of this toy, what would we need to know? We would need to know the surface area of the toy, which consists of the CSA of the hemisphere and the CSA of the cone. So, we can say: Total surface area of the toy = CSA of hemisphere + CSA of cone

Formative Worksheet 1.

2.

3.

4. 5.

6. 7.

8.

A tent is in the shape of a right circular cylinder upto a height of 3m and conical above it. The total height of the tent is 13.5 m and radius of base is 14m. Find the cost of cloth required to make the tent at the rate of Rs. 80 per sq.m. A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and slant height of the conical portion is 53 m, find the area of the canvas needed to make the tent. 22    Use    7   A circus tent is in the shape of a cylinder surrounded by a cone. The diameter of the cylindrical part is 24 m and its height is 11 m. If the vertex of the tent is 16 m above the ground, find the area of the canvas required to make the tent. The surface area of a sphere is 616 cm2. Find its radius. The internal and external diameters of a hollow hemispherical vessel are 42 cm and 45.5 cm respectively. Find its capacity and also outer curved surface area. Three cubes of side 5 cm are joined end to end. Find the surface area of the resulting cuboid. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 15 cm and the radius of each of the hemispherical ends is 178.5 cm, find the cost of polishing its surface at the rate of Re 1 per dm2. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

11.

12.

13.

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

14. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.) 15. From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2. 16. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in below figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

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Surface Areas and Volumes

Conceptive Worksheet 1.

2.

147 4.

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see figure). The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. 22    Take    7  

5.

A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs 45 per m2. A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 108 cm and the diameter of each of the hemispherical ends is 36 cm, find the cost of polishing the surface at the rate of 7 paise per sq. m. (Take  = 3.1416). A toy is in the form of a cone mounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Calculate the surface area of the toy. (Use  = 3.14). A tent is in the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent @ Rs 2 per sq. m, if the radius of the base is 14 metres.

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere (figure). The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find 22   the area he has to colour.  Take    7  

The decorative block shown in figure is made of two solids—a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. 22    Take    7  

6.

7.

8. 3.

A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with ea.ch of these colours. (Take  = 3.14)

3. Volume of a Combination of Solids In the previous section, we have discussed how to find the surface area of solids made up of a combination of two basic solids. Here, we shall see how to calculate their volumes. It may be noted that in calculating the surface area, we have not added the surface areas of the two constituents, because some part of the surface area disappeared in the process of joining them. However, this will not be the case when we calculate the volume. The volume of the solid formed by joining two basic solids will actually be the sum of the volumes of the constituents, as we see in the examples below.

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10th Class Mathematics

148 Example 1 A solid iron pole consists of a cylinder of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. Solution: Let H be the height and R the radius of the lower cylindrical part of the iron pole. 24 then, H  220 cm and R  cm  12 cm 2

220 cm

60 cm

8 cm

Formative Worksheet

24 cm Let h be the height and r the radius of the surmounted cylinder. Volume of the iron pole  R 2 H  r 2 h

   R2H  r 2h   3.14[12  12  220  8  8  60] cm3  3.14[31680  3840] cm3  3.14(35520) cm3  111532.8 cm3 Mass of the pole 111532.8  8 kg 1000  892.262 kg (approx) Hence, the mass of the pole is 892.26 kg  111532.8  8g 

10 cm

Example 2 From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed. Find the volume of the remaining solid.

6 cm

Solution: Let r be the radius and h the height of the cylinder. Then, r  6 cm, h  10 cm Radius of the cone = Radius of the cylinder = 6 cm Height of the cone = Height of the cylinder = 10 cm Volume of the remaining solid = Volume of the cylinder – Volume of the cone 1  r 2 h  r 2 h 3 2 2  r h 3 2 22 5280    6  6  10 cm 3  cm 3 3 7 7 Hence, the volume of the remaining solid is 5280 cm 3 7

17. A tent is in the form of a cylinder of diameter 4.2 m and height 4m, surrounded by a cone of equal base and height 2.8 m. Find the capacity of the tent and the cost of canvas for making the tent at Rs. 100 per sq. m. 18. A solid toy is in the form of a hemisphere surrounded by a right circular cone of the height of the cone is 4 cm and diameter of the base is 6 cm, calculate: (i) The volume of the toy (ii) Surface area of the toy. (Use n = 3.14) 19. An 'ice-cream cone' is the union of a right circular cone and a hemisphere that has the same (circular) base as the cone. Find the volume of the ice-cream if the height of the cone is 9 cm and the radius of its base is 2.5 cm. 20. The radius and height of a right circular cine in 2 the ratio 5:12. If its volume is 2514 cu. m, find 7 its slant height and the radius. 21. 100 circular plates, each of radius 7 cm and 1 thickness  cm , are placed one above another 4 to from a solid right circular cylinder. Find the total surface area and the volume of the cylinder so formed. 22. Water in a canal 6 m wide and 1.5 m deep is flowing with a speed of 10 km/h. How much will it irrigate in 30 minutes, if 8 cm of standing water is needed?

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Surface Areas and Volumes 23. A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm× 6.5 cm? 24. The diagonal of a cube is 27 3 cm. find its surface area and volume. 25. Find the curved surface area, total surface area and volume of a cylinder, if the diameter of its base is 7 cm and its height is 80 cm. 26. A semicircular thin sheet of paper of diameter 28 cm is bent and an open conical cup is made. Find the capacity of the cone. 27. A metallic sheet is of rectangular shape with 48 cm × 36 cm dimension. From each one of its corners a square of side 8 cm is off. An open box is made of the remaining sheet. What is the volume if the box? 28. The diameters of two cylinder are in the ratio of 2:1 and their volumes are equal. Find the ratio of their heights. 29. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be 1/27 of the volume of the given cone, at what height above the base is the section made. 30. The surface area of a cube is equal to the surface area of a sphere. Find the ratio of their volumes. 31. The largest possible sphere is carved out from a cube of 7 cm sides. Find the volume of the sphere. 32. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π. 33. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.) 34. A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig).

149

35. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig).

36. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, onefourth of the water flows out. Find the number of lead shots dropped in the vessel. 37. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14) 38. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. 39. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

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150

Conceptive Worksheet 8.

9.

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder (see figure). If the base of the shed is of dimension 7 m x 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold.

Further, suppose the machinery in the shed occupies a total space of 300 m3, and there are 20 workers, each of whom occupy about 0.08 m3 space on an average. Then, how much air is in 22   the shed ?  Take    7   A juice seller was serving his customers using glasses as shown in figure. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use  = 3.14).

10. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take  = 3.14).

11. A circus tent has cylindrical shape surmounted by conical roof. The radius of the cylindrical base is 20 m. The heights of the cylindrical and conical portions are 4.2 m and 2.1 m, respectively. Find the volume of the tent. 12. Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose total length is 2.7 m and the diameter of each hemispherical end is 0.7 m. 13. An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm3 of iron has 8 g mass 355   (approx.).  Use    113   14. A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank. 15. A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at ? 100 per square metre. 16. A boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 m diameter. Find the volume of the boiler. 17. A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The 2 depth of the cylinder is 4 m and the diameter of 3 the hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid. 18. A cylindrical vessel of diameter 14 m and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation proposes. How many cubic centimetres of cork dust will be required?

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Surface Areas and Volumes

4. Conversion of Solid from One Shape to Another In our day to day life, sometimes we need to convert a solid into another solid into another solid of different shape or size or both, e.g. metal spheres are melted to from cylindrical metal wires, iron blocks are melted and recast into hollow cylindrical pipes, the earth taken out by digging a well is spread uniformly around the well to from an embankment in the shape of a cylindrical shell, lead pellets are made from a lead sphere etc. We shall illustrate the calculation of volumes and surface areas in such cases. Our illustrations will be based on an assumption that there is no wastage during the conversions. Example 3 One metal solid of dimensions 16 cm × 8 cm × 5 cm and another solid of the same metal having dimensions 12 cm × 6 cm × 5 cm are melted to from a cube. Find the edge of the cube. Solution: Let x be the side of the cube. Then, volume of the cube = x3 Volume of the cube = Volume of two rectangular solids

 x 3  16cm  8cm  5cm   12cm  6cm  5cm 

 x 3  640 cm3  340 cm3  1000 cm3  x  10 cm Hence, the edge of the cube is 10 cm.

Formative Worksheet 40. The rain water from a roof 22 m × 20 m drains into a cylindrical vessel having diameter of base 2 m and height 3.5 m. If the vessel is just full, find the rain fall in cm. 41. Water is flowing at the rate of 15 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Find the time in which the level of water in the tank will rise by 21 cm. 42. A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are needed to empty the bowl? 43. The base radius and height of a right circular solid cone are 2 cm and 8 cm respectively. It is melted and recast into spheres of diameter 2 cm each. Find the number of sphere formed.

151 44. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the leap. 45. How many balls, each of radius 1 cm, can be made from a sphere of lead of radius 9 cm? 46. A cylindrical container 2 m high and 3.5 m in diameter has a hemispherical lid. Find its volume. 47. A plot of land in the form of a rectangle has a dimension 240 m × 180 m. A drainlet 10 m wide is dug all around it (on the outside) and the earth dug out is evenly spread ` over the plot, increasing its surface level by 25 cm. Find the depth of the drainlet. 48. A well with inner radius 4 m is dug 14 m deep,. The earth taken out is spread evenly around the boundary of the well to make a circular ring platform of 3 m width. Find the height of the platform. 49. There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii. 50. A spherical ball of lead 3 cm in diameter is melted and recast into three spherical balls. The diameter of two balls is 1 cm and 1.5 cm respectively. Find the diameter of the third ball. 51. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. 52. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere. 53. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. 54. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. 55. A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

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10th Class Mathematics

152 56. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm? 57. A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap. 58. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed? 59. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

25.

26.

27. 28.

29.

Conceptive Worksheet 19. A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere. 20. Selvi's house has an overhead tank in the shape of a cylinder. This is filled by pumping water from a sump (an underground tank) which is in the shape of a cuboid. The sump has dimensions 1.57 m × 1.44 m × 95 cm. The overhead tank has its radius 60 cm and height 95 cm. Find the height of the water left in the sump after the overhead tank has been completely filled with water from the sump which had been full. Compare the capacity of the tank with that of the sump. (Use  = 3.14). 21. A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. 22. A hemispherical tank full of water is emptied by 4 a pipe at the rate of 3 litres per second. How 7 much time will it take to empty half the tank, if it 22   is 3 m in diameter?  Take    7   23. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 7 cm and height 3 cm. Find the number of cones so formed. 24. A solid metallic sphere of diameter 28 cm in melted and recast into a number of smaller

2 cm and height 3 cm. 3 Find the number of cones so formed. A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed. A hemispherical bowl of internal diameter 30 cm contains some liquid. This liquid is to be filled into cylindrical shaped bottles each of diameter 5 cm and height 6 cm. Find the number of bottles necessary to empty the bowl. How many balls, each of radius 1 cm can be made from a solid sphere of lead of radius 8 cm? The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having diameter of cross-section as 0.2 cm. Find the length of the wire. A conical flask is full of water. The flask has base-radius r and height h. The water is poured into a cylindrical flask of base-radius mr. Find the height of water in the cylindrical flask. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere.

cones, each of diameter 4

30.

5.

Frustum of a Cone If a right circular cone cut by a plane parallel to its base, through some point on its axis and the portion containing the vertex is removed, then the left out portion between the cutting plane and the base of the cone is called frustum of cone (figure. iii). thus, a function of a right circular cone has two unequal flat circular based and a curved surface, eg., flower pots, bucket, corks are generally of this shape.

A

D

B

Q

P

E

C

(i)

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Surface Areas and Volumes

153

A

 Volume of the function of the cone 1  h  r12  r1r2  r2 2  3 Draw EL  PC . Then, LC  r1  r2

Q D Q

D

B

In right ∆ELC, l = h 2   r1  r2 

E

Pythagoras Theorem] Curved surface area of the frustum of cone  r1l1  r2  l1  l  C

P

(i)

 r1

(ii)

Q

D

2

E

[By

 lr  lr1  r2  1  r1  r2  r1  r2 

 r2 r2   l  1  2   l  r1  r2   r1  r2 r1  r2  Hence, Curve surface area of the frustum

E

 l  r1  r2  where l  h 2   r1  r2 

B

C

P (iii)

In fig. (i), let ABC represent the right circular cone which is cut by a plane DQE parallel to its circular base BPC. Solid DQECPB forms the frustum of the right circular cone ABC. The circular faces DQE and BPC are called the circular bases of the frustum. QP, the line segment joining the centers of two bases represents the perpendicular distance between the bases and is known as the height of the frustum. Each of the line segments DB and EC is called its slant height. Volume and Surface Area of a Frustum of a Right Circular Cone Let r1 and r2 be the radii (r1 > r2), h the right and l the slant height of frustum of a cone. Then, BP = PC r1, BD = CE = l, DQ = QE = r2 and PQ = h r1

B

Lr1 -r2 C

P

h1

h

h

l

l1 Q

D r2

B

E

Total surface area 2  l  r1  r2   r1  r2 2

of

2

the

frustum

Formative Worksheet 60. A fez, the cap used by the Trucks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of the material used for making it.

O O O O O O O O O O O O O O O O O OO O O O O O OO O O O O

O

O O

61. The perimeters of the ends of the frustum of a cone are 207.24 cm and 169.56 cm. If the height of the frustum be 8 cm, find the whole surface of the frustum. [Taken  3.14 ] 62. The slant height of a frustum of a cone is 4 cm and perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface area. 63. A metallic right circular cone 20 cm high and whose vertical angle is 600 is cut into two parts of the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn in a 1 wire of diameter cm, find the length of the 16 wire. 64. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

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10th Class Mathematics

154 65. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. 66. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

67. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2. (Take π = 3.14) 68. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into 1 a wire of diameter cm, find the length of the 16 wire.

34. Hanumappa and his wife Gangamma are busy making jaggery out of sugarcane juice. They have processed the sugarcane juice to make the molasses, which is poured into moulds in the shape of a frustum of a cone having the diameters of its two circular faces as 30 cm and 35 cm and the vertical height of the mould is 14 cm (see Fig). If each cm3 of molasses has mass about 1.2 g, find the mass of the molasses that can be poured into each mould.

35. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet (see Fig). The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also, find the volume of water the bucket can hold.

Conceptive Worksheet 31. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. 1 If its volume be of the volume of the given 27 cone, at what height above the base is the section made? 32. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular end are 18 cm and 6 cm. Find the curved surface area of the frustum. 33. The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm (see Fig). Find its volume, the curved surface area and the total surface area

Summative Worksheet 1.

The cost of polishing a closed container, which is of the shape of the frustum of a cone, is Rs 440 at the rate of Re 1 per 100 cm2. The respective radii of the upper and the lower bases of the 22   frustum are 1 m and 40 cm.  Use    7  What is the height of the container? (A) 40 cm (B) 56 cm (C) 80 cm (D) 92 cm

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Surface Areas and Volumes 2.

3.

4.

The given figure shows a bucket.

What is the approximate capacity of the bucket? (A) 153.74 L (B) 174.43 L (C) 187.85 L (D) 192.48 L The height of a glass tumbler is 12 cm. The radius of its circular base is 3 cm and that of its rim is 4 cm. The volume of the tumbler is (A) 473.82 ml (B) 464.95 ml (C) 459.86 ml (D) 454.76 ml The given figure shows a funnel in shape of a cylinder surmounted by the frustum of a cone. The ratio of the areas of cross-section of the inlet and outlet is 16:1 and the ratio of the depths of the frustum of the cone and the depth of the cylindrical part is 2:1.

155 6.

7.

There are two buckets - one of red colour and the other of green colour. For the red bucket, the radius of its bottom, the radius of its top, and its height are in the ratio 2:3:6. For the green bucket, the radius of its bottom, the radius of its top, and its height are in the ratio 2:6:9. It is also known that the radius of the top of the red 3 bucket is times that of the green bucket. 2 What is the ratio of the volumes of water that the red and green buckets can hold? (A) 171: 26 (B) 25: 172 (C) 19: 78 (D) 78: 19 The given figure shows the face of a mannequin. The diameters of the circular bases of the hat on its face are 28 cm and 42 cm. The height of the hat is 24 cm. The owner of the mannequin wants to paste glossy paper on the hat. 22    Use   7 

2 per cm2, then 11 the amount required to paste the glossy paper is (A) Rs 208 (B) Rs 306 (C) Rs 612 (D) Rs 821 The given figure shows a hemispherical solid, surmounted by a solid frustum. The height of the hemispherical solid is half the height of the solid frustum.

If the cost of glossy paper is Rs

5.

What is the ratio of the capacities of the frustum and that of the cylindrical part? (A) 9:2 (B) 7:1 (C) 14:1 (D) 9:1 A glass is in the shape of a frustum of a cone. The diameters of its mouth and the base are 8 cm and 4 cm respectively. The height of the glass is 14 cm. Rohan daily drinks water from such glasses 12 times. 22   3 3  Use   7  and [1 L = 10 cm ] How much water is consumed by Rohan in a day? (A) 3.462 L (B) 4.456 L (C) 4.928 L (D) 5.346 L

8.

What is the ratio of the volumes of the hemispherical solid and the solid frustum? (A) 127: 36 (B) 85: 36 ___________________________________________________________________________________________________________________________________________

10th Class Mathematics

156

9.

(C) 36: 85 (D) 36: 127 The given figure shows two frustums. The height of the smaller frustum is half of that of the bigger frustum.

What is the ratio between the volumes of the frustums? (A) 325:37 (B) 425:41 (C) 850:41 (D) 650:37

(A) 332.5 cm2 (B) 335 cm2 (C) 339.5 cm2 (D) 341 cm2 13. A container opened from top is made of a metal sheet and is in the form of a frustum of a cone of height 20 cm. The radii of the lower end and the upper end are in the ratio 2: 5. If the curved surface area of the container is 4466 cm2, then what is the cost of metal sheet used to make the container at a rate of Rs 8 per 150 cm2? (A) Rs 271.04 (B) Rs 273.08 (C) Rs 275.12 (D) Rs 277.16 14. The given figure shows a frustum. The radius of the upper base, the radius of the lower base, and the height of the frustum are in the ratio 2:5:4. The curved surface area of the frustum is 990 cm2. 22    Use   7 

10.

What is the height of the frustum? (A) 6 cm (B) 9 cm (C) 12 cm (D) 27cm 15. A flower vase, in the shape of a frustum of a cone, is shown in the given figure. It is to be 22   painted from outside.  Use    7  What

is the volume of the given 22   figure?  Use    7   18356   24512  (A)  (B)  L L  7   7   22022   23162  (C)  (D)  L L  7   7  11. The capacity of a tub, which is of the shape of the frustum of a cone, is 16720000 cm3. The respective radii of the upper and lower bases of 22   the tub are 90 cm and 40 cm.  Use    7  What is the slant height of the tub? (A) 60 cm (B) 70 cm (C) 120 cm (D) 130 cm 12. The height of the frustum of a cone is 12 cm. If the difference of the radii of its two circularends is 5 cm and product of radii is 114, what is the total surface area of the frustum?

What is the surface area of the vase that is to be painted?





  (D) 77 1  3 41  cm

(A) 154 1  4 29 cm2 (B) 77 1  3 29 cm2





(C) 54 1  4 41 cm2

2

16. A frustum has a height 8 cm. The diameter of the lower base of the frustum is 6 times the radius of

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Surface Areas and Volumes

157

its upper base. It is also 10 cm more than the 22   height of the frustum.  Use    7  What is the total surface area of the frustum? (A) 128cm2 (B) 186 cm2 2 (C) 210 cm (D) 660 cm2 17. A cone of radius 15 cm is cut parallel to its base and a frustum is formed, such that the height of the frustum is one-third of the height of the original cone. The volume of the original cone is 2700π cm3

What is the total surface area of the frustum? (A) 325π (B) 475π (C) 650π (D) 950π 18. The cost of polishing a closed container, which is of the shape of the frustum of a cone, is Rs 440 at the rate of Re 1 per 100 cm2. The respective radii of the upper and the lower bases of the 22   frustum are 1 m and 40 cm.  Use    7  What is the height of the container? (A) 40 cm (B) 56 cm (C) 80 cm (D) 92 cm 19. A pan is of the shape of the frustum of a cone. The diameters of its circular ends are 42 cm and 28 cm respectively and its vertical height 22   is 51 cm.  Use    7  What is the area of the foil required to be pasted on the walls of the pan? (A) 110 cm2 (B) 55 51cm2 (C) 110 51cm 2 (D) 1100 cm2 20. Ramesh, an ice cream seller, uses the container shown in the given figure to freeze ice creams. He gets an order for 700 ice creams. The density of ice is 0.9167 22   g/cm3.  Use    7  

What quantity of ice cream should he freeze in order to complete the order? (A) 79.2 kg (B) 74.9 kg (C) 68.6 kg (D) 64.5 kg 21. If five solid copper spheres each of radius 6 cm are melted to form cylindrical wire of radius 1 cm, then what is the length of the wire? (A) 11.2 m (B) 12.4 m (C) 13.2 m (D) 14.4 m 22. If three small solid metallic spheres of radii 1 cm, 2 cm, and 3 cm respectively are completely melted to form a single solid sphere, then what is the surface area of the sphere so formed? 1

1

(A) 4  6 3 cm2 1

23.

24.

25.

26.

(B) 8  6 3 cm2 1

(C) 12  6  3 cm2 (D) 24  6  3 cm2 A solid copper sphere of radius 6 cm is melted and recast completely into two cylindrical shapes. The radius and height of first cylinder are respectively 2 cm and 28 cm. If the radius of second cylinder is 4 cm, then what is its height? (A) 11 cm (B) 36 cm (C) 44 cm (D) 52 cm If 125 identical metallic balls are melted to form 8 bigger balls, each of same size, then what is the ratio of the surface areas of a bigger ball to that of a smaller ball? (A) 4:3 (B) 5:2 (C) 16:9 (D) 25:4 A cylindrical candle is melted to form a few conical candles in such a way that the respective height and radius of each conical candle is oneeighth and half of the height and the radius of the cylindrical candle. How many conical candles can be made out of 16 such cylindrical candles? (A) 784 (B) 875 (C) 1152 (D) 1536 Ten cylindrical iron rods, each of radius 4 cm and 14 cm long, are melted to form a thin long wire whose area of cross-section is 2 cm2. 22    Use   7 

___________________________________________________________________________________________________________________________________________

158 What is the length of the resulting wire? (A) 3.52 km (B) 35.2 m (C) 3520 dm (D) 352000 cm 27. Seven hundred twenty nine identical metallic solid balls are melted to form bigger solid balls, each having radius that is 1.5 times that of the original small balls. How many bigger solid balls are formed? (A) 48 (B) 72 (C) 216 (D) 256 28. A cylindrical metallic rod of cross-sectional area 25π cm2 and length 144 cm is melted to form 100 identical solid spheres. In this process, no metal is wasted. What is the radius of each solid sphere? (A) 1 cm (B) 3 cm (C) 5 cm (D) 7 cm 29. A toy is in a shape as shown in the figure. It is broken down into two parts − a cone and a hemisphere (as shown). The volume of the toy was 8624 cm3.

What is the increase in the surface area of the toy after it is broken? (A) 1232 cm2 (B) 1104 cm2 2 (C) 552 cm (D) 16 cm2 30. A hemispherical metallic solid of radius 4.2 cm is melted and recast into a cylindrical solid of radius 1.4 cm. What is the length of the cylinder? (A) 21 cm (B) 25.2 cm (C) 31 cm (D) 36 cm 31. A conical metallic solid is melted to form some cylindrical rods. The ratio of the radius of the conical solid to that of the cylindrical rod is 4:1, while the ratio of the height of the conical solid to that of the cylindrical rod is 3: 2. How many cylindrical rods are formed? (A) 4 (B) 6 (C) 8 (D) 10 32. A steam boiler is in the shape of a cylinder with two hemispheres stuck at either ends. The diameter of the hemispherical part is 2 feet. 88 If the volume of the boiler is cubic feet, then 3 what is the height of the cylindrical portion? (A) 2 feet (B) 4 feet (C) 6 feet (D) 8 feet

10th Class Mathematics 33. Some cylindrical shaped capsules are to be filled with a medicine. The capsules are semi-spherical on each end. The length of the cylindrical part of the capsule is 12 mm and the radius is 2 mm.

The volume of each capsule is (A) 135 mm3 (B) 150 mm3 (C) 185 mm3 (D) 200 mm3 34. The adjacent figure shows a one litre bottle made by putting a cone over a cylinder. The height of the cone is half of the height of the cylinder. The height of the bottle is 30 cm.

The radius of the base of the bottle is (A) 11.6 cm (B) 12.1 cm (C) 12.6 cm (D) 13.1 cm 35. A toy has a hemispherical base and a conical top as shown in the figure. The perpendicular height of the cone is 14 cm and the radius of the hemisphere is 7 cm.

What is the approximate volume of the toy? (A) 1 437 cm2 (B) 1 337 cm2 2 (C) 1 117 cm (D) 1 028 cm2 36. The given figure shows a bottle made up of two cylinders. The dimensions of the bottle are also shown. 22    Use   7 

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Surface Areas and Volumes

159 39. A wooden block is in the shape of a cuboid surmounted by a cylinder as shown in the given figure. The dimensions of the cuboidal part are 16 cm × 7 cm × 10 cm. The diameter and the height of the cylindrical part are 7 cm and 6 22   cm respectively.  Use    7 

What is the volume of the bottle? (A) 4680 cm3 (B) 5830 cm3 3 (C) 6420 cm (D) 7540 cm3 37. The given figure shows a tank, which is completely filled with water. There is a tap at the bottom of the tank which releases water at the rate of 0.008 m3/min. It takes 5 hours to empty the tank completely. The height of the conical part of the tank is half the total height of the tank.

What is the total surface area of the wooden block? (A) 816 cm2 (B) 777.5 cm2 2 (C) 765 cm (D) 686.5 cm2

IIT JEE Worksheet I.

1.

What is the volume of the cylindrical part of the tank? (A) 2.6 m3 (B) 2.4 m3 3 (C) 1.8 m (D) 1.2 m3 38. A metallic solid was formed by making conical depressions on each end of a solid cylinder in such a way that the vertices of the conical parts meet at a point. The height of the solid cylinder is 12 times its radius.

2.

3.

4.

5. What is the ratio of the volume of the metallic solid obtained and the volume of the metal taken out? (A) 2:1 (B) 1:2 (C) 3:1 (D) 1:3

Straight Objective Type This section contains multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Find the side of a cube whose total surface area 2 is 32 cm 2 3 7 6 (A) (B) cm cm 3 5 4 8 (C) cm (D) cm 3 7 If the ratio of surface area of two cubes is 1:9, find the ratio of their volumes. (A) 1:27 (B) 2: 27 (C) 1:24 (D) 1:25 Find the volume of water in cubic metres that will fall on 2 hectares of ground if 5 cm of rainfall takes place. (A) 1400 m3 (B) 1000 m3 3 (C) 1200 m (D) 1100 m3 The curved surface area of a solid cylinder is one-third of its total surface area. Determine its height if its radius is 2.5 cm. (A) 2 cm (B) 1.45 cm (C) 1. 35 cm (D) 1.25 cm The sum of the radius of the base and the height of a solid cylinder is 12 cm, Find its circumference, if its total surface area is 540 cm2. (A) 45 cm (B) 40 cm (C) 35 cm (D) 42 cm

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10th Class Mathematics

160 6.

Volume of a cylindrical wire of radius 1 cm is 440 cm3. It is cut into three unequal segments. If the length of two cut segments is 6 cm and 8 cm, find the length of third segment. (A) 122 (B) 132 cm (C) 126 cm (D) 127 cm 7. Volumes of two solid spheres are in the ratio 125: 64. Determine their radii, if the sum of their radii is 45 cm. (A) 25 , 20 (B) 15, 30 (C) 35, 10 (D) 40, 5 8. The volume of cone is 1570 cm3 and the area of the base is 314 cm2. What is its height? (A) 12 cm (B) 13 cm (C) 14 cm (D) 15 cm 9. The height and base radius of a cone, each is increased by 50%. Determine the ratio between the volume of the given cone and the new cone. (A) 8:27 (B) 8:28 (C) 7:24 (D) 6:19 10. The curved surface area of one cone is twice that of the other cone. The slant height of the latter is twice that of the former. Find the ratio of their radii. (A) 4:1 (B) 3:1 (C) 2:1 (D) 5:1 11. Three cubes of same metal whose edges are 6 cm, 8 cm and 10 cm are melted and formed into a signal cube. Find the diagonal of the larger cube formed. (A) 12 3 cm

15. The height and radius of the cone of the cone of which the frustum is a part are , h1 and r1 respectively. If h2 and r2 are the height and radius of the smaller base of the frustum respectively and h2:h1 is 1:2, determine r2:r1. (A) 1 : 3 (B) 1 : 4 (C) 2 : 1 (D) 1 : 2 II.

16.

17.

18.

(B) 12 2 cm

(C) 13 cm (D) 17 cm 12. If the radius of the base of metallic solid right circular cylinder is ‘r’ and its height 3 cm, and it is melted and recast into a right circular cone of the same radius, find the height of the cone. (A) 8 cm (B) 9 cm (C) 10 cm (D) 11 cm 13. What is the maximum volume of a cone that can be carved out of a solid hemisphere of radius ‘r’.? [Leave your answer in terms of π] 1 1 1 (A) r 3 (B) r 2 (C) r 2 (D) r 3 3 9 3 14. The radii of circular ends of a frustum are 6 cm and 14 cm. Find its vertical height if the slant height is 10 cm. (A) 4 cm (B) 7 cm (C) 6 cm (D) 5 cm

19.

20.

Multiple Correct Type This section contains multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22m by 14m. Find the height of the platform (A) 2.5 m (B) 3.5 m (C) 1.5 m (D) 250 cm A cylindrical pipe has inner diameter of 7 cm and water flows through it at 192.5 litres per minute. Find the rate of flow in kilometers per hour. (A) 4 km (B) 3 km (C) 2 km (D) 300000 cm Water is being pumped out through a circular pipe whose internal diameter is 7 cm. If the flow of water is 72 cm per second, how many litres of water are being pumped out in one hour? (A) 9999.2 litres (B) 9898.2 litres (C) 9979200 cm3 (D) 9979.2 litres The rain water from a roof of 22m × 20m drains into a cylindrical vessel having diameter of base 2 m and height 3ss.5 m. If the vessel is just full, find the rain fall in cm. (A) 3.5 cm (B) 1/40 m (C) 2.5 m (D) 1.5 m Water in a canal, 30 dm wide and 12 dm deep is flowing with velocity of 10 km/hr. How much area will it irrigate in 30 minutes, if 8 cm of standing water is required for irrigation? (A) 250000000 cm2 (B) 25000 m2 (C) 15000 m2 (D) 20000 m2

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161

III. Matrix Match Type Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labelled p, q, r, s and t. Any given statement in Column I can have correct matching with ONE OR MORE statement(s) in Column II. 21. For figure shown, match the column hemisphere 3.5 5cm

(A) (B) (C) (D)

Column I Curved area of hemisphere Height of cone Slant height of cone Surface area of top

A

Column II (p) 3.25 (q) 77/4 (r) 3.7 (s) 39.6

22. For a wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm, and its base of radius 3.5 cm, match the column

Column I Column II (A) Volume of cylinder (p) 616/3 (B) Volume of scoops (q) 374 (C) Total surface area (r) 122.5  (D) Volume of the article (s) 171.5/3  23. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and some diameter is hollowed out then match the column : Column I Column II (A) Area of bottom of cylinder (p) 10.56 (B) Outer curved surface area (q) 1.54 (C) Curved area of conical cavity (r) 5.5 (D) Total surface area (s) 17.6 24. In a right circular cylinder Column I Column II (A) The curved surface area = (p) 2rh (B) The total surface area = (q) r2h (C) Volume = (r) 2rh + r2 IV. Integers Type This section contains questions where the answer to each of the question is a single digit integer, ranging from 0 to 9. 25. Find the result when edge of a cube is divided by a, if its volume is 8a3. 26. If x is the number of cubes of side 2 cm can be cut from a cube of side 6 cm, then what is the value of x 3

162

10th Class Mathematics

27. If k is the number of circular plates each of radius 7 cm and thickness 0.5 cm, that should be placed one above the other to form a solid right circular cylinder of volume 1925 cm3, then what is the value of k  5 . 28. If three cubes each of edge ‘a’ are joined together of form a cuboid. If the surface area of the cuboid is A, what is the value of A  7a 2 29. A cylindrical vessel is full of water. How many cones having same diameter and height as that of the cylinder will be needed to store that water. V. Comprehension Type The following questions are based on the paragraph given below 30. An open metallic bucket is in the shape of a frustum of a cone mounted on hollow cylindrical base made of metallic sheet. The diameter of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 30 cm and that of the cylindrical portion is 6 cm (i) What is the area of the metallic sheet used to make the bucket ? (A) 3522.5 cm2 (B) 3622.5 cm2 (C) 3722.5 cm2 (D) 3822.5 cm2 (ii) What is the volume of the water that bucket can hold ? (A) 21728.57 cm3 (B) 22728.57 cm3 (C) 23728.57 cm3 (D) 24728.57 cm3

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163

Learning Outcomes By the end of this chapter, you will understand  Mean of Groped Data  Mode of Groped Data  Median of Grouped Data

Chapter - 13

Statistics  Graphical Representation of Cumulative Frequency Distribution

 Median of Grouped Frequency Distribution

1.

Introduction We have studied in previous class about classification of given data into ungrouped as well as grouped frequency distributions. We have also learnt how to represent the data with the help of various graphs like bar graphs, histograms and frequency polygons. Now, we will study about certain numerical representatives of the ungrouped data, mean, median and mode. In this chapter we will study mean, median and mode from ungrouped data to that of grouped data. Besides this, we will know the concept of cumulative frequency, the cumulative frequency distribution and also how to draw cumulative frequency curves. Direct Method: Mean of Grouped Data: The mean (or average) of observations is the sum of the values of all the observations dived by total number of observations. If x1, x2, x3, ...... xn are observation and their respectively frequencies are f1, f2, f3, ...... fn, then this mean observation x1 occurs f1 times, x2 occurs f2 times, and so on. Then, the sum of the values of all observations = f1 x1  f 2 x2  f 3 x3  .....  f n xn and the sum of number of observations = . Hence, the mean of the data is represented by f x  f 2 x2  f3 x3  .......  f n xn x = 1 1 f1  f 2  f3  .........  f n So,

x

 fi xi .... (i)  fi

[Where  means summation and i means 1 to n.] Assumed Mean Method: We choose assumed mean a (say) and subtract if from each of the values xi. The reduced value xi – a is called the deviation of xi from a. The deviation are multiplied by corresponding frequencies to get fidi. On adding all fidi we get the sum, i.e.  f i d i . Let x1, x2, x3, ..... xn be values of a variable x, with corresponding frequencies f1, f2, f3, .... fn respectively. di = xi – a; where i = 1, 2, 3, ... n  fidi = f i ( xi  a ) ; where i = 1, 2,3, ... n   f i d i =  f i ( xi  a)   f i d i =  f i xi  a  f i

10th Class Mathematics

164

on dividing by  f i , we get  fi di  fi xi a  fi  =  fi  fi  fi  fi di  fi di  =  fi  fi  fi di  = xa  fi  fi di  x = a ... (ii)  fi Step-Deviation Method Deviation method can be further simplified on dividing the deviation by width of the class interval h. In such a case the arithmetic mean is reduced to a great extent. x a ui = i ; 1, 2, 3, ..., n h  xi = a + hui ; i = 1, 2, 3, ..., n   f i xi = afi + hfiui; i = 1, 2, 3, ..., n   f i xi = a f i  h f i ui   f i xi = a f i  h  f i ui  fi xi f f u = = a i  h i i [Dividing by  f i ]  fi f i fi  fi ui  fi xi h  = a f i  fi  Mean ( x ) = a 

fi ui h f i

Example 1 Find the mean of the following distribution X: 4 6 F: 5 10

9 10

10 7

15 8

Solution: Calculation of Arithmetic Mean by Direct Method:

.xi 4 6 9 10 15  Mean = ( x ) =

fi 5 10 10 7 8 fi = 40

fixi 20 60 90 70 120 fixi =360

f i xi 360 = fi 40

 Mean ( x ) = 9.

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Statistics

165

Example 2 If the mean of the following is 6, find the value of p.

2 3

.x: f:

4 2

6 3

10 1

p+5 2

Solution: Calculation of Mean

.x: 2 4 6 10 p +5  Mean =

fi 3 2 3 1 2 fi = 11

fixi 6 8 18 10 2p + 10 fixi = 2p + 52

f i xi fi

2 p  52  66 = 2p + 52 11  66 - 52 = 2p  14 = 2p  p = 7 6=

Example 3 Calculate the Arithmetic mean of the following frequency distribution

Class Interval

50 – 60

60 – 70

70 – 80

80 – 90

90 – 100

Frequency

8

6

12

11

13

Solution: Let assumed mean be 75, i.e. a = 75 di = xi  75

fidi

8

Midvalue (xi) 55

20

160

60 – 70

6

65

70 – 80

12

75

10 0

60 0

80 – 90

11

85

10

110

90 – 100

13

95

20

260

Class Interval

Frequency

50 – 60

fi = 50

 Mean ( x ) = a  = 75 

fidi = 150

fidi fi

150 = 75 + 3  Mean ( x ) = 75. 50

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166

Formative Worksheet 1.

A survey was conducted by a group of students as a part of their environment awarenessprogramme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house. Number of plants

0 − 2 2 − 4 4 − 6 6 − 8 8 − 10 10 − 12 12 − 14

Number of houses

1

2

1

5

6

2

3

Which method did you use for finding the mean, and why? 2.

Consider the following distribution of daily wages of 50 worker of a factory. 100 − 120 120 − 140 140 −1 60 160 − 180 180 − 200

Daily wages (in Rs)

Number of workers 12

14

8

6

10

Find the mean daily wages of the workers of the factory by using an appropriate method. 3.

The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f. Daily pocket allowance (in Rs) Number of workers

4.

11 − 13 13 − 15 15 −17 17 − 19 19 − 21 21 − 23 23 − 25 7

9

f

13

5

4

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of heart beats per minute

65 − 68 68 − 71 71 −74 74 − 77 77 − 80 80 − 83 83 − 86

Number of women 5.

6

2

4

3

8

7

4

2

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes

50 − 52 53 − 55 56 − 58 59 − 61 62 − 64

Number of boxes

15

110

135

115

25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? 6.

The table below shows the daily expenditure on food of 25 households in a locality. Daily expenditure (in Rs) 100 − 150 150 − 200 200 − 250 250 − 300 300 − 350 Number of households

4

5

12

2

2

Find the mean daily expenditure on food by a suitable method.

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Statistics

167

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

7.

concentration of SO2 (in ppm) Frequency 0.00 − 0.04

4

0.04 − 0.08

9

0.08 − 0.12

9

0.12 − 0.16

2

0.16 − 0.20

4

0.20 − 0.24

2

Find the mean concentration of SO2 in the air. 8.a

8.b

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 0 − 6 6 − 10 10 − 14 14 − 20 20 − 28 28 − 38 38 − 40 Number of days 10 7 4 4 3 1 Number of students 11 The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Literacy rate (in %) 45 − 55 55 − 65 65 − 75 75 − 85 85 − 95 3 10 11 8 3 Number of cities

Conceptive Worksheet 1.

Find the mean marks from the following data by step deviation method: Below Below Below Below Below Below Below Below Below Below 10 20 30 40 50 60 70 80 90 100

Marks Number of Students

2.

5

9

17

29

45

78

83

85

Number of Person 100 90 75 50 25 15 5 0

Calculate the Arithmetic Mean of the following frequency distribution: Class Interval 50 – 60 Frequency

4.

70

Find the mean age of 100 residents of a colony from the following data:

Age in Year Grater Than 0 Grater Than 10 Grater Than 20 Grater Than 30 Grater Than 40 Grater Than 50 Grater Than 60 Grater Than 70 3.

60

8

60 – 70

70 – 80

80 – 90

90 – 100

6

12

11

13

The Arithmetic Mean of the following frequency distribution is 50, find the value of p. Class Interval 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 Frequency

17

p

32

24

19

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10th Class Mathematics

168 5.

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2. 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100 - 120 Total Class 8 50 5 f1 10 f2 7 Frequency

2.

Mode of Grouped Data We can locate a class with maximum frequency, called modal class. The mode is a value inside the modal class, and is given by the following formula:

 f1  f 0  Mode  l    h  2 f1  f0  f 2  Where l = Lower limit of the modal class, h = Size of the class interval (assuming all class sizes to be equal), f1 = Frequency of the modal class, f0 = Frequency of the class preceding the modal class, f2 = Frequency of the class succeeding the modal class. Example 4 The following table shows the ages of the patients admitted in a hospital: Age 5 – 15 15 – 25 25 – 35 35 – 45 45 – 55 55 - 65 (in years) 6 11 21 23 14 5 No. of cases Find the mode of the given data. Solution: The class (35 – 45) has maximum frequency, i.e. 23. Therefore, this is the modal class. Lower limit of the modal class (l) = 35, Size of the class internal (h) = 10 Frequency of the modal class (f1) = 23 Frequency of the class preceding the modal class (f0) = 21 Frequency of the class succeeding the modal class (f2) = 14 We know that,

 f1  f0  Mode = l   h  2 f1  f0  f 2     20  23  21 = 35     10 = 35     2  23  21  14   46  36 

20 = 35 + 1.8181 11  Mode = 36.8181 Hence, the average age for which maximum cases occurred is 36.818. = 35 

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Statistics

169

Formative Worksheet 9.

10.

The following table shows the ages of the patients admitted in a hospital during a year: age (in years)

5 − 15 15 − 25 25 − 35 35 − 45 45 − 55 55 − 65

Number of patients

6

11

21

23

14

5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Lifetimes (in hours) 0 − 20 20 − 40 40 − 60 60 − 80 80 − 100 100 − 120 10

Frequency 11.

35

52

61

38

29

Determine the modal lifetimes of the components. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure. Expenditure (in Rs) Number of families

12.

1000 − 1500

24

1500 − 2000

40

2000 − 2500

33

2500 − 3000

28

3000 − 3500

30

3500 − 4000

22

4000 − 4500

16

4500 − 5000

7

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher Number of states/U.T 15 − 20

3

20 − 25

8

25 − 30

9

30 − 35

10

35 − 40

3

40 − 45

0

45 − 50

0

50 − 55

2

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10th Class Mathematics

170

Conceptive Worksheet 6.

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored

Number of batsmen

3000 − 4000

4

4000 − 5000

18

5000 − 6000

9

6000 − 7000

7

7000 − 8000

6

8000 − 9000

3

9000 − 10000

1

10000 − 11000 1 7.

Find the mode of the data. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data: Number of cars 0 − 10 10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 Frequency

8.

7

13

12

20

11

15

8

The marks in science of 80 students of class X are given below. Find the mode of these marks obtained by the students in science. Marks

Frequency

3.

14

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

3

5

16

12

13

20

5

4

1

1

Median of Grouped Data We first arrange the data values of the observations in succeeding order. th

 n 1 Then, if n is odd, then the median =   observation,  2  th th  1  n  n  And if, n is even, then the median =   observation    1 observation  2  2  2  

4.

Median of Grouped Frequency Distribution Frequency distribution may be cumulative frequency distribution of the less than type or cumulative frequency distribution of the more than type n In order to find the median first we have to find out then the class in which lies. This class will be 2 called the median class. Median lies in this class. Median can be calculated by the following formula. Where l = Lower limit of median class, n = number of observations, cf = cumulative frequency of class preceding the median class, f = frequency of median class, h = class size (assuming class size to equal).

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Statistics

171

POINT TO BE REMEMBERD f x (i) Mean ( x ) = i i f i f d (ii) Mean ( x ) = a  i i f i f u (iii) Mean ( x ) = a  i i  h fi (iv) (v)

 f1  f0  Median = l   h  2 f1  f0  f 2  Median (if n is odd) th

 n 1 =   observation  2  (vi) Median (if n is even) th th   1 n n  =   observation    1 observation  2  2  2   (vii) Median = (viii) 3 Median = Mode + 2 Mean

Formative Worksheet 13.

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units) Number of consumers

14.

65 − 85

4

85 − 105

5

105 − 125

13

125 − 145

20

145 − 165

14

165 − 185

8

185 − 205

4

If the median of the distribution is given below is 28.5, find the values of x and y. Class interval Frequency 0 − 10

5

10 − 20

x

20 − 30

20

30 − 40

15

40 − 50

y

50 − 60

5

Total

60

___________________________________________________________________________________________________________________________________________

172

10th Class Mathematics

15.

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

16.

Age (in years)

Number of policy holders

Below 20

2

Below 25

6

Below 30

24

Below 35

45

Below 40

78

Below 45

89

Below 50

92

Below 55

98

Below 60

100

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table: Length (in mm) Number or leaves fi

17.

118 − 126

3

127 − 135

5

136 − 144

9

145 − 153

12

154 − 162

5

163 − 171

4

172 − 180

2

Find the median length of the leaves. (Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5) Find the following table gives the distribution of the life time of 400 neon lamps: Life time (in hours) Number of lamps 1500 − 2000

14

2000 − 2500

56

2500 − 3000

60

3000 − 3500

86

3500 − 4000

74

4000 − 4500

62

4500 − 5000

48

Find the median life time of a lamp.

___________________________________________________________________________________________________________________________________________

Statistics 18.

19.

173

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: 1 − 4 4 − 7 7 − 10 10 − 13 13 − 16 16 − 19 Number of letters 30 40 6 4 4 Number of surnames 6 Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. 40 − 45 45 − 50 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75 Weight (in kg) 3 8 6 6 3 2 Number of students 2

Conceptive Worksheet 9. 10. 11.

12.

Find the median of 6, 8, 9, 10, 12, 13 and 17 Find the median of 21, 23, 24, 25, 26, 27, 28 and 30. Find the median of the following:

Class Interval

0–8

8 – 16

16 – 24

24 – 32

32 – 40

40 - 48

Frequency

8

10

16

24

15

7

The following table shows the weekly drawn by number of works in a factory: Weekly Wages (in Rs.) 0 – 100 100 – 200 200 – 300 300 – 4000 40 39 34 30 No. of Workers

Find the median wage income of the workers.

5.

Graphical Representation of Cumulative Frequency Distribution A graphical representation helps us in understanding given data at a glance. Now, We have to represent a cumulative frequency distribution graphically. Example: Table for less than type: Class 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40

No. of shops Shops Cumulative frequency less than frequency 2 10 2 12 15 14 2 20 16 4 25 20 3 30 23 4 35 27 3 40 30

Table for more than type: No. of shops Shops Cumulative frequency more than frequency 30 5 – 10 2 5 28 10 – 15 12 10 Class

15 – 20

2

15

16

20 – 25

4

20

14

25 – 30

3

25

10

30 – 35

4

30

7

35 – 40

3

35

3

___________________________________________________________________________________________________________________________________________

10th Class Mathematics

174 Y - axis More than ogive 30 28 26 24

Less than ogive

22

Cumulative frequency

20 18 16 14 12 10 8 6 4 Median (17)

2

X - axis

0

5

10

15 20 25 30 35 Profit (in lakhs Rs.)

40

The abscissa of point of intersection is of two curves 17 hence, median = 17

Formative Worksheet 20.

The following distribution gives the daily income of 50 workers of a factory. Daily income (in Rs) 100 − 120 120 − 140 140 − 160 160 − 180 180 − 200 Number of workers

21.

12

14

8

6

10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive. During the medical check-up of 35 students of a class, their weights were recorded as follows: Weight (in kg) Number of students

22.

Less than 38

0

Less than 40

3

Less than 42

5

Less than 44

9

Less than 46

14

Less than 48

28

Less than 50

32

Less than 52

35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula. The following table gives production yield per hectare of wheat of 100 farms of a village. Production yield (in kg/ha) 50 − 55 55 − 60 60 − 65 65 − 70 70 − 75 75 − 80 Number of farms

2

8

12

24

38

16

Change the distribution to a more than type distribution and draw ogive. ___________________________________________________________________________________________________________________________________________

Statistics

175

Conceptive Worksheet 13.

Draw an ogive for the following frequency distribution by less than method. Marks

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100

No. of Students

14.

No. of Pwesons

16.

8

12

15

18

29

38

45

53

0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100 53

48

45

41

38

35

31

24

15

8

Convert the distribution above to a more than type cumulative frequency distribution and draw its ogive The following observations are related to the height of a group of persons. Draw the cumulative frequency curve (less than type) and hence obtain the median value: Height is cms

Frequency

140 – 145

4

145 – 150

11

150 – 155

30

155 – 160

32

160 – 165

43

165 – 170

65

170 – 175

72

175 – 180

81

180 – 185

20

185 – 190

15

190 – 195

5

195 – 200

2

Following is the age distribution of a group of students. Draw the two types of cumulative frequency curves and determine the median. Age

5-6

6-7

7-8

8-9

9-10 10-11 11-12 12-13 13-14 14-15 15-16 16-17

Frequency 35 51 64 66 74

17.

22

The following distribution gives the age of 338 persons in a village Age (in years)

15.

5

99

91

82

62

46

21

9

The annual profits earned by 30 shops in a locality give rise to the following distribution:

Number of (frequency) More than or equal to 5 30 More than or equal to 10 28 More than or equal to 15 16 More than or equal to 20 14 More than or equal to 25 10 More than or equal to 30 7 More than or equal to 35 3 Draw both ogives for the above data and hence obtain the median. Profit (in lakhs Rs.)

___________________________________________________________________________________________________________________________________________

10th Class Mathematics

176

Summative Worksheet 1.

2.

Find the mean of the following distribution 4 6 x 5 10 f

9 10

(A) 9

(C) 18

(B) 15

10 7

15 8

(D) none of these

Find the mean of the following distribution

x f (A) 45 3.

10 7

30 8

(B) 65

50 10

70 15

89 10

(C) 55

(D) 85

If the median of the following frequency distribution is 46, find the missing frequencies Variable 10-20 20-30 30 Frequency 12

30-40 ?

40-50 50-60 60-70 70-80 65 ? 25 18

(A) 33, 55 (B) 32, 35 (C) 34, 45 Find the mode of the following data 26, 16, 19, 48, 19, 20, 34, 15, 19, 20, 21, 24, 19, 22, 16, 18, 20, 16, 19 (A) 48 (B) 20 (C) 19

4.

5.

Find the value of x, if the mode of the following date is 25 : 15, 20, 25, 18, 14, 15, 25, 15, 18, 16, 20, 25, 20, x, 18 (A) 20 (B) 14 (C) 18 The median of the first ten prime numbers is: (A) 7 (B) 12 (C) 11 Mean of all possible factors of 10 is (A) 6 (B) 2 (C) 4.5

6. 7.

8.

9. 10.

Total 229

(D) none of these

(D) 24

(D) 25 (D) 5 (D) 5

The sum of the deviations of a set of n values measured from 50 is –10 and that of the values from 46 is 70. Then the mean of the data will be (A) 40.5 (B) 46.5 (C) 42.5 (D) 49.5 The mean of 10, 12, 16, 20, p and 26 is 17. Then the value of p is (A) 10 (B) 18 (C) 25 (D) 14 The difference between the highest observation and the lowest observation is known as (A) Class Mark (B) Class Size (C) Range

(D) Limit

IIT JEE Worksheet I. 1.

2.

3.

Multiple Correct Answer Type The following are the marks of 9 students in a class. Median of the following data can not be equal to 34, 32, 48, 38, 24, 30, 27, 21, 35 (A) 35 (B) 24 (C) 48 (D) 32 Find the median of the daily wages of ten workers from the following data will be less than Rs. 20, 25, 17, 18, 8, 15, 22, 11, 9, 14 (A) 22 (B) 18 (C) 16 (D) 25 The empirical formula for mode is given by (A) Mode = 3 median – 2 mean (B) Mode - median = 2(median - mean) (C) Mean = 2 median – 3 mode (D) Mode = 2 median – 3 mean

___________________________________________________________________________________________________________________________________________

Statistics II. 4.

Matrix & Matching Column–I (A) The difference between the maximum and the minimum observations in data is called (B) The mid-point of a class interval is called its (C) Facts or figures, collected with a definite purpose, are called (D) Value of the middle-cost observation(s) is called

5. (A) (B) (C) (D)

III. 6.

177

Column–I The variance is the ..... of the standard deviation Median divides the total frequency into .... equal parts The algebraic sum of the deviations from arithmetic mean is always.... percentile divides the number of items into .... equal parts.

(q)

range

(r)

median

(s)

class-mark

(p)

Column–II zero

(q)

square

(r)

two

(s)

hundred

2 3

4 2

6 3

10 1

p+5 2

Obtain the median for the following frequency distribution x f

8.

Column–II data

Integer Answers Type Find the mean of the following distribution is 6, find the value of p x f

7.

(p)

1 8

2 10

3 11

4 16

5 20

6 25

7 15

8 9

9 6

Calculate the value of (median/10) from the following data: Marks No. of Students

0-10 5

10-30 15

30-60 30

60-80 80-90 8 2

9.

For what value of x, the mode of the data 2, 3, 4, 5, 3, 5, 6, 3, x, 7, 5 will be 5

10.

The mean of 5 number is 18. If one number is excluded, their mean is 16. The unit place of excluded number will be

___________________________________________________________________________________________________________________________________________

178

10th Class Mathematics

___________________________________________________________________________________________________________________________________________

IIT FOUNDATION Class X

MATHEMATICS SOLUTIONS

© USN Edutech Private Limited The moral rights of the author’s have been asserted. This Workbook is for personal and non-commercial use only and must not be sold, lent, hired or given to anyone else.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise without the prior written permission of USN Edutech Private Limited. Any breach will entail legal action and prosecution without further notice.

Utmost care and attention to the details is taken while editing and printing this book. However, USN Edutech Private Limited and the Publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in.

Published by

:

USN Eductech Private Limited Hyderabad, India.

CONTENTS 1.

Real Numbers

..........

179 – 192

2.

Polynomials

.........

193 – 216

3.

Pair of Linear Equations in Two Variables

..........

217 – 252

4.

Quadratic Equations

..........

253 – 282

5.

Progressions

..........

283 – 326

6.

Functions

..........

327 – 342

7.

Coordinate Geometry

..........

343 – 366

8.

Some Applications of Trigonometry

..........

367 – 380

9.

Binomial Theorem

..........

381 – 396

10.

Triangles

..........

397 – 430

11.

Circles

..........

431 – 438

12.

Surface Areas and Volumes

..........

439 – 472

13.

Statistics

..........

473 – 498

1. REAL NUMBERS SOLUTIONS

FORMATIVE WORKSHEET 1. (i) 135 and 225 Since 225 > 135, we apply the division lemma to 225 and 135 to obtain 225 = 135 × 1 + 90 Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain 135 = 90 × 1 + 45 We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 × 45 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 45, Therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 196, Therefore, HCF of 196 and 38220 is 196. (iii) 867 and 255 Since 867 > 255, we apply the division lemma to 867 and 255 to obtain 867 = 255 × 3 + 102 Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain 255 = 102 × 2 + 51 We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain 102 = 51 × 2 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51. 2. Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6. Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5 Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer 6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

3.

4.

5.

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer. Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers. And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5 HCF (616, 32) will give the maximum number of columns in which they can march. We can use Euclid’s algorithm to find the HCF. 616 = 32 × 19 + 8 32 = 8 × 4 + 0 The HCF (616, 32) is 8. Therefore, they can march in 8 columns each. Let a be any positive integer and b = 3. Then a = 3q + r for some integer q ≥ 0 And r = 0, 1, 2 because 0 ≤ r < 3 Therefore, a = 3q or 3q + 1 or 3q + 2 OR 2 2 2 2 a = (3q) or (3q + 1) or (3q + 2) 2 2 2 2 a = (9q) or 9q + 6q + 1 or 9q + 12q + 4 2 2 = 3 × (3q) or 3(3q + 2q) + 1 2 or 3(3q + 4q + 1) + 1 = 3k1 or 3k2 + 1 or 3k3 + 1 Where k1, k2, and k3 are some positive integers Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1. Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3  a = 3q or 3q + 1 or 3q + 2 Therefore, every number can be represented as these three forms. There are three cases. Case 1: When a = 3q, 3 3 3 3 a = (3q) = 27q = 9(3q ) = 9m, 3 Where m is an integer such that m = 3q Case 2: When a = 3q + 1, 3 3 a = (3q +1) 3 3 2 a = 27q + 27q + 9q + 1 3 3 2 a = 9(3q + 3q + q) + 1 3 a = 9m + 1

____________________________________________________________________________________________________ _____________________________________________

Real Numbers Solutions

180

Where m is an integer 3 2 (3q + 3q + q) Case 3: When a = 3q + 2, 3 3 a = (3q +2) 3

3

such

that m =

such

that m =

2

a = 27q + 54q + 36q + 8 3

3

2

a = 9(3q + 6q + 4q) + 8 3

a = 9m + 8 Where m is 3

6.

an

integer

2

(3q + 6q + 4q) Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8. Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3 + 2. Case 1: When x = 3q 3

3

3

3

 x = (3q) = 27q = 9(3q ) = 9m, 3

where m = 9q Case 2: when x = 3q + 1 3

3

2

 x = 2q + 27q + 9q + 1 3

2

 x = 9q (3q + 3q + 1) + 1 3

7.

2

 x = 9m + 1, where m = q (3q + 3q + 1). Let x be any positive’s integer of the form 5q + 1. When x = 5q + 1 2

2

x = 25q + 10q + 1 2

x = 5(5q + 2) + 1 Let m = q (5q + 2). 2

x = 5m + 1. 2

8.

Hence, x is of the same form i.e. 5m + 1. Applying Euclid’s division lemma to 196 and 38318. 38318= 195 × 196 + 98 196= 98 × 2 + 0 The remainder at the second stage is zero. So, the H.C.F. of 38318 and 196 is 98. Applying Euclid’s division lemma on 657 and 963. 963= 657 × 1 + 306 657= 306 × 2 + 45 306= 45 × 6 + 36 45= 36 × 1 + 9 36= 9 × 4 + 0 So, the H.C.F. of 657 and 963 is 9. Given : 657x + 963 × (–15) = H.C.F. of 657 and 963.

657 x + 963 × (– 15) = 9 657 x= 9 + 963 × 15 657 x= 14454 14454 x= = 22 657 9. Clearly, the required number is the H.C.F. of the number 626 – 1 = 625, 3127 – 2 3125 and 15628 – 3 = 15625. 15628 – 3 = 15625. Using Euclid’s division lemma to find the H.C.F. of 625 and 3125. 3125 = 625 × 5 + 0 Clearly, H.C.F. of 625 and 3125 is 625. Now, H.C.F. of 625 and 15625 15625 = 625 × 25 + 0 So, the H.C.F. of 625 and 15625 is 625. Hence, H.C.F. of 625, 3125 and 15625 is 625. Hence, the required number is 625. 10. In order to arrange the cartons of the same drink is the same stack, we have to find the greatest number that divides 144 and 90 exactly. Using Euclid’s algorithm, to find the H.C.F. of 144 and 90. 144= 90 × 1 + 54 90= 54 × 1 + 36 54= 36 × 1 + 18 36= 18 × 2 + 0 So, the H.C.F. of 144 and 90 is 18. Number of cartons in each stack = 18. 2

11. (i) 140 = 2 × 2 × 5 × 7 = 2 × 5 × 7 2

(ii) 156 = 2 × 2 × 3 × 13 = 2 × 3 × 13 2

2

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3 × 5 × 17 (iv) 5005 = 5 × 7 × 11 × 13 (v) 7429 = 17 × 19 × 23 12. (i) 26 and 91 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Product of the two numbers = 26 × 91 = 2366 HCF × LCM = 13 × 182 = 2366 Hence, product of two numbers = HCF × LCM (ii) 510 and 92 510 = 2 × 3 × 5 × 17 92 = 2 × 2 × 23 HCF = 2

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

181

LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460 Product of the two numbers = 510 × 92 = 46920 HCF × LCM = 2 × 23460 = 46920 Hence, product of two numbers = HCF × LCM (iii) 336 and 54 336 = 2 × 2 × 2 × 2 × 3 × 7 4

336 = 2 × 3 × 7 54 = 2 × 3 × 3 × 3  54 = 2 × 3 HCF = 2 × 3 = 6 4

3

3

LCM = 2 × 3 × 7 = 3024 Product of the two numbers = 336 × 54 = 18144 HCF × LCM = 6 × 3024 = 18144 Hence, product of two numbers = HCF × LCM 13. (i) 12, 15 and 21 2

12 = 2 × 3 15 = 3 × 5 21 = 3 × 7 HCF = 3 2

(ii)

(iii)

14.

15.

LCM = 2 × 3 × 5 × 7 = 42 17, 23 and 29 17 = 1 × 17 23 = 1 × 23 29 = 1 × 29 HCF = 1 LCM = 17 × 23 × 29 = 11339 8, 9 and 25 8=2×2×2 9=3×3 25 = 5 × 5 HCF = 1 LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800 HCF (306, 657) = 9 We know that, LCM × HCF = Product of two numbers  LCM × HCF = 306 × 657 306× 657 306× 657 LCM = = HCF 9 LCM = 22338 If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5 n

Prime factorisation of 6n = (2 ×3)

It can be observed that 5 is not in the prime n factorisation of 6 . n Hence, for any value of n, 6 will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n. 16. Numbers are of two types - prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself. It can be observed that 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 ×13 × 6 The given expression has 6 and 13 as its factors. Therefore, it is a composite number. 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 ×1009 1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number. 17. It can be observed that Ravi takes lesser time than Sonia for completing 1 round of the circular path. As they are going in the same direction, they will meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia. And the total time taken for completing this 1 round of circular path will be the LCM of time taken by Sonia and Ravi for completing 1 round of circular path respectively i.e., LCM of 18 minutes and 12 minutes. 18 = 2 × 3 × 3 and 12 = 2 × 2 × 3 LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36 Therefore, Ravi and Sonia will meet together at the starting pointafter 36 minutes. 18.

 45470971 = 72 × 132 × 172 × 19.

____________________________________________________________________________________________________ _____________________________________________

Real Numbers Solutions 2

19. 84 = 2 × 3 × 7, 2 90 = 2 × 3 × and 3 120 = 2 × 3 × 5. 1 1  HCF = 2 × 3 = 6. 3 3 1 1  LCM = 2 × 3 × 5 × 7 = 8 × 9 × 5 × 7= 2520 20. Required minimum distance each should walk so, that they can cover the distance in complete step is the L.C.M. of 80 cm, 85 cm and 90 cm 4 80= 2 × 5 85= 5 + 17 2 90= 2 × 3 × 5  LCM = 24 × 32 × 51 × 171 LCM = 16 × 9 × 5 × 17  LCM = 12240 cm, = 122 m 40 cm. 21. Let 5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that a 5 b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime. a  5b 2

182

1a  5 =  - 3 2 b  1 a  Since a and b are integers  - 3  will also be 2 b  rational and therefore, is rational. This contradicts the fact that 5 is irrational. Hence, our assumption that 3 + 2 5 is rational is false Therefore, 3 + 2 5 is irrational. 1 23. (i) Let is rational. 2 Therefore, we can find two integers a, b (b ≠ 0) such that 1 a = 2 b b 2= a b is rational as a and b are integers. a Therefore. 2 is rational which contradicts to the fact that is irrational.

2

Hence, our assumption is false and

2

a = 5b 2 Therefore, a is divisible by 5 and it can be said that a is divisible by 5. Let a = 5k, where k is an integer 2 2 (5k) = 5b . 2 This means that b is divisible by 5 and hence, b is divisible by 5. 2 2 b = 5k . This implies that a and b have 5 as a common factor. and this is a contradiction to the fact that a and b are co-prime. p Hence, 5 cannot be expressed as or it can q be said that

5 is irrational.

22. Let 3 + 2 5 is rational Therefore, we can find two integers a, b (b ≠ 0) such that a 3+ 2 5 = b a 2 5 = -3 b

1 is 2

irrational. (ii) Let 7 5 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that a 7 5 = for some integers a and b b a  5= 7b a is rational as a and b are integers. 7b Therefore, 5 should be rational. This contradicts the fact that

5 is irrational.

Therefore, our assumption that 7 5 is rational is false. Hence, 7 5 is irrational. (iii) Let 6 + 2 is rational. Therefore, we can find two integers a, b (b ≠ 0) such that a 6+ 2 = b a 2 = -6 b

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

183

a - 6 is also rational b and hence, 2 should be rational. This contradicts the fact that 2 is irrational. Therefore, our assumption is false and hence, 6 + 2 is irrational. 24. Let assume on the contrary that 2 is a rational number. Then, there exists positive integer a and b such that a 2 = where, a and b are co primes i.e. their b HCF is 1. Since a and b are integers



 2



2= 2

2

a =  b

2

a2 b2

2

 a = 2b 2  a is multiple of 2 a is a multiple of 2 ....(i)  a = 2c for some integer c. 2 2  a = 4c 2 2 2 2  2b = 4c  b = 2c 2  b is a multiple of 2 b is a multiple of 2 ....(ii) From (i) and (ii), a and b have at least 2 as a common factor. But this contradicts the fact that a and b are co-prime. This means 2 that is an irrational number. 25. Let assume that on the contrary that 3 - 5 is rational. Then, there exist co-prime positive integers a and b such that, a 3- 5 = b a  3- = 5 b 3b - a  = 5 b  5 is rational 3b - a [ a,b, are integer  is a rational b number] This contradicts the fact that 5 is irrational Hence, 3  5 is an irrational number.

13 3125 5 3125 = 5 m The denominator is of the form 5 . Hence, the decimal expansion 13 terminating. 3125 17  ii  8 m The denominator is of the form 2 . 3 8=2

26.

i

Hence, the decimal expansion of

of

is

17 8

is

terminating. 64  iii  455 455 = 5 × 7 × 13 m n Since the denominator is not in the form 2 × 5 , and it also contains 7 and 13 as its factors, its decimal expansion will be non-terminating repeating. 15  iv  1600 6 2 1600 = 2 × 5 15 Hence, the decimal expansion of is 1600 terminating. 29  v 343 3 343 = 7 Since the denominator is not in the form 2m × 5n, and it has 7 as its factor, the decimal 29 expansion of is non-terminating 343 repeating. 23  vi  3 2 × 52 m n The denominator is of the form 2 × 5 . 23 Hence, the decimal expansion of 3 2 2 ×5 is terminating. 129  vii  2 7 5 2 ×5 ×7 Since the denominator is not of the form 2m × 5n, and it also has 7 as its factor, the decimal 129 expansion of 2 7 5 is non-terminating 2 ×5 × 7 repeating.

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Real Numbers Solutions

 viii 

184

6 2×3 2 = = 15 3× 5 5

 ii  n

The denominator is of the form 5 .

2.125 8) 17 16

16 is 15

Hence, the decimal expansion of

_________

terminating.

 ix 

17 = 2.215 8

10 8

35 7 × 5 7 = =  10 = 2 × 5 50 10 × 5 10

_________

20 m

n

The denominator is of the form 2 × 5 .

16

_________

Hence,

the

decimal

expansion

of

35 is 50

40 40

terminating.

_________

77 11× 7 11 = = x 210 30 × 7 30

_________

0

 iv 

30 = 2 × 3 × 5 Since the denominator is not of the form 2m × 5n, and it also has 3 as its factors, the decimal expansion

of

77 is 210

repeating. 27.

i

13 = 0.00416 3125

0.00416 3125) 13.00000 0 _________

130 0 _________

13000

non-terminating

15 = 0.009375 1600

0.009375 1600) 15.000000 0

_________

150 0 _________ 1500 0 _________ 15000 14400 _________ 6000 4800

_________

12000 11200

_________

8000

12500 _________

8000 _________

5000 3125

0

_________

18750 18750

_________

0

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10th Class Mathematics

185

 vi 

23 = 0.115 2 × 52

28. 43.123456789

3

Since this number has a terminating decimal expansion, it is a rational number of the form

0.115 200) 23.000 0

p and q is of the form q

_________

230 200

i.e., the prime factors of q will be either 2 or 5 or both.

_________

300

(ii) 0.120120012000120000…

200

The decimal expansion is neither terminating

_________

nor recurring.

1000 1000

Therefore, the given number is an irrational

_________

number.

0

_________

(iii) 43.123456789

 viii 

6 2×3 2 = = = 0.4 15 3× 5 5

Since

the

decimal

expansion

is

non-

terminating recurring, the given number is a

0.4 5) 2.0 0

of the form i.e., the prime factors of q will

_________

20 20 _________

0

p and q is not q

rational number of the form

also have a factor other than 2 or 5. 29.

13 13 = 3125 2°× 55

_________

This, shows that the prime factorisation of the

 ix 

35 = 0.7 50

m

n

denominator is of the form 2 × 5 . Hence, it has terminating decimal expansion.

0.7 50) 35.0 0 _________

350 350 _________

0

_________

30. (i) Since, 43.123456789 has terminating decimal, so prime factorisations of the denominator is of the form 2m × 5n, where m, n are non – negative integers. (ii) Since, 43.

123456789 has non-terminating

repeating decimal expansion. So, its denominator has factors other than 2 or 5.

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10th Class Mathematics

186 31 32

33 34 35 36 37

38

39 40

C

A

D

C

C

A

B

A

A

(i) Rational number: q = 23 × 53 (ii) Rational number: q = 23 × 32 × 53 × 11

28.

D

29 30

31 32 33 34 35

36

37 38

B

A

D

D

CONCEPTIVE WORKSHEET 1. 2. 3. 6. 7. 8.

89 13 31 13 = 65M + 117n, where m = 2 and n = –1 19 23

1741740

11. 25478976

12.

(i) 176 = 2 × 2 × 2 × 2 × 11 (ii) 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 (iii) 4825 = 5 × 5 × 193 (iv) 12673 = 19 × 23 × 29

C

A

2.

2 5 1    3 24 4

3.

2 1 0 1 2 3 4 5 6 7 , , , , , , , , and 11 11 11 11 11 11 11 11 11 11 29 28 1 0 1 70 , ,....... , , ,...... 130 130 130 130 130 130 0.875 6. 2.1875

8.

0.18

9. 11.

13. 15.

(a)

13.

14/11

1 10. a = 2/7, b= 1/3, c = 1/7 31 32 33 34 35 36 , , , , , 10 10 10 10 10 10 21/120 14.

12.

17.

www.betoppers.com

D

 -2 < 2 < 6

16.

14. No 15. No 16. No 17. Yes, If n = 4 18. Yes, If n = 4 25. (i) Irrational number (ii) rational number (iii) Irrational number (iv) rational number 26. (i) Irrational number (ii) Irrational number (iii) rational number (iv) rational number 27. (i) Terminating decimal (ii) Non-terminating decimal

D

1.

5. 7. 10.

B

SUMM ATIVE WORKSHEET

4.

9.

D

3 5

98

5185 (b) 0.33 33 (i) 0.12212221…., (ii) 1.123123312333…….

2 3  3 2  30 12

18 19

20 21 22 23 24

25

26 27

A

A

B

A

C

A

D

28 29

30 31 32

D

D

A

D

D

A

D

D

Real Numbers Solutions

187

HOTS WORKSHEET

1.

Q.No

1

2

3

4

5

6

7

8

Key

A

B

C

A

A

D

D

C

Q.No

9

10

11

12

13

14

15

16

Key

C

B

C

A

B

C

D

B

Q.No

17

18

19

20

21

22

23

24

Key

D

A

A

B

C

B

B

A

Q.No

25

26

27

28

29

30

Key

D

C

B

C

A

C

Let the other number be a. Let b be 123. We know that, LCM (a, b) × HCF (a, b) = a × b  39483 = a × b 39483 = 123 × a 38483  321 123 Hence, the other number is 321. The correct answer is A. The numbers 124 and 224 can be primefactorised as 124 = 2 × 2 × 31 = 22 × 31 224 = 2 × 2 × 2 × 2 × 2 × 7 = 25 × 7 HCF is the product of the smallest powers of each common prime factor of the numbers, while LCM is the product of the highest power of all the prime factors of the numbers.  LCM (124, 224) = 25 × 7 × 31 = 6944 HCF (124, 224) = 22 = 4 The correct answer is B. The number 1224 can be prime-factorised as 1224 = 2 × 2 × 2 × 3 × 3 × 17 = 23 × 32 × 17 The correct answer is C. Leta be any positive integer. On applying division algorithm on aand b = 2, we obtain a = 2q + r; 0 d   r < 2 Hence, the possible remainders are 0 or 1. This means a is  of  the  form  2q or  2q +  1, where q is the quotient. a

2.

3.

4.

5.

6.

Thus, all positive integers can be written in the form 2q or 2q + 1. On squaring 2q and 2q + 1, we obtain 4q2 and (4q2 + 4q + 1). We can write them as 4(q2) and [4(q2 + q) +1] or, we can write them as 4k or 4k +1, where k is a constant. Thus, any perfect square can be written in the form 4k or 4k + 1.The correct answer is A. The numbers 24, 32, and 60 can be primefactorised as 24 = 2 × 2 × 2 × 3 = 23 × 31 32 = 2 × 2 × 2 × 2 × 2 = 25 60 = 2 × 2 × 3 × 5 = 22 × 31 × 51 HCF is the product of the smallest powers of each common prime factor of the numbers.  HCF (24, 32, 60) = 22 = 4 The correct answer is A.

p  has a non-terminating q decimal expansion if the prime factorisation of the denominator (q) cannot be expressed as 2n5m, where m and n are non-negative integers. A number of the form

The rational number

45  can  be  written  as 630

45 33 5 1   | | 630 2  3  3  5  7 2  7 Here, the denominator is not of the form 2n5m. Thus, has a non-terminating decimal expansion. The correct answer is D. www.betoppers.com

10th Class Mathematics

188

7.

p A number of the form  has a terminating decimal q expansion if the denominator (q) can be expressed as 2n5m, where m and n are  nonnegative integers. The rational number

12.

420  can  be written  as 240

420 2 235 7 7 7    2 0 240 2  2  2  2  3  5 2  2 2  5 Here, the denominator is of the form 2n5m, where n = 2 and m = 0. 420 has a terminating decimal expansion. 240 The correct answer is D. 8. If a number ends in 0, then it must be divisible by 10 i.e., it must have 2 and 5 as its factors. Thus, (2 × 5)n can  end  in  0  for  any  natural number n. The correct answer is C. 9. The numbers 36, 96, and 72 can be primefactorised as 36 = 2 × 2 × 3 × 3 = 22 × 32 96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32 LCM is the product of the highest powers of all the prime factors of the numbers.  LCM (36, 72, 96) = 25 × 32 = 288 The correct answer is C. 10. Machine A takes 15 minutes and machine B takes 25 minutes to produce an item. Therefore, the time at which they will next produce an item simultaneously will be the LCM of 15 and 25. 15 = 3 × 5 25 = 5 × 5 = 52 LCM (15, 25) = 3 × 5 × 5 = 75 Thus, the two machines will simultaneously produce an item 75 minutes after 12 p.m., i.e., at 1.15 p.m. The correct answer is B. 11. HCF (a, b) = 30 LCM (a, b) = 45 We know that, HCF (a, b) × LCM (a, b) = a × b Thus, a × b = HCF (a, b) × LCM (a, b) = 30 × 45 = 1350 The correct answer is C.

13.

Thus,  

www.betoppers.com

14.

15.

16.

We first need to find the number of bottles that have to be kept in the last crate. Using Euclid’s division lemma, we can write 1051 as 1051 = 28 × 37 + 15 Hence, 15 bottles have to be kept in the last crate. It is given that one crate can hold a maximum of 28 bottles. Hence, number of empty spaces in the last crate = (28 “ 15) = 13 The correct answer is A. The number 20000 can be prime-factorised as 20000 = 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5 = 25 × 54 The number 25600 can be prime-factorised as 25600 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 = 210 × 52 HCF is the product of the smallest powers of each common prime factor of the numbers.  HCF (20000, 25600) = 25 × 52 Since b is greater than a, a = 2 and b = 5. Thus, the respective values of a and b are 2 and 5. The correct answer is B. The number 58500 can be prime-factorised as 58500 = 2 × 2 × 3 × 3 × 5 × 5 × 5 × 13 = 22 × 32 × 53 × 13 The correct answer is C. According to the fundamental theorem of arithmetic, every number can be factorised as a product of its primes in a unique way. Therefore, the number 75600 can be primefactorised as 75600 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 7 = 24 × 33 × 52 × 71 On comparing it to (a1 × b2 × c3 × d4), we obtain a = 7, b = 5, c = 3, d = 2 Thus, the value of (a + c) is (7 + 3) = 10. The correct answer is D. Least common multiple × greatest common divisor = product of the two numbers Let the second number be n.  8 × 336 = n × 48 8  336  56 48 The correct answer is B. n

Real Numbers Solutions

17.

18.

According to the given information, the number of books in each shelf should exactly divide the numbers of English and Mathematics books. Since we require the minimum number of shelves, the number of books in each shelf should be the H.C.F. of the numbers of English and Mathematics books. Thus, we have to find the H.C.F. of 154 and 84. By fundamental theorem of arithmetic, the numbers 154 and 84 can be prime factorised in a unique way, apart from the order in which the prime factors occur. 154 = 2 × 7 × 11 = 21 × 71 × 111 84 = 2 × 2 × 3 × 7 = 22 × 31 × 71   H.C.F. (154, 84) = Product of the smallest powers of all common prime factors in the numbers = 21 × 71 = 14 Hence, the required number of books in each shelf is 14. The correct answer is D. It is given that starting from the first bag, every third bag has a blue ball, every sixth bag has a green ball, and every seventh bag has a red ball. The position of the first bag to contain all the three balls is equal to the L.C.M. of 3, 6, and 7. The positions of other bags which contain all the three balls are multiples of the L.C.M. of 3, 6, and 7. By fundamental theorem of arithmetic, the numbers 3, 6, and 7 can be prime factorised in a unique way, apart from the order in which the prime factors occur. 3 = 31 6 = 21 × 31 7 = 71   L.C.M. (3, 6, 7) = Product of the greatest power of each prime factor involved in the numbers = 21 × 31 × 71 = 42 Thus, starting from the first bag, every 42nd bag will have balls of all three colours. Thus, total number of bags that contains the three 126 3 42 The correct answer is A.

balls 

189

19.

20.

From Mumbai, a bus leaves for Pune every 10 minutes and a bus leaves for Thane every 15 minutes. Thus, the time after which two buses will simultaneously depart from Mumbai for Pune and Thane after 9 a.m. = L.C.M. (10, 15) By fundamental theorem of arithmetic, the numbers 10 and 15 can be prime factorised in a unique way, apart from the order in which the prime factors occur. 10 = 2 × 5 = 21 × 51 15 = 3 × 5 = 31 × 51   L.C.M. (10, 15) = 21 × 31 × 51 = 30 Thus, the buses for Pune and Thane will again simultaneously leave Mumbai after 30 minutes. Therefore, after 9 a.m., two buses will simultaneously depart from Mumbai for Pune and Thane at 9:30 a.m. The correct answer is A. The least number of squares which can be cut from the piece of cloth have to be found. This implies that each square must have the largest possible size. Since no cloth should be wasted, the length of each side of a square should divide the length and breadth of the piece of cloth i.e., it should be a factor of the length and width of the piece of cloth. Since we require the biggest possible square, the length of each side of the square should be the largest possible factor of the length and breadth of the piece of cloth i.e., it should be their H.C.F. By fundamental theorem of arithmetic, the numbers 18 and 8 can be prime factorised in a unique way, apart from the order in which the prime factors occur.   18 = 2 × 3 × 3 = 21 × 32 8 = 2 × 2 × 2 = 23   H.C.F. (18, 8) = Product of the smallest powers of all common prime factors in the numbers = 2 Thus, each square piece is of dimensions 2 m × 2 m. Thus, number of squares along the length of the piece of cloth 

18m 9 2m www.betoppers.com

10th Class Mathematics

190

Number of squares along the breadth of the piece 8m 4 2m Thus, total number of square pieces that can be cut from the rectangular piece of cloth = 9 × 4 = 36 The correct answer is B. Let a be any positive odd integer and let b = 4. Then, by Euclid’s division lemma, whole numbers q and r exist such that a = 4q + r              0    r  x. According to the given information, x

______________

y = 3x

(1) ______________

y – x = 26 (2) On substituting the value of y from equation (1) into equation (2), we obtain 3x – x = 26 ______________

x = 13 (3) Substituting this in equation (1), we obtain y = 39 Hence, the numbers are 13 and 39. (ii) Let the larger angle be x and smaller angle be y. We know that the sum of the measures of angles of a supplementary pair is always 180º. According to the given information, _____________

x + y = 180°

(1) _____________

x – y = 18° From (1), we obtain

(2)

_____________

x = 180º − y (3) Substituting this in equation (2), we obtain 180° – y – y = 18° 162° = 2y ________________

81° = y (4) Putting this in equation (3), we obtain x = 180º − 81º = 99º Hence, the angles are 99º and 81º. (iii) Let the cost of a bat and a ball be x and y respectively. According to the given information, _____________

7x + 6y = 3800

(1) _____________

3x + 5y = 1750 (2) From (1), we obtain 3800  7x ________________ y (3) 6 Substituting this value in equation (2), we obtain  3800  7x  3x  5    1750 6  

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

225

9500 35x   1750 3 6 35x 9500 3x   1750  6 3 Let the cost of a bat and a ball be x and y respectively. 18x  35x 5250  9500  6 3 17x 4250   6 3 17x = 8500 3x 

______________

x = 500 (4) Substituting this in equation (3), we obtain 3800  7  500 y 6 300   50 6 Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50. (iv) Let the fixed charge be Rs x and per km charge be Rs y. According to the given information, _____________

x + 10y = 105

(1) _____________

x + 15y = 155 From (3), we obtain

(2)

_____________

x = 105 – 10y (3) Substituting this in equation (2), we obtain 105 – 10y + 15y = 155 5y = 50 ______________

y = 10 (4) Putting this in equation (3), we obtain x = 105 – 10 × 10 x=5 Hence, fixed charge = Rs 5 And per km charge = Rs 10 Charge for 25 km = x + 25y = 5 + 250 = Rs 255 x (v) Let the fraction be y According to the given information, x2 9  y  2 11 11x + 22 = 9y + 18 ___________

11x – 9y = –4 x3 5  y3 6 6x + 18 = 5y + 15

(1)

____________

6x – 5y = –3

From equation (1), we obtain 4  9y _______________ x (3) 11 Substituting this in equation (2), we obtain  4  9y  6   5y  3  11  –24 + 54y –55y = –33 –y = –9 ______________

y=9 (4) Substituting this in equation (3), we obtain 4  81 x 7 11 7 Hence, the fraction is 9 (vi) Let the age of Jacob be x and the age of his son be y. According to the given information, (x + 5) = 3 (y + 5) _______________

x – 3y = 10 (x – 5) = 7(y – 5)

(1)

_______________

x – 7Y = –30 From (1), we obtain

(2)

____________

x = 3y + 10 (3) Substituting this value in equation (2), we obtain 3y + 10 – 7y = –30 –4y = –40 ___________

y = 10 (4) Substituting this value in equation (3), we obtain x = 3 × 10 + 10 = 40 Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years. 14. (i) By elimination method x + y = 5 --------------- (1) 2x − 3y = 4 --------------- (2) Multiplying equation (1) by 2, we obtain 2x + 2y =10 --------------- (3) Subtracting equation (2) from equation (3), we obtain 5y = 6 6 y  --------------- (4) 5 Substituting the value in equation (1), we obtain 6 19 x 5  5 5 19 6  x , y 5 5

(2)

____________________________________________________________________________________________________ _____________________________________________

Pair of Linear Equations in Two variables Solutions

By substitution method From equation (1), we obtain x = 5 − y --------------- (5) Putting this value in equation (2), we obtain 2(5 − y) – 3y = 4 −5y = −6 6 y 5 Substituting the value in equation (5), we obtain 6 19 x 5  5 5 19 6  x , y 5 5 (ii) By elimination method 3x + 4y = 10 ---------- (1) 2x − 2y = 2 ---------- (2) Multiplying equation (2) by 2, we obtain 4x − 4y = 4 ---------- (3) Adding equation (1) and (3), we obtain 7x = 14 x=2 ---------------- (4) Substituting in equation (1), we obtain 6 + 4y = 10 4y = 4 y=1 Hence, x = 2, y = 1 By substitution method From equation (2), we obtain From equation (2), we obtain Putting this value in equation (1), we obtain 3(1 + y) + 4y = 10 7y = 7 y=1 Substituting the value in equation (5), we obtain x=1+1=2  x = 2, y = 1 (iii) By elimination method 3x − 5y − 4 = 0 ---------------- (1) 9x = 2y + 7 = 0 9x − 2y − 7 = 0 ---------------- (2) Multiplying equation (1) by 3, we obtain 9x − 15y − 12 = 0 ------------- (3) Subtracting equation (3) from equation (2), we obtain 13y = − 5 5 y ---------------- (4) 13 Substituting in equation (1), we obtain 25 3x  40 13

226

27 13 9 x 13 9 5  x , y 13 13 (iii) By substitution method From equation (1), we obtain 5y  4 x -------------- (5) 3 Putting this value in equation (2), we obtain  5y  4  9   2y  7  0  3  13y = – 5 5 y 13 Substituting the value in equation (5), we obtain  5  5   4 13 x   3 9 x 13 9 5  x , y 13 13 (iv) By elimination method x 2y   1 2 3 3x + 4y = – 6 -------------------- (1) y x  3 3 3x – y = 9 -------------------- (2) Subtracting equation (2) from equation (1), we obtain 5y = – 15 y = – 3 -------------------- (3) Substituting this value in equation (1), we obtain 3x – 12 = – 6 3x = 6 x=2 Hence, x = 2, y = −3 (iv) By substitution method From equation (2), we obtain y9 x -------------------- (5) 3 Substituting the value in equation (1)  y9 3   4y  6  3  5y = −15 3x 

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

227

y = −3 Substituting the value in equation (5), we obtain 3  9 x 2 3 ∴ x = 2, y = −3 x 15. Let the fraction be . y According to the given information, x 1 1 y 1  x – y = –2 -------------------- (1) x 1  y 1 2  2x – y = 1 -------------------- (2) Subtracting equation (1) from equation (2), we obtain x = 3 -------------------- (3) Substituting this value in equation (1), we obtain 3–y=–2 –y=–5 y= 5 3 Hence, the fraction is . 5 (ii) Let present age of Nuri = x and present age of Sonu = y According to the given information, (x – 5) = 3(y – 5) x – 3y = – 10 -------------------- (1) (x + 10) = 2(y + 10) x – 2y = 10 -------------------- (2) Subtracting equation (1) from equation (2), we obtain y = 20 -------------------- (3) Substituting it in equation (1), we obtain x – 60 = – 10 x = 50 Hence, age of Nuri = 50 years And, age of Sonu = 20 years (iii) Let the unit digit and tens digits of the number be x and y respectively. Then, number = 10y + x Number after reversing the digits = 10x + y According to the given information, x + y = 9 -------------------- (1) 9(10y + x) = 2(10x + y) 88y − 11x = 0 − x + 8y =0 --------------------(2) Adding equation (1) and (2), we obtain 9y = 9 y = 1 -------------------- (3)

(iv)

(v)

16.

(ii)

Substituting the value in equation (1), we obtain x=8 Hence, the number is 10y + x = 10 × 1 + 8 = 18 Let the number of Rs 50 notes and Rs 100 notes be x and y respectively. According to the given information, x + y = 25 --------------------(1) 50x + 100y = 2000 -------------------- (2) Multiplying equation (1) by 50, we obtain 50x + 50y = 1250 -------------------- (3) Subtracting equation (3) from equation (2), we obtain 50y = 750 y = 15 Substituting in equation (1), we have x = 10 Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100. Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively. According to the given information, x + 4y = 27 -------------------- (1) x + 2y = 21 -------------------- (2) Subtracting equation (2) from equation (1), we obtain 2y = 6 y=3 -------------------- (3) Substituting in equation (1), we obtain x + 12 = 27 x = 15 Hence, fixed charge = Rs 15 and Charge per day = Rs 3 (i) x – 3y – 3 = 0 3x – 9y – 2 = 0 a1 1 b1 3 1 c1 3 3  , = = , = = a 2 3 b2 9 3 c 2 2 2 a1 b1 c1 =  a2 b2 c 2 Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations. 2x +y = 5 3x + 2y = 8 a1 2 b1 1 c1 5  , = , = a 2 3 b2 2 c 2 8 a1 b1  a2 b 2

____________________________________________________________________________________________________ _____________________________________________

Pair of Linear Equations in Two variables Solutions

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication method, x y 1 =  b1c 2  b2c1 c1a 2  c 2 a1 a1b 2  a2b1 x y 1 = = 8   10  15 +16 4  3

x y = =1 2 1 x y = 1,  1 c 2 1  x = 2, y = 1 (iii) 3x – 5y = 20 6x – 10y = 40 a1 3 1 = = , a2 6 2 b1 5 1 = = , b2 10 2 c1 20 1 = = , c 2 40 2 a1 b1 c1   a2 b 2 c 2 Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations. (iv) x – 3y – 7 = 0 3x – 3y – 15 = 0 a1 1 = , a2 3 b1 3 = = 1, b 2 3 c1 7 7 a1 b1 = = ,  c 2 15 15 a2 b 2 Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations. By cross-multiplication, x y 1 = = 45   21 21   15  3   9  x y 1   24 6 6 x 1 y 1  and  24 6 6 6  x = 4 and y = –1

228

17. 2x + 3y – 7 = 0 (a – b) x + (a + b) y – (3a + b – 2) = 0 a1 2 b 3  , 1  , a 2 a  b b2 a  b c1 7 7   c2   3a  b  2   3a  b  2  For infinitely many solutions, a1 b1 c1   a 2 b2 c2 2 7  a  b 3a  b  2 6a + 2b – 4 = 7a – 7b _________

a – 9b = –4 (1) 2 3  a b a b 2a + 2b = 3a – 3b _________

a – 5b = 0 (2) Subtracting (1) from (2), we obtain 4b = 4 b=1 Substituting this in equation (2), we obtain a–5×1=0 a=5 Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions. (ii) 3x + y – 1 = 0 (2k – 1) x + (k – 1)y – 2k – 1 = 0 a1 3  , a 2 2k  1

b1 1  , b2 k  1 c1 1 1   c2 2k  1 2k  1 For no solution, a1 b1 c1   a 2 b 2 c2 3 1 1   2k  1 k  1 2k  1 3 1   2k  1 k  1  3k – 3 = 2k – 1 k=2 Hence, for k = 2, the given equation has no solution.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

229 ______________

18. 8x + 5y = 9

_______________

(i)



3x – y = 3 x 1  y 8 4



4x – y = 8 (2) Subtracting equation (1) from equation (2), we obtain

______________

3x + 2y = 4 (ii) From equation (ii), we obtain 4  2y ______________ x (iii) 3 Substituting this value in equation (i), we obtain  4  2y  8   5y  9  3  32 – 16y + 15y = 27 –y = –5 __________

y=5 (iv) Substituting this value in equation (ii), we obtain 3x + 10 = 4 x = –2 Hence, x = –2, y = 5 Again, by cross-multiplication method, we obtain 8x + 5y – 9 = 0 3x + 2y – 4 = 0 x y 1   20   18 27   32  16  15

x y 1   2 5 1 x y  1 and  1 2 5 x = –2 and y = 5 19. (i) Let x be the fixed charge of the food and y be the charge for food per day. According to the given information, _______________

x + 20y = 1000

(1) _______________

x + 26y = 1180 (2) Subtracting equation (1) from equation (2), we obtain 6y = 180 y = 30 Substituting this value in equation (1), we obtain x + 20 × 30 = 1000 x = 1000 – 600 x = 400 Hence, fixed charge = Rs 400 And charge per day = Rs 30 x (ii) Let the fraction be y According to the given information, x 1 1  y 3

(1)

_______________

_______________

x=5 (3) Putting this value in equation (1), we obtain 15 – y = 3 y = 12 5 Hence, the fraction is 12 (iii) Let the number of right answers and wrong answers be x and y respectively. According to the given information, _______________

3x – y = 40 4x – 2y = 50

(1) _______________



2x – y = 25 (2) Subtracting equation (2) from equation (1), we obtain _______________

x = 15 (3) Substituting this in equation (2), we obtain 30 – y = 25 y=5 Therefore, number of right answers = 15 And number of wrong answers = 5 Total number of questions = 20 st

nd

(iv) Let the speed of 1 car and 2 car be u km/h and v km/h. Respective speed of both cars while they are travelling in same direction = (u – v)km/h Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u +v) km/h According to the given information, 5(u – v) = 100 ____________



u – v = 20

(1) ____________

1(u + v) = 100 (2) Adding both the equations, we obtain 2u = 120 ____________

u = 60 km/h (3) Substituting this value in equation (2), we obtain v = 40 km/h Hence, speed of one car = 60 km/h and speed of other car = 40 km/h

____________________________________________________________________________________________________ _____________________________________________

Pair of Linear Equations in Two variables Solutions

230

(v) Let length and breadth of rectangle be x unit and y unit respectively. Area = xy According to the question, (x – 5) (y + 3) = xy – 9

y

 q in the given

_____________

2p + 3q = 2

______________

(1) _____________



3x – 5y – 6 = 0 (x + 3) (y + 2) = xy + 67



2x + 3y – 61 = 0 (2) By cross-multiplication method, we obtain x y 1   305   18  12   183 9   10 

(1)

4p – 9q = –1 (2) Multiplying equation (1) by 3, we obtain

______________

_____________

x y 1   323 171 19 x = 17, y = 19 Hence, the length and breadth of the rectangle are 17 units and 9 units respectively. 1 1  2 20.  i  2x 3y 1 1 13   3x 2y 6 1 1  q , then the equations Let  p and x y change as follows. p q  2 2 3 _______________



3p + 2q – 12 = 0 p q 13   3 2 6



2p + 3q – 13 = 0 (2) Using cross-multiplication method, we obtain p q 1   26   36  24   39  9  4

(1)

_______________

p q 1   10 15 5 p 1 q 1  and  10 5 15 5 p = 2 and q = 3 1 1  2 and  3 x y 1 1 x  and y  2 3 2 3  2  ii  x y 4 9   1 x y

1

1  p and x equations, we obtain

Putting

6p + 9q = 6 (3) Adding equation (2) and (3), we obtain 10p = 5 1 _______________ p (4) 2 Putting in equation (1), we obtain 1 2   3q  2 2 3q = 1 1 q 3 1 1 p  x 2 x 2 x=4 1 1 and q   y 3 y 3 y=9 Hence, x = 4, y = 9 4 3  3y  14 ,  4y  23  iii  x x 1 Substituting  p in the given equations, we x obtain 4p + 3y = 14 ____________



4p + 3y – 14 = 0 3p – 4y = 23

(1)



3p – 4y – 23 = 0 (2) By cross-multiplication, we obtain p y 1   69  56 42   92  16  9

____________

p y 1   125 50 25 p 1 y 1  and  125 25 50 25 p = 5 and y = –2 1 p 5 x 1 x y = –2 5

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

231

 iv 

p q 1   105   40  35  30 16  49

5 1  2 x 1 y  2 6 3  1 x 1 y  2

1 1  q in the given  p and x 1 y2 equation, we obtain Putting

_____________

5p + q = 2

(1) _____________

6p  3q = 1 (2) Multiplying equation (1) by 3, we obtain _____________

15p + 3q = 6 (3) Adding (2) an (3), we obtain 21p = 7 1 p 3 Putting this value in equation (1), we obtain 1 5  q  2 3 5 1 q 2  3 3 1 1 p  x 1 3 x–1=3 x=4 1 1 q  y2 3 y–2=3 y=5  x = 4, y = 5 7x  2y 5  v xy ____________ 7 2  5 (1) y x

8x  7y  15 xy ____________ 8 7   15 (2) y x 1 1 Putting  p and  q in x y equation, we obtain 2p + 7q = 5

the

given

__________

 2p + 7q  5 = 0 7p + 8q = 15

(3) __________

 7p + 8q  15 = 0 (4) By cross-multiplication method, we obtain

p q 1   65 65 65 p 1 q 1  and  65 65 65 65 p = 1 and q = 1 1 1 p  1 q  1 x y x = 1 and y=1 (vi) 6x + 3y = 6xy ______________ 6 3   6 (1) y x 2x + 4y = 5xy ______________ 2 4  5 (2) y x 1 1 Putting  p and  q in these equations, x y obtain 3p + 6q – 6 = 0 4p + 2q – 5 = 0 By cross-multiplication method, we obtain p q 1   30   12  24   15 6  24

we

p q 1   18 9 18 p 1 q 1  and  18 18 9 18 p = 1 and 1 q 2 1 1 1 p  1 q   x y 2 x=1 y=2 10 2  4  vii  xy xy 15 5   2 xy xy 1 1  p and  q in the given Putting xy xy equations, we obtain 10p + 2q = 4 ________________

 10p + 2q – 4 = 0 15p – 5q = -2

(1) ________________

 15p – 5q + 2 = 0 (2) Using cross-multiplication method, we obtain

____________________________________________________________________________________________________ _____________________________________________

Pair of Linear Equations in Two variables Solutions

232

p q 1   4  20 60   20  50  30

p q 1   16 80 80 p 1 q 1  and  16 80 80 80 1 p  and q  1 5 1 1 1 p  and q  1 xy 5 xy ________________

x+y=5

(3) ________________

and x–y=1 (4) Adding equation (3) and (4), we obtain 2x = 6 ______________

x=3 (5) Substituting in equation (3), we obtain y=2 Hence, x = 3, y = 2 1 1 3    viii  3x  y 3x  y 4 1 1 1   2  3x  y  2  3x  y  8

1 1  p and q 3x  y 3x  y in these equations, we obtain 3 _____________ pq  (1) 4 p q 1   2 2 8 1 _____________ pq  (2) 4 Adding (1) and (2), we obtain 3 1 2p   4 4 1 2p  2 1 p 4 Substituting in (2), we obtain 1 1 q  4 4 1 1 1 q   4 4 2 1 1 p  3x  y 4 Putting

q

1 1  3x  y 2 ____________

3x – y = 2 (4) Adding equations (3) and (4), we obtain 6x = 6 ____________

x=1 (5) Substituting in (3), we obtain 3(1) + y = 4 y=1 Hence, x = 1, y = 1 21. (i) Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively. Speed of Ritu while rowing Speed of Ritu while rowing Upstream = (x – y) km/h Downstream = (x + y) km/h According to question, 2(x + y) = 20 ______________



x + y = 10 2(x - y) = 4

(1)

________________



x–y=2 (2) Adding equation (1) and (2), we obtain 2x = 12  x = 6 Putting this in equation (1), we obtain y=4 Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h. (ii) Let the number of days taken by a woman and a man be x and y respectively. Therefore, work done by a woman in 1 1 day = x 1 Work done by a man in 1 day = y 2 5 According to the question, 4     1 x y 2 5 1   x y 4 3 6 3    1 x y 3 6 1   x y 3 1 1 Putting  p and  q in these equations, we x y 1 obtain 2p  5q  4

____________

3x + y = 4

(3)

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

233



8p + 20q = 1 1 3p  6q  3  9p + 18q = 1 By cross-multiplication, we obtain p q 1   20   18 9  8  144  180

p q 1   2 1 36 p 1 q 1  and  2 36 1 36 1 1 p  and q  18 36 1 1 1 1 p   and q   x 18 y 36 x = 18 y = 36 Hence, number of days taken by a woman= 18 Number of days taken by a man = 36 (iii) Let the speed of train and bus be u km/h and v km/h respectively. According to the given information, ______________ 60 240  distance   4 (1)  t   u v speed  

CONCEPTIVE WORKSHEET 1.

2.

100 200 25   10   ______________   (2)   4   hrs  u v 6   60   1 1 Putting  p and  q in these equations, u v we obtain ______________

60p + 240q = 4 25 100p + 200q  6

(3)

______________

600p + 1200q = 25 (4) Multiplying equation (3) by 10, we obtain

3.

______________

600p + 2400q = 40 (5) Subtracting equation (4) from (5), we obtain 1200 q = 15 15 1 ______________ q  (6) 1200 80 Substituting in equation (3), we obtain 60 p + 3 = 4  60p = 1 1 p 60 1 1 1 1 p   and q   u 60 v 80 u = 60 km/h and v = 80 km/h Hence, speed of train = 60 km/h Speed of bus = 80 km/h ____________________________________________________________________________________________________ _____________________________________________

234 4.

x = –2, y = 3

7 2

 

7. x = 3, y = 1, (1, 0),  , 0 

5.

x = 2, y = –3

8.

x = 1, y = 3,  0,

6.

 

10.

x = 5, y = 0; (5, 0), (0, 3), (0, –4)11.

x

13.

x = 2, y = –1

14.

16.

x = 0.4, y = 0.6

19.

x

22. 25. 28.

x

31.

x = 1, y = 2

10th Class Mathematics x = 3, y = 2

7  , (0, 1) 9. (–4, 2), (1, 3), (2, 5) 2

49 19 , y 29 29

12.

x = 3, y = 1; x = 0, y = 3....

x = 3, y = 4

15.

x = a, y = –b

17.

x = 1, y = –1

18.

x

20.

x = 3.2, y = 2.3

21.

x = a, y = b

x = –2, y = 8 x = 12, y = 15

23. 26.

x = 2, y = –1 x = 3, y = –1

24. 27.

x = –1, y = –2 x = a. y = b

1 , y = –2 5

29.

x = 4, y = 5

30.

x = 1, y = 1

32.

x = 3, y = 2

5b  2a a  10b , y 10ab 10ab

1 2 , y 2 3

SUMMATIVE WORKSHEET Q.no Ans Q.no Ans Q.no Ans 1.

1 B 11 C 21 D

2 C 12 D 22 A

3 D 13 D 23 A

4 D 14 A 24 B

KEY 5 A 15 D 25 C

The given equations are: –3x – 5y = 10 ––––––––– (1) –2x + 5y = 5 –––––––––– (2) Equation (1) can be written as y  Then, for x = 0, y  For, x = –1, y 

10  3x 5

10  3   1 7   1.4 5 5

 Equation (1) passes through the points (0, 2) and (–1, 1.4).

Then, for x = 0, y  For x = 2.5, y  www.betoppers.com

7 B 17 B 27 C

8 C 18 B 28 B

9 A 19 C 29 A

10 C 20 C 30 C

 Equation (2) passes through the points (0, 1)

10  0 2 5

Equation (2) can be written as y 

6 D 16 A 26 D

5  2x 5

5  2 0 1 5

5  2  2.5 10  2 5 5

and (2.5, 2). By plotting and joining the ordered pairs (0, 2) and (–1, 1.4) on a co-ordinate plane, a line for equation (1) is obtained. Similarly, by plotting and joining the ordered pairs (0, 1) and (2.5, 2) on a co-ordinate plane, a line for equation (2) is obtained. The two equations can be plotted on the same graph as:

Pair of Linear Equations in Two Variables Solutions

235 Which is of the form

x y   1 , where (a, 0) and a b

(0, b) are x and y intercepts  respectively.  Equation (1) forms x and y intercepts at (3, 0) and (0, 2) respectively. Consider equation (2): –2x + 3y = 12

2x  3y 1 6 2x 3y  1 6 6 x y  1 3 2 Which is again of the form

x y  1 a b

 The equation (2) forms x and y intercepts at (–

2.

It is seen that the given equations intersect at the point (–5, –1). One must verify if the point (–5, –1) satisfies equations (1) and (2). Verification: In equation (1), Left hand side = – 3x + 5y = –3 × (–5) + 5× (–1) = 15 – 5 = 10 = Right hand side In equation (2), Left hand side = – 2x + 5y = –2 × (–5) + 5× (–1) = 10 – 5 =5 = Right hand side Hence, the point of intersection obtained is correct. The correct answer is B. The given equations are: 4x + 6y = 12 ––––––––– (1) –2x + 3y = 6 ––––––––– (2) Consider equation (1): 4x + 6y = 12

4x  6y 1 12 4x 6y  1 12 12 x y  1 3 2

3, 0) and (0, 2) respectively. By plotting and joining the ordered pairs (3, 0) and (0, 2) on a co-ordinate plane, a line for equation (1) is obtained. Similarly, by plotting and joining the ordered pairs (–3, 0) and (0, 2) on a co-ordinate plane, a line for equation (2) is obtained. The two equations can be plotted on the same graph as:

It is seen that the given equations intersect at the point (0, 2). One must verify if the point (0, 2) satisfies equations (1) and (2). Verification: In equation (1), Left hand side = 4x + 6y = 4 × 0 + 6 × 2= 0 + 12=12 www.betoppers.com

10th Class Mathematics

236

3.

= Right hand side In equation (2), Left hand side = –2x + 3y = –2 × 0 + 3 × 2 =0+6 =6 = Right hand side Hence, the point of intersection is correct. The correct answer is C. The given equations are: 3x – 2y = –11 –––––––––– (1) 2x + y = 2 –––––––––– (2) Consider equation (1): 3x – 2y = –11

 3x  2y   1  11 

4.

3 2 x  y 1 11 11

x y  1  3.6   5.5

 2x  5y   1 16

x y which is of the form   1 , where (a, 0) and a b (0, b) are x and y intercepts  respectively.  Equation (1) forms x and y intercepts at (–3.6, 0) and (0, 5.5) respectively. Consider equation (2): 2x + y = 2

2 5 x  y 1 16 16

x y  1 8  3.2  which is of the form

2x  y 1 2 x y  1 1 2 which is again of the form

equations (1) and (2). Verification: In equation (1), Left hand side = 3x – 2y = 3 × (–1) – 2 × 4 = –3 – 8 = –11 = Right hand side In equation (2), Left hand side = 2x + y = 2 × (–1) + 4 = –2 + 4 =2 = Right hand side Hence, the point of intersection obtained is correct. The correct answer is D. The given equations are: 2x – 5y = 16 ––––––––––– (1) x + 2y = –1? ––––––––––– (2) Consider equation (1): 2x – 5y = 16

x y  1 a b

 Equation (2) forms x and y intercepts at (1, 0) and (0, 2) respectively. By plotting and joining the ordered pairs (–3.6, 0) and (0, 5.5) on a co-ordinate plane, a line for equation (1) is obtained. Similarly, by plotting and joining the ordered pairs (1, 0) and (0, 2) on a co-ordinate plane, a line for equation (2) is obtained. The two equations can be plotted on the same graph as: The given equations intersect at the point (–1, 4). One must verify if the point (–1, 4) satisfies www.betoppers.com

x y   1 , where (a, 0), a b

and (0, b) are the x and y-intercepts respectively.  Equation (1) forms x and y intercepts at (8, 0) and (0, “3.2) respectively. Consider equation (2): x + 2y = –1

x 2  y 1 1  1 x 2  1 1  0.5  which is again of the form

x y  1 a b

 Equation (2) forms x and y-intercepts at (–1, 0) and (0,–0.5) respectively. By plotting and joining the ordered pairs (8, 0) and (0, –3.2) on a co-ordinate plane, a line for equation (1) is obtained.

Pair of Linear Equations in Two Variables Solutions

237

Similarly, by plotting and joining the ordered pairs (–1, 0) and (0,–0.5) on a co-ordinate plane, a line for equation (2) is obtained. The two equations can be plotted on the same graph as:

5.

It is seen that the given equations intersect at the point (3, –2). One must verify if the point (3, –2) satisfies equations (1) and (2). Verification: In equation (1), Left hand side = 2x – 5y = 2 × 3 – 5 × (–2) = 6 + 10 = 16 = Right hand side In equation (2), Left hand side = x + 2y = 3 + 2 × (–2) =3–4 = –1 = Right hand side Hence, the point of intersection obtained is correct. The correct answer is D. x2 + x – 2 = x2 + 2x – x – 2 = x2 + 2x – x – 2 = x(x + 2) – 1(x + 2) = (x – 1) (x + 2) It is given that the polynomial x3 + ax2 + bx – 8 is exactly divisible by x2 + x – 2 = (x – 1) (x + 2)  (x – 1) and (x + 2) are factors of the polynomial f (x) = x3 + ax2 + bx – 8

6.

Therefore, by factor theorem, f (1) = 0 and f (– 2) = 0   f (1) = 0  1+a+b–8=0   a + b = 7 –––––––– (1) f (–2) = 0  –8 + 4a – 2b – 8 = 0  4a – 2b = 16  2a – b = 8 ––––– (2) Adding (1) and (2): 3a = 15   a = 5  b=7–a=7–5=2 It is also given that the polynomial g(x) = x3 + 4x2 – cx + b = x3 + 4x2 – cx + 2 is divisible by x + 1. i.e., (x + 1) is a factor of g(x). Therefore, by factor theorem: g (–1) = 0  (–1)3 + 4(–1)2 – c(–1) + 2 = 0  –1 + 4 + c + 2 = 0   c = –5 Thus, a + b + c = 5 + 2 – 5 = 2 The correct answer is A. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents a pair of parallel lines, if

a1 b1 c1   a 2 b 2 c2 The given system of linear equations is: ax – 4y – 10 = 0 and  2x 

b y5  0 3

Since this system represents a pair of parallel lines, it can be concluded that:

a 4 10   2  b 5    3 a 4  2  b    3 

a 12  2 b

 ab  24 Thus, the value of ab is 24. The correct answer is D. www.betoppers.com

10th Class Mathematics

238 7.

Speed of boat along flow (downstream) = x + y Speed of boat against flow (upstream) = x – y It is given that the boat goes 112 km upstream and 96 km downstream in7 hours.



8.

112 96   7 –––––––– (1) xy xy

i.e.,

It is also given that the boat goes 28 km upstream and 64 km downstream in 3 hours.



k a   k  2   3 b 5

(3)

Taking first and third term:

Then, equations (1) and (2) become: 112u + 96v = 7 ... (4) 28u + 64v = 3 ... (5) Multiplying equation (4) by 2 and equation (5) by 3 and then subtracting: 224u + 192v = 14 84u + 192v = 9 – – – ––––––––––––––– 140u =5



u

5 140



u

1 28

k  3      

9.



64v = 3 – 1



v



i.e.,

1  64v  3 28

v

  k  2 5

5k = 3 (k + 2) 5k = 3 (k + 2) 5k = 3k + 6 5k – 3k = 6 2k = 6 k=6 Thus, the value of k is 3. The correct answer is C. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents a pair of parallel lines, if the system is inconsistent.

Substituting the value of u in equation (5):

25 

a1 b1 c1   a1 b1 c1 ––––––––––– (1)

The given system of linear equations is: kx + ay – (k + 2) = 0 and 3x + by – 5 = 0 Since this system represents coincident lines, it can be concluded that:

28 64   3 –––––––– (2) xy xy

1 1  u and  v –––––––– Let xy xy

It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents coincident lines, if the system is dependent and consistent.

a1 b1 c1   a1 b1 c1

Consider the system of equations given in alternative A: x – 2y = 6 and 3x – 6y = 2

2 64

Here

1 32

a1 1 b1 2 1 c 6  ,   and 1  3 a1 3 b1 6 3 c1 2

a1 b1 c1   a1 b1 c1

Substituting the values of u and v in equation (3):



1 1 1 1   and x  y 28 x  y 32

Thus, the system of linear equations given in alternative A represents a pair of parallel lines. The correct answer is A. Let the ones digit and tens digit of the number be x and y respectively. Then, the number is 10y + x. Sum of the digits = x + y According to the first condition:

 x – y = 28 and x + y = 32 Thus, the pair of equations x + y = 32 and x – y = 28 represents the given situation. The correct answer is B.

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10.

Pair of Linear Equations in Two Variables Solutions

239

10y  x 7  xy 1 10y + x = 7x + 7y 3y – 6x = 0 ––––––––– (1) According to the second condition: y = x + 4 ––––––––– (2) Substituting the value of y from equation (2) in equation (1): 3 (x + 4) – 6x = 0  3x + 12 – 6x = 0  12 – 3x = 0   x = 4 Substituting the value of x in equation (2): y = 4 + 4   y = 8 Therefore, the number is 84 and hence its successor is 85. The correct answer is C. The given equations are: 35x – 10y = 515 and 11x + 21y = 355 These equations can be rewritten as: 7x – 2y = 103 ––––––––– (1) 11x + 21y = 355 ––––––––– (2) Multiplying equation (1) with 21 and equation (2) with 2 and then adding the obtained equations: 147x – 427y = 2163 22x + 42y = 710 ––––––––––––––– 169x = 2873

2 k 2  2k  1   –2  k3 + 2x2 – k k 1  k3 + 2k2 – k + 2  0  (k – 1)(k + 1)(k + 2)  0  k  1 or k  –1 or k  –2

 

11.



x

13.

0 has no solution, if

1 2 6   k 6 0



12.

0 has unique solution, if

a1 b1  a 2 b2

The given system of linear equations is: –2x + (k2 + 2k – 1)y = 6 and kx + y = 2 Since this system has unique solution, it can be concluded that:

a1 b1 c1   a 2 b 2 c2  

The given system of linear equations is: x – 2y = 6 kx = 6y These equations can be rewritten as: x – 2y – 6 = 0 kx – 6y = 0 Since this system has no solution, it can be concluded that:

2873 169

x = 17 Substituting the value of x in equation (1): 7(17) – 2y = 103  119 – 2y = 103  2y = 16   y = 8 Thus, the solution of the given pair of equations is x = 17 and y = 8 The correct answer is C. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 =

Thus, k can take any values except –1, –2, and 1. Hence, 2 can be the value of k so that the given system has unique solution. The correct answer is D. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 =

14.



1 2  k 6



1 1  k 3

 k=3 Thus, the required value of k is 3. The correct answer is D. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents intersecting lines, if

a1 b1  a 2 b2

Consider the system of equations given in alternative A: x + y = 15 and x – y = 8 Here



a1 1 b 1   1, 1   1 a2 1 b 2 1

a1 b1  a 2 b2

Thus, the system of equations given in alternative A represents  intersecting  lines. The correct answer is A. www.betoppers.com

10th Class Mathematics

240 15.

It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has no solution, if

17.

a1 b1 c1   a 2 b 2 c2  

a1 b1 c1   a 2 b 2 c2 .

The system of linear equations is: 3x – 5y – 2 = 0, – 3lx + {3 (l + 1) + m}y + 8 = 0 Since this system has infinitely many solutions, it can be concluded that:

The given system of linear equations is:

3 5 2   3l 3  l  1  m 8

38 = 0 Since this system is inconsistent, it can be concluded that:



15   2x   k 2  6k   y  32  0 and 4x – 3y – 2 

1 5 1   l 3  l  1  m 4

2   4

Taking first and third term:

1 1  l 4

2  4

l=4 Taking second and third term:





5 1  3  l  1  m 4 

5 1  3  4  1  m 4

[ l

15 + m = 20 m=5 Thus, l + m = 4 + 5 = 9 The correct answer is D. It is known that the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 =

 

0 has no solution, if

a1 b1 c1   a 2 b 2 c2

The given system of linear equations is:

4  a  b  2b  2b ab  4(a + b)(a – b) = –4b2  a2 – b2 = –b2  a2 = 0  a=0 4(a + b) x – 2b y – 1 = 0 2bx + (a – b) y + 8 = 0 Since this system has no solution, it can be concluded that: Thus, the value of a is 0. The correct answer is A. www.betoppers.com

15 2  32 3 38

k 2  6k 

k 2  6k 

15 2

3 3 15  k 2  6k  2 2

–3 = 2k2 – 12k + 15 2k2 –13k + 18 = 0 k2 – 6k + = 0 (k – 3)2 = 0 k–3=0 k=0 Thus, the value of k is 3. The correct answer is B. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 =

     

=4]

16.

The system of linear equations a1x + b1y + c1 = 0 and a 2 x + b 2 y + c 2 = 0 is inconsistent, if

18.

0 has a unique solution, if

a1 b1  a 2 b2

Consider the system of equations given in alternative B: 3x + 4y = 11 and 6x – 8y = 7 Here,



a1 1 b 1 1  and 1   a2 2 b 2 2 2

a1 b1  a 2 b2

Thus, the system of linear equations given in alternative B has a unique solution. The correct answer is B.

Pair of Linear Equations in Two Variables Solutions 19.

241

The given equations andcan be respectively rewritten as:

 2x = 10   x = 5   y = 4x – 4 = 4 (5) – 4 = 20 – 4 = 16 x 5 Therefore, required fraction =  y 16

1 1   7 –––––––––––– (1) x y 3 4   25 –––––––––––– (2) y x Let

1 1  u and  v y x

21.

Then, equations (1) and (2) become: u + v = 7 –––––––– (3) 4u + 3v = 25 –––––––– (4) Multiplying equation (3) by 3 and then subtracting from equation (4): 4u + 3v = 25 3u + 3v = 21 – – – –––––––––––– u = 4 Substituting the value of u in equation (3): 4+v=7  v=7–4  v=3

x

1 1 1 1  and y   u 4 v 3

Thus, xy 

uv



x . y

According to the given condition:

x 1 1 x 1  and  y 4 y6 2 x 1 1  y 4  

4(x – 1) = y 4x – 4 = y ––––––––––– (1)

4(x – 1) = y 2x = y – 6 ––––––––––– (2) Substituting the value of y from (1) in (2): 2x = 4x – 4 – 6  4x – 2x = 4 + 6

21 7

u–v=3 ––––––––––– (2) Adding equations (1) and (2): 2u = 10    u = 5 Thus, the speed of rowing in still water is 5 km/hr. The correct answer is D. It is known that the lines of the form x = k are parallel to the y-axis. Now, the x-axis and y-axis are perpendicular to each other. Therefore, the lines of the form x = k are perpendicular to x-axis. Thus, among the given equations, the line x = –4 is perpendicular to the x-axis. The correct answer is A. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 represents a pair of parallel lines, if

22.

23.

a1 b1 c1   a 2 b 2 c2   The given system of equations is: 3x – ky + 11 = 0 and 15x + 10y + 17 = 0 Since this system represents a pair of parallel lines, it can be concluded that:

3 k 11   15 10 17

x 1  y6 2  

––––––––––– (1)



1 1 1   4 3 12

Let the fraction be

35 5

u+v=7

uv 

The correct answer is C. 20.

Thus, sum of the numerator and denominator of the fraction = 5 + 16 = 21 The correct answer is C. Let u and v be  the  speed  of  rowing  (in  stream) and the speed of stream respectively. Therefore, according to the given information:

3 k   15 10

1 k  5 10

–k = 2 k = –2 Thus, the value of k is –2. The correct answer is A.

 

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10th Class Mathematics

242 24.

25.

Let l and b be the length and breadth of each congruent rectangle.

Perimeter of rectangle ABCD = (b + b) + (l + b) + l + (b + l) = 3l + 4b Perimeter of the rectangle is given as 40 cm.  3l + 4b = 40 … (1) It is known that the opposite sides of a rectangle are equal.  AD = BC   b + b = l  2b = l … (2) Substituting the value of l from equation (2) in (1): 3(2b) + 4b = 40 6b + 4b = 40 10b = 40 b = 4   l = 2b = 2 × 4 = 8 Thus, perimeter of each congruent rectangle = 2 (l + b) = 2 (4 + 8) cm = 24 cm The correct answer is B. The given equations are: x –  3y =  7  ––––––––––  (1) 3x – 3y =  15 ––––––––––  (2) x + ky =  3  ––––––––––  (3) Subtracting equation (1) from equation (2): 2x = 8   x = 4 Substituting x = 4 in equation (1): 4 – 3y = 7  3y = 4 – 7  3y = –3   y = –1  (4, –1) is the solution of the equations (1) and (2). It is given that equations (1), (2), and (3) have a common solution.  (4, –1) is also a solution of equation (3). Therefore, (4, –1) satisfies equation (3).  4 + k (–1) = 3  4–k=3   k = 4 – 3   k = 1 Thus, the value of k is 1. The correct answer is C.

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26.

27.

The equation of the first line is x – y = 5 –––––– –––– (1) The equation of the second line is x + y = 11 ––– ––––––– (2) The equation of the x-axis is y = 0 Therefore, the given lines meet x-axis at y = 0 Substituting y = 0 in equation (1): x – 0 = 5   x = 5 Thus, the line x – y = 5 meets the x-axis at (5, 0). Substituting y = 0 in equation (2): x + 0 = 11   x = 111 Thus, the line x + y = 11 meets x-axis at point (11, 0). The correct answer is D. The given pair of equations are 6x + 3y = 6xy and 2x + 4y = 5xy. These two equations can be respectively rewritten as:

6 3   6 ––––––––––– (1) y x 2 4   5 ––––––––––– (2) y x Let  

1 1  u and  v y x ––––––– (3)

Then, equations (1) and (2) become: 3u + 6v = 6 ––––––– (4)  4u + 2v = 5 ––––––– (5)  Multiplying equation (5) with 3: 12u + 6v = 15 ––––––– (6)  Subtracting equation (4) from equation (6): 12u + 6v = 15 3u + 6v = 6 – – – –––––––––––– 9u = 9 u=1 Substituting the value of u in equation (4): 3 (1) + 6v = 6  3 + 6v = 6  6v = 3

 v

1 2

Substituting the values of u and v in equation (3):

Pair of Linear Equations in Two Variables Solutions x = 1 and y = 2 Thus, x + y = 1 + 2 = 3 The correct answer is C. 28. The equations of the given lines are: 2x + y =  5  –––––––––  (1) 3x + 2y = 8 –––––––– (2) Equation (1) can be written as: y =  5 –  2x –––––––––  (3) Substituting the value of y from equation (3) in equation (2): 3x + 2(5 – 2x) = 8  3x + 10 – 4x = 8  –x = 8 – 10   x = 2 Substituting x = 2 in equation (3): y = 5 – 2(2) = 1 Therefore, the point of intersection of the given lines is (2, 1). Thus, the x-coordinate of the point of intersection is 2. The correct answer is B. 29. It is known that the sum of all the interior angles of a triangle is 180°.  15x° + 15(x + y)° + 25y° = 180°  30x + 40y = 180°  3x + 4y = 18 ––––––––– (1) It is given that 5x – 4y = –2 –––––– (2) Adding equations (1) and (2): 8x = 16 x = 2 Substituting the value of x in equation (1): 3 (2) + 4y = 18  4y = 18 – 6  4y = 12   y = 3 The angles of the triangle can be obtained by substituting the values of x and y. 15x° = 15 (2)° = 30° 15 (x + y)° = 15 (2 + 3)° = 75° 25y° = 25° × 3° = 75° Thus, the measure of smallest angle of the triangle is 30°. The correct answer is A. 30. The given system of linear equations is: x + ky = 7 and –3x + 8y = 10 It is known that if the system of equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is inconsistent, then

a1 b1 c1   a 2 b 2 c2  

243 it can be concluded that:

1 k  7     3 8  10 

1 k   3 8 

8 = –35



k

8 3 8 3

Thus, the required value of k is  . 31

The correct answer is C. The difference between the ages of Biju and Ani is 3 years. Either Biju is 3 years older than Ani or Ani is 3 years older than Biju. However, it is obvious that in both cases, Ani’s father’s age will be 30 years more than that of Cathy’s age. Let the age of Ani and Biju be x and y years respectively. Therefore, age of Ani’s father, Dharam = 2 × x = 2x years y Cathy years 2 By using the information given in the question, Case (I) When Ani is older than Biju by 3 years, x – y = 3 —————— (i)

And age of Biju’s sister 

y  30 2 4x – y = 60———————— (ii) Subtracting (i) from (ii), we obtain 3x = 60 –3 = 57 2x 

57  19 3 Therefore, age of Ani = 19 years And age of Biju = 19 “ 3 = 16 years Case (II) When Biju is older than Ani, y – x = 3 —————————(i) x

y  30 2 4x – y = 60—————— (ii) Adding (i) and (ii), we obtain 3x = 63 x = 21 Therefore, age of Ani = 21 years And age of Biju = 21 + 3 = 24 years 2x 

Since the given system of equations is inconsistent, www.betoppers.com

10th Class Mathematics

244 32.

33.

Let those friends were having Rs x and y with them. Using the information given in the question, we obtain x + 100 = 2(y – 100) x + 100 = 2y – 200 x – 2y = –300 —————————(i) And, 6(x – 10) = (y + 10) 6x –60 = y + 10 6x – y = 70 ——————————(ii) Multiplying equation (ii) by 2, we obtain 12x – 2y = 140——————————— (iii) Subtracting equation (i) from equation (iii), we obtain 11x = 140 + 300 11x = 440 x = 40 Using this in equation (i), we obtain 40 – 2y = –300 40 + 300 = 2y 2y = 340 y = 170 Therefore, those friends had Rs 40 and Rs 170 with them respectively. Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that, Speed 

34.

Dis tan ce travelled Time taken to travel that distnace

d t Or, d = xt—————————————— (i) Using the information given in the question, we obtain x

 x  10  

d  t  2

(x + 10) (t – 2) = d xt+10t – 2x – 20 = d By using equation (i), we obtain – 2x + 10t = 20————————— (ii)

 x  10  

d  t  3

(x – 10) (t + 2) = d xt – 10t + 3x – 30 = d By using equation (i), we obtain 3x – 10t = 30—————————— (iii) Adding equations (ii) and (iii), we obtain www.betoppers.com

35.

x = 50 Using equation (ii), we obtain (–2) × (50) + 10t = 20 –100 + 10t = 20 10t = 120 t = 12 hours From equation (i), we obtain Distance to travel = d = xt = 50 × 12 = 600 km Hence, the distance covered by the train is 600 km Let the number of rows be x and number of students in a row be y. Total students of the class = Number of rows × Number of students in a row = xy Using the information given in the question, Condition 1 Total number of students = (x – 1) (y + 3) xy = (x – 1) (y + 3) = xy – y + 3x – 3 3x – y – 3= 0 3x – y= 3 —————(i) Condition 2 Total number of students = (x + 2) (y – 3) xy = xy + 2y – 3x – 6 3x – 2y = –6 ——————————(ii) Subtracting equation (ii) from (i), (3x – y) – (3x – 2y) = 3 – (–6) – y + 2y =3+6 y= 9 By using equation (i), we obtain 3x – 9= 3 3x = 9 + 3= 12 x= 4 Number of rows = x = 4 Number of students in a row = y = 9 Number of total students in a class = xy = 4 × 9 = 36 Given that, 3 B = 2(A + B) 3 B = 2 A + 2 B B = 2 A We know that the sum of the measures of all angles of a triangle is 180°. Therefore, A + B + C = 180° A + B + 3 B = 180° A + 4 B = 180°—— (ii) Multiplying equation (i) by 4, we obtain

Pair of Linear Equations in Two Variables Solutions

245

8 A – 4 B = 0——— (iii) Adding equations (ii) and (iii), we obtain 9 A = 180° A = 20° From equation (ii), we obtain 20° + 4 B = 180° 4 B = 160° B = 40° C = 3 B = 3 × 40° = 120° Therefore, A, B,  C are 20°, 40°, and 120° respectively. 36.

It can be observed that the required triangle is ABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, – 3), C (0, – 5).

HOTS WORKSHEET Q.no Ans Q.no Ans Q.no Ans

1 D 11 B 21 A

2 B 12 C 22 C

3 A 13 D 23 A

4 A 14 C 24 B

5 B 15 C 25 D

6 A 16 C 26 C

7 C 17 B 27 A

8 C 18 C 28 C

9 A 19 A 29 C

10 A 20 B 30 D

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10th Class Mathematics

246 1.

Let the sum of the weights of the 10 passengers be x kg  and  the  weight  of  the  original  boatman be y kg. Sum of the weights of the 11 people = x + y It is given that the average weight of the 11 people was 61 kg.

xy  61 11   x + y = 671 ... (1)



x  52  86 11   x + 52 = 649   x = 597

50 x x  100 2

It is also given that Gaurav failed by 10 marks.

x   10  y 2   x + 20 = 2y ... (1) Samik scored 75% of the total marks.

75 3x x   Samik’s score = 75% of x  100 4 Also, Samik scored 12.5% more than the passing marks.

3x  y  12.5% of y 4



3x 125  y y 4 1000



3x y  y 4 8

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x

9y  4 8 3



x

3y 2

3y  20 2



2y 



y  20 2



y = 40

x

Substituting this value of x in equation (1): 597 + y = 671   y = 671 – 597   y = 74 Thus, the weight of the former boatman is 74 kg. The correct answer is D. Let the total marks and the passing marks in the exam be x and y respectively. It is given that Gaurav scored 50% of the total marks.





3y  20  2y 2



 Gaurav’s score = 50% of x 

3x 9y  4 8

Substituting this value of x in equation (1):

Weight of the new boatman = 52 kg Weight of the 11 people now = x + 52 It is given that the average weight of the 11 people would dip by 2 kg i.e., it would become: 61 – 2 = 59 kg

2.



3y 3  40   60 2 2

 Gaurav’s score = Samik’s score =

3.

x 60   30 2 2

3x 3  60   45 4 4

Thus, difference between Gaurav’s and Samik’s scores = 45 – 30 = 15 The correct answer is B. It is known that the sum and the product of the zeroes of quadratic polynomial ax 2  + bx + c are  

b a

and

c a

respectively. Therefore, for the quadratic polynomial x 2  + 6x + k, it can be observed that:

    6 ––––––– (1)   k ––––––– (2) It is given that:     8 ––––––– (3) Adding the equations (1) and (3): 2   = 2  =1 From equation (3): 1 –   = 8  = –7 Substituting the values of   and    in  equation (2):

Pair of Linear Equations in Two Variables Solutions

4.

5.

(1) (–7) = k   k = –7 Thus, the value of k is –7. The correct answer is A. Let the cost price of a shirt and a trouser be 100x and 100y respectively. The shopkeeper earns a gain of Rs 92 on selling a shirt at 6% loss and a trouser at 10% profit.  10% of 100y – 6% of 100x = 92  10y – 6x = 92  – 3x + 5y = 46 … (i) It is also given that he earns a gain of Rs 144 on selling a shirt and a trouser at the profit of 8% each.  8% of 100x + 8% of 100y = 144  8x + 8y = 144   x + y = 18 ... (ii) Multiplying (ii) with 3 and then adding with (i): –3x + 5y + 3x + 3y = 46 + 54  8y = 100   y = 12.5 Substituting y = 12.5 in (ii): x + 12.5 = 18   x = 18 – 12.5   x = 5.5 Thus, cost price of a shirt = Rs 100x = Rs (100 × 5.5) = Rs 550 The correct answer is A. The given equations of the lines are: 5x + 4y = 16 ... (1) 7x + 4y = 8 ... (2) The point of intersection of the two lines is the solution of the system of equations. Subtracting equation (1) from equation (2): 2x = –8   x = –4 Substituting x = –4 in equation (1): 5(–4) + 4y = 16  –20 + 4y = 16  4y = 16 + 20 = 36

 y

6.

247

7.

y

8.

9.

18 6 3

From (2): x = 6 + 2 × 6 = 6 + 12 = 18 Thus, the present age of Anju is 18 years. The correct answer is C. Let x be  the  larger number  and y be  the  smaller number. According to the given information: x + y = 306 –––––––––––– (1) x – y =  36  ––––––––––  (2) Adding equations (1) and (2): 2x = 306 + 36  2x = 342



36 9 4

Thus, the point of intersection of the given two lines is (–4, 9). The correct answer is B. The given system of equations is: x + 3y = 4 ––––––––– (1) 4x + 7y = 1 ––––––––– (2) Multiplying equation (1) with 4: 4x + 12y = 16 ––––––––– (3)

Subtracting equation (2) from (3): 5y = 15    y = 3 From equation (1): x + 3 (3) = 4    x + 9 = 4    x = 4 – 9 = –5 Thus, the solution of the given system of equations is x = –5 and y = 3 The correct answer is A. Let the present ages of Anju and Jyoti be x years and y years  respectively. 3 years ago: Anju’s age = x – 3 and Jyoti’s age = y – 3 According to the given information: x – 3 = 5(y – 3)    x – 5y + 12 = 0  ––––––––– (1) 6 years later: Anju’s age = x + 6 and Jyoti’s age = y + 6 According to the given information: x + 6 = 2(y + 6)   x – 2y – 6 = 0  ––––––––– (2) Subtracting equation (1) from equation (2): 3y – 18 = 0

x

342  171 2

Thus, the larger number is 171. The correct answer is C. It is known that the system of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 has infinitely many solutions, if

a1 b1 c1   a 2 b2 c2 .

The given equations are: 5x + ay + 7 = 0 by –12y + 21 = 0 www.betoppers.com

10th Class Mathematics

248 Here, a1 = 5, b1 = a, c1 = 7; a2 = b, b2 = –12,  c2 = 21



 2x = 900   x = 450

5 a 7   b 12 21



5 a 1   b 12 3



b = 5 × 3 = 15, a 

13.

12  4 3

Thus, the values of a and b are respectively –4 and 15. The correct answer is A. 10. 5x – 7y =  14 ––––––––––  (1) 5x + 7y = 56 –––––––––– (2) Adding (i) and (ii): 10x = 70    x = 7 Substituting x = 7 in (i): 5(7) – 7y = 14  35 – 7y = 14  35 –14 = 7y  7y = 21    y = 3   x + y = 7 + 3 = 10 The correct answer is A. 11. The given equations are: 3x –  4y =  –2  ––––––––––  (i) x + y =  –3  ––––––––––  (ii) Multiplying (ii) with 4 and then adding with (i): 7x = –14 x = –2   Substituting x = –2 in (ii): –2 + y = –3   y = –1  2  2 Thus, x y  = (–2)2 × (–1)2 = 4 The correct answer is B. 12. Let the cost price of a shirt be Rs x and that of a trouser be Rs y. According to the given condition: –10% of x + 20% of y = 115  –0.1x + 0.2y = 115  –x + 2y = 1150 –––––––– (i) 20% of x – 10% of y = 10  0.2x – 0.1y = 10  2x – y = 100 –––––––– (ii) Multiplying (i) with 2 and then adding with (ii): 3y = 2400   y = 800 Substituting the value of y in (ii): 2x – 800 = 100 www.betoppers.com

14

Thus, total cost of a shirt and a trouser = Rs (800 + 450) = Rs 1,250 The correct answer is C. The area of the rectangle = 2 000 m2   xy = 2 000 –––––––– (i) The total area of the garden = 6 100 m2   x2 + y2 + xy = 6 100   x2 + y2 = 4 100  (x + y)2 – 2xy = 4 100  (x + y)2 = 4 100 + 4 000   x + y = 90 –––––––– (ii) (x – y)2 + 2xy = 4 100   (x – y)2 = 100   x – y = 10 –––––––– (iii) Adding (ii) and (iii)  2x = 100   x = 50 m   y = 40 m Hence, the lengths of the sides of the rectangle are 40 and 50 meters The correct answer is D. The pair of linear equations a1x + b1y + c1 =  0 and a 2x + b2 y + c2 =  0  has  infinite  solutions  if

a1 b1 c1   a 2 b2 c2 . It is given that the pair of equations x – ay + 6 = 0 and bx + y + 6 = 0 has infinite solutions.



1 a 6   b 1 6



1  a b

ab = –1 The correct answer is C. The given equations are 3x – 4y + 6 = 0 and 4x – 3y +  1  =  0.  To  represent  these  equations graphically, we need to find at least two solutions for each equation. The tables for the given equations are



15.

X

-2 -6 2 and

y

3x  6 4

0 -3 3

Pair of Linear Equations in Two Variables Solutions

249 18.

X

4x  1 y 3

-1 -4 2

-1 -5 3

On plotting these points, we obtain the graph as

16.

It can be seen that these lines intersect at point (2, 3). Thus, the solution of the given pair of equations is (2, 3). The correct answer is C. A pair of coincident lines has infinite solutions. If a1 x + b1 y + c1 =  0  and a 2x + b2 – y + c2  =  0 represent a pair of coincident lines, then

19

a1 b1 c1   a 2 b 2 c2

a1 b1 c1   a 2 b2 c2 . 

For the pair of equations 7x + 2y – 17 = 0 and 21x + 6y – 34 = 0,

a1 7 1   a 2 21 3

3 a a 1   2 a 9 a  3 3 a  2 a 9



b1 2 1   b2 6 3

 a3 = –27  a = –3 17.

b1 17 1 a b c    1  1  1 b 2 34 2 a 2 b 2 c2

Thus, the required value of a is –3. The correct answer is C. A pair of intersecting lines has a unique solution. If a 1 x + b 1 y + c 1  =  0  and a 2 x + b 2 –y + c 2  =  0 represent a pair of intersecting lines, then



a1 b1  a 2 b2 .

3 p   4 4

Let us consider the given equations. 5x + 8y = 59 2x + 3y = 23 In order to eliminate x, the first equation is multiplied with 2 and the second equation is multiplies with5. 5x + 8y = 59 ] × 2 2x + 3y = 23 ] × 5 10x + 16y = 118 10x + 15y = 115 Subtracting these equations: 10x + 16y = 118 10x + 15y = 115 – – – –––––––––––––– y=3 On substituting the value of y in 2x + 3y = 23, we obtain 2x + 3 × (3) = 23 2x + 9 = 23 2x = 14 x = 7 Hence, we can say that statements (i) and (iv) are correct. The correct answer is C. If a 1 x + b 1 y + c 1  =  0  and a 2 x + b 2 ­y + c 2  =  0 represent a pair of parallel lines, then

20.

p  3

The correct answer is B.

Thus, the pair of equations 7x + 2y – 17 = 0 and 21x +  6y –  34  =  0  represents  a  pair  of  parallel lines. The correct answer is A. A pair of intersecting lines has a unique solution. If a 1 x + b 1 y + c 1  =  0  and a 2 x + b 2 –y + c 2  =  0 represent a pair of intersecting lines, then

a1 b1  a 2 b2 www.betoppers.com

10th Class Mathematics

250 For the pair of equations 2x + 3y – 14 = 0 and 7x + 4y – 40 = 0,

16 51  16 35  8 x  3 2   3    17 17 17  17 

a1 2  a2 7

Therefore, the solution of the given pair of equations is x 

b1 3  b2 4 

21.

a1 b1  a 2 b2

Thus, the pair of equations 2x + 3y – 14 = 0 and 7x + 4y – 40 = 0 has a unique solution. The correct answer is B. An inconsistent pair of lines is always parallel. In this case, the pair of equations does not have a solution. For a1x + b1y + c1 = 0 and a2x + b2­y + c2 = 0 to be parallel,

22.

23.

24.

8 y 17 Substituting y in (i): www.betoppers.com

The correct answer is A. Let the units place digit = x Let the ten’s place digit = y  The original number = 10y + x The number obtained by interchanging the digits = 10x + y 10x + y =  

8 (10y + x) 3

 30x + 3y = 80y + 8x  22x = 77y  2x = 7y

a1 b1 c1   a 2 b 2 c2 .

The correct answer is A. Let the costs of 1 banana and 1 orange be x and y respectively. Thus, 3 bananas and 5 oranges cost 3x + 5y.  3x + 5y = 21 –––––––––– (1) Similarly, 4 bananas and 2 oranges cost 4x + 2y.  4x + 2y = 14 –––––––––– (2)  2x + y = 7   y = 7 –  2x –––––––––– (3) Substituting this value of y in (1): 3x + 5(7 – 2x) = 21  3x + 35 – 10x = 21  –7x = 21 – 35  –7x = –14   x = 2 Substituting this value of x in (3): y = 7 – 2 × 2 = 7 – 4 = 3 Thus, one banana and one orange together cost x + y = Rs (2 + 3) = Rs 5 The correct answer is C. Consider the equation x + 2y = 3. x = 3 – 2y Substitute x in 5x – 7y = 7 … (i) 5(3 – 2y) – 7y = 7 15 – 10y – 7y = 7 –17y = –8

35 8 and y  . 17 17

25.

2 is not a multiple of 7 But 2x is a multiple of 7   x is a multiple of 7 But x  0 So the given equation has real roots, given by,

16abcd  34abcd 50abcd 25cd   9ab 18a 2 b2 18a  b 2

x

Hence, the roots are

2 2 2 2  B  D   16abcd   1156a b c d  2A 2  9a 2 b 2 

3x2 – 2x – 1 = 10x2 – 21x + 9  7x2 – 19x + 10 = 0  (7x – 5) (x – 2) = 0 7x – 5 = 0 or x – 2 = 0

5 or x = 2 7

 x=

4.

2  3  

1

2  3 

 The given equation can be expressed as



2+ 3



x 2  2x+1

 b  b2  4ac   7   289  2a 2 4

1

+



7  17 24   3 8 8 x

25cd cd , . 9ab ab

2+ 3



x 2  2x 1



= 2 2+ 3



---- (1)



Let us put y = 2  3



x 2  2x 1

(1) can be expressed as

2  3 y y

2



 2 2 3



, then the equation

10th Class Mathematics

254  1    x 2  2x 1  2 3 2 3 











1



x 2  2x 1



 y2 – 2 2  3 y + 2  3











x 2  2x 1





 x  3 x  2    x  2 





x



 x  5  

 x  3   x  2 

 x  5



 0

 x  3   x  2 



 0

x 62

7.

 x  5

10 in the original 3

2

1  1  x   2 2 x x 

2   1 1   6  x    2   25  x    12  0 x x   

 x=

5 4

For x =

5 , LHS = 4

RHS =

5 5 1   1  1 and 4 4

5 4    1  4  2 but LHS  RHS  4

5 is not a root of the equation. 4  The given equation has no solution.  x=

10 equation, we find that x =  does not satisfy the 3 equation.  x = 2, 6

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x  1  x  1  4x  1

 2 x 2  1 = 1 – 2x Squaring again, we get, 4(x2 – 1) = 1 – 4x + 4x2

 x  3 x  2 

10  x = 6 or x =  3 Since the equation involves radical therefore

2 Put x 

1 or x = 3. 3

+ 1) + (x – 1)– 2 x 2  1 = 4x – 1

Again squaring on both sides x2 + 12x + 36 = 4(x2 + x –6) 3x2 – 8x – 60 = 0 (x – 6)(3x + 10) = 0

6.

 x=

 x + 1  0, x – 1  0, 4x – 1  0 All these inequalities are satisfied when x  1. Squaring both the sides of the equation, we get, (x

x 3 x 2  x 5

substituting x = 2, 6 and 

1 8  x 3

1 1  x =  ,  , 2, 3 3 2

Squaring both sides,

 x  3   x  2   2

and

1 3 1  and x = 2  x=– x 2 2

 x

 x=2

0

1 1 3 , we have x   x x 2

1 8  x 3

 x

 x  5 x  2 



3 8 or y = 2 3

Since y = x 

x 2  7x  10

 x  2    x  3    x  2 

or

=0

 x(x – 2) = 0

x2  x  6  x  2 

 x  2

2

 y = 2 3

 2 3

5.

Either



2

 x2 – 2x + 1 = 1  x = 0 or x = 2



Let x 

 y=

  y  2  3   0 2 3





1 y x  6(y2 + 2) – 25y + 12 = 0  6y2 – 25y + 24 = 0  (2y – 3)(3y – 8) = 0

  2  2 3  

8.

Given equation,

Let

x 1 + 3x

x 1 = y then 3x

1  y+ y =2

3x =2 x 1

1 3x = y x 1

Quadratic Equations Solutions y2 – 2y + 1 = 0 (y – 1)2 = 0  y=1

x 1 i.e., =1 3x Squaring on both the sides x 1 1 3x  x + 1 = 3x  2x = 1 1 2

x=

9.

y 1 6 5x 1 6  x  or   5 y 5 5 5 5 [putting y for 5x] or y2 – 6y + 5 = 0, or (y – 1)(y – 5) = 0, Hence, y = 1 or 5. i.e., 5x = 1 or 5x = 5 i.e., 5x = 50 or 5x = 51  x = 0 or 1. Here, we have

2

x  8x  80 m 2x  8 x2 + 8x + 80 = m (2x + 8) x2 + (8 – 2m) x + 80 – 8m = 0 given that roots are real

10. Let

D  0  b 2  4ac  0 4m2 + 64 – 32m – 320 + 32 m  0 4m2 – 256  0 4 (m2 – 64)  0 m2 – 64  0 (m – 8) (m + 8)  0 [If (x–  ) (x–  )  0 then x does not lie between  and  ]  m does not lie between –8 and 8. 11. Given, x =

2  2  2  2  ....

 x  2  x  x2 = 2 + x  x2 – x – 2 = 0 1 9 1 3   2,  1 2 2  x = 2, –1 ( x > 0)  x=2



x

255 12.

x

2   log 2 x-1 3

= 2

2     log 2 x  1 = log x 2 3  a x  N  log a N  x   



2 1  log2 x  1  log x 2 3 2

2 1  3  log 2 x  1  2log x 2 2 1  3  y  1  2y , where log2x = y

 4y2 – 4y – 3 = 0  y=

3 2

Now log2 x =

or y = 

3 2

1 2

 x = 23 2  x  2 2

1 1 1  x= 22  x = 2 2 13. Let the length of the room be x meters. Its breadth will be (x – 8) m.  Its area = x (x – 8) m2 But by the hypothesis x (x – 8) = 180. i.e., x2 – 8x – 180 = 0  (x – 18) (x + 10) = 0  x = 18 m, (x = –10 is not permissible as the length of a room can’t be negative) So the length x = 18 m. and the breadth x – 8 = 18 – 8 = 10 m. 14. Let the normal rate of walking of the man be x km/ hour.

And log2 x = 

Time taken to walk 48 km 

48 hours x

48 hours. x2 This time is 4 hour less than the usual time. At 2 km an hour faster he will take 

48 48  4 x x2 48 (x + 2) = 48 x + 4x (x + 2)  x2 + 2x – 24 = 0 (x – 4) (x + 6) = 0  x = 4 or –6  Rate of walking is 4 km/hour as (x = –6 is not admissible) www.betoppers.com 

10th Class Mathematics

256 15. (x – 2) (x – 4) (x + 3) (x + 5) = 120  [(x – 2) (x + 3)] [(x – 4) (x + 5)] = 120  (x2 + x – 6) (x2 + x – 20) = 120 Let x2 + x = y Then, (y – 6) (y – 20) = 120  y2 – 26y + 120 = 120  y (y–26) = 0  y = 0 or y = 26 If y = 0, then x2 + x = 0  x(x +1) = 0  x = 0, –1 If y = 26, then x2 + x =26  x2 + x – 26 = 0  x

1  1  4  26  2



1  105 2

1  105 1  105 , 2 2 16. i) x2-15x+56=0 Here, a = 1, b = –15, c = 56

x = 0, –1,

Sum of roots      =  Product of roots    =

1  0  x 2  1  x   i x 2

1 1  2  x   1  0  x  2  3  0 x x 

 x 4  3x 2  1  0 3  9  4 3  5  0 2 2  There is no real root. x2 

Alternatively for any non-zero real x, x 

c 56  = 56 a 1

ii) 2x2+11x+5=0 Here a = 2, b = 11, c = 5

2

19. (i) If  are the roots of ax2 + bx + c = 0, then

b c and    a a

But if  

1     

c 1 a c = a which is the required condition. 

(ii) If  =2  , then     3 and   22

b 11  a 2

c 5  a 2 17. (i) We know that    = p and  = q Product of roots    

2

 2 2       2  = p2 – 2q

Take  

 22 

 

c a

c a

c 2a

(ii) We know that 3

Now, take    

3

3

        3    

 a

3

3

 b 3   a  b   3ab  a  b 



 3 

= p3 – 3q(p) = p3 – 3pq 3

1  1  18.  x     x    0 x  x  2

 1   1    x    x    1  0 x   x  

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1 is also x

1  real and so  x    1  0. x 

  

b  15   = 15 a 1

Sum of roots       

x

3

b a

b a

c b  2a a

9c b 2  2a a 2  2b2 = 9ac which is the required condition. 

Quadratic Equations Solutions 20.

1 1 1   xa xb c

257

can

be

rewritten

as

x a xb 1   x  a  x  b  c

    2  2  2       3

 [2x + (a + b)] c = (x + a) (x + b)  x2 + x (a + b) + ab – 2cx – (a + b)c = 0  x2 + x (a + b – 2c) + (ab – ac – bc) = 0

But since roots are equal in magnitude but opposite in sign, we have   0

(a  b  2c) 0 1  a+b –2c = 0  2c = a + b. Also   = ab – ac – bc = ab – c(a + b) 

But c =

ab 2

 = ab – 21. We have S 

P

 a  b 2 2

=



2 22  aa  b b



22

k  1 k  2 b   and k k 1 a

k 1 k  2 c k2 c   or  or k k 1 a k a

2 c ca 2a  1  .  k k a a ca Now eliminate k. Putting the value of k in 1st relation, we get, k  1 k  2 b   k k 1 a

    



4 8

  3  2  2   3 4 4 4  3 3  3

4

3 = 4 3 4  8 3 2 23. As, a, b are roots of x +px+q=0 and c,d are roots of x2 + rx + s = 0 we have, a+b= –p and ab = q ---------(1) c +d = – r and cd =s --------(2) Now (a – c) (a – d) (b – c) (b – d)

=

2 2 =  a  a  c  d   cd   b  b  c  d   cd  2 2 =  a  ar  s   b  br  s  = a2b2 + a2br + a2s + b2ar + r2ab + ars + b2s + sbr + s2 = (ab)2 + r(ab)(a + b) + s(a2 + b2) + r2(ab) + sr(a + b) + s2

2 = q2 + rq(–p) + s   a  b   2ab  + r2q +

sr(–p) + s2 = q2 – pqr + s(p2 – 2q) + r2q – prs + s2 = (q2 + s2 – 2qs) – qr (p – r) + sp(p – r) = (q – s)2 + (p – r) (sp – qr) 24. If ,  are the roots of ax2 + bx + c = 0, then

  

     (  ) 2  2  2

b  b  c b 2 2c    2       a a2 a a  a

ca 2c b   2a a or c  a 

b b 2  2ac   a a2

 

4 6  2 and   3 3

b c ,    a a

Given that      2  2

2a  c  a 2a  2c  2a  b   2a 2a  c  a a

(a + c)2 + 4ac = –2b(a + c) or (a + c)2 + 2b(a + c) = – 4ac. Add b2 to both sides.  (a + c + b)2 = b2 – 4ac. 22.

   2  2  2   3 =    



b 2  a  b 2  2ac a

 ab  b 2  2ac  2ac = b2 + ab

 2ac = b2 + ab

   1 1        2       3 =    

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10th Class Mathematics

258

Hence, the required equation is

25. Since ,  are the roots of the given equation, we have,

  

q r   and p p

or p3 x 2  pqrx  r 3  0

Let  and  be the roots of the required equation. Then (i)     2  2  2(  )  2q p

  (2) (2)  4 

4r p .

– (  )x  ()  0 . Hence, the required equation is

( 2q) 4r x 0 p p

or px 2  2qx  4r  0 (ii) In this case,      2   2  (  ) 2  2 

q 2 2r q 2  2pr   p2 p p2

   22  () 2 

r2 p2

Hence, the required equation is x2 

(q 2  2pr) r2 x 2 0 2 p p

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ac  b 2 a x   0 or bc c 2 2 bc x + (b + ac)x + ab = 0. 27. For equal roots, D = 0. i.e., b2 – 4ac = 0 But now x2 + p(4x + p – 1) + 2 = 0  x2 + 4px + p2 – p + 2 = 0  x2 + 4px + (p2 – p + 2) = 0 a = 1 , b = 4p , c = (p2 – p + 2) b2 – 4ac = 0  (4p)2 – 4 × 1 × (p2 – p + 2) = 0  16p2 – 4p2 + 4p – 8 =0  12p2 + 4p – 8 = 0  12p2 + 12p – 8p – 8 = 0  12p(p + 1) – 8(p + 1) = 0  (p + 1) (12p – 8) = 0 x2 

r3 p3

and

8 2  12 3

1 be the roots of x2 – ax + b = 0. 

Also  

p  q  x 2    x   0 or rx 2  qx  p  0 r  r  (iv) In this case,

     2  2  (  ) 



1   1 a         c  Equation is x2 – Sx + P = 0 or

Then     p and   q .

1 p   r Hence, the required equation is  





ac  b2 1   a b     Sum =   b c bc

28. Let  and  be the roots of x2 – px + q = 0 and 

(iii) In this case,     1  1      q    r

 

b c      ,   . a a

 p = – 1 or p =

or p2 x 2  (q 2  2pr)x  r 2  0

   2  2  ()3 

26.

Product =

The equation whose roots are ,  is given by x2

x2 

 qr  r3 x2   2  x  3  0 p  p 

qr p2

1   a and  b.   2

Now,

q  b

2

   1        2         2

2

     2 = .             bq  p  a      

Quadratic Equations Solutions

259

29. Given quadratic equation is x 2  px  q  0 .

1600 hr x New speed = (x + 400) km/hr and the new time =

Sum of roots        p

 Usual time taken =

Product of roots     q The required equation is  2  2 2 2 x 2    2   2  0 x       





 2  2 









2x 640000 800   3 x 3 2  4800x = 4800x – 2x + 1920000 – 800x





 1600 = 1600 –

2

 2  2 2  2 2    2  2

  

 2x2 – 1920000 + 800x = 0  x2 – 960000 + 400x = 0  x2 + 400x – 960000 = 0  x2 + 1200x – 800x – 960000 = 0  x(x + 1200) – 800(x + 1200) = 0  (x + 1200) (x – 800) = 0  x = 800 km/hr ( x = – 1200 is inadmissible) 31. Given equation is m2x2 + (2m – m2) x –3 = 0 -------(1) Since  and  are the roots of equation (1)

2 



2

 2  2  2

       2  

2

2

 p  

2



q2

p2 q

2   1 1  2    2  2       2  2          

1600 40  1600 2       hr x 60 3  x  Distance = speed × time  1600 2   1600 = (x + 400)  x  3   

1 1   2    2 2   2

2 1 1 2     2 2     2

30. Let the usual speed of plane be x km/hr.

2 2  3 3   

 2m  m   m  2     2

m2

and   

----(2)

3 ---- (3) m2

  4 2  2 4  Given,   or   3  3

2 2 22  2 2 2         3   

2

m

2        2a  2   p   2q          q    2

 3   2   2   4  2  3       2   4  

2



2  p 2  2q  q

2

2



2p  4q q2

 p2   2p 2  4q  x2   2  x   0 2 q   q 





 q 2 x 2  p 2 x  2p 2  4q  0

 3       10   0  m  2 2 1   3   10. 2   0 from  2  and  3   m   m 

 3  m2  4  4m   10   0 -----(4)  m 2  4m  6  0 Since m1 and m2 are the two values of m  m1  m 2  4 and m1m 2  6

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10th Class Mathematics

260 m12 m22 m13  m32   Now, m 2 m1 m1m 2 3

 m  m2   1

 3m1m 2  m1  m2 

absolute minimum at x = 

m1m 2

43  3  6  4 64  72 136 68     6 6 6 3

32.

34. i) 3x2 + 4x + 1 Here a = 3, b = 4 and c = 1, Since a = 3 > 0, the expression 3x2 + 4x + 1 has

Given, ,  are roots of x2 + px + q = 0

    p,   q Now the quadratic equation whose roots are

2 1 4ac  b2 4(3)(1)  (4) = = . is 4(3) 3 4a

ii) x2 – 2x + 10 = (x – 1)2 + 9  0  At x = 1 the given expression have minimum value and the minimum values is 9. 35. Let f(x) = (2x + 3) (x – 2) (x – 1)

  and is  

Then, f  x   0  x 

    x2    x  .  0    

When x 

  2  2   x2    x 1 0     x2 

x

2

   

2

 2



 p  

2

q

 2q

x 1 0

x 1  0

 qx 2   p 2  2q  x  q  0 2

33. i) 4x – x – 10. Here a = –1, b = 4 and c = –10, Since a = –1 < 0, the expression –x2 + 4x – 10 has absolute maximum at x = 2 and the maximum value is 2 4ac  b 2 4( 1)(10)  (4)   6. 4(1) 4a

ii)

x  xR . x - 5x + 9 2

x  yx2 – 5xy + 9y = x x  5x + 9  yx2 –x(5y + 1) + 9y = 0 Given, x is real  (5y +1)2 – 4y(9y)  0  25y2 + 10y + 1 – 36y2  0  –11y2 + 10y + 1  0  (11y + 1) (y – 1)  0 Let y =

2

1    y 1 11 Maximum value of y is 1.  www.betoppers.com

2 and the minimum value 3

3  or  x  2  or  x  1 2

3 , then all the three factors are 2 negative and therefore, f(x) is negative. Then f(x) changes sign alternately. And, f(x) = 0 at

 3  x    , x  1 and x  2 .  2  Using the sign scheme, we have x:

3 2

1

f(x): – 0

f(x) > 0 when

+

0

2 –

0

3 < x < 1 and when x > 2 2

 3   ,1   2,    2  36. Given, x2 + 6x – 27 > 0   x  3 x  9   0

 x  9, x  3 -----(1) and  x 2  3x  4  0  x 2  3x  4  0  x   1, 4 

-----(2)

From (1) and (2) x   3, 4   3  x  4

37. Given expression is

3.7.log7

x2

2log

3

+

32

x

 2log2 3  3x 2  x  3

 Minimum value 

4  3 3  1 35  4  3 12

Quadratic Equations Solutions

261

8x 2 +16x  51 38. We have, 2x  3 x + 4 > 3   

CONCEPTIVE WORKSHEET i) The given equation is abx2 – (a + b) x + 1= 0

1.

8x 2 +16x  51   2x  3 x + 4   3  0

 (ax – 1) (bx – 1) = 0  x 

1 1 , . a b ii) The given equation is 3x2 – 11x + 10 = 0  3x2 – 6x – 5x + 10 = 0  3x (x – 2) – 5(x – 2) = 0 Hence, the roots are

2



8x +16x  51  3  2x  3  x  4  0  2x  3 x + 4 



2x 2  x  15 0  2x  3 x + 4 



 2x  5  x  3  2x  3 x + 4 

 (3x – 5) (x – 2) = 0  x 

0

Hence, the roots are 2.

Critical points are, x = – 4, – 3, 3/2, 5/2

1 1 , a b

5 or x = 2 3

5 and 2. 3

i) 3a 2 x 2  8abx  4b 2  0 Let the standard quadratic equation is given by Ax 2  Bx  C  0 , then Clearly,,

 2x  5  x  3   2x  3 x  4   0

A  3a 2 , B  8ab,C  4b 2 2



x   ,  4   3, 3



+

– –4

39.

+ –3

 

 5 , 2 2

–

2 2 2 Now, D  B  4AC  8ab   4  3a  4b 



 64a 2 b 2  48a 2 b 2  16a 2 b2

+

3/2

x

5/2

8ab  4ab 4ab 2b  2  2 6a 6a 3a

x 2  2x + 5 1 x 2  2x + 5 1 >  > 0  3x 2  2x  5 2 3x 2  2x  5 2



 x  5  x  3  0 x  2x  15   0   x  1 3x  5 3x 2  2x  5

x

2

2



 x  5 x  3 x  1 3x  5   x  1 3x  5 

– –5

+ –1

0



 –5 < x < –1 or

– 5/3

 B  D 8ab  4ab 12ab 2b    2 2 2A 6a a 2  3a 

The roots of the given equation are 

2

 (x + 5) (x – 3) (x + 1) (3x – 5) < 0 +

 B  D 8ab  16a  b2  2A 2  3a 2 

2b and 3a

2b . a

2 2 2 ii) 9x  9  a  b  x   2a  5ab  2b   0

+

2 2 2 We have, 9x  9  a  b  x   2a  5ab  2b   0

3

Here, A = 9, B = –9(a + b), C = 2a2 + 5ab + 2b2  B  B2  4AC  x    We know that 2A  

5 0 x2 + 9x – 3x – 27 > 0 (x + 9) (x – 3) > 0  x > 3 or x < – 9 ...(i) Again, –x2 + 3x + 4 > 0 x2 – 3x – 4 < 0 x2 – 4x + x – 4 < 0 (x + 1) (x – 4) < 0  x < 4 – 1 < x < 4 ... (ii) from (i) and (ii) x  (3, 4)  17. Since  ,  are the roots of ax2 + bx + c = 0 then     

b a

... (i)

c a Now sum of new roots and  

... (ii)

  2      2     4     

 b  4a  b  4   a  a product of new roots = (2 +  ) (2 +  ) = 4 + 2      

 b c = 4 +2      a a 4a  2b  c a The required equation is x2 – (sum of roots ) + product of roots = 0 

4a  2b  c  4a  b  2 0  x   x a  a   ax2 – (4a – b)x + 4a – 2b + c = 0  ax2 + (b –4a)x + 4a – 2b + c = 0 18. Let  ,  be the roots of x2 + bx + c = 0 then     b,   c Again let  ,  be the roots of x2 + qx + r = 0 then      q,   r

  

   

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2

 4  b 2  4c

and     Given that

   

2

 4  q 2  4r

    

            

b 2

b  4c



q 2

q  4r

 b2 (q2 – 4r) = q2 (b2 – 4c)  b2q2 – 4b2r = q2b2 – 4q2c  b2 r = q2 c 19. The given equation is x2 – (a –2) x – (a + 1) = 0 Sum of the roots    = (a – 2), product of the roots  = – (a + 1) Now, let 2

y   2   2       2 = (a – 2)2 + 2 (a + 1) = a2 – 4a + 4 + 2a + 2 = a2 – 2a + 6 On differentiating w.r.t. a, we get

dy = 2a – 2 da for maxima and minima, put dy 0 da

 a=1

d2y  2  ve da 2 Thus y is minimum at a = 1.

Then the least value of  2   2 at a = 1

y 2   2 = (1)2 – 2(1) + 6 = 1 – 2 + 6 = 5 20. Consider the quadratic equation is ax2 + bx + c = 0 Given that c = – 6 and a = 1 b  b = – 5a a  b = – 5 (1) = – 5 Thus a = 1, b = – 5, c = –6  The equation is x2 – 5x – 6 = 0 Now, x2 – 5x – 6 = 0 2  x – 6x + x – 6 = 0  x (x – 6) +1 (x – 6) = 0  (x + 1) (x – 6) = 0  x = – 1, 6 Also 2 + 3 =

5. PROGRESSIONS SOLUTIONS annum is the rate of interest, compound annually, then the amount An at the end of n years is

FORMATIVE WORKSHEET 1.

2.

The arithmetic progression with first term a and common difference d is given by a, a + d, a + 2d, a + 3d, ……… i.e., each term is obtained by adding ‘d’ to the preceding term. Here, First term (a) = 4 and common difference (d) = –3 . So, the arithmetic progression is a=4 a+d=4+(–3)=1 a + 2d = 4 + 2( – 3 ) = 4 + ( – 6 ) = – 2 a + 3d = 4 + 3( – 3 ) = 4 + ( – 9 ) = – 5  The required arithmetic progression is is 4, 1, – 2, – 5, – 8, …. i)

Given, A.P. is

1 5 9 13 , , , ,... 3 3 3 3

We have, first term (a) =

1 and common 3

n

r   given by A n  P  1    100  Here, P = Rs 100 and r = 4. n

4.

5 1 4 difference (d) = t2 – t1 = - = 3 3 3 So, the given sequence is an A.P. with first term

3.

1 4 and common difference . 3 3 ii) Given, A.P. is 0.6, 1.7, 2.8, 3.9, ….. We have, first term (a) = 0.6 and common difference (d) = t2 – t1 = 1.7 – 0.6= 1.1 So, the given sequence is an A.P. with first term 0.6 and common difference 1.1. i) Given, total strength of students in the auditorium = 1000 Number of students left in the auditorium when first batch of 25 students leaves the auditorium = 1000 – 25 = 975 Number of students left in the auditorium when second batch of 25 students leaves the auditorium = 950 – 25 = 925 and so on. Thus, the number of students left in the auditorium at different stages are 1000, 975, 950, 925,…… Clearly, it is an A.P. with first term 1000 and common difference – 25. ii) We know that if P is the principal and r % per

n

4  n   26   A n  100  1    100     100  1.04   100   25  Thus, the amount of money in the account at the end of different years is given by At the end of first year Rs. 100 × 1.04 = Rs. 104 At the end of second year Rs. 100 × (1.04)2 = Rs. 108.16 At the end of third year Rs. 100 × (1.04)3 = Rs.112.49 and so on Money at the end of the different years are Rs 104, Rs 108.16, Rs 112.49, … Clearly, it does not form an A. P.

Given,

2 5k , k, are in A.P.. 3 8

t2 – t1  k  2 3 t3 – t2  5k  k 8 Since, the given sequence is in A.P  t2 – t1 = t3 – t2

 k

2 5k  k 3 8

 kk 

5k 2  8 3

11k 2  8 3

2 8 16  k   3 11 33 16 33 Given, First term (a) = –7 and Common difference (d) = 5 We know that the general term ( tn ) = a + (n – 1)d  t18 = a + (18 – 1)d  k

5.

10th Class Mathematics

284

6.

7.

8.

 t18 = a + 17d = – 7 + 17 × 5 = 78 and, tn = a + (n – 1) × 5 = – 7 + (n – 1) × 5 = – 7 + 5n – 5 tn = 5n – 12 i) Find the 20th term from the end of the A.P. –25 , –20, –15, ….. 100. ii) Find the 6th term from the end of the A.P. 17, 14, 11, …..– 40. i) In the given sequence –25 , –20, –15, ….. 100. First term (a) = –25 Common difference (d) = t2 – t1 = 9 – 4 = 5 Last term (l ) = 100. nth term from the end = l – (n – 1)d 10th term from the end = 100 – (20 – 1) × 5 = 100 – 19 × 5 = 100 – 95 = 5. ii) In the given sequence . 17, 14, 11, …..– 40. First term (a) = 17 Common difference (d) = t2 – t1 = 14 – 17 = –3 Last term (l ) = – 40. nth term from the end = l – (n – 1)d 6th term from the end = l– (n – 1)d = – 40 – (6 – 1) × (– 3) = (– 40 + 15) = – 25 Given sequence is an A.P. –1, 3, 7, 11, …. 95. First term (a) = – 1 Common difference (d) = t2 – t1 = 3 – (– 1) = 4 Let 95 be the nth term of the given A.P. (i.e., tn = 95)  tn = 95 tn = a + (n – 1)d  a + (n – 1)d = 95  –1 + (n – 1) × 4 = 95  –1 + 4n – 4 = 95  4n – 5 = 95  4n = 100  n = 25  Thus, 95 is the 25th term of the given sequence. Given sequence 3, 6, 9, 12, ….. 1111 is an A.P. We have first term (a) = 3 and common difference (d) = 3. Let there be n terms in the given sequence. Then, nth term = 111  a + (n – 1)d = 111  3 + (n – 1) × 3 = 111  3 + 3n – 3 = 111  3n = 111  n = 37 Thus, the given sequence contains 37 terms.

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9.

10.

Since, we know that nth term of an A.P. = tn = a + (n – 1)d i) The first A.P. is given by 23, 25, 27, 29, …… Here, first term (a) = 23, common difference (d) = t2 – t1 = 25 – 23 = 2 tn = 23 + (n – 1) 2 = 2n + 21  nth term of first A.P. (tn)= (2n + 21) ——— ————— (1) ii) The second A.P. is given by –17, –10, –3, 4, …. Here, first term (a) = – 17 , common difference (d) = – 10 – ( – 17 ) = 7  tn = – 17 + (n – 1) 7 = – 17 + 7n – 7 = 7n – 24  nth term of second A.P. (tn)= (7n – 24) —— —————— (2)  nth term of (1) = nth term of (2)  2n + 21 = 7n – 24  7n – 2n = 21 + 24  5n = 45  n=9  9th term of the above two A.P.s are equal. Given, A.P. is 3, 15, 27, 39,……….. Here, first term (a) = 3,Common difference (d) = t2 – t1 = 15 – 3 = 12 and n = 54 We know that tn = a + (n – 1) d t54 = 3 + (54 – 1) (12) = 3 + 53 × 12 = 3 + 636 = 639 Let the required term be tn Let the required term = t54 + 132 tn= 639 + 132 = 771 tn = 771 Again tn = a + ( n – 1 )d 771 = 3 + (n – 1) (12)  771 – 3 = 12 (n – 1)

771  3 768   64 12 12  n = 64 + 1 = 65 Hence, 65th term is greater by 132 than the 54th term of the given A.P. 

11.

 n  1 

1 1 3 The given sequence 20,19 , 18 , 17 ,... is an 4 2 4 A.P. in which first term a = 20 and 3 . Let the nth term of 4 the given A.P. be the first negative term. Then, an < 0  a + (n – 1) d < 0 common difference d = 

Progressions Solutions

 20 + (n – 1) × (  

 

13.

14.

3 4

We have to find out the sum of n terms of the A.P. 105, 112, 119, 126, ….. 994. Here a = 105, d = 112 – 105 = 7, tn = 994, n = ? and Sn = ? We know, tn = a + (n – 1)d  994 = 105 + (n – 1) × 7  7 (n – 1) = 889  n – 1 = 127  n = 128 Sum of n terms of an

) 83

2 Þ n ³ 28 3 [ n is a natural number] Thus, 28th term of the given sequence is the first negative term. Let 1st term of an A.P. be ‘a’ and its common difference ‘d’. Given, t10 = 31 and t20 = 71 We know that tn = a + (n – 1)d  t10 = a + (10 – 1)d = a + 9d.  31 = a + 9d. ———————— (1) Again, t20 = a + (20 – 1)d = a + 19d. ———— (2)  71 = a + 19 d. Subtracting (1) from (2), we get 10 d = 40  d = 4. Substituting the value of d in (1), we have 31 = a + 9 × 4  a = 31 – 36  a = – 5 Now, t30 = a + (30 – 1)d = a + 29d = – 5 + 29 × 4 = – 5 + 116 = 111. Hence, 30th term is 111. Let ‘a’ be the 1st term and ‘d’ the common difference of the A.P. Given, t6 = 5t1 and t11 – 2t5 = 3  a + 5d = 5a ———————— (1)  4a = 5d  a + 10d – 2( a + 4d ) = 3  a + 10d – 2a – 8d = 3  a + 3 = 2d  a = 2d – 3 ———————— (2) Substituting the value of a in (1), we get 4(2d – 3) = 5d  8d – 12 = 5d  3d = 12  d=4 On putting d = 4 in (2), we get a = 2 × 4 – 3 = 5.  Now, t8 = a + 7d = 5 + 7 × 4 = 33. Hence, 8th term is 33. The first number between 100 and 1000, which is a multiple of 7, is 105 and The second number between 100 and 1000, which is a multiple of 7, is 112 The largest number between 100 and 1000, which is a multiple of 7, is 994 The set of all these numbers will form an A.P.



12.

285

n > 27

A.P. Sn 

n  2a   n  1 d  2

128 [ 2 × 105 + (128 – 1) (7) ] 2 = 64[ 210 + 127 × 7] = 64 (210 + 889) = 64 (1099) = 70336 Let a be the first term and d be the common difference of the given A.P. Then, (m times mth term of an A.P.) = (n times nth term of an A.P.)  m tm = n t n  m { a + (m – 1) d } = n { a + (n – 1) d }  m { a + (m – 1) d } – n { a + (n – 1) d } = 0  a (m – n) + {m (m – 1) – n (n – 1)} d = 0  a (m – n) + {(m2 – n2) – (m – n)} d = 0  a (m – n) + (m – n) (m + n – 1)} d = 0  (m – n) { a + (m + n – 1) d } = 0  a + (m + n – 1) d = 0  tm + n = 0 Hence, (m + n)th term of the given A.P. is zero. Given, (m + 1)th term of an A.P. is twice the (n + 1)th term tm + 1 = 2 t n + 1 . We have to prove that, (3m + 1)th term is twice the (m + n + 1)th term Let ‘a’ be the 1 st term and ‘d’ the common difference of the given A.P. We know that tm + 1 = a + {(m + 1) – 1} d = a + md and tn + 1 = a + {(n + 1) – 1} d = a + nd tm + 1 = 2 t n + 1 a + md = 2 (a + nd)  a = md – 2nd = ( m – 2n) d  —————— (1) Now, t3m+1 = a + {(3m + 1) – 1}d = a + 3md —————— (2) tm+n+1 = a + {(m + n + 1) – 1}d = a + ( m + n) d —————— (3)

 S128 

15.

16.

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10th Class Mathematics

286

17.

Substitute, the value of a from (1) in (2), we get t3m + 1 = [(m – 2n) d ] + 3 md = 4md – 2nd = 2d (2m – n).—————— (4) substituting the value of a from (1) in (3), we get tm + n + 1 = [(m – 2n) d ] + (m + n) d = 2md – nd = d (2m – n) —————— (5) From (4) and (5), we get t3m + 1 = 2d (2m – n) = 2 × tm + n + 1  Hence, (3m + 1)th term is twice the (m + n + 1)th term. Given, pth, qth and rth terms of an A.P. are a, b, c respectively tp = a, tq = b and tr = c. Let the 1st term of the given A.P. be ‘x’ and ‘d’ be the common difference. Then, tp = x + (p – 1)d; tq = x + (q – 1)d and tr = x + (r – 1)d. a = x + (p – 1) d — (1)  b = x + (q – 1) d — (2) c = x + (r – 1) d — (3) Subtracting equation (2) from equation (1), we have a – b = (p – 1) d – (q – 1) d = (p – 1 – q + 1) d = (p – q) d.

ab — (4) pq d Again, subtracting (3) from (2), we have b – c = (q – 1)d – (r – 1)d = (q – 1 – r +1)d  (b – c) = (q – r)d

19.



20.

bc —— (5) qr d Again, subtracting (1) from (3), we get (c – a) = (r – 1)d – (p – 1)d = (r – 1 – p +1)d = (r – p) d. 

nth term = 5. tn = a + (n – 1)d   5 = 23 + (n – 1) × (– 2)  5 = 23 – 2n + 2  20 = 2n  n = 10 Hence, there are 10 rows of rose plants. Let the numbers (a – 3d), (a – d), (a + d), (a + 3d) be in A.P. Since, given sum of these four numbers = 20.  (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20.  4a = 20  a = 5. Thus, the numbers are (5 – 3d), (5 – d), (5 + d), (5 + 3d) Again, as sum of squares of the numbers = 180  (5 – 3d)2 + (5 – d)2 + (5 + d)2 + (5 + 3d)2 = 180  (25 – 30d + 9d2) + (25 – 10d + d2) + (25 + 10d + d2) + (25 + 30d + 9d2) = 180  20d2 = 180 – 100 = 80  d2 = 4  d =  2 When d = 2 and when d = – 2 numbers are numbers are 5 – 3d = 5 – 3 × 2 = –1 5 – 3d = 5 – 3 (– 2) = 11 5–d=5–2=3 5 – d = 5 – (– 2) = 7 5+d=5+2=7 5 + d = 5 + (– 2) = 3 5 + 3d = 5 + 3 × 2 = 11 5 + 3d = 5 + 3 (– 2) = – 1 Hence, the required numbers are –1, 3, 7, 11 or 11, 7, 3, –1 Let the four parts be (a – 3d), (a – d), (a + d) and (a + 3d). Then, sum = 32. (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32   4a = 32  a = 8 It is given that

 a  3d  a  3d   7  a  d  a  d  15

ca —— (6) r p d Now, (q – r) a + (r – p) b + (p – q)c 



bc ca   a  b  [ From (4), (5)  a   b   c  d   d   d  and (6) ] 1 [ ab – ac + bc – ab + ac – bc ] = 0 d Hence proved. The number of rose plants in first, second third,…, and last row are respectively. 23, 21, 19, ……, 5. Let the number of rows of rose plants be n. The sequence 23, 21, 19, ……5, is an A.P. with first term a = 23, common difference d= – 2 and 

18.

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21.

a 2  9d 2 7 64  9d2 7    2 2 2 a d 15 64  d 15 ( Since a = 8 )

 960  135d 2  448  7d 2  128d2 = 512  d2 = 4  d =  2 Thus, the four parts are a – 3d, a – d, a + d and a + 3d i.e., 2, 6, 10, 14 5 + 7 + 9 + 10 + 13 + 13 + 17 + 16 + …. = (5 + 9 + 13 + 17 + ….. to 20 terms) + (7+ 10 +13 + 16 + …. to 20 terms )

Progressions Solutions

287

In the first set of 20 terms a = 5, d = 4, n = 20 and in the second set of 20 terms a = 7, d= 3, n = 20.



22.

20 20  2  5   20  1 4   2  7   20  1 3 2 2

n   S  2a   n  1 d   2   [ a = 5, d = 4 in the 1st case and a = 7, d = 3 in the later case] = 10(10 + 19 × 4) + 10(14 + 19 × 3) = (10 × 86) + (10 × 71) = 860 + 710 = 1570 Hence, sum of the series is 1570. Given,first term (a) = 17, last term (l) = 350 and common difference (d) = 9. We know that, l = a + (n – 1)d  350 = 17 + (n – 1)d  333 = (n – 1)9 333  n – 1 = 37 9  n = 37 + 1  n = 38 We also know that

 n  1 



S

n  2a   n  1 d  2

S38 

23.

 Sn = 1 + 2 = 4 + ……. + tn – 1 + tn Subtracting we get 0 = 1 + [1 + 2 + 3 + ……. + to (n – 1) terms] – tn

38 [ 2 × 17 + (38 – 1) 9 ] = 19[ 34 + 37 × 9 2

] = 19[ 34 + 333 ] = 19 × 367 = 6973  n = 38 and S = 6973. Hence, there are 38 terms, and their sum is 6973. Since 1st group contains one number, 2nd group contains 2 numbers, 3rd group contains 3 numbers and so on, therefore, nth group will contain n numbers. Here we observe that numbers in each group are in A.P. whose common difference is 1. Thus, for nth group d = 1, n = n. Sequence of first terms of groups 1, 2, 4, 7 …… First term of this sequence is the first term of the first group. Second term is the first term of the second group. Third term is the first term of the third group and so on. nth term of this sequence i.e., tn will be the first term of then nth group. Let Sn = 1 + 2 + 4 + 7 + ………. + tn

 n  1 n

n2  n  2 2 2 Thus, for the nth group, a (first term)

or t n  1 



n2  n  2 d  1, n  n 2  Sum of numbers in the nth group. 

 24.

 n  n 2  1 n   n 2  n  2  2  n  1 1       2   2  2 

Given A.P. is 63, 60, 57…. Here, first term (a) = 63, common difference (d) = (60 – 63) = – 3, Sn = 693. n = ?

n n 2 2  (129 – 3n)n = 1386  3n2 – 129n + 1386 = 0  n2 – 43n + 462 = 0  n2 – 22n – 21n + 462 = 0  n(n – 21) – 21 (n – 22) = 0  (n – 21) (n – 22) = 0  n = 21 or n = 22. Hence, sum of first 21 terms is 693 and the sum of first 22 terms is also 693. Verification In the given A.P. 63, 60, 57…. Here, a = 63, d = – 3, n = 21

 Sn   2a 693  n  1 d2  63   n  1   3

Sn 

n  2a   n  1 d  2



21  2  63   21  1   3  2 



21 126  20   3 2 



21 126  60 2

21  66  693 2 In the given A.P. 63, 60, 57…. Here, a = 63, d = – 3, n = 22 

Sn 

n  2a   n  1 d  2 www.betoppers.com

10th Class Mathematics

288 i.e., Sp + q = 0

22   2  63   22  1   3  2 22 126  21   3 2  = 11 × [126 – 63] = 11 × 63 = 693  sum of first 21 terms is 693 and the sum of first 22 terms is also 693. Hence, verified Given , nth term = a + nb. 1st term = a + (1)b = a + b 2nd term = a + 2b 3rd term = a + 3b 4th term = a + 4b and so on. The set of all the terms a + b, a + 2b, a + 3b….. a + nb is an A.P. Second term – First term = (a + 2b) – (a + b) = b. Third term – Second term = (a + 3b) – (a + 2b) = b. Fourth term – Third term = (a + 4b) – (a + 3b) = b and so on. Clearly, the difference between two consecutive terms is the same in all cases. The common difference = b Hence, the sequence with nth term a + nb is always an A.P. whatever be the value of a and b. Here, 1st term = a + b, n = 20, d = b

Sp q 



25.

We know that, Sn  n  2a   n  1 d   2

20 2  a  b    20  1 b  2  = 10 [2a + 2b + 19b] = 10 (2a + 21b) = 20a + 210b Given , sum of p terms of an A.P. = Sum of q terms Let a = first term and d = common difference. According to the given condition Sp = Sq 

26.



p q 2a   p  1 d    2a   q  1 d  . 2 2 (p – q)2a + p (p – 1)d – q (q – 1)d = 0. (p – q)2a + d (p2 – q2 – p + q) = 0. (p – q)2a + d [(p + q) (p – q) – (p – q) ]

   = 0. Cancelling (p – q), we get 2a + d [(p + q) – 1 ] = 0.   2a = (1 – p – q)d ______ (1) Now we have to prove that sum of p + q terms of an A.P. is zero. www.betoppers.com

 

p  q 2

p  q 2

1  p  q  d   p  q  1 d 

d  pd  qd  pd  qd  d 

p  q

 0  0 2  Sp + q = 0 Given, sum of first p, q and r terms of an A.P. are a, b, c respectively. Let A = first term and d = common difference. According to the given condition Sp = a, Sq = b and Sr = c Sum of p terms is Þ Sp = a 

27.

pq  2a   p  q  1 d  2 

Sum of p terms 





a

p  2A   p  1 d  2

p  2A   p  1 d  2

a 1   2A   p  1 d  _________ (1) p 2

Sum of q terms is  Sq = b Sum of q terms  q 2A   q  1 d  .  2 q  b   2A   q  1 d  2



b 1  2A   q  1 d  ____ (2) q 2

r Sum of r terms   2A   r  1 d  . 2



c

r c 1 2A   r  1 d    2A   r __1_____d__ 2 r 2

(3) Subtracting (2) from (1), we get

a b 1 1    p  1 d   q  1 d    p  q  d  . p q 2 2 

pq 

2 a b    _________ (4) d p q 

Subtracting (3) from (2), we get

Progressions Solutions

289

b c 1 1    q  1 d   r  1 d    q  r  d. q r 2 2 

qr 

2 b c    _________ (5) d q r 

Subtracting (1) from (3), we get c a 1 2 c a    r  p d  r  p     r p 2 d r p _________ (6) Substituting the values of (q – r), (r – p) and (p – q) from (4), (5) and (6), in

29.

Sum of n terms of first A.P. 5n  4  Sum of n terms of second A.P. 9n  6

a b a c b c  pget  q  r    r  p  q rpq  , rwe    p  q p q p r q r

28.



a  2  b c  b  2  c a  c  2  a b                  p  d  q r  q  d  r p  r  d  p q  



2  ab ac bc ab ac bc  2      0  0    d  pq pr qr pq pr qr  d

40 + 40d = 10 + 35 d.  40d – 35d = 10 – 40  5d = –30  d = –6  We know that tn = a + (n – 1) d. 20th term = a + (20 – 1)d  ( a = 2, d = – 6 ) 2 + 19 (– 6) = 2 – 114 = – 112 .  Hence, proved. Let a1, a2 be the first terms and d1, d2 the common differences of the two arithmetic progressions respectively. According to the given data,



n  2a1   n  1 d1  5n  4 2  n  2a1   n  1 d1  9n  6 2



2a 1   n  1 d1 5n  4  2a 2   n  1 d 2 9n  6



n 1 d1 5n  4 2  n 1 a2  d 2 9n  6 2

Hence, proved. Given, first term (a) = 2 and also given that Sum of 1st five terms 

a1 

1 (sum of next five 4

terms).

n We know that Sn   2a   n  1 d  2 Sum of 1 st

As we want the ratio of 18th term i.e.,

n 1  17 2 n – 1 = 34, n = 35.  ]

five



5 5 terms   2  2   5  1 d    4  4d  = 10 + 2 2 10d Sum of 1 st ten terms 

Sum of 1st five terms 

1 (sum of next five 4

terms).



10  10d 

1 10  35d  4

30.

[  t18 = a + 17d

So,

a1  17d1 5  35  4 179   a 2  17d 2 9  35  6 321

Hence,

Eighteenth term of 1st series 179  Eighteenth term of 2nd series 321

10  2  2  10  1 d   5 4  9d  = 20 + 2

45d Sum of next five terms = sum of 1st ten terms – sum of 1st five terms. = 20 + 45d – 10 – 10d = 10 + 35d According to the question,

a1  17d1 a 2  17d 2

Given, the penalty for each succeeding day is Rs.50 more than for the preceding day, the, amount of penalty for different days forms an A.P. with first term a(= 200) and common difference d (= 50). We have to find how much a delay of 30 days costs the contractor. In other words, we have to find the sum of 30 terms of the A.P. www.betoppers.com

10th Class Mathematics

290

30 { 2 × 200 + (30 – 1) × 50 } 2  Required sum = 15 (400 + 29 × 50)  Required sum = 15 (400 + 1450)  Required sum = 15 × 1850 = 27750 Thus, a delay of 30 days will cost the contractor Rs. 27750. We know that S1 = Sum of ‘n’ natural numbers =

2

Required sum 



31.

1  2  3  ......  n 

n  n  1

2 S2 = Sum of squares of ‘n’ natural numbers = n  n  1 2n  1 6 S3 = Sum of cubes of ‘n’ natural numbers = 12  22  32  ....  n 2 

2

n 2  n  1  n  n  1  1  2  3      n    4  2  We have to prove that 9S22 = S3(1 + 8S1). RHS = S3 [1 + 8S1] 3

3

3

n 2  n  1  4 

n 2  n  1

2

  n  n  1   1  8   2    

2

 n 3  n  1

4

3

2





3

n 2  n  1  4 n 3  n  1

33.

4

ac  b  a + c = 2b ___________ (1). 2 ‘m’ is the A.M. between ‘a’ and ‘b’ 

2

n  n  1 1  4 n  n  1  4 2

ab  m ___________ (2). 2 ‘n’ is the A.M. between ‘b’ and ‘c’ 

2

n 2  n  1 1  4n 2  4n   4 2



n 2  n  1  2 n  1

LHS = 9S2

2

3

2

________

4 2

 n  n  1 2n  1   9  6  

2

 n 2  n  12  2n  12   9  36  

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2

________ (2) 4 From (1) and (2)  LHS = RHS  9S22 = S3(1 + 8S1) Let the first term be –5 and the last term be 13 and A1, A2, A3, A4, A5, A6, A7, A8 be the arithmetic means between –5 and 13. As we are inserting ‘8’ arithmetic means between –5 and 13, the number of terms will be ten. t1 = a = 5 ____________ (1) t10 = a + 9d = 13 __________ (2) Substituting (1) in (2)  –5 + 9d = 13  9d = 18  d=2 The second, third, ………… ninth terms of the A.P. becomes the eight arithmetic means A1, A2, A3, A4, A5, A6, A7, A8.  A1 = t2 = a + d = – 5 + 2 = –3 A2 = t3 = a + 2d = –5 + 2 × 2 = 1 A3 = t4 = a + 3d = –5 + 3 × 2 = –1 A4 = t5 = a + 4d = –5 + 4 × 2 = 3 A5 = t6 = a + 5d = –5 + 5 × 2 = 5 A6 = t7 = a + 6d = –5 + 6 × 2 = 7 A7 = t8 = a + 7d = –5 + 7 × 2 = 9 A8 = t9 = a + 8d = –5 + 8 × 2 = 11  The eight arithmetic means between –5 and 13 are –3, –1, 1, 3, 5, 7, 9, 11. Given that a, b, c are in A.P.



32.

n 2  n  1  2n  1

(1)

bc  n ___________ (3). 2 Now, we have to prove that A.M. between m 

mn b 2 Using the values of ‘m’ and ‘n’ from (2) and (3) we get, and n is b i.e.,

L.H.S 

mn 2

ab bc  2  2 2

Progressions Solutions

291

a  2b  c 4



1 1   c a b c

2b   a  c 





4 [ a + c = 2b, from (1) ]

mn 2 = b = L.H.S L.H.S 



mn b 2  ‘b’ is the A.M. between ‘m’ and ‘n’. 34.



 in

a  b  c b a  c c a  b , , are in A.P. bc ac ab By adding 1 to each term of the above A.P., we get

 a  b  c   b a  c   c a  b   1 ,   1 ,   1   bc   ac   ab  are also in A.P.





35.

abc , we get, ab  bc  ca

c a



b c



 

b a



c a

b a b c



b a



b c

c a a b a b

c a

a b



c a



c b



b c



c b



a b



 

b a 

1 1 1 , , are in A.P.. b c c a a b Given sequence is 128, – 96, 72…… a = 128

Hence,

36.

r 37.

t 2 96 3   t1 128 4

Let a be the first term and r be the common ratio of the given G.P; then

T2 

25 25 ________  ar  (1) 4 4

 ab  ac  bc   ab  bc  ac    ,   , bc ac  ac  bc  ab    are in A.P. ab

Dividing (2) by (1), we get

abc  ab  ac  bc   ,  bc ab  bc  ca  abc  ab  bc  ac   ,  ac ab  bc  ca 

2  r6    5  r = 2/5 Substituting the value of r in (1), we get

abc  ac  bc  ab   are in A.P.  ab ab  bc  ca  Hence, a, b, c are in A.P. Given, a, b, c, are in A.P. We have to prove that 1 1 1 , , is in A.P.. b c c a a b To prove the required, it is enough if we prove



c b



 



b–a=c–b  2b = a + c  Thus, a, b, c, are in A.P.



Multiply each term by







 1 1  1 1  1 1 Since, a    , b    , c    are b c a c a b A.P.

b c c a

1 1  a b c a

and T8 

16 16 ________  ar 7  (2) 625 625 ar 7 16 4   ar 625 25

6

2 25 125 a.  a 5 4 8 125  2   n term t n    8 5

n 1

th

Putting n = 1, 2, 3, 4, …., the required G.P., is

125 25 5 , , , 1... 8 4 2

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10th Class Mathematics

292 38.

41.

2 7 , x, , ..... are in G.P., 7 2 If a, b, c are in G.P. then b  ac

2 7   1  1 7 2 x=1 x

39.

The series is

1 1   1  .... 4 2

Here 1 T3 1 T2  2   2, T  1   2.... T2  T3  .....   2 1 2 T1 T1 T2 2 4

The given series is a G.P. with a = ¼ and r = – 2 Let nth term of the sequence be 256.  tn = 256  arn – 1 = 256

1 n 1  2   256 4  (– 2)n – 1 = 1024  (– 2)n – 1 = 210  (– 2)n – 1 = (– 2)10  n – 1 = 10  n = 10 + 1 = 111 Therefore 11th term of the sequence is 256. 1 series: We have 5 + 10 + 20 + … 

40.

Given a = t1 = – 3 and t4 = t22 Let common ratio be r First term a = – 3 t2 = a.r and t4 = a.r3 From the given data, fourth term of the G.P. is equal to the square of the second term. i.e., t4 = t2 2 a.r3 = (a.r)2 – 3.r3 = (– 3.r)2 [ Since a = – 3 ] 3 2 – 3r = 9r r3 9  r 2 3  r= –3 t7 = a. r7 – 1 = a.r6 = – 3(– 3)6 = – 3 × 729 = – 2187.

42.

Given t 6  24 and t13 

We know t6 = arn–1  t6 = ar6–1= ar5 = 24 _____________ (1) 3  t13 = ar13–1= ar12  16 __________ (2) Equations (1) and (2) are two equations in terms of the unknown variables ‘a’ and ‘r’. Solving of (1) and (2) gives us the values of ‘a’ and ‘r’.

 2   ar12 1 ar5



3 / 16 24

10 20  2 5 10  Tn = a1 r1n – 1 = 5(2)n – 1 2nd series: We have 1280 + 640 + 320 + …

 r7 

640 320 1   Here a2 = 1280 and r2  1280 690 2

1 1  r7     r  2 2

Here a1 = 5 and r1 

n 1

1 Tn = a2 r2 = 1280   2 th Let n terms of given series be equal n–1

n–1

1 = 1280   2



5(2)



2n – 1 = 256 ×

22n – 2 = 28 2n – 2 = 8 2n = 10

10 5 2 www.betoppers.com n

1 2n 1

n 1



(2n – 1)2 = 256

3 16

3 1 3 1 1    4 3 7 16 24 16  3  8 2  2 2 7

We can find the value of ‘a’ by substituting the value of ‘r’ in the equation (1). 5

1 ar = 24  a     24  a = 24 × 25 = 24 2 × 32 = 768 Thus first term of the G.P. = a = 768 and 5

1 2 The G.P is 768, 384, 192, 96 ……. Now let us write the 20th term by substituting a common ratio r 

= 768, r 

1 and n = 20 in tn = arn – 1 2

Progressions Solutions

293 19

 t 20  ar

3 1  768    11 2 2

20 1

 The 20 th term of G.P is 43.

45.

3 211

 Product =

6 2

 3

th

 tn  a r

n 1

 243 2   2   3 

New numbers are n 1

 16  n 1

 243   3 

n 1

  3 

10

1th term of the G.P. is 243 2  The 11 Substituting a  2, r  13 and n = 11 in the formula a  r n  1 r 1 2  S11 



44.



we get the sum of the 11 terms. 11

3



3 1



1

2 35 3  1

3 1 Given 2a = b + c and b, G1, G2, c  G.P.. i.e., b, br, br2, c  G.P..

     

10  10r  9 r 16r – 10 = 10r2 – 9r 16r – 10 = 10r2 – 9r 10r2 – 25r + 10 = 0 2r2 – 5r + 2 = 0 2r2 – 4r – r + 2 = 0 2r(r – 2 ) – 1( r – 2 ) = 0



(r – 2 ) ( 2r – 1 ) = 0  r 

____________

(1)

1/3

c   b

2/3

c   b

46.

2/3

 G13 = b2c __________

2/3

1 or 2 2

10 1 ,10, 10   1/ 2 2 i.e., 20, 10, 5 Case II : If r = 2, then the numbers in G.P. are

c r    b

10 ,10, 10  2  2 i.e., 5, 10, 20. 0.15, 0.015, 0.0015,….given G.P.

a  0.15; r 

(2)

c and G2 = br = b    G23 = c2b ________ (3) b On adding (2) and (3) we get G13 + G23 = b2c + c2b = bc (b + c) = 2abc 2

10  10r  7   16] r

1 Case I : If r  , then the numbers in G.P. are 2

3 1

2  243 3  1

Now G = br = b

10 , 10 + 6, 10r + 7. These are r

 16 

 n  1  10  n  11

Sn 

10 , 10 and 10r.. r

in A.P.

 243 2   3 

a = 10

 The numbers are

Let the n term of the G.P be 243 2

a , a and ar.. r

a . a . ar = 1000 r

 a3 = 1000 i.e., a3 = 103 (or)

In the given G.P., the first term  a  2 and The common ratio r 

Let the numbers in G.P. be

t 2 0.015   0.1 t1 0.15

 1  rn  Formula for Sn  a   [ r < 1]  1 r  1   0.18  S8  0.15    1  0.1 



0.15  1 8 8 1   0.1   1   0.1   6  0.90  www.betoppers.com

10th Class Mathematics

294 1   0.1200.15  1 20 20 S20  0.15    1   0.1 1   0.1      1  0.1   0.90 6

47.

Given, G.P. is 3,

3 3 , ,.... 2 4

Here, a = 3; r 

3/2 3 1 1    1 3 2 3 2

49.



= 0.5 + 0.55 + 0.555 + . . . n terms

5 55 555    ...n terms 10 100 1000

 1 11 111   5    ...n terms   10 100 1000  ( by taking 5 common) Multiplying by 9 and dividing by 9 5  9 99 999       ...n terms  9  10 100 1000 

 1  rn  Sn  a    1 r  But, according to the problem,

Sn 

Sn

5  1  1   1     1     1    1    ...n terms  9  10   100   1000  

3069 512

1  r n  3069 a    1  r  512

 1  1  1  ...n terms    5    1 1 1  9     ...n terms      10 100 1000

  1 n  1     3069 2 3     1  512   1 2   

 1  1  1  ...n terms    5    1 1 1  9     ...n terms      10 100 1000

In the second set of terms

  1 n  3069 6 1        2   512

1 1 1    ...n terms     10 100 1000 

n

a

 1  3069 1 1023  1      512 6 1024 2

1 1 and r  10 10

n

1023  1024  1  1023     1  1024  2  1024

 1   1 5   10n    1  rn   n  .  In G.P. S  a n  9 10 1  1    1 r    10  

n

1 1 1 1       n  10  n = 10 1024 2 2 2

3069 . 512 This is not quite a G.P. However, we can convert it to a G.P. by writing the terms as (10 – 1), (102 – 1), (103 – 1), ……… (10n – 1) If Sn is the sum to n terms, Sn = (10 – 1) + (102 – 1) + (103 – 1) + ………+ (10n – 1) = (10 + 102 +103 + . . . . +10n) – (1 + 1 + 1 + . . . n times )

 10n  1   n 5     n  . 10 9  9 10   10  

 Ten terms are needed to give a sum of

48.



10 10n  1 10  1

 n  Sn 

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10 n 10  1  n 9

n 10n  1  5  10  1 .10  5   n  .    n   9 9  10 9.10 n 9.10 n   

50.

Given sequence is 5, Here, a = 5; r 

20 80 , ... 7 49

4 also (r < 1) 7

  

Progressions Solutions

 S 

51.

a 1 r



5 4 3  5  7  35 1 3 3 7 7 Given the first term of G.P. a = 2 and S¥ = 6 S¥ = 6 

5

295

ab2 a b2



a    S   1 r  

a 6 1 r

  

a b

 

a b



2 6 1 r 6 – 6r = 2 6r = 4



4 2  6 3 Given y = (0.64)

a



1 1 1    .......upto  3 32 33 a = 1/3 and r = 1/3

2



a

a b a b



a b



2 a 2 b



6  4

a

3 2

54.

1/ 3 1  2/3 2

G.M of a, b  ab

log0.251/2

1/ 22

1/2

1 log1/21/2 2

= (0.64)1/2 1 2   0.8  2  0.8   Let the numbers be a, b (a > b).

H.M. between a and b 12  G.M. between a and b 13

2ab 12  2 ab  12  a  b  13 a  b  2 ab 12 a  b 13 ab 13 Applying componendo and dividendo, we get

5 1 5 1

ab 2

S 

y   0.64 

b





a 9  b 4  a:b::9:4 Let the numbers be a and b, a > b > 0, then A.M of a,b 

Given

25 1

25 1

a 1/ 3  1  r 1 1 / 3

  0.64 



2

S 

log

a

2



In the sequence

  0.64 

2

5 b 1

a b

1 1 1  log 0.25   2  3  .......upto   3 3 3 

53.

 b

 b

a b

Again applying componendo and dividendo, we get

r

52.

ab 13  12    ab 13  12 

2ab ab Given, the A.M of two positive numbers exceeds its G.M by 3/2 H.M of a,b 

ab 3  ab   a  b  2 ab  3 ____ (1) 2 2 And also given, the G.M of two positive numbers exceeds its H.M by 6/5 and

ab 

2ab 6  _________ (2) ab 5

Let t  ab then the equations 1 and 2 reduces to a + b = 2t + 3 and t 

2t 2 6  ab 5

a + b = 2t + 3 and t 

2t 2 6  ab 5

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10th Class Mathematics

296 2t 2 6  ( a + b = 2t + 3 ) 2t  3 5 2  (2t + 3t) 5 = 10t2 + 6 (2t + 3)  10t2 + 15t = 10t2 + 12t + 18  3t = 18  t

57.

1 1 1 If x, y, z are in H.P., then x , y , z are in A.P.. On multiplying each term with x + y + z, still the sequence remains in A.P.

18  t  6 3



xyz xyz xyz , , are in A.P. x y z

 ab  6  ab  36 _________ (3) Then from (1), a + b = 2 × 6 + 3 a + b = 15 ________ (4) (a – b)2 = (a + b)2 – 4ab = (15)2 – 4 × 36 = 255 – 144 = 81  a – b = 9 _______ (5) (4) – (5)  2b = 6  b = 6/2 = 3 Substituting b = 3 in equation (4) a + b = 15 a = 15 – 3 = 12  The numbers are 12, 3.



x yz y xz z xy  ,  ,  are in A.P. x x y y z z

 1

th

 1 14d  8  14     8  7  1  2 th  15 term in A.P. is 1. 1 1 1 Since x, y, z are in H.P. therefore  15th term in H.P. is

2zx  y(x + z) = 2zx _________ (1) xz  L.H.S. = log (x + z) + log (x – 2y + z) = log [(x + z)(x – 2y + z)] = log [(x + z)2 – 2y(x + z)] = log [(x + z)2 – 2.2zx] [ y(x + z) = 2zx ] = log (x2 + z2 + 2xz – 4xz) = log (x2 + z2 – 2xz) = log (x – z)2 = 2 log (x – z) = R.H.S Hence, proved y

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yz xz xy , 1 ,1  are in A.P. x y z

Subtracting ‘2’ form each term, still the sequence remains in A.P.

1 1 The 3 and 7 terms of a H.P. are and 7 5 respectively. The 3rd and 7th terms of the corresponding A.P.. are 7 and 5. ________ Therefore, a + 2d = 7 (1) a + 6d = 5 ________ (2) On solving (1) and (2), we get rd

1 d   and a = 8 2  15th term in A.P. t15 = a +

58.

59.



yz xz xy  1  2,  1  2,  1  2 are in A.P. x y z



yz xz xy  1,  1,  1, are in A.P. x y z



yzx xzy x yz , , , are in A.P. x y z

The reciprocals of the above terms will be in H.P.



60.

x y z , , are in H.P. yzx xz y x yz

1 1 1 1 1 1 Let x , x , x , x , x , x be the six 1 2 3 4 5 6 2 2 . and 3 31 Then the corresponding Arithmetic progression will be of the form harmonic means between

3 31 , x1 , x 2 , x 3 , x 4 , x 5 , x 6 , 2 2 Now, a 

3 31 ; Last term t 8  2 2

But t8 = a + 7d =

31 2

3 31  7d  2 2 7d 

31 3 31  3 28     14 2 2 2 2

Progressions Solutions

d

3.

14 2 7 3 2

 x1  a  d   2  7 2

x 2  x1  d   2 

34 7  2 2

7  4 11  2 2

x3  x2  d 

11 11  4 15 2  2 2 2

x4  x3  d 

15 15  4 19 2  2 2 2

x5  x4  d 

19 19  4 23 2  2 2 2

x6  x5  d 

23 23  4 27 2  2 2 2



Arithmetic means between

4.

3 and 2

31 7 11 15 19 23 27 are , , , , , 2 2 2 2 2 2 2 

Harmonic means between

5.

6.

2 and 3

2 2 2 2 2 2 2 are , , , , , . 31 7 11 15 19 23 27 7.

CONCEPTIVE WORKSHEET 1.

2.

Here, a = –1.25 and d = –0.25 So, the arithmetic progression is –1.25, –1.25 + (– 0.25), –1.25 + 2 (–1.25), –1.25 3 + (– 0.25), . . . or, –1.25, –1.50, –1.75, –2, . . . i) 8, 11, 14, 17, 20, ........ Common difference = Second term – First term = 11 – 8 = 3 Check. 8 + 3 = 11 (Second term) 11 + 3 = 14 (Third term) 14 + 3 = 17 (Fourth term)etc. Hence, our result is correct. ii) 2, 0.5, –1, –2.5, –4, ...... Common difference = Second term – First term = 0.5 – 2 = –1.5. Check. 2 + (–1.5) = 0.5 (Second term) 0.5 + (–1.5) = –1 (Third term) –1 + (–1.5) = –2.5 (Fourth term) etc. Hence, our result is correct.

297 18, a, b, –3 are in A.P. Difference between 18 and a = a – 18. Difference between a and b = b – a. Difference between b and –3 = –3 – b. Since, the pattern of numbers is in A.P., So difference between 18 and a = difference between a and b = difference between b and –3.  a –18 = b – a = – 3 – b  a –18 = b – a  2a – b = 18 –– (1) And b – a = – 3 – b  a – 2b = 3 –– (2) On solving (1) and (2), we get a = 11, b = 4. We have, a = First term = –7 and, d = Common difference = 5  a18 = a + (18 – 1)d [ an = a + (n – 1)d]  a18 = a + 17d = –7 + 17 × 5 = 78 and an = a + (n – 1) × d = –7 + (n – 1) × 5  an = –7 + 5n – 5 = 5n – 12 Here, a = 17, d = (14 – 17) = –3, l = – 40 and n=6 6th term from the end = l – (n – 1)d = – 40 – (6 – 1) × (–3) = (– 40 + 15) = – 25 Clearly, the given sequence is an A.P. with first term a = 4 and common difference d = 5. Let there be n terms in the given sequence. Then, nth term = 24 a + (n – 1)d = 24  4 + (n – 1)×5 = 24  n = 25.  Clearly, 1, 7, 13, 19, ..... forms are A.P. with first term 1 and common difference 6. Therefore, its nth term is given by an = 1 + (n – 1) × 6 = 6n – 5 Also, 69, 68, 67, 66, .... forms an A.P. with first term 69 and common difference –1. So, its nth term is given by an = 69 + (n –1) × (–1) = –n + 70 The two A.Ps will have identical nth terms, if a n  an  6n – 5 = – n + 70  7n = 75

75 , which is not a natural number.. 7 Hence there is no natural number such that nth terms of two A.P. are equal Let a be the first term and d be the common difference of the A.P. We have, a8 = 31 and a15 = 16 + a11  a + 7d = 31 and a + 14d = 16 + a + 10d

 n

8.

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298

10th Class Mathematics

 a + 7d = 31 and 4d = 16 3a + 36d = 0   a + 7d = 31 and d = 4 3(a + 12d) = 0   a + 7 × 4 = 31 a + 12d = 0   a + 28 = 31  a=3 a + (13 – 1) d = 0  Hence, the A.P. is a, a + d, a + 2d, a + 3d,.... a13 = 0  i.e., 3, 7, 11, 15, 19, ... Hence, 13th term is zero. 9. Let a be the first term and d be the common 13. Let ‘a’ be the first term and ‘d’ the common difference of the A.P. difference of the given A.P. We have, Given, ptp = qtq a10 = 52 and a17 = a13 + 20  p[a + (p – 1)d] = q[a + (q – 1)d]  a + 9d = 52 and a + 16d = a + 12d + 20  ap + (p2 – p)d = aq + (q2 – q)d  a + 9d = 52 and 4d = 20  a(p – q) + (p2 – q2 – p + q)d = 0 a + 9d = 52 and d = 5   a(p – q) + ((p – q)(p + q) – (p – q).1]d = 0  a + 45 = 52 and d = 5  (p – q) [a + (p + q – 1)d] = 0  a = 7 and d = 5 As p  q, i.e., p – q  0, Hence, the A.P. is a, a + d, a + 2d, a + 3d,.... we have a + (p + q – 1)d = 0  tp + q = 0. i.e., 7, 12, 17, 22, ... Hence (p + q)th term of the given A.P. is zero. 10. Let a be the first term and d be the common 14. Let a be the first term and d be the common difference of the given A.P. difference of the given A.P. Then, Let the A.P. be a1, a2, a3, ....an,... pth term = q  a + (p – 1)d = q ––––– (i) It is given that qth term = p  a + (q – 1)d = p ––––– (ii) a10 = 52 and a16 = 82 Subtracting equation (ii) from equation (i), we get  a + (10 – 1)d = 52 and a + (16 – 1)d = 82 (p – q)d = (q – p)  d = – 1 ––––––– (i)  a + 9d = 52 Putting d = – 1 in equation (i), we get and, a + 15d = 82 ––––––– (ii) a + (p – 1) × (–1) = q  a = (p + q – 1) Subtracting equation (ii) from equation (i), we get  nth term = a + (n – 1)d = (p + q – 1) + (n – 1) – 6d = –30  d = 5 × (–1) = (p + q – n) Putting d = 5 in equation (i), we get 15. Let the three parts of 69, which are in A.P. a + 45 = 52  a = 7 = a – d, a, a + d.  a32 = a + (32 – 1) d = 7 + 31 × 5 = 162 Sum of three parts = 69. (Given) and, an = a + (n – 1)d = 7 + (n – 1)×5 = 5n + 2 Then clearly, (a – d) + (a) + (a + d) = 69. Hence, a32 = 162 and an = 5n + 2  3a = 69  a = 23. 11. First two digit-number divisible by 5, is 10 and last Also, the product of two smaller parts = 483. two digit number divisible by 5 is 95. So, we have to (Given) find out the number of terms in the A.P. –––––– (1)  (a – d) × a = 483 10, 15, 20, 25, ....., 95, Substituting a = 23 in (1), we have Let there be n terms in the above A.P.  (23 – d) × 23 = 483 Clearly, here 1st term = a = 10, d = 5 and last term 483 = 95 = tn (say).  23  d   21  d = 23 – 21 = 2. t = a + (n – 1)d, 23  n  We have, 95 = last term Thus, the three parts of 69 are  95 = 10 + (n – 1) × 5.  5(n – 1) = 85 a – d = 23 – 2 = 21. a = 23. 85  n 1   17  n = 18. a + d = 23 + 2 = 25. 5 Hence, there parts of 69 are 21, 23, 25. 12. Let a1, a2, a3, ...., an,.... be the A.P. with its first 16. Let the numbers be a – d, a, a + d, a + 2d term a and common difference d. a, d  Z, d > 0 where according the hypothesis It is given that (a – d)2 + a2 + (a + d)2 = a + 2d 5 a5 = 8 a8 i.e., 2d2 – 2d + 3a2 – a = 0 5(a + 4d) = 8(a + 7d)  1 5a + 20d = 8a + 56d 2   d  [1  (1  2a  6a )] 2 www.betoppers.com

Progressions Solutions

299

Since, d is positive integer, 1 + 2a – 6a2 > 0

 (n –1).5 = 895  n – 1 = 179  n = 180

 1 7   1  7  a 1 a2    0   a   a  0  3 6 6   6  

We know that Sn =

180 (102 + 997) 2 = 90 × 1099 = 98910. 20. Here Sn = 513, a = 54 and d = (51 – 54) = – 3.

 1 7  1 7      a     6   6  Since, a is integer, a = 0

 The required sum =

1 then d  [1  1]  1 or 0. Since d > 0  d = 1 2 Hence, the numbers are –1, 0, 1, 2 17. Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P. Then a = 1/9 and d = 1/9.  Required sum = S25 =

=

25 [2a + (25 – 1)d] 2

25   1   1   25  26  325 2    24     .    2  9 2  9  9  9 

18. We have,

a = 5, l = 45, s = 400

We know that S 

 400   n

n a  l  2

n  5  45 2

 400 

n  50 2

400  16 25

We also know that, Sn 

n  2a   n  1 d  2

16  2  5  16  1 d  2  400 = 8 [10 + 15d]  400 

 10  15d   d

400 5 8

n (a + l) 2

 15d = 50 – 10 = 40

40 8 2  2 15 3 3

2 Hence, n = 16 and d  2 . 3 19. The three-digit numbers which leave 2 as remainder when divided by 5 are 102, 107, 112, .., with 997 as the last of such numbers. These numbers from an A.P. with a = 102, d = 5 and l = 997. Let ‘n’ be the number of terms. As l = a + (n – 1)d, we have 997 = 102 + (n – 1).5  (n – 1).5 = 997 – 102

n 2

 513   2  54    n  1   3  n 111  3n   n(111 – 3n) = 1026 2  3n2 – 11n + 1026 = 0 n2 – 37n + 342 = 0  n2 – 19n – 18n + 342 =0  n(n – 19) – 18(n – 19) = 0  (n – 19)(n – 18) = 0  n = 18 or n = 19 Hence, sum of 18 terms is 513 and also the sum of 19 terms is 513. 21. Let 1st term be a and d the common difference of the given A.P. Given t5 = 30 and t12 = 65. t5 = a + (5 – 1)d. 30 = a + 4d.  and t12 = a + (12 – 1)d ––––– (1) 65 = a + 11d. ––––– (2)  Subtracting (1) from (2), we get 35 = 7d  d = 5. From (1)a + 4d = 30 ( d = 5)  a = 30 – 4d = 30 – 4 × 5 a = 10  Hence, a = 10, d = 5 Now, sum of first 20 terms =

 513 

20  2a   20  1 d  2   S20 = 10 (2 × 10 +19 × 5) = 10 (20 + 96) = 1150 22. Let a = First term and d = common difference. Sum of m terms, Sm = n (Given) S20 

Then,

Sm 

m 2a   m  1 d  2

m 2a   m  1 d  2 2am + m (m – 1)d = 2n –––– (1)  Also, Sn = m (Given)



n

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10th Class Mathematics

300



Sn 

n  2a   n  1 d  2

n 2a   n  1 d  2 2an + n(n – 1)d = 2m –––– (2)  Subtracting (2) from (1), we get 2a(m – n) + {(m2 – m) – (n2 – n)}d = 2(n – m).  2a(m – n)+{(m2 – n2) – (m – n)}d = 2(n – m).  2a + {(m + n) – 1}d = – 2 [Cancelling (m – n)] –––– (3)  2a + {(m + n – 1}d = – 2.



Now

m

Sm  n

mn  2a   m  n  1 d.  2 

–––– (4) Putting the value of 2a + (m + n – 1)d from (3) in (4), we get

mn   2     m  n  . Proved. 2 23. We have, S1 = Sum of n terms of an A.P. with first term 1 and common difference 1 Sm  n 

n n 2  1   n  1  1  d  n  1  2 2 S2 = Sum of n terms of an A.P. with first term 1 and common difference 2

 S1 

n 2  1   n  1  2  n 2  2 S3 = Sum of n terms of an A.P. with first term 1 and common difference 3

 S2 

 S3 

n n 2  1   n  1  3   3n  1  2 2

n n  n  1   3n  1  2n 2 = 2s2 2 2 Hence, S1 + S3 = 2S2 24. Let A1, A2, .., A6 be the six arithmetic means. Then, 2, A1, A2, ..., A6, 16 are in A.P. Let d be the common difference. Here 16 is the 8th term.  16 = 2 + (8 – 1)d, i.e., 7d = 16 – 2 = 14 or d = 2,  A1 = a + d = 2 + 2 = 4, A2 = a + 2d = 2 + 4 = 6 A3 = a + 3d = 2 + 6 = 8, A4 = a + 4d = 2 + 8 = 10 A5 = a + 5d = 2 + 10 = 12, A6 = a + 6d = 2 + 12 = 14. Also, if A is the A.M. between 2 and 16, then Now, S1  S3 

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A=

2  16  9. 2

6  4  14 = 54 2 = 6.9 = 6A = 6 times the A.M, between 2 and 16. Now, A1 + A2 + ... + A6 =

25. A.M between any two numbers a, b, is Now A.M between

ab 2

xa x a and is x x

x  a x  a x  a  x  a  x x  x 2 2

2x 1  1. x 2 26. Let A1, A2, A3, ...., An be n A.M.’s between 3 and 54. Total number of terms in A.P. are (n + 2). If d is the common difference, then 54 = 3 + (n + 2 – 1)d  (n + 1)d = 51----- (1) We have, A8 = t9 = 3 + (9 – 1)d = 3 + 8d and An – 2 = tn –1 = 3 + (n – 1 – 1)d = 3 + (n – 2)d. 3  8d 3 A8 3   Given : A  3   n  2 d 5 5 n 2

 5(3 + 8d) = 3[3 + (n – 2)d]  15 + 40d = 9 + 3(n – 2)d  6 = [3(n –2) – 40]d = (3n – 46).

51 n 1

17 n 1  2(n + 1) = 51n – 782.  49n = 784 or n = 16 27. (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 are in A.P.  (b + c + a) (b + c – a), (c + a + b) (c + a – b), (a + b + c) (a + b – c) are in A.P.  b + c – a, c + a – b, a + b – c are in A.P.. [Dividing each term by a + b + c]  –2a, –2b, –2c are in A.P.. [Adding –a – b – c to each term]  a, b, c are in A.P.. [Dividing each term by – 2], which is given.  (b + c)2 – a2, (c + a)2 – b2, (a + b)2 – c2 are also in A.P.

 2 = (3n – 46).

Progressions Solutions 28.

301

a b  c b c  a  c a  b , , are in A.P.. bc ca ab

 

b c  a  a  b  c c a  b  b c  a     ca bc ab ca

b2  c  a   a 2  b  c 



c2  a  b   b2  c  a 

abc abc 2 2 2 2  b c + b a – a b – a c = c a + c2b – b2c – b2a  (b2c – a2c) + (b2a – a2b) = (c2a – b2a) + (c2b – b2c)  c(b2 – a2) + ab(b – a) = a(c2 – b2) + bc(c – b)  c(b – a)(b + a) + ab(b – a) = a(c – b) (c + b) + bc(c – b)  b – a = c – b  a, b, c are in A.P., which is given, Hence the result. 2

29.

1 1 1 , , bc ca ab

are in A.P. if and only if

1 1 1 1    , ca bc ab ca ba cb  , abc abc i.e., if and only if b – a = c – b, i.e., if and only if a, b, c are in A.P. 30. Given sequence is 5, 0.5, 0.05,...... a = t1 = 5

 2   ar12 1 ar5 r7 



General term of G.P.(tr) = ar t7 = 5(0.1)7–1 = 5(0.1)6 = 5 × 0.000001 = 0.000005 tn = arn–1 = 5 × (0.1)n–1 31. Given We know 

n–1

3 t6 = 24 and t13  16 t6 = arn–1 t6 = ar6–1 = ar5 = 24 ––––––– (1)

3 –––– (2) 16 Equations (1) and (2) two equations in terms of the unknown variables ‘a’ and ‘r’. Solving of (1) and (2) gives us the values of ‘a’ and ‘r’.



t20 = ar20–1 = ar12 =

3/16 24

3 1 3 1 1    4 3  7 16 24 16  3  8 2  2 2 7

1 1  r7     r 2 2 We can find the value of ‘a’ by substituting the value of ‘r’ in the equation (1). 

5

1   a    24 2 5  a = 24 × 2 = 24 × 32 = 768 Thus first term of the G.P = a = 768 and common

ar5 = 24

1 2 The G.P is 768, 384, 192, 96....... Now let us write the 20th term by substituting ratio r 

a = 768, r 

i.e., if and only if

t 0.5 r 2   0.1 t1 5



1 and n = 20 in tn = arn–1 2 19

3 1 t20 = ar20–1 = 768    11 2 2

3

 The 20th term of G.P is 211 32. Let a be the first term and r, the common ratio of the given G.P., then t2 =

25 , 4

and t8 =

i.e., ar =

25 4

---------- (1)

16 16 , i.e., ar7 = ---------- (2) 625 625

Dividing (2) by (1), we get

ar 7 16 4   ar 625 25

6

2  2  r =    r= . 5 5   Subtracting this value of r in (1), we get a = 125/8. 6

125  2   tn = 8 .  5   

n 1

Putting n = 1, 2, 3, 4, ..., the required G.P. is

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10th Class Mathematics

302 33.

The given sequence is a G.P. with a = 18 and =

2 r=  . 3 Let the nth term of the given sequence be 512  2   18.   then 729  3 

7 [(10 – 1) + (10 – 0.1) + (10 – 0.01) + (10 – 9 0.001) + ... to 50 terms)] =

512 , 729

n 1

7 [(10 + 10 + 10 + ... to 50 terms) – (1 + 0.1 + 9 0.01 + ... to 50 terms)] =

n 1

8

512 256 28  2   2       8    729  18 729  9 3  3  3  n – 1= 8  n = 9  The given term is the 9th term.

34.

50

1   0.1 7  50  10  1. = 9 1  0.1  50

1  1/10  7 = 9 500  9 /10 

a Let the numbers in G.P. be , a and ar.. r

 Product =

a a.ar = 1000 r

i.e., a3 = 103 (or)

7 1  1/1050   500   9 9 /10 

=

7 10  1  500   1  50    9 9  10  

a = 10

 The numbers are

10 , 10 and 10r.. r

10 New numbers are , 10 + 6, 10r + 7. These are r 36. in A.P.

37.

S 

t2 9 t 8.1 9  , 3   t1 10 t 2 9 10

1  r ,2 2 1 10 Case I: r  , the numbers in G.P. are ,10, 2 1/ 2 10(1/2) i.e., 20, 10, 5 10 , 10, 2

10(2) i.e., 5, 10, 20. The given sequence is not a G.P. However, we 38. can relate it to a G.P. Let S50 be the sum to 50 terms of the given sequence, then S50 = 7 + 7.7 + 7.77 + 7.777 + ... to 50 terms

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4 1  28 7

7 98 a 28 28  28     6 3 1 r 1 1  6    7 7 Given progression is 10, –9, 8.1 ... Now, 

10  16   10r  16 r  16r – 10 = 10r2 – 9r  2r2 – 5r + 2 = 0

Case II: r = 2, the numbers in G.P. are

7 10  7 1  4500  10  50  = 4490  49    81  81  10  10  Given a = 28 and ar = 4 =

r

10  16   10r  7   16 r

  

  

=

 a3 = 1000

35.

7 (9 + 9.9 + 9.99 + 9.999 + ... to 50 terms) 9

 The ratio of its any term (except the first) to its preceding term is constant and nonzero. Therefore, the given sequence is a G.P. with first term a = 10 and common ratio r = –9/10 < 1. 

S 

a 10 10   1  r 1   9 /10  1  9 /10

10  10 100  19 19 Let a, ar be the first two terms of an G.P 

 a + ar =

5 ——— (1) 3

Progressions Solutions also

a  3  a = 3 – 3r 1 r



3

–

 3  3r 2  If r 

39.

303

3r

+

(3

–

41. 3r)

r

=

5 3

2 5 4  r2   r =  3 9 3

2  2  a  3  3   1 3  3

2 2 If r    a  3    3  5 3 3  a = 1 (or ) 5 The 12th and the 19th terms of H.P. are respectively 1/5 and 3/22. The 12th and the 19th terms of the corresponding 42. A.P. will be 5 and 22/3. If ‘a’ and ‘d’ be respectively the first term and the common difference of the corresponding A.P. then a + 11d = 5 ———— (1) and a + 18d =

22 ———— (2) 3

on subtracting (2) from (1), 7d =

Let the H.P. be The 7th term =

1 1   a  6d  10 a  6d 10

1 1   a  11d  25 a  11d 25 Solving these two equations, a = – 8, d = 3 Hence the 20th term of H.P. = The 12th term =

1 1 1   and nth term a  19d 8  19.3 49 

1 1 1   a   n  1 d 8   n  1 3 3n  11

Since a, b, c are in A.P.

a c --------(1) 2 Since p, q, r are in H.P.  b=

2pr --------(2) pr Since ap, bq, cr are in G.P.  (bq)2 = (ap) (cr) i.e., b2q2 = acpr --------(3) Putting the values of b and q in (3), we get  q=

1 7  d= . 3 3

11 4  3 3 Hence the 4th term of H.P. = a=5–

40.

1 1 1 , , ,...... a a  d a  2d

2

2

 a  c   2pr    acpr     2   pr

1 1 3   a  3d 4 / 3  3/ 3 7 Let the H.P be



 a  c 2 pr 2 p  r

1 1 1 1 , , , ,..... a a  d a  2d a  3d Then,



a 2  c2  2ac p 2  r 2  2pr  ac pr



a c p r  2   2 c a r p

1 2 1 12 5 7  and   a  and d   a 5 a  d 13 2 12 Now, 1 12 nth term of the H.P = a   n  1 d  37  7n So, the nth term is largest when 37 – 7n has the 43. least value.

12 Clearly, is least for n = 5 37  7n Hence, 5th term is the largest term.

2

 ac 

a  c  ac

2



p  r  pr

p r a c    . r p c a Since a, b, c are in A.P.

Hence,

a c ------------ (1) 2 Since b, c, d are in G.P. ------------ (2)  c2 = bd Also since c, d, e are in H.P.  b

 d=

2ce ce

----------- (3) www.betoppers.com

10th Class Mathematics

304 Substituting the values of ‘b’ and ‘d’ from (1) and (3) respectively in (2), we get ea  c a  c 2ce  c 2 ce ce 2  c + ce = ae + ce  c2 = ae, which show that a, c, e are in G.P.. Let a be the first term and (| r | < 1), the common ratio of the G.P.



g12 g2  a and 2  b g2 g1



g12 g 22  ab g 2 g1

c2 =

44.

 S1 

a 1  r n  1 r

a 1  r

S1  S1  S  





a 1  r n  1 r



1 r

n

a 1  r 2n  1 r

,S

 ,  a 1  r   n

 1  r

ab   ; 2A  a  b   A  2    (1)  2A = a + b A – a = b – A.

a 1 r

a   1 r  

 2A  46.

2

1 1 1    mm 2 2 4 The distance covered by the insect in 1st second, 2 nd second, 3 rd second, ... respectively

a 2 1  r n  r n

1 1 1  1, , ,.... which is a G.P. with a = 1, r  2 4 2 Let time taken by the insect in covering 3mm be n seconds.  Sn of the above G.P. = 3

2

n 2n a  a 1  r  a 1  r    S  S1  S2   ,  1 r  1 r 1 r   

  1 n  1 1       2   1 3 1 1    3 or 1  n  or n  1 2 2 2 2 or 1 2

a a 1  r n  1  r 2n    .  1 r 1 r  







45.

a2

1  r  a

2

1  r n  1  r 2n 

2n = –2 This is impossible because 2n > 0,  n  N  Our supposition is wrong.  There is one n  N , for which the insect could cover 3mm in n seconds.  It will never be able to cover 3mm.

2 2

 r n  r 2n 

2

 r 

1  r  a2

1  r 

n

1  r  2n

 a 2 r n 1  r n 

1  r 

2

Let the numbers be a,b.  a, g1, g2, b are in G.P..

g g b  1 2  a g1 g 2 g12 g2  a and 2  b g2 g1 www.betoppers.com

Distance covered by the insect in the 1st second

1 1  mm 2 2 Distance covered by it in the 3rd second

 r n 

1  r 

g12 g 22  g 2 g1

 1

a 1  r n  1 . 1 r 

a 2 1  r n 

1  r 

, S2 

Also, a, A ,b are in A.P.

47.

1 1 1 We have S   2  3  ..... 3 3 3 

1/ 3 1  1/ 3

1 3 1    3 2 2 1

 y   0.64 

2 log0.25

Progressions Solutions

305 50.

1 log 2 log  0.64  log y  log  0.25 

48.

ab 3  ab  ________ (1) 2 2 2ab 3  ________ (2) ab 2

and

1 1  ra

Here t  ab

 1  ra   ra  1  ra 

1 1  rb

 a + b = 2t + 3 and t 

1 A 1 A

2t 2 6  (Eliminating a + b) 2t  3 5  (2t2 + 3t) 5 = 10t2 + 6 (2t + 3)  10t2 + 15t = 10t2 + 12t + 18  3t = 18  t

A 1 A

 1 rb 

1 B

 t

1  r  1 B

 A 1  and b log r = log  A 

 a log r = log 

 B 1    B 

 A 1 log   A 1    A   log  A  B 1  B 1  B log    B 

H.M. between a and b 

2ab ab

a n 1  b n 1 2ab  a n  bn ab n+2 n+1 + ab + an +1b + bn + 2 a n +1 = 2 a b + 2abn +1  an + 2 + bn + 2 = an + 1. b + abn +1  an + 2 – an +1.b = abn + 1– bn +2  an + 1(a – b) = bn + 1(a – b)

a n 1 a  n 1  1    b b  n+1=0 n=–1

SUMM ATIVE WORKSHEET 1.

 a  b  0

 a n + 1 = bn + 1 n 1

a 1   b

18 6 3

 ab  6  ab = 36 ––––––– (3) Then from (1), a + b = 2 × 6 + 3 a + b = 15 –––––––– (4) (a – b)2 = (a + b)2 – 4ab = (15)2 – 4 × 36 = 225 – 144 = 81  a – b = 9 –––––––– (5) (4) – (5)  2b = 6 b = 6/2 = 3 Substituting b = 3 in equation (4) a + b = 15 a = 15 – 3 = 12  The numbers are 12, 3.

B 1  r  B b

a   b

2t 2 6  ab 5

Substitute t  ab in equation (1) and (2) then the equations can be rewriten as

b

49.

ab 

 log 2  log  0.64   1 log  0.64   log 4 2 = log (0.64)1/2 y = 0.8 A

B

Let the numbers be a and b a > b > 0, then

0

2.

We know that if a is the first term and d is the common difference, then the arithmetic progression is a, a + d, a + 2d, a + 3d,.... Here, a = 10 and d = 3. So, the arithmetic progression is 10, 13, 16, 19, 22, ..... Given that k + 2, 4k – 6, 3k – 2 are in A.P (4k – 6) – (k +2) = (3k – 2) – (4k +6) 4k – 6 – k – 2 = 3k – 2 – 4k + 6 3k – 8 = – k + 4 4k = 12 k = 3. www.betoppers.com

10th Class Mathematics

306 3.

Let ‘a’ be the first term and ‘d’ the common difference of the A.P. Given : t2 = 7

d

3 31  a + d = --------- (1) 4 4

Putting d  

1 1 --------- (2)  a + 30d = 2 2

and t31 =

Subtracting (1) from (2), 29d =

1 31 29    2 4 4

1 13  tn Given last term  6  2 2  Last term = tn = a + (n – 1)d



1 31 31 1 32 a   a   8 4 4 4 4 4 13 Let the number of terms be n so that tn = . 2

n 1 13  32 – n + 1 = –26   4 4 2  n = 59. Hence, first term = 8 and number of terms = 59. We have, a = 5 and d = 10  a31 = a + 30d = 5 + 30 ×10 = 305 Let nth term of the given A.P. be 130 more than its 31st term. Then, an = 130 + a31  a + (n – 1)d = 130 + 305  5 + 10 (n – 1) = 435 7.  10(n – 1) = 430  n – 1= 43  n = 44 Hence, 44th term of the given A.P. is 130 more than its 31st term. Let ‘a’ be the first term and ‘d’ the common difference of the A.P.

 8

4.

5.

6.

3 31  a  d  –––– (1) 4 4 t29 = 1  a + 28d = 1 ––– (2) Subtracting (1) from (2), we have Given that

27d  1 

t2  7

31 4

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 27d 

4  31 27  4 4

13  1  8   n  1     2  4

n 1 13 16  13 29 8   4 2 2 2  n – 1= 58  n = 59 Hence, the first term = 8 and number of terms = 59. Let a be the first term and d be the common difference of the given A.P. Let the A.P. be a1, a2, a3, ...., an,... It is given that a7 = – 7 and a16 = 17 a + (7 – 1) d = –1 and a + (16 – 1)d = 17 ––––––– (i)  a + 6d = – 1 and, a + 15d = 17 ––––––– (ii) Subtracting equation (i) from equation (ii), we get 9d = 18  d=2 Putting d = 2 in equation (i), we get a + 12 = –1  a = – 13 Hence, General term = an = a + (n – 1) d = –13 + (n – 1) × 2 = 2n – 15 We observe that 56 is the first integer between 50 and 500 which is divisible by 7. Also, when we divide 500 by 7 the remainder is 3. Therefore, 500 – 3 = 497 is the largest integer divisible by 7 and lying between 50 and 500. Thus, we have to find the number of terms in an A.P. with first term = 56, last term = 497 and common difference = 7 (as the numbers are divisible by 7). Let there be n terms in the A.P. Then, tn = 497 a + (n – 1)d = 497   56 + (n – 1) × 7 = 497[ a = 56 and d = 7]  7n + 49 = 497  7n = 449 

13 2

13  1   8 + (n –1)     2  4 

1 in (2), we get 4

 1 a  28      1  a – 7 = 1  a = 8.  4 Now, let the number of terms be n.

1 d 4 Putting the value of d in (1), we get

i.e., a + (n –1)d =

1 4

Progressions Solutions

307

 n = 64 Thus, there are 64 integers between 50 and 500 which are divisible by 7. 8.

1 1 , tm  m n Let ‘a’ be the 1st term and ‘d’ the common difference of the A.P., then we know that

 (a – b)r + (b – c) p + (c – a)q

Here, t n 

10.

1  a   n  1 d –– (1) m And tm = a + (m – 1)d tn = a + (n – 1)d



1  a   m  1 d –––– (2) n Subtracting (2) from (1), we get 

1 1    n  1 d   m  1 d m n 

nm   n  1  m  1 d . nm

11.

nm    n  m d mn Dividing by ( n – m) both sides, we get 1 d mn Putting the value of d in equation (1), we have 1 1  a   n  1 m mn 

1 n 1 a  m mn mn



1 1 1 a  m m mn

12.

1 1  a mn mn Now, tmn = a + (mn – 1)d  0a

9.

1 1 1 1  t mn    mn  1  1  mn mn mn mn  tmn = 1 Let A be the first term and D be the common difference of the given A.P. Then, 13. a = pth term  a = A + (p – 1) D ––––– (i) th a = q term  b = A + (q – 1) D ––––– (ii) a = rth term  c = A + (r – 1) D ––––– (iii) Equation (i) – equation (ii), we get a – b = (p – q) D Equation (ii) – equation (iii), we get b – c = (q – r) D Equation (iii) – equation (i), we get c – a = (r – p) D

= (p – q)D r + (q – r)D p + (r – p) Dq = D {(p – q) r + (q – r)p + (r – p)q} =D×0 =0 Let the numbers be (a – 3d), (a – d), (a + d), (a + 3d). Then, Sum = 20  (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20  4a = 20  a = 5 Now, Sum of the square = 120  (a – 3d)2 + (a – d)2 + (a + d)2 + (a + 3d)2 = 120  4a2 + 20d2 = 120  a2 + 5d2 = 30  25 + 5d2 = 30  5d2 = 5  d  1 If d = 1, then the numbers are 2, 4, 6, 8. If d = –1, then the numbers are 8, 6, 4, 2. Thus, the numbers are 2, 4, 6, 8 or 8, 6, 4, 2. Let D be the common difference of the given A.P. Then, b = a + D, c = a + 2D, d = a + 3D and e = a + 4D  a – 4b + 6c – 4d + e = a – 4 (a + D) + 6 (a + 2D) – 4(a + 3D) + (a + 4D)  a – 4b + 6c – 4d + e = a – 4a – 4D + 6a + 12D – 4a – 12D + a + 4D  a – 4b + 6c – 4d + e = a – 4a + 6a – 4a + + a – 4D + 12D – 12D + 4D  a – 4b + 6c – 4d + e = 0 Here, a = 4, d = 8, tn = 100, S = ? tn = a + (n – 1)d. 100 = 4 + (n – 1)8  96 = (n – 1)8  n – 1 = 12  n = 13 Number of terms = 13 Sum 

n a  l  2

[l = tn = 100]

13 13  4  100  104  676 2 2 Hence, sum of the given series is 676. We know that, Tn = a + (n –1)d We have, T2 = 14  a + (2 – 1)d =14  a + d = 14 –––––––––– (1) and T3 = 18  a + (3 – 1)d = 18 –––––––––– (2)  a + 2d = 18 Solving (1) and (2), we get a = 10 and d = 4 We know that, 

S

n  2a   n  1 d  2

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10th Class Mathematics

308

 

S51 

51 2  10   61  1 4  2 

16.

51 51  2  50  4   20  200 2 2

51  220  51 110  5610 2 Hence, sum of 51 terms is 5610. Let 1st term be a and d the common difference of the given A.P. Given t5 = 30 and t12 = 65. t5 = a + (5 –1)d. ––––– (1)  30 = a + 4d. and t12 = a + (12 – 1)d ––––– (2)  65 = a + 11d. Subtracting (1) from (2), we get 35 = 7d  d = 5. From (1) a + 4d = 30 ( d = 5)  a = 30 – 4d = 30 – 4 × 5  a = 10 Hence, a = 10, d = 5 Now, sum of first 20 terms = 

14.

20  2a   20  1 d  2   S20 = 10 (2 × 10 + 19 × 5) = 10 (20 + 96) = 1150 Let a = first term and d = common difference. Since, tn = a + (n – 1)d t12 = a + (12 – 1)d  –13 = a + 11d ––––––– (1)

We know that Sn 

Again since, Sn 

n  2a   n  1 d  2

 Sum of four terms =

4 2a   4  1 d  2

 24 = 2 (2a + 3d) ––––––– (2)  2a + 3d = 12 Multiplying (1) by 2 and subtracting from (2), we get 3d – 22d = 12 – (– 26)  – 19d = 38  d = – 2 Putting the value of d in (1), we get a = – 13 – 11d = – 13 – 11 × (–2) = – 13 + 22 = 9 Thus, a = 9, d = –2 Now, sum of ten terms 

10  2a  10  1 d  2

 5 2  9  9   2 

[a = 9, d = –2]

= 5 (18 – 18) = 0. www.betoppers.com

n  2a   n  1 d  2

n  2  1600   n  1 500 2 77000 = n[3200 + (n – 1) 500] 770 = n [32 + (n – 1)5] [ Cancelling 100] 770 = 32n + 5n(n – 1) 5n2 + 27n – 770 = 0 5n2 + 77n – 50n – 770 = 0 n(5n + 77) – 10(5n + 77) = 0 Either 5n + 77 = 0 or n – 10 = 0

38500 

      

77 (Not possible) or n = 10 5 Number of years = n = 10 years. Hence in the next 10 years he will save Rs.38500 Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it i.e., House: H1 H2 H3 ... Hx–1 Hx Hx+1 ....H49 House No: 1 2 3 (x – 1) x (x +1) ...49 1 + 2 + 3 + ... + (x –1) = (x + 1) + (x + 2)+ ...+49  1 + 2 + 3 +.... + (x –1) = {1 + 2 + 3 + .... + x + (x + 1)+ ...+49} – (1 + 2 + 3 + ...+ x)

 n

S20 

15.

In first year, Vijay saved Rs. 1600 In second year, he saved Rs. 2100. In third year, he saved Rs. 2600. Pattern of numbers of his savings is Rs. 1600, Rs. 2100, Rs. 2600, ...... Here, 2100 – 1600 = 2600 – 2100 = 500 The above pattern of numbers are in A.P. Here, a = 1600, d = 500, s = 38500

17.



x 1 49 x 1   x  1  1  49   1  x   2 2 2

n    Using: Sn  2  a  l   x  x  1 49  50 x  x  1   2 2 2  x(x – 1) = 49 × 50 – x (x + 1) [Multiplying both sides by 2]  x(x – 1) + (x2 + x) = 49 × 50  2x2 = 49 × 50  x2 = 49 × 25  x2 = 72 × 52  x = 7 × 5 = 35 

Progressions Solutions

18.

Since x is not a fraction. Hence, the value of x satisfying the given condition exists and is equal to 21. 35. Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P. Then a = 1/9 and d = 1/9.

25 [2a + (25 – 1)d] =  Required sum = S25 = 2 25   1   1   25  26  325 2    24     .    2  9 2  9  9  9 

19.

20.

Let the first term be ‘8’ and the last term 33, and A1 , A2 , A3 , A4 be the four arithmetic means between 8 and 33. 22. 8, A1, A2, A3, A4, 33 are in A.P. As we are inserting ‘4’ arithmetic means in between 8 and 33 the number of terms will be six. t1 = a = 8 –––––––––– (1). t6 = a + 5d = 33 –––––––––– (2). Substituting (1) in (2)  8 + 5d = 33  d=5 The second, third, fourth and fifth terms of the A.P. become the four arithmetic means A1, A2 , 23. A3, A4. A1 = t2 = a + b = 8 + 5 = 13 A2 = t3 = a + 2d = 8 + 2 × 5 = 18 A3 = t4 = a + 3d = 8 + 3 × 5 = 23 A4 = t5 = a + 4d = 8 + 4 × 5 = 28  The four arithmetic means between 8 and 33 are 13, 18, 23, 28. (C) When n arithmetic means are inserted between a and 2b, we have

 2b  a  A m  m th mean  a  m   ............. (i)  n 1  When n arithmetic means are inserted between 2a and b, we have  b  2a  A'm = mth mean = 2a  m   . It is given  n 1  that Am = A'm

 2b  a   b  2a   a  m  2a  m     m(2b –  n 1   n 1  a) = a (n + 1) + m(b – 2a)  a(n + 1 – 2m + m) = b(2m – m) 

a m  b n  m 1

309

1 1 1 , , are in A.P., b+c c+a a +b 1 1 1 1    ca bc ab ca





b a cb   b  c  c  a   a  b  c  a 

 b2 – a2 = c2 – b2.  2b2 = a2 + a2 Hence, a2, b2, c2 are in A.P. 1 t y 2 x x a  , r  t  x  x 2 and 1 y y

 tn = arn – 1  x  y   t7 =    2   y  x  

7 1

x y6 y5  12  11 y x x

Let a be the first term and r, the common ratio of the given G.P. Given : t3 = and

25 25 ----------- (1)  ar2 = 4 4

1 4 4 t7 = t  25  ar6 = ----------- (2) 25 3

Dividing (2) by (1), we get

ar 6 4 4   ar 2 25 25

4

2 2  r   r . 5 5 Substituting this value of r in (1), we get 4

a.

4 25 625   a= 25 4 16

Let tn = 1  arn –1 = 1  n 1

625  2  . 16  5  n 1

16 2 2  2          5 625 5     5 n – 1 = 4  n = 5. Therefore 5th term of G.P. is unitity.

n 1

1

4



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10th Class Mathematics

310 24.

25.

Let a be the first term and r, the common ratio of G.P. Given t3 = t1 + 9  ar2 = a + 9 ---------- (1)  a(r2 – 1) = 9 Also, t2 = t4 + 18  ar = ar3 + 18 ---------- (2)  ar(r2 – 1) = –18 Dividing (2) by (1), we have r = –2. Substituting r = –2 in (1), we get a(4 – 1) = 9  3a = 9  a = 3 Hence the four numbers are a, ar, ar2, ar3, i.e., 3, –6, 12, –24. The given series is a G.P. with first term a = and common ratio r =

81 9 729  3  3        32 2 64  2  2  n–1=6  n=7  There are 7 terms in the G.P.. Hence, sum of the given series =

28.

1 1 1 The sequence 1, , , ,.... is an H.P.. 3 5 7 Corresponding A.P. is 1,3,5,7,........... Now for the corresponding A.P., first term a = 1, d = 2 100th term of the corresponding A.P. = a + (100 – 1) d = 1 + (100 – 1) 2= 199 Hence the 100th term of the given sequence = 1 . 199

29.

n 1

6

1 1 1 , , , then a d a a d a – d + a + a + d = 12 3a = 12 a = 12/3 = 4 Let numbers be

and



1 1 1 1 1 a . .   2  2 a  d a a  d 48 a d 48 1 4  2 16  d 48

26.

2 2  2187  2059     1  . 9 1  128  288

Let A1 , A2, A3, ...., An be n A.M.’s between 3 and 54. Total number of terms in A.P. are (n + 2). If d is the common difference, then 54 = 3 + (n + 2 – 1)d  (n + 1)d = 51 ----------- (1) We have, A8 = t9 = 3 + (9 – 1)d = 3 + 8d and An – 2 = tn –1 = 3 + (n – 1 – 1)d = 3 + (n – 2)d.

 d2 = 4  d  2

7

 2 / 9   3/ 2   1  3/ 2   1

45 8

3 45 8 3   1    15  15  8 1  8 45 7  15

1 9 3   . 3 2 2

81 81 81 2  3   tn   ar n 1   32 32 32 9  2 

2 Since 3  3  3  ........ 



2 9

81 Also, last term = . If n be the number of terms, 32 then

n 1

27.

 Numbers are

30.

1 1 1 , , 42 4 42

1 1 1  , , 2 4 6 Given, a,b,c are in H.P., 1 1 1  , , are in A.P.. a b c abc abc abc , , are in A.P.. a b a

3  8d 3 A8 3   Given : A  3   n  2 d 5 5 n 2



 5(3 + 8d) = 3[3 + (n – 2)d]  15 + 40d = 9 + 3(n – 2)d

 1

51  6 = [3(n –2) – 40]d = (3n – 46). n 1



bc ca a b , , are in A.P.. a b c

17  2 = (3n – 46).  2(n + 1) = 51n – 782. n 1  49n = 784 or n = 16



a b c , , are in H.P.. bc ca a b

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bc ca ab ,1  ,1  are in A.P.. a b c

Progressions Solutions 31.

311

2 1 1  y  xz

32.

Again apply componendo and dividendo and square

Since a, b, c are in G.P., b2 = ac ---------- (1) Let ax = by = cz = k (say)  a = (k)1/x, b = (k)1/y, c = (k)1/z Substituting the values of a, b, c in (1), we have (k)2/y = (k)1/x (k)1/z = (k)(1/x) + (1/z) 1 1 1  x , y , z are in A.P..

34.

ac ------ (1) 2 and since b, c, d are in H.P.



35.

ab





 



 

m 1 m 1 2

a b    m 1 m 1 a b  

a b a b



 m  1 m  1  m  1 m  1

a : b  m  m 2  1 : m  2 m2  1 Let d be the common difference of the A.P., then d = a2 – a1 – a2 = ..... = an – an – 1

1 d d d      .....   d  a1a 2 a 2 a 3 a n 1a n  1 a  a a  a2 a  a n 1    2 1 3  .....  n  d  a1a 2 a 2a 3 a n 1a n  

m 1

ab m  2 ab 1 Apply componendo and dividendo

    

2

1 1 1 1 Now a a  a a  a a  .....  a a 1 2 2 3 3 4 n 1 n

1  1 1   d  a 1 a 2





a  b  2 ab

m 1 

  1 1   1 1        .....     a2 a3   a n 1 a n

    

1 1 1      d  a1 a n 

2ab

a  b  2 ab

2

 

 

2ab ab

a  b

2

 m  1

m 1  m 1

2 a 2 m  m 1  b 2 m  2 m2  1

48 = 3.  a= 16 Hence, the required G.P. is 3, 6, 12, 24, ... Given a and b are two positive integers and also given that G.M = m. H.M ab  m.

a b a 2

2bd ------ (2)  c= bd multiplying (1) and (2), we get bc =

33.

  b 

a b a b

m 1 m 1 2 2 a    m 1 m 1 2 2 b

a, b, c are in A.P.  b 

 a  c  bd a  c 2bd  2 bd bd  b2c + bcd = abd + bcd  b2c = abd  bc = ad  ad = bc. We have, t7 = 8t4  ar6 = 8ar3, where a and r are the first term and common ratio respectively of the G.P.  r 3 = 8 = 23  r = 2 Also, t5 = 48  ar4 = 48  a .24 = 48  16a = 48

2

 

m 1 m 1

a n  a1 a1   n  1 d  a1 n  1   da1a n d a1a n a1a n

HOTS WORKSHEET 1.

Since, given that nth term of the pattern of numbers is a linear expression  tn = An + B  let it be An + B Hence, t1 = A × 1 + B = A + B t2 = A × 2 + B = 2A + B t3 = A × 3 + B = 3A + B t4 = A × 4 + B = 4A + B t5 = A × 5 + B = 5A + B ..................................................... ..................................................... Here, tn+1 = A (n + 1) + B www.betoppers.com

10th Class Mathematics

312

2.

3.

4.

t2 – t1 = [(2A + B) – (A + B)] = A t3 – t2 = [(3A + B) – (2A + B)] = A t4 – t3 = [(4A + B) – (3A + B)] = A t5 – t4 = [(5A + B) – (4A + B)] = A ..................................................... ..................................................... tn + 1 – tn = [A (n + 1) + B] – [An + B] = A Clearly, the differences between each pair of consecutive terms are the same and equal to A. Hence, the pattern of numbers so formed, is an A.P. and common difference = A = coefficient of n. Let ‘a’ be the first term and ‘d’ the common difference of the given A.P. last term = l kth term from the beginning = tk ––––(1) = a + (k – 1)d ––––(2) kth term from the end = l – (k – 1)d On adding (1) and (2), we get (kth term from the beginning) + (kth term from the end) = [a + (k – 1)d] + [l – (k – 1)d] = a + l = First term + last term Let the same common difference of two A.P. be d and their first terms be a and A respectively. If t100 and T100 are their 100th terms, then t100 = a + (100 – 1)d ––––––– (1)  t100 = a + 99d and T100 = A + (100 – 1)d ––––––– (2)  T100 = A + 99d Taking difference of (1) and (2), we have t100 – T100 = (a + 99d) – (A + 99d)  100 = a – A –– (3) [ t100 – T100 =100] Now, t1000 = a + (1000 – 1)d  t1000 = a+ 999d T1000 = A + (1000 – 1)d  T1000 = A + 999d  t1000 – T1000 = (a + 999d) – (A + 999d)  t1000 – T1000 = a – A [ a – A = 100]  t1000 – T1000 = 100 Hence, the difference between their 1000th terms is 100. Let the A.P. be a, a + d, a + 2d, a + 3d,.... Hence, t2r3 = (a + d)(a + 2d). t1t4 = a(a + 3d). t2t3 – t1t4 = (a + d)(a + 2d) – a(a + 3d) = a2 + 3ad + ad2 – (a2 + 3ad) = 2d2 –––––––––––– (1) 2(t1 – t2)2 = 2 [a – (a + d)]2 = 2[a – a – d]2 = 2d2 –––– (2) From (1) and (2), we get t2t3 – t1t4 = 2 (t1 – t2)2.

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5.

6.

7.

Let the two A.P’s be a1, a2, a3, ...., an, ....and b1, b2, b3,...., bn,.... Let d be the common difference of two A.P’s. Then, an = a1 + (n – 1) d and bn = b1 + (n –1)d  an – bn = {a1+(n – 1)d} – {b1+(n – 1) d}  a n – bn = a 1 – b1 Clearly, an – bn is independent of n and is equal to a1 – b1. In other words an – bn = a1 – b1 for all n  N .  a100 – b100 = a1 – b1 and, ak – bk = a1 – b1, where k = 10,00,000. But, a100 – b100 = 111 222 333  a1 – b1 = 111 222 333  ak – bk = a1 – b1 = 111 222 333, where k = 10, 00, 000. Hence, the difference between millionth terms is same as the difference between 100th terms i.e., 111 222 333. Let a be the first term and d be the common difference of the given A.P. Then, pth term = q  a + (p – 1)d = q ––––– (i) qth term = p  a + (q – 1)d = p ––––– (ii) Subtracting equation (ii) from equation (i), we get (p – q)d = (q – p)  d = – 1 Putting d = – 1 in equation (i), we get a + (p – 1) × (–1) = q  a = (p + q – 1)  nth term = a + (n – 1)d = (p + q – 1) + (n – 1) × (–1) = (p + q – n) Let ‘d’ be the common difference of the given A.P. a2 = a1 + d, a3 = a1 + 2d, a4 = a1 + 3d ........ an–1 = a1 + (n – 2)d, an = a1 + (n – 1)d N o w , 1 1 1 1 1 1     ......   . a1a 2 a 2 a 3 a 3 a 4 a 4 a 5 a n  2 a n 1 a n 1a n



1 1   a1  a1  d   a1  d  a1  2d  1



1

 a1  2d  a1  3d   a1  3d  a1  4d  ......... 



1   a1   n  3 d   a1   n  2  d 

1  a1   n  2  d   a1   n  1 d 



11 1  1 1 1        d  a1 a1  d  d  a1  d a1  2d 

Progressions Solutions

1 1 1  1 1 1        d  a1  2d a1  3d  d  a1  3d a1  4d   1 1 1 ......     d  a1   n  3 d a1   n  2  d   1 1 1    d  a1   n  2  d a1   n  1 d 



11 1 1 1 1       d  a1 a1  d a1  d a1  2d a1  2d

1 1 1    a1  3d a1  3d a1  4d  1 1 1 1 .........      a1   n  3 d a1   n  2  d a1   n  2  d a 1   n  1 d 



 1  a1   n  1 d  a1  11 1      d  a1 a1   n  1 d  d  a1  a1   n  1 d  

 n  1 d

313

 x 8 10. Given 1, log92 (3x + 48) log9  3    A.P. 3  1 log9 (3x + 48) log9(3x – 8/3)  A.P. 2  9, (3x + 48)1/2, (3x – 8/3)  G.P.  3x + 48 = 9 (3x – 8/3) 8.3x = 72 3x = 9 3x = 3 2 x=2 11. Let ‘a’ be the 1st term and ‘d’ the common difference of an A.P. S1 = sum of n terms of the A.P.

 log99,

n 2a   n  1 d  2 S2 = sum of 2n terms of the A.P. =

2n  2a   2n  1 d   n  2a   2n  1 d  2  S3 = sum of 3n terms of the A.P. =

n 1 da1 a n a1 a n 3n  S3   2a   3n  1 d  ––––––––– (1) 8. i) The first number between 100 and 1000 divisible 2 by 7 = 105 and last number = 994. Now, R.H.S = 3(S2 – S1) So, the numbers are 105, 112, 119, ..., 994. n   Here a = 105, d = 7, tn = 994, n = ?  3 n 2a   2n  1 d  2a   n  1 d 2   Now, tn = a + (n – 1)d  994 = 105 + (n – 1).7  7n = 896 n  3  4a  2  2n  1 d  2a   n  1 d  n = 128 2 ii) Number of integers not divisible by 7 = Total number of integers between 100 and 1000 3n   4a   4n  2  d  2a   n  1 d  – number of integers divisible by 7 2 = 899 – 128 = 771. 3n 9. Let ‘S’ be the required sum. Then,  2a   4n  2  n  1 d  2 S = 1 + 3 + 7 + 9 + 11 + 13 + 17 + 19 + 21 + 23 + 27 + 29 + ... + 991 + 993 + 997 + 999. 3n   2a   3n  1 d   S3 [From (1)] = (1 + 3 + 7 + 9) + (11 + 13 + 17 + 19) + (21 + 23 2 + 27 + 29) + ... + (991 + 993 + 997 + 999) = L.H.S. [ making groups of four terms each] Therefore S3 = 3(S2 – S1) = 20 + 60 + 100 + ... + 3980, which is an A.P., whose first term is 20, common difference is 40 12. Let first term = a and the common difference = d. and the last term is 3980. Let the number of terms be ‘n’, then 1 1 Given, q  t p and p  t q 3980 = 20 + (n – 1).40  40(n – 1) = 3960  n – 1 = 99  n = 100  tp = a + (p – 1)d 100 1  20  3980  = 50 × 4000  The required sum =   a   p  1 d –––––– (1) 2 q n    tq = a + (q – 1)d = 200000.  Sn   a  l   2   www.betoppers.com 



10th Class Mathematics

314



travelled by the policeman.

1  a   q  1 d –––––– (2) p

100  n  1 

Subtracting (1) from (2), we get

1 1    q  1 d   p  1 d p q 

q  p pq

Putting d 

 q  p d

 a

1 pq

 d

1 pq –––– (3)

 a

1 1 1   q q pq

–––––––– (4)

Now, sum of the first pq terms



pq  2a   pq  1 d  . 2 

 Spq 

pq  2 1    pq  1  .  2  pq pq 

[From (3) and (4), a  d 

 Spq 

1 ] pq

pq  2  pq  1  1     pq  1 2  pq  2

13. Let the policeman catch the thief in n minutes. Since the thief ran one minute before the police, therefore the time taken by the thief before being cought = (n + 1) minutes Distance travelled by the thief in (n +1) minutes = 100 (n + 1) metres In first minute, speed of policeman = 100 m/minute. In second minute, speed of policeman = 110 m/minute. In third minute, speed of policeman = 120 m/ minute and so on. Speeds: 100, 110, 120, ...... form an A.P. Total distance travelled by the policeman in n minutes

n  2  100   n  110 2 on catching the thief by policeman. Distance travelled by the thief = Distance 

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n  n  110 2  100 = n(n – 1)5  n2 – n – 20 = 0  (n – 5) (n + 4) = 0  Either n – 5 = 0  n = 5 or n + 4 = 0  n = – 4 not possible. Hence, time taken by the policeman to catch the thief = 5 minutes. 14. We have, IMAGE d1 = Distance run by the competitor to pick up first potato = 2 × 5m d2 = Distance run by the competitor to pick up second potato = 2 (5 + 3)m d3 = Distance run by the competitor to pick up third potato = 2 (5 + 2 × 3)m dn = Distance run by the competitor to pick up nth potato = 2 (5 + (n – 1) × 3)m  Total distance run by the competitor to pick up n potatoes = d1 + d2 + d3 + ......+ dn = 2 × 5 + 2(5 + 3) + 2(5 + 2 × 3) + 2(5 + 3 × 3) + ...... + 2{5 + (n – 1) × 3} metres = 2[ 5 + {5 + 3} + {5 + (2 × 3)} + {5 + (3×3)}+ ... +{5 + (n – 1) × 3}] = 2[ (5 + 5 + ...... + 5) + {3 + (2 × 3) + (3 × 3)+ ..... + (n – 1) × 3}n – 1times] = 2[5n + 3 {1 + 2 + 3 + ....+ (n – 1)}]  100 n  100  100 n

1 from (3) in (1), we get pq

1 1  a   p  1 . q pq

n  2  100   n  110  2

   n 1   2 5n  3   1   n  1  2   

n    Usin g : Sn  2  a  l   3n  n  1   2 5n   2   = [10n + 3n (n – 1)] = 3n2 + 7n = n(3n + 7) metres 15.  A is the single A.M between a and b.

ab 2 Let A1, A2, ....... An be n A.Ms between a and b.  a, A1, A2 ......... An, b is an A.P with common A 

difference d 

ba . n 1

Progressions Solutions

315

Now, S = A1 + A2 + ....... + An

n = (A1  A n ) ––––––––– (1) 2 A1 is the 2nd term  A1 = A + d ––––––––– (2) An is (n + 1)th term An = a + [(n + 1) – 1]d = a + (n + 1)d – d

18.



1 d 2  ............... (1) 2 ,  are roots of Bx – 6x + 1 = 0



n n (a  d  b  d)  (a  b) 2 2

n [2(a + b)2 + (n – 1)(–2ab)] 2

n = .2[a2 + b2 + 2ab – (n –1)ab] 2 = n(a2 + b2) + nab(3 – n)

   1 1 6/ B     6 or    1/ B

1  2d  3 ......... (2)  From (1) and (2), on solving, we get 1  1, d  1  

 b  2a  A'm = mth mean = 2a  m   . It is given that  n 1  Am = A'm

=

or

or

 2b  a  A m  m th mean  a  m   ............. (i)  n 1  When n arithmetic means are inserted between 2a and b, we have

a m   b n  m 1 17. The given series is an A.P. with first term = (a + b)2 and common difference = (a2 + b2) – (a + b)2 = –2ab  Sum of n terms

1 1   2d  4  

1 1  d  3d  6  

S ab S  n    nA   n A  2  16. (C) When n arithmetic means are inserted between a and 2b, we have

 2b  a   b  2a   a  m  2a  m     n 1   n 1   m(2b –a) = a (n + 1) + m(b – 2a)  a(n + 1 – 2m + m) = b(2m – m)

   4/A 1 1   4 or  4  1/ A  

i.e.,

 ba  = a +  n +1  d  n +1  =a+b–a–d  An = b – d ––––––––– (3) Substitue (2) and (3) in (1) we get 

1 1 1 1 , , , are in A.P     Let d be the common difference of this A.P Now, ,  are roots of Ax2 – 4x + 1 = 0

1 1 1 1  1,  2,  3,  4    

Since,

1 1  A  A = 3, Also,  B,  B = 8.   Hence A = 3 and B = 8

19. 2

2

1  1   1   Sn   x     x 2  2    x 3  3  x x x       1   ...   x n  n  x  

2

2

1 1 1         x2  2  2    x4  4  2    x6  6  2  x x x       1   ...   x 2n  2n  2  x  

1 1 1   1   x 2  x 4  x 6  .....  x 2n    2  4  6  ....  2n  x x x  x   2  2  2  ...n times 

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10th Class Mathematics

316







x2 1   x2 

n



1  x2

x 2 1  x 2n  1 x2



21. Let G.P. be a, ar, ar2, ar3,......... since, the G.P. is infinite and decreasing 0 < r < 1 and a > 0. According to the hypothesis ar2, 3a.ar3, ar are in

n 1   1   1      x 2   x 2     2n 1 1 2 x

1 A.P. with common difference . 8

x 2n  1  2n x 2n  x 2  1

1  x 2n  2 1   x  2n   2n 2  1 x  x  20. Let a and r be the first term and common ratio of the G.P. respectively.

 Sn 

a 1  r 2  1 r



a2

1  r 

2

10

.......(1)

2 and ar  ar  2.

1 8

.......(2)



a 2  a  2d 

From (2), we get a 

2

 1 + r 20 

 r10  r10  1   a 2 r 20  r10  1

L.H.S. 

2

1 or a  4r 1  r 







a 1  r

1  r  10

1 r a 1  r



4 r  r2



3r 3 16r 2 1  r 

2

.......(3)

From (1) and (3) we have

2

2

2



r2 1  4r 1  r  8

1 which gives r  , 2 but 0  r  1 2 From (3), we have a = 1

_______

1 r



a 2 1  r10 

1  r 

2

1 r



2



b c d   r. a b c  b = ar, c = br = ar.r = ar 2, d = cr = ar2.r = ar3

22. As a, b, c, d are in G.P., we get

From (1) and (2) we get (S10 – S20)2 = S10(S30 – S20).

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b  c ar  ar 2 ar 1  r  a) Now, a  b  a  ar  a 1  r  r.  

a 2 r 20  r10  1

1  r 

1 2

Hence, S  2

r 20 1  r 30  2

1 2

1 1

a 1  r 30  1  r 20   1 r

a 2 r 30 1  r10 

1  r 

20

1 r



10

 S 

  a 1  r   a 1  r   30

 r

1 1 1 Hence, G.P. is 1, , , ,...... 2 4 8

(1)

R.H.S. = S10 (S30 – S20)



1

 2r 2  3r  2  0

a2

1  r 

1  r

1 8

, nN

 a 1  r10  a 1  r 20   L.H.S. = (S10 – S20)2 =  1  r  1  r   

=

 3a 2r 3  ar 2 

2

2

_______

(2)

c  d br  cr r  b  c    r bc bc bc

bc cd  .P.  a + b, b + c, c + d are in G.P. ab bc b) L.H.S. = a(ar – ar2)3 = a4r3(1 – r)3 R.H.S = ar3(a – ar)3 = a4r3(1 – r)3  L.H.S. = R.H.S. 

Progressions Solutions

317

23. Let the numbers be a and b. Then a + b = 100 ---------- (1) ab 5  and 2 ab 4

25. Since a,b,c,d are in H.P.,

1 1 1 1 , , , are in A.P. Let D be its common a b c d difference. 

------------- (2)

From (2), 2(a + b) = 5 ab

Now D =

 2 × 100 = 5 a 100  a 

 40 =

Thus ab 

a 100  a 

[ a + b = 100, b = 100 – a] Squaring both sides, we get 1600 = a(100 – a)  1600 = 100a – a2  a2 – 100a + 1600 = 0  (a – 80) (a – 20) = 0  a = 80 or a = 20 when a = 80, b = 100 – 80 = 20 and when a = 20, b = 100 – 20 = 80 Hence, the numbers are 80 and 20. 24. Let a be the first term and d the common difference of the corresponding A.P. Given, pth and qth terms of H.P. are respectively, 'qr' and ‘rp’. Then the p th and qth term of the 1 1 and corresponding A.P. will be , respectively.. qr pr

For A.P. the nth term = a   n  1 d

1 1  b  c  and cd   c  d  D D Adding (1),(2) and (3), we get ab  bc  cd 



3ad  a  d  1  3ad a  d   a d  a  d  /  3ad 

26. Since a,b,c are in H.P.

2ac a c

1  qth term of A.P.  a   q  1 d ––– (2) and pr subtracting (1) from (2), we get

Now,

———— (3)

1 1  a  d   1/ d  1/ a  a  d  D     2 1

ba     d  n 1   

b

1 1 q p    q  p  d or  q  p d pr qr pqr

1 a  b D

Similarly bc 

1   p th term A.P.  a   p  1 d —— (1) qr

1 d  p  q pqr

1 1 ab 1    D  ab   a  b  b a ab D



1 1 1 1    2ac 2ac bc a b c a ac a c

ac a c a c ac  2   2 ac  c a  ac c  a  c  a  a  c 

 a  c  1 1  a  c a  c        a  c  c a  a  c ca 2

1 1 p 1 From (1), a    p  1 d   qr qr pqr p  p 1 1  ———— (4) pqr pqr Now the rth term of the corresponding A.P. a

1 r 1 1  r 1 1    pqr pqr pqr pq th  r term of H.P. = pq.  a   r  1 d 

 a  c   1 . a 2  c2  2ac 1   a c ca a c ca 

1 a c 1  a c      2    2   4    a c c a  a  c  c a  

2  1  a c 4       4   a  c  c a  a c   2   a  c     0    c a    

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10th Class Mathematics 1 , are in A.P..  a a b 1 1   c a b c a

318 27. We have, 2y = x + z --------- (1) [ x, y, z are in A.P.] b2y2 = ax.cz ------ (2) [ ax, by, cz are in G.P.]

29.

2ac ------ (3) a c [ a, b, c are in H.P.] Substituting for ‘y’ and ‘b’ from (1) and (3) in (2), b=

 x  z 2

4a 2 c 2



  c  a    c  a   a  b a  b  c   a  b  c  a 

b c 



c

x z a c x z a c  2   2     z x c a z x c a The given series is a geometric series with

b a c b   b – a = c – b. b a a b  a, b, c are in A.P., which is given. Hence the result. No. of letters in the 1st set = 4 (These are letters sent by Karthik) No.of letters in the 2nd set = 4 + 4 + 4 + 4 = 16 No.of letters in the 3rd set=4+4+....16 terms =64 ––––––––––––––––––––––––––––––––– The number of letters sent in the 1st set, 2nd set, 3rd set ...... are respectively 4, 16, 64.....wich is a G.P. with

a  3 and r  3   1 Let n be the number of terms whose sum

a = 4, r 

we get

 

 a  c

 x  z 2 xz

2

4

 axcz

 a  c 2



30.

ac

x 2  z 2  2xz a 2  c 2  2ac  xz ac



28.

1 1 , b c c 1 1  c a b 

16 4 4  Total no.of letters written in all the first 8 sets = S8 of the above G.P.

 39  13 3,

39  13 3  3



 39  13 3



 3

n

1



3 1

3 1  3  



 3

 39 3  39  39  13 3  3    26 3  3  

 3

 3

n



 3

n





     3

n

n

n

 3

 1 

 1  26

n

 27  33



 3   3

3

2 n

6

n=6  The required number of terms = 6

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 1  n

 1 

4  48  1 4 1

 87, 380

25 100 = Rs. 21, 845

Total money spent on letters = 87, 380 

31.

Let the edges be

a , a, ar, where r > 1 r

From the data

a .a.ar  216 r a3 = 216 a 3 = 63 a=6

a  a 2  .a  a.ar  .ar   252 r  r 1 7 1  r ,r2 r 2 2  a = 6, r = 2 So the longest side = ar = 6.2 = 12.

  r 1

Progressions Solutions 32.

Since x, y, z are in H.P. therefore

33.

2zx  y (x + z) = 2zx –––––– (1) xz  L.H.S. = log(x + z) + log (x – 2y + z) = log[(x + z) (x – 2y + z)] = log[(x + z)2 – 2y(x + z)] = log[(x + z)2 – 2.2zx] = log(x2 + z2 + 2xz – 4xz) = log(x2 + z2 – 2xz) = log(x – z)2 = 2 log (x – z) The given equation is Ax2 + Bx + C = 0, Where A = a(a – b), B = b(c – a), C = c(a – b)  A+ B + C = 0 Discriminant = B2 – 4AC = (– A –C)2 – 4AC = (A – C)2 Roots are equal  (A – C)2 = 0  A=C  a(b – c) = c (a – b)  2ac = ab + bc

319 a b 1 H1  a H n  b H1 Hn    a b Now, H1  a H n  b 1 1 H1 Hn 1

y

35.

1.

Thus,

1 1 1 1   d and  d H1 a Hn b



a b  1  ad and  1  bd H1 Hn

The two quantities be a and b. Then a, A1, A2, b are in A.P.  A1 – a = b – A2 [Each = common difference]  A1 + A2 = a + b Also a, G1, G2, b are in G.P.

G1 b  a G2

 G1G2 = ab Again a, H1, H2, b are in H.P. 1 1 1 1 , , a H1 H1 b are in A.P..

 , 

1 1 1 1    H1 a b H 2

[Each = common difference]

Let d be the common difference of this A.P., then

1 ab . n  1 ab

2a  abd  2b  abd 2  a  b   abd   abd abd



1 1 1 1 1 , , ,......, , a H1 H 2 H n b are in A.P..

 d



IIT JEE WORKSHEET

2ac ac a,b,c are in H.P. Let the four consecutive terms of the A.P. be a – 3d, a – d, a + d, a + 3d So that the common difference is 2d. Now, (2d)4 + (a – 3d)(a – d)(a + d)(a + 3d) = 16d4 + (a2 – 9d2) (a2 – d2) = 16d4 + a4 – 10d2 a2 + 9d4 = 25 d4 – 10d2a2 + a4 = (5d2 – a2)2 = square of an integer As a, H1, H2,......, Hn, b are in H.P.,

1 1    n  1 d b a

1  1  ad 1  1  bd 2  ad 2  bd    1  1  ad 1  1  bd ab bd

2  n  1 dab  abd     2n abd

b

34.



2.



1 1 1 1    H1 H 2 a b



H1  H 2 A1  A 2 GG A  A2   1 2  1 H1H 2 G1G 2 H1H 2 H1  H 2



H1  H 2 a  b  H1H 2 ab

Given a1 = 2  a = 2 –––––––––– (1) a10 = 3  a + 9d = 3 –––––––––– (2) Subtracting (1) from (2), we get 9d = 1

 d

1 9

1  

 a4 = a + 3d = 2 + 3  9  2

1 7  = h1, h2, h3, .......... h10 are in H.P.. 3 3 www.betoppers.com

10th Class Mathematics

320

 a

h1 = 2

1 –––––––– (A) 2

4.

1  3 –––––––– (B) a  9d Subtracting (A) from (B), we get,

 x 7  2 log3 (2x – 5) = log3 2 + log3  2   2 

1 1 1 9d    3 2 6

 x 7  log3 (2x – 5)2 = log3 2  2   2 



h10 = 3

h7 

 d

1 54

1 1  a  6d 1  1   6   2  54 



 x 7  (2x – 5)2 = 2  2   2  Let 2x = a

1 1 18   1 1 7 17  2 9 18

7 18 6 3 7

 a 4h7   3.

 x 7 Given log3 2, log3(2x – 5), log3  2   are in A.P.. 2 

Tm 

1 n

 a   m  1 d  Tn 

1 –––––––– (1) n

1 m

1 –––––––– (2) m Subtracting (1) from (2) we get,  a   n  1 d 

 n  m d   d

 a 

1 mn

1 1   m  1 n mn

1 1 1 1    n n mn mn

 Tmn  

1 1 nm   m n mn

1 1   mn  1 mn mn

1 1 1 1 mn mn

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5.

7   (a – 5)2 = 2  a   2   a2 + 25 – 10a = 2a – 7  a2 – 12a + 32 = 0  a2 – 8a – 4a + 32 = 0  (a – 8) (a – 4) = 0  a = 4 or 8 If a = 8 If a = 4 x 3 2 = 8 = 2   2x = 4 = 2 2  x=3  x=2 x = 2 does not satisfy becasue in the second term if we subsititute we get log(–1) as log of negative value does not exist.  x = 3. Sum of the interior angles of a polygon of ‘n’ sides = (2n – 4) 90° = (n – 2) 180° Also a = 120° d = 5° 

n [2(120°) + (n – 1)(5°)] = (n – 2)180° 2

n (240° + 5n – 5) = (n – 2)180° 2  n[5n + 235] = (n – 2) 360°  5n2 + 235n – 360n + 720 = 0  5n2 – 125n + 720 = 0  n2 – 25n + 144 = 0  (n – 9)(n – 16) = 0  n = 9 or 16 When n = 16 Tn = a + (16 – 1)d = 120 + 15 × 5 = 195 This is not possible as interior angle cannot be greater that 180°.  n=9 

Progressions Solutions 6.

321 Here first term = a – 57 Common difference, d = 2

Let the two numbers be ‘a’ and ‘b’

ab 2  a + b = 2A  A

 G2 = ab

G  ab Now, H 

Sum to n terms =



2ab 4 ab

27 9  6 2

ab 9   2 2 G2 = 4A = 4×

7.

8.

or

n n  2  57    n  1 2  114  2n  2  2 2

n (2n + 112) = n(n + 56) ––––– (2) 2 Given (1) = (2)  n(6n + 1) = n(n + 56)  6n + 1 = n + 56  5n = 55  n = 111 Given a, b, c, d are in A.P.  d, c, b, a are in A.P.. 

G2 = AH Gives G2 = 4A Also 2A + G2 = 27  2A + 4A = 27

 A

n 2a   n  1 d  2

9. a+b=9

9 =18 2



d c b a , , , are in A.P.. abcd abcd abcd abcd

1 1 1 1  , , , are in A.P..  ab = 18 abc abd acd bcd a+b=9 ab = 18  abc, abd, acd, bcd are in H.P.. The equation formed is 10. Let t1 = log a, t2 = log b x2 – x(9) + 18 = 0  d = t2 – t1 = log b – log a = log(b/a)  x2 – 6x – 3x + 18 = 0 n  x(x – 6) – 3(x – 6) = 0 Sn   2a   n  1 d  2  x = 3, 6  numbers are 3, 6. n b Given for an A.P. and G.P. pth term = x   2a   n  1 log  p–1 2 a  a + (p – 1)d = x = AR th q term = y n b   2log a   n  1 log   a + (q – 1)d = y = ARq – 1 2 a rth term = z  a + (r – 1)d = z = ARr – 1  n 1  n b 2 Now, y – z = (q – r)d   log a  log     2  a z – x = (r – p)d  x – y = (p – q)d n xy – z . yz – x . zx – y n a 2 b n 1    log n 1   log  a 3n bn 1  2 p – 1 (q – r)d q – 1 (r – p)d r – 1 (p – q)d = (AR ) × (AR ) × (AR ) 2 a  d(q – r + r – p + p – q) ((p – 1)(q – r) + (r – 1)(p –q) + (q – 1)(r – p)) =A ×R = A0 × R pq – rp – q + r + rp – pq – r + p + rp – rq – p + q 11. Let ‘d’ be the common difference of the given A.P. = A0 × R0 = 1  d = a2 – a1, d = a3 – a2, ..... d = an – an – 1 Given A.P. is 2, 5, 8, ....... first term = a = 2 common difference = d = 3 a a  d n 1 n 1 2n  2a   2n  1 d  Sum to 2n terms = 2   1 1  .... = n[2(2) + (2n – 1)(3)]   Now a n 1  a n   a1  a 2 = n(4 + 6n – 3) = n(6n + 1) –––––– (1) Another A.P. is 57, 59, 61,......... www.betoppers.com

10th Class Mathematics

322  a 2  a1 a  a n 1  .... n  a1  a 2 a n 1  a n  d

=



 

 a n  a1 





a1 + a 2 +

a 3 - a 2 +...+

14. Given x = 1+ y + y2 + ......... 

  



 x

a n + a n-1



2mr mr

m  rn

 m  rn

 2n  m  r  a   



ab  G  ab = G2

Now, harmonic mean = 2G2 

2

  n2  d 

a n  m  r  2n  n    d 2  2n  m  r  2

common difference 2 d 2    first term n a n www.betoppers.com

2ab ab

G2 2A A 16. Let ‘d’ be the common difference of A.P. and ‘r’ be the common ratio of G.P., then the A.P. is (r) + (r + d) + (r + 2d) + ........ and the G.P. is d + dr + dr2 + ........ Given S10 of A.P. = 155 



10 (2r + 9d) = 155  2r + 9d = 31 2 and also given that S2 of G.P. = 9  d + dr = 9  d(1 + r) = 9 On solving these, we get r = 2 and d = 3 

25 2 d and 2 3 A.P. is 2 + 5 + 8 + .......  and G.P. is 3 + 6 + 12 + ........ r

or A.P. is

2 Substituting this value of mr in (1), we get



ab A 2  a + b = 2A Geometric mean = G 

12. Let the angles of a quadrelateral be a, a + d, a + 2d, a + 3d. Difference between (a + 3d) and ‘a’ is 90.  3d = 90°  d = 30° Also sum of angles is 360°.  a + a + d + a + 2d + a + 3d = 360°  4a + 6d = 360°  4a = 360° – 6d  4a = 360° – 6(30°) = 180°  a = 45°  angles of a quadrelateral are a = 45° a + d = 45° + 30° = 75° a + 3d = 45° + 60° = 105° a + 3d = 45° + 90° = 135° 13. Let ‘a’ be the first term and ‘d’ be the common difference of the A.P. then, Tm + 1 = a + md Tn + 1 = a + nd Tr + 1 = a + rd Since Tm + 1, Tn + 1, Tr + 1 are in G.P.  T 2n + 1 = T m + 1. T r + 1  (a + nd)2 = (a + md)(a + rd)  2and +n2d2 = ard + amd + mrd2  (2n – m – r)a = (mr – n2)d ––––––– (1) Given m, n, r are in H.P.



1 x

 y  1

a n  a1

 mr 

 1 y 

1 x 1  x x 15. Let the two numbers be ‘a’ and ‘b’ Given Arithmetic mean = A

a n  a1 a n  a1

 n  1 d

 n

1 1 y

25 79 83    ....... 2 6 6

2 25 625    ....... 3 3 6 17. Given pth term = q  a + (p – 1)d = q ––––––––– (1) qth term = p  a + (q – 1)d = q ––––––––– (2) Subtracting (1) from (2), we get, (q – 1– p + 1)d = p – q  (q – p)d = p – q and G.P. is

Progressions Solutions

 d=–1  a = q – (p – 1)(– 1) = q – (– p + 1) =q+p–1  rth term = a + (r – 1)d = q + p – 1 + (r – 1)(–1) = q + p – 1– r + 1 = q + p – r 18. Odd numbers are 1, 3, 5, .......... n terms first term = a = 1 Common difference d = 3 – 1 = 2 n  sum to n terms = 2a   n  1 d  2 n   2 1   n  1 2  2 

n  2  2n  d  2

n  2n   n 2 2 19. Given 5th term of G.P. = 2  ar4 = 2 –––––––– (1) Product of first 9 terms is (a) × (ar) × (ar2) × (ar3) × (ar4) × (ar5) × (ar6) × (ar7) × (ar8) = a9r36 = (ar4)9 =(2)9 = 512 20. Given third term = 12  a + 2d = 12 –––––– (1) and seventh term is 24  a + 6d = 24 –––––– (2) Subtracting (1) from (2), we get, 4d = 12  d=3  a = 12 – 2d = 12 – 2(3) = 6  10th term = a + 9d = 6 + 9(3) = 6 + 27 = 33 21. Given fourth term = 4  a + 3d = 4 Sum of first 7 terms 



7  2a   7  1 d  2

7   2  a  3d  2 

7  2  4   28 2

323 22. Given p, q, r are in A.P.  2q = p + r ––––––– (1) Given x, y, z are in G.P.  y  xz 1

1

 y  x 2 .z 2 ––––––– (2)

Using (2), we get xq – t.yr – p.zp – q.

x

q r

.x

r p 2

.z

r p 2

2q  2r  r  p 2

.zp q

r  p  2p  2q 2

2q  r  p 2

x .z x = x°.z° =1 23. Let the roots be ‘a’ and ‘b’ Given A.M. is 8

.z

p  r  2q 2

ab 8 2  a + b = 16 –––––– (1) G.M. is 5 

 ab  25  ab = 25 –––––– (2)  sum of the roots = 16 product of the roots = 25  equation is x2 – x(16) + 25 = 0 24. Given a, b, c, d, e, f are in A.P.  b–a=c–b=d–c=e–d=f–e Now e – c = (e – d) + (d – c) = (d – c) + d – c = 2(d – c) 25. Given 7th term = 34  a + 6d = 34 –––––––– (1) 13th term = 64  a + 12d = 64 –––––––– (2) Substracting (1) from (2) we get, 6d = 30  d=5 a = 34 – 6d = 34 – 6(5) = 4  18th term = a + 17d = 4 + 17(5) = 89 26. Let the two numbers be ‘a’ and ‘b’. Given A.M. is 5

ab  5  a + b = 10 2 G.M. is 4 



ab  4  ab = 16

Now H.M. 

2ab 2 16  16   ab 10 5 www.betoppers.com

10th Class Mathematics

324 27. Let ‘a’ be the first term and ‘d’ be the common difference. Given first term = 1  a=1 last term = 11  a + (n – 1)d = 111 Sum of the n terms 

 36 

n  2a   n  1 d  2

n 2a   n  1 d  2

n a  a   n  1 d  2  72 = n(1 + 11)  72 = 12n

1 q

 a    p  1 d 

1 1   p  1 q pq



1 1 1 1    q q pq pq

 (pq)th term 

 36 

72 6 12  number of terms are 6, 28. Let first term be ‘a’ nad common ratio be ‘r’. Given second term = 2

1 a   pq  1 d

1 1 1   pq  1 pq pq



 n

 ar = 2

2  r a

Sum of its infinite terms =

a 8 1 r

 a = 8 – 8r 

 2 a = 8 – 8  a

16 a  a2 = 8a – 16  a2 – 8a + 16 = 0  (a – 4)2 = 0  first term is 4. 29. In H.P. pth term = q

1 1 1 1 pq pq

1

30. Given 3rd term of a G.P. = 3  ar2 = 3 Product of first five terms = (a)(ar)(ar2)(ar 3)(ar4 ) = a5r10 = (ar2)5 = (3)5 = 243  The product of first 5 terms = 243. 31. Given fourth term of H.P. = 5



 a 8





1 5 a  3d

 a  3d 

1 ––––––– (1) 5

5th term = 4

 a=4

1 q a   p  1 d



1 4 a  4d

1 ––––––– (2) 4 Subtracting (1) from (2) we get,  a  4d 

1 1 1   4 5 20

1  a   p  1 d  ––––––––– (1) q

d

1 Similarly  a   q  1 d  p ––––––––– (2)

 a

Subtracting (1) from (2), we get,

 20th term 

 q  p d 

1 1 qp   p q pq

1 pq www.betoppers.com

 d



1 1 3 1  3d    5 5 20 20 1 a  19d

1 1 19  20  1  20 20 20

Progressions Solutions 32. H is harmonic mean between ‘a’ and ‘b’.

 H

2ab ab

H 2b H 2a   and a ab b ab Applying componendo and dividendo, we get, 

Ha 2b  a  b 3b  a   H  a 2b  (a  b) b  a and

H  b 2a   a  a  3a  b   H  b 2a  (a  b) a  b

H  a H  b 3b  a 3a  b    Now, Ha Hb ba ab 3b  a  3a  b 2a  2b  2 ab ab 33. Let the two geometric means be g1 and g2 1, g1, g2, 64 are in G.P.  a=1 a r3 = 64  1.r3 = 64 = 43  r=4 g1 = ar = 1(4) = 4 g2 = ar2 = 1(4)2 = 16 The two geometric means are 4 and 16. 

325 34. Given A.P. is 3, 10, 17, ....... nth term = 3 + (n – 1)7 = 3 + 7n – 7 = 7n – 4 Another A.P. is 63, 65, 67, .......... nth term = 63 + (n – 1)2 = 63 + 2n – 2 = 2n + 61 Given 7n – 4 = 2n + 61  5n = 65  n = 13 35. Given ‘c’ is the harmonic mean between ‘a’ and ‘b’.

 c

2ab ab



c a  b 2 ab



ca cb  2 ab ab



c c  2 b a

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326

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10th Class Mathematics

6. FUNCTIONS SOLUTIONS

1.

Given, n(A) = 5, n(B) = 2 Number of relations from A to B n(a) × n(B)

=2 5×2

=2 10

2.

=2 = 1024. Given, n(A) = 5 and n(B) =3. Then, number of functions from A to B

5.

n(A)

= n(B)

1 2  3  x  0

5

3.

b 0 2a b  x > – 1 and x  2a  b   x   1,       2a   b    x  R         ,  1    2a   To define f(x), 3  x  0, 2  3  x  0, and  x + 1 > 0 and x +

FORMATIVE WORKSHEET

=3 = 243 If we write the above relation in terms of an ordered pair, we get

( square root of negative is imaginary)

3  x, 2  3  x and 1  2  3  x 3  x, 4  3  x and 1  2  3  x 3  x, x  1 and 3  x 1 3  x, x  1 and 3  x 1 3  x, x  1 and x  2 The domain of the real function f  x   1  2  3  x is –1  x  2.

f = {(2, 5), (3, 4), (4, 3), (5, 2)} 4

6.

The set of ordered pairs will satisfy f ={(x, y): x + y = 7} 4

4.

For domain of the given function, 3

2

we must have, ax + (a + b)x + (b + c)x + c > 0. 2

 (ax + bx + c) (x+1) > 0 ( logarithm of negative number does not exist) b c   a  x 2 + x +   x +1 > 0 a a  b2 By adding and subtracting 4a 2 2  b   c b2      x +  +  - 2    x +1 > 0  2a   a 4a    2  b   4ac  b2    x + +    x +1 > 0  2a   4a2    2

2 b   x +   x +1 > 0 ( a > 0 and b – 4ac = 0) 2a  

7.

5x  1 ( y = f(x)) 2x  7 5x  1 then y  2x  7 y (2x + 7) = 5x – 1 x (2y – 5) = – 1 – 7y 7y  1 x 5  2y Clearly, the above fraction does not exist when 5 y= 2 5  The range of f = R    2 f (x)  R  5 + x > 0, 5 – x > 0  x + 5 > 0, x – 5 < 0  x > – 5, x < 5  x  (– 5, 5) Given f(x) 

n

8. i) P  R r

 n > 0, r  0 and n – r > 0. Also, n, r  N.  7 – x > 0, (x – 3)  0 and (7 – x) – (x – 3) > 0  x – 7 < 0, x  3 and 7 – x – x + 3 > 0

____________________________________________________________________________________________________ _____________________________________________

Functions Solutions

328

 x < 7, x  3, and 10 – 2x > 0  x < 7, x  3, and x – 5 < 0  x < 7, x  3, and x < 5  x  3 and x < 5  x = {3, 4, 5} ii) We have domain A of f(x) = {3, 4, 5} Now range of f(A) = {f(3), f(4), f(5) 



7  3

P3 3 

7 4

P4  3 

7 5 

P5 3



 n!    4 P0 3 P1 2 P2   n Pr    n  r !  

= {1, 3, 2}  4 P0  1, 3 P1  3 and 2 P2  2  9. 1)

Let f(x) =

 x  x 2  1    x   x 2  1  g  x   1  x  1  x  x   x 2  1 Domain of f(x) =  0,   Domain of g(x) =  0, 1

2)

3)

 Functions are not equivalent ( Domain of ‘f’  Domain of ‘g’ ) Let h(x) = 2 x3  x x + x, u(x) = x Domain of ‘h’ =R , Domain of ‘u’ =R – {0}  functions are not equivalent. Let v(x) = x 2  1  x and w(x) = x2 1  1 x  x  1  x Domain of ‘v’ =  0,  

2

2

 x = y(1 + x ) 2

 x (1 – y) = y y  x2  1 y y is a real number. 1 y Thus, for every y  [0, 1), there exist real numbers such that f(x) = y.  y belongs to the range of f(x).  [0, 1) is a subset of range of f(x). ------- (2) From (1) and (2), range of f(x) = [0, 1) 1 f  1  f 1  2  f(x) is not one-one . 12. Here, n(A) = 4, n(B) = 6  Number of one-one functions  x

=

 v, w are not equivalent. 10. To find whether the function is one-one or not, we should see that no two elements have same image. Without loss of generality, let us randomly assume that f(x) = 2. x 2  4x  30  2 2 x  8x  18 2

 x + 4x + 30 = 2x – 16x + 36 2

10  94 and 10  94 have the same image 2  ‘f’ is not one-one. x2 11. f  x   is defined for all real values of x. 1  x2  Domain of f = R. f(x) is non-negative and f(x) = 1 x2 1  x2 1   1 1 2 2 1 x 1 x 1  x2  Range of f(x) is a subset of [0, 1) ------ (1) Let y  [0, 1) and f(x) = y x2  y 1 x2

n(B)

Domain of ‘w’ =  0, 1

2

 f  10  94   2  f 10  94 

2

 2x – 16x + 36 – x – 4x – 30 = 0 2

 x – 20x + 6 = 0 20  400  24  x 2  10  94

P

n(A)

6

= P

4

= 6 × 5 × 4 ×3 = 360 13. If m, n are the number of elements in first and second set respectively, then n = 4 and m = 3. Number of many-one functions m

n

=n – P 3

m

4

=4 – P

3

= 64 – 24 = 40 14. Here, n (A) = a and n (B) = 2. Number of onto functions from A to B = n(A)

n(B)

–n(B)

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

329 5

15.

16.

17. 18.

=2 –2 = 32 – 2 = 30 Here, n(A) = 4 and n(B) = 3.  n(A) ≠ n(B). The number of bijections from A to B is zero when n(A)  n(B).  Number of bijections from A to B is 0. n 1 Given f  n   2 When n is odd Put n =1  f(1) = 0 Put n =3  f(3) = 1 Put n = 5  f(5) = 2 n f n  2 When n is even Put n =2  f(2) = –1 Put n = 4 f(4) = –2 Put n = 6  f(6) = –3 Here, all the elements have different images in the domain and all the elements of co-domain have preimages.  f: N  Z is one-one onto. Number of identity functions from A to A is always one. Number of constant functions from A to B is n(B) = 5. 2

19. Given, f(x) = (x + 1)  2  x  2



2

g  f  x    g  x  1  2

 2 

 x  1

2

– 2 and



22

 2  x  1

g(x)

21. f(x) = 2x + 1 y = 2x + 1 y 1 x 2 –1

f( x ) = y  x = f (y) y 1 f 1  y   2 x 1 f 1  x   2 1  hof   x   h  f 1  x  

 x 1  h   2   x 1 x 1    2  2

 hof   2   2 2 1  

1 1   2 2 1  2 2  1  22. f 1    3x  2  x 1  1  f  3x  2   x 1

a2 3 3 3 1 Now f(a)    f a  a2 a5 1 a  5 3 3 Hence, f  x   x5 Let a = 3x – 2. Then x 

2x + 3

Since we have x  1, x  1 = – (x + 1) Therefore, g(f(x)) = –2 – (x + 1) =–3–x 20. f(g(x)) =f(x + k)

23. Given, f(x) = e Let y = f(x) –1

 x=f

(y) 2x + 3

But y = e Applying log on both the sides,

2

2x + 3

= (x + k) – 2( x + k) + 3 2

2 1  2

1

2

= x + (2k – 2)x + (k – 2k + 3) But it is given that f(g(x)) = 2

x + 10x + c 2

By comparing coefficients of x , x, 2

we have 2k – 2 = 10 and k – 2k + 3 = c k = 6 and hence, c = 36 – 12 + 3 = 27.

 log y = log e  log y = (2x + 3) loge  log y = 2x + 3 ( loge = 1)

log y  3 2 log y  3 f 1  y   2 log 1 3 0  3 3 f 1 1    2 2 2 x

____________________________________________________________________________________________________ _____________________________________________

Functions Solutions

330

x (x –1)

24. Let y = 2 Applying log2 on both the sides, we get  log 2 y  log 2 2x  x 1  log 2 y  x  x  1 ( log 2 = 1) 2

 x 2  x  log 2 y  0  x 

1  1  4log 2 y

2 ( x  1)

 x  f 1  y  

 f x  1

1  1  4log 2 y

2 1  1  4log 2 x 2

10x  10 x  y ------- (1) 10x  10 x x 1 Consider 10 = t  10 x  t Then equation (1) becomes 1 t 2 t  y  t 1  y 1 t2 1 t t Applying componendo and dividendo, we get t 2  1 t 2  1 y  1 2t 2 y  1    t 2  1 t 2  1 y  1  2 y 1 1 y  t2  1 y

25. Let f (x) 

1

1 y  1  y 2 t2   t  1 y 1 y  1 2

x 1 y  10x   [  10 =t ]   1 y  Applying log on both sides, we get 10

1 y  1 x log10 10  log10   2 1 y  1 y  1 x  log   2 1 y  1 y  1  f 1  y   log   2 1 y 

1 1 x  f  x   log   2  1 x  28. If x  0, then f(x) = 2x + |x| = 3x If x < 0, then f(x) = 2x + |x| = x If x < 0, then f(x) = 2x + |x| = x Then f(2x) + f(–x) – f(x) 1

= 3(2x) – x – 3x = 2x = 2 |x| Suppose x < 0. Then f(2x) + f(–x) – f(x) = 2x + 3 (– x) – x = – 2x = 2 |x| 30. Since f(x) is even, f(x) = f(– x) ax 1 a x  1  n x n  a x  1   x   a  x  1 Multiplying

the

numerator

and

the

x

denominator of R.H.S by a , we get  1 ax 1 1 ax   1  n n n x x  a  1  1 x n 1  a x   1 Multiplying

the

numerator

and

the

n

denominator of R.H.S by (–1) , we get n+1

1 = (– 1)  (n + 1) is even. Therefore n  {1, 3, 5, 7, ……..} 31. i) f  x   1  x  x 2  1  x  x 2 f  x   1  x  x 2  1  x  x 2 



1  x  x2  1  x  x2



 f  x 

 f (– x) = – f(x) Hence, it is an odd function.  ax 1 ii) f  x   x  x   a 1   ax  1  f (  x)    x    x   a 1  1  ax    x   x   1 a   ax 1   x x   a 1   f(–x) = f(x) Hence, it is an even function.  1  x2  iii) f  x   log  2   1 x   1  x2   f   x   log  2  1 x   f x

 f(–x) = f(x) Hence, it is an even function.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

331

iv)

f(x) = k  f(– x) = k = f(x)  f( –x) = f(x) Hence, it is an even function. 33. f(2) = f(2×1) = f(2+1) = f(3) But f(3) = f(3×1) = f(3+1) = f(4)  f(3) = f(4) In this way we can write; 2006 = f(2) = f(2×1) = f(2+1) = f(3) = f(3×1) = f(3+1) = f(4) = f(4×1) = f(4+1) = f(5) = f(5×1) = f(5+1) = 2006

f(6) =.... = f(2006

)

2006

 f(2006

) = 2006. n

n–1

n–2

34. Let f(x) = a x +a x +a x +.....a x+a . 0

1

2

n–1

n

Then,

1 1 f xf    f x  f   x x n

n–1

1  f 3  x   10 3 1  x 10 10    3  2   10   10 3 3 3 3  x 10 10 10     10 34 33 32 3 Continuing in this manner, we obtain, x 10 10 10 f n  x   n  n 1  n 2  .......  10 3 3 3 3 1   1 n   x x 1   n  10  3   n  15 1  n  1 3  3   1  3 3   x  15 x  15  n  15  f 2006  x   2006  15 3 3 a 37. a n 1  n 1 an 

(a x +a x +......+a x+a ) × 0

1

n–1

1

n

a1 a n 1  a0   x n  x n 1  ........  x  a n    n

a n 1

= (a x +a x +......+a x+a ) 1

n–1

1 an 1  1 an an

Substituting n = 1, 2, 3, ................... in the above equation, we get 1 1  1 a2 2

n–1

0



n

a a a    0n  n11  ........  n 1  a n  x x x  On comparing the coefficients of like powers of ‘x’ on both sides, we get a = 1, a = 1 and a = a = ....= a = 0, 0 n 1 2 n–1 n

 f(x) = x +1,

1 1 1   1    1  1 a3 a2 2  

2

1 2

1 1 1    1    2  1 a 4 a3 2 

3

1 2

n

or f(x) = –x +1. 35. f (x) = f (f (x)), n  2 n

1

1 1 1  1   1    n  2   1  n  1 an an1 2 2 

n–1

f (x) = f (f (x)) 2

1

1

1  f1  x   10 3 1 x     10   10 3 3  

x 10   10 32 3

1 f (x) = f (f (x))  f 2  x   10 3 1 2 3 1  x 10    2   10   10 3 3 3  x 10 10  3  2   10 3 3 3 f (X) = f (f (x)) 4 1 3



1 1 3457 2  1728    a1729  a1729 2 2 3457

38.

f(1) = 1 f(2) = 2 f(1) + 1 = 3 = 22 – 1. f(3) = 2 f(2) + 1 = 7 = 23 – 1. f(4) = 2 f(3) + 1 = 15 = 24 – 1.

f(100) = 2f(99) + 1

= 2100 – 1.

____________________________________________________________________________________________________ _____________________________________________

Functions Solutions

332 y

f(1) + f(2) + f(3) +...+ f(100) 2

3

4

100

= 1 + (2 – 1) + (2 – 1) + (2 – 1) + ....+ (2 2

3

4

100

x

= 1 + 2 + 2 + 2 + ......+ 2 2

3

– 1)

4

– 99

2f(x) = f(x) + [f(1)]

– 100

 f(x) = 2 ( f(1) = 2)

100

x

= 2 + 2 + 2 + 2 + ......+ 2 2  2100  1   100 2 1

r

 f(r) = 2 50

50

Now  f  r    2 r

100

= 2(2

x

43. Given 2f(xy) = [f(x)] + [f(y)] x, y  R] Putting y = 1, we get,

– 1) –100

r 1

40.

2

r 1 3

 2  2  2  ......  250 2

Let f(0) = k f(–1) = f(0) + 4

50 –1

= 2[1 + 2 + 2 + ...........+ 2 ]  a.  r n  1  1  a  a 2  a 3  .....    r 1     t where r  2   t1  

=k + 4

f(–2) = f(–1) + 9 = k + 4 + 9 f(–3) = f(–2) + 14 = k + 4 + 9 + 14



2  250  1 2 1

51

=2 –2

50

 f  r   251  2 r 1

4x ------------- (1) 4x  2 4 2 41x  f 1  x   1 x   ------ (2) x 4  2 4  2.4 2  4x Clearly, f(x) + f(1 – x) 4x 2 4x  2  x   =1 4  2 2  4x 4x  2 2001  r  Now,  f   r 1  2002   1   2   3   2001  f f f   ....  f    2002   2002   2002   2002   1   2   3   2001  f f f   ....  f    2002   2002   2002   2002    1   2001    f  f   2002     2002 

44.

f(–25) = f(–24) + 124 = k + 4 + 9 + 14 + ....... + 124. = k + 4 + 9 + 14 + ....... + 124. 24  k   4  124  2 = k + 12 × 128 = k + 1536 It is given that f(– 25) = 1600 k + 1536 = 1600 k = 64  f(0) = 64

1

42. Given f  x    3  x n  n , x  0 1

1 1 n n      f f  x    f  3  x n  n   3   3  x n  n        

1 n

 3   3  x     3  3  x   1  1 Similarly, f  f       x  x n

n

1 n

  x  n

1 n

=x

1   1  1 f  f  x    f  f     x   2 x.  2 x x   x  Minimum value is 2.

Given f  x  

  2   2000    f  f  + . . .   2002   2002     1000   1002    1001   f  f   f    2002    2002   2002  

1 = 1 + 1 + 1 + ........ to 1000 terms  f   [ f(x) + 2 f(1 – x) = 1] 4  1000  42 1  1000  = 1000.5 2

____________________________________________________________________________________________________ _____________________________________________

Functions Solutions

333

CONCEPTIVE WORKSHEET 1.

2.

Number of relations which are not functions is 2mn – nm = 25 × 2 – 25 = 1024 – 32 = 992 f  x   R  x  6x  5  0 2

  x  1 x  5  0

 x  1, x  5  R – {1, 5} 1 3.

f  x

 x2 – 2x > 0  x (x – 2) > 0  ( x – 0) (x – 2) > 0  [x < 0 or x > 2]

4.

  , 0    2,  

f  x   R  f  x  0  x 3 0  x 3

 x  3,  

5.

f  x 

 D  0 for x  R 2

 1 4y  0 1  4y 1  y  4  y

2

1 2

7.

f is not one-one is f(0) = 0 = f (2n+1) for all n  Z . However it is onto as for any i  Z , there

2i i 2

f  0   0, f  1  0, f  2   1, f  3  0, f  4   2.........

Range = Z and f  1  0  f 8.

x 1 1  x  1  and y  f   3  3 

 x 1 g 1  x   f 1    3  11. (fog)(x) = 1 + x2 – 2 x3 + x4. f(g(x)) = 1 + x2 – 2 x3 + x4. 1 + (g(x))2 = 1 + x2 – 2 x3 + x4. (g(x)) 2 = x 2 – 2 x 3 + x 4 = x 2 (x 2 – 2x +1) = (x(x –1))2  g(x) = x(x –1) or x(1 – x) i.e

y

y 1 Interchanging x and y, 

x 1 x 1 1 , f  x  But it is given that  

f–1(x) = f(x) Therefore 1  x 



2

 1 x     1  0

x 1 

   1

ax  b then, cx  d

 b  dx  2x  5 f 1  x     f  x  then,  3x  4  a  cx 

1 1

is 2i  Z such that f  2i  

 a 1   g  4.  1  g  a  1  1 4   3 = g[a] = a + 2 10. Let y = g–1(x) so x = g(y) = 3f(y) + 1 Solving for

13. We know that If f  x  

 y   ,   2 2 6.

a 1    a  1  gf    4    4 

 gof  

12. Let y  1   x , x 

x  y (say) 1  x2  x2y – x + y = 0

2

Given f(x) = 4x – 1 and g(x) = x3 + 2.

f(y), f  y  

 R  f  x  0

(or)

9.

is onto but not one – one. Given f(x) = 4x – 1 and g(x) = x3 + 2. (fog) (–5) = f[g(–5)]= f[(–5)3 + 2] = f[–125 + 2] = f(–123) = 4(–123) –1 = – 492 – 1 = – 493

 5  4x   4x  5  f 1  x        2  3 x   2  3x  14. Let f–1(21) = y  f(y) = 21  y2 + 5 = 21

 y2 = 16

 y  4

15. Range = [0, a] for

a2  x2

= 0, 2  for 2  x 2 16. Take cos 3x =1 and cos 3x = – 1 since they are maximum, minimum values.  1 1  1  Range:  2  1 , 2  1    3 ,1        

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10th Class Mathematics

334 17. Minimum value is

C  2 ab  25  2 9  4  25  12  13 Maximum value is C  2 ab  25  2 9  4  25  12  37 Range = [13, 37] 18.

f  x   x  1  x  2  x  3 when 2 < x < 3

f  x 

1  R  x  0, 1  x  0, x  1 log x

 x  0, x  1  Domain is R – {–1, 0, 1}

22. f(–3) is defined only for equation (3). f(–3) = 3x + 2 = 3 (–3) + 2 = – 9 + 2 = – 7 23. Range of

a 2  x 2 is [0, a]

Range of

32  x 2 is [0, 3]

24. We have, f  x   x 

20. Let 100  8  13x  y

 100  y  0

 100  y  8  13x

f(– x) = e–2x + sin(– x) = e–2x – sinx   f  x 

 f is neither even nor odd.

 f(x) = x –1 + x – 2 + 3 – x = x  f is an identity function. 19.

x 1  e x 

x =–x+x=0 ex  1  f(– x) = f(x) Hence it is an even function. 4) f(x) = e2x + sin x 

1 1  f  x3   x3  3 x x 3

1 1  3 1   Now, [f(x)] =  x     x  3   3  x   x x   x   3

 y  100  0

 y  100

 y    , 100

1 = f(x3 ) + 3f(x) = f(x3 ) + 3 f    x

 1 x  21. 1) f  x   log    1 x  1 x  1 x    log     f  x0 1 x  1 x 

 f   x   log 



It is an odd function.



2 2) f  x   log x  x  1





 f   x   log  x  x 2  1





1 1  3 f    f  x   4log e   x    x



log x  x 2  1  log  x  x2  1





 log x  x 2  1  x  x 2  1





= log (x2 + 1 – x2) = log (1) = 0  f(–x) = – f(x)  It is an odd function. 3) f  x  

1 25. We have, 3 f  x   f    log e x 4 x 1  3 f  x   f    4.loge x –––––––– (i)  x

 f  x   f   x 



  1   f ( x )  f  x     

x x  1 e 1 2

1  3 f    f  x   4loge x –––––––– (ii)  x Solving (i) and (ii), we get 8f(x) = 8logex  f(x) = logex  f(ex) = x 26. We have, 2f(xy) = {f(x)}y + {f(x)}x for all x, y  x  2f(x) = f(x) + [f(1)]x [Replacing y by 1]  f(x) = ax

x

x x  x.e x x f x  x  1  1 e 1 2 1  ex 2 x    x.e x x   x f  x  f x   x   1     1 x 2   e  1 2  1  e

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n

 (a –1)  f(r) = an+1 – a r 1

n

 r 1

 a n  1 f r  a    a 1 

Functions Solutions 27. F(1 + 1) + F(1 – 1) = F (1) F(1) F(2) + F(0) = 5 × 5 F (2) = 25 – 1 = 24 F(2 + 1) + F(2 – 1) = F (2) F(1) F (3) + F(1) = 24 × 5 F (3) = 120 – 5 = 115 F(3 + 1) + F(3 – 1) = F (3) F(1) F (4) + F(2) = 115 × 5 F(4) = 575 – 24 = 551 F(4 + 1) + F(4 – 1) = F (4) F(1) F (5) + F(3) = 551 × 5 F (5) = 2755 – 115 = 2640 28. f(0) = f(–3 + 3)= f(–3) + g(3) + 7 = – 3 + 21 + 7 = 25 29. Put x1 = x2 = x3 = x4 = ............x101 = 0 Then f(0) = f(0) + f(0) + f(0) + .......+ f(0) – 25 f(0) = 101 f(0) – 25 100 f(0) = 25

335 33. F(n + 2) = 2F(n) – F(n +1) For n = 0, F(2) = 2 f(0) – F(1) =2×2–3=4–3=1 For n = 1, F(3) = 2F(1) – F(2) =2×3–1=5 For n = 2, F(4) = 2F(2) – F(3) = 2 × 1 – 5 = –3 For n = 3, F(5) = 2F(3) – F(4) = 2 × 5 + 3 = 10 + 3 =13 34. We know that f(x + y) = f(x) f(y)  f(x) = ax Also f(3) = 27  a3 = 27  a = 3 Now f(x) = 3x  f(4) = 34 = 81 35. If a polynomial function satisfies

1 1 f  x. f    f  x   f   x  x then, f(x) = xn + 1 f(12) = 12n + 1 1729 = (12)n + 1 1728 = (12)n (12)3 = (12)n  n=3 f(x) = x3 +1  f(10) = 103 + 1 = 1001

1 4 30. Let f(x) = ax + b Then f(1) = a + b, f(2) = 2a + b, f(3) = 3a + b and f(4) = 4za + b f  0 

f 1  f  2   a  0 ----------(1)

(property)

f  3  f  4   a  0 ----------(2)

From (1) and (2) a = 0 Then f(x) = b, a constant function. Also given that f(5) = 5. Hence f(0) = 5 is the correct answer. 31. Let f(x) = g(x) – 6 --------------(1) Note that g(x) is an odd function. Therefore g(– x) = – g(x) or g(x) = – g(– x) -------------(2) Now g(– x) = f(– x) + 6 (from equation (1)) g(– 1) = f(– 1) + 6 = 12 + 6 = 18 g(1) = –g(– 1) = – 18 (from equation (2)) f(1) = g(1) – 6 (from equation (1)) = – 18 – 6 = – 24 32. f(0) = f(–2 + 2) = f(–2) [f(x + 2) = f(x) for all x] = – f(2), [f (– x) = – f(x)] = – f(2 + 0) = – f(0) 2f(0) = 0 f(0) = 0 f(100) = f(2 + 98 ) = f(98) = f(2 + 96) = f(96) ----------------------= f(2) = f(0) = 0

SUMM ATIVE WORKSHEET 1.

f  x 

x x 1

a a f ( a) a2  a 1  a  1  2  f  a2  a 1 a 1 a 1 f (a  1) a 11 a

2.

3.

f  x If g x  R  

 g  x  0

 5x  3  0

 x

3 5

 3   R  5

log f  x   R  f  x   0

( Logarithms are defined for only positive values)

 3x  2  0

 x

2 3

 x

2 3

2   x  ,  3  www.betoppers.com

10th Class Mathematics

336 4.

log f  x   R

 f  x  0

11.

 x2  3x  2  0   x  1 x  2   0  x  1, x  2 5.

 R – {1, 2}

f  x   R  5  x  0, 5  x  1

 x  5  0, x  4  x  5, x  4    , 5  4 6.

 7. 8.

 f 1  0   1

3   3 , , , ........ 2 2 2 2

 3 x   2n  1

 2

 x   2n  1

 6

   R   2n  1 , n  Z  6  

f(x) = 2x + 3 and A = {–2, –1, 0, 1, 2}.  f(–2) = 2(–2) + 3 = –4 + 3 = – 1 f(–1) = 2(–1) + 3 = –2 + 3 = 1 f (0) = 2.0 + 3 = 0 + 3 = 3  A f (1) = 2.1 + 3 = 2 + 3 = 5  A

9.

10.

f (2) = 2.2 + 3 = 4 + 3 = 7  A But 0, 1 and 2 of domain have no images. f is not a function.

x 1 Given f  x   x 1 If the number of f s are even, then answer is x. 

 1 x  3x  x3 f  x   log  g x     and  1 x  1  3x2  3x  x 3  fog(x) = f[g(x)]  f  2   1  3x   3x  x 3   1  3x 2  3x  x 3  1     2  1  3x 2  log  1  3x 3   log  2 3  1  3 x  x   1  3x  3x  x  2 1  3x 2  1  3x    3

 1  x    log    log  1  x  3  1  x   1 x 

 1 x   3log   = 3 f(x) 1 x  www.betoppers.com

13. Since log x is defined x  0  f(x) = log4(log5 (log3(18x – x2 – 77))) log5 (log3(18x – x2 – 77)) > 0  log3(18x – x2 – 77)) > 5°  18x – x2 – 77 > 31  (x – 8) (x – 10) < 0  8 < x < 10 

 x  8, 10 

f  x   cos  3x  5   R  x  R

 

y

1 x  y  3x  f  x   3 12. Let f–1(0) = y  f(y) = 0  y3 + y2 – y – 1 = 0  y2(y +1) –1 (y +1) = 0  (y +1)(y2 –1) = 0 y  1

sec 3x is defined if cos3 x  0

 3x  ....

f 1  x   y  x  f  y   log 3

3

14.

x  0, f  x   x

 f(– x) = x

 –x–1=x 1 2 15. putting x = 2 and y = 1 in the given relation, we obtain g(2)g(1) = g(2) + g(1) + g(2) – 2 5g(1) = 5 +g(1) + 5 – 2  g(1) = 2 x

Putting y 

1 in the given relation, x

1 1 we get, g  x  g    g  x   g    g 1  2 x x

1 1  g  x  g    g  x   g    g (1)  2   x x n  g(x) = x +1 n=2  g(2) = 2n + 1  5 = 2n +1 2 Thus, g(x) = x +1  g(3) = 9 +1 = 10 16. we have, f  x 

4x 4x  2

f 1  x  

41 x 4 2   1 x x 4  2 4  2 4 2  4x

 f  x   f 1  x  

4x 2 4x  2   1 4x  2 2  4x 4x  2

Functions Solutions

337

17. 321 = f (123) = f(123 × 1) = ff(123 + 1) = f(124) = f(124 × 1) = f(125) = f(125 × 1) = f(125 + 1) = f(126) ----------------------------------------------------------------= f(2012)

21. f(6) = f(2 + 4) = f(5) f(4) = f(1 + 4) f(4) = [f(4)f(3)] f(4) = 2 × 2 × 2 = 8 22.

    

 1  a 2   1  a 2  2a   log     log  2 2 1  a  2a   1  a  

1   1  3

1 Let f    x and f(3) = y then we get two 3 linear equations in x and y. x – 5y = 9 ---------------(1) –5x + y = 1 ---------------(2)

1 a 1 a

2a  1  1  a 2  2a  f  log  2  1 a  1  2a 2  1  a

1 18. Put r = 3, then f    5 f  3  9  3 1 Again put r  , then f  3  5 f 3

f  a   log

2

1 a   1 a   log    2log    2 f a 1  a    1 a 

23.

e

f  x



10  x 10  x 10  x

23 Solving equations (1) and (2) we get y  12

200 x  10    200 x  100  x 2 f  log e  2   100  x  10  200 x 2  100  x

23 f  3  12 2 19. [f(xy)] = x [f(y)]2 Put x 

100 and y = 9 in the above equation, then 9 2

2   100  9   100  f  9    9  f  9   for positive    values of y.

100 2   f 100     21 for positive values of y. 9 2

10  21 3 Hence f(100) = 70. 20. g(0) = f(1) – f(0) g(1) = f(2) – f(1) g(2) = f(3) – f(2) g(3) = f(4) – f(3) ------------------------------------------------------------g(999) = f(1000) – f(999)  f 100  

    

 100  x 2  20 x    2  loge  100 2 x   100  x  20 x   100  x 2   10  x 2   log e   2  10  x  

 10  x   loge    10  x 

2

 2. f  x   k 

 2loge

10  x 10  x

1  0.5 2

24. y = max {(x + 3), (7 – 2x) } 3  3 3    For x  , y  max   3  ,  7  2    2 2     2

Adding, g(0) + g(1) + .......+ g(999) = f(1000) – f(0) = 1000000 Average of the 1000 numbers 

f  x   log e10  x

9  9  max  , 4   4.5 2  2  Minimum value of y is 4.

1000000  1000 1000 www.betoppers.com

10th Class Mathematics

338

1 25. We know that f  x  f    f  x    x  f(x) = xn + 1 Also f(2) = 17  2n +1 = 17  2n = 16  2n = 2 4  n=4  f(x) = x4 +1  f(10) = 104 + 1 = 10001

1 f   x 4.

2 

 b say



2

b 1

= C say

4

2 c 



2

c 1 4

2

  2 b 1   2    1 2  2   2 b 2      4 4  

20 3

5 4

2

  2 a 1   2    2   2    2 a 3  4 4



17 16  x   2x  1  0  x  3 20  3x  4 x  5  0  x 

25 7

2

2

1 1  1 3    3x    1    3x    2 4  2 4  2

 x 1 Now  3    0 2 

 x 1 3 3   3  2   4  4



2

2

  2 x 1   2    3  2 x 4  2      4 4



1 4 1  x  3  4 7  x = 2, 3 When x = 2, y = 14C3 + 14C3 = 728 x = 3, y = 13C5 + 11C7 = 1617 Let f(x) = y = 9 x – 3 x + 1 = 3 2x – 3 x +1 = (3x)2 – 3x +1

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2

4

gofogofogofogof = (gof)

1 2x  1  0  x  2

3.



a 1

2 (gofogofogof)(x) = (gof)  b  

16  x  0  x  16

4x  5  0  x 



4( x  x )  1 (2 x  1) 2   a (say) 4 4 Now (gofogof)(x) = (gof{ (gof)x} = (gof) (a)

Given f(x) = a + bx ff(x) = f(f(x)) = f(a + bx) = a + b(a + bx) = a + ab + b2 x Likewise fff(x) = a + ab + ab2 + b3 x  fr (x) = a + ab + ab2 +..........abr–1 + br x =

20  3x  0  x 

4x  1 4



 br  1  2 r–1 r a(1+ b+b +.....+b )+ b x = a  b  1  + br x.   2.

f  x   x  x and g  x  

Consider (gof) (x) = g  f ( x ) = g x + x

HOTS WORKSHEET 1.

3  Required range  ,   4 

3

 f  x  4

 (gofogofogofogof) (3) =





2



32





x 2



2

2

74 3

1  a + b = 11 1

5.

Given f  x    a  x n  n 1

 f  f  x   a   f  x 



 a   a  xn 



1 n

n

 1

1 n

1 n n   n n   a   a  x        

  xn n  x

Functions Solutions 6.

7.

339

We have, f(g(x)) = 1+x2 –2x3 +x4 = 1+x2 (1–2x+x2 ) = 1+x2 (x–1)2 = 1+{x(x – 1)}2  g(x) =  x(x–1) f(x) = 3x – 4

x4  f 1  x   3

g(x) = 2 + 3x

x2  g 1  x   3

f

1

8.

replacing x by

Given f  x  

32 1  3 3

3x  2 5x  3

Let f  x   y  x  f 1  y 

3y  2 3x  2 x  y; 5y  3 5x  3 f 1  y   9.

3y  2 3x  2  f 1  x    f  x 5y  3 5x  3

sin2x + cos4x = 1 – cos2 x + cos4x = 1 – cos2x (1 – cos2x) = 1 – sin2x cos2x

1  1  sin 2 2 x 4 1  3   1 Range: 1  1 , 1   0     , 1 4  4   4 10.

f  g  x    sin x  sin 2 x _________ (1) and

g  f  x    sin 2 x _______ (2) From (1) it is clear that when f operates, result is the positive square root of a number and from (2) it is clear that g operates, result is of the sin 2  . Thus f  x   x and g(x) = sin2x 11. First we prove by induction that f(k) = 2k by induction. f(a) = 2a  f  a  k   f  a  f  k   2a 2k .

Simplifyinga = 3.

1 , we get x

1 af    bf  x   x  5  x

5 4 9  3  5  3 3

(g–1 o f–1) (5) = g–1 [f–1 (5)] = g–1[3] 

1 1 12. we have, af  x   bf     5 ----------(1) on x x

1  bf  x   af    x  5 -------------(2)  x Multiplying (i) by a, ( ii) by b and then subtracting, a  2 2 we get,  a  b  f  x     bx   5  a  b  x  f  x 

1 a 5   bx   2  a b  x  a b 2

13. Case I : when x  N , In this case, we have f(x) = f(1+1+1+.....+1) x – times = f(1)+f(1)+f(1)+....+f(1) x – times = xf(1) Case II : when x is a negative integer : Let x   y y  N , Now, f(x+y) = f(x)+f(y) for all x y  R .  f(0+0) = f(0)+f(0) [putting x = y = 0] f(0) = f(0)+f(0) = f(0) = 0 Again, f(x+y) = f(x)+f(y) for all x, y

 f

  1  1

= f(–1)+f(1)

[putting x =–1 and y = 1]  f(0) = f(–1)+f(1)  0 = f(–1)+f(1) (  f(0) = 0)  f(–1)= –f(1) Now, f(x)= f(–y) =

  ( 1)  ( 1)  ( 1)  ...( 1)  f    y times   = f(–1) +f(–1)+f(–1)+......+f(–1) y – times = y f (–1) = – yf (1) (  f(–1) = –f(1)) = – yf(1) = xf(1) 14. Let x 

m where m, n  z, n  0 . n

1 1 1 1  we have, f 1   n       ....   n   n n n  n  times

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10th Class Mathematics

340

1  f   n

1 f    ....  n

1 f    n. f n

2000

1   n

  1   1  ........ 1   2   f  i   = 

 f  1   1 f 1 n



1000 times

i 1

 2000

f(2) + f(3) + .......f(2000) + f(1) = – 1000 2  f  i 

n

i 1

1 m  Now, f  n    m  n      

(  f(2001) = f(1))

  1 1 1 f    .......   n n  n   m  times  

2000

2000

 f i   2  f i   1000 , i 1

i 1

2000

1 1 1 1  f    f    ....  f    m. f    n  n n n  1  m   1 1 m   .f (1)    f    f (1)     n n  n  n

f(1) = xf(1). 

15.

  1

n n 1 2

x n  1 – x – x2 +x3 + x4 – x5 – x6 + ......

n0

i 1

3.

4.

5.

= (1 – x2 + x4 – x6 + ....) – (x – x3 + x5 – x7 + .....)

1 x  2 1  x 1  x2 (Both are infinite geometric series with common 

2 2 ratio = – x2 and  x  x  1 )



6.

1 x 1  x2

IIT JEE WORKSHEET f(2.4) – f(2) = f(1.4) – f(1) = f(0.4) – f(0) = f   0.6   0  0.6  0  0.6 2.

f(m + 1) = m(–1)m + 1 – 2f(m) for m  1 m = 1 f(2) = 1 (–1)2 – 2f(1)  f(2) = 1 – 2f(1) m = 2 f(3) = 2 (–1)3 – 2f(2)  f(3) = – 2 – 2f(2) m = 3 f(4) = 3 – 2f(3) -----------------------f(2000) = 1999 – 2f(1999) f(2001) = – 2000 – 2f(2000) Adding all the equalities f(2) + f(3) + ...... f(2001) = (1 – 2 + 3 – 4 + ....– 2000) – 2 (f(1) + ....f(2000)) f(2) + f(3) + .......f(2000) + f(2001))

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7.

 f i   i 1

1000 3

f(x + f(x)) = 4f(x) For x = 1, f(1 + f(1)) = 4f(1)  f(1 + 4) = 4.4  f(5) = 16. i. Yes it is a function. Because no vertical line cuts the graph in two points. ii. Yes. Because different values of x has different images. f(a + b) = f(ab) f(a – a) = f(– a2)  f(0) = f(– a2)  1    1  f    0   f     x 0   f 0  2   2   1  311   1 2  f   515   f a  f  0   f   2    2 f(n) – f(n – 1) = 2n – 1 f(1) – f(0) = 2(1) – 1 f(2) – f(1) = 2(2) – 1 f(3) – f(2) = 2(3) – 1 ------------------------------------------f(n) – f(n – 1) = 2n – 1 Adding all the equations f(n) – f(0) 2n n  1 = 2(1 + 2 + 3 ...+ n) – n   n 2 = n2 + n – n = n2  f(n) = n2 + f(0) Each element of B. Should have some pre-image in A. f : A  B is onto in the following way ‘1’ can have 3 pre – images, ‘2’ can have 2 pre-images, 3 can have 1 pre-image. Number of ways of dividing ‘6’ elements into 3 sections are i) 3, 2, 1 ii) 4, 1, 1 iii) 2, 2, 2 Number of onto functions possible in case (1). i.e first element of co-domain can have 3 pre6! image ways. Second element of co3!2!1! domain has ‘2’ pre-images. Third element have only-one pre-image. Pre-images of elements ‘3’, ‘2’, ‘1’ can be interchanged in group wise in 3! ways. (i.e in the beginning element ‘3’ can have two pre-images,



P  x 1  x  Q  x  1  x2

1.

2000

 3  f  i   1000 



Functions Solutions

8. 9.

later an element ‘1’ can have two pre-images.)  Number of possible onto functions possible in  6!  case (i) =   3!  360  3!2!1!  Similarly number of all possible onto functions in  6!  3! case (ii) =     90  4!1!1!  2! Similarly number of all possible onto functions in  6!  case (iii) =    90  2!2!2!   All possible onto functions = 540. f is both one – one and onto function Consider option (1) 2 Let f  x   x   x  1

g  x   1  x  1  x  x   x 2  1

Domain of f  x    0,   Domain of g(x) = [0, 1]  functions are not equivalent. (  Domain of ‘f’  Domain of ‘g’) Consider option (2) x3  x Let h(x) = x2 + x, u  x   x Domain of f = R, Domain of u = R – {0}  functions are not equivalent. Consider option (3) Let v  x   x 2  1  x , w x   x2  1  1  x  x  1  x

Domain of v =  0,   Domain of w = [0, 1]  v, w are not equivalent. Consider option (4) 1 1 l  x   log 1 sinx  , m  x   sinx  2 8cos 2 x 8cos 2 x 1 1 l  x   log 1 sin x  m  x   sinx  2 8cos 2 x 8 cos 2 x 2 for cos x 

1 , m(x) is defined where as same 8

1 ‘l(x)’ is not defined. 8  Domain of ‘l’  Domain of m.  ‘l’ and m are not equal 10. Let f(x) = x2 + b x + c, f(x + k) = f(– x)  (x + k)2 + b(x + k) + c = x2 – bx + c  x2 + k2 + 2xk + bx + bk + c = x2 – bx + c  k2 + 2xk + bx + bk = – bx  k2 + 2xk + 2bx + bk = 0  k2 + bk + 2xk + 2bk = 0  k (k + b) + 2x(k + b) = 0  (k + 2x)(k + b) = 0  (k + b) = 0  k is constant k  2 x  k = – b cos 2 x 

341 hence f(x) = x2 – kx + c ( c is arbitrary) 11. Case 1: Assume ‘1’ is true and ‘2’, ‘3’ are false. f(x) = a, f(y) = a, f(z) = b  f is not one-one function. ( f(x) = f(y) = a)  It is a contrary to the statement that f is a bijection Case 2: Assume ‘2’ is true and ‘1’, ‘3’ are false. f  y   a, f  x   a, f  z   b

f  y   a, f  x   a  either f(y) = b, f(x) = c (or) f(y) = c, f(x) = b f is not one one function in both the cases.  Case 3: Assume ‘3’ is true and ‘1’, ‘2’ are false. f  x   a, f  y   a, f  z   b f  x   a, f  z   b  f(x) = b, f(z) = c  f is one-one function.  f–1(a) = y 12. If x = – 2 then f(x) = –3, g(x) = x –1 = –3, h(x) = – 3 then f(x) + g(x) – 2h(x) = –3 –3 –2(–3) = 0 If x = – 1 then f(x) = –2, g(x) = – 2, h(x) = – 2 then f(x) + g(x) – 2h(x) = –2 –2 –2(–2) = 0. If x  1, 2 then f(x) = x – 1, g(x) = x – 1, (x + 1)(x + 2)(x  1) h(x ) = = (x  1) (x + 1)(x + 2) then f(x) + g(x) –2h(x) = (x –1) + (x –1)– 2(x –1) = 0  f(x) + g(x) – 2h(x) is a polynomial of degree zero.

13. We know that

1 f  x f    f  x   x

1 f   x

 f  x    x n  1 Also f(3) = – 242

   



f(x) = – xn +1 = – 3n +1 = – 242 –3n = – 243 3 n = 35 n=5 f(x) = – x5 + 1

2006  i  14. Given  f   i=1  2007   1   2   3  =f  + f  2007  + f  2007  + 2007      

 2006  ..... + f    2007    1   2006   +f But  f    +  2007     2007 

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10th Class Mathematics

342 15.

  2   2005    f  2007  + f  2007   + ......     

Let the sum be S and set f(0) = 0. 

Then S   n 1

 f  n 1 f  n  f 1   n n 2 2 n 2 2

  1003   1004   + f  +f    2007     2007 



1  f  n  1  f  n  2   2 n 2 2n

  1  1   = f  + f  1  +  2007     2007 



1  f  n  1  f  n  2    2 n 2 2 n 2n n 2

  2  2    f  2007  + f  1- 2007   + ......     



1 1  f  n  1 1  f  n  2    f 0     2 2 n1 2n 1 4 n 2 2 n 2



1 1 1   S  0  S 2 2 4

  1003   1003   + f  + f 1    2007     2007  Each of the above brackets is of the form, f(x) +f(1–x) each bracket is 1. 2006

Hence,

 

i





i



 f  2007  + f 1- 2007   i =1



2006

 1 = 1003 i =1

Therefore, There are1003 pairs

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

3 1 S S 4 2 1 1 S 4 2 

 n1

S=2

f  n 2 2n

7. COORDINATE GEOMETRY SOLUTIONS = 9  256  265 Since AB + BC  CA,

FORMATIVE WORKSHEET 1.

(i) Distance between the two points is given by

 x1  x 2 

2

  y1  y2 

4.

2

Therefore, distance between (2, 3) and (4, 1) is given by l=

 2  4

2

2

  3  1 

 2 

2

  2

 5   1    7  3 2

 5   1    7  3

2

2



 4   4 



2

2

 4

 4



2.

2

 a   a     b    b   2

 2a    2b 

2

2

  4   2   

2

= 1296  225  1521  39

5.

2

 6

2

 3  6

2

2

  4  7  2

 6  9   4  7 

2

 3

2

  3

 3



2

2

  3

2

 9  9  18  3 2

CB =

Yes, we can find the distance between the given towns A and B. Assume town A at origin point (0, 0). Therefore, town B will be at point (36, 15) with respect to town A. And hence, as calculated above, the distance between town A and B will be 39 km. Let the points (1, 5), (2, 3), and (−2, −11) be representing the vertices A, B, and C of the given triangle respectively. Let A = (1, 5), B= (2, 3), C = (2, 11) 2

2

 1

As two sides are equal in length, therefore, ABC is an isosceles triangle. It can be observed that A (3, 4), B (6, 7), C (9, 4), and D (6, 1) are the positions of these 4 friends. AB =

 15  0   362  152

1  2 

2

Therefore, AB = BC

BC =

 AB =

  6 

2

Distance between points (0, 0) and (36, 15) 2

2

2

2

 9  9  18  3 2

 36  0 

 1

CA=  5  7    2   2     2   02

2

2

2

= 4a 2  4b 2 = 2 a 2  b 2

=

3.

2

  2  4  

= 1  36  37

2

(iii) Distance between (a, b) and (a,  b) is given by l=

6  7 

BC =

2

2

= 1  36  37

2

(ii) Distance between (-5, 7) and (-1, 3) is given by

=

 5  6

AB =

= 44  8 2 2

l=

Therefore, the points (1, 5), (2, 3), and (−2, −11) are not collinear. Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of the given triangle respectively.

2

 9  6    4  1

2

 3



2

  3

2

 9  9  18  3 2

 3  6

AD =

2

2

  4  1 

 3

2

  3

2

 9  9  18  3 2 Diagonal AC = 2

 3  9   4  4

2



2

 6    0 

2

6

Diagonal BD=

 6  6

2

2

  7  1 

 0

2

2

  6  6

2

  5  3  5

2

 2   2    3   11 

2

 4 2  14 2

= 16  196  212 CA =

2

1   2    5   11 

2

 32  16 2

____________________________________________________________________________________________________ _____________________________________________

Coordinate Geometry Solutions

344

 36  16  52  2 13 2

 3  0   1  3

BC = 

 3

2

2

  2   9  4  13 2

 0   1    3   4  

CD = 

2

1   7 

It can be observed that all sides of this quadrilateral ABCD are of the same length and also the diagonals are of the same length. Therefore, ABCD is a square and hence, Champa was correct 6.

 1  1

2

  2  0  

 2 

2

  2 

2



2

2

2

 2    2 

1   3    2  0 

2



 2

2

 5  6  2

2

 9  1  10

2

2



2

 3   3

2

1   3    2  0 

  2

2



2

2

2

2

 3  2 

 4  1

2

  2 

Diagonal AC=

2



 44  8 2 2 2

 1   1    2  2 

 3

2

 1

2

2

 3  2 

 3

2

 1

2

 7  1

 02   4   16  4 2

1   3    0  0 

2

 4 2  0 2  16  4 It can be observed that all sides of this quadrilateral are of the same length and also, the diagonals are of the same length. Therefore, the given points are the vertices of a square. (ii)Let the points (− 3, 5), (3, 1), (0, 3), and (−1, −4) be representing the vertices A, B, C, and D of the given quadrilateral respectively. 2



 6 

2

  4

2

 0   2

2

 4  4

2

  5  3

2

 0 4  2

Diagonal CD=

2

2

 3  3   5  1

 4  1

 9  1  10

2

2

2

 7  4    6  3

AD =

2

 44  8 2 2

AB =

 4  81  85

 9  1  10

2

Diagonal BD =

 4  7 2

CD=

2

 44  8 2 2

Diagonal AC =

2

 9  9  18

1   1    0  2  

AD=

2

 3   1

BC=

 44  8 2 2

CD =

2

 2   9 

AB =

BC = 2

 1  49  85

It can be observed that all sides of this quadrilateral are of different lengths. Therefore, it can be said that it is only a general quadrilateral, and not specific such as square, rectangle, etc. (iii)Let the points (4, 5), (7, 6), (4, 3), and (1, 2) be representing the vertices A, B, C, and D of the given quadrilateral respectively.

(i) Let the points (−1, −2), (1, 0), (−1, 2), and (−3, 0) be representing the vertices A, B, C, and D of the given quadrilateral respectively.  AB = 2

2

2

 3   1    5   4  

AD= 

2

7.

2

2

  6  2 

2

 6   4

2

 36  16  52  13 2 It can be observed that opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram. We have to find a point on x-axis. Therefore, its y-coordinate will be 0. Let the point on x-axis be (x, 0). Distance between (x, 0) and (2, 5) =

 x  2

2

2

  0  5 

 x  2

2

 5

2

2

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

345

Distance between (x,0) and ( 2, 9) = 2

 x   2     0   9  

2

 x  2



2

 9

2

By the given condition, these distances are equal in measure. 2

2

2

 x  2   5 

 x  2

2

 x  2  9 2

Therefore, the point is (− 7, 0). It is given that the distance between (2, −3) and (10, y) is 10.

 2  10 

 8

2

2

2

  3  y   10

5

2

2

2

  3  1  2

  4  =

 x 

0  x  2

  5 

2

 1  6 

2

2

25  16  x 2  25

41=x2+25 16 = x2 x = ±4

PR =

 5  4    3  6 

2

2

2

2

2

2

x2 + 96x+ y2 + 36  12y = x2 + 9+ 6x + y2 +168y 36 16 = 6x + 6x + 12y  8y 20 = 12 x + 4y 3x + y = 5 3x + y  5 = 0

AB  ( a  a)2  ( a  a)2  ( 2a)2  ( 2a)2

Therefore, point R is (4, 6) or (−4, 6). When point R is (4, 6), 2

 x   3     y  4 

= (y – a – b)2 + (y – b + a)2  (x – a – b + x – a + b) (x – a – b – x + a – b) = (y – a – b + y – b + a) (y – a – b – y + b – a)  (2x – 2a) (–2b) = (2y – 2b) (–2a)  –4bx + 4ab = –4ay + 4ab  –4bx = –4ay  bx = ay. 12. A = (a, a) ; B = (–a, –a), C( a 3, a 3)

PQ = QR

 5  0



11. Let P(x, y), A(a + b, b – a) and B(a – b, a + b) be the given points. Since PA = PB,  PA2 = PB2  (x – a – b)2 + (y – b + a)2 = (x – a + b)2 + (y – a – b)2  (x – a – b)2 – (x – a + b)2

2

  3  y  =10

64 + (y + 3)2 = 100 (y + 3)2 = 36 y + 3 = ±6 y + 3 = 6 or y  3 = 6 Therefore, y =3 or  9 9.

  y  6

2

 x  3 2

 25   x  2  81

Therefore,

2



 x  3   y  6    x  3   y  4  2 2 2 2  x  3   y  6   x  3   y  4

2

x2 + 4 4x + 25 = x2 + 4 + 4x + 81 8x = 25  81 8x =  56 x=7 8.

10. Point (x, y) is equidistant from (3, 6) and (−3, 4).

 4a2  4a2  8a2  2 2a

 12   9 

2

BC  (a 3  a)2  (a 3  a)2

 1  81  82

QR =

 0  4

2

2

 1  6  

 4 

2

  5 

2

 3a 2  a 2  2 3a 2  3a 2  a 2  2 3a2

 8a 2  2 2a

 16  25  41 When point R is (−4, 6),

PR=

2

 5   4     3  6 

AC  ( a 3  a)2  (a 3  a)2 2



9

2

  9 

2

 3a 2  a 2  2 3a 2  3a 2  a 2  2 3a2

 81  81  9 2

QR =

2

 0   4    1  6 

2



 4

2

  5 

2

 8a 2  2 2a Since AB = BC = AC  ABC is equilateral.

 16  25  41

____________________________________________________________________________________________________ _____________________________________________

Coordinate Geometry Solutions

13. A = (3, –4) ; B = (4, 2) ; C = (5, –4) ; D = (4, –10). AB  (4  3)2  (2  4)2  1  36  37 BC  (5  4)2  ( 4  2)2  1  36  37 CD  (4  5)2  ( 10  4)2  1  36  37 DA  (3  4)2  ( 4  10)2  1  36  37  AB = BC = CD = DA Diagonal AC  (5  3)2  ( 4  4)2  4  10  2 Diagonal BD  (4  4)2  ( 10  2)2  0  144  12  Diagonal AC  Diagonal BD.  All four sides are equal but diagonals are unequal. Hence ABCD is a rhombus. 14. Co-ordinates of mid-point of the line segment joining the points (5, 7) and (3, 9) are  5  3 7  9   8 16   2 , 2    2 , 2   (4,8)     Also the co-ordinates of mid-point of the linesegment joining the points (8, –6) and (–0, 10) are  8  0 6  10   8 16   2 , 2    2 , 2   (4,8)     Since the co-ordinates of the mid-point of two line segments are same.  The mid-point of the line segment joining the points (5, 7) and (3, 9) is also the midpoint of the line segment joining the points (8, 6) and (–0, 10). 15. Let the co-ordinates of R be (x, y). mx 2  nx1 my 2  ny1 Then, x  and y  mn mn

Here x1 = 1, y1 = 2; x2 = 2, y2 = 3 and m = 4, n = 3. 4(2)  3(1) 11  x  43 7 4(3)  3(2) 18 y  43 7  11 18  Hence,  ,  are the co-ordinates of R. 7 7  16. Let the co-ordinates of R be (x, y). Then 4(4)  7( 2) 16  14 2 x   47 11 11

346

4(7)  7(3) 28  21 49   47 11 11  2 49  Hence  ,  are the co-ordinates of R.  11 11  17. Let the co-ordinates of P be (x, y). Then 4(1)  3(5) 4  15 19 x   34 7 7 4(4)  3(2) 16  6 22 y   34 7 7  19 22  Hence  ,  are the co-ordinates of P.  7 7  18. Let P(x, y) be the required point. It divides A (–3, –4) and B(–8, 7) internally in the ratio 7 : 5. 5( 3)  7( 8) 15  56 17 Then x    75 12 12 5( 4)  7(7) 20  49 29 y   75 12 12  71 29   Co-ordinates of P are   ,  .  12 12  19. Let the given point be A, B, C and D respectively. Then, Coordinates of the mid-point of AC are  2  4 1  3  ,    1, 1 2   2 Coordinates of the mid-point of BD are  11 0  2  ,    1, 1 2   2 Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, we shall see whether ABCD is a rectangle or not. We have, y

AC 

2

 4   2  3   1

and BD 

1  1

2

2

 2 13

2

 0  2  2

Clearly, AC  BD . So ABCD is not a rectangle. 20. Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other.  Coordinates of the mid-point of AC = Coordinates of the mid-point of BD  1  2 0  2   3  x 1  y   , ,    2   2 2   2 3 x 1 y 1   and 1 2 2 2  x  2 and y  1

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

347

Hence, the fourth vertex of the parallelogram is (–2, 1) 21. We know that the diagonals of a parallelogram bisect each other. So, coordinates of the midpoint of diagonal AC are same as the coordinates of the mid-point of diagonal BD.  6  9 1 4   8  p 2  3   , ,    2   2 2   2  15 5   8  p 5   ,   ,   2 2  2 2 15 8  p    15  8  p  p  7 2 2 22. We know that the diagonals of a parallelogram bisect each other. Therefore, the coordinates of the mid-point of AC are same as the coordinates of the mid-point of BD i.e.,  2  4 1  b   a  1 0  2  , ,    2   2 2   2 a 1 b 1   1 and 1 2 2 Hence, a = 1 and b = 3 23. Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of ABC . Let D(1, 2), E(0, –1), and F(2, –1) be the mid-points of sides BC, CA and AB respectively. Since D is the mid-point of BC. x2  x3 y  y3   1 and 2 2 2 2  x 2  x3  2 and y2  y3  4 ...(i) Similarly, E and F are the mid-point of CA and AB respectively. x1  x 3 y  y3   0 and 1  1 2 2  x 2  x3  2 and y2  y3  4 ...(ii)

 x1  x3  0 and y1  y3  2 ...(iii) From (i), (ii) and (iii), we get

A(x1, y1 )

F(2, –1)

B(x2, y2)

E(0, –1)

D(1, 2)

C(x3 , y3)

From (ii) and (iv), we get x2 + 0 = 3 and y2 – 2 = 0 So, coordinates of B are (3, 2) x3 + 4 = 3 and y3 – 2 = 0  x 2  3 and y2  2

 x3  1 and y3  2 So, coordinates of C are (–1, 2) Hence, the vertices of the triangle ABC are A(1, –4), B(3, 2) and C(–1, 2) 24. Co-ordinates of mid-point of the line-segment joining the points (5, 7) and (3, 9) are  5  3 7  9   8 16   2 , 2    2 , 2   (4,8)     Also the co-ordinates of mid-point of the linesegment joining the points (8, –6) and (–0, 10) are Since the co-ordinates of the mid-point of two line segments are same.  The mid-point of the line segment joining the points (5, 7) and (3, 9) is also the midpoint of the line segment joining the points (8, 6) and (–0, 10). 25. Let the centroid be G(x, y), then abc x 3 (b  c)  (c  a)  (a  b) 0 and y   0 3 3 Since the ordinate of the centroid is zero, it lies on the x-axis.

 x 2  x 3    x1  x 3    x1  x 2   2  0  4 and  y2  y3    y1  y3    y1  y 2   4  2  2 

2  x1  x 2  x 3   6

and 2  y1  y 2  y 3   0

 x1  x 2  4and y1  y2  2 ...(iv) from (i) and (iv), we get x1 + 2 = 3 and y1 + 4 = 0  x1  1 and y1  4 So, the coordinates of A are (1, –4)

26.

k1 k a  2 b  c  0 and the line ax + by +C = k3 k3

k k  0 Clearly line passes through  1 ,  2  .  k3 k 3  1 1  27. Slope of diagonal BD  62 2 53 1 eq. of line y – b   x  5   x  2y  17 . 2

____________________________________________________________________________________________________ _____________________________________________

Coordinate Geometry Solutions

348

28. Let AB be a diameter of the circle having its centre at C(1, –3) such that the coordinates of one end A are (4, –1) Let the coordinates of B be (x, y). Since C is the mid-point of AB. Therefore, the  x  4 y 1 coordinates of C are  ,  2   2 But, the coordinates of C are given to be (1, –3) x4 y 1   1 and  3 2 2  x  4  2 and y  1  6 B(x, y) C(1, –3) A(4, –1)

Hence, the coordinates of B are (–2, –5) 29. Let D, E, F be the mid-point of the sides BC, CA and AB respectively. Then, the coordinates of D, E and F are  5  3 3 1   3  7 1  3  D , ,   D  4,1 , E   2  2   2  2  E  5,  2   7  5 3  3  and F  ,   F  6, 0  2   2 A(7, –3)

E(5, –2)

F(6, 0)

B(5.3)



D(4, 1)

AD 

 7  4

2

C(3, –1)

  3  1

2

 9  16  5 units and

CF 

 6  3

2

2

  0  1  9  1  10 units

30. Let AD be the median through the vertex A of ABC . Then, D is the mid-point of BC. So, the coordinates of D are  3  1  2  8  ,   i.e.,  2,3 2   2



AD 

 5  2

2

  1  3

2

A(5, –1)

B (–3,–2)

D (–2, 3)

C (–1,8)

and divides it in the ratio 2 : 1. So, coordinates of G are  2  2  1  5 2  3  1  1  ,   2 1 2 1    4  5 6  1 1 3      ,  ,  3  3 5  3 31. Let P(x, y) be the required point. Using the section formula, we obtain 2  4  3   1 8  3 x  1 23 5 2   3  3  7 6  21 15 y   3 23 5 5 Therefore, the point is (1, 3). 32.

Let P (x1, y1) and Q (x2, y2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB Therefore, point P divides AB internally in the ratio 1:2. 1  2   2  4 1   3  2   1 x1  , y1  1 2 1 2 1  2   2  4 1   3  2   1 x1  , y1  1 2 1 2 5 Therefore, P(x1, y1) =  2,   3  Point Q divides AB internally in the ratio 2:1. 2   2   1  4 2   3  1   1 x2  , y1  2 1 2 1 4  4 6  1 7 x2   0, y2   3 3 3 7 Q(x2, y2) =  0,   3 

 49  16  65 units

Let G be the centroid of ABC . Then, G lies on median AD ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

349

33. It can be observed that Niharika posted the 1 green flag at of the distance AD 4 1   i.e.,   100  m  25 m from the starting point 4   of 2nd line. Therefore, the coordinates of this point G is (2, 25). 1 Similarly, Preet posted red flag at of the 5 1  distance AD i.e.,   100  m  20 m from the 5  Distance between these flags by using distance formula = GR =

2

 8  2    25  20 

2

 36  25  61m

The point at which Rashmi should post her blue flag is the mid-point of the line joining these points. Let this point be A (x, y). 28 25  20 x ,y  2 2 10 45 x   5, y   22.5 2 2

Hence, A(x, y) = (5.22.5) Therefore, Rashmi should post her blue flag at 22.5m on 5th line. 34. Let the ratio in which the line segment joining (−3, 10) and (6, −8) is divided by point (−1, 6) be k : 1. 6k  3 Therefore, 1 = k 1

k1 = 6k  3 7k = 2 k=

2 7

Therefore, the required ratio is 2: 7. 35. Let the ratio in which the line segment joining A (1, −5) and B (−4, 5) is divided by x-axis be k : 1. Therefore, the coordinates of the point of  4k  1 5k  5  division is  , .  k 1 k 1  We know that y-coordinate of any point on xaxis is 0. 5k  5  0 k 1

Division point =

 4 1  1 5 1  5   4  1 5  5   3  , ,      ,0  11   2 2   2   11 36.

starting

point of 8th line. Therefore, the coordin

Let (1, 2), (4, y), (x, 6), and (3, 5) are the coordinates of A, B, C, D vertices of a parallelogram ABCD. Intersection point O of diagonal AC and BD also divides these diagonals. Therefore, O is the mid-point of AC and BD. If O is the mid-point of AC, then the coordinates of O are 1 x 2  6   x 1  , ,4   2   2  2  If O is the mid-point of BD, then the coordinates of O are  43 5 y 7 5 y ,   ,  2  2 2   2 Since both the coordinates are of the same point O, x 1 7 5 y   and 4  2 2 2

 x + 1 = 7 and 5 + y = 8  x = 6 and y = 3 37. Let the coordinates of point A be (x, y). Mid-point of AB is (2, −3), which is the center of the circle.  x 1 y  4   (2,  3) =  ,  2   2 x 1 y4   2 and  3 2 2

 x + 1 = 4 and y + 4 =  6  x = 3 and y = 10 Therefore, the coordinates of A are (3, 10)

k=1 Therefore, x-axis divides it in the ratio 1:1.

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Coordinate Geometry Solutions

38.

350

40.

The coordinates of point A and B are (−2, −2) and (2, −4) respectively. 3 Since AP  AB , 7 Therefore, AP: PB = 3:4 Point P divides the line segment AB in the ratio 3:4. Coordinates of P =

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) are the vertices A, B, C, D of a rhombus ABCD. Length of diagonal AC = 2

3   1    0  4 

= 16  16  4 2 Length of diagonal BD=

 3  2  4   2  3   4   4   2   ,   3 4 34  

2

 4   2    5   1 

6  8 12  8  ,  7   7

= 

2  7

=   , 

2

2

= 36  36  6 2 Therefore, area of rhombus ABCD =

20   7 

1  4 2 6 2 2

=24 square units 41.

39.

(i) Area of a triangle is given by Area of a triangle = 1 x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2 1 = 2  0   4     1  4    3  2  3  0  2 1 21 8  7  6  squareunits 2 2 (ii) Area of given triangle =



From the figure, it can be observed that points P, Q, R are dividing the line segment in a ratio 1:3, 1:1, 3:1 respectively. Coordinates of P =

 1 2  3   2  1 8  3  2  ,   1 3 1 3   7 =  1,  

2

 2   2  2  8  ,  =(0, 5) 2 2  

Coordinates of Q =  Coordinates of R =

 3  2  1  2  3  8  1 2   13  ,   = 1, 31 3  1   2  



1  5    5   2    3  2   1   5  1   5    2 1 35  9  20  32square units 2 42. (i) For collinear points, area of triangle formed by them is zero. Therefore, for points (7, −2) (5, 1), and (3, k), area = 0 1  7 1  k   5  k   2    3   2   1   0 2

7 7k +5k + 10 9 = 0 2k +8 = 0 k=4 (ii) For collinear points, area of triangle formed by them is zero. Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0

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10th Class Mathematics

351

Area of a triangle = 1 x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2

1 8  4   5   k   5  1   2 1   4     0 2

8 6k + 10 = 0 6k = 18 k=3 43.

Area of  ABC = 1  4   5    2    3 2    2   3 2    5  2 1 21 = 12  0  9   square units 2 2

Area of ACD = 1  4   2    3  3 3   2   2 2    2  2 1 35 = 20  15  0  square units 2 2

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3). Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by  0  2 1  1  D ,   1, 0  2   2  0  0 3 1  E ,    0,1 2   2  2  0 1 3  F ,   1, 2  2   2

Area of ABCD = Area of ABC + Area of ACD 21 35    square units = 28 square units 2   2

=  45.

Area of a triangle = 1 x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2

Area of DEF = 1 1 2  1  11  0   0  0  2   2 1 = 1  1  1 square units 2

Area of ABC = 1  0 1  3  2  3   1   0  1  1  2 1 = 8  4square units 2

Therefore, required ration = 1: 4 44.

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2). Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC. Coordinates of point D =  3  5 2  2  ,     4, 0  2   2

Area of a triangle = 1 x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2

Area of  ABD =

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

1  4  2    0    3  0    6    4  6    2  2 1 =  8  18  16   3square units 2 However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

____________________________________________________________________________________________________ _____________________________________________

Coordinate Geometry Solutions

352

So, ADBC is a parallelogram. C(–2,3) B(6,7)

Area of ADC = 1   4 0   2    4 2   6    5  6    0  2 1  8  32  30   3 Square units. 2 However, area cannot be negative. Therefore, area of ΔADC is 3 square units. Clearly, median AD has divided ΔABC in two triangles of equal areas.

A(0,–1)

Now, AB =

 x  2

=

 x  2

4

2

2

2

2

=  x  2    9 

BC=

 5  1 5  9

And AC=

2

2

2

4.





let B(x, y) be the third vertex of equilateral  OAB . Then, OA = OB = AB  OA2 = OB2 = AB2 We have, OA2 = (3 –0)2 +

2

2

2

2

  1  5  = 64  36 =

2

  3  1 = 64  16 = 80



2

= 9 + 3 = 12 Y-axis A(3,

X -axis

2

2

  3  1 = 4  16

= 20 = 2 5

8  6 

2

2

 3  7  =

= 20 = 2 5  AD = BC and AC = BD

4  16

X-axis

O (0,0) B(x,y) Y -axis

OB2 = x2 + y2





2

And

AB2 = (x – 3)2 + y  3



AB2 = x2 – 6x + 9 + y2 2 3y + 3

2

  7  3 = 64  16 = 80

 2  0  =

3 0

2

8  0 

And BD



  2  5 = 16  9 = 25 =5

= 4 5 AC =



O (0,0) and A 3, 3 be the given points and

2

1  9 

 6  2

2

  3  3 = 100  0

  2  1 = 16  9 = 25 =5

=4 5 BC =

2

5  2 5 sq.units = 40 sq.units.

100 = 10 Now, it is clear that AC = AB + BC Hence, A, B, C are collinear points. Let A(0, –1), B(6, 7), C(–2, 3) and D(8, 3) be the given points. Then

AD =

2

  7  1 = 36  64 =

= 100 = 10 It is clear that: AB2 = AD2 + DB2 and Hence, ADBC is a rectangle. Now, Area of rectangle ABCD = AD  BD =

2

[Squaring both sides]  x2 – 4x + 4 + 25= x2 + 4x + 4 + 81  –4x – 4x = 85 – 29  –8x = 56  x = –7 Hence, the required point is (–7, 0). .Let A(1, –1), B(5, 2) and C(9, 5) be the given points. Then, we have AB =

3.

  0  9

  0  5

 x  2    5



2

2

2

8  2

And CD =

We know that a point on x-axis is of the form (x, 0). So, Let P(x, 0) be the point equidistant from A(2, –5) and B(–2, 9). Then, PA = PB



 6  0

100 =10

CONCEPTIVE WORKSHEET 1.

D(8,3) 2

 AB2 = x2 + y2 – 6x 2 3y + 12  OA2 = OB2 = AB2  OA2 = OB2 and OB2 = AB2  12 = x2 + y2 and x 2  y 2  x 2  y 2  6x  2 3y  12 

x 2  y 2  12

and  6x  2 3y  12

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10th Class Mathematics

353

x 2  y 2  12 and 3x  3y  6



2

  2  3 = 1  25

2

  2  1 = 25  1

2

  3  1 = 16  16 =

 3  2

RS=

2

2

 6  3x  x2     12  3   3x  3y  6    6  3x   y  3 

=



x2 



 6  3x 

     

3x 2  36  36x  9x 2  36 12x 2  36x  0 x = 0 and x = 3 x=0 3y  6 23

2 3  3

2 3 3 3 3 [Putting x = 0 in 3x  3y  0 ] and x=3  9  3y  6 69  y  3 3 [Putting x = 3 in 3x  3y  6 ] Hence, the co-ordinates of the third vertex B





 

are 0, 2 3 or 3,  3 5.



The given points are P(2, –1), Q(3, 4), R(–2, 3) and S(–3, –2). We have S(–3,–2) R(–2,3)

Q(3,4)

P(2,–1) PQ=

 3  2

2

2

  4  1 = 1  25 = 26

units. QR=

 2  3

= 26 units.

2

 3  3

2

  2  4 

2

=

36  36 = 72 = 6 2 units.

2

3x   6  3x   36



And QS =

 12



6

2

2

3

y

 2  2 

PR=

32 = 4 2 units.

3

2

2

= 26 units.

 12

3x   6  3x 



 3  2 

SP =

2

2

26 units.

2

  3  4  = 25  1

7.

 PQ = QR = RS = SP = 26 units. and PR  QS i.e., PQRS is a quadrilateral in which all sides are equal but diagonals are not equal. Now, area of rhombus 1 PQRS =  (Products of length of diagonals) 2 1 = (PR  QS) 2 1 =  4 2  6 2 sq.units. 2 = 24 sq. units. Let us apply the distance formula to find the distances PQ, QR and PR, where P(3, 2), Q(–2, –3) and R(2, 3) are the given points. We have 2

2

  2  3   52  5 2  50  7.07(approx.)

PQ 

3  2

QR 

 2  2 

2

  3  3  

PR 

 3  2

2

  2  3   12   1  2  1.41(approx.)

2

 4 

2

2

  6   52  7.21(approx.)

2

2

Since the sum of any two of these distances is greater than the third distance, therefore, the points P, Q and R form a triangle. Also, PQ2 + PR2 = QR2, by the converse of Pythagoras theorem, we have P = 90°. Therefore, PQR is a right triangle. 8. Let A(1, 7), B(4, 2), C(–1, –1) and D(– 4, 4) be the given points. One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its digonals should also be equal. Now, 2

2

AB 

1  4 

  7  2   9  25  34

BC 

 4  1   2  1

CD 

 1  4    1  4 

DA 

1  4    7  4 

2

2

 25  9  34

2

2

2

2

 9  25  34

 25  9  34

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Coordinate Geometry Solutions 2

2

AC 

1  1

  7  1  4  64  68

BD 

 4  4    2  4

2

2

 64  4  68

Since, AB = BC = CD = DA and AC = BD, all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal. Thereore, ABCD is a square. 9. Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5). We are given that AP = BP. So, AP2 = BP2 i.e., (x – 7)2 + (y – 1)2 = (x – 3)2 + (y – 5)2 i.e., x2 – 14x + 49 + y2 – 2y + 1 = x2 – 6x + 9 + y2 – 10y + 25 i.e., x – y = 2 which is the required relation. 10. We know that a point on the y axis is of the form (0, y). So, let the point P(0, y) be equidistant from A and B. Then (6 – 0)2 + (5 – y)2 = (– 4 – 0)2 + (3 – y)2 i.e., 36 + 25 + y2 – 10y = 16 + 9 + y2 – 6y i.e., 4y = 36 i.e., y = 9 So, the required point is (0, 9). Let us check our solution: 2

AP =

 6  0   5  9

BP =

 4  0 

2

2

 36  162  52 2

  3  9   16  362  52

11. Let the co-ordinates of B be  ,   in figure. It is given that AC : BC = 3 : 4. So, the coordinates of C are : 3 4

A(2,5) C(–1,2) B(, )  3  4  2 3  4  5   3  8 3  20   3 4 , 3 4  =  7 , 7      But the co-ordinates of C are (–1, 2) 3  8 3  20   1 and =2 7 7  3  8 = –7 ;  3  20  14  3 = –7 –8 ;  3 = 14 – 20 = –6   = –5 ;   = –2 Hence, the co-ordinates of B are (–5, –2). 12. Let the point P(–3, p) divide the line segment joining points A(–5, –4) and B(–2, 3) in the ratio K : 1. Then, the co-ordinates of P are  2K  5 3K  4  ,    K 1 K 1  But, the co-ordinates of P are given as (–3, p). 2K  5 3K  4   3 and p K 1 K 1

354

 2K  5 = –3K –3 and

3K  4 p K 1

3K  4 K 1 2  K = 2 and p = . 3

 K = 2 and p =

2 . 3 13. Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other :  Co-ordinates of the mid-point of AC = Coordinates of the mid-point of BD :  1  2 0  2   3  x 1  y   , , =   2   2 2   2 1   3  x y 1   ,1  =  ,  2  2   2 3 x 1 y 1   and 1 2 2 2  x = –2 and y = 1 Hence, the fourth vertex of the parallelogram is (–2, 1). 14. We know that diagonals of a parallelogram bisect each other. So, the co-ordinates of the mid-point of diagonal AC are same as the coordinates of the mid-point of diagonal BD. 6  9 1 4   8  p 2  3    , , =   2   2 2   2  15 5   8  p 5    , =  ,   2 2  2 2 15 8  p  = 2 2  15 = 8 + p  p = 15 – 8 = 7 15. We know that the diagonals of a parallelogram bisect each other. So, the co-ordinates of the mid-point of AC are same as the co-ordinates of the mid-point of BD.  2  4 1  b   a  1 0  2  i.e.,  , ,  =  2   2 2   2  1  b   a  1    1, ,1   = 2   2   a 1 1  b 1 = and 1 2 2  a + 1 = 2 and b –1 = 2  a = 1 and b = 3 Hence, a = 1 and b = 3.

Hence, the ratio is 2 : 1 and p 

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355

16. Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of triangle ABC. Let D(1, 2), E(0, –1) and F(2, –1) be them idpoints of sides BC, CA and AB respectively. In figure. A(x 1,y 1)

17. Let D, E, F be the mid-points of the sides BC, CA and AB respectively in figure. A(7,–3)

E(5,–2)

F(6,0)

F(2,–1)

E(0,–1) B(5,3)

B(x2,y2) D(1,2) C(x3,y3)  D is the mid-point of BC x  x3 y  y3  2  1 and 2 2 2 2  x2 + x3 = 2 and y2 + y3 = 4 … (i) Similarly, E and F are the mid-points of CA and AB respectively. x x y1  y3  1 3 0 and  1  2 2 x1  x 3  0 and y1 + y3 = –2 … (ii) x  x2 y  y2 and 1  2 and 1  1  x1 + x2 = 2 2 4 and y1 + y2 = –2 … (iii) From (i), (ii) and (iii), we get (x2 + x3) + (x1 + x3) + x1 + x2) = 2 + 0 + 4 and (y2 + y3) + (y1 + y3) + (y1 + y2) = 4 – 2 – 2  2(x1 + x2 + x3) = 6 and 2(y1 + y2 + y3) = 0  x1 + x2 + x3 = 3 and y1 + y2 + y3 = 0 … (iv) From (i) and (iii), we get x1 + 2 = 3 and y1 + 4 = 0  x1 = 1 and y1 = –4  Co-ordinates of A are (1, –4) From (ii), and (iv), we get x2 + 0 = 3 and y2 –2 = 0  x2 = 3 and y2 =2  Co-ordinates of B are (3, 2) From (iii) and (iv), we get x3 + 4 = 3 and y3 = 2  x3 = –1 and y3 = 2 And, So co-ordinates of C are (–1, 2) Hence, the vertices of the triangle ABC are A(1, –4), B(3, 2) and C(–1, 2).

C(3,–1)

D(4,1)

Then, co-ordinates of D, E and F are :  5  3 3 1 D ,  = D(4, 1); 2   2  3  7 1  3  E ,  = E(5, –2) 2   2  7  5 3  3  and F ,  = F(6, 0) 2   2

 AD

2

2

  3  1 = 9  16 =

= 5 units.

25

BE=

 7  4

=

5  5

2

2

  2  3 = 0  25 = 25

= 5 units. AndCF =

 6  3

2

2

  0  1 = 9  1 = 10

= 10 units. 18. Let AD be the median through the vertex A of  ABC . Then, D is the mid-point of BC. So,  3  1 2  8  the co-ordinates of D are  ,  2   2 i.e., (–2, 3) A(5,–1)

B(–3,2)

 AD=

5  2 

D(–2,3) 2

C(–1,8) 2

  1  3 = 49  16 =

65 units. Let G be the centroid of  ABC , G lies on median AD and divides it in the ratio 2 : 1. So, co-ordinates of G are :  2  2  1  5 2  3  1  1   4  5 6  1  , ,  =   2 1 2 1 3     3 1 5 =  ,  3 3

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Coordinate Geometry Solutions

19. Let P(x, y) be the required point. Using the section formula, we get 3  8   1 4  3  5   1 3  x  7, y  3 31 3 1 Therefore, (7, 3) is the required point. 20. Let (– 4, 6) divide AB internally in the ratio m1 : m2. Using the section formula, we get  3m  3m 2 8m1  10m 2  ,  4,6    1  …(i) m1  m 2   m1  m 2 Recall that if (x, y) = (a, b) then x = a and y = b. 3m1  6m 2 8m1  10m 2 So, 4  and  6  m1  m 2 m1  m 2 3m1  6m 2 Now, 4  gives m1  m 2 – 4m1 – 4m2 = 3m1 – 6m2 i.e., 7m1 = 2m2 i.e., m1 : m2 = 2 : 7 You should verify that the ratio satisfies the y-coordinate also. m 8 1  10 8m1  10m 2 m2  m1 m1  m 2 1 m2 (Dividing throughout by m2) 2 8   10 7 6 2 1 7 Therefore, the point (– 4, 6) divides the line segment joining the points A(– 6, 10) and B(3, – 8) in the ratio 2 : 7. 21. Let P and Q be the points of trisection of AB i.e., AP = PQ = QB (see Fig).

Therefore, P divides AB internally in the ratio 1: 2. Therefore, the coordinates of P, by applying the section formula, are  1 7   2  2  1 4   2  2   ,   ,i.e.,  1, 0  1 2 1 2   Now, Q also divides AB internally in the ratio 2: 1. So, the coordinates of Q are  2  7   1 2  2  4   1 2   ,   ,i.e.,  4, 2  2 1 2 1   Therefore, the coordinates of the points of trisection of the line segment joining A and B are (–1, 0) and (– 4, 2).

356

22. Let the ratio be k : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are  k  5 4k  6  ,   k 1   k 1 This point lies on the y-axis, and we know that on the y-axis the abscissa is 0. k  5 Therefore, 0 k 1 So, k = 5 That is, the ratio is 5 : 1. Putting the value of k = 5, we get the point of intersection as  13   0,  3   23. We know that diagonals of a parallelogram bisect each other. So, the coordinates of the mid-point of AC = coordinates of the mid-point of BD  6  9 1 4   8  p 2  3  , ,    2   2 2   2  15 15   8  p 5  ,   ,   2 2   2 2 15 8  p  2 2 p=2 24. Here, x1 = 5, y1 = 2, x2 = 4, y2 = 7, x3 = 7 and y3 = –4.  Area of  ABC 1 = x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2 1 = 5  7  4   4  4  2   7  2  7  2 1 1 = 55  24  35 = 4 2 2 Hence, area of  ABC = 2 sq. units. 25. We have, A(4,6) 1 D

1 E

3

3

B(1,5)



C(7,2)

AD AE 1   AB AC 4 AB AC  4 AD AE

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357

AD  DB AE  EC  4 AD AE DB EC  1 1 4 AD AE DB EC  = =3 AD AE [Subtracting 1 from each side] AD AE 1  = = DB EC 3  AD : DB = AE : EC = 1 : 3  D and E divide AB and AC respectively in the ratio 1 : 3. So, the co-ordinates of D and E are  1  12 5  18   13 23  and ,    ,   1 3 1 3   4 4   7  12 2  18   19  ,   =  , 5  respectively.  1 3 1 3   4  We have : 13 19 4 4 4 4 23 5 6 6 4  Area of  ADE =

23 13 19    4   5   6 4 4 4 1  2  13 19 23    6    4 5 4 4 4    Area of  ADE = 1  92 65 114   78 437    20       2 4 4 4   4 16  1 271 1069  Area of  ADE =  2 4 16 1 15 =  2 16 15 = sq.units. 16 Also, we have 4 4 4 7

5 2 6 6  Area of  ABC = 1  4  5  1 2  7  6   1 6  7  5  4  2  2  Area of  ABC = 1  20  2  42    6  35  8  2 1  Area of  ABC = 64  49 2

15 sq.units. 2 Area of  ADE 15 / 32 1    Area of  ABC 15 / 2 16 Hence, Area of  ADE : Area of  ABC = 1: 16. The area of the triangle formed by the vertices A(1, –1), B(– 4, 6) and C (–3, –5), by using the formula above, is given by 1 1 6  5    4  5  1   3 1  6  2 1 11  16  21  24 2 So, the area of the triangle is 24 square units. The area of the triangle formed by the vertices A(5, 2), B(4, 7) and C (7, – 4) is given by 1 5  7  4   4  4  2   7  2  7   2 1 4 55  24  35   2 2 2 Since area is a measure, which cannot be negative, we will take the numerical value of – 2, i.e., 2. Therefore, the area of the triangle = 2 square units. The area of the triangle formed by the given points is equal to 1 1.5  2  4   6  4  3   3 3  2   2 1 9  6  15  0 2 If the area of a triangle is 0 square units, then its vertices will be collinear. Since the given points are collinear, the area of the triangle formed by them must be 0, i.e., 1  2  k  3  4  3  3  6  3  k    0 2 1  4k   0 2 Therefore, k = 0 By joining B to D, you will get two triangles ABD and BCD. Now the area of ABD = 1 5  5  5    4  5  7   4  7  5  2 1 106 50  8  48   53square units 2 2 Also, the area of BCD 1 4  6  5  1 5  5  4  5  6  2

 Area of  ABC =



26.

27.

28.

29.

30.

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Coordinate Geometry Solutions

358

If we join points B and D, we obtain two triangles ΔABD and ΔBCD. The area of the triangle formed by the vertices (x1, y1), (x2, y2), and (x3, y3) is given by the numerical value of 1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 In ΔABD, the vertices are A (–1, –2), B (–3, – 4), and D (8, 4). 1 ∴ Area of ΔABD = [(–1)(– 4 – 4) + (–3) 2 {4 – (–2)} + 8{–2 – (–4)}] 1 = [(–1) × (–8) – 3 × (4 + 2) + 8 × (–2 + 4)] 2 1 = [8 – 3 × 6 + 8 × 2] 2 1 = [8 – 18 + 16] 2 1 = × 6 = 3 square units 2 ∴ Area of ΔABD = 3 square units In ΔBCD, the vertices are B (–3, – 4), C (–7, –3), and D (8, 4). 1 ∴ Area of ΔBCD = [(–3) (–3 – 4) + (–7) 2 {4 – (– 4)} + 8{– 4 – (–3)}] 1 = [–3 × (–7) –7 × 8 + 8 (– 4 + 3)] 2 1 = [21 – 56 – 8] 2 1 43 = [21 – 64] = 2 2 43 ∴ Area ΔBCD = square units 2 Thus, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD 43 49 3  square units 2 2 Hence, the area of the given quadrilateral 49 ABCD is square units. 2 The correct answer is C.

1  44  10  4  19square units 2 So, the area of quadrilateral ABCD = 53 + 19 = 72 square units.

SUMMATIVE WORKSHEET 1.

Using distance formula, we obtain

BC 

 6  6

2



  3  13  16

Area of ABC = 1 1  BC  AD  0 13  3  6  3  5   6  5  13  2 2  16 × AD = 96  16AD = 96 (Area is always non negative)  AD = 6 Thus, the length of AD is 6 units. The correct answer is A 2. The vertices of the triangle are given as (−3, 1), (0, h), and (−1, 0). Area of the triangle formed by the given points 9 is given as square units. 2 1   3 h  0    0  0  1   11  h   2 9 = 2  3h 1 + h = 9  2h  1 = 9   2h = 10  h=5 Thus, the value of h is −5. The correct answer is A. 3. The vertices of the quadrilateral ABCD are given as A (–1, –2), B (–3, –4), C (–7, –3), and D (8, 4).

4.

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10th Class Mathematics

359

5.

Let ABCD be the given quadrilateral whose vertices are A (3, 4), B (1, 6), C (–3, 3), and D (2, – 1). Then, area of ABCD = area of ΔABD + area of ΔBCD The area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is given by the numerical value of the following expression. 1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 Therefore, area of ΔABD 1 3  6   1   1 1  4   2  4  6   2 1 = 3  7  1  5   2   2   2 1 =  12  6 squareunit 2 Similarly, area of ΔBCD 1 1 3   1   3  1  6   2  6  3 2 1 = 1  4  3   7   2  3 2 1 =  4  21  6 2 1 = 31  15.5square unit 2 ∴ area of ABCD = (6 + 15.5) square unit = 21.5 square unit Thus, the area of quadrilateral ABCD is 21.5 square unit. The correct answer is B. The area of a triangle formed by the vertices (x1, y1), (x2, y2), (x3, y3) is given by the expression 1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 Since (4, 7), (6, ), and (2, 13) lie on a straight line, the area of the triangle formed by these points is 0. 1 ⇒  4    13  6 13  7   2  7      0 2  4   52 + 36 + 14  2  = 0  2   52 + 36 + 14 = 0  2   52 + 50 = 0  2 = 2 =1 Thus, the required value of  is 1. The correct answer is C.

6.

7.

The area of a triangle formed by the points (x1, y1), (x2, y2), and (x3, y3) is given by, 1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 ∴ Area of ΔABC 1 7  5  9   x  9  3  4  3  5   2 1  18  7  4   6x  4  2  2 1   28  6x  8 2 1   6x  36 2  18 = 3x  18 ⇒ 3x = 36 1 ⇒ x =  36  12 units 3 Hence, the required value of x is 12 units. The three points are A (5, 2), B (x, y), and C (4, 3). The distance between two points P (x1, y1) and Q (x2, y2) is given by PQ = ∴ AB 

BC 

 x 2  x1  5  x  2

2

2

  y 2  y1 

2

2

  2  y  and

 x  4    y  3

2

It is also given that AB = BC. ⇒ (AB) 2 = (BC) 2 ⇒ (5 – x)2 + (2 – y)2 = (x – 4)2 + (y – 3)2 ⇒ 25 + x2 – 10x + 4 + y2 – 4y = x2 + 16 – 8x + y2 + 9 – 6y ⇒ 29 – 10x – 4y = 25 – 8x – 6y ⇒ 29 – 10x – 4y – 25 + 8x + 6y = 0 ⇒ –2x + 2y + 4 = 0 ⇒ 2x – 2y – 4 = 0 ⇒ x – y = 2 … (1) 7 Area of ΔABC = square units 2 It is known that the area of the triangle formed by the vertices (x1, y1), (x2, y2), and (x3, y3) is 1 given by the expression [x1(y2 –y3) 2 + x2 (y3 –y1) + x3 (y1 – y2)]. 1 7 ⇒ [5 (y – 3) + x (3 – 2) + 4 (2 – y)] = 2 2 ⇒ 5y – 15 + x + 8 – 4y = 7 ⇒ y + x – 15 + 8 – 7 = 0 ⇒ x + y – 14 = 0 ⇒ x + y = 14 … (2) Adding equations (1) and (2), we obtain

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Coordinate Geometry Solutions

360

xy2 x  y  14

8.

9.

2x  16 ⇒x=8 Putting x = 8 in x – y = 2, 8–y=2 ⇒y=6 Thus, the required coordinates of B are (8, 6). The given points A (3, 2), B (a, 7), and C (–4, 3) are collinear. Therefore, the area of the triangle formed by these points is 0. The area of the triangle formed by the vertices (x1, y1), (x2, y2) and (x3, y3) is given by 1 [x1(y2–y3) + x2 (y3 –y1) + x3 (y1 – y2)]. 2 1 ∴ [3(7 – 3) + a (3 – 2) – 4(2 – 7)] = 0 2 ⇒ 3 × 4 + a – 4 × (–5) = 0 ⇒ 12 + a + 20 = 0 ⇒ a + 32 = 0 ⇒ a = –32 Thus, the required value of a is –32. The area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is given by the numerical value of the expression 1 [x1(y2–y3) + x2 (y3 –y1) + x3 (y1 – y2)]. 2 Therefore, area of ΔPQR 1 = 4  5   7     2   7   3  7  3  5   2 square units 1 = 4  12   2    10   7   2   square 2 units 1 =  48  20  14 square units 2 1 =  54squareunits 2 = 27 square unit 1 Area of ΔPQR  QR  PS 2 By distance formula, the distance between Q(−2, 5) and R (7, −7) is

QR 

2

2

7   2   7   5 units 2

= 92   12  units = 81  144units = 225units

=15 units 1 ∴27 unit2 =  15unit  PS 2 27  2  PS =   units  15  18 PS = units 5

18 units. 5 10. If point (x, y) lies on the line joining the points (3, 5) and (−2, −3), then the area of the triangle formed by these vertices is zero. The area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is given by the numerical value of the expression 1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 Thus, the length of PS is

1   x  5   3  3  3  7    2  y  5   0 2 ⇒ 8x − 9 − 3y − 2y + 10 = 0 ⇒ 8x − 5y + 1 = 0 Thus, the relation between x and y is given by the algebraic relation 11. In ΔADE and ΔABC, AD AE  [Given] AB AC ∠DAE = ∠BAC [Common] ∴ADE ~ ΔABC [by SAS similarity criterion] 2 Area  ADE   AD     Area  ABC   AB 

Area  ADE 

2

1 1     Area  ABC   3  9 ⇒ Area (ΔABC) = 9area (Δ ADE) Now, area (BDEC) = area (ΔABC) − area (ΔADE) = 9area (ΔADE) − area (ΔADE) = 8area (ΔADE) We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is given by the numerical value of the expression

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10th Class Mathematics

361

1 [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]. 2 Therefore, area of ΔADE 1  2  4   3  5 3  1  2 1  4   unit 2 2 1 =  2   7  5   4   2   3  unit 2 2 1 =  14   20    6   unit 2  2 1 =   40  unit 2 2 =20 unit2 Since the area of triangle cannot be negative, the area of ΔADE is 20 unit2. Thus, area (BDEC) = 8area (ΔADE) = 8 × 20 unit2= 160 unit2 12. It is known that the coordinates of the point P(x, y) which divides the line segment with end points A (x1, y1) and B (x2, y2) internally in the  m x  m 2 x 2 m1 y 2  m 2 y1  ratio m1: m2 are  1 2 , . m1  m 2   m1  m 2 Here, m1 = 1, m2 = 3; x1 = –2, y1 = 8; x2 = 10, y2 = 4



x



m1 x 2  m 2 x 2 1  10  3   2  10  6 4 =   1 m1  m 2 1 3 4 4

m1 y 2  m 2 y1 1  4  3  8 4  24 28 =   7 m1  m 2 1 3 4 4 Thus, the coordinates of the required point are (1, 7). The correct answer is A. 13. Let A (3, 2) and B (8, 7) be the end points of the line segment AB. Let P (5, 4) divide the line segment AB in the ratio m: n. Then, by section formula, 8m  3n 7m  2n  ,  5, 4     mn   mn 8m  3n  5 . mn  5(m + n) = 8m + 3n  5m + 5n = 8m + 3n  3m = 2n m 2   n 3 It can be verified that the y-coordinate also satisfies this ratio. Thus, the point (5, 4) divides the line segment joining the points (3, 2) and (8, 7) in the ratio 2: 3.

y=

The correct answer is B.

14. Let the x-axis divide the line segment obtained by joining the points (–6, 3) and (5, –7) in the ratio k: 1. Using the section formula, we obtain the  5k  6 7k  3  coordinates of that point as  , .  k 1 k 1  Since this point lies on the x-axis, its ycoordinate will be zero. 7k  3  0 k 1   7k + 3 = 0  7k = 3 3 k= 7 Thus, the x-axis divides the line segment 3 joining the given points in the ratio :1 , i.e., 7 3:7. The correct answer is A. 15. Let the line 2x + y – 8 = 0 divide the line segment joining the points (2, 3) and (5, 7) internally, in the ratio of k: 1. Let the point of division be P(x, y). By section formula,  5k  2 7k  3  (x, y) =  ,   k 1 k 1  The point P (x, y) lies on the line 2x + y – 8 = 0.  5k  2  7k  3  2 8  0   k 1  k 1 2(5k + 2) + 7k + 3 – 8 (k + 1) = 0 10k + 4 + 7k + 3 – 8k – 8 = 0 9k – 1 = 0 9k = 1 1 ∴k = 9 Thus, the line 2x + y – 8 = 0 divides the line segment joining the points (2, 5) and (5, 7) internally in ratio of 1:9. The correct answer is C. 16. Let the vertices of ΔABC be A (6, 12), B (6,





18) and C 6  3 3,15 . Then, by distance formula, we have

AB 

 6  6

BC 

6  3

2

2

 18  12   6units



2

2

3  6  15  18   6units

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Coordinate Geometry Solutions

AC 

6  3



2

362 2

3  6  15  12   6units

AB = BC = AC So, ΔABC is an equilateral triangle. We know that in an equilateral triangle, the circumcentre and the centroid coincide. Let O be the centre of the circumcircle and AD be the median corresponding to side BC. Then, AD passes through the centre O and O divides AD in the ratio 2:1.

Now, coordinates of D are  6  6  3 3 18  15   12  3 3 33  , ,  .   2 2   2 2  ∴ AD = 3 3units Since O divides AD in the ratio 2:1, we have OD 1  AO 2 OD 1  1  1 AO 2 AD 3   AO 2 2 2  AO   AD   3 3units  2 3units 3 3 Thus, the measure of the radius of the circle is 2 3units . The correct answer is C.

Therefore, co-ordinates of P are  19k  1 23k  7  ,  . k 1   k 1 It can be observed that P is a point on the line, 2x + 7y − 4 = 0  19k  1   23k  7  2   7 40  k 1   k 1   2 ( 19k + 1) + 7(23 k 7) 4(k+1) = 0  119 k 51 = 0  119 k = 51 51 3 k=  119 7 Thus, the line segment joining the points (1, −7) and (−19, 23) is divided by the line, 2x + 7y − 4 = 0, in the ratio 3:7. The correct answer is B. PA 3 18. It is given that  PB 4 Thus, point P divides the line segment joining the points A(−5, 6) and B (9, 6) in the ratio 3: 4. Using section formula, we obtain Coordinates of point  3  9   4  5 3  6   4  6   , P =  3 4 3 4    27  20 18  24  = ,  7   7  7 42  = ,  7 7  =(1, 6) Thus, the coordinates of point P are (1, 6). The correct answer is C. 19. Let the coordinates of the other end of the diameter be (a, b).

17. Let the line segment joining the points (1, −7) and (−19, 23) is divided by the line, 2x + 7y − 4 = 0, in the ratio k : 1.



AP k  PB 1

It is known that the centre of a circle is the mid-point of any diameter of the circle. Therefore, (−1, 2) is the mid-point of the line joining the points A(a, b) and B(5, 9). a 5 b9   ,   1, 2   2  2 

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10th Class Mathematics

363

a 5 b9  1 and 2 2 2  a + 5 = 2 and b + 9 = 4  a = 2  5 and b + 9 = 4  a =  7 and b = 5 Thus, the coordinates of the other end-point of the diameter are (−7, −5). The correct answer is A. 20. For ΔABC, we obtain 

AB 

3  2

2

2

 6  4 

 5

2

  2

2

 25  4  29units

BC 

 3  1

2

2

 2

2

 3

  6  1 

2

  5

2

2

  3

 4  25  29units

CA 

1  2 

2

 1  4  

2

 9  9  3 2units

It can be seen that, AB = BC = 29 units Thus, ABC is an isosceles triangle. The correct answer is C. 21. Any point on the x axis is of the form (x, 0) Let A(x, 0) been a point on the x axis ⇒ PA = 5 ⇒ (5 − x)2 + (−3 − 0) = 25 ⇒ x2 − 10x + 9 = 0 ⇒ (x − 1)(x − 9) = 0 ⇒ x = 1 or x = 9 Hence, the required coordinates are (1, 0) and (9, 0) The correct answer is A. 22. The nature of a quadrilateral can be found by finding the lengths of its sides. AB =



 4 

BC =



 1  3

 3  0 

2

 3  16  9  5units 2

2

 0  3 2

2

  4  0 

 7 

2



 4  3



122   4 

2

2

2

  6  2 

2

2

  8  64  64  8 2units

 1  3

2

2

  6  6 

 4 

2

 122

= 8 10  8 2  8





10  2 units



 3  5

 2

2

2

  2  4 2

  2   4  4  8units 2

2

 4  3  3   3  3  2   1  3   1  3   8units CA = 5   4  3   4   3  3   1  3   1  3   8 units 2

  1  0 

2

BC =

2

2

2

2

2

2

2

2

 16  144  160  4 10units ∴ Perimeter of the given triangle = AB + BC + CA = 4 10  8 2  4 10



  1  49  1  50units

Diagonal BD =

 3  11

 8

CA

2

2

Diagonal AC =



BC =

AB =

  4   9  16  5units 2

  2  6

Then, by distance formula,

  4   9  16  5units 2

2

 144  16  160  4 10units

2

0   4  4   1

 3

11   1



2



=

are A (5, 4), B (3, 2), and C 4  3,3  3 .

2

 42   3  16  9  5units AD =

∴ AB = BC = CD = DA. Also, diagonal AC = BD. The quadrilateral ABCD has equal sides and equal diagonals. Therefore, ABCD is a square. The correct answer is B. 23. Let P (x, y) be equidistant from the points A (–2, 8) and B (7, 12). ∴AP = BP or AP2 = BP2 ∴ By distance formula, we obtain (x – (– 2)) 2 + (y – 8)2 = (x – 7)2 + (y – 12)2 ⇒ (x + 2)2 + (y – 8)2 = (x – 7)2 + (y – 12)2 ⇒ x2 + 4 + 4x + y2 + 64 – 16y = x2 + 49 – 14x + y2 – 24y + 144 ⇒ 18x + 8y = 125 The correct answer is B. 24. Let the triangle be ABC, such that A (–1, 6), B (11, 2), and C (3, –6). ∴AB

The correct answer is A. 25. Let ΔABC be the given triangle. Its vertices

2

4  1 1   1  3

 3

CD =

2

2

2

 12   7   50units

0   1   4  3

2

2

2

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Coordinate Geometry Solutions

364

As AB = BC = CA, the given triangle is equilateral. The correct answer is B.

26. The vertices of rectangle ABCD are given as A (2, 6), B (6, 2), C (3, −1), and D (−1, 3)  AB=

 6  2

2

2

  2  6   42   4 

BC =

 3  6

2

  1  2  

 3

2

  3

(a sin   acos  )2  ( bcos   bsin  )2

2

2 2   a2 cos2   2a2 sin  cos   a sin b2 cos2   b2 sin2   2b2 sin  cos 

 9  9  18  3 2units Thus, perimeter of rectangle ABCD = 2[AB + BC] =  2  4 2  3 2  units  2  7 2units  14 2units

The correct answer is B. 27. If the points P (1, 3) and Q (−2, 6) lie on the circle with the centre O (5, y), then OP and OQ will be the radii of the circle. ∴OP = OQ ⇒ OP2 = OQ2 ⇒ (1 − 5)2 + (3 − y)2 = (−2 − 5)2 + (6 − y)2 [Using distance formula] ⇒ (−4)2 + 9 + y2 − 6y = (−7)2 + 36 + y2 − 12y ⇒ 16 + 9 − 6y = 49 + 36 − 12y ⇒ 25 − 6y = 85 − 12y ⇒ 6y = 85 − 25 ⇒ 6y = 60 60 y  10 10 Thus, the required value of y is 10. The correct answer is C. 28. Let O (x, y) be equidistant from points P (4, 2) and Q (−2, 3). ∴OP = OQ or OP2 = OQ2 ∴ By distance formula, we obtain (4 x)2 + (2 y)2 = (2 x) 2 + (3 y)2  (4 x)2 + (2 y) 2 = (2 + x)2 + (3 y)2 ⇒ 16 + x2 − 8x + 4 + y2 − 4y = 4 + x2 + 4x + 9 + y2 − 6y ⇒ 20 − 8x − 4y = 13 + 4x − 6y ⇒ 12x − 2y = 7 ⇒ 12x − 2y − 7 = 0 The correct answer is A.

IIT JEE WORKSHEET 1.

0 6m  2n  1 mn  6m + 2n = 0

m 1  n 3

(a sin ,  cos  )( a cos ,b sin )

2.

2

 16  16  32  4 2units 2



 (a2  b2 )  (a2  b2 )sin  a 2  b 2 | sin   cos  |

3.

4.

   2 a2  b2 cos     4  6m  3n 0  6m  3n  0 mn m 3 1    n 6 2 5m  2n 0  5m  2n mn m 2   n 5 since the given points (–2, –5), (2, –2), (8, a) are collinear. 2 5

1 2 2 0 2 8 a 2 5  4 + 10 + 21 + 16 – 40 + 2a = 0 5  4a – 10 = 0  a  2 Hence (B) is the correct answer. 

8.

Use area formula for triangle.

9.

0

10.

5m  3n  mn m 3   n 5

5m  3n

9  (k  2) 2  (k  1) 2  9 also k – 2 = – k – 1 k – 2 = ± (k + 1)  2k = 1 k–2=k+1

k

1 2

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10th Class Mathematics

365



11.

14.

3p  2 4  2q 2  2p  2 2 2 3p – 2 = 4  p=2  4  2q 6  2 2q = 8  q=4 Let G(x, y) be the centroid of the triangle whose vertices are (4, –8), (–9, 7) and (8, 13). x  x2  x3 4  9  8  x 1  1 3 3 y1  y 2  y 3 8  7  13  4 3 3  Centroid of the triangle is (1, 4). Let point on y-axis is Q(0, k), P(2, 3) and R(–4, 1), QP = QR. y

15.

Again let Q be the mid-point of line segment AP joining the points A(–4, 0) and P(–2, 3).

–2 = 1 contraction.

(2  0) 2  (k  3)2  (4  0) 2  (k  1) 2



Co-ordinates of Q are



 4  2 0  3   6 3   2 , 2    2 , 2   ( 3,1.5)     Also let R be the mid-point of line segment PB joining the points P(–2, 3) and B(0, 6). Co-ordinates of R are

20.



 k = –1 So Ans. is (0, –1) 16.

4  

17.

2m  6n mn 4m + 4n = –2m + 6n 6m = 2n



m 2 1   n 6 3

26.

Let A(4, 0), B(–1, –1), C(3, 5) be the given points AB  25  1  26

2  2  0 3  6    2 , 2    1, 2   ( 1,4.5)     Hence the co-ordinates of the points which divide the line segment joining the points (–4, 0) and (0, 6) in four equal parts are : Q(–3, 1.5), P(–2, 3) and R(–1, 4.5). Let the required ratio be k : 1. Then co-ordinates of the point of division are  2k  1 7k  3   k 1 , k 1    This point lies on the line 3x + y – 9 = 0  2k  1   7k  3  3  90  k 1   k 1   6k + 3 + 7k + 3 – 9k = 0 3  4k = 3,  k 4 The line joining (1, 3) and (2, 7) is divided by 3x + y = 9 in ratio 3 : 4 internally. Let C is a mid-point of AB then, coordinate of C will be 62 84  ,  2   2 C = (5, 6) (6, 8) A

BC  16  36  52

(2, 4) B

C

CA  1  25  26 

Also

19.

AB  AC  26 AB2 + CA2 = BC2

  ABC is isosceles and right angled. Hence (A) is the correct answer. Let the points (–4, 0) and (0, 6) be denoted by A and B respectively. Let P be the mid-point of line segment AB.  Co-ordinates of P are  4  0 0  6   2 , 2   ( 2,3)  

D

New,

 27.

 1

(1, 2)

CD 

1  2 

2

2

  2 

2

  2  4

2

 1  4  5  5m A(2, 4) B(2, 6) C(3, 5) Area= 1  x1  y1  y3   x 2  y3  y1   x 3  y1  y 2   2

____________________________________________________________________________________________________ _____________________________________________

Coordinate Geometry Solutions

28.

366

1   2  6  5   2  5  4   3  4  6   2 We know that in circumcenter case, AP = PB = PC 2 2    4   2   2  2   2   6    2

1  2  2  6  1 2 1   2  6  14 2 22    11 2 

A(4, 0)

(–2, –2) A

(5, 1) B

D (–1, 5)

C (2, 4)

P B (0, 0)

again

29.

30.

C (0, 6) 2

2

2

Area is not negative, So, area will be 11m2 Area of ABC

2

      36    12 3  2 so, abcissa of coordinates of circumcenter is 2 2(–1 –3) + p(3 – 1) + (–1)(1 + 1) = 0 2 × (–4) + 2p – 2 = 0 –8 + 2p – 2 = 0 2p = 10 p=5 (i) Area ACD

1   x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2   2 1   2  5  4   1 4   2   2  2  5    2

(ii) 1   x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2   2 1   2 1  4   5  4  2   2  2  1  2 1  6  30  6 = 15 2

(iii)

Ratio of ABC and quadrilateral ABCD 15 15    15 : 26 11  15 26

____________________________________________________________________________________________________ _____________________________________________

8. SOME APPLICATIONS OF TRIGONOMETRY SOLUTIONS

1.

Let AB be the surface of the lake and D be a point of observation such that AD = hm. Let C be the position of the cloud and F be its reflection in the lake. Then CB = FB. Let DE be perpendicular from D on CB. Then CDE   and EDF   . Let CE = x in figure  CB  CE  EB  x  h Now, in right CDE, we have CE x tan    tan    DE  AB DE AB x  DE   x cot ....  i  tan  EF Again, in right DEF, we have tan   DE

 tan  

x  2h  EF  EB  BF  x  h  h  AB

h tan   h tan  tan   tan  Hence, height of the cloud (CB) h  tan   tan   = tan   tan  Let PQ = x metres denote the tower so that PAQ  30o 

FORMATIVE WORKSHEET

2.

B

0

60

P h

x Tower 300

A

3.

600 Q

h  tan 600  3 AQ h  PQ   h cot 600 3 Let OP be a vertical tower P

 AB 

x  2h tan 

 AB   x  2h  cot  ….(ii) From (i) and (ii), we get x cot    x  2h  cot   x cot   x cot   2h cot   x cot   x cot   2h cot 

h

 x  cot   cot    2h cot  2h cot  ….(iii) cot   cot  Hence, the height CB of the cloud is given by CB = x + h 2h cot   CB  h cot   cot  1 2h 2h  tan  tan   CB  h h 1 1 tan   tan   tan  tan  tan .ta 2h tan   h tan   tan  2h tan   h tan   h tan   CB  tan   tan  

4.

3

2

 A

B

O

C

APB  2      PAB OP d OBP,  sin 2 BP  OP  BPsin 2  a sin 2 Let QBP  , QAP   Q 

O

N



P 



M B

b

A

a

C

QN = (CA + AM) tan

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

368

7.

Let AB be the tower of height h meter and BC be the height of flag surmounted on the tower. In DABD, AB 1 h tan 30     x  3h ... (1) BD 3 x

ba  b    CA  MB  tan    a  tan   tan   2  2 

5.

 PQ  2NQ   a  b  tan  Let DE be a clock tower standing at the midpoint D of BC.

C 5m

E

B

h

h C

60° 30°

m 100

90

 A

In DADC, AC 5 h 5 h tan 60   3 x ... (2) AD x 3 From (1) and (2), 5 h 3h   3h  5  h  2h  5 3 5  h   2.5m 2 So, that height of tower = 2.5 m Let AB be the multistoried building of height h and let the distance between two buildings be metres.  XAC =  ACB = 45° (Alternative angels)  XAD =  ADE = 30° (Alternative angles) AE 1 h 8 In ADE, tan 30    ED x 3 [ CB  DE  x]

D 0

 B

100m

AD AD cot    3.2  h h BD cot    BD  h cot   h cos ec2  1 h  BD  h (2.6) 2  1  h 6.76  1

6.

h 5.76  2.4h ABD (AB)2 = (AD)2 + (BD)2 (100)2 = (3.2h)2 + (2.4h)2  h = 25m Let x be the distance of hill from man and h + 8 be height of hill which is required.

A

x

D 900

8.

A

…… (1)

 x  3(h  8)

A

h

B

60° 30°

30°

45°

C

8

h–8 D

30° E h

x

D

AC h h In rt. DACB tan 60    3 BC x x In rt. DBCD, CD 8 1 8 tan 30      x 8 3 BC x 3 x  Height of hill = h  8 3x  8  ( 3)(8 3)  8  32m

8m 45° C

x

B

In ACB, h h tan 45   1   x  h ... (2) x x

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

369

From (1) and (2), 3(h  8)  h

 3h  8 3  h

 3h  h  8 3

 h( 3  1)  8 3

h

8 3



( 3  1)

h

8 3( 3  1) 2

3 1 3 1  h  4 3( 3  1)  h  4(3  3) metres From (2), x = h So, x  4(3  3) metres Hence, height of multostired building = 4(3  3) metres distance between two 9.

building = 4(3  3) metres. Let the point ion the ground is E which is y metres from point B and left after 15 sec. fight it covers x metres distance In DAEB, AB 3000 tan 45  1  y  3000m EB y …… (1) In DCED, CD 1 3000 tan 30    ED 3 xy ( AB  CD) …. (2)

 x  y  3000 3 From eq. (1) and (2) x  3000  3000 3

 x  3000 3  3000  x  3000( 3  1)  x  3000  (1.732  1)  x = 3000 × 0.732  x = 2196m Distance covered Speed of aeroplane = Time taken A

C

3000 m

10. Let PX be the horizontal ground. Let A be the initial position of the aeroplane. Let P be the point of observation on the ground. Draw AC  PX.

Then,  ACP = 90º And AC = 1.5 km. The angle of elevation of the point A at point P is 60º i.e.,  APC = 60º. After 15 seconds, let B be the position of the aeroplane. Draw BD  PX. Then,  BDP = 90º As the plane is flying horizontally, BD = 1.5 km The angle of elevation of the point B at point P is 30º, i.e.,  BPD = 30º Let PC = x km and CD = y km Then, PD = PC + CD = (x + y) km … (i) In right  triangle ACP, we have AC 1.5 tan 60o   3 PC x 1.5 x …..(ii) 3 In right triangle BDP, we have BD 1 1.5   tan 30o  [Using (i)] PD 3 xy 1.5  x  y  1.5 3  y  1.5 3  3 [Using (ii)] 1    y  1.5  3   3   3  1  1.5  2 3  y  1.5    3 …(iii)  3 3  3 

Distnace y km 3 km   time 15 sec onds 15 seconds 45° [Using (iii)] 30° B D x y 3   3600 km / hour  415.68 km / hour 15 2196 2196 18 = m / sec  146.4m / sec   km / hr  527.04 kmThus, / hr the speed of the aeroplane is 415.68 km/hour. 15 15 5 Speed 

E

Hence, the speed of aeroplane is 527.04 km/hr ____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

CONCEPTIVE WORKSHEET 1.

370



Let BC be the light house. 60 In ABC, tan15o  d 15°

3 12   h  8 3m 2 h

In AEC, tan 60o  C

500 500  d1  m d1 3 ....(i) C

60 m A

2.

15°

  d  60cot15o  60   h In ABD, tan    2h

A 60° d1

B

d

3 1 m 3  1 

and in BEC, tan 30o 

1 2 C

p D

h A

B

2h

hp 2h 2 tan  hp   2 2h 1  tan  1 2  hp 2     2 2h 1 1   2 4 hp    8h  3h  3p 3 2h 5h  5h  3p  p  m 3 6 6 In ABD, cos 60o   BD   12 BD 1/ 2 BD In DBE, sin 60o  BE In ABC, tan 2 

3.

E

D

A

90°

E

C

5.

30° B d2

500 d2

 d 2  500 3m Required diameter, 500 500 AB  d1  d 2   500 3  1  3 3 3 2000  m 3 1 Given tan   0.5  2 10 10 In ABC, tan    AB  AB tan  D

20 m C

10 m A

B

30 In ABD, tan       AB tan   tan  30 tan    1  tan  tan  10 1 3  tan    3tan   tan 2  2 2 2  3 tan   4 tan   1  0   3tan   1 tan   1  0 1  tan   ,1 3  tan   1

60°

6

B

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

371

6.

DE h a  tan   …(i) CE CE a a In ABC, tan 2   …..(ii) AB CE [ CE = AB] In DEC, tan  

8.

Given tan  

1 9

In ABC, tan  

3h h 3 D

D

h C

3h

h

C

E

A

a





 a tan  1  tan 2   2  h  a  tan   0



 3  27h  36h 2  4h 2



 tan  a  a tan 2   2h  2a  0

 tan   0, tan 2  

2h  3a a

2h  3a 2h    tan  tan 1 3  a a  PQ = 150 m PQ 150 In APQ, tan 60o   AP  ... (i) AP 3 1

7.

PQ AB  AP 150 150  AB   150  AB  3 1 3 3

9.

 4h 2  9h  3  0 9  81  48 9  33 h  24 8 1 In ADF, tan 60o  AF 1 o AF  cot 60  3 E

D

and in BPQ, tan 45o 



Q 45° 60° 150 45°

60°

B

A

P

The speed of boad AB 1 150    3  1  60m / h 2 2 3





4500 3





3 1 m / h



4h 3

tan   tan  4h  1  tan  tan  3 1 h 4h 1  9h 4h 9    h 3 9  h 3 1 9

From eqs. (i) and (ii), we get tan  tan   h a a 2 tan   a tan    h  a   1  tan 2 



B

In ABD, tan      

B

A

3 km



d

C

1 km 60° 30° A

F

B

1 In ABC, tan 30o   AB  cot30o AB  AF  FB  3 1 2 d 3  3 3 distancefrom D to C  speed  time taken 2  3  60  60  240 3km / h 10 h In ABC, tan 60o  BC o ...(i) BC  h cot60

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

372

A

and QP = AQ –PA  l cos   l cos   l  cos   cos  

30° 60° h 30° D

60° B

C

d

In ABD, tan 30o 

h BD



CB l  sin   sin    QP l  cos   cos 

       2sin  cos    y  2   2   y  cot          x x         2  2sin  cos     2   2 

 BD  h cot 30o   x  y tan    2 

 BC  CD  h cot 30o

 CD  h cot 30o  BC  d  h cot 30o  h cot 60o [from (i)] distance from D to C d  Speed   time taken 3 o



12. ln ABD, tan  

d  cos  ...(i)] D

o

h cot 30  h cot 60 3

Time taken from C to B 

distance speed

E



h cot 30o  cot 60o

A

3

3 

3  1.5 min 2

 1   3  3  11. Let l be the length of the ladder i.e., BP = CQ = 1 PA AB ln PAB, cos   and sin   PB PB  PA  l cos  and AB  l sin  ... (i) AQ ln QAC, cos   QC  AQ  l cos  and AC  lsin B

 DC  d tan  [ BC=EA=h] [from (i)]

 cos  sin    h  60  1  .   sin  cos   60 sin      h cos  sin  60 sin      60 sin        (given) x cos  sin   x  cos  sin  13. Since QPC     QPB  BPC  2 Q

l

60

B

DC ln DEC, tan   EC  60  h  d tan   60  h  60cot  tan 

3 1





C

d

h

h cot 60o 

60 d

r

B

C

l

C

l

h

x

Now, CB = AB – CA = l sin   l sin  = l  sin   sin  

...(ii) [From (i) and (ii)]

O

P

 r ln PQB, sin  2 l

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

373

 l sec

 2

..(i)

and in POB, sin  

h l

 h  l sin 

  h  r cos ec sin  [from (i)] 2 h 14. ln CDE, tan 3  x  x  h cot 3 ...(i) h EBD, tan   100  x  100  x  h cot 2 ...(ii) h EAD, tan   300  x  300  x  h cot  ...(iii) From eqs. (i) and (ii) we get 100  h cot 3  h cot 2

15. Let BD is the surface of the lake and A be the point of observation such that AB = 60 m. Let C be the position of the cloud and E be its reflection in the lake. Then CD = ED. Let AF be perpendicular from A on CD. Then CAF  30o and EAF  60o . Let CF = hm. Now, CD = (60 + h) m. Similarly ED = (60 + h)m. Now, In right  CFA , we have CF 1 h tan 30o    AF AF 3  AF  3 h EF Again, in right  EFA we have tan 60o  AF

A

B 200

C



D

 4sin 3   sin   0  sin   0 1    or sin 2    sin 2      4 6 6 hence h  200sin

 3  200  100 3 3 2

 AF 

120  h 3

 3AF  60  h  60

… (ii) 3h 

120  h 3

 3h = 120 + h 3h – h = 120 h = 60 m Now, CD = CF + FD = h + 60 = 60 + 60 CD = 120 m Hence, the height of the cloud from the surface of the lake = CD = 120 m

SUMMATIVE WORKSHEET

100

 cos2 sin 3  sin 2 cos3   100  h   sin 3 sin 2   h sin   100  ...(iv) sin 3 sin 2 From eqs. (ii) and (iii)  sin   200  h    sin 2 sin   ...(v) From eqs. (iv) and (v), 100 sin    sin 3  2sin  200 sin 3  3sin   4sin3   2sin 

ED  DF AF

From (i) and (ii), we get

E

h

 3

1.

Let B and D be the positions of the two ships and let AC be the lighthouse. It is given that ∠XAB = 60°, ∠YAD = 30°, and AC = 60 m. Then, ∠ABC = 60° and ∠ADC = 30°

In right ΔABC: Oppositeside  tan ABC Adjacent side AC  tan 60 BC

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

60m  3 BC 60m 60 3 BC    20 3m 3 3 In right ΔACD: Oppositeside  tan ADC Adjacent side AC  tan 30 CD 60m 1  CD 3 CD  60 3 Thus, BD = BC + CD = 20 3m  60 3m

2.

374

4.5m BC 4.5m _______  BC  (2) 3 From equations (1) and (2), we obtain 4.5m DC  3 4.5m Thus, Sohan is sitting at a height of 3 from the ground. The correct answer is C. Let A be the position of the bird and B be the point at which the shooter is holding the gun. Let CD be ground level. ∴ BC = 1 m ⇒ AE = 11 m − 1 m = 10 m  3

3.

= 80 3m =(80×1.732) m  138.56 Hence, the distance between the two ships is 138.56 m. The correct answer is C. The given information can be diagrammatically represented as

In right ΔAEB, AE tan 30  EB 1 10m   3 EB Let A be Rohan’s position and D be Sohan’s position. Point B denotes the boy on the road. ∴AC = 4.5 m. Now, ∠BAX = ∠ABC = 60° (Alternate angles) Similarly, ∠BDY = ∠DBC = 45° In right ΔDBC, DC tan 45  BC DC 1= BC _______  BC = DC (1) In right ΔABC, AC tan 60° = BC

 EB = 10 3m Thus, the horizontal distance between the bird and the shooter is 10 3m . The correct answer is A. 4.

Let AB and CD be the poles and let point P be the point of observation on the road. Then BD = 20 m. Let the height of the poles be h. Then, AB = CD = h

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

375

Let PD = x ⇒ BP = 20 – x In ΔABP, ∠APB = 45° AB   tan 45 BP h  1 20  x  h = 20  x ________  x = 20  h (1) In ΔCDP,∠CPD = 60° CD   tan 60 PD h   3 x  h =x 3

7000 3m  3 SR 7000 3  SR  m 3 =7000 m  QR = QS + SR = (21000 + 7000) = 28000 m = 28 km Thus, the distance between the two observation points is 28 km. The correct answer is B. The given information can be diagrammatically represented as



6.

3 [from (1)]

 h = (20 h)

 h  20 3  h 3  h  h 3  20 3





 h 1  3  20 3 

20 3 1 3

= 10 3

5.





20 3 1 3



 

Let AD be the building and B and C be the respective positions of Aman Suman. ∴ BD = x1 and CD = x2 In ΔABD, AD tan 45  BD 30m 1= x1  x1= 30 m In ΔADC, AD tan 30° = DC 1 30m   x2 3

  20 3  3  1 3 1 3  1 3 1



3  1  10  3  10 3  12.68m

Thus, the height of each pole is 12.68 m. The correct answer is B. The given situation can be represented as

Here, points Q and R represent the two observation points, and point P is the position of the aeroplane. S is a point on QR such that PS ⊥ QR.  PS = 7000 3 PS In right ΔPQS,  tan 30 QS

7.

 x2 = 30 3m  51.9m Thus, the respective values of x1 and x2 are 30 and 51.9. The given information can be diagrammatically represented as

7000 3m 1  QS 3





 QS = 7000 3  3 m =21000 m In right ΔPRS, PS  tan 60 SR

Let AB be the traffic signal and C be Vicky’s position. ∴ AB = 7.8 m

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

In right ΔABC, AB tan 45  BC 7.8m 1= BC  BC= 7.8 M Thus, the horizontal distance between the traffic signal and Vicky is 7.8 m.

IIT JEE WORKSHEET 1.

Let C and D be the two positions of the ship. Let BD = x m be the distance traveled by the ship during the period of observation, i.e., BD = x m. Let observer be at A, the top of the light house AB = 100 m. The angles of depression from A of C and D are 30o and 45o respectively i.e., C  30o and D  45o in figure. Now, in right  ABD , we have AB 100 tan 45o  1  BD BD  BD (x) = 100 m Again, in right  ABC , we have

tan 30o 



AB 1 100 1 100     BC 3 BD  DC 3 xy

1 100   100  y  100 3  y  100 3 100  y

 y  100

2.





3  1  y  100 1.732  1

 y  100  .732  y  7.32m Hence, the distance traveled by the ship from C to D is 7.32 m Let AB be the building and its height is 60 m, and CD is a tower. Let CD = h m and CE be horizontal from C. The angles of depression of the top C and the bottom D of the tower CD are 30o and 60o respectively. Let DB = CE = x m. AE Now, In right  AEC , we have tan 30º = CE 1 60  h   x 3  x   60  h  3 m … (i)

376

Again, in right  ABD , we have tan 60º = AB BD 60  3 x 60 x  m … (ii) 3 From (i) and (ii), 60 we get  60  h  3  m 3   60  h   3  60  60 – h = 20  60 – 20 = h  h  40m Hence, the height of the tower (CD) = h = 40 m. 3. Let the mass who is standing on the deck of a ship at point A and CD be the hill. It is also given that the angle of depression of the base D of the hill DC observed from A is 30o and the angle of elevation of the top C of the hill CD observed from A is 60o, i.e. EAC  60o , BDA  30o and AB = ED = 10 m in figure. CE Now, in right  AEC , we have tan 60º = AE h  3 x  h  3x … (i) Again, in right  ABD , we have tan 30º = 3  100 AB BD 1 10   3 x

 x  10 3 … (ii) Now, substituting the value of x from equation (ii) in equation (i), we get h  3x  h  3  10 3  h  30m  h  CE   30m and CD = CE + ED = 30 m + 10 m  CD  40m Hence, the distance of the hill from the ship BD = xm = 10 3 m and the height of the hill (CD) = 40 m

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

377

4.

h  tan 450  1  h  x x ha 1  tan 300  x 3

7. Q h-a

300

A

B tower

x a

a 0

45

P

O

x h  3 3 8.  1  3a  h 1  a  h 3 3 1  Let PQ be given tower of height h. If A, B be given points then suppose  PAQ = a and  PBQ = b  Then a  b  2 h Now in PAQ, tan a  …(1) a h In PBQ, tan b  …(2) b (1),(2) 2 h h   h  tan a tan b  .  tan a tan       h  ab a b 2  ab ha 

5.

6.

hx 1 2h  x h 3 3 Let P be position of the aeroplane and A,B, given km stones on the road. Then as given PAB = a and PBA = b If height of the aeroplane from the road be PQ = h, then h / AQ = tan a and h / QB = tan b If height of the aeroplane from the road be PQ = h, then h / AQ = tan a and h / QB = tan b

 2   1 

Let AB and PQ be given cliff and tower respectively. If PQ = x, then as given from adjoining diagram APC  30 o and AQB  60o

9.

 1 1   AQ + QB =h    tan  tan   tan  tan   1 1   1  h    h tan   tan   tan  tan  

Let BD be horizontal position of the lake and A be the given point at height h above it. If position of the cloud and its reflection be P and Q respectively, then from diagram PAC = a and QAC = b Let h be the height of the cloud, then from triangles PCA and QCA ha  tan  ........ (1) AC ha and  tan  ........ (2) AC h  a tan    2   1  h  a tan  2h tan   tan  sin  cos   sin  cos     2a tan   tan  sin  cos   sin  cos 

 ha

sin      sin     

h  500 500 3  tan 300  h   m 500 3 3 3

A h

30° 300 500

h P

30° C

x 60°

Q

B

Now in triangles ABQ and ACP hx h  tan 60  3 and  tan 30 QB PC hx 1   QB 3

10. Let O be the centre of the equilateral triangle ABC and OP the tower of height h. Then each of the triangle PAB, PBC and PCA equilateral. Thus, PA = PB = PC = a. Therefore from right-angle triangle POA, we have PA2 = PO2 + OA2 2

a2 4 a2 a   a 2  h 2   sec30   h 2  .  h 2  4 3 3 2  2  a 2  h 2 or 2a2 = 3h2 3

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

11. Let OP be the tower of height x. A the point on the same level as the foot O of the tower and B be the point h m above A (see fig.). Then  AOB = 60° and  PAO = 30° . From right-angled triangle AOP, we have OA = x cot 30° and from right-angled triangle OAB, we have OA = h cot 60° Therefore, from (1) and (2), we get x cot 30° = h cot 60° 3 

1 3

h x 

1 hm 3

12. Let the tower OP subtend an angle of 30° at A. a point on the same level as the foot of the tower. From a point B at the height h, vertically above A, from where the angle of depression of foot O of the tower is 60° B

378

14.

6 1  sin 30o   l  12m l 2

15. h1  x tan 600  3x x h 2  x tan 300  3

h1 h2 0

0

A

60°

30°

O

i.e.  AOP = 60°. From DOAP, OA = h cot 60°. Also from DOAP, OA = OP cot 30°.  OA = h cot 60° = OP cot 30°  OP = h/3.  OA = (h/3) cot 30° 13. Let AD be the building of heigth h and BP be the hill. Then hx x tan q  and tan p  y y

x

p

C

q A

 tan q 

y

B

x

x 1  h1 : h 2  3x :  3:  3:1 3 3 16. AB = 15m CAB  60o B

h 60° A

A BC sin 60  AC 3 h   2 15 15 3  h m 2 17. T = 2 minutes B

B

100m

hx  x cot p   h  x  cot q x cot p

h cot q h cot q x hx  h cot p  cot q cot p  cot q h cot p  cot p  cot q

x

o

P

D

30

60

A

P Tower

h

D

45° x C

60°

In BDA tan 45o 



1

A

y

AB AD

100 xy

____________________________________________________________________________________________________ _____________________________________________

C

10th Class Mathematics

379

x  y  100 Now, In BCA AB tan 60o  AC 100 3 y

.....(i)

AD CD h  20 3 CD  h  20 tan 60o 

3 y  100 ......(ii) from equation (i) and equation (ii), we get xy 3y

xy





C

x  y 3 1 from equation (i) x + y = 100 x  x  100 3 1  3  1  1  x   100 3  1  

100



x



Speed  5







3 1 6 3



5 3 3 

18 18. AB = hm BD = 20m In BDC,

D

1

E



3





Distance 100 3  1  time 3  120

3 3

 m/s

hm

20m 45°

21. 22. 23. 25.

B

45° xm

B

h x

 xh In right triangle AEC, we have AE tan 30o  EC 1 AB  EB   EC 3 1 h  100   x 3

3 1

BD tan 45o  CD 20  1 CD  CD  20m In right ADC ,

30°

hm

A

C



3 1 m

19. Let AB = (h metres) be the rock Let CD = (100 m) be the tower Now, In right triangle ABD, tan 45° = AB A

3 1





......(i)

 x  3 h  100 3 .......(ii) On equating the values of x from equation (i) and equation (ii) we get h  3 h  100 3  h  236.5 m A - q, B - p, C - q, D – r A - s, B - r, C - p, D – q A - q, B - p, C - r, D – s AC sin 30o  AB 1 AC   2 60 A

D

60m



30° B  

so,

C AC  30 K  30 K 30   2m 15 15

____________________________________________________________________________________________________ _____________________________________________

Some Applications of Trigonometry Solutions

380

A

26. AB = hm AD = 1.46m In DBC, D

x 30°

D

1.46

E

10

16

C

B

A

AE AED, sin 30  DA 14  2x  x  8m x 8 then   2m 4m 4 29. In PQR o

hm 60°

45° B C DB o tan 60  BC DA  AB tan 60o  BC 1.464  h 3 BC 1.464  h  BC  m 3 In right triangle ABC, AB h tan 45o   1 BC BC  BC  hm ....(ii) from equation (i) and equation (ii), we get 1.464  h h 3 1.464 h 2 3 1





27. BC = 1m

tan 45o 

AB BC 45°

45° C

AB 1 1 AB 1    1m  1m 1 28. AB = 16 m DC = 10 m AE = 4m New tan right triangle

A

PQR  45o  QR  PR  20cm  PS  13cm  SR  PR  PS  20  13  7cm SR 7cm   1 7cm 7cm 30. (i) In triangle BDC, A 30°

B

210m

30° D

C

BC tan 30o  DC 1 200  3 DC DC  200 3 m  346.4 m (ii) Let give DC = 200 m, BC  tan   DC 200 so, tan   1 200    45o so, angle of depression = 45°.

B

AB  1m

____________________________________________________________________________________________________ _____________________________________________

9. BINOMIAL THEOREM SOLUTIONS

FORMATIVE WORKSHEET 1.

i) If n is odd, then the expansion of

 n 1 (x + a)n + (x –a)n contains   terms. So,  2  9



the expansion of 1  5 2x

  1  5



2x has

 9 1    5 terms.  2  ii) We have, (2x + 3y – 4z)n = {2x + (3y – 4z)}n = nC0(2x)n(3y – 4z)0 + nC1(2x)n–1(3y – 4z)1 + nC2(2x)n–2(3y – 4z)2 +...+ nCn–1(2x)1(3y – 4z)n–1 + nCn(3y – 4z)n Clearly, the first term in the above expansion gives one term, second term gives two terms, third term gives three therms and so on. So, Total number of terms = 1 + 2 + 3 + ...+ n + (n + 1)  2.

 n  1 n  2  2

4:

9

Putting x  2 and y = 1, we get



5

5

 



2 1 

 2 

 2

2 1

5

3

 10

 2

4  5 2 1  

 2 4 2  20 2  5 2   58 2 4

5.

4

2

.

3

 2x 3  4  2x   2x   3  4     C0    C1      3 2x   3   3   2x  4

2

 2x   3   2x  3  C 2      4 C3     3   2x   3  2x 

3

2 6

(x + y ) 6 6 6 5 2 1 6 4 2 2 = C x + C x (y ) + C x (y ) 0 1 2 6 3 2 3 6 2 2 4 6 2 5 + C x (y ) + C x (y ) + C x (y ) 4 5 6 3 2 6 + C (y ) 66 5 2 4 4 3 6 2 8 = x + 6x y + 15x y + 20x y + 15 x y 10 12 + 6xy + y 6

Here, C = 2

 3   4C 4    2x 

6

C = 3

6

6! 6! =  6 – 2 !2! 4!2!

 2x   27   81  4     3   1 4   3   8x   16x  4.3  4  4 4 4 4  6  C4  C0  1, C3  C1  4, C2  1.2  

6! 6! 4 .5. 6 = = = 20  6 – 3!3! 3!3! 1. 2 .3 6

4

 16x 4   8x 3   3   4x 2   9   1  4  6        2  81   27   2x   9   4x 

1. 2 . 3. 4 . 5. 6 = . . . = 15 1 2 3 4 × 1. 2

3.

= 1 – 4x + 4x2 + 6x2 (1 – 2x + x2) – 4x3 (1 – 3x + 3x2 – x3) + x4 (1 – 4x + 6x2 – 4x3 + x4) = 1 – 4x + 4x2 + 6x2 – 12x3 + 6x4 – 4x3 + 12x4 12x5 + 4x6 + x4 – 4x5 + 6x6 – 4x7 + x8 = 1 – 4x + 10x2 – 16x3 + 19x4 – 16x5 + 10x6 – 4x7 + x8 We have (x + y)5 + (x – y)5 = 2[5C0x5 + 5C2x3y2 + 5C4x1y4] = 2(x5 + 10x3y2 + 5xy4)

n

4

n

C = C = 15 (since C = C ) r n–r 6 4 6 2 C = C =6 5 1 Let y = –x + x2. Then, (1 – x + x2)4 = (1 + y)4 = 4C0 + 4C1 + 4C2 y2 + 4C3y3 + 4C4y4 = 1 + 4y + 6y2 + 4y3 + y4 = 1 + 4(– x + x2) + 6(– x + x2)2 + 4(– x + x2)3 +(– x + x2)4 = 1 – 4x(1 – x) + 6x2(1 – x)2 – 4x3(1 – x)3 + x4 (1–x)4

6.

2x 3 16 16 9 81      x 4  x 2  6  2  9 x 16x 4  3 2x  81 Now (x + y)6 + (x – y)6 = {6C0x6 + 6C1x5y1 + 6C2x4y2 + 6C3x3y3 + 6 C4x2y4 + 6C5x1y5 + 6C6y6} + {6C0x6 – 6C1x5y1 + 6 C2x4y2 – 6C3x3y3 + 6C4x2y4 – 6C5x1y5 + 6C6y6} = 2{6C0x6 + 6C2x4y2 + 6C4x2y4 + 6C6y6} = 2{x6 + 15x4y2 + 15x2y4 + y6}

(6C0 = 6C6 = 1 and 6C2 = 6C4= = 2x6 + 30x4y2 + 30x2y4 + 2y6.

65  15 ) 1 2

10th Class Mathematics

382 Substituting x  3 and y = 1 in the above result, we obtain 6



 

3 1 

2

 3



6

12

3 1

6

 30

3



6

 3 1

2

2

 30

 3 1

4

6

 2 1

 33  27,

 3

4

 32  9



n! n Cr   n  r !r! –––––––– (1) n

Cn–r can be obtained by substituting r = (n – r) in (1). n

Cn r

=

n!   n   n  r   ! n  r !

n! n! =  n - n + r ! n - r ! r! n - r ! –––– (2)

By observing (1) and (2), we see that their R.H.S is the same.  (1) = (2) Hence, nCr = nCn–r 8.

12  8

x T9 = T8+1 = C8   a

i) 6 !  2 !

4

 x   3a   12 C8    2  a  x 

a4   12 C 4 x 12 a 4  38 12 x 11. Since n = 10, the number of terms in the expansion is 11 and hence the middle term is the sixth term. 12 C 4  38

10

3 x  

2 10–5

T = C (2x )



6

5

10

5

5

5

T = C (6) (x) 6 5 12. Here n = 20, which is an even number.

 20  So,   1 th term i.e., 11th term is the middle  2  term. Hence, the middle term = T11 = T10+1 = 10

 3  = 20C x10. C10  2 x 2  10   3   2x  13. Here total number of terms is 9+1 = 10 (even).  There are two middle terms given by T9 1 2

10  9  7  6! 2! 21    6! 5  4  3  2! 2 iii) 3 ! 2 ! 4 ! 3 !

and

T9  3 2

 3! 2!  4  3! 8 = 6 – 8 = –2 We have, n n–r r T = C x y

, i.e., T5 and T6.

We know that Tr+1 = nCrxn–r ar,  x 2   T5 = T4+1 = 9C4(2x)9 – 4    4 

4!  3! 2!  4  3 !  3! 2!  3! 3!



3 Here r = 3, x is to be replaced by and n = 7 x 73

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3

7

.y = C

3

34 3 .y x4

9 5

and T6  T51  C 5  2x  9

9

4

 C4 2 x



4

4

  8  7  6 5 5 x 8 63 13 .2 x . 4  x 1 2  3  4 4 4

r

7 3 T = C   4 3 x

8

20 10

10 ! 2 ! ii) 6 !  5 !

r+1

8

20

=6×5×4×3×2×1×2×1 = 1440

9.

 3a   2   x 

12

= 54 + 270 + 90 + 2 = 416. 7.

 x 3a  In the expansion of   2  , we have a x 

 4

= 2 (27) + 30 (9) + 30 (3) + 2

 

10. We know that the (r + 1)th term in the expansion of (x + a)n is given by Tr+1 = nCr xn–r ar

 1 .

5

 x 2     4 

5

x10

45

 x10  87 6 63 . 16x 4    5    x14 1 2  3  4 4 32  

Binomial Theorem Solutions 14. i) Let (r + 1)th term be independent of x in the given expression.  1 Now, Tr+1 = 10Cr (2x)10–r     x

  1

r 10

r

Cr 210 r x10 2r .......... (i)

For this term to be independent of x, we must have 10 – 2r = 0  r = 5. So, (5 + 1)th i.e., 6th term is independent of x. Putting r = 5 in (i), we get T6 = (–1)5 10C5  210 5  10 C5  25

10  9  8  7  6  32  8064. 5  4  3  2 1 Hence, required term = –8064. ii) Let the general term Tr+1 be the term independent of x. 

11 r

383

 ( 1)2 15 CS x 60 7S ...... (ii) If this term contains x–17, we must have 60 – 7S = –17  S = 11 1 th th So, (11 + 1) i.e., 12 term contains x–17. Putting S = 11 in (ii), we get T12 = (–1)11 15C11 x–17 = –15C11 x–17 [ since  n C r  n C n  r ]  Coefficient of x–17 = – 15C4 = – 1365. 16. In the expansion of (1+a)m+n, the general term is = – 15C4 x–17

Tr 1  m n Cr a r  Coeff. of ar = m + nCr Taking r = m, n; we obtain ocefficient of a m  m n Cm  

r

2 x   1     i.e., C r   , be the term  5   2x x  independent of ‘x’ 11

x

11 r 2

.x

3r 2

 x0

11 r 3r 11   0 r  , 2 2 2 4 i.e., the value of r is not an integer according to the definition of Tr+1 Hence, no term of the expansion is independent of x. 15. Suppose (r + 1)th term involves x32 in the 15

 4 1  expansion of  x  3  . x   4 15–r

Now, Tr+1 = Cr (x )

 1   3   x 

r

 ( 1) r 15 Cr x 607r .............. (i) For this term to contain x32, we must have 60 – 7r = 32  r = 4. So, (4 + 1)th i.e., 5th term contains x32. Putting r = 4 in (i), we get T5 = (–1)4 15C4 x(60 – 28) = 15C4 x32.  Coefficient x32 = 15C4 = 1365 Suppose (S + 1)th term in the binomial expansion

.......... (i)

n m n and coeff. of a  C n 

mn mn n n

mn ............(ii) mn From (i) and (ii), we find that the coefficients of am and an in the expansion of (1+a)m+n are equal. 17. We shall write 9n+1 as (8+1)n+1 and use Binomial Theorem for positive for positive integral index so that 9n+1 – 8n – 9 = (1 + 8)n+1 – 8n – 9 = n + 1C0 + n + 1 C18 + n + 1C2 82 +.....+ n + 1 Cn + 8n + 1 – 8n – 9 1 = 1 + (n + 1)8 + 82{n + 1C2 + n + 1C3 8+..... + n + 1C n + 1 8 n – 1} –8n – 9 = 82 {n+1C2 +n+1C3 8+......+n+1Cn+18n-1}, which is a multiple of 82=64. Hence 9n+1-8n-9 is divisible by 64 for all n  N . 18. Third term in the expansion of (x+xlog10x)5 is 5 C2x5-2 (xlog10x)2. So we must have 5 C2 x3 (xlog10x)2 = 1,000,000  10x3 (x2log10x) = 1,000,000  x3 (x2log10x) = 100000 Taking log on both sides we get log10x3 + log10 (x2log10x) = log10100000  3log10x + 2log10xlog10x = 5  2t2 + 3t – 5 = 0, where t = log10x

5 2 Now t = 1  log10x = 1  x = 101

15

 4 1  of  x  3  contains x–17. x   15

mn nm





15

mn mnmm

4 15–S

Now, TS+1= CS (x )

 1   3   x 

 t  1, 

S

 x = 10 and t = 

5 2 www.betoppers.com

10th Class Mathematics

384



5  x = 10– 5/2 2 A note on Arithmetic Progression Numbers a1, a2, a3, ..., an, ... are said to be in arithmetic progression (abbreviated A.P.) if a2–a1 = a3–a2 = a4 – a3 = ..... In particular, three numbers a1, a2, a3 are in A.P. iff a2-a1 = a3-a2, i.e., iff 2a2=a1+a3. 19. Here, the index = 2n is an even number. Hence,

 log10x = 

 ...11C11 x 22 (1  x)11  = 3 (3159 – 160C13158 +...160C159) + 1 = 3m + 1 32

Now, 3232 = 323(5m+1).22 = 4. (8)5m+1 = 4. (7+1)5m+1 4

= (n + 1)th term  C n 1 

nn

1 3  5  ...   2n  1 n

xn

T2  nc1  2x 



4

968

1

 3 2 2

1

2





500



 9

1

4

8

1

500

6



500  r

  . 2 

 Generalk term  500C r 3

1

2

1

1

2

1

250  r

2

2

r

11 2 (1+ x + x2 x3)11 =  1  x   x 1  x  

11

11

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2

 1  x n4  x   nc2 (2 ) 4 

2

1  1  1  1 

.2

r

7

2

For r = 0, 2, 4, 6, 8, 10, 12....500; powers of 3 and 2 are integers. Hence the integer terms = 251. 21. We have,

 1  x  1  x 2 

x n2



1

 1  x n 3  x   nc1 (2 ) and, 4 

Sum of even powers of x 

r

r

  . 2   500C 3

 500C r 3

n 1

n  n  1  36  n2 + – 72 = 0 2  (n + 9)(n – 8) = 0  n = 8 and T3 = 7T2  nc2 (2x)n–4 = 7.nc1(2x)n–3  8C2(2x)4 = 7 × 8c1(2x)5  2x = 2 –1  x = –1 23. Let f(x) = (1 – x + x2 – x3)7

500

500  r

5m 1

 5m 1 C 2  7 

 n

n



5m

 5m 1 C1  7 

It is given that, T3 = 3 T2 and nc1 + nc2 = 36 Now, nc1 + nc2 = 36

1.3.5....  2n  1 2n x n

20. Since

T3  nc 2  2

2  2  2  ...  2x n

n factors 

5m 1

C0  7 

5m 1 C0 75m  5m1 C1 75m1  .....  1 =  = 4 [7n + 1] = 28n + 4

1 3  5  ...   2n  1  2  4  6  ...   2n   n   x n   1 2  3  ...  n 



5m 1

This shows that there 323232 is divided by 7, then remainder is 4. 22. We have,

2n n x  x nn n

1 2  3  4  5  ...x  2n  1   2n 



......  5m 1 C5m 7  5m 1 C5m 1 

 2n   The required middle term   2  1 th term   2n  n

11

C 0 11 C1 x 2 11 C 2 x 4  ... 11 C11 x 22  1  x 

11C0 (1  x)11  11C1 x 2 (1  x)11  11C2 x 4 (1  x)11

 2n   1 th there is only one middle term namely   2  term, i.e., (n+1)th term.

2n

11

=2 24.

f 1  f  1 2 7

 1  1  1  1 2



0  47 214  2 2

13 r

 1  .  2  = nCr(–1)r x3x – 5r x  For coefficients of x5, 3n – 5r = 5 Tr 1  n C r  x 3 

nr

3n  5  p  say  5 For coefficients of x10, 3n – 5r = 10 r

Binomial Theorem Solutions

385 = 64x12 + 576 x9 + 2160 x6 + 4320 x3 + 4860

3n  10  q  say  5 Note p – q = 1 Since sum of coefficients = 0  nCp (–1)p + nCq (–1)q = 0  nCp (–1)p + nCp – 1 (–1)q – 1 = 0  nCp = nCp–1  n – p = p – 1 r

 n = 2p – 1  P 

3.

n 1 2

4.

n  1 3n  5  2 5  5n + 5 = 6n – 10.  n = 15 n 25. Sum of the coefficients in (p + q) = 1024 n 10  2 = 2  n = 10 10 Which is even so greatest coefficient in (p + q) is coefficient of most middle term which is

n! n!  5! n  5 ! 10! n  10 !



n

1 5! n  5  n  6  n  7  n  8  n  9  n  10 ! 

C n  10 C10  10 C5 2

 n  8 n  9 

Using binomial theorem, we have (2x – 3y)4 = [2x + (– 3y)]4 = 4C0(2x)4 (– 3y)0 + 4C1(2x)3 (– 3y) + 4C2(2x)2 (– 3y)2 + 4C3(2x)1 (– 3y)3 + 4C4(– 3y)4 = 16x4 + 4(8x3)(–3y) + 6(4x2)(9y2) + 4(2x)(–27y3) + 81y4 = 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4

 2 3 2.  2x   x 

6

5.

6 2 6 6 2 53 6 2 43 = C0  2 x   C1  2 x     C2  2 x    x  x

 C3  2 x 6

2 3



3

4

3 6 2 23    C4  2 x     x x 5

=1× 64 × x12 + 6 × 32x10 ×

3 9 + 15.16x8 × 2 x x

27 81 +20 × 8x6 × 3 + 15 × 4x4 × 4 + 6×2x2 x x 243 729 1 6 5 + x x



6.

5

3 3 C5  2 x 2     6C6   x  x [  6C0 = 1,6C1= 6, 6C2 = 15, 6C3 = 20, 6C4 = 15, 6C5 = 6, 6C6 = 1] 6

×

1 10  9  8  7  6  5! n  10 !

10  9  8  7  6   n  5  n  6  n  7 

2

CONCEPTIVE WORKSHEET 1.

2916 729  6 x3 x 5 (1 – 2y) = 5C0 (1)5 + 5C1(1)4 (–2y) + 5C2(1)3 (–2y)2 + 5 C3(1)2 (–2y)3 + 5C4(1) (–2y)4 + 5C5 (–2y)5 = 1× 1 + 5×1× (–2y) + 10 × 1 × (4y2) + 10 × 1 × (–8y3) + 5 × 1 × (–16y4) + 1× (–32y5) =1–10y + 40y2 – 80y2 + 80y4 – 32y5. Given, nC5 = nC10 +

7.

If n – 5 = 10,  n = 15 then n – 6 = 15 – 6 = 9 n – 7 = 15 – 7 = 8 n – 8 = 15 – 8 = 7 n – 9 = 15 – 9 = 6  n = 15 n C15 = 15C15 = 1. ( nCn = 1) R.H.S. = 8C6 = 7+1C6 =7C6 + 7C5 = 6+1C6 + 7C5 (n+1Cr = nCr + nCr–1) = 6 C6 + 6 C5 + 7 C5 = 1 + 6 C5 + 7 C5 = 7C5 + 6C5 + 5C5 = L.H.S. ( nCn = 1) th The (r + 1) term in the binomial expansion of n n n–r r (x + y) is T = C x .y r+1 r Therefore, the fifth term, 14 14–4 . 2 4 14 10 4 8 T = C .2 (3x ) = C (2) (3) x 5 4 4 Clearly, the given expansion contains 26 terms. So, 11th term from the end = (26 – 11 + 1)th term from the beginning i.e., 16th term from the beginning.  Required term = T16 = T15+1 15

=

25

25–15

C15 (2x)

25 C15  210  x10

 1   2   x 

(1)15 210 25   C  15 x 30 x 20

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10th Class Mathematics

386 r

8.

 1  Tr 1  Cr  9 x      3 x For the 13th term, we put, r + 1 = 13 i.e., r = 12 in the above term. 18 r

18

6

1    3 x

 1   C6 9 x   312 x  6

6

  12   



 

n

Cr  n Cn  r 

2 6

18  17  16  15  14  13  3  x  . 6  5  4  3  2 1 3 x  9.

6



 Middle term =  + 1 2 

T =T 4

3+1

2 6–3

= C . (x ) 3

13 15 Hence, the middle terms are  105 x and 35 x 8 48 11. Number of terms in the expansion is 10 + 1 = 11 (odd).

There is only one middle term given by

 2 x2  Now, T6  T5 1  10 C5    3 

 T

r 1

th

3

7

th

 73  7 1 So,   and  2  i.e., 4th and 5th terms    2  are two middle terms.   x3   Now, T4 = T3+1 = 7C3 (3x)7–3   6 

= –35 × 81 x4 × 

7

105 x13 and T5 = T4+1 8 7–4

= C4 (3x)

 x3     6 

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5

be Tr+1.

3

 1 Tr 1  Cr .  2 x 2    10

10

10 r

 Cr .2

r 10

1    .  3 x 3   

r

r

10 r 2

10  r

r

10  r r  2 3

.  1 .3 .x

  1 . Cr .2

.3 .x

.x



r

r 3

.

If the term is independent of x, then the power of x equals ‘0’.



10  r r  0 2 3

3

x9 216

4

 3   2  2x 

10

 x  10. The given expression   3x   . 6   Here, n = 7, which is an odd number.

 ( 1)

, i.e.,

10  9  8  7  6 25 x10 35 . .  252 1 2  3  4  5 35 25 x10 12. Let the term independent of x in the expression

10  r

 x3  C3 (3x )    6

1

 n Cr x n  r a1 

1    2 3 of  2 x  3 x   

3

4

2



–20 –5 6  –1  = = 20 . x .  6  = 8 2  8x 

3 7

T10

th

term i.e., 4 term

 –1   2  2x 

x12 35 x15  1296 48

10  5

= 18564

n=6

6

 35  27 x3 

T6 .

6

1/ 2 12

12

4

3

12



 T13  18C12  9 x   

18

 x3   C4 (3x )     6  7

30  3r  2r 0 6  30  5r  0 

30 6 5 Hence, the term independent of x is, r 

6

T6 1  T7   1 . 10 C6 .210 6.36

10  9  8  7 4 6 .2 .3 4 3 2  210  16  729  2449440. 

Binomial Theorem Solutions

387 Now,

10

x 3  13. Given :   2  2 x 

6

T5  T4 1 10  r

 x 10 General term, Tr 1  Cr   2 r

r

 3  .  2  10 Cr x 

r

x10 r  3 10  3 10 3r .  Cr 10 r .x 210 r x 2 r 2 For the power of x to be 4. 10  3r  4  3r  6  r  2 4 10  coefficient of x  C2

 3

2

210 2

10  9 9 405   2  1 28 256 r r+1 14. If the coefficient of x , x in the expansion of

10! 2 1  10.9.8.7 2. 4.3.2.1 4!6! 22 = 10 × 3 × 7 = 210 17. Number of terms in the expansion of n n+r–1 (x + x + ....x ) is given by C 1 2 r r–1 Required number of terms in the given expansion is (here, n = 15, r = 4) 18 17 16 . . = 816 4–1 3 3 2 1 n 18. The number of terms in the expansion of (x+a) 15 + 4 – 1

n

1

  16. Last term of  2 3  1   Tn 1 2  

n

n

1

n

  1 .

n

22 Also, we have

  1  53 3



   

n

1



log 3 2 5

3

 1  n

 53  3   

3log 3 2



1 5log 3 2

2

6 1  3 1  4 2 120

1  1    4 3  54  19. Given, 4  5   Number of rational terms



3

4

120



  n 120   1     1  10  1  11.  12   L.C.M of 3, 4   Total number of terms = 120 + 1 = 121 Number of rational terms = 11 Number of irrational terms = 121 – 11 = 110 2 3 8 20. Let f(x) = (1 + x + x + x ) Sum of othe coefficients 8 = f(1) = (1+1+1+1) 8 2 8 16 = 4 = (2 ) = 2 2 3 9 21. Let f(x) = (1 – x + x – x ) Sum of the coefficients of odd powers of x

0  49 218 17  = –2 2 2 2 3 7 22. Let f(x) = (1 – x +x – x )

3



1 x  25  a log a  x  5   2 5

n  1 if n is even 2

9

1

n



18 = C 

 1  n

22

n   5  n  10 2

22

n

9

f 1  f  1 f 1  1  1  1  1  1  1  1   2 2

log3 8



C

+ (x – a) is

n

x   a   are equal, then b  n = (r + 1) (ab + 1) –1 = (7 + 1) (2.3 +1) –1 = 8 (7) –1 = 56 – 1 = 55 15. For the sum of binomial coefficients put x y z u p = q = r = s = 1. 1001 = (1 + 2 – 5 + 10)  Required sum 1001 3 1001 3003 = (8) = (2 ) =2

4



=



 1   1   Cn  2 3       2

 1   1   C4  2 3       2 10

10

 1  25

Sum of even powers of x  7

f 1  f  1 2

1  1  1  1  1  1  1  1 

7

2



0  47 214 13  =2 2 2

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10th Class Mathematics

388 n

10

ii) 8 is even  There is only one middle term in the expansion.

23. Sum of binomial coefficients = 2 = 2 = 1024 2 3 50 3 50 150 24. (1 + 3x + 3x + x ) = [(1 + x) ] = (1 + x) Middle term

Tn 2

1

 T150 2

1

 T75  1

1 log x  25. Given, T = 1,000 in   x 10  3 x  T = 1,000

8

x  8 8–4 4 4   9 y  namely, T5 = T4 + 1 = C4(x/3) 9 y 3  4 4 = 5670 x y

150

= 150 x

5

5.

i)

3

3

x 1 5 C2   x log10 x





7! 3! 7  6  5! 3  2  1 21 ii) 4! 5!  4  3  2  1  5!  2

2

 1,000

x

iii) 5! 3!  5  4  3  2  1  3  2  1 = 120 – 6 = 114

10 x 3 .x 2 log10  1,000

x

 2log 3 x 10

 102 6.

 2log   3   log  102 x

x 10

[Taking logarithm on both sides at the x] x 10

x 10

7.

x 10

2 x where t  log10 . t (2t + 1)(t – 2) = 0 2t  3 

1   t = 2 and  t   2  

 log10x  2 2  x = 10 = 100

8.

SUMM ATIVE WORKSHEET Number of terms = n + 1 = 99 + 1 = 100 3 2 100 3 100 300 (x + 3x + 3x + 1) = [(x + 1) ] = (x+1)  Number of terms = 300 + 1 = 301 10

3.

5! 5  4  3! = 10  3! 2! 3! 2  1

8! 8  7  6  5  4!   70 4!4! 4! 4  3  2  1 21 21– r 2 r i) T = C (2x) (1/3x ) 21r+1 21– r r 21 –r r –2r = C2 (x) (1/3) (x ) 21 r 21– r r 21 – 3r = C2 (1/3) (x) r For independent term 21 – 3r = 0  3r = 21  r=7 21 14 14 Value of the independent term C 2 (1/3) 7 21 21– r r 21 – 3r ii) T = C2 (1/3) (x) r+1 r 3 For coefficient of x 21 – 3r = 3  3r = 18  r = 6 3 21 15 6 Value of coefficient of x is C 2 (1/3) 6 a) 2 x 4  24 x 2  8 8 iii) C4 

 2log   3  log2 x 10

i) 5 C3 

10! 10  4  8  7  6! ii) 10C6 = 6! 4!  = 210 6! 4  3  2  1

 2log   3  log2

1. 2.

8! 8  7  6  5  4  3!   560 3! 4! 3! 4  3  2  1

10 2 20  1  1  2 1    x  2  2    x      x   x x   x     

b) 140 2 4 2 2 4 c) 2 x 16 x  20 x a  5a  .

9.

Since n = 9, the number of terms in the expansion is 10 and hence the middle terms will be the fifth and the sixth terms. 9 5 4 9 4 5  T = C (3x) (4y) and T = C (3x) (4y) 5 4 6 5 are the middle terms of the expansion.

 Number of terms = 20 + 1 = 21 4.

i) 15 is odd  There are two middle terms namely, T8 and T. 9 15 15 – r 2 r T in = C (3x) (–2/x ) .........(1) r+1 15r 15 – 7 7 14 T =T = C (3x) (–2) / x 815 7 8+18 77 14 15 8 –6 7 = C 3 x (–2) /x = C 3 x (–2) 7 7 15 15–8 8 16 T = T = C (3x) (–2) / x 915 8+1 7 7 8 16 15 7 –9 8 = C 3 x (–2) / x = C 3 x (–2) 8

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8

10. T = T 3

10

2+1

 10 C2 x16

2 10–2

= C (x ) 2

 1   2 x 

2

1 10 12 Cx 4 = 2 x

Co-efficient of third term 

10  9  45 2

Binomial Theorem Solutions

11.

r

389

np  m p  q when, n = 16,

HOTS WORKSHEET

p = 2, q = 3, m = – 18 1.

32  18 50   10 5 5 16 16–10 10 16 6 = C (5) (–1) = C 5 T 

10+1

12.

r

10

General Term : Cr  ax

p nr



 1  . q   bx 

r

 n r pn  pr 1  qr  . r .x  = x pn  pr  qr  x 0 . = Cr  a x b  

10

np 1 p  , q = 2, n = 10 p  q where, 2

That is, pn-pr-qr = 0 . Hence, r 

1 2  10  2  2  1 2 2 5 2

pn . pq

10 

2.

C2 x n 2 a 2  ....  n Cn 1 xa n 1  n Cn a n 10  2

1 10 Independent of x  C2   3

 45 

 3

2

1 45 5 3  3  34 3 3

13. 5 term in T = T 5

2

 n C4 a 3

 23   C4  a    n

th

4+1

n 4

1 a  

4

 n 4  4

Since it does not contain a :

2  n  4  4  0 3

2n – 8 – 12 = 0 2n = 20 n = 10 14. (x + 1) (x + 4) (x + 9) (x + 16) ..... (x + 400) 2 2 2 2 = (x + 1) (x + 2 ) (x + 3 ) (x + 4 ) ..... (x + 20 ) 19 2 2 2 2 Coefficient x = 1 + 2 + 3 + ...... + 20 n  n  1 2n  1 20  20  1 2.20  1   6 6



15.

n

n n 0 n n 1 1 n i)  x  a   C0 x a  C1 x a 

3.

4.

2 128 128 128 17 256 = (17 ) = (289) = (290 – 1)

= 128C0 (290)128 – 128C1 (290)129 + 128C2 (290)126 = 128C0 (290)128 – 128C1 (290)129 + 128C2 (290)126 –............– 128C127 (290) + 1 = 1000 m + 128C2 (290)2 – 128C1 (290) + 1 (128)(127) (290) 2  128  290  1 = 1000 m + 2

20  21  41  2870 6

np 9  2 18 r   6 p  q 2 1 3

 Trth+ 1 = T6 + 1 = T7  7 term is independent of x.

(x + a)n = (nC0 xna0a0 + nC2 xn-2 a2 +........) = nC1 xn-1 a1 = nC3 xn-3 a3 +.......)  (x + a)n = O + E and ii) Likewise, (x - a)n = O – E From (i) and (ii) we get (x + a)n (x – a)n = (O + E) (O – E) (xz – az) = O2 – E2 We have, 4 OE = (O + E)2 – (O – E)2 4 OE = (x + a)2n – (x – a)2n (x + a)2n + (x – a)2n = (O + E)2 + (O – E)2 = 2(O2 + E2) 50 50 (101) – 99 = (100 + 1)50 – (100 – 1)50 = 2 [50C1 (100)49 + 50C3 (100)47 + ....... + 50C49 1001] = {(100)50 + a positive number } > 10050  (101)50 >9950 + 10050

5.

= 1000 m+ 683527680 + 1 = 1000 m +683527000 + 680+ 1 = 1000 (m + 683527) + 681 Hence the answer. 32 = 25 (32)32 = (25)32 = 2160 (32)32 = (3 – 1)160 = 160C0 3160 – 160C13159 + ..... – 160 C159.3 + 160C160.1 159 = 3(3 – 160C13158 + ..... – 160 C159) + 1 = 3m + 1 www.betoppers.com

10th Class Mathematics

390 32

Now, 3232  323 m1 = 25 3m1 = 215m = 23 (5m + 1) .22 = 4. (8)5m + 1 = 4(7+1)5m+1 = 4[7n + 1] = 28n + 4 Hence, remainder is 4. 6.

(x + y + z + w)n =

1  2 2   Coefficient of x3 in 1  x  2 x   2 x   3x  

+5

9 9 r 9  r  r 18 3 r   = Coefficient of x3 in  Cr (1) 2 3 .x  r 0  = 9 C5×(–1)5 29–53–5

n

 x  y    2  w 

= –9C5×24×

= (x + y)n + nc1(x + y)n-1(z + w) + nc2(x + y)n 2 (z + w)2+...+ncn(2 + w)n  (n + 1) + n.2 + (n – 1).3 + ...+ 1. (n + 1)   ( n  r  1)(r  1)

9.

n

  ( n  1)  nr  r 2 r o

n

n

 ( n  1) .1  n r   r 2 = (n + 1).(n + 1) r o

7.

8.

r o

We have, = 1  x  2 x

2



 2 1   2x   3x  

9

9



9

Cr (1) r 29  r 3 r .x18 3 r

r 0

9



9

Cr ( 1) r 29  r x18 3 r +

r 0

9



9

Cr ( 1) r 29 r 3 r .x19  3r

r 0

9 9 r 9  r  r 20 3 r  2  +  Cr ( 1) 2 3 .x  r 0  Since 19–3r = 3 and 20 – 3r = 3 do not have integral soltions, the last two expressions do not contain x3. Only the first expression contains x3. For which, we must have 18–3r = 3  r = 5 www.betoppers.com

2



9

is

n(n  1) n( n  1)( n  2)  3.  .....nCn ] 2! 2!

n( n  1)(n  1)  ...  1] 2! = 2n + n [n–1C0 + n–1C1 + n–1C2 + .....+ n–1Cn–1) = 2n + n. 2n–1 = (n + 2)2n–1 10. We have, nC0 + nC1 + nC2 = 46 n = 2  n [1  ( n  1)

n  n  1  46 2  2 + 2n + n2 – n = 92  n2 + n – 90 = 0  n = 9 as n > 0  1 n 



= 1  x  2 x 2  =

2n  [ n  2.

r o

n( n  1) n( n  1)(2n  1)  + n. 2 6 = (n + 1) (n + 2)(n + 3)/62. The given expression = 23n+3- 7n - 8 = 23n.23 - 7n - 8 = 8n.8 - 7n - 8 = 8 (1+7)n - 7n - 8 = 8{1 + nc17+nc272 + ....+ ncn7n} - 7n - 8 = 8 + 56n + 8 (nc2.72 +....+ ncn7n)- 7n - 8 = 49n + 8 (nc2.72 +....+ ncn7n) = 49 {n + 8 (nc2 +.....+ ncn7n-2)} Hence, 23n+3 - 7n-8 is divisible by 49.

1   3x 

224 27 2 3 C0 + C1 + C2 + ..... (n + 1) Cn = (C0 + C1 + ....... + Cn) + (C1 + 2C2 + ...... nCn) = 2n +

1  x  2 x   2 x 2

n

1 224 = 5 3 27

 Coefficient of x3 in

n r o

9

r

np 9  2 18   6 p  q 2 1 3

Independent of x is 9C6 = 84 11. (x + a)n = (t1 + t3 + t5 +....) + (t2 + t4 + t6 +........)  (x + a)n = P + Q .............(A) and (x – a)n = (t1 + t3 + t5 +....) – (t2 + t4 + t6 +........) (x – a)n = P – Q ...........(B) Squaring and subtracting (A) and (B), we get, (x + a)2n – (x – a)2n = (P + Q)2 – (P – Q)2 = 4 PQ 12. Given (1 + x – 2x2)10 = 1 + a1 x + a2x2 + ........+ a20x20 Putting x = 1, –1, we get 0 = 1 + a1 + a2 + a3 + a4 + .....+ a20 .......(i) 210 = 1 – a1 + a2 – a3 + a4 + ......+ a20 ............(ii) Adding (i) and (ii), we get, 210 = 2 [1 + a2 + a4 + a6+ .....+ a20] 29 = 1 + a2 + a4 + a6+ +.....+ a20 29 – 1 = a2 + a4 + a6+ .....+ a20 511 = a2 + a4 + a6+ .....+ a20

Binomial Theorem Solutions

391 n

13. C02  C12  C22  C32  ...... 1 Cn2  0 if n is odd Here, n = 15 is odd, so evaluated value of S = 0 14. Clearly (n + 1)th term is the middle term in the expansion of (1 + x)2n and its Coefficient is 2nCn. In the expansion of (1 + x)2n–1 , nth and (n + 1)th terms 2n–1 are middle terms and their coeff. are C n–1 2n–1 and Cn respectively.  Sum of the coeff. of two middle terms in the expansion of (1 + x)2n–1 = 2n–1C n–1 + 2n–1Cn = 2nCn th 15. It is given that 6 term in the expansion of

17. It is given that T3 = 84, T4 = 280 and T5 =560. Now, T3 = 84  T3 = n C2 xn  2 a 2  84 ......(i)

T4  280  T4  n C3 xn 3a 3  280 ...... (ii) and T5  560  T5  n C4 xn  4 a 4  560 ..... (iii) In order to eliminate x and a, we multiply (i) and (iii) and then divide the product by the square of (ii)

T3  T5 84  560 3   (T4 ) 2 (280)2 5



8

 1  2  x8 / 3  x log10 x  is 5600, therefore,  



3

8

2 5  1  C (x log x) .  8 / 3  = 5600 5 10 x  10

 56 . x

10

2

 Tr 1 n  r  1 a      r x  Tr  5n – 15 = 4n – 8  n = 7

1 = 5600 x8

5

(log x) . 5

 x (log x) = 100  x = 10 10

n

4 1  th 16. In the expansion of  2  4  , 5 term from 3  the beginning is T5  T4 1  n C4  21/ 4   n C4

2

 1   1/ 4  3 

n4

3 x n3 a 3 T3 T5 3        n2 a 4 x 5 T4 T4 5

T n2 a Now, Tr 1  n  r  1  a  4   T3 3 x Tr r x 

280  7  2  a   84  3  x

4

n  7 

n4 / 4

–––––––– (1)

3 th Also, the 5 term from the end in the expansion



T3 = 84



n

C2 xn  2 a 2  84 

4 1  beginning in the expansion of  2  4  3   1  n . T5  C4  4   3

n 4

 2 4

4

 n C4

2 3

n 4 / 4

n

–––– (2)

7

C2 x5 a 2  84

n  7

n

4 1  th of  2  4  and is equal to 5 term from the 3 

a 10 5  a      x 2 3 3 x 

a  2

5

 x  a / 2

 21  (a 2 )  84

 a 7  27  a  2  x = 1  x  a / 2

Hence, x = 1, a = 2 and n = 7.

 x 1 x 1   1 1 18.  2  x 3  x3  1 x  x 2

10

   

We are given that. T5 : T5 :: 6 :1

T 1  5  T5 6  6

 n4 / 4



2

 n4 / 4

3 1

6 6 6

1 2

 n4 / 4

.

3

2

 6

n4 3   2n – 8 = 12 4 2  2n = 20  n = 10. 



 n 4  / 4

6 1

6

3/ 2

1   13  23    x  1 x  x 3  1     1   23  3   x  x  1   

 1   x 3  1  

10

x 1  x 



 x 1 x 1   x x 1   





10





10

 1 1    x3  11  x 

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10th Class Mathematics

392

 1 1   x3  1  x2 

3.

10

   

1 1 n = 10, p  , q  3 2

 r

np pq

10 r 3 4 1 1  2 3 4

Term independent of ' x '  10C4  1  210 19.

2 2  1  1   x     x    x  x    

4.

108

108

1    4. x .  x 

 2200

(only one) 20. 3400 = (34)100 = (81)100 = (80 + 1)100 = 100C0(80)100 + 100C1 (80)99 + .... + 100C97(80)3 + 100C98 (80)2 + 100C99 (80)1 + 100C100(1) = 102[(100C0(8)100 (10)98 + 100C1(8)99 (10)97 +...+ 100 C98 82) + 100C99(80) + 1] = 100 m + 100C1 (80) + 1, m is a + ve integer = 100 m + 100 × 80 + 1 = 100 (m + 80) + 1 = 100 m1 + 1 (m1 = m + 80 is an intger)  Last two digits = 01.

 a5  b5  c 5   Divided by c on both sides    c   = 5a4 + 10a3c+10a2c2 + 5ac3 –––––– (1) Consdier RHS, as it contains only b & c, we take (b + c)5. (b + c)5 = b5 + 5b4c + 10b3c2 + 10b2c3 + 5bc4 + c5  (–a)5 – b5 – c5 = 5b4c +10b3c2 + 10b2c3 + 5bc4 ( a + b + c = 0  b + c = – a) divided by c on both sides

IIT JEE WORKSHEET 1.

2.

(1 + x + x2)25 = a0 + a1x + a2x2 + ..... a50x50 Put x = 1  (1 + 1 + 1)25 = a0 + a1 + a2 + ..... a50  a1 + a2 + ...... a50 = (3)25 – a0 = (1 + 2)25 – nC0 where, a0 = nC0 = 1 = (2 + 1)25 – 1 = 25C0(2)25 + 25C1(2)24 + ...... 25C25(2)0 – 1 = 25C0 225 + 25C1 224 + ......+ 25C2421 + (1) – 1 = 2(M) is even. General term in p

x y x p     Tr 1  Cr    y x  y

Tr 1  p Cr

p r

y   x

General term is independent of x if p – 2r = 0 But no term of expansion is independent of x  p  2r  0  p  2r  p is odd.

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5.

p

x p r y r , , Tr +1 = nCr xp – 2r . y2r – p y p r x r

1999 – 9919 = (20 – 1)99 – (100 – 1)19 =99C0(20)99 – 99C1(20)98 + 99C2(20)97 – ........... + 99C98(20)1 – 99C99(20)0 – {19C0(100)19 – 19C1 (100)18 + 19C2(100)17 – ......– 19C19(100)0} = (20)99 – 99C1(20)98+99C2 (20)97 –...99C98(20)–1 – {(100)19 – 19C1(100)18 + 19C2(100)17 –..... + 19C18 (100)–1} = 202[2097 – 99C1(20)96 + 99C2(20)95–....99C97 (20)2] + 99 ×20 – 1 – (100)2 {(100)17 – 19C1 (100)16 + 19 C2 (100)15 – ....... 19C17} – 19×100+1 202 = M(80), (100)2 = 80M1 = 80m – 80m1 + 1980 –1900 = 80(m – m1 + 1) is divisible by 80. where, m, m1 are integers. Consider LHS. As it contains only a & c, we take (a + c)5 (a + c)5 = a5 + 5a4c + 10a3c2 + 10a2c3 + 5ac4 + c5  (– b)5 – a5 – c5 = 5a4c + 10a3c2 + 10a2c3 +5ac4 ( a + b + c = 0  a + c = – b)

6.

 a5  b5  c 5     c   = 5b4 + 10b3c + 10b2c2 +5bc3 ––––––––– (2) From (1) & (2), we have LHS = RHS. Hence proved. (a0+ a1x + a2x2 +...... an xn)2 = (a 0 + a 1 x + a 2 x 2 + ...a n x n )(a 0 +a 1 x+a 2 x 2 + ...an x n ) Coefficient of xn+1 in the product is a1an + a2an–1 + .....an a 1 Given that (1 + x)15 = a0 + a1 x + a2 x2 + ... + a15 x15  15 C0 + 15 C1 x + 15 C2 x2 + ... + 15 C15 x15 = a0 + a1 x + a2x2 +...+ a15 x15 Equating the coefficient of various powers of x, we get a0 = 15 C0, a1 = 15 C1, a2= 15 C2 , ..., a15 = 15 C15 15



r r 1

15 15 ar C   r 15 r ar 1 r 1 Cr 1

Binomial Theorem Solutions

393 2 3  1  x  x  x  1           ... 3   3   3   3  

15! 15 r !15  r !  r 15! r 1  r  1!15  r  11 15

 r 1

 Coefficient of x3 in

r  r  1!15  r  1! r !15  r !

3

 1 11        2  3 3

15 r 1

15 15  1 =120 2



1  2 x 

2

 2 k 10. General term in the expansion of  x   is x 

–2

= (1 + 2x) (1–2x)

2.3  2  1  2 x  1   2 x    2 x 2  ... 2!  1! 

2.3.4...  r  1 2 x  2.3...r r 1 2x  r!  r  1!

r

   

2

 r  1! 2r r! 2r 1  r!  r  1!

= r 2r + (r + 1) 2r = 2r (2r + 1) The series of binomial coefficient is 15C , 15C , 15C 0 1 2

…15C7,

decreasing value

15C , 15C 8 9

greatest value

…15C14, 15C15

decreasing value

From the above discussion, we can say that decreasing series is 15C7, 15C6, 15C5 therefore option (4) is correct. 9.

x4 x4  x  5 x  6  x  2  x  3 2



2 1  –1 –1  x  2   x  3 = 2 (x – 2) – (x – 3) 1

x x 1  1   2  2   1     3 1    2  3

  x   x  2  x 3    1           ...   2   2   2  

1

r

k  Tr 1  Cr  x    x 5 10–3r r = Cr x .k Let this term contains x then, 10 – 3r = 1  3r = 9  r = 3, then coefficient of x = 5 C3 k3 = 10k3 Given that 10k3 = 270  k3 = 27  k = 3 11. We have, (1 + x + x2 )n Put x =1 = (1 + 1 + 12)n = 3n 12. We have, Tp = n Cp –1 = p T p + 1 = n Cp = q 2 5 r

5

The coefficient of xr

8.

73 648 5

1  2x

7.

3

1 1 81  8    8 81 648

  [15 – r + 1] = 15 + 14 +13 + ... + 2 + 1 

x4 x  5x  6 2

n C p 1 p  n q Cp





p p  q n  p 1

 p+q=n+1

13. Let S = C0 + 2C1 + 3C2 + ... + (n+1) Cn ...(i) S = Cn + 2Cn–1 +3Cn–2+ ... + (n+1) C0 or S = (n + 1) C0 + nC1 + ... + Cn ...(ii) On adding equations (i) and (ii), we have 2S = C0 [1 + n + 1] + C1 [2 + n] + ... + Cn [n + 1 + 1] 2S = (n + 2) (C0 + C1 + ... + Cn ) S

n  2 2

2n = (n + 2) 2n–1 = n2n–1 + 2n

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10th Class Mathematics

394

6

 2 1 19. The rth term of  x   is x 

14. We have, 2

1  3x  1  2 x 

= (1 + 9x2 – 6x) (1 – 2x)–1 6

= (1 + 9x2 –6x) (1 + (2x) + (2x)2 + (2x)3 + ...) Coefficient of x4 = 16 + 36 – 48 = 52 – 48 = 4 15. Coefficient of rth term, T r = 3n Cr–1 Coefficient of (r + 1)th term, Tr + 1 = 3nCr Given that,

Tr 1  Tr 1 2 3n

Cr 1 1  Cr 2

3n

 2 3n Cr–1 = 3nC r 2



3n ! 3n!   r  1! 3n  r  1! r ! 3n  r !

2  r  1!. 3n  r  1 3n  r !

1 (3n + 1) 3

6! = 15 4!2!

20. We have, n C0 + nC1 x + n C2 x2 + n C3 x3 + ... +n Cn xn = (1 + x)n On differentiating w.r.to x, we get 0 + nC1 + 2nC2x + 3nC3x2 + ... + n nCn xn–1 = n(1+x)n–1 Put x = 1. n C1 + 2n C2 + 3n C3 + ... + n n Cn = n.(2)n–1 15

 2 3 21. We have,  2x  3  x   The rth term is

2 15–r

2

3

1 k  k  1  1  k  k  1 k  2   1  1 k        ... 3 2!  3  3! 3

16. We have, 2n

1  x  1 n 1  x  1    xn  x =x–n [2nC0 + 2nC1 x + .... + 2n C2nx2n ] Coefficient of x–n = 2n C0 =1 17. The middle term is n 40 1   1 = 21th term 2 2 The coefficient of 21th term 

 40 ! 2  20!

18. We have, T8 = T7 + 1 = 10 C7(1)3 (x)7 Coefficient = 10 C 7 = 10 C10–7 = 10 C 3 www.betoppers.com

T4+1 = 6C4 (–1)4 =

r

2 1  3n  r  1 r  2r = 3n – r + 1  3r = 3n + 1

T20 + 1 =40 C20 =

r

= 6 Cr (-1)r x12–3r Let this term is indpendent of x, then 12 – 3r = 0  r=4

15



n

Tr + 1 = Cr (x )

 1    x

 3 Tr + 1 = Cr (2x ) .   3   x  15 15–r = Cr 2 .(–3)r . x30–5r This term is independent of x, if 30 – 5r = 0  r = 6  T 6 + 1 =15C6 215–6 (–3)6 = 15 C6 29 (3)6 = 15C9 29 . 36 [  15 C6 = 15 C9 ] 22. We have

1  r  r  1! 3n  r !

 r=

2 6–r

k

k

k

 1 2  3  1          3 3  2 23. We have (3 + 2x)44 = 44 C0 (3)44 (2x)0 + 44 C1 (3)43 (2x)1 + ... + 44 C44 (3)0 (2x)44 It is clear from the expansion of (3 + 2x)44. The 1st term is independent of x. 24. Given that (1 + x + x2)n = a0 + a1x + a2x2 + ..... + a2n x2n On differentiating w.r.t. x, we get n(1 + x +x2)n–1 (1+2x) =a1 + 2a2x + .... + 2na 2n x 2n–1 Put x = –1, n(1 – 1 + 1)n–1 (1–2) = a1 – 2a2 + ..... + (–1)2n–1 2na2n  a1 – 2a2 + ....+ (–1)2n–1 2n.a2n = –n

Binomial Theorem Solutions 25. Coefficient of x5 in (1 + x2)5 (1 + x)4 = Coefficient of x5 in (5 C0 + 5 C1 x2 + 5C2x4 + ...) (1 + 4 C1x + 4 C2x2 + 4C3 x3 + 4 C4x4 ) 5 = C 1. 4 C 3 + 5 C 2. 4 C 1 = 5.4 + 10.4 = 20 + 40 = 60 26. We have 11

1 T12 = C11 x    x 23

12

and 12

1 T 13 = 23 C12 x11    x

11



12

 T12 = x2 T 13 27. We have (1 + x)9 = 9C0 + 9 C1 x + ... + 9 C9 x9 Put x =1 9 C0 + 9C1 + 9C2 + ... + 9C8 + 9C9 = 29 ... (i) Agains put x = –1 9 C0 – 9C1 + 9 C2 – ... + 9C8 – 9 C9 = 0 ...(ii) Adding equations (i) & (ii) 2(9C0 + 9C2 + 9 C4 + 9C4 + 9C6 + 9C8) = 29 C0 + C2 + C4 + C6 + C8 = 28 = 256 28. Since n C4 , nC5 and n C6 are in AP  2 nC 5 = nC 4 + n C 6 2  n ! n! n!   5! n  5 ! 4! n  4 ! 6! n  6 !



29. We have 2n

2  n  5 .5.4! n  6 !

(1 + x) =



2 1 1   5  n  5   n  4  n  5 5.6



 n  4  n  5  2 n  4  1 5 30 n 2  9n  20 6 12 (n – 4) = 30 + n2 – 9n + 20 12n – 48 = 30 + n2 – 9n + 20 n2 – 21n + 98 = 0 n2 – 14n – 7n + 98 = 0 n (n – 14) –7 (n – 14) = 0 (n – 14) (n – 7) = 0  n = 7, 14

 2  n  4  5 

     

r

r

r 0

The rth term in the expansion of (1 + x)2n is Tr + 1 = 2n Cr.xr The coefficient of xr i.e., ar = 2n Cr 30. The rth term in the expansion of (x5/2 + 3x–3/2 )6 is T r+1 = 6 Cr (x5/2 ) 6 – r (3x –3/2 ) r 30 5 r 2

= 6 Cr 3 r .  x 

. 3r .  x 

3 r 2

30 8 r 2

Let rth term contains the coefficient of x3 .

30  8r 3 2  30 – 8r = 6  8r = 24  r =3 The coefficient of x3 = 6C3 33 Now, put

6.5.4 1.2.3 = 27 . 20 = 540  27.

31. We have (1 + x)n = C0 + C1x + C2x2 + ... + Cnxn Put x = i, we get (1 + i)n = C0 + C1 i – C2 – C3i + ... Cnin ...(i) Again put x = – i, we get (1 –i)n = C0 – C1 i – C2 + C3i + ... + Cn (–1)n in ...(ii) On adding equations (i) and (ii), we get (1+i)n + (1 –i)n = 2(C0 –C2 + C4 ...) 

1 1   4! n  4  n  5  n  6 !  6.5  4! n  6 !

a x

2n

= 6 Cr  x 

1 C11 x   T12  x   x2  12 T13 23 1 C12 x11   x 23

395

1 2 



2e i  / 4

n

 

n 2e  i  / 4  



1  .2n / 2 ein  / 4  e in / 4  2 1 n  .2n / 2 2cos 2 4  2n / 2 cos

n 4

32. We know that n C 0 – nC 1 x + n C 2 x 2 – ... + (–1)n n Cn xn = (1–x)n Integrating from 0 to 1, we get 1

n x 2 n x3 x n 1  n n n C  C  C  ...   1 C   0 1 2 n   2 3 n  10 

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10th Class Mathematics

396 34. We have

1

 1  x n 1    1    n  1  0

(1 + 3 2x )9 = 9C0 + 9C1 ( 3 2x ) n

 n C 0 – n C1 .

1 n 1 Cn + C2 . 1 – ... + (–1)n 2 3 n 1

1 0 n 1 

33. We have n

r 1

n

+ 9C5 ( 3 2x )5 + 9C6 ( 3 2x )6 + 9 C7 ( 3 2x )7 + 9C8 ( 3 2x )8 + 9C9 ( 3 2x )9 Again

...(i)

(1 – 3 2x )9 = 9C0 – 9C1 ( 3 2x )

C0 C1 C2 1 n Cn    ...   1  1 2 3 n 1 n 1



+ 9C2 ( 3 2x )2 + 9C3 ( 3 2x )3 +9 C4 ( 3 2x )4

2

C r  = ( nC ) 2 + ( n C ) 2 + ( nC ) 2 1 2 3

+ ... + (nCn)2

...(i)

Now, (1 + x)n = nC0 + n C1 x + nC2 x2 + n C3 x3 + ... + n Cn xn ...(ii) (x + 1)n = n C0 xn + n C1 xn–1 + n C2 xn–2 + n C3 xn–3 + ... + n Cn ...(iii) From equations (ii) & (iii), we get (1 + x)2n = (nC0 + nC1 x + nC2 x2 + n C3 x3 + ... + nCn xn).(n C0 xn + n C1 xn–1 + n C2 xn–2 + n C3 xn–3 + ... + n Cn ) Equating the coefficient of xn on both sides, we get 2n Cn = (nC0 )2 + (n C1)2 + (nC2)2 + ... + (nCn )2 (n C1)2 + (nC2)2 + (nC3 )2 + ... + (nCn )2

 

2n ! 1 n !n ! 2n!

 n!

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2

1

+ 9 C2 ( 3 2x )2 – 9 C3 ( 3 2x )3 +9 C4 ( 3 2x )4 – 9 C5 ( 3 2x )5 + 9 C6 ( 3 2x )6 – 9C7 ( 3 2x )7 + 9 C8 ( 3 2x )8 – 9 C9 ( 3 2x )9 Adding equations (i) and (ii), we get

...(ii)

(1 + 3 2x )9 + (1 – 3 2x )9 = 2 9 C0 + 29C2 ( 3 2x )2 + 2 9 C4 ( 3 2x )4 + 2 9C6 ( 3 2x )6+ 2 9 C8 ( 3 2x )8 There are 5 terms in the expansion of (1 + 3 2x )9 + (1 – 3 2x )9 . 35. We know that (1 + x)n = C0 + C1x + C2x2 + C3x3 + ... + Cn xn Integrating, on both sides, we get

1  x 

n 1

n 1

 C0 x  C1

x2 x3 x4  C2  C3 ... 2 3 4  Cn .

 C0 

C1 C C C x  2 x 2  3 x 3  ...  n x n 2 3 4 n 1

1  x  1  . x  n  1

n 1

x n 1 n 1

10. TRIANGLES SOLUTIONS Let EC = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain AD AE 1.5 1 3×1   = x= DB EC 3 x 1.5 x=2  EC = 2cm

FORMATIVE WORKSHEET 1.

(i) Similar (ii) Similar (iii) Equilateral (iv) (a) Equal (b) Proportional (i) Two equilateral triangles with sides 1 cm and 2 cm

2.

5.

(ii)

Two squares with sides 1 cm and 2 cm Trapezium and square

Let AD = x cm It is given that DE || BC. By using basic proportionality theorem, we obtain AD AE x 1.8 1.8×7.2 =   x= DB EC 7.2 5.4 5.4 x = 2.4  AD = 2.4 cm 6.

Triangle and parallelogram

3.

4.

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.

Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm PE 3.9 = =1.3 EQ 3 PF 3.6 = = 1.5 FR 2.4 PE PF Hence,  EQ FR Therefore, EF is not parallel to QR.

____________________________________________________________________________________________________ _____________________________________________

Triangles Solutions

398

AN AL __________ = (ii) AD AC From (i) and (ii), we obtain AM AN = AB AD In ABC, DE || AC BD BE  = (Basic Proportionality Theorem) DA EC _______ (i) 

8.

PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm PE 4 8 = = EQ 4.5 9 PF 8 = FR 9 PE PF Hence,  EQ FR Therefore, EF is parallel to QR.

7.

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm PE 0.18 18 9 = = = PQ 1.28 128 64 PF 0.36 9 = = FR 2.56 64 PE PF Hence,  PQ PR Therefore, EF is parallel to QR. In the given figure, LM || CB By using basic proportionality theorem, we obtain AM AL = AB AC Similarly, LN || CD

In ABC, DE || AE BD BF  = (Basic Proportionality Theorem) DA FE _______ (ii) From (i) and (ii), we obtain BE BF = EC FE

8.

In  POQ, DE || OQ PE PD  = (Basic Proportionality Theorem) EQ DO _______

(i)

In  POQ, DF || OR PE PD  = (Basic Proportionality Theorem) FR DO _______

(ii)

From (i) and (ii), we obtain ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

399

PE PF = EF || QR (Converse of basic EQ FR proportionality theorem) In  POQ, DF || OR

10.

PE PD = FR DO _______ (Basic Proportionality Theorem) (ii) From (i) and (ii), we obtain PE PF = EQ FR EF || QR (Converse of basic proportionality theorem) In  POQ, AB || PQ OA OB  = (Basic Proportionality Theorem) AP BQ

From (i) and (ii), we obtain OB OC = BQ CR  BC || QR (By the converse of basic proportionality theorem) Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ ||BC.



9.

_______

By using basic proportionality theorem, we obtain

(i)

11.

AQ AP = QC PB AQ 1 = (P is the mid-point of AB,  AP = QC 1 PB)  AQ = QC or, Q is the mid-point of AC. Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

In  POQ, AC || PR

OA OC = AP CR (Basic Proportionality Theorem) 

_______

(ii)

i.e., AP = PB and AQ = QC It can be observed that AP 1 AQ 1 = and = PB 1 QC 1 AP AQ  = PB QC Hence, by using basic proportionality theorem, we obtain PQ || BC.

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Triangles Solutions

12.

400

Draw a line EF through point O, such that EF || CD.

In ADC, EO || CD By using basic proportionality theorem, we obtain AE AO _________ = (1) ED OC In ABD, OE || AB

⇒

By using basic proportionality theorem, we obtain AE BO _________ = (1) ED OD However, it is given that AO OB _________ = (2) OC OD From equations (1) and (2), we obtain AE AO = ED OC EO || DC [By the converse of basic proportionality theorem] ⇒ AB || OE || DC ⇒ AB || CD  ABCD is a trapezium.

13.

So, by using basic proportionality theorem, we obtain ED OD = AE BO AE BO _________ =  (2) ED OD From equations (1) and (2), we obtain AO BO = OC OD AO OC =  BO OD Let us consider the following figure for the given question.

14. Since the ratio of the areas of two similar triangle is equal to the ratio of the squares of their corresponding altitudes and is also equal to the ratio of the squares of their corresponding medians Hence, ratio of altitudes = 9: 7 = ratio of medians 15. 5. We have, XY || AC and, Area(BXY) = Area (quad. XYCA)  Area  ABC   2 Area  BXY  ….(i) Now, XY || AC and BA is a transversal ….(ii)  BXY  BAC Thus, in ’s BAC and BXY, we have XBY  ABC [Common] [From (ii)] BXY  BAC Therefore, AA-criterion of similarity, we have BAC ~ BXY Area  BAC  BA 2   Area  BXY  BX 2



BA 2 BX 2 BA  2 BX



BA  2  BA  AX 







AX 2 1  AB 2



Draw a line OE || AB In ABD, OE || AB

2

[Using (i)]



2  1 BA  2 AX

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10th Class Mathematics

401

16. AE = 3cm BE = 4cm AF = 15cm FC = ? In ABD, D is bisector, so AD AE  …(i) BD BE AD AF   …(ii) [since, BD = DC] DC FC AE AF  so, BE FC AE AF  BE FC 3 15  4 FC FC = 20 cm 17. (i) A = P = 60° B = Q = 80° C = R = 40° Therefore, ABC ∼ PQR [By AAA similarity criterion] AB BC CA = = QR RP PQ  ABC ~ QRP [By SSS similarity criterion] (iii) The given triangles are not similar as the corresponding sides are not proportional. (iv) The given triangles are not similar as the corresponding sides are not proportional. (v) The given triangles are not similar as the corresponding sides are not proportional. (vi) In DEF, D + E + F = 180º (Sum of the measures of the angles of a triangle is 180º.) 70º + 80º +F = 180º F = 30º Similarly, in PQR, P +Q +R = 180º (Sum of the measures of the angles of a triangle is 180º.) P + 80º + 30º = 180º P = 70º In DEF and PQR, D = P (Each 70°) E = Q (Each 80°) F = R (Each 30°)  DEF ∼ PQR [By AAA similarity criterion]

18.

DOB is a straight line.  DOC + COB = 180°  DOC = 180° − 125° = 55° In DOC, DCO + CDO + DOC = 180° (Sum of the measures of the angles of a triangle is 180º.)  DCO + 70º + 55º = 180°  DCO = 55° It is given that ODC ∼ OBA.  OAB = OCD [Corresponding angles are equal in similar triangles.]  OAB = 55° 19. DOB is a straight line.

In DOC and BOA, CDO = ABO [Alternate interior angles as AB || CD] DCO = BAO [Alternate interior angles as AB || CD] DOC = BOA [Vertically opposite angles]  DOC ∼ BOA [AAA similarity criterion] DO OC  = [Corresponding sides are BO OA proportional] OA OB  = OC OD 20. In PQR, PQR = PRQ ___________  PQ = PR (i) Given, QR QT = QS PR Using (i), we obtain QR QT _____________ = (ii) QS QP In PQR, TQR, QR QT = [Using (ii)] QS QP Q = Q  PQS ~ TQR [SAS similarity criterion]

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Triangles Solutions

21. In RPQ and RST, RTS = QPS (Given) R = R (Common angle)  RPQ ∼ RTS (By AA similarity criterion) 22. It is given that ABE ≅ ACD. _____________  AB = AC [By CPCT] (1) _____________ and, AD = AE [By CPCT] (2) In ADE and ABC, AD AE = [Dividing equation (2) by (1)] AB AC A = A [Common angle]  ADE ∼ ABC [By SAS similarity criterion]

402

In AEP and ADB, AEP = ADB (Each 90°) PAE = DAB (Common) Hence, by using AA similarity criterion, AEP ∼ ADB (iv)

23. (i)

In AEP and CDP, AEP = CDP (Each 90°) APE = CPD (Vertically opposite angles) Hence, by using AA similarity criterion, AEP ∼ CDP (ii)

In ABD and CBE, ADB = CEB (Each 90°) ABD = CBE (Common) Hence, by using AA similarity criterion, ABD ∼ CBE

In PDC and BEC, PDC = BEC (Each 90°) PCD = BCE (Common angle) Hence, by using AA similarity criterion, PDC ∼ BEC 24. In ABE and CFB, A = C (Opposite angles of a parallelogram) AEB = CBF (Alternate interior angles as AE || BC) ABE ∼ CFB (By AA similarity criterion) 25. In ABC and AMP, ABC = AMP (Each 90°) A = A (Common)  ABC ∼ AMP (By AA similarity criterion) CA BC =  PA MP (Corresponding sides of similar triangles are proportional) 26. It is given that ABC ∼ FEG.

(iii)

 A = F, B = E, and ACB = FGE ACB = FGE ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

403

 ACD = FGH (Angle bisector) and, DCB = HGE (Angle bisector) In ACD and FGH,

A = F (Proved above) ACD = FGH (Proved above)  ACD ∼ FGH (By AA similarity criterion) CD AC = GH FG In DCB and HGE, DCB = HGE (Proved above) B = E (Proved above)  DCB ∼ HGE (By AA similarity criterion) In DCA and HGF, ACD = FGH (Proved above) A = F (Proved above)  DCA ∼ HGF (By AA similarity criterion) 27. It is given that ABC is an isosceles triangle.  AB = AC  ABD = ECF In ABD and ECF, ADB = EFC (Each 90°) BAD = CEF (Proved above)  ABD ∼ ECF (By using AA similarity criterion) 28. Median divides the opposite side.

BC QR and QM = 2 2 Given that, AB BC AD = = PQ QR PM 1 BC AB 2 AD  = = PQ 1 QR PM 2 AB BD AD  = = PQ QM PM BD =

In ABD and PQM, AB BD AD = = (Proved above) PQ QM PM  ABD ∼ PQM (By SSS similarity criterion)  ABD = PQM (Corresponding angles of similar triangles) In ABC and PQR, ABD = PQM (Proved above) AB BC = PQ QR  ABC ∼ PQR (By SAS similarity criterion) 29. In ADC and BAC,

ADC = BAC (Given) ACD = BCA (Common angle)  ADC ∼ BAC (By AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. CA CD   CB CA 2  CA = CB × CD AB AC AD 30. Given that,   PQ PR PM

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

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Triangles Solutions

We know that medians divide opposite sides. Therefore, BD = DC and QM = MR Also, AD = DE (By construction) And, PM = ML (By construction) Therefore, quadrilateral ABEC is a parallelogram.  AC = BE and AB = EC (Opposite sides of a parallelogram are equal) Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR It was given that AB AC AD   PQ PR PM AB BE 2AD    PQ QL 2PM AB BE AE    PQ QL PL  ABE ∼ PQL (By SSS similarity criterion) We know that corresponding angles of similar triangles are equal. _____________  BAE = QPL (1) Similarly, it can be proved that AEC ∼ __________ PLR and CAE = RPL (2) Adding equation (1) and (2), we obtain BAE + CAE = QPL + RPL ____________  CAB = RPQ (3) AB AC In ABC and PQR,  (Given) PQ PR CAB = RPQ [Using equation (3)]  ABC ∼ PQR (By SAS similarity criterion) 31. Let AB and CD be a tower and a pole respectively.

404

Let the shadow of BE and DF be the shadow of AB and CD respectively. At the same time, the light rays from the sun will fall on the tower and the pole at the same angle. Therefore, DCF = BAE and DFC = BEA CDF = ABE (Tower and pole are vertical to the ground)  ABE ∼ CDF (AAA similarity criterion) AB BE   CD DF AB 28   6cm 4  AB = 42cm Therefore, the height of the tower will be 42 metres. 32. It is given that ABC ∼ PQR We know that the corresponding sides of similar triangles are in proportion.

AB AC BC ______________  = (1) PQ PR QR Also, A = P, B = Q, C = R 

______________

(2) BD QR ______________ BD = and QM = (3) 2 2 From equations (1) and (3), we obtain AB BD  PQ QM In ABD and PQM, B = Q [Using equation (2)] AB BD  [Using equation (4)] PQ QM  ABD ∼ PQM (By SAS similarity criterion)

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10th Class Mathematics

405

AB BD AD   PQ QM PM 33. We have Area  AOB  4  Area  COD  1





84 4  Area  COD  1

 Area  COD   21cm2 34. We have, Area(ADE) = Area (trapezium BCED)  Area  ADE   Area  ADE   Area  trapezium BCED   Area  ADE  

2Area  ADE   Area  ABC

AB AC  and A  D DE DF [Given] So, by SAS-similarity criterion, we have ABC ~ DEF Area  ABC  AL2   Area  DEF  DM 2 16 AL2  25 DM 2 [Using (i)] AL 4   DM 5 36. In ABC, we have DE || BC 



In ADE and ABC, we have ADE  B [ DE || BC  ADE  B (Corresponding angles] and, A  A  ADE ~ ABC Area  ADE  AD 2   Area  ABC  AB2 





Area  ADE  2Area  ADE 

1  AD    2  AB  AD 1  AB 2



AD 2 AB2

2

BD 2 1 2  2   AB 2 2 35. Let ABC and DEF be the given triangles such that AB = AC and DE = DF, A  D Area  ABC  16 and,  Area  DEF  25 Draw AL  BC and DM  EF Now, AB  AC, DE  DF AB DE   1 and 1 AC DF AB DE   AC DF AB AC   DE DF Thus, in triangles ABC and DEF, we have 

ADE  ABC and AED  ACB [Corresponding angles] Thus, in triangle ADE and ABC, we have A  A [Common] ADE  ABC and, AED  ACB  AED ~ ABC [By AAA similarity] AD DE   AB BC We have, AD 5  DB 4 DB 4   AD 5 DB 4  1  1 AD 5 DB  AD 9   AD 5 AB 9 AD 5     AD 5 AB 9 DE 5   BC 9 In DFE and CFB, we have [Alternate interior angles] 1  3 2  4 [Vertically opposite angles] Therefore, by AA-similarity criterion, we have DFE ~ CFB Area  DFE  DE 2   Area  CFB  BC 2 2

Area  DFE   5  25     Area  CFB  9  81 37. In ABC, AD is a bisector of A

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Triangles Solutions

406

AB BD  AC DC From A, draw AL  BC A 

B 

 

L

C

D

1  BD  AL 2 Area  ACD  1  DC  AL 2 Area of  ABD  BD  Area of  ACD  DC Area  ABD 

Area of  ABD  Area of  ACD 



OBA = ODC (Alternate interior angles)  AOB ∼ COD (By AAA similarity criterion) 2 ar  ΔAOB   AB   = ar  ΔCOD   CD  Since AB = 2 CD, 2 ar  ΔAOB   2CD   = ar  ΔCOD   CD  4 = = 4 :1 1 40. Let us draw two perpendiculars AP and DM on line BC.

AB AC

38. It is given that ABC ~ DEF. 2 2 2 ar  ΔABC   AB   BC   AC  = = = ar  ΔDEF   DE   EF   DF  Given that, EF = 15.4 cm, 2 ar (ABC) = 64 cm , 2 ar (DEF) = 121 cm , 2 ar  ABC   BC   = ar  DEF   EF 

 64cm2  BC2  = 2 2   121cm  15.4cm  BC  8     cm 15.4  11   8 ×15.4   BC =   cm  11  = (8 × 1.4)cm = 11.2cm 39. Since AB || CD,



We know that area of a triangle = 1 ×Base ×Height 2 1 ar  ΔABC  2 BC× AP AP = = ar  ΔDBC  1 BC×DM DM 2 In APO and DMO, APO = DMO (Each = 90°) AOP = DOM (Vertically opposite angles)  APO ∼ DMO (By AA similarity criterion) AP AO =  DM DO ar  ΔABC  AO  = ar  ΔDBC  DO 41. Let us assume two similar triangles as ABC ∼ PQR. 2 2 2 ar  ABC   AB   BC   AC  = = = ar  PQR   PQ   QR   PR  _______________

 OAB = OCD and OBA = ODC (Alternate interior angles) In AOB and COD, AOB = COD (Vertically opposite angles) OAB = OCD (Alternate interior angles)

(1) Give that, ar (ABC) = ar (PQR) ar  ΔABC   =1 ar  ΔPQR  Putting this value in equation (1), we obtain

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10th Class Mathematics

407 2

2

2

 AB   BC   AC  1=   =  =   PQ   QR   PR   AB = PQ, BC = QR and AC = PR  ABC = PQR (By SSS congruence criterion) 42. D and E are the mid-points of ABC.  ABC ∼ PQR 

AB BC AC = = PQ QR PR ________



1 DE || AC and DE = AC 2 In BED and BCA, BED = BCA (Corresponding angles) BDE = BAC (Corresponding angles) EBD = CBA (Common angles)  BED ~ BCA (AAA similarity criterion) 2 ar  BED   DE   ar  BCA   AC  

ar  BED  ar  BCA 



1 4

1  ar  BED = ar  BCA  4 Similarity, 1 ar (CEF) = ar  CBA  and ar  ADF  4 1 = ar  ABC  4 Also, ar (DEF) = ar(ABC) = [ar (BED) + ar(CFE) + ar(ADF)]  ar(DEF) = ar(ABC) 3 1  ar  ABC = ar  ABC 4 4 ar  DEF  1  = ar  ABC  4

A = P, B = Q, C = R (2) Since AD and PS are medians, BC  BD = DC = 2 QR and, QS =SR = 2 Equation (1) becomes AB BD AC = = PQ QS PR In ABD and PQS, B = Q [Using equation (2)] and, AB BD = [Using equation (3)] PQ QS  ABD ∼ PQS (SAS similarity criterion) Therefore, it can be said that AB BD AD _____________ = = (4) PQ QS PS ar  ΔABC 

2

2

2

 AB   BC   AC  =  =  =  ar  ΔPQR   PQ   QR   PR  From equations (1) and (4), we may find that AB BC AC AD = = = PQ QR PR PS and hence, 2 ar  ΔABC   AD  = ar  ΔPQR   PS  44. Let ABCD be a square of side a.

43. Let us assume two similar triangles as ABC ∼ PQR. Let AD and PS be the medians of these triangles.

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Triangles Solutions

408

Therefore, side of BDE =

x 2

2

  area  ABC   x  4    = area  BDE   x  1 2 Hence, the correct answer is (C). 46. If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles. It is given that the sides are in the ratio 4:9. Therefore, ratio between areas of these Therefore, its diagonal = 2a Two desired equilateral triangles are formed as ABE and DBF. Side of an equilateral triangle, ΔABE, described on one of its sides = a Side of an equilateral triangle, ΔDBF, described on one of its diagonals = 2a We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. Area of ABE  a  =  Area of DBF  2a  1 = 2

2

45. We know that equilateral triangles have all its angles as 60º and all its sides of the same length.

Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles. Let side of ABC = x

2

 4  16 triangles =   =  9  81 Hence, the correct answer is (D). 47. Let ABC be an equilateral triangle and let AD  BC. In ADB and ADC, we have AB = AC [Given]  B  C [Each equal to 60°] and ADB  ADC [Each equal to 90°]  ADB  ADC  BD = DC BC  BD = DC = 2 Since ADB is a right triangle right angle  AB2 = AD2 + BD2   

 BC  AB2  AD 2     2  BC 2 AB2  AD 2  4

AB2  AD2 

AB2 4

2

 BC  AB

3 AB2  AD2  3AB2  4AD 2 4 48. From Pythagoras theorem, BE = 10cm So, area of rectangle BCDE = BC × BE = 12 × 10 = 120 cm2 49. Draw a line BE perpendicular on DC. D 

B

12cm

8cm A

5cm E 13cm

12cm

C

____________________________________________________________________________________________________ _____________________________________________

B

10th Class Mathematics

409

Then, DE = 5 cm

12 

New, BD =

2

Squaring the lengths of these sides, we will obtain 9, 64, and 36. However, 9 + 36 ≠ 64 Or, 32 + 62 ≠ 82 Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. Therefore, the given triangle is not satisfying Pythagoras theorem. Hence, it is not a right triangle. (iii) Given that sides are 50 cm, 80 cm, and 100 cm. Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000. However, 2500 + 6400 ≠ 10000 Or, 502 + 802 ≠ 1002 Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side. Therefore, the given triangle is not satisfying Pythagoras theorem. Hence, it is not a right triangle. (iv)Given that sides are 13 cm, 12 cm, and 5 cm. Squaring the lengths of these sides, we will obtain 169, 144, and 25. Clearly, 144 +25 = 169 Or, 122+52=132 The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a right triangle. We know that the longest side of a right triangle is the hypotenuse. Therefore, the length of the hypotenuse of this triangle is 13 cm.

2

  5   169

BD = 13 cm 50. LA = ? AB2 + AC2 = BC2 A 4 3 cm

4cm B

 4

2



C

8cm

 4 3



2

  8

2

so, A = 90° 51. If AB2 + AC2 = BC2 then, A = 90° A

2a

a B

2a

52. AB = 8 cm BC = 6 cm We know that A

C

C

B Angle subtended by diameter On circle = 90° So ABC = right angle So AC  53.

 8

2

2

  6   10cm

(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm. Squaring the lengths of these sides, we will obtain 49, 576, and 625. 49 + 576 = 625 Or, 72+242=252 The sides of the given triangle are satisfying Pythagoras theorem. Therefore, it is a right triangle. We know that the longest side of a right triangle is the hypotenuse. Therefore, the length of the hypotenuse of this triangle is 25 cm. (ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

54.

Let MPR = x In  MRP = 180°-90°-x MRP =90°-x Similarly, in MPQ, MPQ = 90°-MPR = 90°-x MQP = 180°-90°-(90°-X) MQP = x

____________________________________________________________________________________________________ _____________________________________________

Triangles Solutions

In QMP and  PMR, MPQ = MRP PMQ = RMP MQP= MPR QMP PMR (By AAA similarity criterion) QM MP   PM MR

410

∴ AC = CB Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain AC2 + CB2 = AB2  AC2 +AC2 = AB2 (AC = CB)  2 AC2 = AB2 57.

PM2 = QM ×MR 55.

(i) In ADB and CAB, DAB = ACB (Each 90°) ABD = CBA (Common angle) ADB CAB (AA similarity criterion) AB BD   CB AB

AB2 = CB×BD (ii) Let CAB = x In CBA, CBA = 180°-90°-x CBA = 90°-x Similarly, in CAD, CAD = 90°-CAB = 90°-x CDA = 180°-90°-(90°-x) CDA = x In  CBA and CAD, CBA = CAD, CAB = CDA ACB = DCA (Each 90º) CBA CAD (By AAA rule) AC BC   DC AC

Given that, AB2 = 2AC2  AB2 = AC2 + AC2  AB2 = AC2 + BC2 (As AC = BC) The triangle is satisfying the Pythagoras theorem. Therefore, the given triangle is a right-angled triangle. 58.

AC2 = DC × BC (iii)

Let AD be the altitude in the given equilateral triangle, ΔABC. We know that altitude bisects the opposite side. ∴ BD = DC = a In ADB, ADB = 90° Applying Pythagoras theorem, we obtain AD2 + DB2 = AB2  AD2 + a2 = (2a)2  AD2 + a2 = 4a2  AD2 = 3a2  AD = a 3 In an equilateral triangle, all the altitudes are equal in length. Therefore, the length of each altitude will be 3a .

In DCA and DAD ∠DCA = ∠ DAB (Each 90º) ∠CDA = ∠ ADB (Common angle) DCADAB (AA Similarity criterion) DC DA   DA DB

 AD2 = BD × CD 56.

59. Given that ΔABC is an isosceles triangle. ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

411

AF2 +BD2 +EC2 = AE2 + CD2 +BF2 61.

In ΔAOB, ΔBOC, ΔCOD, ΔAOD, Applying Pythagoras theorem, we obtain AB2 = AO2 + OB2____ (1) BC2 = BO2 + OC2____ (2) CD2 = CO2 + OD2____ (3) AD2 = AO2 + OD2____ (4) Adding all these equations, we obtain AB2 + BC2+ CD2 +AD2 = 2(AO2 + OB2 + OC2 + OD2)   AC  2  BD  2  AC  2  BD  2  = 2       2   2   2   2    

(Diagonals bisect each other)   AC  2  BD  2  2     2   2    

Let OA be the wall and AB be the ladder. Therefore, by Pythagoras theorem, AB2 =OA2 +BO2 (10m)2 = (8m)2 +OB2 100m2 = 64m2 + OB2 OB2 = 36m2 OB = 6m Therefore, the distance of the foot of the ladder from the base of the wall is 6 m. 62.

=(AC)2 + (BD)2 60. Join OA, OB, and OC.

Let OB be the pole and AB be the wire. By Pythagoras theorem, AB2 = OB2 + OA2 (24 m)2 = (18m)2 + OA2 OA2 = (578324)m2 = 252 m2 OA = 252m  6  6  7m  6 7m Therefore, the distance from the base is 6 7m m. (i) Applying Pythagoras theorem in ΔAOF, we obtain OA2 = OF2 = AF2 Similarly, in ΔBOD, OB2 = OD2 + BD2 Similarly, in ΔCOE, OC2 = OE2 +EC2 Adding these equation, OA2 +OB2 + OC2 = OF2 + AF2 + OD2 + BD2 + OE2 + EC2 OA2 +OB2 + OC2 OD2  OE2 OF2 = AF2 + BD2 + EC2 (ii) From the above result, AF2 + BD2 +EC2 = (OA2 OE2) + (OC2 OD2) + (OB2 OF2)

63.

Distance travelled by the plane flying towards 1 1 north in 1 hrs  1000  1  1500km 2 2

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Triangles Solutions

412

Similarly, distance travelled by the plane 1 flying towards west in 1 hrs 2 1 = 1200 1  1800km 2 Let these distances be represented by OA and OB respectively. Applying Pythagoras theorem, 1 Distance between these planes after 1 hrs, 2

Applying Pythagoras theorem in ΔACE, we obtain AC2 = CE2 = AE2 Applying Pythagoras theorem in BCD, we AB = OA 2  OB obtain = BC2 + CD2 = BD2 2 2 Using equation (1) and equation (2), we obtain 1500   1800  km  2250000  3240000 km AC2 +CE2 + BC2 +CD2 = AE2 +BD2 _________ (3) Applying Pythagoras theorem in  CDE, we = obtain 5490000 km  9  610000 km  300 61km DE2 =CD2 +CE2 Applying Pythagoras theorem in CDE, we Therefore, the distance between these planes obtain DE2 = CD2 +CE2 1 will be 300 61km after 1 hrs . Applying Pythagoras theorem in ABC, we 2 obtain 64. AB2 =AC2 +CB2 Putting the values in equation (3), we obtain DE2 + AB2 = AE2 + BD2 66. Applying Pythagoras theorem for ΔACD, we obtain AC2 =AD2 +DC2 AD2 = AC2 DC2 _______ (1) Applying Pythagoras theorem in ΔABD, we obtain Let CD and AB be the poles of height 11 m AB2 =AD2 +DB2 and 6 m. AD2 =AB2 DB2 _______ (2) Therefore, CP = 11 − 6 = 5 m From equation (1) and equation (2), we obtain From the figure, it can be observed that AP = AC2 DC2 = AB2 DB2_______ (3) 12m It is given that 3DC =DB Applying Pythagoras theorem for ΔAPC, we BC 3BC obtain  DC = and DB  4 4 AP2 +PC2 = AC2 2 2 2 Putting these values in equation (3), we (12m) +(5m) = AC obtain AC2 = (144 + 25)m2 = 169 m2 AC = 13m BC2 9BC 2 2 2 AC   AB  Therefore, the distance between their tops is 16 16 13 m. 16AC2 BC2 = 16AB2 9BC2 65. 2 2 2

















16 AB 16AC = 8BC 2AB2 = 2AC2 + BC2

67.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

413 2

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC. BC a  ∴ BE = EC = 2 2 a 3 And, AE = 2 1 Given that, BD = BC 3 a ∴ BD = 3 a a a DE = BE − BD =   2 3 6 Applying Pythagoras theorem in ΔADE, we obtain AD2 = AE2 + DE2

a a 2  AE 2    2 a2 AE 2  a 2  4 2 3a AE 2  4 4AE2 = 3a2 ⇒ 4 × (Square of altitude) = 3 × (Square of one side)

69.

Given that, AB = 6 3 cm, AC = 12 cm, and BC = 6 cm It can be observed that AB2 = 108 AC2 = 144 And, BC2 = 36 AB2 +BC2 = AC2 The given triangle, ΔABC, is satisfying Pythagoras theorem. Therefore, the triangle is a right triangle, rightangled at B. ∴ ∠B = 90° Hence, the correct answer is (C).

2

 a 3   a 2 AD        2  6  3a 2   a 2  =    4   36  2

28a 2 36 7 = AB2 9 ⇒ 9 AD2 = 7 AB2

=

CONCEPTIVE WORKSHEET 1.

In ABC, DE || BC A

68. 4x–3 D 3x–1

8x–3 E 5x–3

B

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC. BC a  ∴ BE = EC = 2 2 Applying Pythagoras theorem in ΔABE, we obtain AB2 = AE2 + BE2

C



AD AB  DB EC 4x  3 8x  7  3x  1 5x  3  4x  3 5x  3   8x  7 3x  1

  

20x 2  27x  9  24x 2  29x  7 4x 2  x  1  0 2x 2  x  1 = 0

 

____________________________________________________________________________________________________ _____________________________________________

Triangles Solutions

  

414

P

 x  1 2x  1  0 Either x  1  0 or 2x  1  0 1 x  1 or x  2 1 1    x   2  rejected, as x = 2 , makes the   line segments negative]

Q

PQ QM  PR MR M is a mid-point of QR Then, QPR  60o PM bisect of QPR and

Hence, x  1 2.

Through C draw CG || DF, meeting AB at a point, say G. B

4.

G

R

M

F

So, QPM  30o Given In PQR, PQ QM  PR MR P

A

C

E

Q

M So, PM bisect of QPR  QPM  MPR

D

In AEF,



AF  AE

 

   

AEF  AFE

[given]

…(i)

[sides opp to equal s as  are equal] In  ACG, EF || CG [ By construction] AE AE  [By BPT] CE  GF CE GF ….(ii) CE = GF [using (1) and (ii)] …..(iii) In BDF, CG || DF [ By construction] BC BG  [ By BPT] CD GF BC BG 1   1 [Adding 1 to both sides] CD GF BC  CD BG  GF  CD FG BD BF  CD GF



BD BF  CD CE

3.

Q  75o R  45o

[using (iii)]

5.

R

New,

Q  75o , R  45o

Then,

QPR  60o



QPM  30o

BAC  180o  110o  70o so, BAD  35o AB = 5cm, and AC = 7cm BC = 3cm 110° A 7cm

5cm B

D 3cm

C

By angle bisect BD AB  DC AC BD 5   BC  BD 7 BD 5   3  BD 7

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

415

5 4 so, DC = BC – BD 



BD 



35 7   1.75cm 4 4



AB BC AC   [corresponding DA AC DC sides of similar triangles are proportional] CA CB  CD CA Draw AL  BC and DM  BC

8.

6.

A A 2.5 cm D

4 cm 3 .5

cm

M

E

B

9.5 cm

L

O

C

3.5 cm

D B

Now, in ALO and DMO, [each equal to 90] ALO  DMO [Vertically AOL  DOM opposite angles] [By AA criterion ALO ~ DMO of similarity] AL AO  [Corresponding sides of DM DO similar triangles are proportional] ….(i) Now, 1 ar  ABC  2 BC  AL AL AO    ar  DBC  1 BC  DM DM DO 2 [using (i)] ar  ABC  AO Hence,  ar  DBC  DO

C

AD 2.5cm 2.5cm 1    AC  4  3.5 cm 7.5cm 3 AE 4cm 4cm 1    AB  2.5  9.5  cm 12cm 3





AD AE 1   ….(i) AC AB 3 In ADE and ACB, AD AE  [from (i)] AC AB and A  A [common]  ADE ~ ACB [By SAS criterion of similarity] AD AE DE    [Corresponding AC AB CB sides of similar triangles are proportional]



1 3.5 cm   [from (i)] 3 x cm  x  10.5 Hence, x = 10.5

9.

and

AB BC AD   [given] PQ QR PM AB 2BD AD    PQ 2QM PM [ AD and PM are the medians of ABC and PQR respectively] We have,

7. A

P A

B



D

C

In ABC and DAC, we have [given] BAC  ADC and [common] C  C [By AA criterion ABC ~ DAC of similarity]

B

D

C

Q

M

AB BD AD   PQ QM PM ABD ~ PQM [By SAS criterion of similarity] 



____________________________________________________________________________________________________ _____________________________________________

R

Triangles Solutions

416



and AP and DQ are the bisectors of A and D respectively.

B  Q ….(i) Now in ABC and PQR, we have AB BC  [given] PQ QR and B  Q [from (1)] Hence, ABC ~ PQR [By SAS criterion of similarity] 10. (i) Given : Two triangles ABC and DEF in which A  D, B  E and C   F . AM and DN are the medians of ABC and DEF respectively. A

B

(ii)

B

D

M

C

E

N

F

BC AM  Then EF DN Proof : Since equiangular triangles are similar, we have ABC ~ DEF AB BC   [Corresponding sides of DE EF similar triangles are proportional ] ….(i) AB 2BM   [ M and N are the DE 2EN midpoints of BC and EF respectively] AB BM   ….(ii) DE EN Now, in ABM and DEN, we have AB BM  [from (ii)] DE EN and B  E [given] [By SAS criterion ABM ~ DEN of similarity] AB AM   …..(iii) DE DN [Corresponding sides of similar triangles are proportional] From equation (i) and equation (iii), we get AB AM  DE DN Hence, the ratio of the corresponding sides of the equiangular triangles is same as the ratio of their corresponding medians.



A

Given : Two triangles AC and DEF in which A  D, B  E and C   F





D

P

C

E

F

Also, since equiangular triangles are similar, ABC ~ DEF AB BC   ….(iii) DE EF [Corresponding sides of similar triangles are proportional] From equations (ii) and (iii), we get BC AP  EF DQ Hence, the ratio of the corresponding sides of equiangular triangles is same as the ratio of the their corresponding angle bisectors. (iii) Given : Two triangles ABC and DEF in which A  D, B  E, C  F and AL  BC, DM  EF A

B



Q

BC AP then  EF DQ Proof : A  D 1 1 A  D  2 2  BAP  EDQ ……(i) [ AP and DQ are bisectors of A ad D respectively] In ABP and DEQ BAP  EDQ [from (i)] and B  E [given] ABP ~ DEQ [By AA criterion of similarity] AB AP  ….(ii) [Corresponding DE DQ sides of similar triangles are proportional]

P

D

C

E

M

F

BC AL  Then EF DM Prove : Since equiangular triangles are similar. ABC ~ DEF

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

417



 

11.

  



12.



  

AB BC  ….(i) DE EF [Corresponding sides of similar triangles are proportional] In ALB and DME, we have ALB  DME [Each equal to 90] and B  E [given] ALB ~ DME [ By AA criterion of similarity] AB AL  ….(ii) DE DM [Corresponding sides of similar triangles are proportional] From equation (i) and (ii), we get BC AL  EF DM Hence, the ratio of the corresponding sides of the equiangular triangles is the same as the ratio of their corresponding altitudes. Given: ABCD is a trapezium in which AB || DC. OA OB  . then OC OD AB || DC [Given] [Alternate  S ] CAB  ACB OAB  OCD …..(i) [same angels] Again, AB || DC [Given] [Alternate  S ] DBA  BDC  OBA  ODC ….(ii) [Same angles] [AA similarity, OAB ~ OCD OA OB  using (i) and (ii)]  OC OD In AEB and CED, we have [Alternate interior EAB  ECD angles] [Vertically AEB  CED opposite angles] [By AA criterion AEB ~ CED of similarity] AE BE  CE DE [Corresponding sides of similar] BE BE DE  ….(i) AE AE CE AED ~ BEX [given] AE DE AD   ….(ii) [corresponding BE CE BC sides of similar triangles are proportional]

 

 

 13.

From equation (i) and equation (ii), we get BE AE  AE BE BE 2  AE 2 BE = AE From equation (i) and equation (ii), we get BE AE  AE BE BE 2  AE 2 BE  AE Putting BE = AE in (ii), we get AD AE  1 BC AE AD = BC Two triangles ABC and EF such that ABC ~ DEF, AP  BC and DQ  EF D A

B

P

C

E

ar  ABC 

Q

F

2

AP ar  DEF  AQ 2 Proof : Since the ratio of the area of the two similar triangles is equal to the ratio of the squares of any two corresponding ar  ABC  AB2 Therefore,  ….(i) ar  DEF  DE 2

Then

 



14.



Now, in APB and DQE, we have APB = DQE [each equal to 90] B = E [corresponding angles of similar triangles are equal] APE ~ DQE [By AA criterion of similarity] AB AP  [corresponding DE DQ angles of similar triangles are proportional] AB2 AP 2  ….(ii) DE 2 DQ 2 From equation, (i) and (ii), we get ar  ABC  AP 2  ar  DEF  AQ 2 Two triangles ABC and PQR such that ABC ~ PQR, Al and PM are the medians of ABC and PQR respectively. ar  ABC  AL2 Then  ar  PQR  PM 2

____________________________________________________________________________________________________ _____________________________________________

Triangles Solutions

418

Proof: ABC ~ PQR [given] AB BC  [corresponding sides of PQ QR similar triangles are proportional] AB 2BL  [QAL and PM are the PQ 2QM medians of DABC and DPQR respectively] AB BL  …(i) PQ QM Now, in ABL and PQM, we have AB BL  [from (i)] PQ QM B  Q [corresponding s of similar triangles] ABL ~ PQM [By SAS criterion of similarity] BL AL  ….(ii) [Corresponding QM PM sides of similar triangles are proportional] AB2 AL2  ….(iii) PQ 2 PM 2 Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. ar  ABC  AB2  ….(iv) ar  PQR  PQ 2 From equation (iii) and equation (iv), we get ar  ABC  AL2  ar  PQR  PM 2







 





15.

Given : ABC ~ DEF and AX and DY are the bisectors of A and D respectively, ar  ABC  AX 2 Then  ar  DEF  DY 2

A  D [corresponding angles of similar triangles] 1 1 A  D.  2 2  BAX  EDY …..(ii) [AX and DY are the bisectors of A and D respectively] In ABX and DEY, we have BAX = EDY [from (ii)] B = E [corresponding angles of similar triangles] ABX ~ DEY [By AA criterion of similarity] AB AX  [corresponding DE DY sides of similar triangles are proportional] AB2 AX 2  ….(iii) DE 2 DY 2 From equation (i) and (iii) we get ar  ABC  AX 2  ar  DEF  DY 2 Two triangles ABC and DEF such that ABC ~ DEF and ar  ABC   ar  DEF

 



16.

A

B

C

E

Y

F

1 ar  DEF  since the ratio of the areas of two similar triangles is equal of the ratio of the squares of their corresponding sides. ar  ABC  AB2 BC2 AC 2    ar  dEF  DE 2 EF2 DF2



From equation (i) and equation (ii), we get AB2 BC2 AC2   1 DE 2 EF2 DF2 AB BC AC   1 DE EF DF AB = DE, BC =EF and AC = DF ABC  DEF [By SSS congruence]

F

Proof : Since the ratio of the area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have ar  ABC  AX 2  ...(i) ar  DEF  DY 2

E

ar  ABC 

A

X

C

Then ABC  DEF Proof : ar  ABC   ar  DEF

D

B

D

   17.

Let the length of the third side be x cm. 2

p  q2  x2

[by Pythagoras Theorem]

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

419

x 2  p2  q 2   p  q  p  q   p  q



[ p – q = 1, given]

x  p  q  2q  1  p  1  1  p  1  q 



Hence, the length of the third is 18.

2q  1 cm.

  19. 20.

A PQR in which QM  PR and PR2–PQ2 = QR2 2

Then QM  PM  MR

PR 2  PQ2  QR 2

Proof :

 PR 2  QR 2  PQ2

2PR 2  2  PQ2  PQ.QN  RQ.QM 

[given ]

Now, in right triangle QMP, we have

PQ2  QM 2  PM 2 ….(ii) [By Pythagoras Theorem] and in right triangle QMR, we have



  21.

BD AB  DC AC

Then

Construction : Draw CE || DA to meet BA produced at E. E

A

B



D

C

Proof : Since CE || DA DAC  ECA [Alternate angles]….(i) and BAD  AEC [corresponding angles] …(ii) BAD  DAC …..(iii) [given] From (i), (ii) and (iii), we get ECA  AEC ….(iv) Thus, in ECA  AEC [from (iv)]

PR 2  PQ 2  QR 2  PQ  PN  PQ   RQ  RM  RQ 



PR2 = PQ2 + QR2 + PQ.PN – PQ2 + RQ.RM – RQ2  PR2 = PQ.PN + RQ.RM.



AE  AC [sides of triangle opposite to the equal angles are equal ] ….…(v) In BCE, DA || CE [By construction]

QR 2  QM 2  MR 2 …(iii) [By Pythagoras Theorem] Adding equations (ii) and (iii), we get PQ2 + QR2 = 2QM2 + PM2 + MR2 PR2 = 2QM2 + PM2 + MR2 [Using (i)] 2 2 2 2 (PM + MR) = 2QM + PM + MR Given : A triangle ABC in which AD, the internal bisector of A meets BC and D.

PM2 + MR2 + 2PM. MR = 2QM2 + PM2 + MR2 PM.MR=QM2 Hence, QM2 = PM  MR. See Formative Q.no.68 In PQR, Q is obtuse angle  PR2 = PQ2 +QR2 +2PQ.QN ..(i) in PQR, Q is obtuse angle  PR2 = PQ2 +QR2 +2PQ.QM ….(ii) Adding equations (i) and (ii), we get

BD BA  [by basic proportionality theorem] DC AE BD BA   [using (5)] DC AC BD AB   [ BA = AB] DC AC 22. In ABC,AD is the bisector of BAC BD AB   [ By the angle-bisector theorem] DC AC BD 10cm    6BD  10BC  10BD BC  BD 6cm  6BD  120 cm  10 BD  BC  12cm  16BD  120 cm  BD  7.5 cm DC  12  7.5  cm  4.5cm. 

Hence, BD = 7.5 cm and DC = 4.5 cm.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

420 2

2  CD  AC 3

PQT is right-angled at Q.

Applying Pythagoras theorem in PQT : PT2 = PQ2 + QT2 1   PT 2  PQ 2   QR  3  

1  PT 2  PQ2  QR 2 9

2 a  CD   3 2

2

 CD 

_____________

2

BD  = BC  + CD ... (2) It is given that point D trisects AC.

SUMM ATIVE WORKSHEET 1.

2

a 3

From equations (1), (2), and (3):

(1)

3a 2 a 2  4 9

QRS is right-angled at Q.

BD 2 

Applying Pythagoras theorem in QRS : RS2 = QS2 + QR2

 BD2 

27a 2  4a 2 36

1   RS   PQ   QR 2 3 

 BD2 

31a 2 36

1  RS2  PQ 2  QR 2 _____________ (2) 9



2

2

PQR is right-angled at Q.

Applying Pythagoras theorem in PQR : PR2 = PQ2 + QR2 _____________ (3) Adding equations (1) and (2): 1

 

 1  

3.  

2 2 2 2 PT2 + RS2   PQ  QR    PQ  QR  9 9

 PT 2  RS2 

10 2 10 PQ  QR 2 9 9

 PT 2  RS2 

10 2 PR 9

[Using equation

2

49  x    cm    26  169

10 2 PR 9



 18  PT 2  RS2   20PR 2

2.

 BC2 

a2 4

3a 2 4

Again applying Pythagoras theorem in BCD : www.betoppers.com

x 7  26cm 13

26  7 cm 13  x  14cm x

Thus, the value of á is 20. The correct answer is C. Applying Pythagoras theorem in ABC : AB2 = BC2 + AC2  a 2  BC 2 

The correct answer is D. Let the length of the shortest side of the smaller triangle be x. It is known that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.  Shortest side of smaller triangle  49    Shortest side of larger triangle  169

(3)] 18  PT 2  RS2   18 

AB2 a2 36    36 : 31 2 2 BD  31a  31    36 

4.

Thus, the length of the shortest side of the smaller triangle is 14 cm. The correct answer is D. It is given that: ABC ~ LMN It is known that the corresponding sides of similar triangles are proportional. AB AC  LM LN

Triangles Solutions

3LM AC   LM LN

421

= (7 cm)2 + (24 cm)2 = 49 cm2 + 576 cm2 = 625 cm2  AC = 25 cm Therefore, from equation (1):

[AB = 3LM]

18cm LN  LN  6cm 3

5.

Thus, the measure of LN is 6 cm. The correct answer is A. It is given that: ABC ~ YZX It is known that if two triangles are similar, then corresponding angles are equal.

 24cm  LC  25cm

8.

A  Y, B  Z, C  X It is given that A  53 and Z  39 B  Z  39

Applying angle sum property of triangles in ABC :

6.

7.

LC BC   BC AC  LC 

BC 2 AC

_________________

(1)

Applying Pythagoras theorem in ABC : AC2  = AB2 + BC2



576 cm  23.04cm 25

Thus, the measure of LC is 23.04 cm. The correct answer is D. By basic proportionality theorem, it is known that if a line is drawn parallel to a side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 

A  B  C  180  53  39  C  180  92  C  180  C  88 Thus, the measure of C is 88°.

The correct answer is C. In the given figure, AE  BC AEB and AEC are right-angled triangles. Applying Pythagoras theorem in AEB an AEC (D) AB2 = AE2 + BE2 ... (1) AC2 = AE2 + EC2 ... (2) Subtracting (1) from (2): AC2 – AB2 = EC2 – BE2  AC2 + BE2 = EC2 + AB2  AB2 + CE2 = AC2 + BE2 The correct answer is C. Comparing ABC an BLC (D) ABC  BLC  90 (Given) ACB  BCL (Common)  ABC ~ BLC (AA similarity criterion) It is known that the corresponding sides of similar triangles are proportional.

2

AB AC  BD EC



AB AC  AD  AB EC



AB 4cm  3AB  AB EC



AB 4cm  2AB EC

1 4cm  2 EC  EC  8cm 

9.

Thus, the length of EC is 8 cm. The correct answer is B. Comparing  C an DEC (D) ABC  DEC  90 (Given) ACB  DCE (Common) C ~ DEC (AA similarity criterion) It is known that the corresponding sides of similar triangles are proportional. 

AC AB  CD ED

AC 5cm  5cm ED

__________________

(1)

[CD = 5cm, AB = 5cm] It is observed that: In DEC , CD = 5 cm and CE = 4 cm Applying Pythagoras theorem in ÄCED: CD2 = EC2 + ED2  52 = 42 + ED2 www.betoppers.com

10th Class Mathematics

422

 25 = 16 + ED2  ED2 = 9  ED = 3

13.

Substituting ED = 3 in equation (1): AC 5cm  5cm 3cm  AC 

25 cm 3

ar  AOD 

 AB     ar  COB   BC 

25 Thus, the measure of AC is cm 3

10.

The correct answer is B. Since the triangles ABC and DEF have the same angles, they are similar. It is known that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.  Rat io of areas of ABC and DEF

11.

12.

2

2

x 1 1     ar  BOC   3  9  ar  BOC   9x

It is observed that ABD and ACD are lying on the same base AD and between same parallels AD and BC.  ar ( ABD ) = ar ( ACD )

2

9 3     9 :16  4  16

Let ar ( AOD ) be x. In AOD an COB (D) AOD  COB (Vertically opposite angles) ADO  CBO (Alternate interior angles) AOD ~ COB (AA similarity criterion) It is known that the ratio of two similar triangles is equal to the square of the ratio of corresponding sides.

 ar  AOD   ar  AOB  ar  AOD   ar  COD 

It is also known that the ratio of the areas of two similar triangles is equal to the square of the ratio of corresponding altitudes. Thus, ratio of the corresponding heights of

It is given that: ar ( AOB ) = 3 ar ( AOD ) = 3x

9 3   3: 4 ABC and DEF  16 4

ar  ABD  ar  AOD   ar  AOB  ar  BCD  ar  COD   ar  BOC 

 ar  AOB   ar  COD   ar  COD   3x

The correct answer is A. Let the height of the triangle be x. According to Pythagoras theorem: (Hypotenuse)2 = (Base)2 + (Altitude)2 2 2 2  (3a + 1)  = (3a – 1)  + x  9a2 + 1 + 6a = 9a2 + 1 – 6a + x2   x2 = 12a



ar  ABD  x  3x  ar  BCD  3x  9x



ar  ABD  4x  ar  BCD  12x

 x  2 3a



Since the length cannot be negative, x  2 3a The correct answer is D. Comparing ABC an ADB (D) ABC  ADB (Given) BAC  BAD (Common) Therefore, by AA similarity criterion,

ar  ABD  1  ar  BCD  3

Thus, the ratio of the areas of ABD and BCD is 1:3. The correct answer is B. Comparing APU an ABC (D)

14.

C ~ ADB

AP AU 1   AB AC 3

It is known that corresponding sides of similar triangles are proportional. The correct answer is B.

and points T and U trisect AC] PAU  BAC (Common) APU ~ ABC (SAS similarity criterion)

www.betoppers.com

[Points P and Q trisect AB

Triangles Solutions

423

It is known that the ratio of two similar triangles is equal to the square of the ratio of corresponding sides. 

ar  APU 

 AP    ar  ABC   AB 



ar  APU  1296cm 2

HOTS WORKSHEET 1.

2

Since XYZ is an equilateral triangle, XY = YZ = ZX ... (1) Applying Pythagoras theorem in XLY : XY3 = XL2 + YL2

2

1 1    3 9

 YZ   XY 2  XL2     2 

1  ar  APU    1296cm 2  144cm2 9 Similarity, it can be shown that: ar ( BQR ) = ar

2

[Altitude of equilateral

triangle bisects the base] YZ  XY  XL  4 2

( CST ) = 144 cm2 It is given that: area of ABC = 1296 cm2

2

2

 XY 2  XL2 

XY 4

2

[Using equation (1)]

2

2 XY  ar  APU   ar  BQR   ar  CST   ar  PQRSTU   1296cm  XY 2   XL2 2 2 2 4  144 cm  +  144  cm  +  144  cm  +  ar

(PQRSTU) = 1296 cm2   ar  PQRSTU   1296cm 2  432cm 2  864cm 2

 3XY 2  4XL2

2

15.

Thus, the area of hexagon PQRSTU is 864 cm . The correct answer is C. Applying Pythagoras theorem in BDC : BD2 = BC2 + DC2 (89)2 = (2x)2 + (x • 1)2  7921  4x 2  x 2  1  2x  5x 2  2x  7920  0  5x 2  200x  198x  7920  0  5x  x  40   198  x  40   0   5x  198 x  40   0

x

198 (or) 0 5

Since x cannot be negative, x = 40 Applying Pythagoras theorem in ADE : AE2 = AD2 + DE2  (82)2 = (2x)2 + DE2  (82)2 = (80)2 + y2  y2 = 6724 – 6400  y2 = 324  y = 18 Thus, the values of x and y are 40 and 18 respectively. The correct answer is B.

3XY 2  XL2 4

2.

Thus, 3XY2  r and let h be its height. Then, 2R  207.24 cm 

P R

A

Radius (r2) of circular end of cylindrical tank 10 = 5m 2 Depth (h2) of cylindrical tank = 2 m Let the tank be filled completely in t minutes. Volume of water filled in tank in t minutes is equal to the volume of water flowed in t minutes from the pipe. Volume of water that flows in t minutes from pipe = Volume of water in tank t × 0.5π = π ×(r2)2 ×h2 t × 0.5 = 52 ×2 t = 100 Therefore, the cylindrical tank will be filled in 100 minutes. 60. Let r be the radius of the upper base (closed) and R the radius of the lower base (open) of the cap. Then, r  4cm and R  10cm. Let l be the slant height of the frustum shaped cap. Then, l = 15 cm Area of the material used = Curved surface area + Area of closed base

B

8 cm

D

O r

C

207.24 cm  33 cm 2  3.14 and 2r  169.56 cm 169.56  r cm  27 cm 2  3.14 2 l 2  h2   R  r 

R

…. (1)

…..(2)

2

 82   33  27   64  62  64  36  100  l = 10 cm ....(3) Whole surface area of the frustum    R 2  r 2   R  r  l  2

2

 3.14  33   27    33  27 10  cm 2    3.14 1089  729  600  cm2

 3.14  2418 cm2  7592.52 cm2 Hence, the whole surface area of the frustum is 7592.52 cm 2 ____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

454

62. Here slant height = l = 4 cm Let the two radii of frustum be r1 and r2 9  2r1  18  r1   3 And 2r2  6  r2   Now   h 2   r1  r2 

64.

2

Curved surface area of frustum    r1  r2   

22 22  9 3   9  3  4   4  7   7 22 7   12   4  48 sq.cm 7 22 

63. Height of cone  h  20cm A

4  2cm 2 2 Radius (r2) of lower base of glass =  1cm 2 Capacity of glass = Volume of frustum of cone 1 = h  r12  r22  r1 r2  3 1 2 2 = h  2   1   2 1 3 1 22 =   14  4  1  2  3 7 308 2 =  102 cm3 3 3 Therefore, the capacity of the glass is 2 102 cm 3 . 3

Radius (r1) of upper base of glass = D

r1

G

E H

B

r2

C

F

 AG  GF  10 cm Vertical angle of cone = 600  BAC  600  ABC  ACB  60



0

ABF  ADG  600 1 Volume of frustum  h r12  r2 2  r1r2 3 10  r1  10cot 600  3 20 and r2  20cot 600  3 Volume of frustum 1  100 400 10 20     10      3 3 3 3  3 Let the length of wire be = H cm 1 1 1 Result of wire    cm 16 2 32 Now volume of wire = Volume of frustum 1  r 2 H  h r12  r2 2  r1r2 3 









65.



2

1  1   100 400 200      H   10     3 3 3   32   3 32  32  0  700   H    796444.44cm  7964.44 m 3  3 

Perimeter of upper circular end of frustum = 18 cm 2πr1 =18 9 r1  

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

455

Perimeter of lower end of frustum = 6 cm 2πr2 = 6 3 r2   Slant height (l) of frustum = 4 cm CSA of frustum = π (r1 + r2) l 9      4   =12×4 =48 cm2 Therefore, the curved surface area of the frustum is 48 cm2.

Slant height (l) of frustum =

 20  8  16

=

12   16

2

2

2

 h2

2

 144  256

= 20 cm Capacity of container = Volume of frustum 1 = h  r12  r22  r1 r2  3 1 2 2 =  3.14  16   20   8    20  8     3 1 =  3.14  16   400  64  160  3 1 =  3.14  16  624 3 =10449.92 cm3 =10.45 litres. Cost of 1 litre milk = Rs 20 Cost of 10.45 litre milk = 10.45 × 20 = Rs 209 Area of metal sheet used to make the container =   r1  r2  l  r22 = π (20 + 8) 20 + π (8)2 = 560 π + 64 π = 624 π cm2 Cost of 100 cm2 metal sheet = Rs 8 624  3.14  8 Cost of 624  cm2 metal sheet = 100 =153.75 Therefore, the cost of the milk which can completely fill the container is Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.

66.

Radius (r2) at upper circular end = 4 cm Radius (r1) at lower circular end = 10 cm Slant height (l) of frustum = 15 cm Area of material used for making the fez = CSA of frustum + Area of upper circular end = (r1 + r2 ) l + r22 = π (10 + 4) 15 + π (4)2 = π (14) 15 + 16 π 226  22 = 210  16   7 2 2 = 710 cm 7 Therefore, the area of material used for 2 making it is 710 cm 2 . 7

2

=

 r1  r2 

68.

67.

Radius (r1) of upper end of container = 20 cm Radius (r2) of lower end of container = 8 cm Height (h) of container = 16 cm

In ΔAEG, EG  tan 30 AG

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

EG=

10

cm =

3 In ΔABD, BD  tan 30 AD

BD 

20 3



456

10 3 3

CONCEPTIVE WORKSHEET 1.

20 3 cm 3

Radius (r1) of upper end of frustum =

10 3 3

cm Radius (r2) of lower end of container 20 3 cm 3 Height (h) of container = 10 cm 1 Volume of frustum h  r12  r22  r1 r2  3 =

=



2

2  3.5  r h      3.25  cm  3.7cm(approx)  2  2





10  100 400 200     3  3 3 3 

10 22 700 22000 3    cm 3 7 3 9 1 1 1 Radius (r) of wire =   cm 16 2 32 Let the length of wire be l. Volume of wire = Area of cross-section × Length = (πr2) (l)

22 3.5 11   3.5  3.7  cm2    3.5  3.7  cm 2 7 2 2  39.6cm 2 (approx) You may note that ‘total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere. The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2. Note that the part of the cube where the hemisphere is attached is not included in the surface area. So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – r2 + 2 r2 = (150 + r2) cm2  22 4.2 4.2  2  150cm 2      cm 2 2   7 = (150 + 13.86) cm2 = 163.86 cm2 Denote radius of cone by r, slant height of cone by l, height of cone by h, radius of cylinder by r’ and height of cylinder by h’. Then r = 2.5 cm, h = 6 cm, r’ = 1.5 cm, h' = 26 – 6 = 20 cm and 

=

2

2

 1  =    l  32  Volume of frustum = Volume of wire 2

22000 22  1     l 9 7  32  7000  1024  l 9 l =796444.44 cm =7964.44 metres

2

Therefore, CSA of cone = rl =  22 3.5   3.7  cm 2   7 3   This gives the surface area of the top as 22 3.5 3.5  2  22 3.5     3.5  cm 2 2   cm    7 2 2  2   7 

2 2  10 3 20 3  1 10 3   20 3       10        3   3   3 3 3  

=

TSA of the toy = CSA of hemisphere + CSA of cone Now, the curved surface area of the 1 hemisphere=  4r 2   2r 2 2 22 3.5 3.5  2   2    cm 7 2 2   Also, the height of the cone = height of the top – height (radius) of the hemispherical part 3.5   5   cm  3.25cm 2   So, the slant height of the cone (l )=

3.

l  r 2  h 2  2.52  62 cm  6.5cm ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

457

Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted. So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder = rl + r2 – (r’)2 =  [(2.5 × 6.5) + (2.5)2 – (1.5)2] cm2 =  [20.25] cm2 = 3.14 × 20.25 cm2 = 63.585 cm2 Now, the area to be painted yellow = CSA of the cylinder + area of one base of the cylinder = 2r’h’ +  (r’)2 = r’ (2h’ + r’) = (3.14 × 1.5) (2 × 20 + 1.5) cm2 = 4.71 × 41.5 cm2 = 195.465 cm2 4. Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere = 2rh + 2r2 = 2 r (h + r) 22  2   30 145  30  cm2 7 = 33000 cm2 = 3.3 m2 5. Rs. 55687.50 6. Rs. 855.02 7. 103.62 cm2 8a. Rs. 2068. 8b. The volume of air inside the shed (when there are no people or machinery) is given by the volume of air inside the cuboid and inside the half cylinder, taken together. Now, the length, breadth and height of the cuboid are 15 m, 7 m and 8 m, respectively. Also, the diameter of the half cylinder is 7 m and its height is 15 m. So, the required volume = volume of the 1 cuboid + volume of the cylinder 2 1 22 7 7   3 15  7  8  2  7  2  2  15  m   3  1128.75m Next, the total space occupied by the machinery = 300 m3 And the total space occupied by the workers = 20 × 0.08 m3 = 1.6 m3 Therefore, the volume of the air, when there are machinery and workers = 1128.75 – (300.00 + 1.60) = 827.15 m3

9.

10.

11. 12. 13. 14. 15. 16. 17 18. 19.

Since the inner diameter of the glass = 5 cm and height = 10 cm, the apparent capacity of the glass = r2h = 3.14 × 2.5 × 2.5 × 10 cm3 = 196.25 cm3 But the actual capacity of the glass is less by the volume of the hemisphere at the base of the glass. i.e., it is less by 2 3 2 r   3.14  2.5  2.5  2.5cm3 3 3  32.71cm3 So, the actual capacity of the glass = apparent capacity of glass – volume of the hemisphere = (196.25 – 32.71) cm3 = 163.54 cm3 Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere. The radius BO of the hemisphere (as well as 1 of the cone) = × 4 cm = 2 cm. 2 2 1 So, volume of the toy = r 3  r 2 h 3 3 2 1 3 3      3.14   2    3.14   2   2  cm 3 3 3    25.12cm3 Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm So, the volume required = volume of the right circular cylinder – volume of the toy = (3.14 × 22 × 4 – 25.12) cm3 = 25.12 cm3 Hence, the required difference of the two volumes = 25.12 cm3. 6160 m3 0.95 m3 102.24 kg 8316 cm3 500  m3, Rs. 55,000 10 3 10 m 21 7 2 56.15 m3, 7 m 12 1980 cm3. 1 Volume of cone =   6  6  24cm3 3 If r is the radius of the sphere, then its volume 4 is r 3 3

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

Since, the volume of clay in the form of the cone and the sphere remains the same, we have 4 3 1 r    6  6  24 3 3 i.e., r3 = 3 × 3 × 24 = 33 × 23 So, r = 3 × 2 = 6 Therefore, the radius of the sphere is 6 cm. 20. The volume of water in the overhead tank equals the volume of the water removed from the sump. Now, the volume of water in the overhead tank (cylinder) = r2h = 3.14 × 0.6 × 0.6 × 0.95 m3 The volume of water in the sump when full = l × b ×h = 1.57 × 1.44 × 0.95 m3 The volume of water left in the sump after filling the tank = [(1.57 × 1.44 × 0.95) – (3.14 × 0.6 × 0.6 × 0.95)] m3 = (1.57 × 0.6 × 0.6 × 0.95 × 2) m3 So, the height of the water left in the sump = volume of water left in thesump lb 1.57  0.6  0.6  0.95  2 = m = 0.475 m = 47.5 1.57  1.44 cm capcity of tan k Also, = capacity of sump

3.14  0.6  0.6  0.95 1  1.57  1.44  0.95 2 Therefore, the capacity of the tank is half the capacity of the sump. 21. The volume of the rod =

458

22. Radius of the hemispherical tank =

3 m 2

Volume of the tank = 3

23. 24. 25. 26. 27. 28. 29. 30. 31.

2 22  3  3 99 3    m  m 3 7 2 14 So, the volume of the water to be emptied = 1 99 3 99  m   1000litres 2 14 28 99000  litres 28 25 Since, litres of water is emptied in 1 7 99000 second, litres of water will be emptied 28 99000 7 in  seconds, i.e., in 16.5 minutes. 28 25 126 672 504 60 512 36 m h units 3m2 539 cm3. 3 Let the radius of the base of the given cone (say ABC) be R. Its height AG = H (say) = 30 cm [Given] …..(1) A

2

1     8cm3  2cm3  2 The length of the new wire of the same volume = 18 m = 1800 cm× If r is the radius (in cm) of cross-section of the wire, its volume =  × r2 × 1800 cm3 Therefore,  × r2 × 1800 = 2 1 1 i.e., r2= r= 900 30 So, the diameter of the cross section, i.e., the 1 thickness of the wire is cm, 15 i.e., 0.67mm (approx.)

F

D

H

x B

E

G

C

Let the height above the base where the section (say DEF) is made be x cm., i.e., FG = x cm Let the radius of the smaller cone  h  AF  AG  FG   30  x  cm .....(2) In ∆AFE and ∆AGC, we have A is common and AFE  AGC  right angle  AFE AGC [by AA criterion of similarity]

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

459

AF FE  AG GC 30  x r   [Using (2)] ……(3) H R Given, Volume of smaller cone 1  (Volume of the given cone) 27 1 2 1 1   r h   R 2 H  3 27  3  1  r2h  R2H 27 r2 1 H     2 R 27  h  2  30  x  1  H      H2 27   30  x   1 3 3   30  x    30  27 3 3  30   30  x       3 

We have : r1 = 28 cm, r2 = 7 cm and the height of frustum (h) = 45 cm. Also, h1 = 45 + h2 _________ (1) We first need to determine the respective heights h1 and h2 of the cone OAB and OCD. Since the triangles OPB and OQD are similar h 28 4 _________ we have 1   (2) h2 7 1 From (1) and (2), we get h2 = 15 and h1 = 60. Now, the volume of the frustum = volume of the cone OAB – volume of the cone OCD 1 22 2 2  1 22  3  3  7   28   60  3  7   7   15   cm   =48510 cm3 The respective slant height l2 and l1 of the cones OCD and OAB are given by



3

3

  30  x   10   30  x  10  x  20 . Thus, the height of the section from the base is 20 cm. 32. Let R and r be the radii of the circular ends of the frustum such that R > r and let l be its slant height.

4 cm

R

r

Then, 2R  18 cm  R 

9 cm 

3 cm and l = 4cm .  Curved surface area    R  r  l

2R  6 cm  r 

9 3      4 cm 2  48 cm 2 . Hence, the   curved surface area of the frustum is 48 cm2. 33. The frustum can be viewed as a difference of two right circular cones OAB and OCD. Let the height (in cm) of the cone OAB be h1 and its slant height l1, i.e., OP = h1 and OA = OB = l1. Let h2 be the height of cone OCD and l2 its slant height.

7

l2  l1 

 28 

2

2

2

 15   16.55cm(approx) 2

  60   4

7

2

2

 15   4  16.55cm

=66.20 cm Thus, the curved surface area of the frustum = r1l1 – r2l2 22 22  28  66.20   7 16.55  5461.5cm2 7 7 Now, the total surface area of the frustum = the curved surface area + r12  r22 22 22 2 2 = 5461.5cm2   28 cm 2   7  cm 2 7 7 = 5461.5 cm2 + 2464 cm2 + 154 cm2 = 8079.5 cm2 1 (i) Volume of the frustum= h  r12  r22  r1r2  3 1 22 2 2   45  28    7    28  7   cm2   3 7 (ii) we have 2

l  h 2   r1  r2   2

 45

2

2

  28  7  cm

2

 3 15    7   49.65cm

So, the curved surface area of the frustum 22   r1  r2  l   28  7  49.65  5461.5cm2 7 (iii) Total curved surface area of the frustum   r1  r2  l  r12  r22 22 22 2  2 2   5461.5  7  28   7  7   cm   2 = 8079.5cm

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

34. Since the mould is in the shape of a frustum of a cone, the quantity (volume) of molasses  that can be poured into it h  r12  r22  r1r2  3 where r1 is the radius of the larger base and r2 is the radius of the smaller base.

460

SUMMATIVE WORKSHEET 1.

 35  2  30  2  35 30   1 22   14           cm 3 3 7  2   2   2 2  

=11641.7 cm3 It is given that 1 cm3 of molasses has mass 1.2g. So, the mass of the molasses that can be poured into each mould = (11641.7 × 1.2) g = 13970.04 g = 13.97 kg = 14 kg (approx.) 35. The total height of the bucket = 40 cm, which includes the height of the base. So, the height of the frustum of the cone = (40 – 6) cm = 34 cm. Therefore, the slant height of the frustum, l = h 2   r1  r2 

 40000    r1  r2  l .

2.

2

where r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm. So,

2

34 2   22.5  12.5  cm

342  102  35.44cm The area of metallic sheet used = curved surface area of frustum of cone + area of circular base + curved surface area of cylinder = [× 35.44 (22.5 + 12.5) +  × (12.5)2 + 2 × 12.5 × 6] cm2 22 = (1240.4 + 156.25 + 150) cm2 7 = 4860.9 cm2 Now, the volume of water that the bucket can hold (also, known as the capacity of the bucket)  h   r12  r22  r1r2  = 3 22 34  2 2    22.5   12.5   22.5  12.5 cm3   7 3 22 34   943.75  33615.48cm3 7 3 = 33.62 litres (approx.)

Let l be the slant height of container and let h be its height. Let r1 = 1 m = 100 cm; r2 = 40 cm The cost of polishing 100 cm2 of the container is Re 1. Rs 440 is the cost of polishing (440 × 100) cm2 = 44000 cm2 of the container. ∴ Total surface area of the container = 44000 cm2 The total surface area of the frustum of a cone is given by   r1  r2  l 44000 =(100+40)l 22 44000 =  140  l 7  l = 100 The given bucket is in the shape of the frustum of a cone. Radius of upper end, r1 = 40 cm Radius of lower end, r2 = 30 cm Height of bucket, h = 45 cm It is known that the volume of the frustum of a 1 cone is given by h  r12  r22  r1r2  . 3 ∴Capacity of the bucket 1 22 2 2   45  40    30   40  30 cm3 3 7 330 = 1600  900  1200 cm3 7 330  3700 3 = cm 7 =174428.57 cm3 =174.43L (1L = 1000 cm3] Thus, the capacity of the bucket is approximately 174.43 L. The correct answer is B.





3.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

461

From the similar triangles COA and CPB CO OA 4   CB PB 3  3 CO = 4 CB  3CB + 3 ×12 = 4 CB  CB = 36 ⇒ The volume of the glass tumbler = 1   42  48  32  36  3 =464.95 cm3 =464.95 mL The correct answer is B. 4.

2 =  21Hr22 3 =14H r22

Also, volume of the cylindrical part = r 2 h  r22 H

5.

Letr1 and r2be the radii of the inlet and the outlet respectively. As the ratio between the depths of frustum and cylindrical part is 2:1, their depths can be assumed to be 2Hand H respectively. The ratio of the areas of cross-section of the inlet and outlet is 16:1. r 2 16  12  r2 1  r1 2  16    r2  1 r 4  1  r2 1  r1 = 4r2 ....(1)

1 Volume of frustum  h r12  r22  r1r2  3 1 2 =   2H  4r2   r22  4r1r2  from (1)   3 2 = H 16r22  r22  4r1r2  3

14r22 H  14 :1 r22 H Thus, the ratio of the capacities of the frustum and the cylindrical part is 14: 1. The correct answer is C. It is given that the diameters of the mouth and the base of the glass are 8 cm and 4 cm respectively. Height of the glass = 14 cm The glass is in the shape of a frustum of a cone. It is known that volume of frustum of cone 1 h  r12  r22  r1r2  , where r1 and r2 are the 3 radii of the ends of the frustum, and h is its height. r1 = 4 cm, r2 = 2 cm, h = 14 cm Therefore, volume of water contained in one glass  1 22      14   4 2  2 2  2  4   cm 2 3 7  44 44 =  16  4  8  cm3 =  28cm3 3 3 Therefore, volume of water contained in 12 44 such glasses 12   28cm3  4928cm3 3 Thus, water consumed by Rohan in a day 4928 is L = 4.928 L. 1000 The correct answer is C. As the radii of the top and bottom of the buckets are different, the buckets are frustum-shaped. For the red bucket The radius of the bottom, the radius of the mouth, and the height of the red bucket are in the ratio 2: 3: 6. Let the radius (r1) of the bottom of the bucket be 2x. Thus, radius (r2) of the top of the bucket = 3x Thus, height (h) of the bucket = 6x Thus, capacity of the red bucket= 1   r12  r22  r1r2  h 3

∴ Required ratio =

6.

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

1 2 2 =   2x    3x    2x  3x  6x 3  = 4x 2  9x 2  6x 2  6x 3 =38x3 For the green bucket, The radius of the bottom, the radius of the mouth, and the height of the green bucket are in the ratio 2:6:9. Let the radius (r1) of the bottom of the bucket be 2y. Thus, radius (r2) of the top of the bucket = 6y Thus, height (h) of the bucket = 9y Thus, capacity of the green bucket 1   r12  r22  r1r2  h 3 1 2 2 =   2y    6y    2y  6y   9y   3  = 4y 2  36y2  12y 2  9y 3 =156 y3 3 Radius of the top of the red bucket =  2 radius of the top of green bucket 3  3x =  6y 2  x = 3y Thus, capacity of the red bucket = 38πx3 = 38π(3y)3 = 38π × 27y3 = 1026πy3 The ratio of the volumes of water that the red and green buckets can hold equals the ratio of the capacity of the buckets. Thus, required ratio capacity red bucket 1026y3 171    171: 26 capacity of green bucket 156y3 26



7.

462



Thus, the ratio of the volumes of water that the red and green buckets can hold is 171:26. The correct answer is A. Let us analyse the figure of the hat on the mannequin.

Let r1 and r2 be the radii of the circular bases of the hat. Let l and h be its slant height and vertical height respectively.

Then, r1 

42 28  21cm,r2   14cm and 2 2

h = 24 cm  Slant height(l) 

=

 21  14 

2

 r1  r2 

2

 h2

 24 2

= 7 2  242  25cm The area that is to be pasted with glossy paper is the curved surface area of the frustum and the area of the upper circular base. ∴ Required area =   r1  r2  l  r22 2 =    21  14  25   14   cm2   22  22  =   35  25   14  14  cm 2 7 7  2 = 110  25  616 cm

=  2750  616 cm2  3366cm2 Thus, amount required to paste glossy paper = 2 Rs  3366 = Rs 612 11 8.

Let the radius of the hemispherical solid be r and let the height of the solid frustum be h. It is given that the height (radius) of the hemispherical solid is half the height of the solid frustum. h r  2 The total height of the solid (18 m) is the sum of the heights of the hemisphere and the frustum. h   h  18m 2 3h   18m 2  h = 12 m

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

463

12 m  6m 2 Volume of the hemisphere= 2 3 2 3 r    6m   144m3 3 3 1 Volume of the frustum   r12  r22  r1r2  h , 3 where r1 and r2 are the radii of the ends of the frustum Thus, volume of the frustum  1  14 2        62  7  6   12 m3  3  2    =  4  49  36  42   m3 r=

=  4  127  m2  508m3

144 36   36 :127 508 127 Thus, the ratio of the volumes of the hemispherical solid and the solid frustum is 36:127. The volume of a frustum with radii r1, r2, and 1 height h is   r12  r22  r1r2  h . 3   152  52  15  5 h 3  325h 3 ___ =   225  25  75 h  cm (1) 3 3 Thus, required ratio

9.

Let r1 and r2 be the radii of the ends of the smaller frustum. It is given that the height of the smaller frustum is half of that of the bigger h one i.e., height of the smaller frustum is . 2 ∴Volume of the smaller 1 h frustum  32  42  3  4 3 2 1 h =  9  16  12 3 2 37h 3 = cm ____ (2) 6 From (1) and (2), the required ratio is obtained 325h 3  650 = 650:37 as 37h 37 6 Thus, the ratio of the volumes of the given frustums is 650: 37.

Height of the frustum of the cone = 315 cm − 35 cm − 70 cm = 210 cm The given solid consists of a hemisphere, the frustum of the cone, and the cone. 2 Volume of hemisphere = r 3 3 3  2 22 =     70   cm 3 3 7  2  =   22  49000  cm3 3  2156000 3 = cm 3 Volume of the frustum of the 1 cone   r12  r22  r1r2  h 3 1 22 2 2 =   210 120    70   8400 cm3   3 7 = 220 (14400 + 4900 + 8400) cm3 = 220 (27700) cm3 = 6094000 cm3 1 Volume of cone = r 2 h 3 2  1 22  =    120   35  cm 3 3 7   22  =   14400  5  cm3 3  = 110  4800  cm3 =528000 cm3 ∴ Volume of given  2156000  solid   6094000  528000  cm 3 3    2156  =  6094  528  L  3  2156   =  6622  L 3   22022 = L 3 Thus, the volume of the given solid is  22022   L .  3  11. The volume of the frustum of a cone is given 1 by   r12  r22  r1r2  h . 3 1 ∴ Volume of the tub   r12  r22  r1r2  h 3

10. According to the given figure, ____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

464

1 22 1672000cm 3   3 7  2 2  h  40    90    40  90   cm 3   22  1672000cm3  h 1600  8100  3600 7 22  1672000cm3  h 13300 7  h = 120 Slant height (l) of the frustum of cone =

 r1  r2 

2

 h2

=

 90  40 

=

 50 

2

2

 120 

2

2

 120   130cm

Thus, the slant height of the tub is 130 cm. 12. D 13. A 14. It is given that the radius of the upper base, the radius of the lower base, and the height of the frustum are in the ratio 2:5:4. Let the radius (r1) of the upper baseof the frustum be 2x. Thus, radius (r2) of the lower base of the frustum = 5x Thus, height (h)of the frustum = 4x Slant height (l) of the given frustum

 r2  r1  =

 3x 

2

2

 h2 

 5x  2x 

2

  4x 

2

2

  4x   9x 2  16x 2  25x 2  5x

Curved surface area (A) of the frustum 990cm2    r1  r2  l

22   2x  5x  5x 7 22  990=  7x  5x 7 990 = 110x2  x2 = 9 x=3 Thus, height of the given frustum = 4x = 4 × 3 cm = 12 cm The correct answer is C. 15. It can be easily seen that the total surface area of the vase that is to be painted is the sum of C.S.A of the frustum of cone and the area of the circular base. C.S.A of frustum of cone = π(r1 + r2)l, where r1 and r2 are the radii of the lower end and the upper end of the frustum respectively, and l is its slant height. r1 = 7 cm, and r2 = 21 cm  990 =

h 2   r2  r1 

l= 

 35cm 

2

2

  21cm  7cm 

2

= 1225  142 cm2  1225  196cm2  l = 1421cm  7 29cm Therefore, C.S.A. of the frustum of cone =   r1  r2  l

22   21  7   7 29cm 2 7 22   28  7 29cm2 7 =22×28× 29cm 2 =616 29cm 2 Area of the circular base 22  r12   7  7cm 2  154cm2 7 Thus, total surface area of the vase that is to be painted = 616 29  154 cm 2  154 1  4 29 cm 2 









The correct answer is A. 16. Height (h) of the frustum = 8 cm Diameter of the lower base = 10 cm more than the height = (10 + 8) cm = 18 cm Thus, radius (r2) of the lower base of the 18 frustum cm  9cm 2 Let the radius of the upper base of the circle be r1 cm. It is also given that the diameter of the lower base is six times the radius of the upper base. ∴ 6r1 = 18 cm ⇒ r1 = 3 cm Thus, radius (r1) of the upper base of the frustum = 3 cm Slant height (l) of the frustum 

 r2  r 1  

2

 h2

 9cm  3cm 

2

  8cm 

2

 62  82 cm  100cm  10cm Thus, curved surface area of the frustum   r1  r2  l =   3cm  9cm  l0cm  120cm3 2

Area of the upper base r12    3cm   9cm2 Area of the lower base 2

r22    9cm   81cm2 Thus, total surface area of the frustum

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465

= C.S.A. of the given frustum + Area of the upper base + Area of the lower base = (120π + 81π + 9π) cm2 = 210π cm2 22     210   cm   660cm 2 7   Hence, the total surface area of the given frustum is 660 cm2. The correct answer is D. 17. The radius and volume of the given cone is 15 cm and 2700π respectively. 1 ∴Volume of a cone r 2 h = 2700π 3 1 2   15  h  2700 3 2700  3 h =  36cm 15  15 The height of the frustum is one-third of the height of the original cone. 1 ⇒ Height of the frustum, DG =  36 cm = 12 3 cm

Slant height =

15  10 

2

 12 2  25  144  169cm  13cm

Thus, total surface area of the frustum =   r1  r2  l  r12  r22 2 2 =   15  10   13   15    10   cm2   2 = 325  225  100 cm

= 325  225  100 cm 2 =650 cm2 The correct answer is C. 18. Let l be the slant height of container and let h be its height. Let r1 = 1 m = 100 cm; r2 = 40 cm The cost of polishing 100 cm2 of the container is Re 1. Rs 440 is the cost of polishing (440 × 100) cm2 = 44000 cm2 of the container. ∴ Total surface area of the container = 44000 cm2 The total surface area of the frustum of a cone is given by   r1  r 2  l . 44000 = (r1 + r2 )l 22 44000 = (100+40)l 7  l =100 Slant height (l) =

∴AG = AD – DG = (36 – 12) cm = 24 cm ΔAGF and ΔADC are similar. AG GF   AD DC 24 GF   36 15 24 GF =  15cm  10cm 36

 r1  r2 

2

 h2

⇒ (100 cm)2 = (100 cm − 40 cm)2 + h2 ⇒ 10000 cm2 = 3600 cm2 + h2 ⇒ h2 = 6400 cm2 ∴ h = 80 cm Thus, the height of the container is 80 cm. The correct answer is C. 19.

Let r1 = 21 cm, r2 = 14 cm, and h = 51 cm. The area of the walls of the pan is equal to the curved surface area of the frustum of the cone. Slant height of the pan, l  h 2   r1  r2  Slant height (l) of the frustum is given by

 r1  r2 

2

 h12 .

=





2

51cm   21cm  14cm 

2

2

= 51cm2  49cm2 =10 cm

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Surface Areas and Volumes Solutions

Curved surface area of the pan =   r1  r2  l = π (21 cm + 14 cm) × 10 cm 22 =  35cm  10cm 7 = (22 × 5 × 10) cm2 = 1100 cm2 Thus, the area of the foil required to be pasted on the walls of the pan is 1100 cm2. The correct answer is D. 20. It can be seen that the given container consists of a cone and a cylinder. ∴ Volume of one ice cream in the container = 1 r 2 h1  r 2 h 2 3 = 2 2  22  3  1 22  3  7   2    7   cm   3  7   2   3 cm     88   =  88   cm 3 7    616  88  3 =  cm 7   704 3 = cm 7 Volume of ice cream to be frozen for 704 3 completing the order = 700  cm 7 = 70400 cm3 Therefore, quantity of ice cream to be frozen = 70400 × 0.9167 ≈ 64.5 kg The correct answer is D. 21. Radius of each sphere = 6 cm It is known that the volume of a sphere of 4 radius r is given by r 3 . 3 Volume of each sphere= 4 4 3    6cm     6  6  6cm3  288cm3 3 3

466

22. Let r be the radius of the solid sphere so formed. It is known that the volume of solids remains the same in case of conversion of solids from one shape into another. ∴Volume of three smaller spheres = Volume of big sphere 4 4 3 3 3 4 3  3  1  3   2   3   3  cm =   4 3  3 r  volume of sphere    Radius   3 3   4 4   1  8  27  cm3  r 3 3 3  r3 = 36 cm3 1

 r =  36  3 cm ∴Surface area of the sphere = 4πr2 2

 1 =   36 3  cm 2   2

=   36  3 cm 2 1

=   362  3 cm 2 1

= 24   6  3 cm 2 Thus, the surface area of the sphere so formed 1

is 24   6  3 cm 2 . The correct answer is D. 23. Radius of the copper sphere, r = 6 cm ∴ Volume of the copper 4 4 3 sphere r 3    6cm   288cm3 3 3 Radius of first cylinder, R1 = 2 cm Height of first cylinder, H1 = 28 cm ∴ Volume of first cylinder = R12 H1 = π (2 cm)2 (28 cm) = 112π cm3 Radius of the second cylinder, R2 = 4 cm Let the height of the second cylinder be H2. ∴ Volume of the second 2

Volume of five spheres = 5 × 288 π cm3 = 1440 π cm3 It is known that the volume remains same in conversion from one shape into another. ∴Volume of cylindrical wire = Volume of five spheres ⇒ π (1 cm)2 × h = 1440 π cm3 [Volume of cylinder = π(Radius)2 × Height] ⇒ h = 1440 cm ⇒ h = 14.4 m Thus, the length of the wire is 14.4 m. The correct answer is D.

cylinder R 22 H 2    4cm  H 2  16H 2 cm3 It is given that the copper sphere is melted to form two cylindrical shapes. Therefore, the volume of the copper sphere will be equal to the sum of the volumes of the two cylinders. 288 = 112 + 16H2  16 H2 =288  112 176  16 H2 = 176  H2 =  11 16 Thus, the height of the second cylinder is 11 cm. The correct answer is A.

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10th Class Mathematics

467

24. Let the radius of smaller metallic ball be r while that of bigger ball be R. It is given that 125 smaller metallic balls are melted to form 8 bigger balls. Therefore, the volume of 125 smaller metallic balls will be equal to the volume of the 8 bigger balls. It is known that the volume of a sphere 4 3 =   Radius  3 4 Volume of 125 smaller balls 125  r 3 3 4 Volume of 8 bigger balls = 8  R 3 3 4 3 4  125  r = 8  R 3 3 3 125  R  r3 8 5 R  r 2 It is also known that the surface area of a sphere = 4π(Radius)2 ∴ Surface area of a bigger ball = 4πR2 = 2

 5r  4   2 Surface area of a smaller ball = 4πr2 Surface area of a bigger ball  Surface area of a smaller ball 2

 5r  4   25 2   2   25 : 4 4r 4 Thus, the required ratio is 25:4. The correct answer is D. 25. Let R and H be the respective radius and height of the cylindrical candle. Let r and h be the respective radius and height of the conical candle. R H Then, by given condition, r  and h  2 8 Volume of the cylindrical candle = πR2H Volume of the conical candle 2

1 2 1 R H 1 r h      R 2 H 3 3  2  8 96 Let n number of conical candles be formed out of 16 cylindrical candles. Then, volume of n conical candles = volume of 16 cylindrical candles 1  n × R 2 H  16R 2 H 96

⇒ n = 16 × 96 = 1536 Hence, 1536 conical candles can be made out of 16 cylindrical candles. The correct answer is D. 26. The radius (r) and height (h) of the rod are 4 cm and 14 cm respectively. As each rod is cylindrical, the volume is given by πr2h. 22 Volume  42  14  22  4  4  2  704cm3 7 Therefore, volume of 10 rods = 704 × 10 = 7040 cm3 Let the length of the resulting wire be l cm. The area of cross-section of the wire is given as 2 cm2. Thus, volume of the wire = Area of crosssection × length = l × 2 cm3 As 10 cylindrical rods are melted to form the wire, Volume of 10 rods = Volume of the resulting wire ⇒ 7040 = 2l ⇒ l = 3520 cm = 35.2 m Thus, the length of the resulting wire is 35.2 m. The correct answer is B. 27. Let the radius of each small metallic solid ball be r and that of the bigger metallic solid ball be R. 3 It is given that R = 1.5 r = r __ (1) 2 Let n be the number of bigger balls formed. Then, volume of n bigger balls = volume of 729 smaller balls 4 4  n  R 3  729  r 3 3 3 3

4 3  4  n   r   r 3  729 from (1) 3 2  3 22   n  r 3  r 3  729 8 22   n  729 8 8  729  n  216 27 Thus, 216 bigger solid balls are formed. The correct answer is C. 28. Cross-sectional area of the metallic rod = 25π cm2 Length of the rod = 144 cm The metallic rod is cylindrical in shape. ∴Volume of the metallic rod = Cross-sectional area × Length

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

= 25π × 144 cm3 = 3600 π cm3 Let the radius of each sphere be r cm. The rod is melted to form 100 solid spheres. ∴Volume of the rod = Volume of 100 solid spheres 4  3600  = 100 ×  r 3 3 3600  3 3  r 100  4  r3 =27  r = 3 cm Therefore, the radius of each solid sphere is 3 cm. The correct answer is B. 29. Volume of toy = 8624 cm3 Radius of hemisphere = Height of hemispherical portion = r ∴Height of conical part = 2r − r = r Radius of cone = Radius of hemisphere = r Now, Volume of conical part + Volume of hemispherical part = 8624 cm3 1 2  r 2 h  r 3  8624cm 3 3 1 2 2  r  r   r 3  8624cm  3 3 3  r = 8624 cm3 8624  7 3  r3 = cm 22  r3 = 2744 cm3  r = 14 cm Now, original surface area of toy = C.S.A of conical part + C.S.A of hemispherical part = πrl + 2πr2 Surface area of toy after breaking = T.S.A of cone + T.S.A of hemisphere = πr (l + r) + 3πr2 = πrl + 4πr2 Increase in area = (4πr2 + πrl) − (πrl + 2πr2) = 22 2πr2  2   14  14  1232cm 2 7 Thus, the surface area of the toy increased by 1232 cm2 after it was broken. The correct answer is A. 30. Radius of hemispherical metallic solid = r1= 4.2 cm Radius of cylinder = r2 = 1.4 cm Let h be the length of the cylinder. Since the hemispherical solid is converted into the shape of a cylinder, we obtain Volume of the hemisphere = Volume of the cylinder

468

2  r13  r22 h 3 2 3 2   4.2   1.4  h 3 3 2   4.2  h =  25.2cm 2 3  1.4  Thus, the length of the cylinder is 25.2 cm. The correct answer is B. 31. Let the radii of the conical solid and the cylindrical solid be 4r and r respectively. Let the heights of the conical solid and the cylindrical solid be 3h and 2h respectively. ∴ Volume of conical metallic solid = 1 2   4r   3h   16r 2 h 3 Volume of cylindrical rods = 2

  r   2h   2r 2 h Let n be the number of rods so formed. ∴ 16r 2 h  n  2r 2 h ⇒n=8 Thus, 8 cylindrical rods can be formed from the conical block. The correct answer is C. 32. Diameter of the hemispherical portion = 2 m ∴Radius of the hemispherical portion = 1 m Let h feet be the height of the cylindrical portion.

Since the two hemispheres are stuck at either ends of cylinder, their radii are same. Capacity of the boiler = Volume of the cylinder + Volume of the two hemispheres 2 = r 2 h  2  r 3 3 4   = r 2  h  r  3   4 2 =  1  h   ft 3 3  4  =   h   ft 3 3  The volume of the boiler is given as

88 3 ft . 3

4  88    h    3 3 

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10th Class Mathematics

469

4 88  7  3 22  3 28 4 h   3 3 24 h  8 3 Thus, the height of the cylindrical portion of the boiler is 8 feet. The correct answer is D. 33. The shape of the capsule is:

 22  =   82  28  32  7   cm3 7   22  =   64  28  9  7  cm 3 7 

h+

 22  =  1792  63  cm3 7   22  =  1855   cm 3 7  =5830 cm3 Thus, the volume of the bottle is 5830 cm3. The correct answer is B.

37.

The volume of the cylinder = πr2h = 150.796 4 mm3 ≈ 151 mm3 2 The volume of the semi-sphere = r 3 = 3 16.7619 mm3 ≈ 17 mm3 Therefore, the estimated volume of the capsule = 151 + 17 + 17 = 185 mm3 The correct answer is C. 34. The volume of the bottle = 1 litre = 1 000 cm3 Let the height of the conical part be h. ⇒ The height of the cylindrical part = 2h But (h + 2h) = 30 cm ⇒ h = 10 cm Let the radius of the base be r. 1 r2 × 20 + r 2  10  1000 3 ⇒ r = 11.6 cm The correct answer is A. 35. The volume of the toy = the volume of the cone + the volume of the hemisphere 1 2 1 2 = r 2 h  r 3     7  14  2  7  3 3 3 1437 cm3 The correct answer is A. 36. Radius (R) of the bigger cylinder = 8 cm Height (H) of the bigger cylinder = 28 cm Radius (r) of the smaller cylinder = 3 cm Height (h) of the smaller cylinder = 7 cm Volume of the given bottle = volume of the bigger cylinder + volume of the smaller cylinder = R 2 H  r 2 h =  R 2H  r 2h 

It is given that the height of the conical part is half of the total height of the tank. Therefore, the heights of conical part and cylindrical part are equal. Let the height be h. Also, radii of conical part and cylindrical part are equal. Let the radii be r. 1 Therefore, volume of the conical part = r 2 h 3 Volume of the cylindrical part = πr2h Volume of the tank = volume of cylinder + volume of cone 1 4 = r 2 h  r 2 h  r 2 h 3 3 The tap is releasing water at the rate of 0.008 m3/min. Thus, amount of water released in 5 hours = (0.008 × 60 × 5) m3 = 2.4 m3 The volume of water emptied by the tap in 5 hours is equal to the total volume of the tank. Therefore, total volume of the tank = 2.4 m3 4 Thus, r 2 h  2.4m3 3 3  r 2 h   2.4  1.8m3 4 Thus, the volume of the cylindrical part of the tank is 1.8 m3. The correct answer is C.

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

38. Let the radius of the original solid cylinder be r. Then, height of the original solid cylinder = 12r Since the vertices of the two conical parts meet at a point, the height of each conical part is 6r.

Volume of each conical part = 1 1 2   radius   height   r 2  6r   2r 3 3 3 Therefore, volume of the metal taken out = Volume of the two cones = 2 × 2πr3 = 4πr3 Volume of the original solid cylinder = π ×(radius)2× height = πr2(12r) = 12πr3 Volume of the metallic solid obtained = Volume of original solid cylinder – Volume of the metal taken out = 12πr3 – 4πr3 = 8πr3 Thus, required ratio = volume of the metallicsolid obtained 8r 3 2   volume of the metal taken out 4 r 3 1 = 2:1 Thus, the ratio of the volume of the metallic solid obtained and the volume of the metal taken out is 2:1. The correct answer is A. 39. T.S.A of the wooden block = T.S.A of cuboid + C.S.A of cylinder + Area of the base of cylinder – Area of base of cylinder = T.S.A of cuboid + C.S.A of cylinder The dimensions of the cuboidal part are 16 cm × 7 cm × 10 cm. T.S.A of the cuboid = 2(lb + bh + hl) = 2(16 × 7 + 7 × 10 + 10 × 16) cm2 = 2(112 + 70 + 160) cm2 = 2 × 342 cm2 = 684 cm2 Diameter of the cylindrical part = 7 cm 7 Radius of the cylindrical part = cm 2 Height of the cylindrical part = 6 cm C.S.A of cylindrical part = 2πrh 22 7 2    6cm 2  132cm 2 7 2

470

Therefore, T.S.A of the of the wooden block = 684 cm2 + 132 cm2 = 816 cm2 The correct answer is A.

IIT JEE WORKSHEET KEY 1. (A) 2. (A) 4. (D) 5. (A) 7. (A) 8. (D) 10. (A) 11. (A) 13.(A) 14. (C) 16.(AD) 17. (BD) 19. (BC) 20. (AB) 21. A-q, B-p, C-r, D-s 22. A-r, B-s, C-q, D-p 23. A-q, B-p, C-r, D-s 24. A-p, B-r, C-q 25. 2 26. 9 27. 5 28. 2 29. 3 30. (I-D), (ii-C)

3. 6. 9. 12. 15. 18.

(B) (C) (A) (B) (D) (CD)

HITS / SOLUTIONS TO THE SELECTED QUESTIONS 2 1. Total surface area of a cube = 32 cm2 3 98 6a 2  3 7 a  cm 3 2. Let sides of a cube are a1 and a2 6a 2 1 then, 12  6a 2 9 a1 1  a2 3  volume of cube = 1 : 27 3.

4.

Volume of water 5  20000m2  m  1000m3 100 2rh  2r 2 2r  h  r  Given 2rh =  3 3 hr  h 3  2h  r

5.

 2h  2.5  h  1.25cm r  h  12cm 2r  h  r   540

…….(i)

from (i) 2r × 12 = 540 540 r 24

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

471

7.

then, find 2r Let radii are R1 and R2 4 3 R1 125 So, 3  4 3 64 R 2 3 R1 5  R2 4

5 R1  R 2 …..(i) 4 Given, R 1  R 2  45 5 From (i) R 2  R 2  45 4 9R 2  45 4 R2 = 20 , R1 = 25 2 r h 8.  1570 3 Area = 314, r2 = 314 1570  3 h 314 9. Let radius of one cone is r1 and radius of other cone is r2  r l  2r1l1 r l  2 r2 l r1 1  r2 2 10. Let radius and height of cone are r and h. then after increasing 50% in r  50 r r  100 2 r 3r New radius = r   2 2 h 3h New height = h   2 2 2

 3r  3h    2 2 And value of new cone =   3 r 2 h  8 : 27  required ratio = 9r 2 3h   4 2

11. 63  83  103 = Total value a3 1728 = a3  a = 12 cm  diagonal = 3 a  3 12  12 3 cm 13. Let height of the cone is hcm then value of right circular cylinder = value of right circular cone r 2  h r 2  3  3  h  9cm 19. Let L be the length and B the breadth of the roof Let the rainfall in meter be x  volume of rain water drain from the roof into the cylindrical vessel = volume of the cylindrical vessel 22  22  20  x   1 3.5 7  x  2.5cm 22. Total surface area of the article = CSA of the cylinder + CSA of the two hemispheres  2rh  2  2r 2  2r  h  2r 

22  3.5 10  2  3.5  7  22  17  374cm 2 27. Let n cones have same diameter and height so, volume of cylindrical vessel = n(volume of cone) n  r 2 h  2  r h  3  n 3 28. After joining sides = 39 so total surface area  2

2

 6   3a   6  9a 2 A = 6× 9a2 6  9a 2 a  27a 2  2 27a 2

____________________________________________________________________________________________________ _____________________________________________

Surface Areas and Volumes Solutions

472

29. Let n cones volume of cylinder = r 2 h r 2 h volume of cone = 3  n  volume of cone = volume of cylinder r 2 h n  r 2 h 3 n 3 30. Let height ‘h’ of the frustum of cone = (30-6) = 24 cm O'

B'

A'

24cm B

A

6cm

Radii of circular ends are R1 = 22.5 cm and r2 = 12.5 cm   slant height of the frustum   h 2   r1  r2 



 24 

2

2

  22.5  12.5 

2

 26cm  (i) Area of metallic sheet used = curved surface area of the frustum of cone + area A circular base + curved surface area of cylinder    r1  r2   r22  2r2 h 2 2

   22.5  12.5   26  12.5  2  12.5  6 cm2   2  3822.5cm (ii) volume of water that the bucket can hold 1 =   r12  r22  r1r2   h 3 1 22 2 2    22.5  12.5  22.5  12.5  24 cm3 3 7 1 22 2     2.5   92  52  9  5 24 3 7  23.728 litres .





____________________________________________________________________________________________________ _____________________________________________

13. STATISTICS SOLUTIONS

FORMATIVE WORKSHEET 1.

To find the class mark (xi) for each interval, the following relation is used. Upper class limit  Lower class limit Class mark (xi) = 2 xi and fixi can be calculated as follows. Number of plants Number of houses xi fixi (fi) 0−2 1 1 1×1=1 2−4 2 3 2×3=6 4−6 1 5 1×5=5 6−8 5 7 5 × 7 = 35 8 − 10 6 9 6 × 9 = 54 10 − 12 2 11 2 ×11 = 22 12 − 14 3 13 3 × 13 = 39 Total 20 162 From the table, it can be observed that  f i  20

 f i x i  162

Mean, x 

 fi xi  fi

162  81 20 Therefore, mean number of plants per house is 8.1. Here, direct method has been used as the values of class marks (xi) and fi are small. To find the class mark for each interval, the following relation is used. Upper class limit  Lower class limit xi= 2 Class size (h) of this data = 20 Taking 150 as assured mean (a), di, ui, and fiui can be calculated as follows. Daily wages Number of workers (fi) xi di = xi − 150 d ui  i (in Rs) 20 =

2.

100 −120 120 − 140 140 − 160 160 −180 180 − 200 Total

12 14 8 6 10 50

110 130 150 170 190

− 40 − 20 0 20 40

−2 −1 0 1 2

fiui

− 24 − 14 0 6 20 − 12

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

474

From the table, it can be observed that  f i  50

 f i u i  12

3.

  fi ui  Mean x  a   h   fi   12  = 150    20  50  24 = 150  5 =150 4.8 =145.2 Therefore, the mean daily wage of the workers of the factory is Rs 145.20. To find the class mark (xi) for each interval, the following relation is used. Upper class limit  Lower class limit xi= 2 Given that, mean pocket allowance, x  Rs18 Taking 18 as assured mean (a), di and fidi are calculated as follows. Daily pocket allowance Number of children Class mark xi di = xi − 18 (in Rs) fi

11 −13 13 − 15 15 − 17 17 −19 19 − 21 21 − 23 23 − 25 Total

7 6 9 13 f 5 4 f   i 44  f

12 14 16 18 20 22 24

−6 −4 −2 0 2 4 6

fidi

− 42 − 24 − 18 0 2f 20 24 2f − 40

From the table, we obtain  fi  44  f

 f i u i  2f  40   fidi  x a     fi   2f  40  0   44  f  2f 40 = 0 2f = 40 f = 20 Hence, the missing frequency, f, is 20

4.

To find the class mark of each interval (xi), the following relation is used. Upper class limit  Lower class limit xi = 2 Class size, h, of this data = 3 Taking 75.5 as assumed mean (a), di, ui, fiui are calculated as follows.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

475

Number of heart beats per minute 65 − 68 68 − 71 71 − 74 74 − 77 77 − 80 80 − 83 83 − 86 Total From the table, we obtain  f i  30

Number of women fi

xi

di = xi − 75.5

2 4 3 8 7 4 2 30

66.5 69.5 72.5 75.5 78.5 81.5 84.5

−9 −6 −3 0 3 6 9

ui 

di 3

−3 −2 −1 0 1 2 3

fiui

−6 −8 −3 0 7 8 6 4

 fi u i  4   fi ui  Mean x  a   h   fi   4  = 75.5     3  30  =75.5 + 0.4 =75.9 Therefore, mean hear beats per minute for these women are 75.9 beats per minute.

5. Number of mangoes Number of boxes fi 50 − 52 15 53 − 55 110 56 − 58 135 59 − 61 115 62 − 64 25 It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. 1 1 Therefore, has to be added to the upper class limit and has to be subtracted from thelower class limit 2 2 of each interval. Class mark (xi) can be obtained by using the following relation. Upper class limit  Lower class limit xi = 2 Class size (h) of this data = 3 Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows. Class interval fi xi di = xi − 57 fiui d ui  i 3 49.5 − 52.5 15 51 −6 −2 − 30 52.5 − 55.5 110 54 −3 −1 − 110 55.5 − 58.5 135 57 0 0 0 58.5 − 61.5 115 60 3 1 115 61.5 − 64.5 25 63 6 2 50 Total 400 25 It can be observed that  f i  400

 f i u i  25

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

6.

476

  fi ui  Mean x  a   h   fi   25  = 57    3  400  3 = 57   57  0.1875 16 =57.1875 = 57.19 Mean number of mangoes kept in a packing box is 57.19. Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di. To find the class mark (xi) for each interval, the following relation is used. Upper class limit  Lower class limit xi = 2 Class size = 50 Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows. Daily expenditure (in Rs) fi xi di = xi − 225 fiui d ui  i 50 100 − 150 4 125 − 100 −2 −8 150 − 200 5 175 − 50 −1 −5 200 − 250 12 225 0 0 0 250 − 300 2 275 50 1 2 300 − 350 2 325 100 2 4 Total 25 −7 From the table, we obtain  f i  25

 f i u i  7

7.

  fi ui  Mean x  a   h   fi   7  = 225     50  25  =225-14 =211 Therefore, mean daily expenditure on food is Rs 211. To find the class marks for each interval, the following relation is used. Upper class limit  Lower class limit xi = 2 Class size of this data = 0.04 Taking 0.14 as assumed mean (a), di, ui, fiui are calculated as follows. Concentration of SO2 Frequency Class mark di = xi − 0.14 (in ppm) fi xi

0.00 − 0.04 0.04 − 0.08 0.08 − 0.12 0.12 − 0.16 0.16 − 0.20 0.20 − 0.24 Total

4 9 9 2 4 2 30

0.02 0.06 0.10 0.14 0.18 0.22

− 0.12 − 0.08 − 0.04 0 0.04 0.08

di 0.04 −3 −2 −1 0 1 2

ui 

fiui − 12 − 18 −9 0 4 4 − 31

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

477

From the table, we obtain  f i  30

 f i u i  31   fi ui  Mean x  a   h   fi   31   0.14     0.04   30  =0.14 0.04133 =0.09867  0.099 ppm Therefore, mean concentration of SO2 in the air is 0.099 ppm. 8a. To find the class mark of each interval, the following relation is used. Upper class limit  Lower class limit xi = 2 Taking 17 as assumed mean (a), di and fidi are calculated as follows. Number of days Number of students xi di = xi − 17 fidi fi 0−6 11 3 − 14 − 154 6 − 10 10 8 −9 − 90 10 − 14 7 12 −5 − 35 14 − 20 4 17 0 0 20 − 28 4 24 7 28 28 − 38 3 33 16 48 38 − 40 1 39 22 22 Total 40 − 181 From the table, we obtain  f i  40

 f i d i  181   fidi  Mean x  a      fi   181  = 17     40  =174.525 =12.475  12.48 Therefore, the mean number of days is 12.48 days for which a student was absent. 8b. To find the class marks, the following relation is used. Upper class limit  Lower class limit xi = 2 Class size (h) for this data = 10 Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows. Literacy rate (in %) Number of cities xi di = xi − 70 d ui  i fi 10 45 − 55 3 50 − 20 −2 55 − 65 10 60 − 10 −1 65 − 75 11 70 0 0 75 − 85 8 80 10 1 85 − 95 3 90 20 2 Total 35

fiui −6 − 10 0 8 6 −2

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

478

From the table, we obtain  fi  35

 f i u i  2

9.

  fi ui  Mean x  a   h   fi   2  = 70     10   35  20 = 70  35 4 = 70  7 =700.57 =69.43 Therefore, mean literacy rate is 69.43%. To find the class marks (xi), the following relation is used. Upper class limit  Lower class limit xi = 2 Taking 30 as assumed mean (a), di and fidiare calculated as follows. Age (in years) Number of patients Class mark di = xi − 30 fidi fi xi 5 − 15 6 10 − 20 − 120 15 − 25 11 20 − 10 − 110 25 − 35 21 30 0 0 35 − 45 23 40 10 230 45 − 55 14 50 20 280 55 − 65 5 60 30 150 Total 80 430 From the table, we obtain  fi  80

 f i d i  430   fidi  Mean x  a      fi   430  = 30    =30+ 5.375  80  =35.375  35.38 Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45. Modal class = 35 − 45 Lower limit (l) of modal class = 35 Frequency (f1) of modal class = 23 Class size (h) = 10 Frequency (f0) of class preceding the modal class = 21 Frequency (f2) of class succeeding the modal class = 14  f1  f 0  Mode = l   h  2f1  f o  f 2 

Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years. ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

479

10. From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80. Therefore, modal class = 60 − 80 Lower class limit (l) of modal class = 60 Frequency (f1) of modal class = 61 Frequency (f0) of class preceding the modal class = 52 Frequency (f2) of class succeeding the modal class = 38 Class size (h) = 20  f1  f 0  Mode = l   h  2f1  f o  f 2    61  52 = 60   20  2  61  52  38      9   = 60     20  122  90   9  20   = 60     132  90 = 60   60  5.625 16 =65.625 Therefore, modal lifetime of electrical components is 65.625 hours. 11. It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals. Therefore, modal class = 1500 − 2000 Lower limit (l) of modal class = 1500 Frequency (f1) of modal class = 40 Frequency (f0) of class preceding modal class = 24 Frequency (f2) of class succeeding modal class = 33 Class size (h) = 500

 f1  f 0 Mode = l    2f1  f o  f 2

 h 

  40  24 = 1500    500  2  40   24  33     16  = 1500     500  80  57  8000 = 1500  23 =1500 + 347.826 =1847.826  1847.83 Therefore, modal monthly expenditure was Rs 1847.83. To find the class mark, the following relation is used. Upper class limit  Lower class limit Class mark = 2 Class size (h) of the given data = 500 Taking 2750 as assumed mean (a), di, ui, and fiuiare calculated as follows. ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

480

Expenditure Number of families (in Rs) fi 1000 − 1500 1500 − 2000 2000 − 2500 2500 − 3000 3000 − 3500 3500 − 4000 4000 − 4500 4500 − 5000 Total From the table, we obtain  f i  200

24 40 33 28 30 22 16 7 200

xi

di = xi − 2750

1250 1750 2250 2750 3250 3750 4250 4750

− 1500 − 1000 − 500 0 500 1000 1500 2000

di 500 −3 −2 −1 0 1 2 3 4

ui 

fiui − 72 − 80 − 33 0 30 44 48 28 − 35

 f i u i  35   fi ui  Mean x  a   h   fi   35  = 2750     500  200  =2750  87.5 =2662.5 Therefore, mean monthly expenditure was Rs 2662.50. 12. It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35. Therefore, modal class = 30 − 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f1) of modal class = 10 Frequency (f0) of class preceding modal class = 9 Frequency (f2) of class succeeding modal class = 3  10  9   f1  f 0  Mode = l      5   h = 30    2f1  f o  f 2   2 10   9  3  5  1  = 30   =30.625 =30.6  5 = 30   30.625 8  20  12  It represents that most of the states/U.T have a teacher-student ratio as 30.6. To find the class marks, the following relation is used. Upper class limit  Lower class limit Class mark = 2 Taking 32.5 as assumed mean (a), di, ui, and fiui are calculated as follows. Number of students Number of states/U.T xi di = xi − 32.5 fiui di u  i per teacher (fi) 5 15 − 20 3 17.5 − 15 −3 −9 20 − 25 8 22.5 − 10 −2 − 16 25 − 30 9 27.5 −5 −1 −9 30 − 35 10 32.5 0 0 0 35 − 40 3 37.5 5 1 3 40 − 45 0 42.5 10 2 0 45 − 50 0 47.5 15 3 0 50 − 55 2 52.5 20 4 8 Total 35 − 23 ____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

481

  fi ui  Mean x  a   h   fi   23  = 32.5   5  35  23 = 32.5   32.5  3.28 7 =29.22 Therefore, mean of the data is 29.2. It represents that on an average, teacher−student ratio was 29.2. 13. To find the class marks, the following relation is used. Upper class limit  Lower class limit Class mark = 2 Taking 135 as assumed mean (a), di, ui, fiui are calculated according to step deviation method as follows. Monthly Number of xi class mark di= xi− 135 fi u i d ui  i consumption (in units) consumers (f i) 20 65 − 85 4 75 − 60 −3 − 12 85 − 105 5 95 − 40 −2 − 10 105 − 125 13 115 − 20 −1 − 13 125 − 145 20 135 0 0 0 145 − 165 14 155 20 1 14 165 − 185 8 175 40 2 16 185 − 205 4 195 60 3 12 Total 68 7 From the table, we obtain  fi u i  7

 f i  68 h = 20   fi ui  Mean x  a   h   fi  7 = 135   20 68 140 = 135  68 =137.058 From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145. Modal class = 125 − 145 Lower limit (l) of modal class = 125 Class size (h) = 20 Frequency (f1) of modal class = 20 Frequency (f0) of class preceding modal class = 13 Frequency (f2) of class succeeding the modal class = 14]  f1  f 0  Mode = l   h  2f1  f o  f 2    20  13 = 125     20  2  20   13  14 

= 125 

7 140  20 = 125   135.76 13 13

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

482

To find the median of the given data, cumulative frequency is calculated as follows. Monthly consumption (in units) Number of consumers Cumulative frequency 65 − 85 4 4 85 − 105 5 4+5=9 105 − 125 13 9 + 13 = 22 125 − 145 20 22 + 20 = 42 145 − 165 14 42 + 14 = 56 165 − 185 8 56 + 8 = 64 185 − 205 4 64 + 4 = 68 From the table, we obtain n = 68 n 68  Cumulative frequency (cf) just greater than  i.e.,  34  is 42, belonging to interval 125 − 145. 2 2  Therefore, median class = 125 − 145 Lower limit (l) of median class = 125 Class size (h) = 20 Frequency (f) of median class = 20 Cumulative frequency (cf) of class preceding median class = 22 n   2  cf  Median  l    h f      34  22  = 125     20  20  =125 +12 =137 Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively. The three measures are approximately the same in this case. 14. The cumulative frequency for the given data is calculated as follows. Class interval Frequency Cumulative frequency 0 − 10 5 5 10 − 20 x 5+ x 20 − 30 20 25 + x 30 − 40 15 40 + x 40 − 50 y 40+ x + y 50 − 60 5 45 + x + y Total (n) 60 From the table, it can be observed that n = 60 45 + x + y = 60 x + y = 15 (1) Median of the data is given as 28.5 which lies in interval 20 − 30. Therefore, median class = 20 − 30 Lower limit (l) of median class = 20 Cumulative frequency (cf) of class preceding the median class = 5 + x Frequency (f) of median class = 20 Class size (h) = 10 n   60   2  cf   2  5  x   Median  l     h  28.5  20     10 20  f         25  x  8.5     2  ____________________________________________________________________________________________________ _____________________________________________

483

10th Class Mathematics

17=25x x=8 From equation (1), 8 + y = 15 y=7 Hence, the values of x and y are 8 and 7 respectively. 15. Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below. Age (in years) Number of policy Cumulative holders (fi) frequency (cf) 18 − 20 2 2 20 − 25 6−2=4 6 25 − 30 24 − 6 = 18 24 30 − 35 45 − 24 = 21 45 35 − 40 78 − 45 = 33 78 40 − 45 89 − 78 = 11 89 45 − 50 92 − 89 = 3 92 50 − 55 98 − 92 = 6 98 55 − 60 100 − 98 = 2 100 Total (n) From the table, it can be observed that n = 100. n 100  Cumulative frequency (cf) just greater than  i.e.,  50  is 78, belonging to interval 35 − 40. 2 2  Therefore, median class = 35 − 40 Lower limit (l) of median class = 35 Class size (h) = 5 Frequency (f) of median class = 33 Cumulative frequency (cf) of class preceding median class = 45 n   2  cf  Median  l    h  f     50  45  = 35   5  33  25 = 35  =35.76 33 Therefore, median age is 35.76 years. 16. The given data does not have continuous class intervals. It can be observed that the difference between 1 two class intervals is 1. Therefore,  0.5 has to be added and subtracted to upper classlimits and lower 2 class limits respectively. Continuous class intervals with respective cumulative frequencies can be represented as follows. Length (in mm) Number or leaves fi Cumulative frequency 117.5 − 126.5 3 3 126.5 − 135.5 5 3+5=8 135.5 − 144.5 9 8 + 9 = 17 144.5 − 153.5 12 17 + 12 = 29 153.5 − 162.5 5 29 + 5 = 34 162.5 − 171.5 4 34 + 4 = 38 171.5 − 180.5 2 38 + 2 = 40 ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

484

From the table, it can be observed that the cumulative frequency just greater than

n 40   i.e.,  20  is 29, 2 2 

belonging to class interval 144.5 − 153.5. Median class = 144.5 − 153.5 Lower limit (l) of median class = 144.5 Class size (h) = 9 Frequency (f) of median class = 12 Cumulative frequency (cf) of class preceding median class = 17 n   2  cf  Median  l    h  f     20  17  = 144.5   9  12  9 = 144.5   146.75 4 Therefore, median length of leaves is 146.75 mm. 17. The cumulative frequencies with their respective class intervals are as follows. Life time Number of lamps (fi) Cumulative frequency 1500 − 2000 14 14 2000 − 2500 56 14 + 56 = 70 2500 − 3000 60 70 + 60 = 130 3000 − 3500 86 130 + 86 = 216 3500 − 4000 74 216 + 74 = 290 4000 − 4500 62 290 + 62 = 352 4500 − 5000 48 352 + 48 = 400 Total (n) 400 It can be observed that the cumulative frequency just greater than

n 400   200  is 216, belonging to  i.e., 2 2 

class interval 3000 − 3500. Median class = 3000 − 3500 Lower limit (l) of median class = 3000 Frequency (f) of median class = 86 Cumulative frequency (cf) of class preceding median class = 130 Class size (h) = 500 n   2  cf  Median  l    h  f     200  130   3000     500 86   70  500  3000  86 = 3406.976 Therefore, median life time of lamps is 3406.98 hours.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

485

18. The cumulative frequencies with their respective class intervals are as follows. Number of letters Frequency (fi) Cumulative frequency 1−4 6 6 4−7 30 30 + 6 = 36 7 − 10 40 36 + 40 = 76 10 − 13 16 76 + 16 = 92 13 − 16 4 92 + 4 = 96 16 − 19 4 96 + 4 = 100 Total (n) 100 n 100  It can be observed that the cumulative frequency just greater than  i.e.,  50  is 76, belonging to 2 2  class interval 7 − 10. Median class = 7 − 10 Lower limit (l) of median class = 7 Cumulative frequency (cf) of class preceding median class = 36 Frequency (f) of median class = 40 Class size (h) = 3 n   2  cf  Median  l    h  f     50  36  7 3  40  14  3 7 40 = 8.05 To find the class marks of the given class intervals, the following relation is used. Upper class limit  Lower class limit Class mark = 2 Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows. di = xi− 11.5 Number of letters Number of surnames xi fiui d ui  i fi 3 1−4 4−7 7 − 10 10 − 13 13 − 16 16 − 19 Total From the table, we obtain ∑fiui = −106 ∑fi = 100   fi ui  Mean x  a   h   fi 

6 30 40 16 4 4 100

2.5 5.5 8.5 11.5 14.5 17.5

−9 −6 −3 0 3 6

−3 −2 −1 0 1 2

− 18 − 60 − 40 0 4 8 − 106

 106   11.5   3  100 

= 11.5 − 3.18 = 8.32 ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

486

The data in the given table can be written as Number of letters Frequency (fi) 1−4 6 4−7 30 7 − 10 40 10 − 13 16 13 − 16 4 16 − 19 4 Total (n) 100 From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10. Modal class = 7 − 10 Lower limit (l) of modal class = 7 Class size (h) = 3 Frequency (f1) of modal class = 40 Frequency (f0) of class preceding the modal class = 30 Frequency (f2) of class succeeding the modal class = 16  f1  f 0  Mode = l   h  2f1  f o  f 2    30 40  30 7  7.88  3  7  34  2  40   30  16  Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88. 19. The cumulative frequencies with their respective class intervals are as follows. Weight (in kg) Frequency (fi) Cumulative frequency 40 − 45 2 2 45 − 50 3 2+3=5 50 − 55 8 5 + 8 = 13 55 − 60 6 13 + 6 = 19 60 − 65 6 19 + 6 = 25 65 − 70 3 25 + 3 = 28 70 − 75 2 28 + 2 = 30 Total (n) 30 n 30  Cumulative frequency just greater than  i.e.,  15  is 19, belonging to class interval 55 − 60. 2 2  Median class = 55 − 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cf) of median class = 13 Class size (h) = 5 n   2  cf  Median  l    h  f     15  13  = 55   5  6  10 = 55  = 56.67 6 Therefore, median weight is 56.67 kg.

____________________________________________________________________________________________________ _____________________________________________

487

10th Class Mathematics

20. The frequency distribution table of less than type is as follows. Daily income (in Rs) Cumulative frequency (upper class limits) Less than 120 12 Less than 140 12 + 14 = 26 Less than 160 26 + 8 = 34 Less than 180 34 + 6 = 40 Less than 200 40 + 10 = 50 Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

21. The given cumulative frequency distributions of less than type are Weight (in kg) Number of students upper class limits (cumulative frequency) Less than 38 0 Less than 40 3 Less than 42 5 Less than 44 9 Less than 46 14 Less than 48 28 Less than 50 32 Less than 52 35 Taking upper class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be drawn as follows.

Here, n = 35 n So, = 17.5 2 Mark the point A whose ordinate is 17.5 and its x-coordinate is 46.5. Therefore, median of this data is 46.5. ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

488

It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below. Weight (in kg) Frequency (f) Cumulative frequency Less than 38 0 0 38 − 40 3−0=3 3 40 − 42 5−3=2 5 42 − 44 9−5=4 9 44 − 46 14 − 9 = 5 14 46 − 48 28 − 14 = 14 28 48 − 50 32 − 28 = 4 32 50 − 52 35 − 32 = 3 35 Total (n) 35 n 35  The cumulative frequency just greater than  i.e.,  17.5  is 28, belonging to class interval 46 − 48. 2 2  Median class = 46 − 48 Lower class limit (l) of median class = 46 Frequency (f) of median class = 14 Cumulative frequency (cf) of class preceding median class = 14 Class size (h) = 2 n   2  cf  3.5  17.5  14  Median  l    46.5   h = 46     2 = 46  7  14   f    Therefore, median of this data is 46.5. Hence, the value of median is verified. 22. The cumulative frequency distribution of more than type can be obtained as follows. Production yield Cumulative frequency (lower class limits) more than or equal to 50 100 more than or equal to 55 100 − 2 = 98 more than or equal to 60 98 − 8 = 90 more than or equal to 65 90 − 12 = 78 more than or equal to 70 78 − 24 = 54 more than or equal to 75 54 − 38 = 16 Taking the lower class limits on x-axis and their respective cumulative frequencies on y-axis, its ogive can be obtained as follows.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

489

CONCEPTIVE WORKSHEET 1.

Let assumed mean (a) be 55 Class Interval

Frequency (fi)

Mid-Value (xi)

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100 Total

5 4 8 12 16 15 10 8 5 2  fi = 85

5 15 25 35 45 55 65 75 85 95

ui 

xi  55 10 5 4 3 2 1 0 1 2 3 4

fiui 25 160 24 24 16 0 10 16 15 8 fiui = 56

We know that, Mean ( x ) = a 

f i ui h f i

56  10 = 55 – 6.59 85 Hence, Mean ( x ) = 48.41. Let assumed mean = a = 35. = 55 

2.

Age (in years) 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Total

Number of Midpersons (fi) Value (xi) 10 15 25 25 10 10 5 fi = 100

5 15 25 35 45 55 65

ui 

xi  35 10 3 2 1 0 1 2 3

fiui 30 30 25 0 10 20 15 fiui = 40

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

490

We know that f i u i  40   h = 35     10 = 35 + 4 f i  100  Hence, Mean ( x ) = 31.

mean ( x ) = a 

3.

ui 

Mid Point (xi) Frequency (fi)

Class 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

55 65 75 85 95

8 6 12 11 13

xi  75 10

fiui

2 1 0 +1 +2

16 6 0 +11 +26 fiui = 3622=14

fi = 50

Let assumed mean = a = 75 Class interval = h = 10 Sum of frequencies =  f i = 50 We know that, 14  f u  Mean ( x ) = a   i i   h = 75   10 = 75 + 2.8 50  N  Hence, Mean ( x ) = 78 (approx).

= 77.8

4. Class Interval Frequency (fi) 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100

17 p 32 24 19

Mid Point (xi) 10 30 50 70 90

fi = 92 + p

fixi 170 30p 1600 1680 1710  fixi = 51600 + 30p

We know that, f x 51600  30 p Mean ( x ) = i i  50 =  50 (92 + p) = 51600 + 30p f i 92  p  50p – 30p = 5160 – 4600  20p

= 560  p =

560  p = 28. 20

5. .Class Interval 0 – 20

Frequency Mid Value (fi) (xi) 5 10

fi x i 50

20 – 40

f1

30

30f1

40 – 60

10

50

500

60 – 80

f2

70

70 f2

80 – 100

7

90

630

100 – 120

8 fi = 30 + f1 + f2

100

880

fixi = 2060 + 30 f1 + 70f2

.

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

491

fi = 30 + f1 + f2  50= 30 + f1 + f2  50 – 30 = f1 + f2  f1 + f2 = 20  f1 = 20 – f2 ...(i) f i xi f i 2060  30 f1  70 f 2  62.8 = 50  62.8 × 50 = 2060 + 30f1 + 70f2

Now, Mean ( x ) =

 3140 = 2060 + 30f1 + 70f2  314 = 206 + 3f1 + 7f2  314 = 206 + 3(20 – f2) + 7f2 [from (i) f1 = 20 – f2]  314 = 206 + 60 – 3f2 + 7f2  314 – 266 = 4f2  48= 4f2  f2= 48/4 = 12 ... (ii) Substituting the value of f2 from equation (ii) to equation (i), we get f1 = 20 – f2  f1 = 20 – 12

6.

7.

 f1 = 8 Hence, f1 = 8 f2 = 12 From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000. Therefore, modal class = 4000 − 5000 Lower limit (l) of modal class = 4000 Frequency (f1) of modal class = 18 Frequency (f0) of class preceding modal class = 4 Frequency (f2) of class succeeding modal class = 9 Class size (h) = 1000  f1  f 0  Mode = l   h  2f1  f o  f 2   18  4  = 4000    1000  2 18   4  9     14000  = 4000     23  = 4000 +608.695 =4608.695 Therefore, mode of the given data is 4608.7 runs. From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals. Therefore, modal class = 40 − 50 Lower limit (l) of modal class = 40 Frequency (f1) of modal class = 20 Frequency (f0) of class preceding modal class = 12 Frequency (f2) of class succeeding modal class = 11 Class size = 10 ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

492

 f1  f 0  Mode = l   h  2f1  f o  f 2    20  12 = 40     10  2  20   12  11

8.

 80  = 40     40  23  80 = 40  17 =40+4.7 =44.7 Therefore, mode of this data is 44.7 cars. The class (50-60) has maximum frequency, i.e. 20. Therefore it is the modal class. Lower limit of the modal class l = 50 Size of the class interval = h = 10 Frequency of the modal class = f1 = 20 Frequency of the class preceding the modal class = f0 = 13 Frequency of the class succeeding the modal class = f2 = 5 We know that,   f1  f0 Mode = l   h  2 f1  f 0  f 2 

  20  13 = 50     10  2  20  13  5 

9.

 7  = 50     10  40  18  10  7 = 50  22 = 50 + 3.1818  Mode = 53.18 Hence, the mode of given marks obtained by the students in science is 53.18 Total number of terms = 7 th

th

 n  1 8 th The middle terms =     4  2   2 Median = Value of the 4th term = 10 Hence, the median of the given series is 10. 10. Total number of terms =n=8 Median th th  1  n  n  =   observation    1 observation  2  2  2   th th  1  8  8  =   observation    1 observation  2  2  2   =

1 th  4 observation  5th observation   2

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

493

1 1  24  25 =  49 2 2  Median = 24.5 = 11. Class Interval

Frequency

Less than type cumulative frequency

0–8

8

8

0 – 16

10

18

16 – 24

16

34 = cf

24 – 32

24 = f

58

32 – 40

15

73

40 – 48

7

n = 80

n 80   40 lies in the cumulative frequency of class interval 24 - 32. So, 24 - 32 belongs to median 2 2 class interval. Lower limit of median class interval= l = 24 Frequency of median class = f = 24 Number of observations = n = 80 Class size =h=8 Cumulative frequency of class preceding the median class = cf = 34. 

n   2  cf  Now, Median = l   h  f     40  34  = 24   8  24  6 = 24  8 24 = 24 + 2 = 26  Median = 26 Hence, the median of the given frequency distribution = 26 12. Find the median wage income of the workers. Weekly Wages (in Rs.)

No. of Workers

0 – 100

40

Less then type cumulative frequency 40

100 – 200

39

79 = cf

200 – 300

34 = f

113

300 – 400

30

143

400 – 500

45

N = 188

n 188   94 lies in the cumulative frequency of class interval (200 - 300), So 200 - 300 is the median 2 2 class Lower limit of median class = l = 200 

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

494

Number of observations = n = 188 Cumulative frequency of class preceding the median class = cf = 79 Frequency of median class = f = 24 Class size = h = 100 n   2  cf  Now, Median = l   h  f     188   2  79  = 200     100  34    15  94  79  = 200     100 = 200   100 35  34  300 = 200  = 200 + 44.117 7  Median = 244.117 = 244.12 (approx)

13.

Marks

Number of student

Marks less than

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 80 – 90 90 – 100

5 8 12 15 18 22 29 38 45 53

10 20 30 40 50 60 70 80 90 100

Cumulative frequency 5 13 25 40 58 80 109 147 132 245

Y - axis 260 240

Less than ogive

220

Cumulative frequency

200 180 160 140 120 100 80 60 40 20 0

10

20

30

40

50 60 70 Upper limit

80

90 100

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

495

14. Age more Cumulative frequency than

Age (In years)

Number of Persons

0 – 10

53

0

338

10 – 20

48

10

285

20 – 30

45

20

237

30 – 40

41

30

192

40 – 50

38

40

151

50 – 60

35

50

113

60 – 70

31

60

78

70 – 80

24

70

47

80 – 90

15

80

23

90 – 100

8

90

8

Y - axis 340 320 300 280 260 240

‘More than’ ogive

220

Cumulative frequency

200 180 160 140 120 100 80 60 40 20 0

10

20

30

40

50

60

70

80

90

X-axis

Lower Limits

____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

496

15. Height in cms 140 – 145 145 – 150 150 – 155 155 – 160 160 – 165 165 – 170 170 – 175 175 – 180 180 – 185 185– 190 190 – 195 195 – 200

Frequency 4 11 30 32 43 65 72 81 20 15 5 2

Height less than 145 150 155 160 165 170 175 180 185 190 195 200

Cumulative frequency 4 15 45 77 120 185 257 358 358 373 378 380

N 380   190 2 2 Locate 190 on the y-axis from this point, drawn a line parallel to x-axis. The point of intersection of this perpendicular with the x-axis is the median of the data hence median = 171 

Y - axis 375 350 325 300 275

Cumulative frequency

250 225 200 175 150 125 100 75 50 25 0

Median (171)

X - axis

145 150 155 160 165 170 175 180 185 190 195 200 Height less than (in cm)

____________________________________________________________________________________________________ _____________________________________________

10th Class Mathematics

497

16. Table for less than cumulative curve Age

Frequency Age less than

5–6 6–7 7–8 8–9 9 – 10 10 – 11 11 – 12 12 – 13 13 – 14 14 – 15 15 – 16 16 – 17

35 51 64 66 74 99 91 82 62 46 21 9

Cumulative frequency

6 7 8 9 10 11 12 13 14 15 16 17

35 86 150 216 290 389 480 562 624 670 691 700

Total for more than cumulative curve: Age

Frequency

5–6 6–7 7–8 8–9 9 – 10 10 – 11 11 – 12 12 – 13 13 – 14 14 – 15 15 – 16 16 – 17

35 51 64 66 74 99 91 82 62 46 21 9

Age More Cumulative than frequency 5 700 6 665 7 614 8 550 9 484 10 410 11 311 12 220 13 138 14 76 15 30 16 9

17. Y - axis

700 650 600 550

Cumulative frequency

500 450 400 350 300 250 200 150 100 50 0

Median (10.6) 5

6

7

8

9

10

11

12

13 14

X - axis 15

16

17

Age

 The two ogives will intersect each other at a point. From this point if we draw a perpendicular on the xaxis, the point at which it cuts the x-axis, is the median of this data. Hence, Median = 10.6 ____________________________________________________________________________________________________ _____________________________________________

Statistics Solutions

498

SUMMATIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

A

C

C

C

D

B

C

D

B

C

9 5

10 6

IIT JEE WORKSHEET KEY

4. 5.

1

2

3

ABC

ABD

AB

4 *

5 *

6 7

7 5

8 4

(A-q), (B-s), (C-p), (D-r) (A-q), (B-r), (C-p), (D-s)



____________________________________________________________________________________________________ _____________________________________________