Class 7 Mathematics - BeTOPPERS IIT Foundation Series - 2022 Edition

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IIT FOUNDATION Class VII

MATHEMATICS

© 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.

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Published by

:

USN Eductech Private Limited Hyderabad, India.

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.

Number System

..........

01 - 32

2.

Exponents and Powers

..........

33 - 40

3.

Algebraic Expressions

..........

41 - 50

4.

Simple Equations

..........

51 - 56

5.

Identities and Special Products

..........

57 - 64

6.

Arithmetic

..........

65 - 78

7.

Lines and angles

..........

79 - 90

8.

Triangles

..........

91 - 106

9.

Quadrilaterals

..........

107 - 116

10.

Symmetry

..........

117 - 124

11.

Mensuration

..........

125 - 136

12.

Statistics

..........

137 - 144

13.

Factorisation

..........

145 - 150

14.

Visualising Solid Shapes

..........

151 - 160

15.

Key and Answers

..........

161 - 317

Number System

Learning Outcomes • • • • • • • •

Integers Representing Integers on Number Lines Properties of Integers Types of Fractions Inter conversion of Fractions Multiplication of Fractions Division of Fractions Decimals

• • • • • • •

Chapter -1

By the end of this chapter, you will understand Comparison and conversion of decimals Types of Decimals Operations on Decimals Rational Numbers Comparison of Rational Numbers Operations on Rational Numbers Properties of Rational Numbers

Integers 2. Representing Integers on Number Lines

1. Introduction Natural Numbers The numbers 1, 2, 3...... are used for counting objects is called ‘Natural Numbers’. Set of natural numbers are represented by ‘N’. Set of natural numbers N = { 1, 2, 3 ...... }

Integers can be represented on a number line. The number line shows that every integer has an opposite number except ‘0’. Negative integers are to the left of zero.

Whole Numbers Including ‘0’ in the collection of natural numbers is called whole numbers. Set of whole numbers are represented by ‘W’. Set of whole numbers W = {0, 1, 2, 3 ......} Numbers greater than 0 are called positive numbers. Extending the number line to the left of 0 allows us to picture negative numbers, numbers that are less than 0. Negative numbers

5 4

3 2

zero

1

Positive numbers

0

1

2

3

4

5

When a single + sign or no sign is in front of a number, the number is a positive number. When a single – sign is in front of a number, the number is a negative number. – 5 indicates “negative five”. 5 and + 5 indicate “positive five”. The number 0 is neither positive nor negative. Reading and writing integers The sign of an integer is read first before the  number. Example : – 5 is read as ‘negative five’. + 9 is read as ‘positive 9’ or simply ‘nine’. 0 is an integer but it is nothing positive nor  negative.

4

3

2

1

Positive integers are to the right of zero. 1

0

2

3

4

5

Comparing the Values of Two Integers Number line can be used to compare the values of two integers. 1. Horizontal number line (A) On a horizontal number line, an integer is greater than the integer on its left. (B) On a horizontal number line, an integer is less than the integer on its right. 4 lies to the right of 1. 4>1

4

3

2

1

0

1

2

3

4

2 lies to the left of 1.  2 < 1

2.

Vertical number line (A) On a vertical number line, an integer is greater than the integer below it. (B) On a vertical number line, an integer is less than the integer above it.

7th Class Mathematics

2

5 4

5 lies above 2.  5>2

3 2 1 0

3 lies below 2.  3 < 2

1 2 3 4 5

Arranging Integers in Order 1. Number lines can be used to arrange order, integers in increasing or decreasing order. 2. The value of integers on a horizontal number line increases from left to right and decreases from right to left. Ascending Order Values increasing Largest

Smallest 3

2

1

0

1

2

3

4

5

Descending Order Values Decreasing

Writing Positive and Negative Integers to Represent Word Descriptions A positive or negative number is used to denote: (A) An increase or a decrease in value Example: (i) Rs. 70 withdrawn is denoted by –Rs. 70. (ii) Rs.70 deposited is denoted by + Rs.70. (B) Values more than zero values less than zero (i) – 18 oC denotes a temperature that is 18 o C below 0 oC. (ii) +18 oC denotes a temperature that is 18 o C above 0 oC. (C) A positive direction or a negative direction (opposite direction) Example: (i) –20 oC denotes an anticlockwise rotation of 20o. (ii) +20o denotes a clockwise rotation of 20o. www.betoppers.com

(iii) +5 m denotes a direction 5 m to the right. (iv) –5 m denotes a direction 5 m to the left. (D) Position above or below sea level (i) The sea level is taken as 0 m. (ii) Anything above sea level is taken as positive. For example if a bird is flying at 50m above, then we say +50m. (iii) Simlarly if a submarine lies 150 m below sea level we write it as –150m. Comparison of Integers: If we represent two integers on the number line, then the integer occurring on the right is greater than that occurring on the left. Note: • 0 is less than every positive integer • Every negative integer is less than every positive integer. • The greater is the integer, the lesser is its negative. i.e., a  b   a   b

3. Properties of Integers addition and subtraction

on

(A) If a and b are two integers then a+b=c where c is also an integer. (B) For any two integers a and b a+b=b+a Which means that if we change the order of the integers, even then their sum does not change. (C) For any three integers a, b and c (a + b) + c = a + (b + c) This means that even if we rearrange the integers their sum does not change. (D) If a is any integer then a + 0 = a This means that the sum of any integer and zero is the integer itself. Example: – 10 + 0 = – 10 6+0=6 – 15 + 0 = – 15 (e) For every integer a (which is not zero) there is another integer – a such that a + (– a) = 0 Example: 3 + (– 3) = 0 5 + (– 5) = 0 6 + (– 6) = 0 (f) The difference of any two integers is an integer i.e. If a and b are two integers then a – b = c, where c is also an integer.

Number System (g) In the whole numbers, 0 has no predecessor. But in integers – 1 is the predecessor of 0, – 2 is the predecessor – 1 and so on. Thus if a is any integer, then a – 1 is its predecessor. (h) If a is any integer then a–0=a Like signs Unlike signs + (+y) = +y + (– y) = – y – (–y) = + y – (+ y) = – y

Formative Worksheet 1.

2.

3.

4.

5.

On a particular day, the temperature in Alabama, U.S., was recorded as –10°C in the morning. Later, it dropped by 7°C at night. If rise in temperature is denoted by positive integers, then which expression represents the temperature in Alabama at night on the particular day? (A) (–10 –7)°C (B) (10 + 7)°C (C) (–10 + 7)°C (4) (10 – 7)°C Meenakshi goes 10 units West. Then, she goes 15 units East and finally, she goes 8 units West. If the distance towards the East direction is represented by a positive integer and her initial position is regarded as zero point of reference, then which of the following expression represents her final position? (A) 10 + (–15) + 8 (B) (–10) +15 + (–8) (C) (–10) + (–15) + 8 (D) 10 + 15 + (–8) For what values of a and b does the expression [14 + (–5)] + a = b + [(–5) + (–9)] show the associative property of integers under addition? (A) –5 and –9 (B) –9 and 14 (C) 14 and –5 (D) 9 and 5 Which alternative correctly shows the associative property of integers under addition? (A) {(–15) × 4} + 1 = –15 × (4 + 1) (B) {(–11) + (–3)} + (–2) = –11+{(–3) + (–2)} (C) (–3) + (5) = 5 + (–3) (D) (–1 – 6) + 4 = –1 – (6 + 4) Which of the following statements is correct? (A) Multiplication of an integer with its additive identity gives 1. (B) Addition of an integer with its multiplicative identity gives the successor of that integer. (C) Subtraction of additive identity of an integer from that integer gives the predecessor of that integer. (D) The product of two integers may or may not be an integer.

3 6.

Which of the following pairs of number lines correctly shows the commutative property of integers under addition?

(A) (B)

(C)

(D) 7.

8.

–5 –4 –3 –2 –1 0

–5 –4 –3 –2 –1

0

0 1 2 3 4 5 6

0 1 2 3 4 5 6

0 1 2 3 4 5 6

0 1 2 3 4 5 6

–5 –4 –3 –2 –1 0

–5 –4 –3 –2 –1

0

The numbers a and b are negative in nature. The sign of which of the following expressions cannot be determined with certainty? (A) a + b (B) a – b (C) a × b (D) a  b Which of the following statements is incorrect? (A) Subtraction is commutative for integers. (B) Integers are closed under subtraction. (C) Integers are closed under addition. (D) Addition is associative for integers.

Conceptive Worksheet 1.

2.

Which of the following statements is correct? (A) Integers are closed under addition but not under subtraction. (B) Integers are commutative under subtraction. (C) Integers are associative under subtraction. (D) Integers are commutative under addition. The information in which alternative is incorrectly matched? Property

Example

(A) Addit ion for integers (–3) + 5 = 5 + (–3) is commutative.

Property

Example

(B) Integers are closed The sum of 6 and

under addition. Property

(C) Addition is associative for integers. Property

(D)

4 is positive. Example 8 + [(–7) + 6] = [8 + (–7)] + 6 Example

Integers are closed The difference between under subtraction. 12 and (–5) is an integer.

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7th Class Mathematics

4 3.

4.

Which of the following expressions is correct? (A) (–15) × (–12)×(100) b × c, then

a c  b d

•

If a × d < b × c, then

a c  b d

•

If a × d = b × c, then

a c  b d

5 7 , 6 8

5 7  6 8 We have 5 × 8 = 40 and 6 × 7 = 42 42 > 40 Cross multiply as shown

7 5  8 6 Method2: Convert given fractions in to like fractions and then compare numerators. 

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Formative Worksheet (i) 2 

21 25 , are like fractions. 35 35

Example: compare

7 5  8 6

26. Solve:

3  7 21  5  7 35 5  5 25  7  5 35

Now

5 7 , 6 8

L.C.M of 6,8 is 48

Conversion of Unlike to Like Fractions

3 5 , in to like fractions 5 7 Take L.C.M of denominators L.C.M of 5, 7 is 35.

Compare

Example:

3 5

9 4  11 15

(ii) 4  (v)

7 8

(iii)

3 2  5 7

(iv)

7 2 3   10 5 2

2 1 1 5 (vi) 2  3 (vii) 8  3 3 2 2 8 27. Arrange the following is descending order: 2 2 8 1 3 7 , , (ii) , , 9 3 21 5 7 10 28. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same.Is this a magic square? 4 9 2 11 11 11 3 5 7 11 11 11 8 1 6 11 11 11 (i)

4 9 2 15 + + = ) 11 11 11 11 1 29. A rectangular sheet of paper is 12 cm long 2 (Along the first row

2 10 cm wide. Find its perimeter.. 3 30. Arrange the following is ascending order: (i)

4 4 6 , , 9 3 28

(ii)

2 5 9 , , 5 7 10

Number System

9

31. Salil wants to put a picture in a frame. The picture

3 is 7 cm wide. To fir in the frame the picture 5 3 cm. wide. How much 10 should the picture be trimmed? cannot be more than 7

3. Multiplications of fractions You know how to find the area of a rectangle. It is equal to length × breadth. If the length and breadth of a rectangle are 7 cm and 4 cm respectively, then what will be its area? Its area would be 7 × 4 = 28 cm2. What will be the area of the rectangle if its length

3 32. Ritu ate part of an apple and the remaining apple 5 was eaten by her brother somu. How much part of apple did somu eat? Who had the larger share? By how much? 33. Michael finished colouring a picture in

7 hour.. 12

3 4 hour. Who worked longer? By what fraction was it longer? Vaibhav finished colouring the same picture in

23. Ramesh solved solved

1 1 15 7 You will say it will be 7  3   cm 2 . The 2 2 2 2 15 7 and are fractions. 2 2 To calculate the area of the given rectangle, we need to know how to multiply fractions. We shall learn that now. numbers

Multiplication of a Fraction by a Whole Number

Conceptive Worksheet 22. Write five equivalent fractions of

1 1 and breadth are 7 cm and 3 cm respectively? 2 2

3 . 5

2 part of an exercise while Seema 7

4 of it. solved lesser part? 5

3 1 24. Sameera purchased 3 kg apples and 4 kg 4 2 oranges. What is the total weight of fruits purchased by her? 25. Suman studies for 5

2 hours daily. She devotes 3

4 2 hours of her time for Science and 5 Mathematics. How much time does she devote for other subjects?

Observe the pictures at the left. Each shaded part

1 part of a circle. How much will the two 4 shaded parts represent together? is

They will represent

1 1 1   2 . 4 4 4

To multiply a mixed fraction to a whole number, first convert the mixed fraction to an improper fraction and then multiply.

5 19 57 1 8 . Therefore, 3  2  3   7 7 7 7 ________Multiplication of a Fraction by a Fraction Farida had a 9 cm long strip of ribbon. She cut this strip into four equal parts. How did she do it? She folded the strip twice. What fraction of the total length will each part represent? www.betoppers.com

7th Class Mathematics

10

Formative Worksheet 34. Multiply and reduce to lowest form and convert into a mixed fraction:

3 (i) 7  5 (iv) 5 

1 (ii) 4  3

2 9

(vii) 11  (x) 15 

(v)

4 7

6 (iii) 2  7

2 4 3

(viii) 20 

(vi)

4 5

5 6 2

(ix) 13 

1 3

3 5

4. Multiplication of a Fraction by a Fraction Farida had a 9 cm long strip of ribbon. She cut this strip into four equal parts. How did she do it? She folded the strip twice. What fraction of the total length will each part represent?

9 of the strip. She took one part 4 and divided it in two equal parts by folding the part once. What will one of the pieces represent? It Each part will be

will represent

35. Find

1 9 1 9 of or  . 2 4 2 4

1 (A) of 2

(i) 24

(ii) 46

Each part will be 9/4 of the strip. She took one part and divided it in two equal parts by folding the part once. What will one of the pieces represent? It

(B)

2 of 3

(i) 18

(ii) 27

will represent

(C)

3 of 4

(i) 16

(ii) 36

Formative Worksheet 37. Find

4 of (i) 20 (ii) 35 5 36. Multiply and express as a mixed fraction: (D)

(a) 3  5

1 5

1 (d) 4  6 3

(b) 5  6

3 4

1 (e) 3  6 4

(c) 7  2

1 4

2 (f) 3  8 5

(i)

1 1 of (a) 4 4

(b)

3 5

(c)

4 3

(ii)

1 2 of (a) 7 9

(b)

6 5

(c)

3 10

38. Multiply and reduce to lowest form (if possible): (i)

Conceptive Worksheet 1 of 2

(i) 36

(ii) 54

5 of (i) 64 (ii) 24 8 27. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of (b)

2 water. Vidya consumed of the water. Pratap 5 consumed the remaining 5 water. (i) How much water did Vidya drink ? (ii) What fraction of the total quantity of water did Pratap drink ? ________________________________________________________________________________ www.betoppers.com

2 2 2 3 3

(ii)

2 7  7 9

(iii)

3 6  8 4

9 3  5 5

(v)

1 15  3 8

(vi)

11 3  2 10

2 7 (ii) 6  5 9

(iii)

3 1 5 2 3

2 4 (v) 3  5 7

3 (vi) 2  3 5

(iv)

26. Find: (a)

1 9 1 9 of or  . 2 4 2 4

4 12  5 7 39. Multiply the following fractions. (vii)

(i)

2 1 5 5 4

(iv)

5 3 2 6 7

4 3 (vii) 3  7 5 40. Which is greater: (i)

2 3 3 5 1 6 2 3 of or of (ii) of or of 7 4 5 8 2 7 3 7

Number System

11

Conceptive Worksheet 28. Saili plants 4 saplings, in a row, in her garden. The

3 m. 4 Find the distance between the first and the last sapling.

So, we are required to divide a whole number by a fraction or a fraction by another fraction. Let us see how to do that. Reciprocal of a Fraction

distance between two adjacent saplings is

3 29. Lipika reads a book for 1 hours everyday. She 4 reads the entire book in 6 days. How many hours in all were required by her to read the book ? 30. A car runs 16km using 1 litre of petrol. How much distance will it cover using 2 31. (A) (i)

3 litres of petrol. 4

Provide the number in the box

,

2 10   3 30 (ii) The simplest form of the number such that

obtained in (B) (i)

The reciprocal of a non-zero fraction

a b is b a

2 5 is 5 2 Division of Whole Number by a Fraction Example: Reciprocal of

1 Let us find 1  . 2 We divide a whole into a number of equal parts such that each part is half of the whole.

1 The number of such half   parts would be  2 Observe the figure. How many half parts do you see? There are two half parts. So, 1 

1 2  2 . Also 1  1 2  2 2 1

is __________.

Provide the number in the box

,

3 24  such that  5 75 (ii) The simplest form of the number obtained in

is __________.

________________________________________________________________________________

5. Division of Fractions

Thus, 1 

1 2  1 2 1

3  3? 4 Based on our earlier observations we have: What will be

John has a paper strip of length 6 cm. He cuts this strip in smaller strips of length 2 cm each. You know that he would get 6  2 =3 strips. John cuts another strip of length 6 cm into smaller

3 3 3 3 1 3 1 3      4 4 1 4 3 12 4 Division of a Fraction by Another Fraction

3 strips of length cm each. 2 How many strips will he get now? He will get

We can now find

6

3 strips. 2

A paper strip of length smaller strips of length

15 cm can be cut into 2

3 cm each to give 2

1 5  . 3 6

1 5 1 5 1 6 2    reciprocalof    . 3 6 3 6 3 5 5 A Fraction Lying between two given Fractions If

a c and are two fractions, then the fraction b d

ac a c a ac c  lies between and , thus  bd b d b bd d

15 3  pieces. 2 2 www.betoppers.com

7th Class Mathematics

12 •

Formative Worksheet 41. Find the reciprocal of each of the following fractions. Classify the reciprocal as proper fractions, improper fractions and whole numbers. (i)

3 7

(ii)

12 7 42. Find (v)

(i)

(vi)

5 8

(iii)

9 7

1 8

(vii)

(iv)

6 5

Decimal Places: The number of digits contained in the decimal part of a decimal gives the number of its decimal places. Example: • 5.89 has 2 decimal places • 23.758 has 3 decimal places Like Decimals: Decimals having the same number of decimal places are called like decimals. Example:• 5.1, 77.3, 109.1, 1009.0 are like decimals; each having one decimal place • 0. 85, 23.58, 134.72, 1000.89 are like decimals; each having two decimal places Unlike Decimals: Some given decimals, all not having the same number of decimal places are called unlike decimals. Example:• 5.75, 57.5 are unlike decimals

1 11

7 4 6 1  2 (ii)  5 (iii)  7 (iv) 4  3 3 9 13 3

1 (v) 3  4 2

3 (vi) 4  7 7

Conceptive Worksheet 32. Find: (i) 12 

3 4

(ii) 14 

(iv) 4 

8 3

(v) 3  2

5 6

(iii) 8 

7 3

1 3

(vi) 5  3

4 7

2. Comparison of Decimals

33. Find (i)

•

2 1  5 2

1 3 2  3 5

•

(ii)

4 2  9 3

1 8 (v) 3  2 3

1 2 (vii) 3  1 5 3

(iii) (vi)

3 8  7 7

(iv)

2 1 1 5 2

1 1 (viii) 2  1 5 5

Decimals 1. Fundamentals of Decimals A fraction whose denominator is 10 or some integral power of 10, is called a decimal fraction.

2 23 234 2345 , , , etc., 10 100 1000 10000 These decimal fractions can be written in the decimal form as 0.2, 0.23, 0.234, 0.2345 etc Decimals: The number written in the decimal form are called decimal numbers or decimals. Thus each of the numbers 0.2, 0.23, 0.234, 0.2345 is a decimal. • A decimal has two parts. Whole number part and decimal part. These parts are separated by a dot (.) is called decimal point. Example:

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The digits lying to the left of the decimal point form the whole number part. The decimal point together with digits lying to its right form the decimal part. Eg: In the decimal number 23.574; Whole number part = 23 & Decimal part = 574

•

If two decimals having greatest Integral part is greater. Example: 45.85, 54.75 Since 54 > 45; hence 54.75 > 45.85 If two decimals having same integral part, then compare the decimal number part which is having greatest decimal part is greater. Example: 23.58, 23.78 58, 78 are in decimal part of these two In this 78 > 58; hence 23.78 > 23.58

3. Conversion of Unlike to Like Decimals Annexing zeroes to the extreme right of the decimal part of a decimal does not change its value. Example: 0.7 

7 70   0.70 10 100

70 700   0.7000 100 1000 0. 7 = 0. 70 = 0. 700 etc. Decimal to Fraction Conversion Write the given decimal without the decimal point as numerator. Take ‘1’ annexed with as many zeroes as is the number of decimal places in the given decimal as denominator. 0.70 

Number System

Example:

13

23.4

=

234 10

2345 23.45 = 100 23.456 =

23456 1000

Conversion of Fraction into a Decimal by Division Method: Procedure: Step1. Divide the numerator by the denominator Step2. Complete the division. Let a non-zero remainder be left Step3. Insert a decimal point in the dividend and the quotient. Step4. Part a zero on the right of the decimal point in the dividend as well as on the right of the remainder. Divide again just as whole numbers. Step5. Repeat step4 still either the remainder is zero or requisite number of decimal places have been obtained.

4. Types of Decimals Terminating Decimals: In the process of converting a fraction into a decimal by the division method. If we obtain a zero remainder after a certain number of steps, then the decimal obtained is a terminating decimal. Example: 3.175 is a terminating decimal. However, there are situations where the division process continues indefinitely and zero remainder is never obtained. Such decimals are known as nonterminating decimals.

1  0.5263157.... 19 Repeating or Recurring Decimals: If in a decimal, a digit or a set of digits in the decimal part is repeated continuously, then such a number is called a recurring or repeating decimal. In a recurring decimal, if a single digit is repeated, then it is expressed by put a dot on it. If a set of digits is repeated, it is expressed by putting a bar on the set. Example:

Example: • •

4  0.444....  0.4 9 22 = 3.142857142857 ...... 7  3.142857

Pure Recurring Decimal: Is a decimal in which all the digits in the decimal part are repeated. Example: 0.2, 0.59 etc. Mixed Recurring Decimal: Is a decimal in which some of the digits in the decimal part are repeated and the rest are not repeated. Example: 0.63, 0.576 , etc. Zero of the Decimal: The value of a decimal number is not altered when zeros are replaced at the end of a number. Example: 8 can be written as 8, 8.0, 8.00, 8.000……… The number 34.5 can be written as 34.50, 34.500,34.5000………

Formative Worksheet 43. Which is greater ? (i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7 (iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88 44. Express as rupees using decimals : (i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise (iv) 50 paise (v) 235 paise. 45. Write the place value of 2 in the following decimal numbers : (i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352 46. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much ? 47. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits ?

Conceptive Worksheet 34. (i) Express 5 cm in metre and kilometre (ii) Express 35 mm in cm, m and km 35. Express in kg : (i) 200 g (ii) 3470 g (iii) 4kg 8g (iv) 2598 mg 36. Write the following decimal numbers in the expanded form : (i) 2.03 (ii) 2.03 (iii) 200.03 (iv) 2.034 37. How much less is 28 km than 42.6 km? ________________________________________________________________________________ www.betoppers.com

7th Class Mathematics

14

5. Operation on Decimals Addition of Decimals • •

• •

Convert all the given decimals into like decimals. Write the decimals one under the other with decimal points of all the addends in the same column. Add as in the case of whole numbers. In the sum, put decimal point directly under the decimal points of the addends. Example: Solve 5.7 + 8.78 + 0.352 5.700 + 8.780 + 0.352 5.700 8.780 0.352 _________ 14.832

Step2: When the division of whole number part of the dividend is complete, put the decimal point in the quotient and proceed with the division as in case of whole numbers. Example: 2.345 10 2.345 10 = 0.2345 (shifting decimal point one place to the left) Dividing a Decimal by a Decimal Step 1: Convert the divisor into a whole number by multiplying the dividend and divisor by a suitable power of 10. Step 2: Divide the new dividend by the whole number obtained above Example: Divide 55.50 by 5.55

55.50  100 5550   10 5.55  100 555

_________

Subtraction of Decimals • Convert the given decimals into like decimals. • Write the smaller number under the larger one in such a way that the decimal points of both the numbers are in the same column. • Subtract as in the case of whole numbers. • In the difference, put the decimal point directly under the decimal points of the given numbers. Example: Solve 100.08 – 98.8 100.08 – 98.80 100.08 97.80  ________ 1.28 ________ Multiplication of Decimals: Multiplication of a decimal by 10,100,1000 etc., when a decimal is multiplied by some power of 10, then the decimal point is shifted to the right by as many digits as there are zeroes in the multiplier. Example: 10.999 10 = 109.99 (shifted decimal point 1 place to the right) 10.999 100 = 1099.9 (shifted decimal point 2 places to the right) Division of Decimals: To divide a decimal by 10,100,1000 etc., shift the decimal point to the left as many places as is the number of zeroes in the divisor. Example: Divide 15.3 10 15.3 10 = 1.53 (shifting decimal point one place to the left) Dividing a Decimal by a Whole Number Step1: Perform the division by considering the dividend a whole number www.betoppers.com

Formative Worksheet 48. Find (i) 4.8 10 (iii) 0.7 10 (v) 272.23  10 (vii) 3.97 10 49. Find (i) 2.7  100 (iii) 0.78  100 50. Find (i) 7.9 1000 (iii) 38.53 1000 (v) 0.5  1000 51. Find (i) 7  3.5

(ii) 52.5  10 (iv) 33.1  10 (vi) 0.56 10

(ii) 0.3  100 (iv) 432.6 100 (ii) 26.3  1000 (iv) 128.9 1000

(iii) 3.25  0.5

(ii) 36  0.2 (iv) 30.94  0.7

(v) 0.5  0.25 (vii) 76.5  0.15

(vi) 7.75  0.25 (viii) 37.8 1.4

(ix) 2.73  1.3

Conceptive Worksheet 38. Find (i) 0.2  6 (ii) 8  4.6 (iii) 2.71  5 (iv) 20.1  4 (v) 0.05  7 (vi) 211.02  4 (vii) 2  0.86 39. Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.

Number System 40. Find (i) 1.3  10 (ii) 36.8  10 (iii) 153.7  10 (iv) 168.07  10 (v) 31.1  100 (vi) 156.1  100 (vii) 3.62  100 (viii) 43.07  100 (ix) 0.5  10 (x) 0.08  10 (xi) 0.9  100 (xii) 0.03  1000 41. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol? 42. Find (i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1 (v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02 (viii) 10.05 × 1.05 (ix) 101.01 × 0.01 43. Find: (i) 0.4  2 (ii) 0.35  5 (iii) 2.48  4 (iv) 65.4  6 (v) 651.2  4 (vi) 14.49  7 (vii) 3.96  4 (viii) 0.80  5 44. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance wil it cover in one litre of petrol ?

Formative Worksheet ADDITIONAL PROBLEMS ON FRACTIONS & DECIMALS 52. As a part of his Science assignment, Warren has to experiment with prisms. For this, he goes to the store and buys five prisms. The cost of nine prisms is Rs 6.84. The cost of five prisms is (A) Rs 1.20 (B) Rs 2.40 (C) Rs 3.80 (D) Rs 4.60 53. Connie earns Rs3.68 for delivering milk to 16 houses. She delivers one packet of milk at each house. How much does Connie earn for delivering each packet of milk? (A) Rs 0.23 (B) Rs 0.37 (C) Rs 0.41 (D) Rs 0.59 54. At a mall's parking lot, the charges are Rs5 per vehicle and an additional Rs1.20 per person in the vehicle. How much money will be charged for a vehicle with five persons? (A) Rs 8 (B) Rs 9 (C) Rs 10 (D) Rs 11

15 55. A train, 150 m long, running with a speed of 50 km per hour, passes a platform, 600 m long. How much time will the train take to pass the platform? (A) 20 sec (B) 28 sec (C) 40 sec (D) 54 sec 56. The height of Eiffel Tower is 324 m while the height of Leaning Tower of Pisa is 55.9 m. The difference between their heights, in centimeters, is (A) 26.81 (B) 268.1 (C) 2,681 (D) 26,810 57. There are four members in Linda's family. They eat a total of 25 kg of vegetables in a month. In a particular month, Linda's father was out of town on an official tour. The quantity of vegetables eaten by Linda's family during that month was (A) 18 825 g (B) 18 750 g (C) 18 675 g (D) 18 300 g 58. At the beginning of a particular month of 30 days, a shopkeeper had 200 m of cloth. At the end of the month, he discovered that on an average, he sold 5.2 m of cloth everyday. How much cloth was left with the shopkeeper at the end of the month? (A) 3 600 cm (B) 3 800 cm (C) 4 200 cm (D) 4 400 cm 59. Two friends Angel and Lily together bought some candies. The amount paid by Angel was Rs4.50. The amount paid by Lily was twice the amount paid by Angel. How much money did the two friends spend on the candies? (A) Rs6 (B) Rs9 (C) Rs13.50 (D) Rs17.50 60. Clint purchased six packets of cookies and paid a Rs5 bill to the cashier. The cost of each packet of cookies was Rs0.75. How much change did he receive from the cashier? (A) Rs0.50 (B) Rs0.75 (C) Rs1.00 (D) Rs1.25 61. If 1 kg of cashew nuts costs Rs7.75, then what is the cost of 3 kg of cashew nuts? (A) Rs21.25 (B) Rs22.25 (C) Rs23.25 (D) Rs24.25 62. Bob along with his three friends goes to a restaurant. They order four milk shakes. If each milk shake costs Rs1.25, then what is the total amount spent by them? (A) Rs3 (B) Rs4 (C) Rs5 (D) Rs6 www.betoppers.com

7th Class Mathematics

16 63. What is the result of the expression

1   84  63  1.5  21  42  ? (A) 121.5 (B) 123.5 (C) 125.5 (D) 127.5 64. What is the result of the expression [19.25 – 7 + 16.2 × 15]? (A) 255.25 (B) 255.75 (C) 260.25 (D) 260.75 65. What is the result of the expression 3 4 3   12.5  5  7  14  2 ? (A) 8 (B) 8.5 (C) 9 (D) 9.5 66. What is the result of the expression  17 75 ?  25  17  6.4  5  15 (A) 10 (B) 15 (C) 20

(D) 25

Conceptive Worksheet 45. What is the value of the expression [26.2 + 38.05 – 6.05 × 5]? (A) 33 (B) 33.50 (C) 34 (D) 34.50 46. What is the value of the expression

6  17  51  12  2.5  8 ? (A) 5 (B) 8 (C) 10 (D) 13 47. Each of the 34 students of a class contributed Rs1.25 each to buy a Christmas gift for their class teacher. What was the total amount contributed by the students? (A) Rs 40 (B) Rs 42.50 (C) Rs 45 (D) Rs 47.50 48. Five friends went to a cafeteria. Each of them ordered a hamburger, each hamburger costing Rs 1.50. What was the total cost of their meal? (A) Rs 6.50 (B) Rs 7.50 (C) Rs 8.50 (D) Rs 9.50 49. Max and Robin went to market to buy clothes for themselves. Max bought two shirts for Rs 10.50 each and a pair of jeans for Rs 12.75. Robin bought a shirt for Rs 8.65, a pair of jeans for Rs 15.30 and two pairs of socks for Rs 2.75 each. How much more did Max spend than Robin? (A) Rs 5.70 (B) Rs 5.30 (C) Rs 4.70 (D) Rs 4.30 www.betoppers.com

50. Jimmy has two long sticks. One of the sticks measures 12 m and the other measures 12 yards. The lengths of the two sticks differ by (A) 40.40 inches (B) 40.44 inches (C) 44.40 inches (D) 44.44 inches 51. The weight of a 4 lb fish in grams is (A) 1,523.4 (B) 1,675.8 (C) 1,714.6 (D) 1,814.4 52. Four friends - Cindy, Pam, Alicia, and Nancy purchased suit pieces of lengths 200 cm, 4 m, 3 feet, and 1 yard respectively. The maximum quantity of cloth was purchased by (A) Cindy (B) Pam (C) Alicia (D) Nancy 53. Liz prepared tomato sauce to be used at a party at her place. She bought 9 kg of tomatoes. Each kilogram of tomatoes cost her Rs 0.58. How much money did Liz spend on buying the tomatoes? (A) Rs 5.22 (B) Rs 6.22 (C) Rs 7.22 (D) Rs 8.22 54. Kate, along with her four friends, went out to eat ice-cream. They ordered two scoops of ice cream for each of them. The cost of each scoop of ice cream was Rs 1.50. How much money did they spend in all? (A) Rs 10 (B) Rs 15 (C) Rs 20 (D) Rs 25 55. What is the value of the expression

 1  5 3  35.5  3.5   17  8 ? (A) 36 (B) 48 (C) 60 (D) 72 56. A group of nine friends went to an amusement park. Five friends rode the roller coaster while the other four went to see the house of horrors. Each ticket for the roller coaster ride costs Rs 1.50 and each ticket to the house of horrors costs Rs 2.75. The total money spent by the friends on the tickets for the two rides was (A) Rs 12.50 (B) Rs 18.50 (C) Rs 24.50 (D) Rs 26.50 57. A school purchased baseball items for the school team. A baseball bat costs Rs 3.75 and a baseball costs Rs 1.25. They purchased 11 bats and 12 balls. How much money was spent on the whole purchase? (A) Rs 50.75 (B) Rs 56.25 (C) Rs 60.75 (D) Rs 66.25

Number System

17

58. A courier company charges Rs 0.50 for each unit of the package and an additional Rs 1.50 as handling charges. What is the cost of mailing a package of seven units? (A) Rs 5 (B) Rs 8 (C) Rs 12 (D) Rs 15 59. What is the value of the expression 3.25 × 2.32? (A) 5.54 (B) 6.54 (C) 7.54 (D) 8.54

n In general, any integer n can be written as n  , 1 which is a rational number. But rational numbers like 4.

numbers. Since every natural number is an integer, therefore, p and q are integers.

1. Introduction The word ‘rational’ is derived from the word ‘ratio. A rational number is any number that can be named

p Thus, the fraction q is the quotient of two integers

a where a and b are integers and b

such that q  0 .

b0.

Thus, each of the numbers

1.

p Hence, q is a rational number..

5 6 13 6 , , , is a 6 11 9 17

Numerator and Denominator

rational number. Note: Every natural number is a rational number but a rational number need not be a natural number.

p Let q ( q  0 ) be a rational number. It has two

1 2 3 We can write 1  ,2  ,3  and so on. 1 1 1 This shown that every natural number n can be written as

n which is a rational number.. 1

But none of the rational numbers like 2.

1.

5 3 1 , , , etc., 6 8 3

is a natural number. Zero is a rational number. We can write 0 in anyone of the forms

3.

2.

p q , where p = 0 and q is any non-

zero integer. Hence, 0 is a rational number. Every integer is a rational number but a rational number need not be an integer. We know that

1 2 3 1 2 3 1  , 2  ,3  , 1  , 2  , 3  and 1 1 1 1 1 1 so on.

terms. One is p above the line ‘____’ and the other is q below the line. p is called the numerator of the rational number and q is called the denominator. Positive and Negative Rational Numbers A rational number is said to be positive if its numerator and denominator are either both positive or both negative.

5 29 6 27 , , , are all positive 7 18 19 83 rational numbers. A rational number is said to be negative if its numerator and denominator are such that one of them is a positive integer and the other is a negative integer. For example, Each of the numbers For example,

0 0 0 0 0 0 , , , , , and so on. Thus, 0 can be 1 1 2 2 3 3 expressed as

integers. Every fraction is a rational number but a rational number need not be a fraction.

p Let q by any fraction. Then p and q are natural

Rational Numbers

in the form

5 7 11 , , are not 7 8 6

1.

7 28 37 56 , , , is a negative rational 9 59 11 217 numbers. Note: Every negative integer is a negative rational number.

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7th Class Mathematics

18 Example., –1, –2, –3 and so on, which may be

1 2 3 , 2  , 3  .... are 1 1 1 all negative rational numbers. expressed as 1 

2.

The rational number 0 is neither positive nor negative.

Two Important Properties of Rational Numbers Property 1 Equivalent rational numbers

p If q is a rational number and m is a non-zero p p× m integer, then q = q× m

Thus, 

3  3   2  3   3   4 4 2 43

 3   4  3   5

i.e.,

4 4



45

 ......

2. Rational Numbers in Standard form A rational number is said to be in standard form if its denominator is positive and it is in the lowest terms. To express a given rational number in standard form, proceed as under: Step1. If not already so, make the denominator of the given rational number positive. Step2. Divide both the numerator and the denominator by their HCF.

Formative Worksheet 67. Name each of these rational numbers in the form

p q where p and q are integers. (i) five-sevenths (ii) four and one-half (iii) nineteen (iv) opposite of two-thirds (v) opposite of two (vi) zero (vii) opposite of one 68. Write down the denominator of each of the following rational numbers:

A

3 6 9 12 15      ....... 4 8 12 16 20

All these rational numbers are equal to one another and are called equivalent rational numbers. Property 2 Reducing to a simpler form

p If q is a rational number and m is a common divisor of p and q then

p p÷m = . q q÷m

48 48  2 24 48 48  6 8   ,   , 60 60  2 30 60 60  6 10

5 6

 B

16 12

1 19  D 9 16 69. In each case write down the rational number whose numerator and denominator are respectively as under: (i) (–3) × 2 and 15 – 8 (ii) 5 × 6 + 3 and 3 + 2 × 5. 70. Write the following rational numbers as integers:

 C  4

8 13 25 0 , , , 1 1 1 1 71. Write the following integers as rational numbers: 6, –18, 23, 0 72. Find five rational numbers equivalent to each of the following rational numbers: 6 3 6 8  B C D   11 5 11 15 73. Write each of the following rational numbers with a positive denominator:

A

48 48  12  4   . 60 60  12 5 The rational number

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4 is in lowest terms. 5

A

3 5

 B

8 17

C

21 25

 D

40 59

Number System

19

74. Write each of the following rational numbers with a positive numerator:

i

9 7

 ii 

31 40

 iii 

16 19  iv  33 50

60. Which of the following rational numbers is not positive?

A

 v  4 15 as a rational number with: 16 (i) numerator = –30 (ii) numerator = 75 (iii) denominator = 48 (iv) denominator = –96

75. Express

19 as a rational number with: 5 (i) numerator = –38 (ii) numerator = –95 (iii) denominator = –35 (iv) denominator = 20

76. Express

15 as a rational number with 16 numerator: (i) 96 (ii) –32 (iii) 16 (iv) –8

77. Express

1260 as a rational number with: 1540 (i) Numerator = –63 (ii) Denominator = –154 (iii) Numerator = –90 (iv) Denominator = –22 79. Express the following rational numbers in standard form: 78. Express

6 i 15

Conceptive Worksheet

18  ii  24

36 1302  v 84 1953 80. Fill in the blanks:

 iv  a 

3 ...... ......   4 12 28

 b

5 ..... 25   8 16 ......

 c

7 14 35   9 ...... ......

 d

8 4 .......   ..... 13 65

21  iii  35

28 41

 B

13 19

 C

39 58

D

0 5

3 is written as a rational number with 6 numerator 12, then the denominator is (A) 18 (B) –18 (C) –24 (D) 24 62. Which of the following is not equal to the others? 61. When

40 25 5 15  B  C  D 56 35 7 21 63. Which of the following is in the standard form?

A

9 3 3 3  B  C D 12 4 4 4 64. Which of the following cannot be written as a rational number with denominator 4?

A

7 20 28 1  B  C D 8 16 32 4 65. If a rational number x/y < 1, where x and y are both positive integers, then which of the following is greater than 1?

 A

x 2y

x y2

y Dx  y x ________________________________________________________________________________

A

 B

 C

3. Rational Numbers on a Number Line In Class VI, you have learnt how to represent integers on a number line. If you draw any line as shown below, take a point O on it which you may call the zero point, set off equal distances on both sides of O on the line, then these distances will be considered as of unit length. If we name the points on the right as A, B, C, D, E, …… and the corresponding points on the left as A|, B|, C|, D|, E|,….. Then A, B, C, D, E, …. will represent the points 1, 2, 3, 4, 5, …. And A|, B|, C|, D|, E|, …. Will represent the points –1, –2, –3, –4, –5….

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7th Class Mathematics

20 In the same manner as done, we can represent rational numbers on a number line and obtain a rational number line. (i) If we bisect OA, we get the point P which

1 . Similarly,, 2 if we bisect OA|, we get the point P| which represent the rational number

represents the rational number 

1 2

We observe from the above example that: • A positive rational number is always greater than a negative rational number. • Zero is greater than each one of the negative rational numbers and less than each one of the positive rational numbers. Arithmetical Process • Express each rational number with a positive denominator. • Find L.C.M of the positive denominators. • Express each of the given rational numbers with L.C.M as the common denominator. Thus,

2 2 3 4 4 5   ,   5 5 3 3 3 5

|

(ii) If we divide the lengths OA and OA into three equal parts and label the points of division as P, Q, P|, Q| as shown, A|

P|

Q|

O P

Q

A

2 1 1  3 3

1 O 3

2 3

1

then P, Q will represent the rational numbers

1 2 and respectively. Likewise, points P|, Q| 3 3 will represent the rational numbers 1 2  and  respectively.. 3 3

4. Comparison of Rational Numbers

2 4 , 5 3

6 20 , 15 15

• • •

6 20  15 15 The number having greater numerator is greater. A positive rational number is always greater than a negative rational number Zero is greater than each one of the negative rational numbers and less than each one of the positive rational numbers.

5. Rational Numbers between Two Rational Numbers

There are two ways to compare Rational numbers: • Number Line • Arithmetical Process Number Line: One can compare rational numbers by using a number line easily. Arrange the rational numbers on the number line in ascending order from left to right. Example: The Rational numbers having 4 as denominator can be represented on the number line as follows. Of the two rational numbers represented on the number line, the number on the left is smaller than the number to its right, Thus

Can you tell the natural numbers between 1 and 5? They are 2, 3 and 4. How many natural numbers are there between 7 and 9? There is one and it is 8. How many natural numbers are there between 10 and 11? Obviously none. List the integers that lie between –5 and 4. They are – 4, – 3, –2, –1, 0, 1, 2, 3. How many integers are there between –1 and 1? How many integers are there between –9 and – 10? You will find a definite number of natural numbers (integers) between two natural numbers (integers). How many rational numbers are there between

3 7 3 7   ,  4 4 4 4

3 7 and ? You may have thought that they are 10 10

7 5 3 5    ,    .... 4 4 4 4 www.betoppers.com

only

4 5 6 , and . 10 10 10

Number System

21

But you can also write

3 30 7 as and as 10 100 10

70 . 100 Now

You know how to add, subtract, multiply and divide integers as well as fractions. Let us now study these basic operations on rational numbers.

Addition the

numbers,

31 32 33 , , ,..... 100 100 100

68 69 3 7 , , are all between and . 100 100 10 10 The number of these rational numbers is 39. Also

6. Operations on Rational numbers

3 3000 7 can be expressed as and as 10 10000 10

Let us add two rational numbers with same denominators, say

7  5    3  3  On the number line, we have: We find

7000 . 10000 Now, we see that the rational numbers 3001 3002 6998 6999 , ,......, , are 10000 10000 10000 10000

7 5 and 3 3

3 3

2 3

1 3

0 3

1 3

2 3

3 3

4 3

5 3

6 3

7 3

The distance between two consecutive points is between

3 7 and . These are 3999 numbers in all. 10 10 In this way, we can go on inserting more and more 3 7 and . So unlike 10 10 natural numbers and integers, the number of rational numbers between two rational numbers is not definite. Here is one more example. How many rational numbers are there between rational numbers between

1 5 7 to will mean, moving to the left . So adding 3 3 3 of

7 , making 5 jumps. Where do we reach? Wee 3 2 . 3

reach at So,

7  5  2    . 3  3  3

Subtraction

1 3 and ? 10 10

Difference of two rational numbers

0 1 2 , , are r ational numbers 10 10 10 between the given numbers.

this way:

Obviously

If we write

1 10000 3 30000 as and as , 10 100000 10 100000

we get the rational numbers

8 3

9999 9998 , ,..., 100000 100000

1 3 29998 29999 , between and . 100000 100000 10 10 You will f ind that you get countless rational numbers between any two given rational numbers.

5 3 and 7 8

in

5 3 40  21 19    7 8 56 56

Multiplication Let us multiply the rational number we find

3 by 2, i.e., 5

3  2. 5

On the number line, it will mean two jumps of

3 5

to the left. 6 5

5 5

4 5

3 5

2 5

1 0 1  0 5 5 5

2 5

3 5

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7th Class Mathematics

22 Where do we reach? We reach at

6 . Let us 5

find it as we did in fractions.

3 3  2 6 2   5 5 5

Division We have studied reciprocals of a fraction earlier. What is the reciprocal of

2 ? It will be 7

7 . We extend this idea of reciprocals to 2 rational numbers also. The reciprocal of

2 7 7 will be i.e., ; 7 2 2

3 5 . that of would be 5 3 Dividing a rational number rational number

4 9

by another

5 as, 7

4 5 4 7 28     . 9 7 9 5 45

7. Properties of Rational Numbers Addition on rational numbers satisfies the closure property, the commutative law and the associative law. Zero is the identity element for the addition of rational numbers.

p Every rational number q has its additive inverse p q . Mutliplication on rational numbers satisfies the closure property, the commutative law, the associative law and the distributive law over addition. One is the multiplicative identity for rational numbers.

p Every nonzero rational number q has its q multiplicative inverse p . www.betoppers.com

Zero does not have its multiplicative inverse.

Order properties of rational numbers Property 1: For each rational number x, one and only one of the following is true. (i) x > 0 (ii) x = 0 (iii) x < 0 Property 2: For any two rational numbers x and y, one and only one of the following is true. (i) x > y (ii) x = y (iii) x < y Property 3: If x, y, z be any three rational numbers such that x > y and y > z ; then x > z.

The role of zero (0) Look at the following. 2+0=0+2=2 (Addition of 0 to a whole number) – 5 + 0 = ... + ... = – 5 (Addition of 0 to an integer)

2  2  2  ...  0    7  7  7 (Addition of 0 to a rational number) You have done such additions earlier also. Do a few more such additions. What do you observe? You will find that when you add 0 to a whole number, the sum is again that whole number. This happens for integers and rational numbers also. In general, a + 0 = 0 + a = a, where a is a whole number b + 0 = 0 + b = b, where b is an integer c + 0 = 0 + c = c, where c is a rational number Zero is called the identity for the addition of rational numbers. It is the additive identity for integers and whole numbers as well.

The role of 1 We have, 5×1=5=1×5 (Multiplication of 1 with a whole number)

2 2  1  ...  ...  7 7 3 3 3  ...  1   8 8 8 What do you find? You will find that when you multiply any rational number with 1, you get back that rational number as the product. Check this for a few more rational numbers. You will find that, a × 1 = 1 × a = a for any rational number a. Negative of a number While studying integers you have come across negatives of integers. What is the negative of 1?

Number System

23

It is – 1 because 1 + (– 1) = (–1) + 1 = 0 So, what will be the negative of (–1)? It will be 1. Also, 2 + (–2) = (–2) + 2 = 0, so we say 2 is the negative or additive inverse of –2 and vice-versa. In general, for an integer a, we have, a + (– a) = (– a) + a = 0; so, a is the negative of – a and – a is the negative of a.

Can you say what is the reciprocal of 0 (zero)? Is there a rational number which when multiplied by 0 gives 1? Thus, zero has no reciprocal.

c is called the d reciprocal or multiplicative inverse of another We say that a rational number

2  2  2   2     0 3  3 3

a a c if   1 . b b d Distributivity of multiplication over addition for rational numbers To understand this, consider the rational numbers

Also,

 2 2     0  3 3

3 2 5 , and . 4 3 6

Similarly,

8  8   ...  ...     0 9  9 

3  2  5   3   4    5          4  3  6  4  6 

For the rational number

2 , we have, 3

rational number

 11   11  ...      ...  0  7   7 



3  1  3 1     4  6  24 8

Also

3 2 3  2 6 1     4 3 4  3 12 2

a  a  a a        0 b  b  b b

And

3 5 5   4 6 8

a a We say that  is the additive inverse of b b

 3 2   3 5  1 5 1   Therefore          4 3  4 6  2 8 8

In general, for a rational number

and

a , we have, b

 a a is the additive inverse of    . b  b

Reciprocal 8 By which rational number would you multiply , 21 21 to get the product 1? Obviously by , since 8 8 21  1. 21 8

3  2 5   3 2   3 5            4 3 6   4 3   4 6 

Formative Worksheet 81. Using appropriate properties find.

2 3 5 3 1 (i)      3 5 2 5 6

2  3 1 3 1 2       5  7  6 2 14 5 82. Write the additive inverse of each of the following. (ii)

5 7 must be multiplied by so as to 7 5 get the product 1. Similarly,

21 8 7 We say that is the reciprocal of and 8 21 5 is the reciprocal of

Thus,

5 . 7

(i)

2 8

(ii)

5 9

(iii)

6 5

(iv)

2 19 (v) 9 6 83. Verify that – (– x) = x for. (i) x 

11 15

(ii) x  

13 17 www.betoppers.com

7th Class Mathematics

24 84. Find the multiplicative inverse of the following.

13 (ii) 19

(i) –13

5 3 2  (v) 1  8 7 5

1 (iii) 5

(iv)

1  4 1  4   6   as   6   . 3  3 3  3

(vi) –1

85. Name the property under multiplication used in each of the following. (i)

4 4 4 1  1  5 5 5

(ii) 

(iii)

13 2 2 13    17 7 7 17

19 29  1 29 19

86. Multiply

6 7 by the reciprocal of . 13 16

87. Represent these numbers on the number line. (i)

7 4

(ii)

88. Represent

5 6

89. Write five rational numbers which are smaller than 2. 90. Which of the following relations exhibits the commutative property of rational numbers under multiplication?

 21  6 6  21        5 7 7  5 

(B)

 3 8  6 3  8 6          5 7  11 5  7 11  1 2

(C) 7  1  7

9 8   1 8 9

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1 2

8 1 the multiplicative inverse of 1 ? Why or 9 8 why not?

67. Is

1 68. Is 0.3 the multiplicative inverse of 3 ? Why or 3 why not? 69. Write. (i) The rational number that does not have a reciprocal. (ii) The rational numbers that are equal to their reciprocals. (iii) The rational number that is equal to its negative. 70. Fill in the blanks. (i) Zero has ________ reciprocal. (ii) The numbers ________ and ________ are their own reciprocals (iii) The reciprocal of – 5 is ________. 1 , where x  0 is ________. x (v) The product of two rational numbers is always a _______. (vi) The reciprocal of a positive rational number is ________. (iv) Reciprocal of

2 5 9 , , on the number line. 11 11 11

(A)  

Conceptive Worksheet 66. Tell what property allows you to compute

2 1 71. Find ten rational numbers between 5 and . 2 72. Find five rational numbers between. 2 4 3 5 1 1 and (ii) and (iii) and 3 5 2 3 4 2 73. Write five rational numbers greater than –2. (i)

3 3 and . 5 4 75. Which property is illustrated by the relation 74. Find ten rational numbers between

(D)

p p  0  , where p and q   0 are integers? q q (A) Commutative property of rational numbers under addition (B) Associative property of rational numbers under addition (C) Multiplicative identity of rational numbers (D) Additive identity of rational numbers

Number System

25

Summative Worksheet 1.

2.

3.

4.

5.

6.

7.

8.

9. 10. 11. 12. 13. 14.

The integer – 18 is read as (A) minus 18 (B) negative 18 (C) subtract 18 (D) negative 18 units A glass of water is at its freezing point of 0 oC. When salt is added, its temperature drops to 5 oC below zero. What is the temperature of the water? (A) 5 oC (B) +5 oC (C) –5 oC (D) 5 oC below zero 28 21

S

7

0

T

The integers represented by S and T on the above number line are: S T (A) – 14 7 (B) –14 2 (C) –8 14 (D) – 14 14 Which of the following is incorrect? (A) 4 > 0 (B) 5 < – 3 (C) 7 > – 3 (D) 8 < 9 Which set of integers below is not in descending order? (A) 0, 1, 2, 3, 4 (B) 3, 2, 1, 0, – 1 (C) 7, 6, 5, 4, 3 (D) –5, –6, –7, –8, –9 Which of the following integers has the greatest value? (A) –11 (B) – 10 (C) – 9 (D) – 8 If +8 means 8 units forward, then – 8 means 8 units (A) forward (B) to the right (C) backward (D) to the left If a profit of Rs. 2180 is written as +Rs. 2180, then a loss of Rs. 2180 should be written as (A) Rs.0 (B) Rs. 180 (C) Rs. –2180 (D) –Rs. 2180 145 + (–200) = (A) 65 (B) –55 (C) 55 (D) 65 If –30 + (–y) + 20 = 0, then the value of y is (A) 20 (B) 10 (C) –10 (D) –20 –40 – 32 + 40 = (A) – 80 (B) – 32 (C) 32 (D) 80 0 – (–3) – (–4) = (A) – 7 (B) – 1 (C) 1 (D) 7 – 8 – (–3) – (–5) = (A) 16 (B) 6 (C) 0 (D) – 16 – 10 – (–2) – 3 = (A) – 10 (B) – 11 (C) – 12 (D) – 15

15. In a quiz, each student is required to answer 40 questions. 5 marks are given for every correct answer and 3 marks are deducted for every wrong answer. If Maheshanswered 35 questions correctly and Suresh answered 32 questions correctly, what is the difference in the total marks obtained by them? (A) 15 (B) 24 (C) 160 (D) 175 16. If the temperature of City A is –20 oC and the temperature of City B is 10 oC, the difference in temperature between the two cities is (A) –30 oC (B) – 10 oC o (C) 10 C (D) 30 oC 17. – 40 + (–5) + (–3) = (A) –48 (B) – 32 (C) 32 (D) 48 18. The diagram shows a number line. x 8

19.

20.

21.

22.

y

0

The value of x + y is (A) – 32 (B) – 8 (C) 8 (D) 32 A lady parked her car on the 6th floor and took a lift up 17 floors to the Finance Department. She then went down 9 floors to the Tax Department. On which floor would you find the lady? (A) 14th (B) 15th (C) 23rd (D) 26th The number of negative integers between –6 and 6 is (A) 5 (B) 6 (C) 11 (D) 12 – 52 + (–7) – (– 42) = (A) – 101 (B) – 59 (C) – 17 (D) – 7 Which of the following represents the number line for operation 0 – (–2) – (–3)?

(A) 5

4

3

2

0

1

2

3

2

1

0

1

1

0

1

2

0

1

(B) 4

5

(C)

(D) www.betoppers.com

7th Class Mathematics

26 23. The diagram below is a number line. 15

p

0

What is the value of p + q? (A) –9 (B) – 6 (C) –3 24. The result of the expression

25.

26.

27.

28.

 1 7 14   3    4  1.5 is  2 4

q

(D) 3

8   42.05  12.15  is 5 (A) 37.48 (B) 47.84 (C) 57.48 (D) 67.84 Henry filled the fuel tank of his car with 12 gallons of gasoline. Each gallon of gasoline costs Rs 2.50. If Smith paid for the gas with a Rs 100 bill, then the amount he got back was (A) Rs 30 (B) Rs 50 (C) Rs 70 (D) Rs 90 There are 27 students in Mrs. Thomas' class. All of them were supposed to go on a tour to a museum but at the last moment, six students could not go on the trip. The total entry fee paid by the students was Rs 6.30. The amount paid by each student toward the entry fee of the museum was (A) Rs 0.15 (B) Rs 0.20 (C) Rs 0.25 (D) Rs 0.30 Dennis bought two baseballs priced at Rs 4.25 each. He gave a Rs 10 bill to the store owner. What was the change that was returned to him by the store owner? (A) Rs 1 (B) Rs 1.50 (C) Rs 2 (D) Rs 2.50 Which of the following statements is true about the result of the expression 0.002 × 0.003? (A) The result is less than 0.002 (B) The result is greater than 0.003 (C) The result is less than 0.003 and greater than 0.002 (D) The result is less than 0.002 and greater than 0.003

 2.8  4.7  4 29. The value of the expression 17    1.5  is (A) –3 (B) –2 (C) 2 (D) 3 30. The value of the expression

 1 7 14  3    4  1.5 is  2 4 (A) 12 (B) 14 (C) 16

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31. The value of the expression

(D) 18

(A) 12 (B) 14 (C) 16 32. The value of the expression

(D) 18

2 1 3   1   6  2.5  10 is 3 5 4 (A) 2.5 (B) 4.0 (C) 5.5 (D) 7.0 33. Barbara had Rs 100. She purchased 8 pounds of cookies at the cost of Rs 0.50 per pound. She also bought 3 pizzas at the cost of Rs 2.50 each and two dresses at the cost of Rs 20 each. How much money was left with Barbara? (A) Rs 48.50 (B) Rs 49.50 (C) Rs 50.50 (D) Rs 51.50 34. The value of the expression

4 25 5  1.5  2.5  6.4  10   4  is 3 3 (A) 2.5

(B) 3.5

(C)

19 3

(D)

35 3

35. A stationery shopkeeper purchased 17 pencil packets and 23 eraser packets for Rs 5.5 each. What is the total cost of the stationery bought? (A) Rs 220 (B) Rs 240 (C) Rs 260 (D) Rs 280 36. How many pieces of length

2 mm can be cut 5

from a page of length 10 cm? (A) 25 (B) 50 (C) 250 (D) 500 37. A group of fifteen people went to a zoo. There were five children and ten adults. A child ticket cost Rs 1.50 and an adult ticket cost Rs 2.50. The total cost of all the tickets was (A) Rs 22.5 (B) Rs 32.5 (C) Rs 42.5 (D) Rs 52.5 38. John's aquarium comprises five fishes. Two are goldfish and three are guppies. The cost of one gold fish is 85 cents and that of one guppy is 80 cents. What is the total worth of the five fishes in John's aquarium? (A) Rs 2.1 (B) Rs 3.1 (C) Rs 4.1 (D) Rs 5.1

Number System

27

39. Stella bought 4 apples, 3 oranges, and 5 pears from a fruit vendor. The cost of one apple was Rs 2, one orange was Rs 1.50, and one pear was Rs 2.50. The total money paid by Stella was (A) Rs 20 (B) Rs 22 (C) Rs 25 (D) Rs 27 40. If a is  a  non-zero  rational  number,  then  which expression shows the existence of multiplicative inverse of a?

1 (A) a   1 (B) a ÷ a = 1 a (C) a ÷ 1 = a (D) a × 1 = a 41. If a, b, and c are  rational  numbers,  then  which property is shown by the expression a ×  (b + c) = a × b + a × c? (A) Distributive property of multiplication over addition (B) Associative property of multiplication (C) Commutative property of addition (D) Existence of additive inverse 42. What is the multiplicative inverse of the

ab number   ? ab ba (A) ab

(A)

3 7

 1 3  2  1 2        1       5 7  3  7 3

Step II:

1 1 3 2     1   7 5 7 3

a b (B) ab

3 ? 7

(A) 66

3 7

7 7 (D)  3 3 44. What is the product of the multiplicative inverse (C)

3 and the additive inverse of the number  ? 2 2 2 (A) 1 (B) –1 (C) (D) 3 3 45. The first two steps applied by a student solve the 2  1 3 1 2 expression,    1          , are 3  5 7 7 3 given as follows.

2 3

(C) 0.015

(B) –1 (D) 66

2 3

47. Which of the following statements is correct?

2 5 is the additive inverse of  . 5 2

(B) The multiplicative inverse of 

3 7 is . 7 3

(C) Integers are commutative for subtraction. (D) Rational numbers are commutative for multiplication. 48. Which of the following statements is correct? (A)

(B)

 

Which property is used in the second step by the student? (A) Identity property (B) Distributive property (C) Associative property (D) Commutative property 46. What is the multiplicative inverse of –0.015?

(A)

ab ab (C) (D) ab ba 43. What is the additive inverse of the rational number

Step I:

2 5 is the additive inverse of  . 5 2

(B) The multiplicative inverse of 

3 7 is . 7 3

(C) Integers are commutative for subtraction. (D) Rational numbers are commutative for multiplication. 49. What is the additive inverse of the value of the  2  9   4  6  expression          ?  5  7   11  5 

(A)

616 19 61 8 (B) (C)  (D)  385 385 55 5

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7th Class Mathematics

28

HOTS Worksheet 1.

The stock market average fell 317 points in one day. Represent this quantity by an integer. 2. The lowest elevation in India is found at Nilgir Valley at an elevation of 282 feet below sea level. Represent this elevation with an integer. 3. The temperature on one January day in Delhi was – 10 degrees Celsius. Tell whether this temperature is cooler or warmer than – 5 degrees Celsius. 4. The temperature at 4 P.M. on February 2nd was – 10o Celsius. By 11 P.M. the temperature had risen 12 degrees. Find the temperature at 11 P.M. 5. Scores in golf can be positive or negative integers. For example, a score of 3 over par can be represented by + 3 and a score of 5 under par can be represented by – 5. If a Couple had scores of 3 over par, 6 under par, and 7 under par for three games of golf, what was the total score? 6. Suppose a deep-sea diver dives from the surface to 165 feet below the surface. He then dives down 16 more feet. Use positive and negative numbers to represent this situation. Then find the diver’s present depth. 7. Suppose a diver dives from the surface to 248 meters below the surface and then swims up 6 meters, down 17 meters, down another 24 meters, and then up 23 meters. Use positive and negative numbers to represent this situation. Then find the diver’s depth after these movements. 8. Arun has Rs. 125 in his saving account. He writes a cheque for Rs. 117, makes a deposit of Rs. 45, and then write another cheque for Rs. 69. Find the amount left in his account. (Write the amount as an integer). 9. In a test, it is possible to have a negative score. If Krishna’s score is 15, what is his new score if he loses 20 points? 10. The temperature on a February morning is – 6o Celsius at 6 A.M. If the temperature drops 3 degrees by 7 A.M., rises 4 degrees between 7 A.M. and 8 A.M., and then drops 7 degrees between 8 A.M. and 9 A.M,. find the temperature at 9 A.M. 11. A mountain peak in Himalayas has an elevation of 14,393 feet above sea level. A deep-sea trench in the Pacific Ocean has an elevation of 12,456 feet below sea level. Find the difference in elevation between these two points. 12. A football team lost 4 yards on each of three consecutive plays. Represent the total loss as a product of integers, and find the total loss. www.betoppers.com

13. Abhi lost Rs. 400 on each of seven consecutive days in the stock market. Represent his total loss as a product of integers and find his total loss. 14. A deep-sea diver must move up or down in the water in short steps in order to keep from getting a physical condition called the bends. Suppose the diver moves down from the surface in five steps of 20 feet each. Represent his total movement as a product of integers, and find the product. 15. A weather forecaster predicts that the temperature will drop 5 degrees each hour for the next six hours. Represent this drop as a product of integers and find the total drop in temperature. 16. Simplify: (i) |8 – 24| . (–2)  (–2) (ii) 3(–10)  [5(–3) – 7(–2)] (iii)

(7)( 3)  (4)(3) 3[7  (3  10)]

10( 1)  (2)(3) (iv) 2[ 8  (2  2)] 17. Samantha went to a florist. She bought five roses, two lilies, and three lotuses. The cost of each rose was Rs 0.50, that of each lily was Rs 0.30, and that of each lotus was Rs 0.75. The amount spent by Samantha on the flowers was (A) Rs 4.35 (B) Rs 4.65 (C) Rs 5.35 (D) Rs 5.65 18. Mike, along with his wife and three kids, went to an amusement park. The cost of the ticket for an adult was Rs 5 and that for a kid was Rs 2.50. How much money did Mike spend on the tickets? (A) Rs 13.50 (B) Rs 15.50 (C) Rs 17.50 (D) Rs 19.50 19. Martha went with her friends to a restaurant. They ordered two hamburgers for Rs 2.50 each, four packs of fries for Rs 0.75 each, and two soft drinks for Rs 1.25 each. She paid for this with a twenty dollar bill. How much money did she get back? (A) Rs 9 (B) Rs 9.50 (C) Rs 11 (D) Rs 11.50 20. Smith had 6 puppies. When the puppies were old enough, he approached a store to sell the puppies. The puppies were sold for Rs 66 each, but only 23% of the money was given to Smith. The amount of money earned by Smith after all six puppies were sold was (A) Rs 15.18 (B) Rs 42.27 (C) Rs 67.41 (D) Rs 91.08

Number System 21. The pantry boys at a pizza store begin their work at 5:17 a.m. and finish their work for the day at 12:17 p.m. For every hour, they are paid Rs 9.23. If they work 5 days every week, then the weekly amount they earn is (A) Rs 452.27 (B) Rs 323.35 (C) Rs 291.61 (D) Rs 142.13 22. Linda is decorating a flower basket. She needs 2.98 feet of pink ribbon, 1.2 feet of white ribbon, and 3 inches of green ribbon. The total length of ribbon needed by Linda is (A) 7.13feet (B) 6.91 feet (C) 5.21 feet (D) 4.43 feet 23. Three friends went to an amusement park. They spent Rs 9.85 to purchase the tickets and Rs 12.50 for food. The money that they spent was equally shared among them. How much money did each person spend? (A) Rs 7.25 (B) Rs 7.35 (C) Rs 7.45 (D) Rs 7.55 24. Mike went to a grocery store. He purchased 500 g of coffee, which costs Rs 2 per 100 g, and two packs of fruit juice which costs Rs 1.25 per pack. He paid a total of Rs 20 to the cashier. How much money did the cashier return to Mike? (A) Rs 7.50 (B) Rs 8.50 (C) Rs 11.50 (D) Rs 12.50 25. Lisa, along with two of her friends, went to a restaurant. They ordered three burgers costing Rs 1.25 each and two glasses of juice costing Rs 1.20 each. How much money did they pay for the meal? (A) Rs 9.30 (B) Rs 8.25 (C) Rs 7.20 (D) Rs 6.15 26. Teddy has Rs 5 with him. He buys some cakes for Rs 2.48 and gives half the remaining money to his younger brother. How much money is left with Teddy? (A) Rs 1.25 (B) Rs 1.26 (C) Rs 1.27 (D) Rs 1.28 27. Matt purchases two kilograms of apples at Rs 1.45 per kg and three packets of strawberries at Rs 1.50 per packet. He gives Rs 10 to the cashier. How much money will be returned to Matt by the cashier? (A) Rs 3.75 (B) Rs 3.65 (C) Rs 2.80 (D) Rs 2.60

29 28. Lisa bought cream and fruits for preparing fruit cream. She bought cream worth Rs 3.50, strawberries worth Rs 5 and eight apples, each apple costing Rs 0.50. The total amount spent on purchasing the ingredients of fruit cream was (A) Rs 12.50 (B) Rs 14.50 (C) Rs 16.50 (D) Rs 18.50 29. Molly bought two cans of grape juice and four cans of apple juice. The cost of each can was Rs 1.50, be it a can of grape juice or apple juice. The amount spent on the fruit juices by Molly was (A) Rs 13 (B) Rs 11 (C) Rs 9 (D) Rs 7 30. Alfred bought bus tickets to another city for his entire family. He booked a total of four tickets, two for adults and two for children. Each ticket for an adult costs Rs 60 and each ticket for a child costs Rs 35.50. The total amount spent by Alfred on the tickets was (A) Rs 161 (B) Rs 171 (C) Rs 181 (D) Rs 191 31. Four friends went to a circus. The entry ticket for each person was for Rs 3.50. Each of them also bought a packet of popcorn, each packet costing Rs 0.90. The total amount spent by them was (A) Rs 13.60 (B) Rs 15.60 (C) Rs 17.60 (D) Rs 19.60

 11 8   2 8  expression         is an  3 7 3 7 equivalent form of which of the following expressions?

32. The

(A)

8  5 11     3 2 7 

(B)

1  8  2      3   7  5 

(C)

8  2 11     3 5 7 

(D)

2  8  11      7  3  7  

 11 8   2 8  expression         is an  3 7 3 7 equivalent form of which of the following expressions?

33. The

(A)

8  5 11     3 2 7 

(B)

1  8  2      3   7  5 

(C)

8  2 11     3 5 7 

(D)

2  8  11      7  3  7  

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7th Class Mathematics

30 34. What is the additive inverse of the multiplicative

4 7

inverse of the number 3 ?

40. Which of the following relations illustrates the associative property of rational numbers under addition?

19 (A)  4

7 (B)  25

(A)

1 7 7 1    2 5 5 2

7 (C) 25

19 (D) n 4

(B)

1  2 11   1 2  11      3 5 8  3 5 8

6 6 (C)   0   7 7

35. What is the additive inverse of the expression

 3 4 11 4  5 10  15  6  45   4 ? (A) 2

8 9

(B)

9 26

(C)

9 26

 8 8 (D)      0  9 9 (D) 2

8 9

4 36. What is the multiplicative inverse of 8 ? 9 76 (A) 9

68 (B) 9

6  6 3  38. The expression      4    is  an 11   11 2   equivalent form of which of the following expressions?

(C)

6 3     4 11  2 

3 6  (B)  4    2  11  (D)

(B) 5

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Single Correct Answer Type

1.

The value of is: (A) 17.26 (C) 58.75

(C)

1 4

(D) 

1 5

2980 correct to two decimal place (B) 54.59 (D) 74.29

2.

Given that (5.142)2 = 26.44, find

3.

(A) 5 142 (B) 514.2 (C) 51.42 (D) 5.142 Which of the following values is not equal to 5? (A)

4.

52

 5

(D) – (5) 2

Given that

45 = 6.708 and

(A) 882.9 (C) 88.29 5.

(B)

( 5) 2

the value of

Given that

4,500 +

264 400 .

2

( C )

45 = 2.121, find

450 .

(B) 458.7 (D) 45.87

64.52 = 8.032, what is the value of

2

(80.32) ? (A) 64,520 (C) 645.2

(B) 6,452 (D) 64.52

6.

Given that

7.

640  64 . (A) 33.30 (B) 27.83 (C) 10.53 (D) 3.33 Given that (3.9268)2 = 15.42, find the value of

6 3     4 11  2 

39. What is the product of the additive inverse of – 0.8 and the multiplicative inverse of 0.2? (A) 4

I.

9 9 (C)  (D)  76 68

37. Rational numbers are closed under addition because (A) Any two rational numbers can be added in any order (B) The sum of any two rational numbers is again a rational number (C) The sum of any three rational numbers is independent of the way the three numbers are grouped together (D) The sum of 0 and any rational number is the same rational number

3 6  (A)    4  2  11 

IIT JEE Worksheet

6.4 = 2.53, find the value of

154 200 .

(A) 3926.8 (C) 39.268

(B) 392.68 (D) 3.9268

Number System 8.

31

III. Paragraph Type

The value of 6.9 100 lies between (A) 800 and 900 (B) 200 and 300 (C) 80 and 90 (D) 20 and 30

II. Multiple Correct Answer Type 9.

10.

x 2  y 2  z 2  _________

(A)

x 2  y2  z 2 (B) x  y  z

(C)

 xyz 

2

(D) None of these

0.87  __________

87 87 87 (C) (D) 10 100 10 11. The square of a number ‘n’ is: (A) (n)2 (B) 2n (C) (n  n) (D) none of these (A) 87

12.

(B)

14.

15.

5 = ____________ (A) 2.3 (upto 1st decimal) (B) 2.24 (upto 2nd decimal) (C) 2.2 (upto 1st decimal) (D) 2.23 (upto 2nd decimal)

13. Which of the following one false for

16.

17.

94

Finding Square root of a Number by Prime Factorization Method The square root of a number can be found by prime factorization method using the following steps. Step-1: Write the prime factorisation of the given number. Step-2: Pair the factors such that primes in each pair are equal. Step-3: Choose one prime from each pair and multiply all such primes. Step-4: The product thus obtained is the square root of the given number. Answer the following questions Find the square root of 9216 (A) 26 (B) 56 (C) 76 (D) 96 Find the square root of 8100 (A) 30 (B) 60 (C) 90 (D) 120 Find the square root of 729 (A) 13 (B) 17 (C) 27 (D) 37 Find the square root of 24336. (A) 56 (B) 156 (C) 256 (D) 356

(A)

9+ 4

(B) 3 + 2

IV. Integer Type

(C)

13

(D) None of these

18. The square root of 16 19. 9 9 = 20. Given that (y – 4)2 = 16, y = 21.

132  36  4 =

V. Matrix Matching Column I (A) Value of 144 + 16 × 4 (B) (C)

7, 456 lies between

Column II (p) 2 5 (q) 0.75

1 52

(r)

(D) Calculate the value of 1 – (0.5)2.

(s)

169  69 ×

80 and 90

75 100 (t) 20 (u) 50 and 60

 

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32

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7th Class Mathematics

Exponents & Powers

Learning Outcomes • • • •

Chapter -2

By the end of this chapter, you will understand Exponents Laws of Exponents Powers with Zero and negative exponents Expressing Large Numbers in the Standard Form

Law 2

1. Exponents In general if ‘x’ is any number and ‘n’ is any natural number, then we have xn = x × x × x ........n times. The number ‘x’ is called the base and ‘n’ is called the exponent (or) the index of the exponential expression, xn. Eg: In exponential form, we write 2 × 2 × 2 as 23 read as 2 raised to the power 3. We have, base = 2 and exponent = 3

Rule p If q is any fractional number, then for any m

p pm positive integer ‘m’, we have    m q q

Reciprocal

Power of a power. i.e.,  a

m n



 a mn for all positive

integers

Law 3 Power of a product (ab) m = a m × b m, where a  0, b  0 and ‘m’ is a positive integer.. Repeated application of this gives a more general result namely (abc..z)m = ambmcm.......zm

Law 4 Quotient of powers of the same base a mn if m  n am   1 a n  mn if n  m a

The reciprocal of a non zero integer ‘x’ is 1 denoted by x–1 and defined as x 

fractional number

1 . For a x

p q (where p  0; q  0 ) we

1

m

am a power of a Quotient i.e.,    m where b b

a  0, b  0 and ‘m’ is a positive integer..

3. Powers with Zero and negative exponents

p q have    . p q p The reciprocal of   q

Law 5

m

q is given by   p

m

2. Laws of Exponents Law 1 The prodcut of the two powers of the same base is a power of the same base with the index equal to the sum of the indices. i.e., if a  0 be any rational number and m,n be positive integers, then am × an = am + n

i)

a0 = 1 for every non zero real number ‘a’

ii)

a m  n am   1 a n  m n a

if m  n if n  m

where m,n are

positive integers and a  0 . If we denote the n multiplicative inverse of an by a-n. a 

1 an

7th Class Mathematics

34

4. Expressing Large Numbers in the Standard Form In earlier classes, we have learnt to write a number in the expanded form, as shown below: 473 = 4 ×100 + 7 × 10 + 3 3758 = 3 × 1000 + 7 × 100 + 5 × 10 + 8 30739 = 3 × 10000 + 0 × 1000 + 7 × 100 + 3 × 10 +9 We can express these using powers of 10 in the exponential form: 473 = 4 ×102 + 7 × 101 + 3 ×100 3758 = 3 × 103 + 7 × 102 + 5 × 101 + 8 × 100 30739 = 3 × 104 + 0 × 103 + 7 × 102 + 3 × 101 + 9 × 100 In this section, we shall learn to write large numbers, using powers of 10 as shown above. Standard Form Any number can b written as a number between 1 and 10 multiplied by a power of 10. This is called standard form of the number. To write a number in standard form we split it into two parts multiplied together. The first part must be a number between 1 and 10 and the second, a power of 10. For the number 56750, we start with 5.6750 in order to have a number between 1 and 10; and then we move the decimal point to the right until it is in its correct place (i.e., 56750.0). Here , it needs to move 4 places. The number 4 is the power of 10. So, 56750 = 5.6750 × 104

Formative Worksheet 1. 2.

3.

4.

5.

Find the value of: (i) 26 (ii) 9 (iii) 112 (iv)54 Express the following in exponential form: (i) 6 × 6 × 6 × 6 (ii) t × t (iii) b × b × b × b (iv) 5 × 5 × 7 ×7 × 7 (v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × d Express the following numbers using exponential notation: (i) 512 (ii) 343 (iii) 729 (iv) 3125 Identify the greater number, wherever possible, in each of the following? (i) 43 or 34 (ii) 53 or 35 8 2 (iii) 2  or 8 (iv) 1002 or 2100 (v) 210 or 102 Express each of the following as product of powers of their prime factors: (i) 648 (ii) 405 (iii) 540 (iv) 3,600

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6.

7.

8.

9.

Simplify: (i) 2 × 103 (ii) 72 × 22 (iii) 23 × 5 (iv) 3 × 44 2 (v) 0 × 10 ­­­­­ (vi) 52 × 33 4  2 (vii) 2 × 3 (viii) 32 × 104 Simplify: (i) (– 4)3 (ii) (– 3) × (– 2)3 2 2 (iii) (– 3)  × (– 5) (iv)(– 2)3 × (–10)3 Compare the following numbers: (i) 2.7 × 1012; 1.5 × 108 (ii) 4 × 1014; 3 × 1017 Using laws of exponents, simplify and write the answer in exponential form: (i) 32 × 34 × 38 (ii) 615 ÷ 610 3 2 (iii) a  × a (iv) 7x× 72 (v) 523  53 (vii) a4 × b4

(vi) 25 × 55 (viii) (34)3

20 15 3 (ix)  2  2   2 (x) 8t ÷ 82

10. Simplify and express each of the following in exponential form: 23  34  4 (i) 3  32

2 4 7 (ii)  5  5   5

(iii) 25  5

3  7 2  118 (iv) 21  113

4

3

37 (v) 4 3 3 3 (vii) 20 × 30 × 40 28  a 5 (ix) 3 3 4 a

3

(vi) 20 + 30 + 40 (viii) (30 + 20) × 50

 a5  8 (x)  a 3   a  

45  a 8 b3 (xii) (23 × 2)2 45  a 5 b 2 11. Say true or false and justify your answer: (i) 10 × 1011 = 10011 (ii) 23 > 52 (iii) 23 × 32 = 65 (iv) 30 = (1000)0 12. Express each of the following as a product of prime factors only in exponential form: (i) 108 × 192 (ii) 270 (iii) 729 × 64 (iv) 768 13. Simplify:

(xi)

5 2

2  (i)

 73

83  7

35  105  25 (iii) 57  65

(ii)

25  52  t 8 103  t 4

Exponents and Powers

35

14. Write the following numbers in the expanded forms: 279404, 3006194, 2806196, 120719, 20068 15. Find the number from each of the following expanded forms: (a) 8 × 104 + 6 × 103 + 0 × 102 + 4 ×101+5 × 100 (b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100 (c) 3 × 104 + 7 × 102 + 5 × 100 (d) 9 × 105 + 2 × 102 + 3 × 101 16. Express the following numbers in standard form: (i) 5, 00, 00, 000 (ii) 70, 00, 000 (iii) 3, 18, 65, 00, 000 (iv) 3, 90, 878 (v) 39087.8 (vi) 3908.78 17. Express the number appearing in the following statements in standard form. (a) The distance between Earth and Moon is 384, 000, 000 m. (b) Speed of light in vacuum is 300, 000, 000 m/s. (c) Diameter of the Earth is 1, 27, 56, 000 m. (d) Diameter of the Sun is 1, 400, 000, 000 m. (e) In a galaxy there are on an average 100, 000, 000, 000 stars. (f) The universe is estimated to be about 12, 000, 000, 000 years old. (g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300, 000, 000, 000, 000, 000, 000 m. (h) 60, 230, 000, 000, 000, 000, 000, 000 molecules are contained in a drop of water weighing 1.8 gm. (i) The earth has 1, 353, 000, 000 cubic km of sea water. (j) The population of India was about 1, 027, 000, 000 in March, 2001.

Conceptive Worksheet 1.

Express the following in exponential notation? (i) 3125

2.

(ii)

(iii) a3 + b3 + 3a2b + 3ab2 Simplify  256  (ii)    325 

2/6

(i) (729) (iii) 3.

343 1331

3



4

64



1 3

1/ 4

?

64, 5 3125, 3 729, 4 1296 in ascending order and decending order?

4.

5.

Express the following numbers as a product of prime factors in the exponential notation (i) 14553 (ii) 8281 (iii) 3456? Simplify  xa   b  x 

6.

a

2

 ab  b 2



 xa    c  x 

1 1 x

 q p

 x

r p

1  pq 

7.

 r p

b

2

 bc  c2



 c  ac a  2

 xc    a  x 

2

 

1 q r 

1 x x 1 x  x If 3n = 729, find the values of

p r 

?

42

(i) 3n-3 8. 9.

(ii) 3-n+8

(iii) 3n/2

(iv)  3n  36 ?

find ‘x’ and ‘y’ if 64x + 2 = 16x + 1 = 4y? Simplify (i) (–1)500 (ii) (–1)2001 (iii)

25 120  100  50 5

1001 (iv)

0

 10020   10030 

20010  20020  20030

(v)

5 1

2 6

4 5

6 3

6 3

2   2   2  3   3   3 

3/ 2 2

10. Express the number appearing in the following statements in standard form? (i) The distance between earth and moon is 384,000,000m (ii) The distance of the sun from the centre of milky way galaxy is estimated to be 300,000,000,0000m (iii) The earth has 1,353,000,000 cubic km of sea water (iv) Thickness of a thick paper is 0,0007 mm (v) Charge of electron is 0.000,000,000,000,000,000,16 columb 3 11. a – 3a2b + 3ab2 – b3 exponential form? (1) (a + b)3 (2) (a – b)3 3 3 (3) a + b (4) a3 – b3 12. a2 + b2 + c2 + 2ab + 2bc + 2ca in exponential form? (1)(a + b + c)2 (2) (a – b – c)2 2 (3) (a +b – c) (4) (ab + bc + ca)2 13. 500o + 501o + 502o + 504o + 505o + 506o + 507o + 508o + 509o? (1)5045 (2) 0 (3) 10 (4) None www.betoppers.com

7th Class Mathematics

36 14

30 + 3-1 + 3-2 + 3-3 + 3-4?

121 (1) 81 15.

(2) 123

Summative Worksheet

2

5

3

3

3

3n  814 , then the vlaue of ‘n’ is?

4 16 (2) 12 (3) (4) 48 3 3 18. If 343x + 1 = 49x + 21 then the value of ‘x’ ? (1)

39 (3) 9 (4) 39 5 a c b  x a   x b   xc  19.  b  ×  c  ×  a  ? x  x  x  (1) 20

(2)

(1) xabc (2) xab + bc + ca (3) x (4) 1 -5 -3 -4 -2 20. Arrange 2 , 3 , 4 ,5 in decending order? (1) 5-2,3-3, 2-5,4-4 (2)2-5,3-3, 4-4,5-2 -2 -4 -3 -5 (3)5 , 4 , 3 ,2 (4)2-5,4-4, 3-3,5-2 21. Match the following Column I Column II (i) (-1)1001 ( ) (a) 32



25 (ii)  2 

(iii)

2 0



1 25

(

)

(b) –1

(

)

(c) 1 (d) 0 (e)

22. Column I (i)

Chang is studying exponents. The solution of 21 =? is to be represented on a number line.

243 is exponential form? 3125

3  3   7   7  (1)   (2)   (3)   (4)   5  5   25   25  16. 60.2 × 2160.6 = ? (1) 6 (2) 36 (3) 0.36 (4) 60.36

17.

1.

(3) –117 (4) 121

225  224  251  250  23  35 ( 325  324  351  350  33  210 x 1

2 x 10

(a) 11

11

 3  3 9   (ii)     2 2      4 then the value of ‘x’

(iii) 3x + 3 = 81x -2, then ‘x’

)

1 32 Column II

(

) (b)

(

) (c) (d)

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Which of the following letters given on the number line represents the solution correctly? (A) E (B) F (C) G (D) H 6 4 3 2. Among the exponents 2 , 3 , 4 , and 62, which is the greatest? (A) 26 (B) 34 (C) 43 (D) 62 3. Which is the greatest prime factor of 2710 –2435? (A) 2 (B) 3 (C) 11 (D) 13 4. If a 1  × b2  × c 3  =  6000,  where a, b, and c are natural numbers, then what is the value of a × b × c? (A) 30 (B) 60 (C) 100 (D) 120 5. Which of the following expressions is an equivalent form of the expression 3x + 1 × 4x? (A) 3 × 12x (B) 3 × 6x + 1 (C) 4 × 9x (D) 4 × 18x – 1 6. If 523 + 523 + 523 + 523 + 523 = 5x, then what is the value of x? (A) 24 (B) 25 (C) 115 (D) 116 7. How can the number 10800 be expressed in the form of exponents? (A) 23 × 32 × 52 (B) 22 × 33 × 54 3 2 (C) 2  × 3  × 5 (D) 24 × 33 × 52 8. What is the prime factorisation of the number 360? (A) 22 × 32 × 52 (B) 22 × 33 × 5 3 2 (C) 2  × 3  × 5 (D) 22 × 32 × 52 9. The distance between the Sun and Jupiter is 77,80,00,000 km. What is the scientific notation (standard form) of this distance? (A) 7.78 ×108 (B) 77.8 ×107 (C) 0.778 ×109 (D) 778 ×106 10. How can the number 90252 be expanded? (A) 9 × 104 + 2 × 102 + 5 × 101 + 2 × 100 (B) 9 × 104 + 2 × 102 + 5 × 101 + 2 + 100 (C) 9 × 103 +10 × 102 +2 × 101 +5 × 100 +2 + 1 (D) 9 × 104 +10×103 +2 ×102 +5 × 101 +2 +100 11. What is the value of the expression {– (–3) × (–3) × (–3) × (–3) – (–3) × (–3) × (–3) × (–3) – (–3) × (–3) × (–3) × (–3)}? (A) (–3)5 (B) (–3)4 (C) (–3)12 (D) 12 (–3)3 12. Which expression is equivalent to the expression (32 + 33 + 34 + 35)? (A) 33 × 2 × 5 (B) 32 × 22 × 5 (C) 32 × 23 × 5 (D) 33 × 2 × 52

Exponents and Powers

37

13. What is the value of (0.6)3? (A) 0.216 (B) 21.6 (C) 2.16 (D) 0.0216 14. Which of the following expressions is equivalent to 7 0

3  the expression 15.

16.

17.

18.

19.

20.

21.

22.

 35

? 32 (A) 0 (B) 33 (C) 310 (D) 333 Which of the following expressions is equivalent to the expression (–2)6? (A) 43 (B) 62 (C) (–6)–2 (D) (–4)–3 Which expression is equivalent to the expression (25 + 3 × 25 + 27)? (A) 28 (B) 82 (C) 215 (D) 43 Which expression is equivalent to the expression {24 ×(2 × 12) × (4 × 6) × (8 × 3) × (3 × 4 × 2)}? (A) 4 × 24 (B) 5 × 24 (C) 244 (D) 245 The value of ‘ y’ that satisfies the equation 5w × 7x × 2y × 3z = 363 × 352 × 302 ×21 is (A) 4 (B) 5 (C) 6 (D) 8 What will be the value of z satisfying the equation 7x × 9y × 4z = 144 × 68? (A) 4 (B) 5 (C) 6 (D) 7 Mike decided to carry out a calculation. He would hop a pawn across all 64 squares of a chess board. Starting with 1 for the first square, he would multiply the previous result with 2, for every square hopped, i.e. after hopping the second square, the result becomes 1 × 2 = 2; after hopping the third square, the result becomes 2 × 2 = 4; and so on. The result obtained by dividing the result after the 63rd hop by the result after the 61st hop is (A) 2 (B) 4 (C) 6 (D) 8 A class of bacteria multiplies by a factor of 10 everyday. The count of bacteria at the end of the first day was 3.27 × 103. The count of the bacteria at the end of the fifth day will be (A) 32,700 (B) 3,270,000 (C) 32,700,000 (D) 327,000,000 The number 9.387 2 × 107 can be written in standard form as (A) 0.000 000 938 72 (B) 0.009 387 2 (C) 9,387,200 (D) 93,872,000

1 1 23. If the value of x is   and the value of y is  , then 2 2 the value of the expression(4x3 + 5y2)2  is (A)

49 32

(B)

49 16

(C)

49 8

(D)

49 4

24. The number of prime factors in [(6)12 ×  (35)28 × (15)16]/[(3)16 × (5)43 × (14)12 × (21)11 × (7)4] is (A) 15 (B) 7 (C) 5 (D) 3 25. If a =  5  then  [a(K +  4) – a×aK]/[a×a(K +  3)]+(1/a)3 (A) 1 (B) 2 (C) 52 (D) (1/5)2 x  26. If 64 × 256 × 1024 = 4 , then x is (A) 16 (B) 14 (C) 12 (D) 11 27. If a=3 and   4 , then the value of a   a is (A) 337 (B) 145 (C) 91 (D) 24 28. What is the value of the expression 213  104 ?? 152  8  343 (A) 35 (B) 75

(C) 150

(D) 180

HOTS Worksheet 1.

2.

The mass of the planet Venus is approximately 4,869,000,000,000,000,000,000,000 kg. The mass of Venus can be expressed in the standard form as (A) 486.9 × 1024 kg (B) 486.9 × 1021 kg (C) 4.869 × 1024 kg (D) 4.869 × 1021 kg The information given in which alternative is incorrectly matched? (A)

Number Exponent ial form - 144

(B)

Number Exponent ial form 1296

(C)

(- 6)4

Number Exponent ial form - 15625

(D)

- 42 × 32

(- 5)6

Number Exponent ial form 83

512 3.

What is the value of the 2

3

2 2  5   1  expression 12  5      ?  13   5 

1 1 (D) 5 13 What expression will be obtained when (2p)5 × (3q)3 × 62 × 5 is divided by 10p ×(3 pq)2 ×36? (A) 16pq (B) 24pq (C) 30p2q (D) 48 p2q (A) 25

4.

(B) 5

(C)

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7th Class Mathematics

38 5.

The number 21384 can be written in an exponential form as x ×  2y × y5. What are the respective values of x and y? (A) 4 and 11 (B) 7 and 5 (C) 9 and 7 (D) 11 and 3 2

6.

The expression

2

18a   5b  2b  2  5a   32  36b 

can be

simplified as

7.

8.

4 4b2 (B) 10b2 (C) b (D) 10b 5 5 What is the expanded form of 5020016.308? (A) 5×106+2×104+1×101+6×100+3×10–1+8×10–3 (B) 5×106+2×104+1×101+ 6×100+3×10–1+8×10–2 (C) 5×106+2×104+1×102+6×101+3×10–1+8×10–3 (D) 5×106+2×104+1×102+6×100+3×10–1+8 ×10–

12. Which of the following relations is incorrect? (A) 2 × (25)2 = 410 (B) [(–2)2]5 =  (25)2 (C) (241)59 =  (259)41 (D) [(–2)5]3 =  [(­2)3]5 13. How can the expression 3

18ab2   3a    4b  12a 2   ab2 

 be simplified?

2

(A) 16ab2 (C) 216 3

2

(B) 24a2b (D) 648 3

4

(A)

 25  42  24  10x  y| , then what are 14. If    2 140  36  the respective values of x and y? (A) 2 and 84 (B) 2 and 21 (C) 4 and 42 (D) 4 and 14 15. What is the value of the expression

3

278  96  27  811 ? 3616  25  32  18 (A) 1 (B) 2 (C) 4 (D) 8 16. What is the value of the expression

What is the simplified form of the expression 242  94  8  a 5 b6 ? 122  162  81  a 3 b 2

243 2 4 a b (A) 16

2 4

(B) 243a b

972 4 2 a b (D) 3888 a4b2 17 9. The number 945 × 100 can be further expanded as (A) 22 × 33 × 53 × 7 (B) 23 × 32 × 53 × 72 (C) 22 × 32 × 5 × 72 (D) 23 × 32 × 52 × 7 10. Which of the following statements is incorrect? (A) The value of (–x)n  is  always  negative, where x is  a  natural  number  and n is  an odd number. (B) The value of x n  is  always  positive, where x and n are  natural  numbers. (C) The value of the expression (xn × yn) is always negative, where x and y are negative integers and n is  a  natural number. (D) The value of the expression (xn × yn) is always positive, where x, y, and n are natural numbers. 11. What is the value of the expression (C)

3

204  54  65  2    ? 1004  43  32  3  (A) 3 (B) 4 (C) 6 www.betoppers.com

(D) 9

184  63  25 ? 3 4  65  2 4 (A) 27 (B) 36 (C) 72 (D) 84 17. What is the value of the expression 984  213  242  43 ? 710  482  123 (A) 14 (B) 24 (C) 28 (D) 38 18. What is the value of the expression 3610 6320 × 3× 239 + ? 99 ×164 2819 × 340 (A) 270 (B) 186 (C) 114 (D) 56 19. What is the value of the expression 812  99 259 169   ? 417  314 1256 224  322 (A) 329 (B) 113 (C) 41 (D) 17 6 20. How can the numbers 2 × 10 , 1.19 × 104, 3.2623 × 105, and 1.41 × 106 be written in descending order? (A) 2×106 > 1.41×106 >3.2623×105>1.19 × 104 (B) 2 × 106 >1.19×104 >3.2623×105>1.41 × 106 (C) 1.41×106 >  3.2623×105 >  2×106>1.19× 104 (D) 1.41×106>2×106>3.2623×105 > 1.19 × 104

Exponents and Powers

39

II. Multiple Correct Answer Type

IIT JEE Worksheet

9.

I.

Single Correct Answer Type

1.

If a,b use whole numbers such that ab = 169 then the value of (a - 3)b - 1? (A) 16 (B) 10 (C) 26 (D) Cannot be determined Which of the following statement is true ?

2.

3

2

3

1 1 1 (A)          3  3  3

6

2 6    3 1 (B)  2  2  3     3 5

3.

  4 2   4 6   5 0   (C)   5    3  (D)      1       6   If 4x + 3 – 4x + 2 = 3, then the value of x3 is?

1 (B) –2 (C) –8 (D) Cannot be determined 16 If abc = 1, then (A)

4.

1 1 1      1 1 1  = ? 1 b  c 1 c  a   1 a  b

5.

1 (A) ab (B) 0 (C) (D) 1 ab If x,y,z are real numbers, then the value of x 2 y2 . y 2 z 2 . z 2 x 2 ?

(A) xyz

1 xyz (C) xyz (D) 1

(B)

0

6.

The value of

(A) 5 7.

(B)

 0.8  1

3

 5  5  1        33   3   7 

11 18

(C)

9 32

1

(D)

?

1 2

l m n l m n The population of india was about 1,027,000,000 in march,2001.In scientific notation the population in (A) 0.1027 × 1010 (B) 1.02 × 109 (C) 102.7 × 107 (D) 1027 × 106

(C) l m – n (D)

Expressed as power of 4 - 32 -8 16

Number (A) (B) (C) (D)

1024 - 1024 - 512 1

Exponential Form 5 4 (- 32)2 (- 8)3 0 16

11. Which of the following relations is incorrect? (A) 25 < 52 (B) 34 < 43 (C) 24 > 42 (D) 32 > 23 12. Which of the following relations is correct? (A) 24 = 42 (B) 36 = 93 (C) 16 = 12 (D) 26 = 62 13. The information in which alternative is correctly matched?

Expression

Value

(A)

25

32

(B)

4

81

(C)

2

5

10

(D)

43

64

3

III. Integer Type 14. What is the exponent of (–11)5 ? 2

2

4 3 15. Value of       3 2 16. What is the base of (10)100 17. Value of (–1)120 =

IV. Matrix Matching (Match the following) 11. Match the following Colunm-I (i) (–1)77 – (–1)74 (ii) If ax = 1 then

Column-II (a) 25×32×53 (b) 26×33×54

the value of ‘x’

1 1  ? n m 1 l 1  l m n (A) l m + n (B) l

8.

  0.1

1

Which of the following relations is correct? (A) (–2)4 = 24 (B) b × a × b × a ×a = a3b2 5 5 (C) (–3) = 3 (D) x4y3 = y3x4 10. The information in which alternative is correctly matched?

(iii) 36,000 is product of powers of thier prime factors (iv) 5x – 1 + 5x + 1 then

(c) –2

(d) –151

the value of ‘x’ (v)

1024  2 x , then the value of’ x’ 4

(e) 2 (f) 25/2 (g) 0 (h) 1

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40

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7th Class Mathematics

Chapter -3

Algebraic Expressions

Learning Outcomes

By the end of this chapter, you will understand • Understanding an expression • Basic definitions related to Algebraic ExpressionsRational numbers • Identifying coefficient and variable • Classification Of Algebraic Expressions • Addition And Subtraction Of Algebraic Expressions • Finding Values Of Algebraic Expressions At Given Points • Formulae And Their Uses

2. Understanding an expression

1. Introduction This is where your work really begins to look and feel like algebra. You are about to begin working with letters of the alphabet as well as numbers and signs of operation. From now on, letters such as x, y and z are just as common as numbers 1, 2 and 3. So what do letters in algebra mean? Until now–-in basic arithmetic–-you work with numbers, each number having a specific value or meaning. A “2” is a 2, for example. A “2” is always a 2–-it is never a 6 and it is never a 10. It’s like that for all numbers. We still use regular arithmetic numbers in algebra, but we also use terms expressed in letters. In algebra, the letter x, for example, can stand for a lot of different values. We can set x equal to 2 in one problem, but then set it equal to 6 in another. The letters in algebra can stand for an endless variety of values and combinations of values. Letters in algebra can even represent other letters. Compare these two expressions ·

Arithmetic expression:  2 + 1

·

Algebraic expression:

x+1

The arithmetic expressions tells us to add 1 to the value of 2. The algebraic expression, however, covers a lot more territory by telling us to add a 1 to any value we choose. So if we let x = 2 in the algebraic expression, it becomes 2 + 1. If we let x = 5, it becomes the same as 5 + 1. You can see that the algebraic version is a lot more flexible than the version.

i)

ii)

A specific numerical value ( such as 2, 4, –6, ¾ ) is called a constant. The value is constant ... unchanging. An algebraic term (such as x, y, a, b, and so on) is called a variable. The value can be varied. Expression

5+y

Constant

iii)

Variable

When constants or variables are connected by operations (such as +, –, x , or ÷ ), you have an expression. Forming an expression We combine constants and variables to make algebraic expressions for this we use four basic operations of mathematics such as addition, subtraction, multiplication and division. The expression 5x + 4 is obtained from the variable x, first by multiplying x by the constant 5 and then adding constant 4 to the product. Let’s see how the following expressions are formed. 4x2 – 5x, 7xy + 5, x2y In 4x2 – 5x, we first obtain x2 by multiplying the variable x by itself and then multiply x2 by 4 to get 4x2. From 4x2 we subtract the product of 5 and x to finally arrive at 4x2 – 5x. In 7xy + 5, we first obtain xy, multiply it by 7 to get 7xy. We add 5 to 7xy to get the expression 7xy + 5 . In x2y, we first obtain x2 by multiplying the variable x by itself and then multiply x2 by y to obtain the expression x2y.

7th Class Mathematics Term Coefficient variable –4x3 –4 x3 2 x 1 x2 –x –1 x –7 –7 none Identifying terms of an expression – Tree Diagram Method Consider an expression : 7mn – 4, 2x2y. Let’s now identify the terms through a tree diagram . We can represent the terms and factors of the terms of an expression by a tree diagram

42

3. Basic definitions related to Algebraic Expressions I.

Term Algebraic expressions contain one or more terms. Terms contain letters and numbers connected by multiplication, division or powers. We combine terms using addition to make expressions. For example, 4x 2 – xy is an expression with two terms, 4x2 and (–xy).

II.

Factors of a Term When numbers and literals are multiplied to form a product then each quantity multiplied is called a factor of the product. • A constant factor is called a numerical factor. • A variable factor is called a literal factor. Example: • In 7ab, the numerical factor is 7 and literal factors are a, b and ab. • In –9x2y, the numerical factor is –9 and the literal factors are x, x2, xy and x2y

II.

(7mn – 4)

III. Coefficients Any of the factors of a term is called the coefficient of the product of other factors. In particular, the constant part is called the numerical. Coefficient of the term and the remaining part is called the literal coefficient of the term. Example:In the term –3xyz Numerical coefficient = – 3 Literal coefficient = xyz Coefficient of x = – 3yz Coefficient of y = – 3xz Coefficient of z = – 3xy

and

Every term in an algebraic expression has a coefficient. The coefficient is the signed numerical part of a term in an algebraic expression that is, the number and the sign (+ or –) that goes with that term. For example, suppose you are working with the following algebraic expression: –4x3 + x2 – x – 7 The table below shows the four terms of this expression, with each term’s coefficient. www.betoppers.com

–4

Factors

7mn

–4

Terms 2x2 y Factors

A term of the expression having no literal factor is called the constant term. Example: In the expression 3x – 7y + 5; the constant term is 5.

coefficient

7mn

2x2 y

Constant Term

4. Identifying variable

Terms

2

x

x

y

Formative Worksheet 1.

2.

3. 4.

5.

Get the algebraic expressions in the following cases using variables, constants and arithmetic operations. (i) The sum of the numbers x and y. (ii) The number which is 7 more than x. (iii) The number which 5 less than y. (iv) The number which is 3 times x. (v) The number which is 1 less than twice of x. Find the terms in the algebraic expression

 xy  2    14xy  3  . 7   Find the factors of (–3x2yz3). Represent the terms and factors of the algebraic expression 7x 2 y –  2xy +  5  through  a  tree diagram. Find the like terms in the algebraic expression x2y  31  6xy  x 2 y . 2 Find the coefficients of pq in  the  following terms. 51x 2 y  21x 2 y 2 

6.

pq 2 , 3pq,15p 2 q 2 , 

31 2 pq 5

Algebraic Expressions

43

III. Trinomial

Conceptive Worksheet 1.

2. 3. 4.

5.

6.

7.

An algebraic expression which contains three terms is called a trinomial. Example: Each one of the expression 7x + 5y – 8z, 5 – abc – a2b , a2 + b2 – 2a2b is a trinomial.

Get the algebraic expression in the following cases using variables, constants and arithmatic operations. (i) Subtraction of z from y. (ii) One half of the sum of numbers x and y. (iii) The number z multiplied by itself. (iv) One fourth of the product of numbers p and q. (v) Numbers x and y both squared and added. (vi) Number 5 added to three times the product of numbers m and n. (vii) Product of numbers y and z subtracted from 10. (viii) Sum of numbers a and b subtracted from their product. Write a number which is half of the sum of x and y. Write the number which is 3 less than the product of x and y. Identify the terms and their factors in the following expressions. Show the terms and factors by the diagram. (A) 4x + 5 (B) 4x2 – 3xy 2 (C) 5x + 10 (D) 2x2 + 5x + 6 Identify, in the following expressions, terms which are not constants. Give their numerical coefficients. (A) xy + 4 (B) 13 – y2 (C) 13 – y + 5y2 (D) 4p2q – 3pq2 + 5. What are the coefficient of x in the following expressions ? (A) 4x – 3y (B) 8 – x + y 2 (C) y x – y (D) 2z – 5xz. What are the coefficients of y in the following expressions ? (A) 4x – 3y (B) – 8 + yz (C) yz2 + 5 (D) my + n

5. Classification Expressions I.

of

IV. Multinomial An algebraic expression containing two or more terms is called a multinomial. Example: Each one of the expression x + 2y , x + 2y + 3z , x + 2y + 3z + 4w is a Multinomial. Note: In general, an expression with one or more terms is called a polynomial. Thus, each of these is a polynomial. Like and Unlike Terms : The terms which have the same algebraic factors, are called like terms and the terms which have different algebraic factors are called unlike terms. For example, the terms 3xy, – 5xy, 2xy etc. are called like terms whereas the terms 5px, 3yz, 4 etc. are called unlike terms. We should follow the following simple steps to decide whether the given terms are like terms (i) Concentrate on the literal or algebraic part of the terms. (ii) Check the variables in the terms. The terms should contain the same variables. (iii) Check the powers of the variables. The powers of the variables must be the same. Note: In deciding like terms, two things do not matter : (A) the numerical coefficients and (B) the order in which the variables occur in the terms.

Algebraic

Formative Worksheet 7.

Monomial An algebraic expression which contains one term is called a monomial. Example: Each one of the expression 5x, 12xy, 3ab2c,–5z2 is a monomial.

II.

Binomial In algebraic expression which contains two terms is called a binomial. Example: Each one of the expression 5 + 2x, 1– 7x2yz , a2 + y2, b 

1 is a binomial. b

8. 9.

Separate monomials, binomials, and trinomials from the following polynomials. (i) x + 7 (ii) 16 2 2 2 (iii) – 21x y z (iv) x2 – 3 (v) 7x + 7y –  6xy (vi) 6xy2 – 2x2y (vii) 4x + xy (viii) 15xy2 – 7 – 2x2y Is the polynomial (5x + 6xy2 –  2x)  a  trinomial? Justify your answer. Group the like terms together from the following: 2xy, – 3x, 5yx, 6x, 15, x, 12.

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7th Class Mathematics

44

Conceptive Worksheet 8.

9.

Classify the following expressions as a monomial, a binomial or a trinomial : (i) 5m (ii) 7xy – 2z 2 (iii) 3z (iv) x + y (v) m – 5n (vi) 4 (vii) mn + 4m (viii) x + y + 1 2 2 (ix) a – b (x) m + n + 10 (xi) ab + a + b (xii) 12x + 3y (xiii) 3x2 – 5x + 2 State with reasons which of the following pairs of terms are not pairs of like terms : (i) 7x, 12y (ii) 15x, – 21x (iii) –4ab, 7ba (iv) 3xy, 3x 2 2 (v) 6xy , 9x y (vi) pq2, – 4pq2 2 (vii) mn , 10mn.

6. Addition and Subtraction of Algebraic Expressions I.

Addition Add similar terms by adding their coefficients and keeping the same variable part. For example, suppose you have the expression 2x + 3x. Remember that 2x is just shorthand for x + x, and 3x means simply x + x + x. So when you add them up, you get the following: = x + x + x + x + x = 5x As you can see, when the variable parts of two terms are the same, you add these terms by adding their coefficients: 2x + 3x = (2 + 3)x. The idea here is roughly similar to the idea that 2 apples + 3 apples = 5 apples. You cannot add non-similar terms. Here are some cases in which the variables or their exponents are different. 2x + 3y 2yz + 3y 2x2 + 3x In these cases, you cannot perform the addition. You are faced with a situation that is similar to 2 apples + 3 oranges. Because the units (apples and oranges) are different, you cannot solve the problem.

II.

Subtraction

Subtraction works much the same as addition. Subtract similar terms by finding the difference between their coefficients and keeping the same variable part. For example, suppose you have 3x – x. Recall that 3x is simply shorthand for x+ x + x. So www.betoppers.com

doing this subtraction gives you the following. x + x + x – x = 2x No big surpr ises here. You simply find (3 – 1)x. This time, the idea roughly parallel the idea that Rs. 3 – Rs. 1 = Rs. 2. Here is another example. 2x – 5x Again, no problem, as long as you know how to work with negative numbers. Just find the difference between the coefficients: = (2 – 5)x = –3x In this case, recall that Rs. 2 – Rs. 5 = Rs. –3 (that is, a debt of Rs. 3). You cannot subtract non-similar terms. For example, you cannot subtract either of the following: 7x – 4y 7x2y – 4xy2 As with addition, you cannot do subtraction with different variables.

Formative Worksheet 10. Add (i) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5 (ii) 4x2y, – 3xy2, – 5xy2, 5x2y (iii) 3p2q2 – 4pq + 5, – 10p2q2, 15 + 9pq + 7p2q2 (iv) ab – 4a, 4b – ab, 4a – 4b (v) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2 11. Which expression when subtracted from the expression (7x – 3y + 45xy + 7) gives (2x – 21y – 42xy)? 12. Subtract the sum of (4y2–6y) and (–2y2+3y– 3) from the sum of (5y + 7)and (3y2 – 9y + 2).

Conceptive Worksheet 10. Add (i) (ii) (iii) (iv) (v)

3mn, – 5mn, 8mn, – 4mn t – 8 tz, 3 tz – z, z – t – 7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3 a + b – 3, b – a + 3, a – b + 3 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy 11. Subtract : (i) – 5y2 from y2 (ii) 6xy from – 12xy (iii) (a – b) from (a + b) (iv) a (b – 5) from b (5 – a) (v) – m2 + 5mn from 4m2 – 3mn + 8 (vi) – x2 + 10x – 5 from 5x – 10 (vii) 5a2 – 7ab + 5b2 from 3ab – 2a2 – 2b2 (viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq

Algebraic Expressions

45

7. Finding Values Of Algebraic Expressions At Given Points We know that the algebraic numbers represent numbers. Thus if the values of algebraic numbers is given, we can find the value of the given expression. For example, if x = 3, y = 5, then to find the value of the expression 2x + 3y, we put x = 3 and y = 5 in 2x + 3y.  The required value of given expression = 2 (C) + 3 (5) = 6 + 15 = 21

Formative Worksheet 13. Find the value of the following expressions. (i)

x3 

x2  8 at x = 5 25

1 (ii) 4k – 2k + 1 at k  2 (iii) 2x 3 y 2 – 4x – 3xy – 5(x 2 y 2 – x 2 ) when x = 5, y = – 2 (iv) At p = – 2, the value of the expression is 12. Find the value of k. Simplify the expressions and find the value if x is equal to 2. (i) x + 7 + 4 (x – 5) (ii) 3 (x + 2) + 5x – 7 (iii) 6x + 5 (x – 2) (iv) 4 (2x – 1) + 3x + 11 Simplify these expressions and find their value if x = 3, a = – 1, b = – 2. (i) 3x –5 – x + 9 (ii) 2 – 8x + 4x + 4 (iii) 3a + 5 – 8a + 1 (iv) 10 – 3b – 4 – 5b (v) 2a – 2b – 4 –5 + a (A) If z = 10, find the value of z3 – 3 (z – 10). (B) If p = – 10, find the value of p2 – 2p –100. What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0 ? Simplify the expression and find its value when a = 5 and b = – 3. 2 (a2 + ab) + 3 – ab. 2

14.

15.

16. 17. 18.

Conceptive Worksheet 12. If m = 2, find the value of (i) m – 2 (ii) 3m – 5 (iv) 3m2 – 2m – 7

(iii) 9 – 5m (v)

5m 4 2

13. If p = – 2, find the value of (i) 4p + 7 (ii) – 3p2 + 4p + 7 (iii) – 2p3 – 3p2 + 4p + 7 14. Find the value of the following expression, when x =–1 (i) 2x – 7 (ii) – x + 2 2 (iii) x + 2x + 1 (iv) 2x2 – x – 2

15. If a = 2, b = – 2, find the value of : (i) a2 + b2 (ii) a2 + ab + b2 2 2 (iii) a – b 16. When a = 0, b = – 1, find the value of the given expressions : (i) 2a + 2b (ii) 2a2 + b2 + 1 (iii) 2a2b + 2ab2 + ab (iv) a2 + ab + 2.

8. Formulae and Their uses Formulae We observe that equation A = S2 is useful in easily solving many problems of the same type. Equation which are used frequently to solve problems are called formulae. Formulae state that relationship existing between two (or) more variables.

Subject of the Formulae Formulae are frequently given with one variable standing alone on one side of the equality sign and the value of the variable given in terms of the others. This variable standing alone on one side its called “The subjects of the formula”

Expressing the Principles Stated in words in terms of Symbols Formulae are time saving devices in this age of computerisation and quick decisions. Formulae that you have already learnt give the meaning of long statements, precisely in symbols. Example: The statement that the area of a rectangular field is the product of its length and breadth exprerssed in symbol as A= l×b Where ‘A’ is unit areas. ‘l’ and ‘b’ are units of length and breadth respectively

Stating Formula given Symbolically in Word Form The ability to read a formula given in symbols and state in words, so that they can have proper understanding of the formula and apply it in correct situation. Example: (i) State the formula V = lbh in words. Volume of a rectangular parallelopiped is equal to the product of its length, breadth and height. (ii)

1 A  bh state the formula in words. Area 2 of a triangle is equal to half the product of its base and height

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7th Class Mathematics

46

Changing the Subject of the Formula Average of a finite number of quantities is obtained by dividing the sum of the quantities by their number. This is symbolically written in the form of formula

s . n

as a 

Where a = average, s = Sum of quantities, n = number of quantities. In a 

s , ‘a’ is the subject of the formula. n

This formula is useful to find the value of ‘a’, when the values of ‘s’ and ‘n’ are known. Suppose the values of ‘a’ and ‘n’ are given and ‘s’ is to be found. Then we have to change the formula in such a way that ‘s’ is given in terms of ‘a’ and ‘n’ (or) ‘s’ is the subject of the formula.

a

s n

Multiplying bothsides by ‘n’

an 

s  n ; a × n = s (or) s = an n

This new formula is deduced from the original

s formula. a  n I.e., the subject ‘a’ is changed (or) transformed to subject ‘s’. This type of transformation is useful in deriving other forms (or) variants of the same formula. These new formula are derived (or) auxilary formulae. The following characterstics of subject in a formula have to be noted; • The subject symbol occurs on the left side of the equality sign. • It is written independently without being linked with any other quantities (or) variables • Its coefficient is always one. • All properties used in solving simple equations are also used in transforming the subject of a formula

Framing a Formula To express a given statement in the form of an equation by using literals and mathematical symbols is called framing the formula www.betoppers.com

Example: (i) The rule that the perimeter (p) of a rectangle is twice, the sum of its length (l) and breadth (b), is given by the for mula p = 2(l + b). (ii) The rule that the volume (v) of a cube is equal to the cube of the length of its edge(a), is given by the formula; V = a3 (iii) The rule that the centrigrade tempertature (C) is equal to five-ninths of the difference between fahrenheit (F) and 32, is given by the formula.

5  F  32  9 (iv) A number decreased by 3, is divided by 2 to get 4. This fact may be expressed in the form C

of an equation as

x 3  4 , where ‘x’ is the 2

number. (v) If ‘5’ is added to both the numerator and the denominator of a certain fraction, its value is

x 5 1 1 .This fact is given by the formula y  5  4 . 4 Where ‘x’ is the numerator and ‘y’ is the denominator of the fraction

Formative Worksheet 19. In a right angled triangle the square of the hypotenus is equal to the sum of the square of the other two sides ‘x’ and ‘y’, is given by the formula 20. In the formula Tn = arn - 1, make r as the subject? 21. In the formula s  

a , make ‘r’ as the subject? 1 r

22. In the formula,

1   x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  , where 2 (x1y1)=(1,2), (x2,y2)=(4,5),(x3y3)=(6,8) ? N F 23. In the formula, median (m) = L  2  c , where f L = 50, N = 30, F = 25, f = 24, C = 20?

n  2a   n  1 d  , find the 2 sum of 100 natural numbers? 25. In the formula, mode = 3 median - 2mean, where median = 100, mode = 50, find the mean? 24. In the formula, Sn 

Algebraic Expressions 26. In the formula, A  R 2  r 2 , where  

22 ,R 7

47 23. Volume of a rectangular parallelopiped having length (l), breadth (b), and height (h) is

= 4mr = 3m? 27. In the formula, b 

(A) V =

2ac , derive all the auxilary ac

formulas 28. Match the following Column-I (i)

In C 

Column-II

a ; ac

(a)

c 1 subject bc

make ‘a’ as the (ii) In M 

N P  R  ; 40

(b)

bc as the subject c 1

make ‘R’ (iii) In the formula

1 1 1   , f u v

(c) 0.133 find u, when v = 15 and f = 5 (d) 7.5

lbh

(B) V = 2lbh

(C) V = 3lbh (D) V = lbh 24. The area of a circle, having radius (r) is (A) A = 2 r (B) A = 2 rh (C) A = 2 r2 (D) A = 4 rh 25. A =

d (a +b) in word form

(A) Area of triangle (B) Area of trapezium (C) Area of rhombus (D) Area of rectangle 26. A + B + C + D = 3600, in word form (A) Sum of the angles of a triangle (B) Sum of the angles of a circle (C) Sum of the angles of a cube (D) Sum of the angles of a quadrilateral 27. In the formula Tn = a + (n - 1)d, when Tn = 100, a = 2nd , d = 2, then the value of ‘n’ is (A) 48 (B) 50 (C) 26 (D) 52 28. In the formula t  2

(e)

Np  40m N

(f)

N NP  40M

Conceptive Worksheet 17. The basic unit of length is (A) Square meter (B) Meter (C) Cubic meter (D) Kilometer 18. The basic unit of area is (A) Square meter (B) Meter (C) Cubic meter (D) Kilometer 19. The basic unit of volume is (A) Square meter (B) Meter (C) Cubic meter (D) Cube 20. The area of a square is 28 sq.units, then length of the side is (A) 625 units (B) 5 units (C) 5 Sq.units (D) 625 Sq.units 21. 1 Meter = _____. (A) 10 Centimeters (B) 1000 Centimeters (C) 100 Centimeters(D) 100 Kilometers 22. Area of a triangle (A), having base (b) and height (h) is (A) A = bh (B) A =2bh

and L = 128, then the value of ‘g’ is (A) 8 (B) 128 (C) 512 (D) 32

Summative Worksheet 1.

2. 3.

4.

State the unknown for each of the following situations. (A) Chinnu has a few cats. (B) A sum of money was donated to help the flood victims. Car W is priced at Rs. p. Identify the object and the unknown. Determine whether the following are algebraic terms with one unknown. (A) 5xy (B) 8r (C) 26 Complete the following table: Terms Coefficients Unknown

f 6 Determine if the following pairs are like terms or unlike terms. –11x

5.

h 10 Identify the like terms from the following list. 2a, (A) 0.6 j and 2 g

6.

5b, 7c, 6d, –c, –0.2b, (C) A =

bh

(D) A  bh

1 , when t = 12.56,   3.14 g

(B) 4.5 h and

3 a , 7.2d 4 www.betoppers.com

7th Class Mathematics

48 7.

State any 3 like terms for the following terms.

2 k 3 8. Determine whether the following are algebraic expressions. (A) 2u + 3u (B) 5p – 7 (C) 0.8s – 6t + 4 (D) 11u + 13v – 17w 9. Determine the number of terms in the expressions below. (A) 15a – 7b + 45 (B) 19t + 15t 10. Simplify the following expressions: (A) 6m – (–2m) + 4m (B) 12x + 7y (C) 5p + 9 – 3p – 6q 11. Determine the unknowns in the following algebraic term. –5rstu 12. State any two like terms for the following algebraic terms. (A) 2xy (B) –3p2qs3 (A) 4p

13.

(B)

x cm

y cm

14.

15.

16.

17. 18.

19. 20.

Write an algebraic expression to represent the perimeter of the triangle given in the diagram. State the number of terms in the following alegebraic expressions. (A) 8xy (B) ab + 2bc + 7 Find the sum of each of the following. (A) 6pq and 3pq (B) –2x2y and 5x2y (C) 4ab and (–3hk) Simplify each of the following expressions. (A) 7r2s2 – (–4r2s2) – 3rs2 (B) 6mn + 2xy – 3mn + 5xy – 4 Given that a = 1, b = –2, x = –1 and y = 3, find the value of 2a2b + xy2 – ax. Ravi bought x cans of water at Rs. 1.50 each and y packets of biscuits at Rs. 2.70 each. Which of the following expressions shows the total amount Ravi paid, in Rs.? (A) 150 x + 270y (B) 1.5x + 2.7y (C) 15x + 27y (D) 0.15x + 0.27y Simplify: (5xy + 3z2) + (2xy – 4z2) Simplify: (3p2k – 2mn) –(rs – 3mn –p2k)

21. l  x  2r , in word form? 360 (A) Length of a rectangle (B) Length of a parallelogram (C) Length of a curve in a sector www.betoppers.com

22. One third of a number incresed by 11 is 25. Then it may be expressed in the form of an equation is ? (A) 3x + 11 = 25 (B) 3x – 11 = 25

x x  11  25 (D)  11  25 3 3 23. In the formula, p = 2(l + b), make ‘b’ as the subject? (C)

(A) b = p – 2l

p  2l 2

p p l l (D) b  2 2 24. In the formula y = mx + c, makes ‘m’ as the subject? (C) b 

(A) m 

mx c

yc x

(B) m 

y c x 25. In the formula x2 + y2 = r2 , make ‘y’ as the subject? (C) m = (y – c)x

(D) m 

(A) y  r 2  x 2

(B) y = r2 - x2

(C) y  r 2  x 2 (D) y = r2 + x2 26. In the formula y2 = 4ax , make ‘a’ as the subject? (A) a 

y2 4x

(B) a 

y2 x

(C) a 

4y 2 4x

(D) a 

4x y2

HOTS Worksheet 1. 2.

3. 4.

5. 6.

(D) None

(B) b 

7.

If y = 2, z = – 1 and k = 3, evaluate the following: (A) 3yz + kz – 2yk (B) kyz + 2kz2 – yz + 2 A fruit-seller bought m apples at Rs. 30 each and n mangoes at Rs. 55 each. He sold the apples at Rs. 50 each and the mangoes at Rs. 70 each. Write an expression for his total profit. A rectangle has length 2y + z units and breadth 3y – 2z units. Find the perimeter of the rectangle. Anitha is x years old. Her sister is twice her age. What is the sum of their ages?

1 + 2 – 11x 3 A muffin costs Rs. x and a curry puff costs Rs. y. Sunitha buys 4 muffins and 6 curry puffs. If she pays Rs. 10 to the cashier, write an expression for change that she gets. Mallika is x years old. Her sister is (2x + 4) years old. Find the difference in their age. Simplify: 4x –

Algebraic Expressions

49

8.

An adult ticket costs Rs. x and a child ticket cost Rs. y. Ramani buys 3 adult tickets and 5 child tickets. If she pays with a Rs. 50 note, find the change that she gets. 9. Vaishnavi bought x apples at Rs. 0.50 each and y pineapple at Rs. 0.80 each. She sold the apples at Rs. 0.85 each and the pineapple Rs. 1.00 each. Write an expression that correctly shows the profit earned by Vaishnavi? 10. The figure shows a rectangle ABCD. A

16. In vanderwalls equation;  p  n2a     v  nb   nRt; then a =? 2  v  v 2 nRt  Pv  pnb (A)  v  nb  v 2 nRt  Pv  pnb (B)  v  nb 

C

v 2 nRt  Pv  pnb (C)  v  nb  n 2

2x cm 4y cm y cm B D 6x cm Find the perimeter of the unshaded part.

PTR , I = 1000, P = 5000, T = 4years, 100 then the value of R is? (A) 25% (B) 5% (C) 40% (D) 50%

v 2 nRt  Pnb  pnb (D)  v  nb 

IIT JEE Worksheet

11. Given I 

I.

Single Correct Answer Type

1.

Leela spent a few weeks preparing for her final year examination. Based on the given statement, state the unknown quantity involved. (A) Making preparations (B) The number of weeks (C) Final year examination (D) Leela Kiran bought a kg of bananas and b kg of apples from stall P. Which of the letters represent the objects? (A) a, b (B) P, Kiran (C) P, a (D) Kiran, b Bookshop K has c books. Select the letter which represents the object. (A) K (B) c (C) c, K (D) none Factory W produces x units of mobile phones in y days. What are represented by W, x and y? W x y (A) object unknown object (B) object object unknown (C) object unknown unknown (D) unknown object unknown State the unknown and coefficient of – 26f. Unknown Coefficient (A) f –26 (B) –f 26 (C) 26 –f (D) –26 f www.betoppers.com

9 12. In the formula f  C  32 , when C = 30, then 5 the value of ‘f’ is? (A) 113 (B) 59 (C) 76 (D) 86 13. In S 

n [2a + (n – 1)d, make d as the subject? 2

s  an (A) n n  1  

n  n  1 (B) s  an

2  s  an  (C) n n  1  

n  n  1 (D) 2 s  an  

2.

3.

xb 1  , make ‘x’ as the subject? 14. In ab c (A)

a  b  bc c

(B)

c a  b  bc

(C)

c  ab  b2 ab

(D)

ab c  ab  b 2

4.

1 2 15. In S  gt , the value of ‘t’ in the formula, when g 2 = 9.8; and s = 40? (A)

10 7

(B)

20 7

(C)

30 7

(D)

40 7

5.

50 6.

7.

8.

7th Class Mathematics Plantation G has q durian trees of grade r, s durian trees of grade t and u durian trees of grade v. The total number of durian trees in plantation G is : (A) p + q + r + s + t+ u + v (B) p + r + u + v (C) q + s + u (D) r + t + v Which of the following are like terms? –2m, 3n, –0.4p, 6n, 0.4n (A) 0.4n, –0.4p (B) –0.4p, –2m (C) –2m, 3n, 6n (D) 0.4n, 3n, 6n Which of the following is a pair of unlike terms? (A) –5x, 9x

9.

10.

11. 12. 13.

14.

15.

3

(B)  p, 6p 4

(C) 8r, – 3s (D) –8s, 9s State the number of terms in the expression 3x – 4y + 5. (A) 5 (B) 4 (C) 3 (D) 2 Calculate –45m + 20m. (A) m25 (B) – m25 (C) 25m (D) –25m –7f – 6f + 4f = ____________ (A) 9f (B) 17f (C) –9f (D) –17f The sum of –6p and 3p is: (A) –3p (B) 3p (C) –9p (D) 9p 3d subtracted from 7d is: (A) 7d – 3d (B) 7d + 3d (C) 4d + d (D) –7d + 3d –15x + 5x = ____________ (A) –10x (B) –5x (C) 5x (D) 10x 20t – 9 + 9t – 20 = ____________ (A) 11t – 11 (B) 29t – 9 (C) 29t – 29 (D) 29t + 29

II. Multiple Correct Answer Type 16. Which of the following are like algebraic terms? (A) 5y2x, 3xy2, –4y2x (B) –4pq, 0.7pq (C) 6ab, 9ba (D) 5xy2, 3xy 17. Which of the following is a pair of unlike algebraic terms? (A) –pqr, 0.8qrp (B) a2bc, –6ba3c (C) 1.5x2zy, 3xyz (D) –kmn, 48kml 18. Which of the following are monomials? (A) 2x3 (B) 5x + 4 (C) 5y (D) –9x2 + 5y + 4 19. Which of the following are binomials ? (A) 2x3 (B) 5x + 4 (C) 5y3 + 5y (D) –9x2 + 5y + 4

III. Paragraph Type .

Arun distributed ‘x’ apples to his three children. Amar gets ‘a’ apples ;  Suneel gets 3 apples more than Amar;  Prem gets 5 apples less than Suneel.  Based on this information answer the questions given below. 20. How many apples did Suneel get? 21. How many apples did Prem get? 22. How many apples did Prem get less than that of Amar? 23. Express ‘x’ interms of ‘a’. 24. What is the coefficient of ‘a’ in the total number of apples?

IV. Integer Type 25. The coefficient for the term 3x2 is _______.

5m – 4. 2 27. If a = 2, b = – 2, find the value of a2 + b2 28. When a = 0, b = – 1, find the value of the given expressions 2a2 + b2 + 1 26. If m = 2, find the value of

V. Matrix Matching (Match the following) 29. (A) (B) (C) (D)

Column - I x(y – 5) – (x + y) (xy) + (x + y) (x2 + xy + y2) + (2x2 + 3xy) (2x2 – 3xy) – (x2 – xy + 2)

30. Match the following Column-I

n  2a   n  1 d  2 when s = 320; n = 16 and a = 5 (B) If ‘s’ is added to a certain number and the sum is multiplied by 4; the result is 56 (C) 8 years ago, the father was thrice as old as her daughter at that time (A) The value of ‘d’ in S 

Column - II (p) 3x + 4xy + y2 (q) 2x2 + 3xy – y2 (r) x2 – 2xy – 2 (s) 2x (t) –6x + xy – y 2

Column-II (p) x – 8 = 3(y – 8) (q) 8 + x = 56 (r) 4(8 + x) = 56 (s) 2

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Simple Equations Chapter -4

Learning Outcomes

By the end of this chapter, you will understand

•

Simple Linear Equations in one unknown

•

Equations involving Brackets

•

Simple Fractional Equations

•

Forming linear equations to solve problems

1. Simple Linear Equations in one unknown

2. Equations involving Brackets We apply the distributive law of multiplication over addition to help us solve equations involving brackets. Example: Solve the equation 9(x + 1) = 2(3x + 8) Sol. 9(x + 1) = 2(3x + 8) 9x + 9 = 6x + 16 9x + 9 – 6x = 16 3x + 9 = 16 3x = 16 – 9 3x = 7

In our previous classes, we have learnt some number games like + 3 = 8, where we have to fill in the circle with any possible number to make the sentence true. Suppose we replace the circle by the letter x. x+3=8 x + 3 = 8 is called an equation in one unknown x. A solution or root of an equation is a value of the unknown that will make the equation true. For example, x = 5 is a solution of the above equation but x = 1 is not. To solve an equation means to find the solution(s) to the equation. A simple equation the form ax + b = c, where a, b and c are constants and a  0 is called a linear equation. We can relate the idea of balance to a linear equation that help us solve it. Let us illustrate how this can be done by using the equation 2x + 3 = 9. 2x + 3 = 9 2x = 9 – 3 2x = 6

x

6 2

x=3 Hence x = 3 is a solution of the equation 2x + 3 = 9. In general, an equation remains unchanged when both sides are added, subtracted, multiplied or divided by the same number.

x

7 3

Formative Worksheet 1.

Solve the equation x – 5 = 11.

2.

Solve the equation

3.

Solve the equation 8  3 x  2

4.

Solve the equation 3x  2  4 x  7 .

5. 6.

Solve 5 (x - 3) - 7 (6 - x) = 24 - 3 (8 - x) - 3. Solve 5x – (4x – 7) (3x – 5) = 6 – 3 (4x – 9) (x – 1).

7.

Solve 4 

8.

Solve

x 7. 4 5

5

6

x 9 x 1   . 8 22 2

x  4 2x  3 5x  32 x  9    3 35 9 28

9. Solve 0.375x – 1.875 = 0.12x + 1.185. 10. Solve the following equations. (i) 3x + 4 = 2x + 7 (ii) 5x + 8 = 3x – 2 (iii) 2(8x + 5) = 4(3x + 1)

7th Class Mathematics

52 (iv) 3(4x – 1) = 7(2x – 5) (v) 5(x + 3) – 4(2x – 9) = 0 (vi) 3 (3x – 1) – 4(5 – 2x) = –10 (vii) 9x – 2(x + 8) = 5x – 11 (viii) 1 – 4(2x + 3) = 5(x – 2) – 3 (x – 1) (ix)

3(5x  6)  2  4x 4

(x)

2(1  4x)  9  3(2  x) 5

(xi)

7x  2 5x  3  2 3

(xii)

2x  3 3x  15  3 11

Solve the equation x + 6 = 13. Solve the equation –6x = 8.

3.

Solve the equation

4.

y y 3 7. Solve the equation  5 2

5.

Solve 7x – 5{x – [7 – 6(x – 3)]} = 3x + 1.

6.

1 1 Solve 6x + 25  x  1.8  0.75x  . 9 3

5(2x  9)  8  2x . 3

Solve the following equations. (i) x + 8 = 9 (ii) x + 36 = -40 (iii) x – 9 = 5 (iv) x – 22 = –15 x 3 2

(vi)

x  4 5

(vii) 4x = 24 (ix) 2x – 3 = 7

(viii) 9x  21 (x) 3x + 8 = –1

(xi) –5x + 2 = –3

(xii)

(xiii)

1 x 9 6 4

x 2  0 3

1 7

(xiv) 1  x  8

3. Simple Fractional Equations When the unknown of an equation is in the denominator of a term, the equation is called a fractional equation. Examples of fractional equations are: 6 3 x2

and

1 2  . x 3 x

We can use multiplication to transform simple fractional equations into linear equations. In solving fractional equations, it is important to check the solutions. They cannot be those values that make a denominator of the original equation zero. www.betoppers.com

Sol.

6 3 x2

6 3 x2

 6  (x  2)    (x  2)(3)  x2

1. 2.

(v)

Solve the equation

Multiplying both sides by x – 2, we have:

Conceptive Worksheet

7.

Example:

6 = 3 (x – 2) Notice that the denominator (x – 2) 6 = 3x – 6does not appear in this equation 3x = 12 x 

12 3

x 4

4. Forming linear equations to solve problems Sometimes we can use linear equations to represent real-life situations. Then by solving the linear equations, we can actually provide solutions to real-life problems. It is therefore useful to learn how to form linear equations from given information. Here, we shall learn how we can use a linear equation to solve for any unknown quantity. Example: The sum of three consecutive integers is 111. Find the integers. Sol. Step 1: We are going to find the three integers. Step 2: Let x be the smallest integer. Step 3: Middle integer = x + 1 Largest integer = x + 2 Step 4: Sum of 3 integers = 111  x + (x + 1) + (x + 2) = 111 3x + 3 = 111 Step 5: 3x = 111 – 3 3x = 108 x

= 108 3

x = 36 Step 6: The three integers are 36, 37 and 38. 

Simple Equations

53

Formative Worksheet 11. Four quantities x, u, v and t are related by the formula s 

Summative Worksheet

1 (u  v) t 2

Find the values of v given that u = 10, s = 15 and t = 3.

13. The price of a table is Rs. 100 less than 6 times the price of a chair. A similar set of one table and 4 chairs is priced at Rs. 1400. Find the price of a chair.

PQR in which PQ = the value of x. (A) 8 (B) 12 2.

t 1 1 t 3 4

P Q 2xo

10. Tommy is 5 kg heavier than Anne. Let Anne’s mass be x kg.

5x

o

M 5xo

Express Tommy’s mass in terms of x.

(ii) If Tommy’s mass is 63 kg, find Anne’s mass. 11. The price of a watch is Rs. 50 more than twice the price of a gold ring. Let the price of the ring be Rs. x. (i)

(D) 24

Q

4x 3 x x    Solve 5 10 5 4

(i)

(C) 16

P

In the triangle PQR above,  P = 2yo,  Q = yo and  R = 60o. Find the value of y.. (A) 20 (B) 40 (C) 60 (D) 80 3.

9.

3 x and PR = 12 cm. Find 4

R

Conceptive Worksheet Solve the equation

R

Q

The diagram above shows an isosceles triangle

12. Mrs. Kumar’s 3 times as old as her daughter. In 5 years’ time, the sum of their ages will be 62 years. Find the daughter’s present age.

8.

P

1.

R

In the figure above, the value of x is (A) 10 (B) 20 (C) 30 (D) 40

(4x - 7) cm

Express the price of the watch in terms of x.

Q

(ii) If the price of the watch is Rs. 208, find the price of the ring. 12. The sum of three consecutive even integers is 66. Find the three integers. 13. Mrs. Kalyan buys 3 identical boxes of chocolates from a supermarket. She uses a Rs. 50 note to pay for them and gets Rs. 8 change. How much does each box of chocolates cost? 14. A computer shop displays a total of 50 type A and type B mice. The cost of a type A mouse is Rs. 6 and that of a type B mouse is Rs. 13. If the total cost of both types of mice is Rs. 433, find the number of type A mice displayed.

P

4.

5.

6.

7.

(2x + 5) cm R

In the diagram above, PQR is an equilateral triangle. Find the length of QR. (A) 6 (B) 12 (C) 15 (D) 17 Divide n by 4 and then add 11. If the result is 13, what is the value of n? (A) 2 (B) 4 (C) 6 (D) 8 Multiply x by 3 and then subtract 3 from it. If the result is 9, what is the value of x? (A) 3 (B) 4 (C) 5 (D) 6 Start with the number y. Double it and then add 5. If the result is y + 8, find the value of y. (A) 5 (B) 4 (C) 3 (D) 2

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7th Class Mathematics

54

9.

10.

11.

12. 13. 14. 15.

The sum of two consecutive whole numbers is 12. If the smaller number is n, which of the following statements is the correct equation? (A) 2n = 12 (B) n + n – 1 = 12 (C) 2n – n = 12 (D) n + n + 1 = 12 The sum of two consecutive numbers is 19. If the smaller number is m, find the two numbers. (A) 5 and 14 (B) 7 and 12 (C) 8 and 11 (D) 9 and 10 The sum of two consecutive numbers is 13. If the larger of the two numbers is n, find the value of n. (A) 5 (B) 6 (C) 7 (D) 8 The sum of three consecutive even numbers is 24. If the smallest of the three numbers is n, what is the largest of the three numbers? (A) 6 (B) 8 (C) 10 (D) 12 Given that 2m – 5 = 10 – m, then m = (A) 2 (B) 3 (C) 4 (D) 5 Given that 25 = 7y – 3, y = (A) 4 (B) 7 (C) 14 (D) 28 Given that 17 = 3 + 2d, d = (A) 5 (B) 7 (C) 8 (D) 14 If 20 = 8 – 4t, then t = (A) – 2 (B) – 3 (C) 3 (D) 4

16.

S

P

R Q

In the figure above, PQR is a straight line. Given that  SQR = xo and  PQS = (x + 80)o find the value of x. (A) 20 (B) 30 (C) 40 (D) 50 17.

19.

Q

x

o

o

o

2x

R

In the figure above, find the value of x. (A) 15 (B) 18 (C) 25 (D) 35 R

S

xo

18.

130 P

o

Q

In the figure above, find the value of x. (A) 40 (B) 50 (C) 70 (D) 100 www.betoppers.com

Q x

4x

S 6

V

R

In the figure above, PQRS and TURV are rectangles. The lengths shown in the figure are in centimetres. If the perimeter of the unshaded rectangle TURV is 32 cm, find the value of x. (A) 2 (B) 4 (C) 6 (D) 8 20.

Q 5x + 1 P 2x + 13 R

In the figure above, the given lengths of  PQR are in centimetres. If PQ = PR, find the length of QR. (A) 3 (B) 4 (C) 5 (D) 6

HOTS Worksheet 1. 2. 3. 4. 5. 6. 7. 8.

Solve 8 – 3(u + 4) = u Solve 20 = 2k – 5(2k – 3) Solve 5d + 4 = 18 – 2(2 – d) Solve 3(2p – 5) – 1 = 2 – 3p Solve 2 – (2y + 1) = 3(1 – y) Solve x – 3(4 + x) = 8 Solve u – 4(u – 1) = 2(u – 3) Give an equation of the form ax + b = c, where a, b and c are constants, such that the solution of the equation is x = 4.

9.

Give an equation of the form x  a  c , where b

a, b and c are constants, such that the solution of the equation is x = –1. 10. What conclusions can you draw about the solutions of the following equations? (a) 2(3x + 8) = 3(2x + 4)

3x  6  x2 3 11. If x  x always equal to 1? If not, when x  x not equal to 1? Explain briefly. 12. Write an application problem such that the equation to be formed for solving the problem is 5x + 4 (x – 10) = 140. (b)

T

8x

U

P 105x

T

P

2x - 3

8.

Simple Equations

55

13.

x 1 4x 2x 2   3 2 9 3

14.

1 1 5 1 x xx  x 8 6 6 2

IIT JEE Worksheet I.

Single Correct Answer Type

1.

If 5 is added to x, the result is 20. The value of x is (A) 5

15. (x + 1)2 – (x2 – 1) = x(2x + 1) – 2(x + 2) (x + 1) + 20. 16. 2x – 5(3x – 7 (4x – 9)} = 66. 17. 20(2 – x) + 3 (x – 7) – 2[x + 9 – 3 [9 – 4(2 – x)]] = 22.

2.

18. (3x + 1) (2x – 7) = 6(x – 3)2 + 7.

(B) 8

(C) 10

(D) 12

(B) 46

(C) 80

(D) 113

20. 4 (x + 5)2 – (2x + 1)2 = 3 (x – 5) + 180.

The length of a rectangle is y cm and its breadth is (y – 5) cm. If its perimeter is 18 cm, find the value of y.

21. (x + 1) (x + 2) (x + 6) = x3 + 9x2 + 4 (7x – 1).

(A) 7

= (x + 2) (x + 3) + x (x + 4) – 9.

22.

5(x  5) 2(x  3) 19  5 . 8 7 28

23. 1 

x 2x 3x 1   4 . 2 3 4 2

4.

(D) 20

Divide x by 4 and then add 13. If the result is 33, find the value of x. (A) 20

19. x (x + 10 + (x + 1) (x + 2)

(C) 15

The sum of x and x + 3 is 15. The value of x is (A) 6

3.

(B) 10

5.

26.

(C) 28

(D) 32

The length of a rectangle is (p + 3) cm and its breadth is (2p – 5) cm. If its perimeter is 26 cm, find the area of the rectangle is cm2. (A) 13

(B) 40

(C) 48

6.

(D) 56

Q

x 1 x  5 1 x 24. 3    4     11   . 4 2 3  6 3 2 25.

(B) 10

S

P

R

In the figure above, PQ, QR and RS are three straight lines. Given that PQ = x cm, QR =

1 4  x x 1 23  x (x  8)   7 . 5 4 7 5

(x + 2) cm and RS =

2x  5  1 5  x   3x    (2x  57)  . 10  6 3 

1 x cm. If the total length of 2

the three straight lines is 27 cm, what is the value of x?

27. 2.25x – .125 = 3x + 3.75.

(A) 8

(B) 10

(C) 11

(D) 12

28. 12[3x – .25(x – 4) – .3(5x + 14)] = 47. 29.

.25(x  3)  .3(x  4)  5x  19. .125

30.

x  0.75 x  0.25   15 0.125 0.25

P

Q

S

R

7.

31. 1.5 

0.36 0.09x  0.18  . 2 0.9

2 3

 

PQRS is a square. If PQ =  x  1  cm and QR = 9 cm, what is the value of x? (A) 8

(B) 10

(C) 12

(D) 18

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7th Class Mathematics

56 Q 2y

P

III. Paragraph Type o

yo

R

o

70

8. 80o S

The figure above shows a quadrilateral. If the sum of all the angles of the quadrilateral is 360o, what is the value of y? (A) 210 (B) 140 (C) 110 (D) 70 4x

P

Q

6

9.

IV. Integer Type

S

R

3x + 10

The figure above shows a trapezium PQRS and the lengths are in centimetres. If the area of PQRS is 72 cm2, find the value of x. (A) 2 (B) 4 (C) 6 (D) 8 Q

R o

2x

10. P

xo S

In the figure above, PQRS is a parallelogram. Find the value of x. (A) 30 (B) 40 (C) 50 (D) 60

II. Multiple Correct Answer Type 11. 2 is a solution of: (A) x – 1 = 1 (B) x + 1 = 2x – 1 (C) 2x – 5 = 1 (D) x + 1 = 4 12. The equation 2x + 1 = 5 is identical to (A) 3x + 1 = 5 (B) 4x + 2 = 10 (C) x = 4 (D) 6x + 3 = 15 13. The equation 5x + 7 = 8 is the same as: (A) 5x + 7 – 7 = 8 – 7 (B) 5x + 7 – 8 = 8 – 3 (C)

5x 7 8   5 7 8

(D) 5x 

8 7

14. Which of the following statements hold good in the process of solving the equation, 15 + 3x = 3. (A) 3x = 3 – 15 (B) 15 – 3 = –3x (C) 15 +

Mrs. Raksha has a certain sum of money to buy fruits. She can buy n mangoes at Rs. 1.60 each and have Rs. 0.80 left. Alternatively, she can buy (n + 10) apples at Rs. 0.70 each and have Rs. 0.10 left. Based on this information answer the questions given below. 16. Find the value of n. 17. How much money does Mrs. Raksha have for buying fruits? 18. If Mrs. Raksha buys 3 mangoes and uses the rest of the money to buy apples, (a) how many apples can she buy? (b) how much money will she have left?

3x 3 3

(D)

15 3x 3   3 3 3

15. (–1) do not satisfy: (A) 2x – 1 = –3 (B) 2x + 1 = –3 (C) 2x + 1 = 3 (D) 2x – 1 = 3 www.betoppers.com

19. If 15 (z + 4) = 75, then z = ? 20. 1 – 2(m + 1) = –5(m – 1), then m = ? 21. x + 5 = 1 – 2(x – 5), then x = ? 22. y + 3 = 4(y – 3), then y = ?

V. Matrix Matching 23. Match the following for the value of x. Column - I Column-II (A) 3x + 4 = 4 p) 4 (B) 2(x + 4) = 12 q) 4 (C) 4(2 – x) = 8 r) no (D) 4 + 5 (x – 1) = 34 s) x = 7

 

Learning Outcomes

By the end of this chapter, you will understand •

Multiplication properties

•

Algebraic Identities

•

Division properties

•

Special Products

1. Multiplication Multiplication in its primary sense signifies repeated addition. Thus 3 × 4 = 3 taken 4 times = 3 + 3 + 3 + 3. Here the multiplier contains four units, and the number of times we take 3 is the same as the number of units in 4. Again a × b = a taken b times = a + a + a + ..., the number of terms being b. Also 3 × 4 = 4 × 3; and so long as a and b denote positive whole numbers, it is easy to show that a × b = b × a. Before taking up the product of algebraic expressions, let us look at simple rules.

I.

Commutative Law for Multiplication When the quantities to be multiplied together are not positive whole numbers, we may define multiplication as an operation performed on one quantity which when performed on unity produces the other. The reasoning is clearly general, and we may now say that a × b = b × a, where a and b are any positive quantities, integral or fractional. In the same way it easily follows that abc = a × b × c = (a × b) × c = (b × a) × c = bac = b × (a × c) = (b × c) × a = bca; that is, the factors of a product may be taken in any order. This is the commutative law for multiplication.

II.

Associative Law for Multiplication The factors of a product may be grouped in any way we please. Thus abcd = a × b × c × d = (ab) × (cd) = a × (bc) × d = a × (bcd). This is the associative law for multiplication.

Chapter -5

Identities & Special products

III. Index law for multiplication By definition, a3 = a.a.a, and a5 = a.a.a.a.a 3 a × a5 = a.a.a × a.a.a.a.a = a.a.a.a.a.a.a.a = a8 = a3 + 5 ; that is, the index of a letter in a product is the sum of its indices in the factors of the product. This is the index law for multiplication. When the expressions to be multiplied together contain, powers of different letters, a similar method is used. Example: 5a3b2 × 8a2bx3 = 5 a.a.a.b.b × 8.a.a.b.x.x.x = 40a5b3x3. Note: The beginner must be careful to observe that in this process of multiplication the indices of one letter cannot combine in any way with those of another. Thus the expression 40a5b3x3 admits of no further simplification.

IV. Multiplication of two simple expressions To multiply two simple expressions together, multiply the coefficients together and prefix their product to the product of the different letters, giving to each letter an index equal to the sum of the indices that letter has in the separate factors. The rule may be extended to cases where more than two expressions are to be multiplied together. By definition, (a + b)m = m + m + m + ... taken a + b times = (m + m + m + ... taken a times), together with (m + m + m + ... taken b times) = am + bm ...(1) Also (a – b)m = m + m + m + ... taken a – b times = (m + m + m + .... taken a times), diminished by (m + m + m + ... taken b times) = am – bm ... (2) Similarly (a – b + c)m = am – bm + cm. Thus it appears that the product of a compound

7th Class Mathematics

58 expression by a single factor is the algebraic sum of the partial products of each term of the compound expression by that factor. This is known as the distributive law for multiplication. Note: It should be observed that for the present, a, b, c denote positive whole numbers, and that a is supposed greater than b.

V.

descending powers of the common letter, and zero coefficients must be used to represent terms corresponding to missing powers of that letter. Example: Multiply 2x3 – 4x2 –5 by 3x2 + 4x – 2. 2–4+0–5 3+4–2

Multiplication of compound expressions We write c + d for m in (1), we have (a + b) (c + d) = a(c + d) + b(c + d) = (c + d)a + (c + d)b = ac + ad + bc + bd ...(3) Again, from (2) (a – b) (c + d) = a(c + d) – b(c + d) = (c + d) a – (c + d)b = ac + ad – (bc + bd) = ac + ad – bc – bd ...(4) Similarly, by writing c – d for m in (1), (a + b) (c – d) = a(c – d) + b(c – d) = (c – d) a + (c – d)b = ac – ad + bc – bd ...(5) Also, from (2) (a – b) (c – d) = a(c – d) – b(c – d) = (c – d) a – (c – d) b = ac – ad – (bc – bd) = ac – ad – bc + bd ...(6) If we consider each term on the right-hand side of (6) and the way in which it arises, we find that (+ a) × (+ c) = + ac (– b) × (– d) = + bd (– b) × (+ c) = – bc (+ a) × (– d) = – ad These results enables us to state what is known as the rule of signs in multiplication.

6 – 12 + 0 – 15 8 – 16 + 0 – 20 – 4 + 8 – 0 + 10 6 – 4 – 20 – 7 – 20 + 10 Here we insert a zero coefficient to represent the power of x which is absent in the multiplicand. In the product the highest power of x is clearly x5, and the others follow in descending order. Thus the product is 6x5 – 4x4 – 20x3 – 7x2 – 20x + 10.

Formative Worksheet

4. 5. 6. 7. 8. 9.

VI. Rule of Signs The product of two terms with like signs is positive, the product of two terms with unlike signs is negative. Multiplication by a negative quantity indicates that we are to proceed just as if the multiplier were positive, and then change to sign of the product.

VII. The Method of Detached Coefficients When two compound expressions contain powers of one letter only, the labour of multiplication may be lessened by using detached coefficients, that is, by writing down the coefficients only, multiplying them together in the ordinary way, and then inserting the successive powers of the letter at the end of the operation. In using this method the expressions must be arranged according to ascending or www.betoppers.com

Multiply 5x2 + 3 – 4x by 5 – 4x. Multiply 3x3 – 7x + 8 – 2x2 by 4x2 + 6 Find the product by arranging the variable a in descending product order of powers. 3a2 – 4ax + x2 by 5x2 – 2ax + 2a2 Multiply a3 – b3 – 3a2b + 3ab2 by a2 + b2 – 2ab. Find the successive product of (x2 + x + 1), (x2 – x + 1) and (x4 – x2 + 1). Multiply 3x2 – 5 + 2x by 7 + 4x. Multiply 5a3 – 4a – 3 by a2 – 4. Multiply 6x3 – y3 + 3x2y by x2 + y2. Multiply 5a3 + 3ab2 – 2b3 by 2a2 – b2 by the method of detached co-efficient and express the product in the descending order of powers of b.

1. 2. 3.

Conceptive Worksheet 1. 3. 5. 7. 8. 9. 10. 11. 12. 13. 14. 15.

Find the value of 8x3y × y5 2. 4x2 y3 × 7x5 4 3 2 2 5p q × r s 4. abc × xyz 4a3bx × 7b2x4 6. 3a3x4y7 × a2 x5 y8 Multiply together 5p + 3q and 2p2 p2 + q2 – r2 and p3q 5x2y + xy2 – 7x2y2 and 3x3 6p3qr – 7pq2 r2 and p2 q2 – 2abx and – 7abx xyz and – 5x2y3z – 2a2b – 4ab2 and – 7a2b2 – 5xy2z + 3x2yz2 – 8xyz and xyz – p2qr + q2rp – r2pq and – pq

Exponents & Powers 16. 17. 18. 19. 20. 21.

59 Note: If we apply the rule to divide any power of a letter by the same power of the letter we are led to a curious conclusion. Thus, by the rule a3  a3 = a3 – 3 = a0;

6 2 2 3 7 3 px – px3 and px 7 2 3 4 5 3 7 4 x y and x3 – y3 7 4 7 x – 17 and x + 18 – x + 21 and x – 21 2pq – 3r and 2pq + 3r 3x – 5y and 3x – 5y

a3 but also a  a = 3 = 1; a a0 = 1.  Division of Compound Expressions To divide one compound expression by another, use the following steps: Arrange divisor and dividend in ascending or descending powers of some common letter. Divide the term on the left of the dividend by the term on the left of the divisor, and put the result in the quotient. Multiply the whole divisor by this quotient, and put the product under the dividend. Subtract and bring down from the dividend as many terms as may be necessary. Repeat these operations till all the terms from the dividend are brought down. 3

2. Division Dividiend and Divisor

1.

When a quantity a is divided by the quantity b, the quotient is defined to be that which when multiplied by b produces a. This operation of division is

2.

a or a/b; in each of these modes b of expression a is called the dividend, and b the divisor. denoted by a  b,

3. 4.

Rules and signs of division Division is thus the inverse of multiplication, and (a  b) × b = a. This statement may also be expressed verbally as follows : quotient × divisor = dividend. Since division is the inverse of multiplication, it follows that the laws of Commutation, Association, and Distribution, which have been established for Multiplication, hold for Division. The rule of sign holds for division. Thus

ab  a =

ab a  b  = b. a a

– ab  a = ab  (–a) =

ab a  ( b)  = – b. a a ab (a)  (  b)  =–b a a

ab (a)  b  – ab  (–a) = = b. a a Hence in division as well as multiplication like signs produce +, unlike signs produce – . The index of each letter in the quotient is obtained by subtracting the index of that letter in the divisor from that in the dividend. To the result so obtained prefix with its proper sign the quotient of the coefficient of the dividend by that of the divisor.

3

Formative Worksheet Divide 10. 6a2 + 2 – 7x by 3x – 2 11. 17x + 4 – 15x2 by 5x + 1 12. 2a3 – a + 2 – 7a2 by a2 – 3a – 2. 13. 81p4 – 1 by 3p – 1. 14. 4p2 – q2 + 9 – 12p by 2p + q – 3. 15. 2a4 – 10a2 – 72 by a – 3 16 12a2 + 14a + 8 by 3a + 2. 17. 10x2 + 15x – 12 by 5x – 3. 18. 5a3 + 2a – 3 – 4a2 by a2 – 1 + 2a. 19. 12a2 – 31ab + 22b2 by 4a – 5b. 20. Find the quotient and the remainder by the method of detached co-efficients.

Conceptive Worksheet Divide 22. x3y3 by x2y 23. p4 x3 by – p2 x3 24. – 42 x9 by – 8x3 25. 35a11 by 7a7 26. x3 – 3x2 + x by x 27. a2 – ab – ac by – a

1 2 1 3 3 ax– abx – acx by ax 4 10 8 8 29. x2 – 7x + 12 by x – 3 30. 3p2 + 10p + 3 by p + 3 31. 15 + 3a – 7a2 – 4a3 by 5 – 4a 28.

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7th Class Mathematics

60

3. Algebraic Identities An equation is a statement that tells us that two algebraic expressions are equal, Example: 5x + 3 = 2x + 15 This equation is true statement only if x takes the value 4. Any other value for x will not satisfy the equation. An identity is an equation consisting of variables which is true for all values of the variables, Example: 3(a + b) + 5 = 3a + 3b + 5

4. Special Products Certain identities, involving the squares and cubes of binomials, are known as special products. These are applied to compute products in an easier and quicker manner. Identity 1: a(b + c) = ab + ac This is based on the distributive property of multiplication of real numbers. Identity 2:(x + a) (x + b) = x2 + (a + b)x + ab (x + a) (x + b) = x(x + b) + a(x + b) = x2 + bx + ax + ab = x2 + x(a + b) + ab Identity 3 : (a + b)2 = a2 + 2ab + b2 (a + b)2 = (a + b) (a + b) = a(a + b) + b(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 2 Identity 4 : (a – b) = a2 – 2ab + b2 (a – b)2 = (a – b) (a – b) = a(a – b) – b(a – b) = a2 – ab – ab + b2 = a2 – 2ab + b2 Identity 5: (a + b) (a – b) = a(a – b) + b(a – b) = a2 – ab + ab – b2 = a 2 – b2 Identity 6 : (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (a + b + c)2 = (a + b + c) (a + b + c) = a(a + b + c) + b(a + b + c) + c(a + b + c) = a2 + ab + ca + ab + b2 + bc + ca + bc + c2 = a2 + b2 + c2 + 2ab + 2bc + 2ca [on rearranging and collecting like terms] Identity 7 : (a + b)3 = a3 + b3 + 3ab(a + b) (a + b)3 = (a + b) (a + b) (a + b) = (a + b) (a2 + 2ab + b2) = a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 = a3 + b3 + 3a2b + 3ab2 = a3 + b3 + 3ab(a + b) www.betoppers.com

Identity 8: (a – b)3 = a3 – b3 – 3ab(a – b) (a – b)3 = (a – b) (a – b) (a – b) = (a – b) (a2 – 2ab + b2) = a(a2 – 2ab + b2) – b(a2 – 2ab + b2) = a3 – 2a2b + ab2 – a2b + 2ab2 – b3 = a3 – b3 – 3a2b + 3ab2 = a3 – b3 – 3ab(a – b) Identity 9: (x + a)(x + b)(x + c) = x3 +(a + b + c)x2 + (ab + bc + ca)x + abc (x + a)(x + b)(x + c) = (x + a) [x2+(b +c)x + bc] = x[x2 + (b + c)x + bc] + a[x2 + (b + c)x + bc] = x3 + (b + c)x2 + bcx + ax2 + (ab + ca)x + abc = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc Identity 10: Consider the following products: 1. (a + b) (a2 – ab + b2) = a(a2 – ab + b2) + b(a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3 = a 3 + b3 2. (a – b) (a2 + ab + b2) = a(a2 + ab + b2) – b(a2 + ab + b2) = a3 + a2b + ab2 – a2b – ab2 – b3 = a 3 – b3 Identity 11: Let us find the following product: (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = a(a2 + b2 + c2 – ab – bc – ca) + b(a2 + b2 + c2 – ab – bc – ca) + c(a2 + b2 + c2 – ab – bc – ca) = a3 + ab2 + ac2 – a2b – abc – a2c – ab2 + a2b – abc + b3 + bc2 – b2c – c2a – abc + a2c – bc2 + b2c + c3 = a3 + b3 + c3 – 3abc Thus, we have the identity (a + b+ c) (a2 + b2 + c2 – ab – bc – ca) = a3 + b3 + c3 – 3abc This identity helps us to calculate the sum a3 + b3 + c3 when a + b + c = 0 If a + b + c is zero, then you can see that the LHS of the identity is zero. Therefore, a3 + b3 + c3 – 3abc = 0  a3 + b3 + c3 = 3abc

Formative Worksheet 21. Simplify: (0.8 m + 1.1 n) (0.8 m – 1.1 n) 22. Find the value of 7122 – 2882 23. If a + b = 11 and ab = 30, then find the value of (i) (a2 + b2) (ii) (a – b) 24. If x +

1 = 5, then find x

(i) x2 +

1 x2

(ii) x4 +

1 x4

Exponents & Powers

61

25. If a + b + c = 22 and ab + bc + ca = 54, find a2 + b2 + c2. 26. Simplify : 2

2

2

b c ab bc ca   a b c  a  2  8  5   4  64  25  16  40  10     3

3

Summative Worksheet 1.

3

27. Simplify: (a – 3b) + (3b – 5c) + (5c – a) 28. Simplify, the expression by removing brackets: a–2b–[4a–6b–{3a – c+(5a–2b– 3a  c  2b )}]. 29. Simplify the expression: – [ – 2x – {3y – (2x – 3y) + (3x – 2y)} + 2x]. 30. Find the value of

2.

84 – 7 [ – 11x – 4 { – 17x + 3 (8 – 9  5x )}]

Conceptive Worksheet 3.

32. Find the square of : 2

(i) (2x – yz)

x 3 (iii)    3 x 2

2

(A) 4m

(B)

1 2 m 2

(C) 2 mn

(D)

1 m 2

24pq 25pr  5p 2 12q = ____________ (A) 10 pqr

(B) 10r

(C) 10pr

(D) p

– 6 mn2  18 mn = (A) – 3n (B) –3mn (C) –

2 2

(iv) (0.5a + 0.2b)2 1   (vi)  3x  y  5  

(D) –

1 n 3

2

(ii) 197 (v) (59)2

5.

(A) 2n2 (B) 2m2 (C) 18 mn (D) 18m2n2 – 18ac2 × 3b  9abc = –––––––––– (A) – 3ac (B) – 6c (C) – 3bc (D) – 6abc

6.

2ab 5bc  = –––––––––– 25c2 4a 2

2

(iii) (703) (vi) (999)2

9.83  9.83  1.17  1.17 9.83  1.17 (iii) 1.05 × 1.05 – 2 × 1.05 × 0.05 + 0.05 × 0.05 (iv) (101)2 – 992 (v) 593 – (80)3 + (21)3 5p2 – 4p(– p + q) = –––––––– (a + b)2 = –––––––– (a + 2b)2 = –––––––– (a – c)2 – 2ac = –––––––––– (a + b)2 – 4ab = –––––––––– a 2 b2  = –––––––– b2 a 2

1 = –––––––––– 3m

6mn2 ×

(A)

(ii)

40.

1 m 3

4.

2

8.63  8.63  1.37  1.37 (i) 72.6

35. 36. 37. 38. 39.

10 r

2

(v) (p q – qr ) 33. Evaluate : (i) 8.3 × 7.7 (iv) (81)2 34. Simplify :

1  (ii)  a   a 

2m  5mn = ____________ 24n

b b b2 b2 (B) (C) (D) 10a 5ac 7ac 10ac 2

7.

 m  6n = ––––––––––     2n  m

(A) 8.

3m 3m 3 3mn (B) (C) (D) n 2n mn 2

In a school threre are 300 pupils. x of them are girls. What fraction of the pupils are boys? (A)

x 300

(B)

300 300  x

(C)

300 x

(D)

300  x 300 www.betoppers.com

7th Class Mathematics

62

HOTS Worksheet

2y + 4 cm 2 cm 9

y + 2 cm 3 cm

10. 11.

12.

13.

The area inside the figure above is: (A) 7y + 8 cm2 (B) 6y + 10 cm2 (C) 9y + 18 cm2 (D) 7y + 14 cm2 2 2 23 – 22 = ––––––– (A) 45 (B) 55 (C) 23 × 22 (D) 45 + 1 2 102 = ––––––––– (A) 10004 + 200 (B) 10004 + 400 (C) 10004 – 200 (D) 10004 – 400 24 × 26 = –––––––– (A) 200 + 24 + 400 (B) 24 × 400 + 200 (C) 40000 + 20 +24 (D) 40,000 + 200 + 24 (60 – 4) (60 – 6) = –––––––– (A) 1200 – 600–24 (B) 1200 + 600 – 24 (C) 3600 + 600–24 (D) 3600 – 600 + 24 3

1 1  14.  a   = a3 + ––––––––– + 3 27 

9a 2  a (B) 9

a (A) 3a2 + 9

a a 2  9a (C) (D) a2 + 3 9 3 3 15. If a + b = 35, a + b = 5 then ab = –––––––– (A) 6 (B) 7 (C) 8 (D) 9 3 3 16. If a – b = 127, a – b = 1, then ab = ––––– (A) 40 (B) 42 (C) 45 (D) 35 2 17. (a + b + c) = 32 ; ab + bc + ca = 10, then a2 + b2 + c2 = –––––––––– (A) 12 (B) 22 (C) 54 (D) 44 2

b b2   18.  a   2c  = a2 + + 4c2 ––––––––– 2 4  

(A) – 2ab – 2bc + 4ac (B) – ab – bc + 2ac (C) – ab – 2bc + 2ac (D) – ab – 2bc + 4ac 19. Simplify : (i) (3a – 4b) (9a2 + 12ab + 16b2) + (4a – 3b) (16a2 + 12ab + 9b2) (ii) (1 – a + a2) (1 + a) – (1 – a) (1 + a + a2)

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I

Multiply together

1. 2. 3. 4. 5. 6.

x3 – 2x2 – 8 and x + 2 a – 2b + c and a + 2b – c – 3a2b2 + 4ab3 + 15a3b and 5a2b2 + ab3 – 3b4 x2 – xy + x + y2 + y + 1 and x + y – 1 – x3y + y4 + x2y2 + x4 + xy3 and x + y – ax2 + 3axy2 – 9ay4 and – ax – 3ay2

7.

3 2 2 3 1 1 x – ax – a2 and x2 – ax + a2 2 3 4 2 3

8.

1 2 2 3 1 2 3 x – x– and x2 + x – 2 3 4 2 3 4

II

Divide

9. 10. 11. 12. 13.

2y3 – 3y2 – 6y – 1 by 2y2 – 5y – 1 a4 – a3 – 8a2 + 12a – 9 by a2 + 2a – 3 30y + 9 – 71y3 + 28y4 – 35y2 by 4y2 – 13y + 6 x5 – 2x4 – 4x8 + 19x2 by x3 – 7x + 5 a12 + 2a6b6 + b12 by a4 + 2a2 b2 + b4

14.

9 4 3 3 7 2 4 16 3 8 a – a – a + a+ by a2 – – a 16 4 4 3 9 2 3

III

Simplify by removing brackets

15. 16. 17. 18. 19. 20.

– (– (– (– c))) – (– (– y)) – (– (– a)) – (– (– (– x))) 3x – [5y – {6z – (4x – 7y)}] –[15x – {14y – (15z + 12y) – (10x – 15z)}] –[p + {p – (p – q) – (p + q) – p} – p] p + q – (r + p – [q + r – (p + q – {r + p – (q + r – p)})]) y + z – (x + y – [z + x – (y + z – {x + y – (z + x – y)}]) – 4 (x + y) + 24 (y – z) – 2 [z + w + x + z{w + x – 4(y – z)}] p – 2 (q – r) – [– {– (4p – q – r – 2 {p + q + r})}] 8(q – r) – [ – {p – q – 3(r – q + p)}]

21. 22. 23. 24. 25.

1 {x – 5(q – x)} – 4 3 2

1  x 2 3 4x      q    x   q   3 9 4 5    3 

IV. Simplify (856  167)2  (856  167) 2 856  856  167  167 27. (i) (3x – y + 2z)2 – (3x – y – 2z)2 (ii) (x + y + z)2 + (x – y + z)2 + (x + y – z)2 (iii) (2x + 3)3 – (2x – 3)3 (iv) (1003)3 (v) (999)3

26.

Exponents & Powers 28. If x +

1 = x

63

7 find the value of x2 +

29. Find the value of x +

1 x2

1 1 , when x2 + 2 = 27 x x

30. (2x – 3y) = 10, and xy = 8, find the value of 4x2 + 9y2 31. If a2 + b2 + c2 = 79 and ab + bc + ca = 45, find the value of a + b + c 32. If x + y + z = 14, x2 + y2 + z2 = 50 find the value of xy + yz + zx. 33. If x + 34. x – 35. x4 +

1 1  3 , find the value of x3 + 3 x x

1 1 = a, find the value of x3 – 3 x x 1 1 3 4 = 194, find the value of x + x x3

1 1 3 2 = 66, find the value of x – x x3 37. Find the product : (i) (x + 2y + 4z) (x2 + 4y2 + 16z2 – 2xy – 8yz – 4zx) 36. x2 +

3   4 2 3 9 2 2  x  y   x  xy  y  4   25 10 16  5 38. Find the product : (i) (0.7x + 0.9y) (0.49x2 – 0.63xy + 0.81y2) (ii) (x2 + x + 1) (x2 – x + 1) (ii)

IIT JEE Worksheet I. 1. 2.

3.

4.

5.

Single Correct Answer Type 5 2 3 2 4 2 1 2 x  x  x  x  x 2 is equal to: 3 4 3 4 2 2 (A) –5x (B) 4x (C) –6x2 (D) –3x2 When x = 1, y = 2, z = 3, the value of the sum of 5x2, –2x3 z and 3y4 is: (A) 40 (B) 10 (C) –37 (D) 47 A boy attempts x + y sums, of which only y – 2z are correct. The number of wrong sums is: (A) –x + 2z (B) –2y – 2z (C) x + 2z (D) 2y + 2z The value of (–1.4 a2b) × (–0.5abc2) is (A) 7a3 b2c2 (B) –0.07 a3b2c3 3 2 2 (C) 0.7 a b c (D) 70 a2b3c2 (a – b) (a – b) – (a + b) (a – b) is equal to : (A) 2a2 – 2b2 (B) 2b2 – 2ab (C) a2 + 2ab (D) 2b2 + 2ab – 2a2

6.

7.

8.

1 2 at y , the quotient is: 4 (A) –25a5t4y2 (B) 25a3 3 (C) –400 a (D) 400 a5 t4 y2 2a (a – b) –b (3b – 2a) is equal to: (A) 3a2 – 2b2 (B) 2a2 – 3b2 (C) –2a2 + 3b2 (D) –3a2 – 2b2 The degree of the polynomial If –100 a4t2y is divided by

1 2 x 5  y  6y3  4x 2 y3 is: 2 (A) 3 (B) 2 (C) 5 (D) 1 9. (18 a4x2 – 24ax5)  (–6ax2) is equal to: (A) 54a3x2 – 4a2x7 (B) 3a3 – 4x3 (C) –3a3 + 4x2 (D) 4a3 – 3x3 10. The quotient when 10 + x2 – 7 divided by x – 2 is (A) 5 – x (B) x – 5 (C) x + 5 (D) –x – 5

II. Multiple Correct Answer Type 11. Which of the following simplifications results in zero? (A) x (y – z) + y(z – x) + z(x – y) (B) 42a2 – 8a(a – 1) – 7a(1 + 5a) + a(a – 1) (C) x2 (3 – 5y2) + x (xy2 – 3x) – 2y (y – 2x2y) (D) 5a + 7a + 2b + 8b – a + 1 12. Which of the following is correct? (A) (12x + 36)  4 = 3x + 9 (B) (3x2 – x)  (–x) = –3x + 1 (C) 35x2y3  5x3 y2 =

7y x

3a 2 c2 13. Which of the following are correct. (A) (3x – 2) (2x2 + 3x – 5) = 6x3 + 5x2 – 21x + 10 2 (B) (x + y) (x – xy + y2) = x3 + y3. (C) (x + 2) (x2 + x + 1) = x3 + 3x2 + 3x + 2 (D) (x – y) (x2 + xy + y2) = x3 – y3

(D) –18a3bc3  –6abc5 =

III. Paragraph Type Identity: (x + a) (x + b) = x2 + (a + b) x + ab Use the above identity to solve the problem 14. (x + 3) (x + 7) 15. (4x + 5) (4x + 1) 16. (4x – 5) (4x – 1) 17. (4x – 5) (4x – 1) 18. (4x + 5) (4x – 1) 19. (2a2 + 9) (2a2 + 5) 20. (xyz – 4) (xyz – 2)

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7th Class Mathematics

64

IV. Integer Type 21. Divide: –33abc2  (–33abc2) 22. Find the remainder when 9x3 + 3x2 – 5x + 7 is divided by (3x – 1) 23. Find the value of the expression 25x2 + 70x + 49 if x = –1.

V. Matrix Matching (Match the following) 24.

Column - I (A) (x + 3) (x + 3) (B) (2x + 3y) (2x + 3y) (C) (3x2 + 5y2) (3x2 + 5y2) (D) (x – 7) (x – 7)

Column - II (p) x2 – 14x + 49 (q) 16x2 + 4xy + 18y2 (r) 9x4 + 30x2 y2 + 25y4 (s) x2 + 9x + 6 (t) 4x2 + 12xy + 9y2 (u) x2 – 49x + 14 (v) x2 + 6x + 9

 

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Arithmetic

Learning Outcomes

Chapter – 76

By the end of this chapter, you will be understand



Ratio



Profit and Loss



Proportion



Simple Interest



Types of Proportion



Time and Work



Percentage and Applications



Time and Distance

1. Ratio The ratio of two quantities of the same kind in the same units is the fraction that one quantity is of the

a b written as a : b. In the ratio a:b; we call ‘a’ as the first term or antecedent and b, the second term or consequent. Eg: In the ratio 4 : 5 We have first term or antecedent = 4 Second term or consequent = 5 The ratio between two quantities of the same kind and in same units is obtained on dividing the first quantity by the second. Eg: Ratio between 100kg and 150kg = 100 : 150 other. Thus, the ratio ‘a’ is to ‘b’ is the fraction

Ratio in Simplest form or in lowest Term: A ratio a : b is said to be in simplest form, if H.C.F of ‘a’ and ‘b’ is 1. Eg: i) The ratio of 2 : 3 in simplest form, Since H.C.F of 2 and 3 is 1 ii) The ratio 15 : 25 is not in simplest form Since H.C.F of 15 and 25 is 5

15 5  3 3    3:5 25 5  5 5 Rule: To convert a ratio a : b in simplest form, divide ‘a’ and ‘b’ by the H.C.F of a and b. Comparison of Ratios: Since ratios are fractions, they can be compared similar to the way we compare fractions. i.e., by converting them into equivalent like fractions or by the cross product method. 15 : 25 

100 2  =2:3 Types of ratios 150 3 • Ratio is a fraction. It has no units. 1) Compound Ratio 2) Duplicate Ratio • The quantities to be compared for a ratio should 3) Triplicate Ratio 4) Sub-duplicate Ratio be of the same kind. We cannot have ratio 5) Sub-triplicate Ratio between 100kg and 2ml. 6) Reciprocal Ratio • To find a ratio between two quantities of the same kind, both the quantities should be taken 1) Compound Ratio in the same unit. The compound ratio of two ratios a : b and c : d Eg: Ratio between 25ml and 1.5ltr is defined as ac : bd. In other words, compound = Ratio between 25ml and 1500 ml ratio is the ratio of the product of antecedents to the product of consequents of the given 25 1   ratios. or 1 : 60 1500 60 Similarly, compound ratio of If each term of a ratio be multiplied or divided a : b, c : d, e : f, g : h.... by the same non-zero number, the ratio remains is a × c × e × g × ... : b × d × f × h × .... or the same. a× c× e× g.... 2 22 4 . b× d × f × h.... 2:3     4:6 Eg: i) 3 3 2 6 Example: 3 33 1 The compound ratio of ratios 2 : 3, 5 : 6 and 1 : 7 is   1: 3 ii) 3 : 9   9 9 3 3 2  5  1 10  3  6  7 126 

7th Class Mathematics

66 2)

Duplicate Ratio The duplicate ratio of a : b is the ratio of the squares of a and b i.e., a2 : b2. Example: The duplicate ratio of 2 : 3 is 22 : 32 = 4 : 9

3)

Sub - Duplicate Ratio The sub-duplicate ratio of a given ratio is equal to the ratio of square roots of the antecedent and the consequent of the given ratio. The sub-duplicate ratio of a : b is

a: b.

Example: The sub-duplicate ratio of 25 : 49 is

25 : 49  5 : 7 . 4)

Triplicate Ratio The triplicate ratio of a given ratio is the ratio of the cubes of the antecedent and the consequent of the given ratio. The triplicate ratio of a : b is a3 : b3 Example: The triplicate ratio of 2 : 3 is 23 : 33 = 8 : 27.

5)

Example: i) 3 : 5 is commensurable ratio. ii) 2 :1 and ratios.

2. Proportion A statement of equality of two ratios is called a proportion. Eg: We know that 16 : 36 = 4 : 9 We write it as 16 : 36 :: 4 : 9, where the symbol :: stands for ‘is as’ We say that 16, 36, 4, 9 are in proportion. Thus, four quantities a,b,c,d are said to be in proportion if a : b = c : d

a c  b d i.e., if ad = bc i.e., if

Facts about Proportion 1.

Sub - triplicate ratio The sub-triplicate ratio of a given ratio is the ratio of the cube roots of the antecedent and the consequent of the given ratio. The sub-triplicate ratio of a : b is 1

2. 3.

1

a : 3 b or a 3 : b 3 . Example: The sub-triplicate ratio of 27 : 125 is 3

3

6)

Reciprocal ratio

16 : 25 is

1 1 : or 25 : 16. 16 25

c  ac Comparison of two Ratios To compare two ratios writing them in the fraction from and equating the denominator and then compare Note: i) A number ‘x’ can be divided into two parts in the ratio m : n ; then two parts are

Commensurable Ratios If the ratio of any two quantities can be expressed exactly by the ratio of two integers, the quantities are said to be commensurable. Otherwise, they are incommensurable. www.betoppers.com

In a proportion: i) The first and fourth terms are called extremes ii) The second and third terms are known as the means iii) Product of means =product of extremes If a : b : : c : d, then ‘d’ is called the fourth proportional to a,b,c If a : b = b : c, then we say that i) a,b,c are in continued proportion i) ‘c’ is the third proportional to ‘a’ and ‘b’ iii) ‘b’ is the mean proportion between ‘a’ and ‘c’ now, we have a : b = b : c

a b  ; ac = b2 ; b  ac b c Hence, mean proportion between a and

27 : 125  3: 5 . 3

1 1 The reciprocal ratio of a : b is : , which is a b same as b : a. Example: The reciprocal ratio of

3 : 5 are incommensurable

ii)

mx nx , mn mn A number ‘x’ can be divided into three parts in the ratio m : n : p; then three parts are mx nx px , , mn p mnp mn p

Arithmetic

3. Types of Proportion i)

67

iv) Fourth Proportional If a : b = c : d, then ‘d’ is called the fourth proportional to a,b,c.

Direct Proportion If two quantities (x, y) are so related that an increase or decrease in one causes a corresponding increase or decrease in other, then they are said to be in direct proportion or direct variation. Eg: Number of pens of same price, their total cost. price of 4 pens = Rs. 12 price of 8 pens = Rs. 24

x If in two variables x and y; y is constant, then

v)

If a,b,c are in continued proportion, then ‘c’ is called the third proportional. Eg: The third proportional of 3, 9 is 27

4.

Various properties of Ratio and Proportion

1.

Invertendo property If a : b :: c : d then b : a :: d : c.

2.

3.

Inverse Proportion If two quantities are so related that an increase in the former causes a corresponding decrease in the later or vice versa; then they are said to be in an inverse variation or inverse proportion. Eg: Number of men engaged in a work and the time they take to finish it. If two quantities x and y change such that

x

Proof a : b :: c : d 

iii) Compound Proportion Some times, change in a quantity depends upon the changes in two or more other quantities in some proportion. The quantity may be in direct proportion with each of the other quantities. The ratio which shows the change in the quantity is the ratio obtained by compounding of the ratios in which the other quantities change. The quantity may be in inverse proportion with all the others. Then the ratio which shows the change in the quantity is the ratio obtained by compounding the inverse ratio of other quantities.

a c a c   1  1 b d b d

ab cd    a  b  : b ::  c  d  : d. b d Dividendo property If a : b :: c : d then (a – b) : b :: (c – d) : d. or

4.

Proof a : b :: c : d 

1 then ‘x’ and ‘y’ are said to be in inverse y

variation. Then x × y = k (constant).

a c b d a b c b     .  . b d a c b c d c or a b   a : c :: b : d. c d Componendo property If a : b :: c : d then (a + b) : b :: (c + d) : d 

So ‘x’ and ‘y’ are in direct variation.

ii)

a c   b : a :: d : c b d Alternendo property If a : b :: c : d then a : c :: b : d. Proof a : b :: c : d Proof a : b :: c : d 

x and y are said to be in direct variation. We write this as x  y and read ‘x is directly proportional to y’ . Eg: Suppose when x = 3, 5, 8,....... y = 9, 15, 24,......

x 3 5 8 1     y 9 15 24 3

Third Proportional

a c a c   1  1 b d b d

a b cd    a  b  : b ::  c  d  : d . b d Convertendo property If a : b :: c : d then a : (a – b) :: c : (c – d). or

5.

Proof a : b :: c : d 

a c  b d

------(1)

a c a b cd  1   1 or  . -------(2) b d b d Dividing (1) by the corresponding sides of (2), or

a c b  d a b cd b d

or

a c   a :  a  b  :: c :  c  d  a b cd www.betoppers.com

7th Class Mathematics

68 6.

Componendo - dividendo property If a : b :: c : d then (a + b) : (a – b) :: (c + d) : (c – d). Proof a : b :: c : d

For example :

a c a c a c     1   1 and  1   1 b d b d b d



ab cd a b cd   and  b d b d Dividing the corresponding sides, ab cd b  d or a  b  c  d a b cd a b cd b d   a  b  :  a  b  ::  c  d  :  c  d  Writing in algebraic expressions, the componendo - dividendo property gives the following.

7.

a c ab cd    b d a b cd This property is frequently used in simplification. Equivalent ratio property If a : b :: c : d then  a  c  :  b  d  :: a : b and

 a  c  :  b  d  :: c : d Proof a : b :: c : d 

a c   k  say  b d

 a  bk,c  dk. Now, 

a  c bk  dk  bd bd

k b  d a c k  . bd b d

  a  c  :  b  d  :: a : b and  a  c  :  b  d  :: c : d. Algebraically, the property gives the following.

a c a c a c a c      b d b d bd bd Similarly, we can prove that a c a c pa  qc     b d b d pb  qd

a c e a c e ace       b d f b d f bdf 

ap  cq  er bp  dq  fr

a c a c 2a  3c 5a  4c      b d b d 2b  3d 5b  4d ab  cd ,etc b2  d 2

a c e a c e a  2c  3e       b d f b d f b  2d  3f 

4a  3c  9e ,etc 4b  3d  9f

5. Percentage and Applications Percent: It means ‘per hundred’ ‘out of hundred’ we abbreviate percent by P.C and denote it by the symbol %; thus 15 percent is written as 15%.S Hundredth Part: Out of 100 equal parts, each part is known as its hundredth part.

5 11  5 hundredths;  11 hundredths 100 100 and so on. Percentage: By a certain percentage, we mean that many hundredths. Thus

7 13 61 , 13%  , 61%  etc.... 100 100 100 Conversion of Fraction into a Percentage: To express a fraction as a percent, multiply it by 100. Thus, 7% 

3 3     100  %   3  25  %  75% 4 4  Conversion of Percentage into a Fraction: To convert a percentage into a fraction. We divide it by 100 and remove the % sign. Eg:

10 1  100 10 Percent as a Ratio: A percentage can be expressed as a ratio with its second term 100 and first term equal to the given percentage. Eg: 10% 

36 9   9 : 25 100 25 Conversion of Ratio into a Percentage: To express a given ratio as a percent, convert the given ratio into a fraction, and then multiply the fraction by 100. Eg: 36% 

3  Eg: 3 : 5    100  %   3  20  %  60% 5   5  5 : 4    100  %   5  25  %  125% 4 

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Arithmetic

69

Conversion of given Percentage into a Decimal : To express a given percentage as a decimal, divide it by 100 and then convert it into decimal form. Eg: 45% 

Formative Worksheet 1.

45  0.45 100 2.

20 Eg: 20% of 100kg = of 100kg; 100

 20   100  kg ; = 20 kg = 100  

3.

4. 5.

6.

Step1: Take both the quantities in the same units Step2: Taking the quantity to be compared to numerator and the quantity with which it is to be compared as denominator, express it as a fraction. Step3: Express this fraction as percent by multiplying with 100 Percentage Change

8.

The change in value is always calculated on the original value.

the purchase of the flower pots was (A) ` 5.00 (B) ` 7.50 (C) ` 12.50 (D) ` 15.00 1 What percent is equivalent to the fraction ? 5 (A) 15% (B) 20% (C) 25% (D) 30% The decimal equivalent of 75% is (A) 0.75 (B) 0.075 (C) 0.007 5 (D) 0.000 75 The fraction equivalent of 25% is

1 3 1 3 (B) (C) (D) 5 5 4 4 A dishonest milkman has 20 L solution containing milk and water in the ratio 3:1. How much more water should he add to the solution so that the ratio of milk and water becomes 1:2? (A) 12.5 L (B) 20 L (C) 25 L (D) 32.5 L In a particular class, one-third of the students comprises of girls. It is also known that 75% of the girls scored more than 75% in the last exam while 50% of the boys scored less than or equal to 75% in the last exam. If 70 students scored more than 75% in the exam, then how many boys are there in the class? (A) 75 (B) 80 (C) 105 (D) 120 Garfield was a naughty cat. On three occasions, while his master was away, he drank some milk from a bottle. Each time, he drank one-fourth of the contents of the bottle. If the bottle was initially full with milk, then what fraction of the bottle is still filled with milk? (A)

7.

 Increase in Value  Increase % =  Orginal Value  100  %  

Diane purchased 10 flower pots. The cost of 2 flower pots is ` 1.50. The amount spent by Diane on

Expressing one quantity as a Percent of the other

 Decrease in Value  Decrease % =  Orginal Value  100  %  

42 1 (B) 7 (C) 0.42 (D) 4,200% 10 6 The perimeter of a rectangle is 120 m and its length is 36 m. The ratio of the width of the rectangle to its length is (A) 2 : 3 (B) 3 : 2 (C) 4 : 3 (D) 5 : 4 (A)

0.15 15 0.15%    0.0015 100 10000 Conversion of given Decimal into a Percentage: To express a given decimal as a percent, multiply it by 100. Eg: 0.4 = (0.4 x 100)% = 40% 0.004 = (0.004 x 100)% = 0.4% Finding certain Percentage of a given Quantity: Convert the percentage into a fraction and multiply the given quantity with the resulting fraction.

Which of the following values is equivalent to the number 42?

9.

1 9 27 81 (B) (C) (D) 4 16 64 256 10. A number x when added to x% of 700 gives 700. What is the value of x? (A) 68 (B) 76 (C) 87.5 (D) 95.4 www.betoppers.com (A)

7th Class Mathematics

70 11. In a football match, there were 12000 spectators. Out of these, there were 2400 children. What percent of the spectators constituted children? (A) 15% (B) 20% (C) 24% (D) 30% 12. If the length of a rectangle is increased by 25% and the area remains unchanged, then the corresponding breadth must be decreased by (A) 10% (B) 15% (C) 20% (D) 25% 13. The management of a baseball ground wishes to purchases tarp material to cover the ground during the rains. The field measures 120 yards by 50 yards, including the end zones. Experts suggest that the material required to cover the field is exactly 12% more than the area of the field. How much material is required to cover the field? (A) 6,020 yards (B) 6,220 yards (C) 6,520 yards (D) 6,720 yards 14. There are two bottles, A and B. Each bottle has a capacity of 1 L and contains a mixture of juice concentrate and water in different ratios. For bottle A, this ratio is 1:3, while for bottle B the ratio is 3:7. The amount of juice concentrate in (A) bottle A is 50 mL more than that in bottle B (B) bottle B is 50 mL more than that in bottle A (C) bottle A is 100 mL more than that in bottle B (D) bottle B is 100 mL more than that in bottle A 15. A company incurred huge loss due to which it sacked 400 of its employees, out of which, 250 were men. What percentage of women employees lost their job? (A) 32% (B) 34% (C) 35% (D) 37.5%

Conceptive Worksheet 1.

Which of the following values is not an equivalent form of 2.5?

5 1 25 (B) 25% (C) 2 (D) 2 2 10 The percent equivalent of the decimal number 0.25 is (A) 25 (B) 35 (C) 45 (D) 55 If x is 25% more than y, then by what percent is y less than x? (A)

2.

3.

2 (A) 16 % (B) 20 % 3 www.betoppers.com

1 (C) 25 % (D) 33 % 3

4.

Singh Sahib kept 8 L of a solution containing 75% milk and remaining water in the open. As a result, 1 L of water evaporated from the solution. Singh Sahib then added 1 L of a solution containing 75% milk and 25% water to the original solution. What percentage of the new solution is constituted by milk

1 (A) 75% (B) 79 % 8 5.

3 (C) 84 % (D) 90% 8

In one test, Appu scored 40 marks out of 80 marks. In the second test, he scored 30 marks out of 50 marks. It is also known that the maximum marks in the third test are 70. How much should Appu score in the third test such that his average percentage score for the three tests becomes 60%? (A) 35

6.

(B) 40

8.

(D) 55

1 By how much is 200 less than 33 % of 720? 3 (A) 20

7.

(C) 50

(B) 30

(C) 40

(D) 50

A bauxite ore contains 45% of aluminium. What amount of bauxite ore contains 333 kg of aluminium? (A) 675 kg

(B) 690 kg

(C) 740 kg

(D) 755 kg

In a school library, there are 600 books of English

2 and Hindi. The number of Hindi books is 66 % of 3 that of English books. How many books of English are therein the library? (A) 240 9.

(B) 280

(C) 360

(D) 390

The length of a rectangle is increased by 70%. If the area of the rectangle is to be maintained, then the width has to be decreased by (A) 20.00 %

(B) 31.00 %

(C) 41.18 %

(D) 54.79 %

10. On a particular day, 60 students were present in class VII of a particular school. During break time, 50% of the students read comics, 30% played indoor games, and the remaining students played outdoor games. How many students of class VII played outdoor games? (A) 30

(B) 20

(C) 18

(D) 12

Arithmetic

71

6. Profit & Loss Cost Price: The price at which an article is purchased is called its cost price, abbreviated as C.P. The overhead expenses like sales tax, transportation etc..... are always included in the cost price. Net C.P of an article = Actual C.P + overhead expenses

7. Simple Interest Some times, in need, we borrow money from a money-lender or a bank. We promise to pay it back after a specified period of time. At the end of this period, we have to pat the money borrowed along with some additional money in lieu of using another man’s money.

Selling Price: The price at which an article is sold is called its selling price, abbreviated as S.P.

Principle: The money borrowed (or) lent out for a certain period is called the principle (or) the sum.

Profit (or) Gain: If (S.P) > (C.P), then the seller has a gain (or) profit given by

Interest: The additional money paid by the borrower for having used the lenders money is called interest.

Gain = S.P – C.P Gain always reckoned on C.P. Gain on ` 100 is called a gain percent.

Amount: The total money paid back to the lender is called amount. Therefore Amount = Principle + Interest

 Gain  Gain %    100   C.P 

Rate: Interest on ` 100 for 1 year is called rate percent per annum. (Abbreviated as rate% p.a).

Loss: If (S.P) < (C.P), then the seller incurs a loss given by

Thus if rate = 5% per annum, then it means that the interest on ` 100 for 1 year is ` 5.

Loss = (C.P.) – (S.P.) If interest is calculated through out the loan period, then the interest is called simple interest, abbreviated as S.I.

Loss is always reckoned on C.P. Loss on ` 100 is called loss percent

 Loss  Loss %    100  C.P  

Finding S.P When C.P. and Gain % (or) loss% are given, S.P is calculated as follows.

Formula for Calculating Simple Interest Let principle = Rs ‘P’ Time = ‘T’ years and Rate = ‘R’% per annum Then, simple interest (S.I.) is given by the formula.

•

 100  Gain %  S.P.     C.P. 100  

 S.I. 

•

 100  Loss %  S.P.     C.P. 100  

Then, P 

Finding C.P When S.P. and Gain % (or) loss% are given, C.P is calculated as follows. C.P. 

•

100 C.P.   S.P. 100  Loss % 

100  S.I 100  S.I 100  S.I ;R ;T  RT PT PR

Note: i)

While calculating the time period between two given dates, the day on which the money is borrowed is not counted for interest while the day on which the money is returned, is counted.

ii)

For converting the time in days into years, we always divide by 365, weather it is an ordinary years (or) a leap year.

100  S.P. 100  Gain %  

•

P R T 100

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7th Class Mathematics

72

8. Time and Work •

If ‘A’ can complete a piece of work in ‘n’ days, then A’s 1 day’s work =

• •

1 n

1 , then time taken by A n to finish the whole work = n days. If a certain number of workers undertake to do a certain job together in a certain number of days for a certain amount, then this amount is distributed among them in the ratio of their one day work. If A’s 1 day’s work =

9. Time and Distance Speed: The distance travelled by a moving body in a unit of time is called its speed.

Speed 

Time 

Distance ; Distance = Speed × Time Time

Distance Speed

Speed in km / hr 

Distance in km Timein hrs

Speed in m / sec 

Distance in metres Timein seconds

Conversion of Speed into other Units i) To convert a speed of km/hr into m/sec;

5 18 To convert a speed of m/sec into km/hr; multiply by

ii)

multiply by

18 5

Uniform and Variable Speeds Uniform Speed: If a moving body covers equal distance in equal intervals of time, its speed is said to be uniform. Eg: Suppose a train covers a 60km in first hour, 60km in second hour, 60km in third hour and so on. We will say that the train is running at a uniform speed of 60km/hr. In this case we assume that the speed is uniform. Variable Speed: if a moving body covers unequal distance in equal intervals of time, its speed is said to be variable. Average Speed:

Average Speed  www.betoppers.com

Total Distance Covered Total Time Taken

Note: i) In crossing a stationary object (like pole (or) a standing man etc) a train has to cover a distance equal to the length of the train. ii) In crossing a platform, a train has to cover a distance equal to the sum of the length of the train and the platform. Average: The term ‘average’ plays a vital role when it comes to comparing the performance of two (or) more groups of individuals with respect to a certain parameter such as number of marks secured by the student of two (or) more classes, number of runs scored by the players of two (or) more terms and so on... We define average as

Average 

Sum of Observations Number of Observations

using the

above formula, we may thus find, a single average value for a set of given individual values. Average thus provides us a single mid-value of the individual scores, that serves to represent the score of the group taken as a whole. From the above formula Sum of given Observations = (Average) × (Number of Observations)

Formative Worksheet 16. A man bought a shirt for ` 700 and sold it at the loss of 15%. At what price was the shirt sold? (A) ` 575 (B) ` 595 (C) ` 620 (D) ` 680 17. Four friends Mohit, Rohit, Sumit, and Amit invested ` 3900, ` 3250, ` 2925, and ` 2925 respectively to run a small business. What percent of total investment is made by Sumit? (A) 20% (B) 22.5% (C) 25% (D) 27.5% 18. Ravi bought a wrist watch for ` 6850 and sold it to Javed at a loss of 20%. Javed then sold the wrist watch to Gopal at a profit of 5%. At what price did Gopal buy the wrist watch from Javed? (A) ` 6480 (B) ` 6210 (C) ` 5754 (D) ` 5480 19. Suman lent some money to her friend for 2 years, for which she received ` 2,500 as interest. If her friend paid back a total of ` 13,750, then what was the rate of interest?

1 1 (A) 9.5% (B) 10 % (C) 10.75% (D) 11 % 7 9

Arithmetic

73

20. Ambar took some loan from a bank for one year at 29. John covers a certain distance cycling at the rate of the rate of 12.5% p.a. If he paid back ` 13,500 to the 10 mph. Alex also takes the same time as John to bank at the end of one year, then what amount did cover a certain distance on his motorcycle at the he borrow? rate of 50 mph. The distances covered by John and (A) ` 10,800 (B) ` 11,200 Alex are in the ratio (C) ` 11,540 (D) ` 12,000 1 1 1 1 (A) (B) (C) (D) 21. A motorcycle dealer bought a second-hand 2 3 4 5 motorcycle and sold it for ` 38,400 at a profit of 20%. At what price did the dealer buy the onceptive orksheet motorcycle? 11. Rajesh bought a scooter at ` 38,000. If he sold it at (A) ` 30,000 (B) ` 31,200 a loss of 20%, then at what price did he sell the (C) ` 32,000 (D) ` 32,400 scooter? 22. At what percentage above the C.P. must an article (A) ` 32650 (B) ` 30400 be marked so as to gain 33% after allowing a (C) ` 29550 (D) ` 28400 customer a discount of 5%? 12. Savita borrowed a sum of ` 12000 at 6% rate of (A) 40% (B) 30% (C) 20% (D) 10% simple interest per annum for 3 years. What is the 23. When a producer allows 36% commission on the amount that she has to pay at the end of the third retail price of his product, he earns a profit of 8.8%. year? What would be his profit percent if the commission (A) ` 12960 (B) ` 13240 is reduced by 24%? (C) ` 14160 (D) ` 15340 (A) 50.6% (B) 49.6% (C) 48.6% (D) 47.6% 24. A sum was put at simple interest at a certain rate for 13. Aditi had two sweaters worth ` 1800 each. She sold one sweater to her friend Simran at a profit of 10% 2 years. Had it been put 3% higher rate, it would and she sold the other sweater to her friend Sanya have fetched ` 72 more. Calculate the sum. at a loss of 20%. Find the total amount of money (A) ` 9 000 (B) ` 1100 that Aditi had after selling both the sweaters? (C) ` 1200 (D) ` 1300 (A) ` 1980 (B) ` 2460 25. The rate of interest on a sum of money is 4% per (C) ` 2840 (D) ` 3420 annum for the first 2 years, 6% per annum for the 14. Appu borrowed some money from Shashank at the next 3 years and 8% per annum for the period rate of 8% p.a. If Appu paid back a total of ` 16,500 beyond 5 years. If the simple interest accrued by the at the end of four years, then what amount did sum for a total period of 8 years is ` 1280. What is Shashank lend to Appu? the sum? (A) ` 12,000 (B) ` 12,500 (A) ` 2500 (B) ` 2560 (C) ` 13,000 (D) ` 13,400 (C) ` 3000 (D) ` 3560 15. Anil bought a washing machine for ` 8,000 and sold it to Dilruba at a profit of 15%. Dilruba, in turn, sold 5 the washing machine to Harsh at a loss of 15%. At 26. Walking of its usual speed, a train is 10 minutes 6 what price did Dilruba sell the washing machine to too late. Find its usual time to cover the journey. Harsh? (A) 50 min (B) 30 min (C) 20 min (D) 10 min (A) ` 7,640 (B) ` 7,820 (C) ` 8,000 (D) ` 8,400 27. If a man walks at the rate of 5 km/hr, he misses a 16. Sunanda buys rice at 10 kg and puts a price tag on ` train by only 7 minutes. However, if he walks at the it to earn a profit of 20%. However, his faulty rate of 6 km/hr, he reaches the station 5 minutes balance shows 1000 gm when it is actually 800 g. before the arrival of the train. Find the distance What is the actual gain percentage? covered by him to reach the station. (A) 50% (B) 40% (C) 30% (D) 20% (A) 2 km (B) 4 km (C) 6 km (D) 8 km 17. An orange seller makes a profit of 20% by selling 28. If a student walks from his house to school at 5 km/ oranges at a certain price. If he charges ` 1.2 higher hr, he is late by 30 minutes. However, if he walks at per orange he would gain 40%. Find the original price 6 km/hr, he is late by 5 minutes only. Find the at which he sold an orange. distance of his school from his house. (A) ` 2 (B) ` 4 (C) ` 4 (D) ` 8 (A) 10.5 km (B) 11.5 km (C) 12.5 km (D) 13.5 km

C

W

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7th Class Mathematics

74 18. A man bought a horse and a carriage for ` 3000. He sold the horse at a gain of 20% and the carriage at a loss of 10%, thereby gaining 2% on the whole. Find the cost of the horse. (A) ` 800 (B) ` 900 (C) ` 1000 (D) ` 1200 19. An uneducated retailer marks all his goods at 50% above the cost price and thinking that he will still make 25% profit, offers a discount of 25% on the marked price. What is his actual profit on the sales? (A) 12.5% (B) 13.5% (C) 14.5% (D) 15.5% 20. A retailer buys 40 pens at the marked price of 36 pens from a wholesaler. If he sells these pens giving a discount of 1%, what is the profit percent? (A) 5% (B) 10% (C) 15% (D) 20%

th

5.

6.

3 of her 5 pocket money. What per cent of her pocket money did she save in that month? (A) 30% (B) 40% (C) 50% (D) 60% What percent of the given figure shaded area in ?

In a particular month, Meena saves

1 1 (A) 33 % (B) 50% (C) 62 % (D) 75% 3 2 7. A person bought an article and sold it at a loss of 10%. If he had bought it for 20% less and sold it for ` 55 more, he would have had a profit of 40%. Find the CP of the article. 22. What annual instalment will discharge a debt of (A) ` 50 (B) ` 200 (C) ` 250 (D) ` 300 1092 due in 3 years at 12% simple interest? ` 8. If a commission of 20% is given on retail price, the (A) ` 300 (B) ` 325 (C) ` 350 (D) ` 400 profit is 60%. Find the profit percentage when the 23. A and B are two stations 350 km apart. A train starts commission is increased by 5% of the retail price. from “A” at 7 a.m. and travels towards B at 40 km/ (A) 25 % (B) 50 % (C) 55 % (D) 70 % hr. Another train starts from B at 8 a.m. and travels towards A at 60 km/hr. At what time do they meet 9. The cost price of an article increases by ` 100. The ? selling price increases by 10%. If the new profit decreases from 15% to 10%, what is the original (A) 9.00 Am (B) 9.20 Am cost price? (C) 9.40 Am (D) 9.54 Am (A) ` 666.66 (B) ` 777.77 24. Excluding stoppages, the speed of bus is 54 km/hr (C) ` 888.88 (D) ` 999.99 and including stoppages, it is 45 km/hr. For how many 10. The profit earned when an article is sold for ` 800 minutes does the bus stop per hour ? is 20 times the loss incurred when it is sold for (A) 5 min (B) 10 min (C) 15 min (D) 20 min ` 275. At what price should the article be sold if it is desired to make a profit of 25%? (A) ` 275 (B) ` 300 ummative orksheet (C) ` 350 (D) ` 375 1. Which fraction is equivalent to 70%? 11. A certain sum of money amounts to ` 1560 in 2 years 3 5 7 9 and ` 2100 in 5 years. Find the rate percentage per (A) (B) (C) (D) annum. 10 10 10 10 (A) 5 % (B) 10 % (C) 15 % (D) 20 % 1 2. The percent equivalent of the fraction is 12. A certain sum of money amounts to ` 1008 in 2 years 2 1 (A) 30 (B) 50 (C) 70 (D) 90 and to ` 1164 in 3 years. Find the sum and the rate 2 3. If x: y = 2: 3 and y: z = 2: 3, then what is the ratio of interest. between x and z? (A) 1 : 1 (B) 2 : 3 (C) 3 : 8 (D) 4 : 9 (A) 10 % (B) 11 % (C) 12 % (D) 13 % 4. If 75% of a number is 24, then what is 125% of the number? (A) 6 (B) 8 (C) 30 (D) 40 21. Rs. 800 amounts to ` 920 in 3 years at simple interest. If the interest is increased by 3%, it would amount to how much? (A) ` 900 (B) ` 910 (C) ` 992 (D) ` 1000

S

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W

Arithmetic

75

13. A man borrowed ` 24000 from two money lenders. For one loan he paid 15% annum and for the other 18% per annum. At the end of one year, he paid simple interest of ` 4050. How much did he borrow at each rate? (A) ` 15000

(B) ` 16000

(C) ` 17000

(D) ` 18000

14. A is twice as fast as B and B is thrice as fast as C. The journey covered by C in 54 minutes will be covered by B in : (A) 18 min

6.

7.

1 1 1 1 (A) 10 % (B) 11 % (C) 12 % (D) 13 % 9 9 9 9 A person incurs 5% loss by selling a watch for ` 1140. At what price should the watch be sold to earn 5% profit?

8.

(A) ` 1060 (B) ` 1160 (C) ` 1260 (D) ` 1360 A book was sold for ` 27.50 with a profit of 10%. If it were sold for ` 25.75, then what would have been the percentage of profit or loss?

(B) 27 min

(C) 38 min (D) 9 min 15. What is the average speed if a person travels at the 9. speed of 20 km/hr and 30 km/hr ? (a) for the equal interval of time. (b) for equal distance. (A) 25 km/hr, 24 km/hr 10. (B) 24 km/hr, 23 km/hr (C) 23 km/hr, 22 km/hr (D) 22 km/hr, 20 km/hr

HOTS Worksheet 1.

Which fraction is equivalent to 40%?

1 2 3 4 (B) (C) (D) 5 5 5 5 In an exam, the passing marks were 40% of the maximum marks. A boy scored 200 marks, which were 20 marks more than the passing marks. What were the maximum marks in the exam? (A) 425 (B) 450 (C) 500 (D) 525 In an examination, a student has to score 33% marks to pass. Raju secured 82 marks and he failed by 83 marks. What is the maximum number of marks in the examination? (A) 500 (B) 450 (C) 300 (D) 250 A company incurred huge loss due to which it sacked 400 of its employees, out of which, 250 were men. What percentage of women employees lost their job? (A) 32% (B) 34.5% (C) 35% (D) 37.5% In a school, a survey was conducted and it was found that 45% of students like to play only football, 35% of students like to play only cricket, remaining students like to play only badminton. How many students like to play only cricket? (A) 350 (B) 450 (C) 700 (D) 900 (A)

2.

3.

4.

5.

Shyam bought a T.V. for ` 4000 and spent ` 500 on its repairs. He later sold it for ` 5000. Find his gain %.

(A) 2 %

(B) 3 %

(C) 4 %

(D) 5 %

By selling 33 metres of cloth, one gains the selling price of 11 metres. Find the gain percent. (A) 25 %

(B) 50 % (C) 75 %

(D) 85 %

A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for ` 10.50 less, he would have gained 30%. Find the cost price of the article. (A) 40

(B) 45

(C) 50

(D) 55

11. A person deposits ` 25000 in a firm who pays an interest at the rate of 20% per annum. Calculate the income he gets, from it annually. (A) ` 5000 (B) ` 5500 (C) ` 6000 (D) ` 6500 12. Find the interest on ` 2400 at 6% per annum for 146 days. (A) ` 54.60 (B) ` 55.60 (C) ` 56.60 (D) ` 57.60 13

3 Calculate the simple interest on ` 7200 at 12 % per 4 annum for 9 months. (A) ` 688.50 (B) ` 788.50 (C) ` 888.50 (D) ` 999.50

14. At what rate per cent per annum will a sum of money double in 8 years? (A) 10.5% (B) 11.5% (C) 12.5% (D) 13.5% 15. The simple interest on a sum of money is 25% of the principal and the rate per annum is equal to number of years. Find the rate per cent. (A) 1 %

(B) 5 %

(C) 10 % (D) 15 %

16. A car covers four successive 3 km stretches at speeds of 10 km/hr, 20 km/hr, 30 km/hr and 60 km/ hr respectively. What is the average speed of the car for the entire journey. (A) 10 km/hr (B) 15 km/hr (C) 20 km/hr (D) 25 km/hr www.betoppers.com

7th Class Mathematics

76 17. A person walks at 4 km/hr for 6 hours and at 6 km/ hr for 8 hours. What is the average speed of the man for the entire journey.

1 (A) 2 km / hr 7

1 (B) 3 km / hr 7

1 1 (C) 4 km / hr (D) 5 km / hr 7 7 18. The average speed in travelling from A to B and back is 24 km/hr. If the speed in travelling from A to B is 30 km/hr, find the speed in travelling from B to A. (A) 20 km/hr (B) 30 km/hr (C) 40 km/hr (D) 50 km/hr

IIT JEE Worksheet I.

Single Correct Answer Type

1.

There are 225 boxes of 8 L capacity and 515 boxes of 27 L capacity. The percentage of the volume of the 8 L boxes with respect to the total volume of all the boxes is (A) 13.64% (B) 11.46% (C) 9.64% (D) 7.46% If 35% of a number is 210, then what is the number? (A) 450 (B) 500 (C) 550 (D) 600 Five litres of petrol costs ` 255. How much does Mani need to pay to fill up his car’s petrol tank, if his petrol tank has a capacity of 24 L? (A) ` 1184 (B) ` 1192 (C) ` 1208 (D) ` 1224 Rajat pays ` 42,000 as house rent for one year. If the rent for each month is same, then what is the rent for three months? (A) ` 9,900 (B) ` 10,200

2. 3.

4.

(C) ` 10,500 5.

(D) ` 10,800

Which of the following expressions is correct?

7 1 (A)  3 4 4

18  3.5 (B) 5

78 (D) 12.5% = 0.125 5 Which of the following values is not equivalent to 304%? (C) 15.2 

6.

1 76 (B) 3.04 (C) (D) 3.4 25 25 7. At what price must Anil sell his cycle, which costs him ` 600, so as to make a profit of 25%. (A) ` 700 (B) ` 725 (C) ` 750 (D) ` 775 www.betoppers.com (A) 3

8.

9.

A man sells his scooter for ` 3000 making a profit of 20%. How much did the scooter cost him? (A) ` 2500 (B) ` 3000 (C) ` 3500 (D) ` 4000 A man borrowed ` 8000 from a bank at 8% per annum. Find the amount he has to pay after 4

1 2

years? (A) ` 10080 (B) ` 10880 (C) ` 11800 (D) ` 11080 10. Neena borrowed ` 900 from her friend at 10% per annum. She returned the amount after six months. How much amount did she pay? (A) ` 900 (B) ` 945 (C) ` 965 (D) ` 985

II. Multi Correct Answer Type 11. Which of the following values is equivalent to the fraction

28 ? 14

10 1 (D) 2 5 14 12. Which of the following expressions correct? (A) 2

(A) 2

(B) 200%

(C)

13 47  17 17

(B)

67 7 8 8 8

9 25 (D) 12.5  200 2 13. Which of the following values is equivalent form of (C) 4.5% 

1 the mixed fraction 25 ? 4 101 404 (D) 4 16 14. Which of the following representation are incorrect of the whole number 22 is (A) 25.25

(B) 25%

264 12

(C)

1 (D) 20% 3 15. Which of the following relations is incorrect? (A) 15% of 1 h = 12 min (A) 21.3

(B)

(C) 7

2 of 160 + 30 5 (C) 15% of one-third of 220 = 12 (D) Three-fourth of 1 km = 37.5% of 2 km (B) 24% of 350 

Arithmetic

77

III. Paragraph Type

16.

17. 18.

19.

A restaurant marks up the items of his menu and allows further discount on each item in the following way Mark up Discount Soups 20% 10% Sizllers 40% 15% Manchurian 30% NIL Fried rice 30% 10% Noodles 10% 10% Based on this information answer the questions given below. A family who has gone to have a dinner at the restaurant paid bill as follows. Soups ( ` 194.40), Sizller ( ` 105), Manchurian (` 98) and Noodles ` 46. Find the profit earned by the restaurant on this bill. In the above question, find the profit percentage. If the payment made by a customer for 2 manchurian after discount is ` 196, then profit earned on this item is what percentage more than the profit earned on 3 soups for ` 194.40? Find the difference in the profit per cent in the following two payments. (i) Soups (` 129.60), Szillers (` 214.20), Noodles (` 99), Fried rice (` 234) (ii) Soups ( ` 302.40), Szillers ( ` 119), Noodles (173.25)

IV. Integer Type 20. Five hundred millilitres of water is mixed with 2 L of juice concentrate containing 5% pure juice and rest of the amount as water. What is the percentage of pure juice in the mixture so obtained? 21. Hardeep went to a bank and deposited ` 80000 at the rate of simple interest of 7% per annum. After some years, he withdrew 91200 and closed the account. For what time period did he deposit the money in the bank? 22. Mayank deposited ` 14,000 in his bank account, which gives him an interest of 9% p.a. After how many years will the deposit in his account amount to ` 21,560? 23. A trader mixes 26 kg of rice at ` 20 per kg with 30 kg of rice of other variety at ` 36 per kg and sells the mixture at ` 30 per kg. His profit percent is –– ––. 24. A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?

V. Matrix Matching (Match the following) 25.

Column-I (A) The ratio of 43.5 : 24 is the same as (B) The mean proportional between 0.32 and 0.02 is

Column-II (p) 18 (q) 0.08

(C) If a man runs at 3m/sec, how many kilometers does he run in 1hour 40min

(r) 8:1 (s) 1:8

26. (A) The average of all prime numbers between 30 and 50

(B) A and B can do a prime of work in 18 days, B and C can

(p) 2

(q)

1 2

do it in 24 days, A and C can do it in 36 days. In how many days can they do it all working together

x (C) 30% of x = 60% of y, then y 

(r) 39.8 (s) 16

  www.betoppers.com

78

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7th Class Mathematics

Lines and Angles

Learning Outcomes 

Angle



Types of Angles



Parallel Lines



Angles formed when a Transversal cuts Two Lines



Condition of Parallelism of Two Lines

1. Angle

Chapter –78

By the end of this chapter, you will be understand

2. Types of Angles

The inclination between two rays having a common initial point is called an angle. The two rays forming an angle are called the arms (or) sides of the angle. The common end point of the two rays forming an angle is called the vertex of the angle. An angle is denoted by the symbol ‘  ’

Acute Angle An angle whose measure is more than 00 but less than 900 is called an acute angle. In the adjoining C

C 50°

A

A

B

In the above figure, ABC  50  90 . Therefore ABC is an acute angle

B

In the above figure, the rays AB and AC form an BAC or CAB having ‘A’ as vertex and AB,AC as its arms. We may also denote this angle by A . When there are more than one angles in a figure, we label them 1,2,3 etc. and call them 1, 2, 3 etc.

Right Angle An angle whose measure is 90 0 is called a right angle. A

D C

C

B 3 O

In the above figure, ABC  90

B 2 1

Therefore ABC is an right angle

A

Thus, in the above figure COB   2 and COD  3

BOA  1 ,

Measuring Angles The magnitude of an angle is the amount of rotation through which one of the arms must be rotated about the vertex to bring it to the position of the other. A complete turn about a point is divided into 360 equal parts, each called a degree. Written as 1°. The standard unit of measuring an angle is degree, to be denoted by ‘0’. One degree is further divided into 60 equal parts, each called one minute, written as 1 . Thus 1  60 . One minute in further divided into 60 equal parts are called one second, written as 1 . Thus 1  60

Obtuse Angle An angle whose measure is more than 900 but less than 1800 is called an obtuse angle A

.

130° B

C

In the above figure, ABC  130 Clearly 90° < 130° < 180° Therefore ABC is an obtuse angle

Straight Line An angle whose measure is 1800 is called a straight line 180° A

O

B

7th Class Mathematics

80

Reflex Angle

Therefore COA is a straight angle.

An angle whose measure is more than 1800 but less than 3600 is called a reflex angle.

So COA  180

360°

Therefore AOB  BOC  180 Therefore the sum of a linear pair of angles is 1800 The sum of the angles at a point on a line on one side of the line is 1800

O

Complementary Angles

C

D

Two angles are said to be complementary, if the sum of their measures is 900. Each one of these angles is called the complement of the other. Eg: • Complement of 400 = (900 – 400) = 500 • Complement of 500 = (900 – 550) = 350

Supplementary Angles Two angles are said to be supplementary, if the sum of their measures is 1800. These angles is called the supplement of the other. Eg: • Supplement of 800 = (1800 – 800) = 1000 • Supplement of 450 = (1800 – 450) = 1350

3 2 1

4

E

A

O

Thus, in the above figure, we have 1  2  3  4  180

Angles at a Point The sum of all the angles at a point is 3600 . C

Adjacent Angles

2

Two angles having a common vertex and a common arm such that the other arms of the angles are on the opposite sides of the common arm, are known as adjacent angles.

B

D

3 O

4

B

1

A

5

E

C

In the above figure, we have

1  2  3  4  5  360

B

Vertically Opposite Angles A

O

In the given figure, AOB and BOC are adjacent angles, since they have a common vertex ‘O’. A common arm of OB and their other arms OA and OC are on the opposite sides of the common arm OB.

Linear Pair of Angles

B

O

D

A

O

B

C

A pair of adjacent angles is said to be a linear pair, if these angles have one arm common and other arms as opposite rays.

C

When two lines intersect, then two angles having no common arm are called vertically opposite angles.

A

In the above figure, AOB and BOC form a linear pair of angles as they have a common arm OB and their other arms OA and OC are opposite rays. www.betoppers.com

Let AB and CD be two lines intersecting at a point ‘O’. Then AOC and BOD form one pair of vertically opposite angles. AOD and BOC form another pair of vertically opposite angles. When two lines intersect, then vertically opposite angles are always equal. AOC = BOD and AOD = BOC .

Lines and Angles

81

3. Parallel Lines

Two pairs of Co-interior or Cojoined or Allied angles

If two lines are drawn in a plane such that they do not intersect, even when extended indefinitely in both the directions, then such line are called parallel lines. If lines ‘ l’ and ‘m’ are parallel, we shown them by marking arrows on the lines pointing in the same directions and we write,

 3, 5  and  4, 6  form

two pairs of

Co-interior angles.

Results when two Parallel lines are cut buy a Transversal

l read as ‘l’ is parallel to m

Let AB and CD be two parallel lines, cut by a transversal ‘t’. Then,

‘m’

t

l m

1 2 3 4

A

B

5 6 7 8

C

D

Transversal 1.

A line which intersects two distinct lines in a plane in exactly two points, is called a transversal. In the above figure, a line ‘t’ cuts two lines ‘l’ and ‘m’ at exactly two points A and B. So, ‘t’ is a transversal.

2.

t

3.

A

l

4. m

B

4. Angles formed when a Transversal cuts Two Lines

If two straight lines are cut by a transversal such that, • A pair of alternate angles are equal. • A pair of corresponding angles are equal • The sum of the interior angles on the same side of the transversal is two right angles, then the two straight lines are parallel to each other.

t

5 6 7 8

4 5 Exterior alternate angles are equal 1  8 and 2  7 Corresponding angles are equal 1  5, 2  6, 3  7 and  4  8 Co-interior angles are supplementary 3 5  180 and 4  6  180

5. Condition of Parallelism of Two Lines

Suppose a transversal cuts two lines ‘l’ and ‘m’. Then the following angled are formed. 1 2 3 4

Interior alternate angles are equal 3 6 and

l m

Two pairs of Interior alternate angles:

 3, 6  and  4, 5  form two pairs of interior

Formative Worksheet 1.

In the given figure, AB||DE. B

alternate angles. Two pairs of Exterior alternate Angles:

 1, 8  and  2, 7  form two pairs of exterior alternate angles. Four pairs of corresponding Angles

 1, 5  2, 6  3, 7  and  7, 8  form

D

90° A

C

E

Which of the following expressions is equal to a? (A) 25° – b (B) 35° – b (C) 55° – b (D) 65° – b

four pairs of corresponding angles. www.betoppers.com

82 2.

7th Class Mathematics In the given figure, AB||DE. A

7.

In the given figure, AOB is a straight line.

D

A C C

x 29°

34°

E

B

3.

What is the measure of BCE ? (A) 43° (B) 63° (C) 146° (D) 151° In the given figure, two intersecting lines AB and CD are shown. C

3x – 19°

D B

What is the measure of AOD ? 8.

A

The ratio of an angle to its complement is 7: 11. What is the measure of the angle? (A) 35°

2x + 11° O

9.

(B) 55°

B

5.

C 92°

6y + 12° 5y – 8°

D

D

F

6.

z C

(D) 154°

B 44° O 35° D

E

What are the respective measures of ABC and ADE ? (A) 44° and 35° (C) 35° and 136°

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G

B D

If x: y = 17: 19, then what is the value of z? (A) 34° (B) 38° (C) 85° (D) 95° 11. P and Q are two complementary angles such that P = Q – 20°. What are the respective values of P and Q? (A) 20° and 70° (B) 35° and 55° (C) 55° and 35° (D) 70° and 20°

In the given figure, AB||CE.

C

F x y

A

(C) 136°

A

(D) 172°

E

B

(B) 92°

E

What is the value of x? (A) 16° (B) 96° (C) 108° 10. In the given figure, AB||CD.

What is the measure of DOE ? (A) 88°

C x

E

O

(D) 110°

A

B

What is the measure of AOC ? (A) 30° (B) 71° (C) 109° (D) 128° What is the complement of the supplement of the angle of measure 118°? (A) 18° (B) 28° (C) 62° (D) 72° In the given figure, OE bisects BOC . A

(C) 70°

In the given figure, AB||CD and BF||AE.

D

4.

4x

O

(B) 44° and 135° (D) 35° and 156°

12. Two complementary angles are in the ratio 2: 3. What is the difference between the angles? (A) 18°

(B) 36°

(C) 54°

(D) 72°

Lines and Angles

83

13. In the given figure, AB||CD. A m

B

(C) 130° and 70° (D) 130° and 60° 18. In the given figure, the lines AB and CD are parallel to each other.

85°

G

X

E

D

x

n

134°

65° A

C

I

K

P

B M

L

Y C

D

O N What are the respective values of m and n? 120° 60° (A) 46° and 95° (B) 46° and 134° J F H (C) 85° and 95° (D) 85° and 134° Find the value of x from the given figure. 14. Which of the following pair of angles is supplementary? (A) 65° (B) 60° (C) 55° (D) 50° (A) 35° and 55° (B) 63° and 37° 19. In the given figure, two parallel lines AB and CD are cut by two lines WX and YZ. (C) 126° and 54° (D) 131° and 39°

15. The ratio of the measures of XOZ and ZOY is 2: 1.

Y

W A

B

P

Z

55° x

Y

X

R

C

Q

D

O

What are the respective measures of XOZ and ZOY ? (A) 30° and 150° (B) 60° and 120° (C) 120° and 60° (D) 150° and 30° 16. In the given figure, CD = CE, AB || CE, and CED  30 . A

Z

X

If the complement of  PQR is 30°, then what are the respective measures of DRZ and APZ ? (A) 65° and 55° (B) 65° and 60° (C) 115° and 55° (D) 115° and 65° 20. In the given figure, AB||CD and PQ||RS.

B

L

A

C

30°

R

P

M

B

E C

Y 85° D

X

D

What is the measure of ABC ? (A) 75° (B) 60° (C) 45° (D) 30° 17. In the given figure, line segment AY || line segment BX. A y

S

What is the measure of  ALX ? (A) 85° (B) 90° (C) 95°

(D) 100°

Y 40°

x

70° B

Q

C

X

What are the respective values of x and y? (A) 140° and 60° (B) 140° and 70° www.betoppers.com

7th Class Mathematics

84

Conceptive Worksheet 1.

6.

In the given figure, POQ is a straight line. The angles x and y are in the ratio 8: 9.

In the given figure, AB||CD||EF. R

A

B

S

92° a

x

D

C

95°

P

y

Q

O b E

F

2.

What are the respective values of a and b? (A) 72° and 98° (B) 88° and 92° (C) 92° and 88° (D) 98° and 72° In the given figure, lines AB, CD, EF intersect at O. C

7.

What are the respective values of x and y? (A) 35° and 50° (B) 40° and 45° (C) 45° and 40° (D) 50° and 35° In the given figure, AB||CD. A 2x + 25°

E

A

100° b

8. B

F

T

C E 84°

54° B

D

4.

(A) 42° and 84° (B) 42° and 126° (C) 96° and 104° (D) 96° and 126° In the given figure, AOB is a straight line. A

O 2p + 16°

Q

P

If a: b = 5: 3, then what are the values of a and c? (A) 10° and 30° (B) 10° and 50° (C) 30° and 40° (D) 50° and 30° In the given figure, what are the respective measures of AOE and AOD ?

O

D

What is the value of x? (A) 25° (B) 45° (C) 115° (D) 135° In the given figure, PQRS is a parallelogram and TQRU is a trapezium.

D

A

F

x + 20°

B

a

O

3.

Q P

E c

C

y

25° 45° x R

S

U

What are the respective values of x and y? (A) 25° and 45° (B) 25° and 135° (C) 45° and 115° (D) 45° and 135° 9. M and N are supplementary angles such that twice the measure of M is 15° more than the measure of N . What are the respective measures of M and N ? (A) 25° and 65° (B) 35° and 55° (C) 65° and 115° (D) 95° and 85° 10. In the given figure, OE is the bisector of BOD . D

B

7p – 16°

E C 7x – 5° 2x

C

5.

What is the value of p? (A) 14° (B) 20° (C) 34° (D) 40° The measure of an angle x is two-third its complementary angle. What is the supplement of the angle x? (A) 36° (B) 80° (C) 100° (D) 144°

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20° A

O

What is the measure of EOB ? (A) 15° (B) 30° (C) 60°

B

(D) 90°

Lines and Angles

85

11. In the given figure, AB||CD and CE||DF. B

A

15. In the given figure, what is the measure of POR ? P

95°

R

144°

.

D

C

130° O

S

E

Q

(A) 84° (B) 86° (C) 94° 16. In the given figure, AB||CD.

F

What is the measure of CDF ? (A) 85° (B) 95° (C) 165°

(D) 96°

P

(D) 175°

A

12. In the given figure, SOQ is the complement of the angle of measure 75°.

B

R 2y + 12°

C 7y – 3° S

R

D

Q S 71° O

P

Q

What is the measure of ROS ? (A) 39° (B) 51° (C) 86° (D) 94° 13. In the given figure, AB is a horizontal line and OX  AB . The rays OC and OD are images of each other with respect to the dotted line OX.

What is the measure of PRB ? (A) 48° (B) 50° (C) 54° (D) 60° 17. In the given figure, what is the complement of EOF in the given figure? C

E

F B 37°

X

79° C

O 84°

D

A D O

A

B

If COD  140 , then what is the measure of BOD ? (A) 10° (B) 20° (C) 70° (D) 80° 14. In the given figure, what is the pair of angles POQ and QOR called? P

(A) 72° (B) 64° (C) 52° (D) 26° 18. In the given figure, DF||BG. What is the measure of AEB in the given figure? A

Q D

R

B

(A) 109°

135° G

64°

O

(A) Adjacent angles (B) Corresponding angles (C) Alternate interior angles (D) Vertically opposite angles

F

E

(B) 106°

C

(C) 128°

(D) 135°

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7th Class Mathematics

86 19. The given figure shows two parallel lines AB and CD that are cut by a transversal, AF. F

6.

In the given figure, OY bisects POB . What is the value of x?

D

A

X

B

44° 108°

E

x

O

P C A

What is the measure of FAB in the given figure? (A) 80° (B) 72° (C) 62° (D) 50° 20. In the given figure, AB||CD||EF.

7.

A B 112°

8.

128°

E

F

C

D

(A) 88° (B) 92° (C) 102° (D) 136° Two complementary angles are in the ratio 7: 8. What are the angles? (A) 40 and 50° (B) 42 and 48° (C) 80 and 100° (D) 84 and 96° In the given figure, OA is the bisector of AOX . What is the value of x?

What is the measure of CAE in the given figure? (A) 52° (B) 56° (C) 60° (D) 70°

A B 2x + 5°

Summative Worksheet 1.

50°

What is the value of x in the given figure?

X

D

C

9.

2.

Y

(D) 65°

a

B

O

O

(A) 35° (B) 45° (C) 55° In the given figure, a and b are

x

25° A

Y

B

(A) 45° (B) 55° (C) 65° (D) 75° What is the value of x in the given figure? b

2x 3x

(A) 29° 3.

4.

5.

(B) 31°

x + 6°

(C) 58°

(D) 62°

If an angle is less than 70°, then its supplement is (A) Equal to 20° (B) Equal to 110° (C) Greater than 20° (D) Greater than 110° If two angles are supplementary and congruent, then what is the measure of each angle? (A) 45° (B) 90° (C) 135° (D) 180° An angle is 12º less than twice its complement. What is the measure of the angle? (A) 34° (B) 38° (C) 42° (D) 56°

(A) Opposite angles (B) Corresponding angles (C) Alternate interior angles (D) Alternate exterior angles 10. What is the value of x in the given figure?

6x + 2°

(A) 18° (B) 20° (C) 22° (D) 24° 11. Find the complement of each of the following angles (i) 4624 (ii) 352452 ? 12. Two complement angles are in the ratio 5 : 4 find the angles? 13. How many degrees are there in (i)

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3x – 2°

1 right angle 5

(ii)

2 straight angles? 5

Lines and Angles

87

14. Write the supplement of (i) 65° (ii) 534525 ? 15. 80% of an angle is the supplement of 1400. Find the angle ? 16. In the given figure AB|| CD and EF is a transversal. If EGB  x and GHC   4x   . value of ‘x’?

HOTS Worksheet 1.

The given figure shows a quadrilateral PQRS where PQ||RS.

Find the

Q

P 2x + 10°

115°

E A

B

x°

60°

G

3y – 10° S

R

4x° H

C

D

2.

F

17. In the figure AB||CD and EF is a transversal find x,y,z? B

3.

D

E z° 130°

G y°

4.

H x° F

not exist? (A) Acute (B) Obtuse (C) Right (D) Straight The complement of the supplement of an obtuse angle is (A) A straight angle (B) An obtuse angle (C) An acute angle (D) A right angle In the given figure, PQ is parallel to RS and PQR  48 . The reason behind SRQ also being equal to 48° is that PQR and SRQ are

C

A

What is the value of (x + y) in the given figure? (A) 60° (B) 70° (C) 80° (D) 90° The complement of supplement of which angle does

18. In the following figure , AB||CD and EF is a transversal. Find the value of ‘x’?

P

Q

E A

B

5x° G

R

3x° H

C

5.

6.

What is the measure of POQ in the given figure?

D

F

19. In the given figure QOR is a straight line find (i) POQ (ii) POR ? P

R

(3x + 55°)

(x + 25°) O

Q

20. In the following figure, two straight lines AB and CD intersect at a point ‘O’. If AOC  70 . Find the measure of each of the angles BOD, AOD, BOC ? A

S

(A) Vertically opposite angles (B) Alternate interior angles (C) Corresponding angles (D) Linear pair The supplement of complement of an angle is (A) 45° more than the angle (B) 90° more than the angle (C) 45° less than the angle (D) 90° less than the angle P Q 2x – 5°

D

x

50° A 70°

C

O

7. B

O

B

(A) 75° (B) 85° (C) 95° (D) 105° If the sum of complement and supplement of an angle is 142°, then what is the measure of the angle? (A) 56° (B) 58° (C) 64° (D) 66° www.betoppers.com

88 8.

In the given figure, the lines x and y are parallel. b

7th Class Mathematics 13. In the given figure, two line segments AB and CD are intersecting at point O. E

x

C

2a + 5° y

40°

A

3a + 45°

B

O

z

9.

What is the value of b? (A) 60° (B) 80° (C) 85° (D) 95° The complement of which of the following angles is 43°?

(A)

(B)

117°

47°

(C)

D

If the measure of the supplement of EOC is 105°, then what are the respective measures of AOE and BOD ? (A) 65° and 140° (B) 65° and 75° (C) 75° and 100° (D) 40° and 75°

(D)

IIT JEE Worksheet

137°

67°

I.

Single Correct Answer Type

1.

What is the value of x in the given figure?

10. In the given figure, BC||DE What is the measure of CDE in the given figure?

30°

A

x

130°

x 70°

70°

E

D

50° B

C

(A) 20° (B) 30° (C) 60° (D) 70° 11. In the given figure, AB is parallel to CD and QX is the angle bisector of CQF .

2.

(A) 60° (B) 65° (C) 70° (D) 75° In the given figure, if OX is the bisector of BOC , then what is the value of x? A

X

O

C

D x B

110°

F C

A Q

X

100° D P

3. B

E

What is the measure of DQX ? 4. (A) 100° (B) 120° (C) 130° (D) 150° 12. In the given figure, DE is a straight line and BD is the angle bisector of ABC .

(A) 40° (B) 45° (C) 50° (D) 55° The supplement of an obtuse angle is (A) A right angle (B) An acute angle (C) An obtuse angle (D) A straight angle OP and OQ are bisectors of AOX and AOY . O

X

Y

80°

A P

D

B C

E

What is the measure of ABE in the figure? (A) 135° (B) 130° (C) 125° (D) 120° www.betoppers.com

Q A

If AOX  80 , then what is the measure of POQ ? (A) 80° (B) 85° (C) 90° (D) 95°

Lines and Angles 5. In the given figure, the measure of  P RQ is onethird that of QRS .

89 13. In the given figure , AB||CD and ‘t’ is a transversal than which of the following are true. t

P

1 2 3 4

A

5 6 7 8

C Q

R S

What is the measure of QRS ? 6.

(A) 120° (B) 125° (C) 135° (D) 140° Which of the following pairs of angles is adjacent with respect to the given figure?

(B) 2 6

(C) 3  5 180

(D) 2  3 180

(D) Reflects angle – 180° <  < 360°

III. Paragraph Type

6

B 3

(A) 1 3

(B) Acute angle – 90° <  < 180° (C) Obtuse angle – 0° <  < 90°

F 5

  In the figure below, FN is a straight line. AR make    an acute angle RAN with FN at A. RN  FN .

4

D

C

R

2

1 A

(A) 1 and 3

D

14. Which of the following are true. (A) Straight angle – 180° <  > 0°

G E

B

F

(B) 4 and 6

(C) 2 and 4 (D) 3 and 6 7. Which is complement to 500 ? (A) 1300 (B) 400 (C) 900 (D) 1800 0 8. Which is supplement to 70 ? (A) 200 (B) 900 (C) 1800 (D) 1100 0 0 9. If x and (x + 60) are complement of each other then = ? (A)150 (B) 600 (C) 1500 (D) 900 0 0 10. If 2y and (5y + 40 ) are supplements of each other then y0 ? (A) 500 (B) 200 (C) 900 (D) 1800

II. Multi Correct Answer Type 11. Which of the following are not acute angles? (A) 1790 (B) 1800 (C) 900 (D) 890 12. Which of the following are obtuse angles? (A) 900 (B) 1400 (C) 1790 (D) 1800

15. 16. 17. 18.

N

A

Using the above information, find each of the following   AR  AN  AF  (exterior RAN)  FN  (interior RAN) (interior RAN)  (exterior RAN)

IV. Integer Type S

10a 120

19.

0

O R

Find the value of ‘a’ in the above figure, given that ROS is a straight line.

20.

y 200

153

0

Find the value of angle y in the above figure. www.betoppers.com

7th Class Mathematics

90 21. How many complete turns is equivalent to 360° ? 22. How many right angles are there in 1 complete turns? 23. How many right angles are equal to two complete turns?

V. Matrix Matching (Match the following) 24.

Column-I (A) In the given figure then x + 2y =

(

)

Column-II (p) 1500

(

)

(q)1200

(

)

(r) 300

(

)

(s) 3000

(A)

(

)

Column-II (p) Acute angle

(B)

(

)

(q) Obtuse angle

(C)

(

)

(r) Straight angle

(

)

(s) Right angle

A

D 120° x

O y

B

C

(B) In the given figure then x = C

D

x 2x

3x

A

B

O

(C) In the given figure , AB||CD and ‘EF’ is a transversal then find x E x

A

y B

30°

C

D

F

Column-I

25.

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Triangles

Learning Outcomes 

Triangle



Congruency of Triangle



Classification of Triangles According to Sides



Conditions for Congruency of Triangles



Classification of Triangle according to Angles



Important Terms related to a Triangle

1. Triangle A closed plane figure bounded by three line segments is called a triangle. Let A,B and C be three noncollinear points A

Chapter – 3- 98 Chapter

By the end of this chapter, you will be understand

In the above figure, points D,E, and F lie in the interior of ABC points P,Q and R lie, in the exterior of ABC , while the points ‘X’ and ‘Y’ lie on ABC Triangular Region The interior of a triangle together with its boundary is called the triangular region Exterior and Interior Angles

B

BCA to be denoted by A, B, C respectively The three sides and three angles of a triangle together are called six parts (or) elements of the triangle In ABC the points A,B, and C are called its vertices ‘A’ is the vertex opposite to the side BC ‘B’ is the vertex opposite to the side CA ‘C’ is the vertex opposite to the side AB

Interior and Exterior of a Triangle •

The part of the plane enclosed by ABC is called the interior of ABC

•

The part of the plane not enclosed by ABC is called the exterior of ABC A P

Q

D

X R E C

A

C

Then the figure formed by the three line segments AB,BC, and CA is called a triangle ABC, denoted by ABC . ABC has i) Three sides namely AB, BC, and CA ii) Three angles namely BAC , ABC ,

F Y

B

B

A

C D

B

A

CB

C

E

•

On producing the side be of ABC to a point D, the exterior angle formed is ACD and its interior opposite angles are CAB and ABC , while BCA is the interior adjacent angle • On the producing the side of ABC to a point E, the exterior angle formed is CBE and its interior adjacent angle CBA • On producing the side CA of ABC to a point F, the exterior angle formed is BAF and its interior opposite angles are ABC and ACB , while BAC is the corresponding interior adjacent angle Note The sum of the lengths of any two sides of a triangle is always greater than the length of the third side. The three line segments can form a triangle only when the sum of the lengths of two smaller line segments is greater than the length of the longest one.

7th Class Mathematics

92

2. Classification of According to Sides

Triangles 3. Classification of Triangle according to Angles

There are three types of triangles with respect to the sides

There are three types of triangles with respect to their angles.

Scalene Triangle

Acute Angled Triangle

If all three sides of a triangle are of different lengths, then the triangles is called a scalene triangle. In the figure below, ABC is a triangle in which AB  BC  CA

A triangle with all the three angles acute is called an acute angled triangle A 60°

A

45°

75°

B B

C

3.2 cm

Therefore ABC is a scalene triangle. All the three angles of a scalene triangle are of different measures i.e A  B  C

Isosceles Triangles A triangle having two sides of equal length is called an isosceles triangle

C

In the figure above, ABC is a triangle in which A  60B  45 and C  75 . Therefore ABC is an acute angled triangle

Obtuse Angled Triangle A triangle having one obtuse angle is called an obtuse angled triangle P R

P

120° Q

Q

4 cm

R

In the above figure, PQR is a triangle in which PQ  PR  QR Therefore PQR is an Isosceles triangle. The angles opposite to the equal sides of an Isosceles triangle are equal. Since PQ = PR; we have Q  R

PQR is an obtuse - angled triangle. In an obtuse - angled triangle one of the angles is obtuse and each of the remaining two angles is acute.

Right Angled Triangle A triangle is called right angled if the measure of one of its angle is 900 L

Equilateral Triangle A triangle having all the three sides of the same length is called an equilateral triangle.

X

90° M

Y Z 3 cm In the figure below, XYZ is a triangle in which XY = YZ = ZX. Therefore XYZ is an equilateral triangle. Each angle of an equilateral triangle measures 60°. www.betoppers.com

N

The side opposite to the right angle is called the hypotenuse of the triangle .In the above figures LMN is right angled at ‘M’ clearly , LN is the hypotenuse of LMN. In a right angle triangle one angle measures 90° and each one of the remaining two angles is acute.

Triangles

93

4. Important Terms related to a Triangle Medians A line segment joining a vertex of a triangle to the middle point of the side opposite to the vertex is called median. All the three medians of a triangle always pass through a point. Centroid of a Triangle The point of intersection of all the three medians of a triangle is called its centroid. A

E

Incentre of a Triangle F

G

B

The points of intersection of all the three perpendicular bisectors of the sides of a triangle is called the circumcentre of the triangle. Then circle passing through the three vertices of a triangle is called the circumcircle. In the given figure, the perpendicular bisectors of the sides BC, CA and AB of ABC intersect at a point ‘O’. ‘O’ is the circumcentre of ABC . With ‘O’ as centre and radius OA, we draw a circle passing through three vertices A,B,C of ABC and hence this circle is the circumcircle of ABC .

D

C

In the given figure, D,E and F are the midpoints of the sides BC, CA and AB respectively of ABC so AD, BE and CF are the medians of ABC intersecting at a points ‘G’ .Therefore ‘’G is the centroid of ABC

The angle bisectors of the triangle always intersects at a point. The point of intersection of all the angle bisectors of the triangle is called its incentre. The circle touching the sides of a triangle is called the incircle of the triangle. A

I

Altitudes A line witch passes through a vertex of a triangle, and is a right angles to the opposite side. A triangle has three altitudes, all the three altitudes of a triangle always pass through a point. Orthocentre of a Triangle The point of intersection of all the altitudes of a triangle is called its orthocentre A N

M

B

T

C

In the above figure, the angle bisectors of its angles intersect at the points I. Therefore ‘I’ is the incentre of ABC . From ‘I’ draw IL BC, with ‘I’ as centre and radius IL, draw a circle, touching the sides of ABC . Hence this circle is incircle of ABC

Angle sum Property of a Triangle The sum of the angles of triangle is 1800

H

Exterior Angle property of a Triangle B

L

C

In the given figure; AL BC, BM CA and CN AB, so AL, BM and CN are the altitudes of ABC

When a side of a triangle is produced, then the exterior angle so formed is equal to the sum of its interior opposite angles.

Circumcentre of a Triangle The perpendicular bisectors of the sides of a triangles always intersects at a point A

F

E O

B

D

C

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7th Class Mathematics

94

Formative Worksheet 1.

5.

The given figure shows a 10 ft long ladder inclined on an 8 ft high wall. In the figure, AC represents the ladder while AB represents the wall. A

6. 10 feet

8 feet

In a  PQR, if the angles of measures 15°, 10°, and 5° are subtracted from angles P, Q, and R respectively, then the ratio of the angles becomes 4: 5: 6. What is the measure of  R in  PQR? (A) 55° (B) 60° (C) 65° (D) 70° The given figure shows an isosceles right triangle ABC. A

C

2.

B

How far is the foot of the ladder from the base of the wall? (A) 4 ft (B) 5 ft (C) 6 ft (D) 7 ft The given figure shows two walls AB and DC of heights 5 m and 9 m respectively. AC and BD are ladders, which are inclined to the walls as shown in figure. The length of the ladder BD is 15 m.

B

D A

7. 9m

5m

8. B

3.

C

What is the length of the ladder AC? (A) 10 m (B) 11 m (C) 12 m (D) 13 m The given figure shows the positions of three houses A, B and C. The measure of ABC in the resulting ABC is 90°.

9.

C

According to the given figure,  A and  C are not (A) Complementary to each other (B) Supplementary to each other (C) Congruent (D) Acute What is the length of the longest pole that can be placed in a cylindrical hall of diameter 12 m and height 5 m? (A) 13 m (B) 15 m (C) 17 m (D) 19 m If the interior angles of a triangle are in the ratio 5:7:8, then the triangle is (A) An isosceles triangle (B) An equilateral triangle (C) A right-angled triangle (D) An acute-angled triangle In the given figure, P is the mid-point of AC. What is the value of (3AC 2 – 4AB2)? A

C

P

3 km

A

4.

4 km

B

What is the distance between the houses A and C? (A) 3 km (B) 4 km (C) 5 km (D) 6 km In the given figure, points B and C represent two persons standing on either side of a 12 m tall building which is represented by AD.

B

C

(A) BP 2 (B) 2BP 2 (C) 3BP 2 (D) 4BP2 10. In the figure, the side BC is extended on both the sides to points X and Y. A

A

12 m

120° Y

B

D

C 21 m

In the  ABC so formed, what is the length of side AC? (A) 16 m (B) 18 m (C) 20 m (D) 22 m www.betoppers.com

B

125° C

What is the measure of A ? (A) 60° (B) 65° (C) 55°

X

(D) 50°

Triangles

95

11. In a triangle, one of the exterior angles is of measure 105° and the interior opposite angles are in the ratio 3: 4. What is the measure of the greatest interior angle of the triangle? (A) 45° (B) 65° (C) 75° (D) 95°

3.

4.

Which of the following lengths can be the sides of a triangle? (A) 2 cm, 4 cm, 7 cm (B) 4 cm, 6 cm, 8 cm (C) 3 cm, 5 cm, 10 cm (D) 4 cm, 7 cm, 12 cm What is the value of y in the given figure?

12. What is the measure of SQR in the given figure? P 35° 3y – 68°

S

Q

R

(A) 65° (B) 55° (C) 45° (D) 35° 13. If two sides of a triangle are of lengths 6.8 cm and 8.2 cm, then the length of third side varies from (A) 1.4 cm to 15 cm (B) 2 cm to 15.6 cm (C) 1 cm to 14 cm (D) 2.4 cm to 13.8 cm 14. If AB||CD and AB = CD, then what is the value of x in the given figure? B

A

y

5.

6.

x O

7.

80°

(A) 28° (B) 34° (C) 56° (D) 68° In a right-angled triangle, the lengths of base and perpendicular are 6 cm and 8 cm. What is the length of the hypotenuse? (A) 9 cm (B) 10 cm (C) 11 cm (D) 12cm In PQR , if 2P   Q  R  , then what is the measure of P ? (A) 50° (B) 55° (C) 60° (D) 65° In the given figure, XY is parallel to PQ. What is the value of x?

40° C

X

D

(A) 40°

(B) 60°

(C) 70°

y

Y

(D) 90°

40° A

A 3x

15.

65° C 100°

x B

x

C

X

B

What is the value of x in the given figure? (A) 15° (B) 20° (C) 25° (D) 30°

55°

A . What type of 2

1.

In ABC , AB = AC and B 

2.

triangle is ABC ? (A) Equilateral triangle (B) Acute-angled triangle (C) Right-angled triangle (D) Obtuse-angled triangle What is the value of x in the given figure?

(A) 58°

(B) 62°

8.

(A) 20° (B) 25° (C) 30° (D) 35° The given figure shows an isosceles right-angled triangle ABC. A

B

C

What is the measure of A ? (A) 35° (B) 45° (C) 55°

55° 2x + 1°

Q

P

Conceptive Worksheet

(D) 65°

70°

(C) 66°

(D) 71° www.betoppers.com

7th Class Mathematics

96 9.

In the given figure, AD is parallel to EF. What is the 15. In the given figure, A and B are in the ratio 3: 5. What is the measure of B ? value of x in the given figure? A

F A

160°

55° 65°

B

x

E

(A) 95° (C) 100°

C

B D

10.

11.

12.

13.

(A) 50° (B) 60° (C) 70° (D) 80° Two trees of heights 9 feet and 14 feet are standing on the ground. If they are 12 feet apart, then what is the distance between their tops? (A) 11 feet (B) 13 feet (C) 15 feet (D) 17 feet Which of the following cannot be the sides of a triangle? (A) 3.2 cm, 4.8 cm, 6.4 cm (B) 2.5 cm, 3.6 cm, 5 cm (C) 2.6 cm, 3.4 cm, 7 cm (D) 2.1 cm, 3.1 cm, 4.2 cm The angles of a triangle are in the ratio 2: 3: 5. What type of triangle is it? (A) Isosceles triangle (B) Equilateral triangle (C) Scalene right triangle (D) Isosceles right triangle In the given figure, AB||DE, ADE  90 , and AB = AC. A

B

C E

D

What is the measure of BCD ? (A) 115° (B) 125° (C) 135° (D) 150° 14. What is the value of x in the given figure? A x 15°

135° B

(A) 30°

C

(B) 45°

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D

E

(C) 60°

(D) 75°

C

D

(B) 98° (D) 105°

5. Congruency of Triangle Congruent Figures Geometrical figures which have exactly the same shape and the same size, are known as congruent figures. For congruence, we use the symbol '  ' read as ‘congruent to’. Thus, two plane figures are congruent if each when superposed on the other, covers it exactly. Eg:

Similar Figures Geometrical figures which have exactly the same shape but not necessarily the same size, are known as similar figures. For similarity, we use the symbol '  ' read as ‘is similar to’ Note Two congruent figures are always similar but two similar figures need not be congruent. Eg:

Triangles •

97 Any two equilateral triangles are always similar, but they are congruent only if they have the same side length. Any two line segments are always similar, but they congruent only if they have the same length.

•

ASA (Angle-Side-Angle)Condition If two triangles have two angles and the included side of them one respectively equal to two angles and the included side of the other. Then the triangles are congruent. Eg

Congruent Triangles

C

Two triangles are said to be congruent, if each one of them can be made to superpose on the other, so as to cover it exactly. Thus, congruent triangles are exactly identical. In congruent triangles, the sides and angles which coincides by superposition are called corresponding sides and angles respectively. Hence, we can also say that two triangles are congruent if pairs of corresponding sides and corresponding angles are equal. A

D

60°

D

50°

60°

5 cm

A

50° 5 cm

B E

F

If two angles of one triangle are respectively equal to two angles of the other, then it is quite clear that their remaining third angles are also equal. Thus even, if the given side is not the one included between the two given angles, then it shall be the one included between one of the given angles and the third angle.

SSS (Side-Side-Side)Condition B

C E

F

Thus, ABC  DEF if AB = DE, BC = EF,, CA = FD A  D, B  E, C  F . We write, ABC  DEF , if means that A,B and C are matched with D,E and F respectively and we write A  D, B  E and C  F and therefore,

ABC  DEF

If two triangles have the three sides of the one respectively equal to the corresponding three sides of the other, then the triangles are congruent. Eg: A

SAS (Side-Angle-Side)Condition If two triangles have two sides and the included angle of the one respectively equal to two sides and the included angle of the other, then the triangles are congruent. Eg: C

60° A

If two right angled triangles have one-side and hypotenuse of the one respectively equal to the corresponding side and the hypotenuse of the other, then the triangles are congruent. Eg: A

B

60° 5 cm

RHS (Right-Angle-Hypotenuse-Side)

5 cm

4 cm

B P

5 cm

ABC  PQR

Q

Z

4 cm

ABC  XYZ

R

4 cm

5 cm

6 cm

C Y

4 cm

6. Conditions for Congruency of Triangles There are four cases as conditions for congruency. In each case, we have a different combination of the three matching parts.

5 cm

6 cm

B

X

L

4 cm C

M

5 cm

4 cm N

ABC  LMN Note: In two triangles, if the three angles of one triangle are respectively equal to the three angles of the other, then the triangles are not necessarily congruent.

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7th Class Mathematics

98

20. In the given figure, AO = OD, BO = OC. The perimeter of AOB is 16 cm and lengths of OC and OD are 8 cm and 5 cm respectively.

Formative Worksheet 16. Bethany has a flower-pot in his room

A

C O D

B

The shape congruent with the shape of the pot is (A)

(B)

(C)

(D)

What is the length of CD? (A) 2 cm (B) 3 cm (C) 4 cm (D) 5 cm 21. In the given figure, C is equal to R and BC = QR. A

17. One night, Jasmine sees a moon crescent in the Sky as shown in the figure.

P

B

R

C Q

By which congruency criterion are ABC and PQR congruent? (A) SSS (B) SAS (C) ASA (D) RHS 22. In the given figure, AB = AC and BE = CD. Which of the following shapes is similar with the shape of moon crescent? (A)

(B)

(C)

(D)

B

18. The given figure shows two triangles. P X

6.8 cm

Y

D

(B) PQR  ZYX by RHS congruence rule (C) RPQ  XYZ by RHS congruence rule (D) RPQ  XYZ by ASA congruence rule 19. In the given figure, ABCD and ABEC are two parallelograms. B

C

B

O

Which of the following statements is correct? (A) PQR  XYZ by SAS congruence rule

D

C

Z

6.8 cm

A

E

By which congruency criterion are ABD and ACE congruent? (A) SSS (B) SAS (C) ASA (D) RHS 23. In the given figure, AB is parallel to CD. A

2.9 cm Q 2.9 cm R

A

E

If DAC = 50° and ADC = 70° , then what is the measure of CEB ? (A) 40° (B) 50° (C) 60° (D) 70° www.betoppers.com

C

D

By which congruency criterion are AOB and DOC congruent? (A) SSS (B) ASA (C) RHS (D) SAS 24. Which of the following statements is incorrect? (A) Square having equal areas are congruent. (B) Rectangles having equal areas are congruent. (C) Circles having equal diameters are congruent. (D) Rhombuses having equal areas are congruent. 25. If ABC  QRP , then which of the following relations is correct? (A) AB = PQ (B) CA = QR (C) BC = RP (D) BC = PQ

Triangles

99

26. In the given figure, AD = 4 cm and BD = 3 cm.

30. In the given figure, AB = EF and AD = 13 cm.

A

A

F

O C

D

B

What is the perimeter of ABC ? (A) 16cm (B) 18cm (C) 21cm (D) 22cm 27. In the given figure, NP = (x + 4) cm, MN = (3x – 4) cm, and PQ = (2x + 1) cm. P

L

B

10 cm

C

2 cm 22 cm

E

D

What is the perimeter of CEF ? (A) 20 cm (B) 24 cm (C) 30 cm (D) 32 cm

Conceptive Worksheet N

16. On his way to school, Joshua spots a sign board showing the way to the school. Q

M

SCHOOL

Find the length of side LM. (A) 6 cm (B) 7 cm (C) 8 cm (D) 9 cm 28. In the given figure, XY bisects WXZ and XW = XZ. W

X

Which of the following shapes is not similar with the arrow in the sign board?

Y

(A) Z

If XWZ  25 and WXZ  60 , then what is the measure of XYZ ? (A) 55° (B) 65° (C) 125° (D) 135° 29. The given figure shows two triangles.

(B)

(C)

(D)

17. The given set of figures shows Jennifer’s collection of erasers (i)

(ii)

(iii)

(iv)

P X

6.8 cm 2.9 cm Q 2.9 cm R Y

6.8 cm

Z

Which of the following statements is correct? (A) PQR  XYZ by SAS congruence rule (B) PQR  ZYX by RHS congruence rule (C) RPQ  XYZ by RHS congruence rule (D) RPQ  XYZ by ASA congruence rule

(A) (i) and (iii) (B) (i) and (iv) (C) (ii) and (iii) (D) (ii) and (iv) 18. The given figure shows two triangles. P X

6.8 cm 2.9 cm Q 2.9 cm R Y

6.8 cm

Z

Which of the following statements is correct? (A) PQR  XYZ by SAS congruence rule (B) PQR  ZYX by RHS congruence rule (C) RPQ  XYZ by RHS congruence rule (D) RPQ  XYZ by ASA congruence rule www.betoppers.com

7th Class Mathematics

100 19. In which of the following alternatives is ABC congruent to DEF by SAS congruence criterion? D

A 3.4 cm

P

3.4 cm

4.2 cm

(A) 4.2 cm A

B

F

C E D

R

3 cm

2 cm 60°

(C)

Q

A

B

(B)

23. In the given figure, BC = PR and the perimeter of PQR is 30 cm.

35° B

3 cm A

60° 2 cm F

C E

B

3 cm

C

A

F

4 cm E

A

P

24 cm

7 cm

(D)

3 cm 50°

3 cm 60°

60°

50°

B

B

D

C

S

Q

40°

D

C

What is the perimeter of ABC ? (A) 24 cm (B) 25 cm (C) 30 cm (D)36 cm 24. In the given figure, ABC is congruent to PQR .

E

3 cm 40° 4 cm

145°

F

20. The given figure represents an equilateral triangle ABC of side 12 cm. The points D and E divide BC into three equal parts. A

C

R

If PC = 20 cm, then what is the length of AR? (A) 25 cm (B) 30 cm (C) 35 cm (D) 40 cm 25. In the given figure, AB = CD and BC = DE. E A

60°

60° B

D

E

C

B

If AD = 9 cm, then what is the perimeter of ADE ? (A) 17cm (B) 18cm (C) 21 cm (D) 22cm 21. In the given figure, ABDC is a parallelogram. A

B

C

E D

Which of the following pairs of triangles is congruent

D

C

What type of a triangle is ACE ? (A) Isosceles triangle (B) Equilateral triangle (C) Scalene right-angled triangle (D) Right-angled isosceles triangle 26. In the given figure, triangles ABC and CDE are right-angled at B and D respectively. The measures of the sides are given (see figure). E

Q

P

B

15 cm C 23 cm

D

What type of a triangle is ACE ? (A) Isosceles triangle (B) Equilateral triangle (C) Scalene right-angled triangle (D) Right-angled isosceles triangle

R

S

What is the measure of QRS ? (A) 40° (B) 80° (C) 100° www.betoppers.com

A

8 cm

(A) ABC and BEA (B) ABC and DCB (C) ABE and ECA (D) ABE and DCB 22. In the given figure, PQ = PS and QR = RS. The measure of QPS is 230° and PQR is 25°.

15 cm

in the given figure?

(D) 130°

Triangles

101

Summative Worksheet

27. In the given figure, AB||CD and AD||BC. A

D

1.

In the given figure, l||m. If AC = BC, then what is the value of x?

5 cm

A

B

C

If the perimeter of ABD  20 cm , then what is the perimeter of quadrilateral ABCD? (A) 25 cm (B) 30 cm (C) 35 cm (D) 40 cm 23. In the given figure, ABF  DEG .

13 cm C

3.

11 cm B

E

F

42°

B

(A) 58° (B) 69° (C) 71° (D) 72° The sides of a triangle are: x cm, 10.4 cm, and 16.2 cm, where x is a natural number. What is the greatest possible value of x? (A) 27 cm (B) 26 cm (C) 25 cm (D) 24 cm Which angle is common to the interior opposite angles of the exterior angles PRX and RQZ of PQR ? Y

G

What is the perimeter of ΔABC? (A) 27 cm

C m

l

2.

D

A

x

.

P

(B) 31 cm

(C) 33 cm (D) 35 cm 29. In the given figure, ABC and ADE are isosceles triangles with AB = AC and AD = AE.

Q Z

B

R

(A) QPR

D

4.

A E

X

(B) PQR

(C) PRQ (D) QPY What is the ratio of B and C in the given figure? A

C

If CE = 2 cm and DE = 6 cm, then what is the length of side BC? (A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm 30. In the given figure, line l is parallel to m. A

l

C O

m D

105°

50° B

5.

C

(A) 1:1 (B) 2:1 (C) 3:2 (D) 3:1 In an isosceles triangle, the measure of one of the equal angles is 42°. What is the measure of the unequal angle? (A) 121° (B) 112° (C) 96° (D) 86°

B

If CD = 8 cm, then what is the length of CO? (A) 2 cm (B) 3 cm (C) 4 cm (D) 5 cm

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7th Class Mathematics

102 6.

Two friends Ishaan and Saurabh are standing at a point A. Both of them want to reach the point C. Ishaan goes directly from A to C whereas Saurabh first goes from A to B and then from B to C.

12. In the given figure, CAB  DEF . D B 12 cm

13 cm

A

A 14 m B

48 m

C

C

F

E

What is the measure of EF in the given figure? (A) 4 cm (B) 5 cm (C) 10 cm (D) 12 cm 13. Which of the following lengths can be the sides of a triangle? (A) 2 cm, 3 cm, and 6 cm

Find how much more distance did Saurabh travel than Ishaan. (A) 12 m (B) 24 m (C) 50 m (D) 52 m 9 5 (B) 2 cm cm, and cm 7. Ruby bought a flat screen television. The length and 2 2 the diagonal of the screen are 2 feet and 2.5 feet (C) 1.5 cm, 3 cm, and 3.5 cm respectively. What is the height of the screen of the 4 7 2 (D) cm, cm and cm television? 3 3 3 (A) 1.2 feet (B) 1.4 feet 14. In the given figure, ΔABC is an isosceles triangle (C) 1.5 feet (D) 1.7 feet with AB = AC. 8. A pole is made to stand on a ground by tying the top A of the pole with a 26 m long rope on the ground. The 40° point on the ground through which the rope is tied is 10 m away from the pole. What is the height of the pole? B C (A) 24 m (B) 25 m D What is the measure of ACD ? (C) 27 m (D) 28 m 9. What is the area of a rectangle whose length is (A) 110° (B) 120° 24 cm and diagonal is 25 cm? (C) 130° (D) 140° (A) 150 cm2 (B) 168 cm2 OTS orksheet (C) 175 cm2 (D) 300 cm2 10. In the given figure, AD is the altitude of ABC 1. What is the measure of MNO ? 60° drawn from the vertex A.

H

W

M

A

125° N

B

D

C

If AB = 10 cm, AD = 8 cm, and BC = 21 cm, then in what ratio does the point D divide the side BC? (A) 2 : 3 (B) 2 : 5 (C) 1 : 3 (D) 1 : 1 11. Which of the following sets of sides can be the sides of a right-angled triangle? (A) 14 cm, 40 cm, and 50 cm (B) 13 cm, 80 cm, and 81 cm (C) 12 cm, 30 cm, and 37 cm (D) 11 cm, 60 cm, and 61 cm www.betoppers.com

2.

O

P

(A) 60° (B) 65° (C) 66° (D) 70° is right-angled at B. If the measures of A ABC and C are in the ratio 7:8, then what is the measure of C ? (A) 34° (B) 42° (C) 48° (D) 56°

Triangles 3.

4.

103

If the measures of the angles of a triangle are in the 7. ratio 1:2:3, then how can the triangle be classified? (A) Right-angled and isosceles (B) Acute-angled and isosceles (C) Right-angled and scalene (D) Acute-angled and scalene The given figure shows ABE . Side BE of the triangle acts as the width of rectangle BCDE. The perimeter of the figure is 38 cm.

The given figure shows two parallel lines AB and CD that are intersected by transversals EF and GH. G

E a A

B

X 140° 75°

C

D

Z

Y

A

F

H 60°

What is the value of a in the given figure? B

60°

(A) 40° (C) 75°

E

10 cm

8.

(B) 65° (D) 140°

If one of the angles of an isosceles triangle measures 40°, then what are the possible differences between

C

5.

D

What is the length of AB? (A) 4 cm (B) 6 cm (C) 10 cm (D) 14 cm The given figure shows an isosceles ABC with AB = BC. Sides BC and AC are extended to D and 9. E respectively, such that CD and CE along with DE form sides of CDE . A 110°

C

the measures of its unequal angles? (A) 20° and 60° (B) 30° and 60° (C) 60° and 100° (D) 70° and 100° The given figure is drawn by attaching an equilateral ABC and an isosceles DEF to the opposite ends of rectangle BCFD. The perimeter of ABC is 24 cm, while that of DEF is 20 cm. D

D

B

55°

B A

E 20 cm

F

E

6.

How can CDE be classified? (A) Acute-angled and scalene (B) Acute-angled and scalene (C) Right-angled and scalene (D) Right-angled and isosceles The given figure shows LMN , which is rightangled at M. MO acts as the altitude from vertex M to side LN. The measure of LMO is 40°.

C

What is the perimeter of the given figure? (A) 64 cm (B) 68 cm (C) 76 cm (D) 84 cm 10. The given figure shows an isosceles ABC with equal sides AB and AC. Side BC of the triangle is extended to point X. A 50°

L

B O

M

C

X

What is the magnitude of ACX ? (A) 50° (B) 65° (C) 115° (D) 130° 11. What is the value of y in the given figure? N

X

What is the magnitude of LNM in the given figure? (A) 40° (B) 45° (C) 50° (D) 55°

A y

145° B

(A) 110°

(B) 125°

C

Y

(C) 130°

(D) 145°

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7th Class Mathematics

104 12. The given figure shows two parallel lines AB and CD, which are intersected by transversal EF.

2.

In the figure, BCD is a straight line and AE = BE. D 120°

A

L

E

x C

C

40° B 50°

r

D

70°

A

N

M F

What is the value of x in the given figure? 3. (A) 30° (B) 45° (C) 70° (D) 80° 13. If two angles of a triangle measure 35° and 68°, then which of the following measures cannot be the measure of one of the exterior angles of the triangle? (A) 103° (B) 112° (C) 136° (D) 145° 4. 14. Sumit wants to construct a triangular kitchen garden whose perimeter is 36 m. Which of the following dimensions are possible for the garden? (A) 13 m, 14 m, and 9 m (B) 18 m, 11 m, and 7 m (C) 6 m, 12 m, and 18 m (D) 19 m, 11 m, and 6 m 15. The given figure shows two buildings AB and CD.

B 0

120

0

25

C

x A

(A) 30° (B) 35° (C) 40° (D) 45° Calculate the value of p in the figure, given that ABC is a straight line. C p 40°

D

B 165°

X is a point between the buildings from which two ropes are tied without slack to the top of the buildings.

B

Find the value of r. (A) 120° (B) 140° (C) 150° (D) 160° Find the value of x in the triangle ABC.

A

5.

A

(A) 120° (B) 125° (C) 130° (D) 135° In the figure, ABC is a straight line. Given that BD = BE, the value of z is A

C

41 m

40 m

17 m B

B

z

15 m

X

C

D

What is the distance between the buildings? (A) 8 m

(B) 9 m

(C) 16 m

0

70 30

(D) 17 m

IIT JEE Worksheet

D

6.

I.

Single Correct Answer Type

1.

In the figure, ABC and FED are straight lines. The value of p is: C 60 B

0

p 0

100

F

1100 E

D

A

(A) 130°

(B) 140°

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E

0

(C) 150°

(D) 160°

7.

(A) 70° (B) 80° (C) 90° (D) 100° Two angles of a triangle measure 52° and 72°. Find the measure of the third angle of the triangle. (A) 36° (B) 46° (C) 56° (D) 66° The angles of a triangle are in the ratio 6 : 5 : 7. Find the measure of each angle of the triangle. (A) 60°, 50°, 70° (B) 50°, 60°, 70° (C) 70°, 60°, 50° (D) 70°, 60°, 50°

Triangles 8.

105

Find the values of the variables in the figures given III. Paragraph Type below. The figure given below has been obtained by using A two equilateral triangles of equal size. A

° 15° 20

G

H

B

F

0

9.

120 D

y

x B

C

(A) x = 25°, y = 40° (B) x = 30°, y = 35° (C) x = 25°, y = 50° (D) x = 40°, y = 25° In  ABC, the angle bisectors of  B and  C meet at O. If  A = 60°, find  BOC. A 0

60

O

B

(A) 120°

(B) 125°

L

I

C

(C) 130°

(D) 135°

II. Multi Correct Answer Type 10. Which of the following can be sets of angles of a triangle? (A) 75°, 45°, 60° (B) 100°, 40°, 50° (C) 110°, 50°, 20° (D) 30°, 60°, 90° 11. Which of the following can be the sets of angles of an acute triangle? (A) 60°, 60°, 60° (B) 45°, 55°, 80° (C) 70°, 70°, 40° (D) 90°, 40°, 50° 12. Which of the following statements is false? (A) If two angles of a triangle are complementary, it is a right triangle. (B) In a right triangle, the side opposite to the right angle is called the hypotenuse (C) A triangle must have three acute angles (D) A triangle can have two obtuse angles 13. Which of the following can be the measures of the sides of a triangle? (A) 3 cm, 3 cm, 5 cm (B) 6 cm, 6 cm, 6 cm (C) 6 cm, 4 cm, 2 cm (D) 10 cm, 6 cm, 3 cm

C

J

K

E

D

Based on this information answer the questions given below.

14 What are the total number of triangles in the figure. 15. What are the number of regular polygons in the above figure? 16. Find the measure of  GHI. 17. Find sum of the angles A, B, C, D, E and F 18. Find the sum of the angles G, H, I, J, K, L

IV. Integer Type 19. Number of vertices of a triangle________ 20. In a right-angled triangle, the lengths of base and perpendicular are 3 cm and 4cm. What is the length of the hypotenuse? 21. ABC is an equilateral triangle of side 3cm. Find the perimeter of the triangle ? 22. Area of a triangle is 16 cm2 and base 8 cm. Find the height of the triangle ? 23. In ABC sum of the two angles equal to 172° degrees, find the another angle ?

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7th Class Mathematics

106

V. Matrix Matching (Match the following) Column – I

24.

Column – II

(A)

(p) R.H.S Axiom

 (B)

(q) A.S.A Axiom

 (C)

(r) S.A.S Axiom

 (D)

(s) S.S.S Axiom

 25.

Column – I

Column – II

(A) The point of intersection of attitude Of triangle (B) The point of intersection of the medians of triangle (C) The point of intersection of (Perpendicular bisectors of the Sides of the triangle (D) The point of intersection of Angular bisector of the angles of triangle

(p) Incenter

(q) Circum centre

(r) Centroid (s) Orthocenter

 

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Chapter – 10 9

Quadrilaterals

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





Quadrilateral



Angle Sum Property of a Quadrilateral



Various Types of Quadrilateral

Diagonals of a Quadrilateral

1. Quadrilateral

The line segment joining the opposite vertices of a quadrilateral is called a diagonal of the quadrilateral. In the given figure, AC and BD are the two diagonals of the quadrilateral.

A simple closed figure bounded by four line segments is called quadrilateral. D

Polygons

C

D

A

B

A quadrilateral ABCD has • Four sides, namely AB, BC, CD, DA • Four angles, namely DAB, ABC, BCD and CDA • Four vertices, namely A, B, C and D

A

Two sides of a quadrilateral which have a common end point are called its adjacent sides. Thus (AB, BC), (BC, CD), (CD, DA) and (DA, AB) are four pairs of adjacent sides (or consecutive sides) of the quadrilateral ABCD.

The sum of the lengths of all the four sides of a quadrilateral is called its perimeter. Perimeter of quadrilateral ABCD = AB + BC + CD + DA

2. Angle Sum Property of a Quadrilateral The sum of all the four angles quadrilateral is 360°.

Opposite Sides Two sides of a quadrilateral which do not have a common end point are called its opposite sides. Thus, (AB, CD) and (AD, BC) are two pairs of opposite sides of the quadrilateral ABCD.

Adjacent Angles Two angles of a quadrilateral having a common side are called its adjacent angles. Thus and  D, A 

are four pairs of adjacent angles of the quadrilateral ABCD.

Opposite Angles Two angles of a quadrilateral which are not adjacent angles are known as the opposite angles of the quadrilateral. Thus,  A, C  and  E, D  are

B

Perimeter of a Quadrilateral

Adjacent Sides

 A, B ,  B, C  ,  C, D 

C

In a quadrilateral ABCD, we have A  B  C  D  360

3

Various Types of Quadrilaterals The quadrilaterals may broadly be classified in to three types. 1. Parallelogram 2. Trapezium 3. Kite

Parallelogram A quadrilateral in which both pairs of opposite sides are parallel, is called a parallelogram. In the given figure, ABCD is a quadrilateral in which AB || DC and AD || BC. Therefore ABCD is a parallelogram. C

D

two pairs of opposite angles of the quadrilateral ABCD.

O A

B

7th Class Mathematics

108 Properties of Parallelogram In a parallelogram ABCD, we have the following properties. i) Its opposite sides are parallel. i.e., AB || DC and AD || BC ii) Its opposite sides are equal i.e., AB = CD and AD = BC iii) Its opposite angles are equal. i.e., A   C and B  D iv) The sum of the measures of all of its angles is 360° i.e., A  B  C  D  360 v) Its adjacent angles are supplementary A  B  180 B  C  180 C  D  180 and D  A  180 vi) Its diagonals bisect each other. i.e., AO = AC and BO = OD vii) Each diagonal of a parallelogram divides it into two congruent triangles. i.e., ABC  CDA and ABD  CDB

Square A square is a parallelogram having all sides equal and each angle measuring 90°. D

C

A

B

In the given figure, ABCD is a parallelogram in which AB = BC = CD = DA and A  B  C  D  90 . Therefore, ABCD is a square Properties of a Square A square ABCD has the following properties D

O

Types of Parallelograms Rectangle A rectangle is a parallelogram each one of whose angle measures 90°. D

A

i)

C

ii) iii) A

B

In the given figure, ABCD is a parallelogram in which A  B  C  D  90 Therefore, ABCD is a rectangle. Properties of Rectangle A rectangle ABCD has the following properties D

C

O A

B

Its opposite sides are parallel i.e., AB||DC and AD||BC Its all sides are equal i.e., AB = BC and AD = BC Each of its angle measures 90°

i.e., A  B  C  D  90 iv) Its diagonals are equal i.e., diagonal AC = diagonal BD v) Its diagonals bisect each other at right angles i.e., AO = OC, BD = OD and AOB  BOC  COD  DOA  90 Rhombus A rhombus is a parallelogram having all sides equal. In the given figure, ABCD is a parallelogram in which AB = BC = CD = DA.

B

Its opposite sides are parallel i.e., AB||DC and AD||BC ii) Its opposite sides are equal i.e., AB = CD and AD = BC iii) Its diagonals are equal. i.e., diagonal AC = diagonal BD iv) Its diagonals bisect each other (not necessarily at right angles) i.e., AO = OC and BO = OD v) Each of its angles measures 90° i.e., A  B  C  D  90 www.betoppers.com

C

A

i)

B

D

C

Therefore , ABCD is a rhombus.

Quadrilaterals

109

Properties of a Rhombus A rhombus ABCD has the following properties. A

5.

B

D

6.

C

i)

Its opposite sides are parallel i.e., AB||DC and AD || BC 7. ii) Its all sides are equal i.e., AB = BC = CD = DA iii) Its diagonals bisect each other at right angles 8. i.e., AO = OC, BD = OD and AOB  BOC  COD  DOA  90 Note: Every rhombus is a parallelogram and every square is a rhombus.

If  DAM = 25° and  ADC = 100º, then what is the measure of  AMB? (A) 90° (B) 70° (C) 55° (D) 35° What is the sum of the interior angles of a regular pentagon? (A) 720° (B) 540° (C) 450° (D) 360° In a rhombus ABCD, if the angle bisectors of  A and  B meet at O, then what is the measure of  AOB? (A) 60° (B) 90°

A  B A  B (D) 4 6 What is the sum of the measures of the external angles of an octagon? (A) 900° (B) 720° (C) 360° (D) 180° The given figure shows rectangle ABCD. The diagonals meet at point O such that DO = 5 m (C)

A

Formative Worksheet 1.

2.

3.

C

A

4.

D

The diagonals AC and BD of parallelogram ABCD intersect at O. What is the ratio of the area of 9. parallelogram ABCD and  AOB? (A) 1 : 1 (B) 2 : 1 (C) 3 : 1 (D) 4 : 1 The three angles of a quadrilateral are in the ratio 2:5:7. The sum of the greatest and the smallest among these angles is 180°. What is the measure of the fourth angle? (A) 90° (B) 80° (C) 65° (D) 45° The given figure shows a parallelogram ABCD where AD = BD and  ABD = 50°. D

B

What are the measures of the adjacent angles of the parallelogram? (A) 50° and 130° (B) 40° and 90° (C) 30° and 90° (D) 20° and 160° The given figure shows a parallelogram ABCD. M is a point on CD such that AM = BM. D

A

M

C

B

C

6m

The area of the rectangle is (A) 24 m2 (B) 48 m2 (C) 60 m2 (D)80 m2 The given figure represents a rhombus WXYZ. W

Z

O X

Y

What are the respective lengths of the diagonals, WY and XZ? (A) 10 units and 14 units (B) 14 units and 18 units (C) 18 units and 10 units (D) 10 units and 20 units 10. What is the number of sides of a regular polygon whose each interior angle measures 150°? (A) 10 (B) 12 (C) 15 (D) 16

Conceptive Worksheet 1.

If the length of the diagonal of a square is 16 2  cm, then what is the perimeter of the square? (A) 64 2cm

(B) 64 cm

(C) 32 2cm

(D) 32 cm

B

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7th Class Mathematics

110 2.

In the given figure, ABCD is a square and PQRS is a rhombus. D

S

C s

A B

R d

r

9. 950

1300

c x

p A

3.

4.

b

a

q B

P

C

D

y

7.

8.

350 E X

D

What is the value of x? (A) 120° (B) 140°

(C) 150°

(D) 160°

Trapezium A quadrilateral having one pair or opposite sides are parallel is called a trapezium. D

C

A

A

6.

1000

C x

5.

900

Q

The relation in which alternative is correct with respect to the given figure? (A) p + r  b, then the sign of the expression will be positive and if a 2

Ritu had the larger share by

3 2 32 1    5 5 5 5

Number System Solutions 33.

165

3 3× 3 9 = = 4 4× 3 12 9>7  9 7  12 12 Vaibhav worked longer by fraction

 

9 7 97 2 22 1      hour.. 12 12 12 12 12  2 6 34.

35.

(ii)

2 2 2  27 54 of 27   27    18 3 3 3 3

(c)

3 of (i) 16 4

(i)

3 3 3  16 48 of 16   16    12 4 4 4 4

(ii)

3 3 3  36 108 of 36   36    27 4 4 4 4

(ii) 36

(i)

3 7  3 21 1 7   4 5 5 5 5

(d)

4 of (i) 20 5

(ii)

1 4 1 4 1 4   1 3 3 3 3

(i)

4 4 4  20 80 of 20   20    16 5 5 5 5

(iii)

2

(ii)

4 4 4  35 140 of 35   35    28 5 5 5 5

(iv)

5

(v)

2 2 4 8 2 4   2 3 3 3 3

(vi)

5 5  6 30 15 6    2 2 2 1

(vii)

11

4 11 4 44 2   6 7 7 7 7

(viii)

20 

4 20  4 80 16    5 5 5 1

(ix)

1 13  1 13 1 13    4 3 3 3 3

(x)

3 15  3 45 9 15     5 5 5 1

6 2  6 12 5   1 7 7 7 7 2 5  2 10 1   1 9 9 9 9

(a)

1 of 2

(i)

1 1 1  24 24 of 24   24    12 2 2 2 2

(ii)

1 1 1  46 46 of 46   46    23 2 2 2 2

(b)

2 of 3

(i)

2 2 2  18 36 of 18   18    12 3 3 3 3

(i) 24 (ii) 46

(i) 18

36.

(a)

1 5

1 26 3  26 78 3 3  5  3    15 5 5 5 5 5 (b)

(c)

(d)

(e)

(ii) 27

35

(ii) 35

3 4 3 27 5  27 135 3 56  5    33 4 4 4 4 4 1 72 4 1 9 7  9 63 3 72  7    15 4 4 4 4 4 1 46 3 1 19 4  19 76 1 4 6  4    25 3 3 3 3 3 1 3 6 4 5 6

1 13 13  6 78 78  2 39 1 3 6  6      19 4 4 4 4 42 2 2 2 3 8 (f) 5 2 17 17  8 136 1 3 8  8    27 5 5 5 5 5

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7th Class Mathematics

166 37.

(i)

1 of 4

(a)

1 1 1 1 1 1 1 of     4 4 4 4 4  4 16

(b)

1 3 1 3 1 3 3 of     4 5 4 5 4  5 20

(c)

1 4 1 4 1 4 4 44 1 of       4 3 4 3 4  3 12 12  4 3

(ii)

1 of 7

(a)

2 9

2 9

(b)

(b)

6 5

6 5

(a)

1 2 1 2 1 2 2 of     7 9 7 9 7  9 63

(b)

1 6 1 6 1 6 6 of     7 5 7 5 7  5 35

(c) 38.

(a)

(i)

(c)

39.

(c)

2 2 2 3 3

(v)

(ii)

2 1 5 5 4

2 7 6  5 9 2 7 32 7 32  7 224 44 6      4 5 9 5 9 5 9 45 45

(iii)

3 1 5 2 3

3 1 3 16 3  16 48 48  6 5      8 2 3 2 3 23 6 66 (iv)

5 3 2 6 7 5 3 5 17 5  17 85  2     2 6 7 6 7 6  7 42 42

(v)

2 4 3  5 7 2 4 17 4 17  4 68 33 3      1 5 7 5 7 5  7 35 35

3 6  8 4 (vi)

3 2 3 5 3 13 13  3 39 4 2 3  3   7 5 5 5 5 5

9 3  5 5

4 3 3  7 7

9 3 9  3 27    5 5 5  5 25

(vii)

1 15  3 8

4 3 25 3 25  3 75 75  5 15 1 3        2 7 7 7 5 7  5 35 35  5 7 7

1 15 1  15 15 15  3 5      3 8 3  8 24 24  3 8 (vi)

4 12  5 7

2 1 2 21 2  21 42 42  2 21 1 5       2 5 4 5 4 5  4 20 20  2 10 10

2 7  7 9

3 6 3  6 18 18  2 9      8 4 8  4 32 32  2 16 (iv)

3 10

(i)

1 3 1 3 1 3 3 of     7 10 7 10 7  10 70

2 7 2  7 14 14  7 2      7 9 7  9 63 63  7 9 (iii)

(vii)

4 12 4  12 48    5 7 5  7 35

2 2 2 8 16 2    3 3 3 3 9 (ii)

3 10

11 3  2 10

11 3 11  3 33    2 10 2  10 20 www.betoppers.com

Number System Solutions 40.

(i)

167

2 3 3 5 of or of 7 4 5 8

(iv)

2 3 2 3 23 6 62 3 of       7 4 7 4 7  4 28 28  2 14

Reciprocal of the fraction

6 5  (an proper 5 6

fraction) 12 (v) 7

3 5 3 5 3  5 15 15  5 3 of       5 8 5 8 5  8 40 40  5 8 LCM (14,8) = 56

Reciprocal of the fraction

3 3  4 12   14 14  4 56

fraction) 1 (vi) 8

3 3  7 21   8 8  7 56 21  12

12 7  (an proper 7 12

Reciprocal of the fraction



21 12  56 56

number) 1 (vii) 11



3 3  8 14

Reciprocal of the fraction number)



3 5 2 3 of  of 5 8 7 4

(ii)

1 6 1 6 1 6 6 62 3 of       2 7 2 7 2  7 14 14  2 7

42.

2 3 2 3 23 6 6 3 2 of       3 7 3 7 3  7 21 21  3 7



3>2



3 2  7 7

 41.

6 5

(i)

(i)

7 2 3 7 7 1 7 1 7 1 2    1 3 3 2 3 2 6 6

(ii)

(iii)

4 5 9

6 7 13 6 6 1 6 1 6 1 7    1 13 13 7 13  7 91 6

3 7  (an improper 7 3

(iv)

5 8 Reciprocal of the fraction  (an improper 8 5 fraction) 9 (iii) 7 9 7 Reciprocal of the fraction  (an improper 7 9 fraction)

(v)

fraction) 5 (ii) 8

1  11 (a whole 11

4 4 1 4 1 4 5    9 9 5 9  5 45

1 6 2 3 of  of 2 7 3 7 3 7

Reciprocal of the fraction

1 8   8 (a whole 8 1

1 4 3 3 1 13 13 1 13 4 4  3  3    1 3 3 3 3 9 9 1 3 4 2 1 7 7 1 7 1 7 3 4 4    2 2 2 4 2 4 8

(vi)

3 4 7 7 3 3 31 1 31  1 31 4 7  4 7     7 7 7 7 7  7 49 www.betoppers.com

7th Class Mathematics

168 43.

(i) 0.5 or 0.05

46. 0.5 > 0.05  0.5  0.50

For Dinesh Distance between place A and place B = 7.5 km

(ii) 0.7 or 0.5 0.7 > 0.5 (iii) 7 or 0.7 Distance between place B and place C = 12.7 km Distance between place A and place C  = Distance between place A and place B + Distance between place B and place C = 7.5 km + 12.7km = 20.2 kg  Distance travelled by Dinesh = 20.2 km 7.5

 7  7.0 7>0.7 (iv) 1.37 or 1.49 1.49 > 1.37 (v) 2.03 or 2.30 2.30 > 2.03 (vi) 0.8 or 0.88

 0.8  0.80 0.88 > 0.8 44.

45.

12.7

(i) 7 paise

20.2

7 paise = Re. 0.07 (ii) 1 rupees 7 paise 7 rupees 7 paise = Rs. 7.07 (iii) 77 rupees 77 paise 7 rupees 77 paise = Rs. 77.77 (iv) 50 paise 50 paise = Re. 0.50 (v) 235 paise 235 paise = Rs. 2.35. (i) 2.56 Place value of 2 in the decimal number 2.56 = 2xl = 2 (ii) 21.37 Place value of 2 are in decimal number 21.37 = 2 x 10 = 20 (iii) 10.25 Place value of 2 in the decimal number

(iv)

(v)

For Ayub Distance between place A and place D = 9.3 km Distance between place D and place C = 11.8 km  Distance between place A and place C = Distance between place A and place D + Distance between place D and place C = 9.3 km + 11.8km = 21.1km 9.3 11.8 21.1 Distance travelled by Ayub = 21.1 km  21.1> 20.2  So, Ayub travelled more by 21.11 km – 20.2 km = 0.91 km 21.11

 1  2 1 2 1 10.25  2        10  10 10 5 9.42 Place value of 2 in the decimal number

1  1  2 1 2 9.42  2       100  100 100 50 63.352 Place value of 2 in the decimal number

2 1  1  2 1 63.352  2       1000  1000 1000 500

20.20 0.91 47.

 20.2  20.20 

For Shyama Apples bought = 5 kg 300 g = 5.300 kg Mangoes bought = 3 kg 250 g = 3.250 kg Fruits bought = Apples bought + Mangoes  bought = 5.300 kg + 3.250 kg = 8.550 kg 5.300

3.250 8.250 www.betoppers.com

Number System Solutions

48.

For Sarala Oranges bought = 4 kg 800 g = 4.800 kg Bananas bought = 4 kg 150 g = 4.150 kg  Fruits bought – Oranges bought + Bananas bought = 4.800 kg+ 4.150 kg = 8.950 kg 8.950 > 8.550  So, Sarala bought more fruits. (i) 4.8  10

4.8  10 

(ii)

4.8 4.8 48 48  4 12      0.48 10 10.0 100 100  4 25

169 50.

(i) (ii) (iii) (iv) (v)

51.

(i)

7  3.5 

52.5  10

(ii)

52.5 52.5 525 525  25 21 52.5  10      10 10.0 100 100  25 4 1  4  5.25 4 (iii) 0.7  10 0.7  10  (iv) (v) (vi) (vii) 49.

(i) (ii) (iii) (iv) (v) (vi)

7.9  1000 7.9  1000  0.0079 26.3  1000 26.3  1000  0.0263 38.53  1000 38.53  1000  0.03853 128.9  1000 128.9  1000  0.1289 0.5  1000 0.5  1000  0.0005 7  3.5

7 7.0 70 2    2 35 3.5 35 1

36  0.2

36  0.2  (iii)

36 36.0 360    180 0.2 0.2 2

3.25  0.5

3.25  0.5  0.7 0.7 7    0.07 10 10.0 100

33.1  10 33.1  10  33.1 272.23  10 272.23  10  27.223 0.56  10 0.56  10  0.056 3.97  10 3.97  10  0.397 2.7  100 2.7  100  0.027 03  100 03  100  0.003 0.78  100 0.78  100  0.0078 432.6  100 432.6  100  4.326 23.6  100 23.6  100  0.236 98.53  100 98.53  100  0.9853

(iv)

30.94  0.7

30.94  0.7 

(v)

30.94 30.94 3094 221     44.2 0.7 0.70 70 5

0.5  0.25

0.5  0.25  (vi)

3.25 3.25 325 13     6.5 0.5 0.50 50 2

0.5 0.50 50   2 0.25 0.25 25

7.75  0.25

7.75  0.25  (vii)

7.75 775 31    31.0 0.25 25 1

76.5  0.15

76.5 76.50 7650 510     510 0.15 0.15 15 1 (viii) 37.8  1.4 76.5  0.15 

37.8  1.4  (ix)

37.8 378 27    27 1.4 14 1

2.73  1.3

2.73  1.3 

2.73 273 21    2.1 1.3 1.30 10

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7th Class Mathematics

170 52.

Cost of nine prisms = Rs 6.84

56.

 Cost of one prism

=

Height of Eiffel Tower = 324 m Height of Leaning Tower of Pisa = 55.9 m Thus, difference between their heights = (324 – 55.9) m = 268.1 m

Rs 6.84  684 1  = Rs  ×  9  9 100 

Now, 1 m = 100 cm

1   = Rs  76×  = Rs0.76 100   57.

 Cost of nine prisms

1    5  Rs0.76  5  Rs  76   100  

25  6.25kg of vegetables 4 Now, Linda's dad was out of town for a month.  In that particular month, the remaining three members eat 6.25 × 3 = 18.75 kg of vegetables = (18.75 × 1000)g ( 1kg = 1000 g)

1    Rs  380    Rs3.80 100   The correct answer is C. 53.

 1 875    1000  g  100 

For 16 packets, Connie earns Rs3.68



For

each

packet,

Connie

earns

Rs3.68  368 1   Rs    16  16 100 

58.

1    Rs  23    Rs0.23 100   The correct answer is A. 54.

Charges for parking 1 vehicle = Rs5 Charges per person in the vehicle = Rs1.20 Charges for five persons in the vehicle = 5 × 1.20 = Rs6.00

59.

Total money = Rs(5 + 6) = Rs11 The correct answer is D. 55.

The distance covered by the train = (150 + 600) m = 750 m

60.

Speed = 50 km/h



50  1000 125 m / sec  m / sec 3600 9

Therefore, the time taken 

The correct answer is D. www.betoppers.com

Thus, difference between their heights = 268.1 × 100 cm = 26,810 cm The correct answer is D. In one month, four members eat 25 kg of vegetables  In one month, each member eats

750  54 125 9

61.

= 18 750 g of vegetables The correct answer is B. Total cloth with the shopkeeper at the beginning of the month = 200 m He sells 5.2 m of cloth everyday for 30 days.  Total cloth sold in the month = (5.2 × 30) m = 156 m  Quantity of cloth left with the shopkeeper = (200 – 156) m = 44 m = (44 × 100) cm (because 1 m = 100 cm). = 4 400 cm The correct answer is D. Amount spent by Angel = Rs4.50 Lily spent twice of Rs4.50. Thus, amount spent by Lily = 2 × Rs 4.50 = Rs 9 Thus, total amount spent by the two friends = Rs 4.50 + Rs 9 = Rs 13.50 The correct answer is C. Cost of 1 packet of cookies = Rs0.75  Cost of 6 packets of cookies = 6 × Rs 0.75 = Rs 4.50 Amount paid to the cashier = Rs5 Thus, change given back by the cashier = Rs 5 – Rs 4.50 = Rs 0.50. The correct answer is A. Cost of 1 kg of cashew nuts = Rs7.75 Thus, cost of 3 kg of cashew nuts = 3 × Rs 7.75 = Rs 23.25 The correct answer is C.

Number System Solutions 62.

63.

171

Total number of people = 1 + 3 = 4 Cost of 1 milk shake = Rs1.25 Cost of 4 milk shakes = 4 × Rs 1.25 = Rs 5. Thus, the four friends spent Rs5 on the milkshakes. The correct answer is C.

84  63  1.5 

1  42 21

1  42 21 = 84 – 4.5 + 42 = 79.5 + 42 = 121.5 The correct answer is A. 19.25 – 7 + 16.2 × 15 = 19.2 – 7 + 243 = 12.25 + 243 = 255.25 The correct answer is A.  84  94.5 

64.

65.

(i)

70.

8, –13, 25, 0

71.

6 18 23 0 , , , 1 1 1 1

72.

(A)



  

6 2 12 6 3   ,  11 2 22 11 3

18 6 4 24 6 5 ,   ,  33 11 4 44 11 5 (C) (same as above) (D) 

8 2 16 8 3 24 8 4 32 8  5   ,   ,   , 15 2 30 15 3 45 15 4 60 15  5

4 3  7 7



40 8 6 48 ,   75 15 6 90

7.5  7 4 3   1 7 7 7

73.

(A)

52.5 4 3   7 7 7

74.

(i)

9 7

(v)

4 1

(i)

15 2 30   16 2 32

(ii)

15 3 75   16 3 48

(iii)

15 3 75   16 3 48

(iv)

15 6 90   16 6 96

52.5  4  3 7

59.5 7 = 8.5 The correct answer is B.

67. 68.

75.

17 75   6.4  5  15 25 17 = 3 + 6.4 × 5 – 15 = 3 + 32 – 15 = 35 – 15 = 20 The correct answer is C. 5 1 (i) (ii) 4 7 2 (vi) 0 (vii) –1 (A) 6 (B) –12

2 (iii) 19(iv)  3

3 5

(B)

8 17

(C)

21 25

(D)

40 59

(ii)

31 40

(iii)

16 33

(iv)

19 50

76. (v) –2

(i)

19 2 38   5 2 10

(ii)

19 5 95   5 5 25

19 7 133   5 7 35

(iv)

19 4 76   5 4 20

(iii) (C) 9 (D) –16

5  6  3 30  3 33   3  2  5 5  5 25

15 3 6 18 ,   25 5 6 30



66.

(ii)

3 2 6 3 3 9 3 4 12 3 5   ,   ,   ,  5 2 10 5 3 15 5 4 20 5 5

(B)

3 4 3 4 3 12.5     2  2.5  3    2 5 7 14 7 14  7.5 

3  2 6  15  8 7

69.

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7th Class Mathematics

172 77.

Note: Express

15 as rational number with 16

denominator (i) 96 (ii) –32

15 6 90   16 6 96

(i) (iii)

15 3 45   16 3 48

(iii) 48

5 3  4 1    2 5 6  5 3 3     2 5 6

(iv) 80

(ii)

15 2 30   16 2 32

(iv)

15 5 75   16 5 80

3 5 3  25 28    10 2 10 10

2  3  1 2 1 3       5  7  14 5 6 2

(ii) 78.

79.

(i)

1260 63  1540 77

(ii)

1260 126 126   1540 154 154

(iii)

1260 90  1540 111

(iv)

1260 18 18   1540 22 22

(i)

6 2  15 5 18 3  24 4

(ii) (iii) (iv)

2  3 1  3    5  7 14  12 2  6  1  1   5  14  4 2  5  1   5  14  4 1 1 4  7 11    7 4 28 28 82.

83.

80.

5 10 25   (b) 8 16 40 (c)

81.

(i)

84.

5 2 3 3 1     2 3 5 5 6

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2 9

19 6

 11  11 (i)     15  15

(i)

1 13

(ii)

19 13

(iii) 5

85.

8 7 5  (v) 1  (vi) –1 5 3 2 (i) Proporty one (Multiplicative identity) (ii) Commutative (iii) Multiplicative inverse.

86.

6 16 96   13 7 91

87.

6 5 6 6

4 6

3 6

2 6

1 6

0 0

5 3 2 1      2 5 5 6 

(iv)

(iv)

7 14 35   9 18 45

8 4 20   (d) 26 13 65

6 5

(iii)

  13    13  13 (ii)           17    17  17

1362 1302 434   (v) 1953 1953 657 3 9 21   (a) 4 12 28

5 9

(ii) (v)

21 3  35 5 36 3  84 7

2 8

(i)

7 1

7 2

7 3

7 4

7 5

7 6

7 7

Number System Solutions 9 11

88.

89. 90.

173 5 11

5.

2 11 0

1 2 3 4 3 , , , , 2 3 2 3 4 a×b=b×a

6.

21 6 6  21      5 7 7  5  7.

CONCEPTIVE WORKSHEET 1.

Integers are commutative under addition. For example, –2 + 3 = 3 – 2 Thus, statement D is correct. The correct answer is D. 2. Integers are closed under addition i.e., the sum of any two integers is also an integer. Thus, for integers a and b, (a + b) is also an integer. The sum of integers may be positive or negative or zero depending upon their values. Thus, the information in alternative B is incorrectly matched. The correct answer is B. 3. Consider the expression given in alternative B. L.H.S = 28 + (–16) + (–54) = 28 + [(–16) + (–54)] = 28 + (–70) = –42 R.H.S = (–46) + (–11) + (67) = [(–46) + (–11)] + (67) = (–57) + 67 = 10 (– 42) < 10  [28 + (– 16) + (– 54)]  2 m > 0.99 m Thus, Pam purchased the maximum quantity of cloth. The correct answer is B. Cost of 1 kg of tomatoes = Rs 0.58 Cost of 9 kg of tomatoes = 9 × Rs 0.58 = Rs 5.22 Hence, money spent by Liz on buying tomatoes = Rs 5.22 The correct answer is A. Total number of people = Kate + four friends = 1 +4=5 Number of scoops of ice cream ordered by each person = 2 Number of scoops of ice cream ordered by five persons = 2 × 5 = 10 Cost of one scoop of ice cream = Rs 1.50 Cost of ten scoops of ice cream = 10 × 1.50 = Rs 15 Thus, total money spent by the five friends = Rs 15 The correct answer is B.

179 58.

59.

 1  The expression is : 5  35.5  3.5   17  8  3 



16  39   136 3

16  13  3  136 3 = 16 × 13 – 136 = 208 – 136 = 72 The correct answer is D. Total money spent by the friends = 5 × Rs 1.50 + 4 × Rs 2.75 = Rs 7.50 + Rs 11.00 = Rs 18.50 The correct answer is B. Cost of 1 baseball bat = Rs 3.75 Cost of 11 baseball bats = 11 × 3.75 = Rs 41.25 Cost of 1 baseball = Rs 1.25 Cost of 12 baseballs = 12 × 1.25 = Rs 15 Total money spent on the whole purchase = Rs 41.25 + Rs 15 = Rs 56.25 The correct answer is B. 

56.

57.

60. 61.

Charges for 1 unit of the package = Rs 0.50 Charges for 7 units of the package = 7 × 0.50 = Rs 3.50 Charges for handling a package = Rs 1.50 Total charges for mailing = 1.50 + 3.50 = 5.00 = Rs 5 The correct answer is A. Firstly multiply the given numbers without the decimal points. Step 1: Write the expanded forms of the numbers which are to be multiplied. This can be done as: 325 = 300 + 20 + 5 232 = 200 + 30 + 2 Step 2: Write the expanded forms of the two numbers on the top and at the side of a table as: 300 20 5 200 30 2 Step3:Write the products of all the pairs of numbers on the top and at the side of the table in each corresponding cell of the table as: 300 20 5 60,000 4,000 1,000 200 9,000 600 150 30 600 40 10 2 Step 4: The required product of the given numbers can now be calculated by adding the numbers in all the cells of the table. Therefore, 325 × 232 = 60,000 + 4,000 + 1,000 + 9,000 + 600 + 150 + 600 + 40 + 10 = 75,400 Now, both the numbers 3.25 and 2.32 have two decimal places each. Thus, the product of the numbers 3.25 and 2.32 will have four decimal places. Thus, the number will be 7.540 0 or 7.54. The correct answer is C.

39 58 –24

62.

15 21

63.

3 4

64.

7 8 www.betoppers.com

7th Class Mathematics

180

65.

x (A) 2y

x (B) y2

66.

Associative property under multiplication

a   b  c  a  b  c 67.

1  1 9 1   1      8  8 8 8  9         1  8  9 It is not multiplicative inverse.

68.

69.

70.

1 10 3  3 3 3 10  1 10 3 So, it is multiplicative inverse (i) 0 does not have reciprocal. Because a number divided by 0 is undifined. (ii) 1 and –1 are equal reciprocal (iii) 0 is the number equal to its negative. (i) no (ii) 1 & –1 1 5 (iv) x (v) Rational number (vi) Positive (iii)

71.

2 1 and 5 2 Equavalent rational numbers of the given rational numbers are having a common denominator are 2 2 4   5 2 10 1 5 5   2 5 10 5 4 , 10 10 2 3 2 1 1 2 3 4 1 , , , ,0, , , , , 5 10 10 10 10 10 10 10 2

72.

2 4 and 3 5 Equavalent rational numbers of the given rational numbers are having a common denominator are (i)

2 5 4 3  and  3 5 5 3 10 12 and 15 15 Dividing both the rational numbers 6 we get, 10 6 12 6  and  15 6 15 6 60 72 and 90 90 61 62 63 71 , , ,.......... are the rational numbers 90 90 90 90 between the

2 4 and . 3 5

3 5 and 2 3 Equavalent rational numbers of the given rational numbers are having a common denominator are (ii)

3 3 5 2  and  2 3 3 2 

9 10 , 6 6



8 7 6 5 9 , , , ....... are the rational numbers 6 6 6 6 6

between

3 5 and . 2 3

1 1 and 4 2 Equavalent rational numbers of the given rational numbers are having a common denominator are (iii)

1 2 1 4  and  4 2 2 4 2 4 and 8 8 Dividing the both rational numbers with 5 we get, 2 5 4 5  and  8 5 8 5

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Number System Solutions

181 25.

10 45 and 40 40

45   11 12 13  , , ........  are the rational number 40   40 40 40 between

1 1 and . 4 2

73.

1 2 3 4 5 , , , , 2 3 4 5 6

74.

3 4 12   5 4 20

26.

3 5 15   4 5 20

27.

12 5 15 5  and  20 5 20 5 60 75 and 100 100

75.

28.

61 62 74 , ........ are the required rational 100 100 100 numbers. D Additive identity of rational numbers.

SUMM ATIVE WORKSHEET 29.

KEY

24.

1

2

3

4

5

6

7

B

C

D

B

A

D

C

8

9

10

11

12

13

14

D

B

C

B

D

C

B

15

16

17

18

19

20

21

B

D

A

C

A

A

C

22

23

B

B

The expression is:

Cost of 1 gallon of gasoline = Rs 2.50  Cost of 12 gallons of gasoline = 12 × Rs 2.50 = Rs 30 Total amount paid by Henry = Rs 100 Thus, amount given back to Henry = Rs (100 ? 30) = Rs 70 The correct answer is C. Total number of students = 27 Number of students who went on the tour = 27 ? 6 = 21 Total entry fee paid by the students = Rs 6.30 Thus, entry fee paid by each student $6.30   $0.30 21 The correct answer is D. Cost of 1 baseball = Rs 4.25  Cost of 2 baseballs = 2 × Rs 4.25 = Rs 8.50 Total money paid by Dennis = Rs 10 Thus, amount given back to Dennis by the store owner = Rs (10 – 8.50) = Rs 1.50 The correct answer is B. The expression is 0.002 × 0.003 0.002 < 1 0.003 < 1 The product of numbers less than 1 is less than both the numbers 0.002 × 0.003 = 0.000 006 The correct answer is A.

 2.8  4.7  17   4  1.5 

 7.5   17   4  1.5  = 17 – 5 × 4 = 17 – 20 = –3 The correct answer is A. 30.

 1 7 14   3    4  1.5  2 4 7 7  14      4  1.5 2 4

8   42.05  12.15  5

8   29.90 5 = 47.84 The correct answer is B.

7 4  14      4  1.5 2 7 = 14 – 2 + 4 × 1.5 = 12 + 4 × 1.5 = 12 + 6 =8 The correct answer is D. www.betoppers.com

7th Class Mathematics

182

 1 7 14   3    4  1.5  2 4 7 7  14      4  1.5 2 4 7 4  14      4  1.5 2 7 = 14 – 2 + 4 × 1.5 = 12 + 4 × 1.5 = 12 + 6 = 18 The correct answer is D. 1 3 2 32.   1   6  2.5  10 5 4 3  1 7  20       2.5  10 5 4 3  4  35  20     2.5  10  20  3 39 20    2.5  10 20 3 = 13 + 2.5 – 10 = 15.5 – 10 = 5.5 The correct answer is C. 33. Total amount that Barbara had = Rs 100 Cost of 8 pounds of cookies = 8 × 0.50 Cost of 3 pizzas = 3 × 2.50 Cost of 2 dresses = 2 × 20 Therefore, amount she was left with = 100 – (8 × 0.50 + 3 × 2.50 + 2 × 20) = 100 – (4 + 7.50 + 40) = 100 – 51.50 = Rs 48.50 The correct answer is A. 4 25 34. 5 × 1.5 + 2.5 × 6.4 + 10   4  3 3 4 55 = 5 × 1.5 + 2.5 × (2 × 3.2) + (5 × 2) ×  4  3 3 4 45  = 5 × 1.5 + (2.5 × 2) × 3.2 + 5   2    5  3 3  8 20 = 5 × 1.5 + 5 × 3.2 × 5   5  3 3 8 20    5 1.5  3.2    3 3   8  20    5  4.7   3   12    5  4.7   3   = 5(4.7 – 4) = 5 × 0.7 = 3.5 The correct answer is B.

35.

31.

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36.

37.

38.

39.

40.

Number of pencil packets = 17 Number of eraser packets = 23 Cost of each packet = Rs 5.5 Therefore, total cost = 5.5 × (17 + 23) = 5.5 × 40 [firstly bracket is solved] = Rs 220 [secondly multiplication is valuated] The correct answer is A. Length of page = 10 cm 2 2 1 Length of pieces to be cut  mm   cm 5 5 10 = 0.04 cm 10  250 Number of pieces  0.04 The correct answer is C. Cost for admission of one child = Rs 1.50 Cost for admission of one adult = Rs 2.50 Total cost of all the tickets = Rs (5 × 1.50 + 10 × 2.50) = Rs 32.5 The correct answer is B. Cost of 1 gold fish = 85 cents Cost of 1 guppy = 80 cents Total worth of all fishes = Rs (2 × 0.85 + 3 × 0.80) = Rs 4.1 (As Rs 1 = 100 cents) The correct answer is C. Number of apples bought = 4 Cost of 4 apples = 2 × 4 = Rs 8 Number of oranges bought = 3 Cost of 3 oranges = 1.50 × 3 = Rs 4.50 Number of pears bought = 5 Cost of 5 pears = 2.50 × 5 = Rs 12.50 Amount paid by Stella = 8 + 4.50 + 12. 50 = Rs 25 The correct answer is C. For any non-zero rational number a, the multiplicative inverse of a is the number which when multiplied by a gives the product as 1. Ther efore, the multiplicative inverse of a is 

 because 

41.

Thus, the expression shows the existence of multiplicative inverse of a. The correct answer is A. The given expression shows the distributive property of multiplication over addition of rational numbers. The correct answer is A.

42.

ba ab

43.

The additive inverse of any rational number is the negative of the rational number. Thus, additive inverse of The correct answer is A.

3  3  3       7  7  7

Number System Solutions 44.

45.

183

3  3  3       Additive inverse of 2  2  2 3 Multiplicative inverse of = Reciprocal of 2 3 2    2 3  Product of the additive inverse and the multi3 3 2    1 plicative inverse of = 2 2 3

48. 49.

 2  9   4  6   5   7    11   5        154  495   140  462    385  385   385  385 154  495  140  462  385 19  385 19 19 The additive inverse of  is . 385 385 

The correct answer is B. The distributive property of rational numbers is given by a × (b + c) = a × b + a× c, where a, b, and c are rational numbers. The second step is given as

The correct answer is B.

1 2  1 2 2  3   1    7  3   3  1  7        Thus, the second step uses the distributive property. The correct answer is B. 46.

Multiplicative inverse of –0.015 



1   0.015

1 0.015

 1  0.015  1 

15 1000

1000 15 200  3 2  66 3  1 

47.

The correct answer is A. Rational numbers are commutative for multiplication. For example,

2 4 8   3 7 21 4 2 8   7 3 21 Thus, the statement given in alternative D is correct. The correct answer is D.

Same as above (repeated problem) L.C.M of 5, 7, and 11 is 385.

HOTS WORKSHEET 1) 2) 3) 4) 5) 6) 7) 8) 9) 10) 11) 12) 13) 14) 15) 16) 17.

–317 –282 cooler 2 0C –10 –181; 181 ft below the surface – 260; 260 m below the surface –16 –5 –12 0C 26,849 ft –12 and a loss of 12 yards –2800 –100 a descent of 100 feet –30 0C (a) 16 (b) 30 (c) –3 (d) –4 Cost of 1 rose = Rs 0.50  Cost of 5 roses = 5 × Rs 0.50 Cost of 1 lily = Rs 0.30  Cost of 2 lilies = 2 × Rs 0.30 Cost of 1 lotus = Rs 0.75  Cost of 3 lotuses = 3 × Rs 0.75 Thus, total amount spent by Samantha = 5 × Rs 0.50 + 2 × Rs 0.30 + 3 × Rs 0.75 = Rs 2.50 + Rs 0.60 + Rs 2.25 [multiplication is done] = Rs 3.10 + Rs 2.25 [addition is done] = Rs 5.35 [addition is done] The correct answer is C. www.betoppers.com

7th Class Mathematics

184 18.

19.

20.

Cost of 1 ticket for an adult = Rs 5 Number of adults = 2  Cost of 2 tickets for adults = 2 × Rs 5 Cost of 1 ticket for a kid = Rs 2.50 Number of kids = 3  Cost of 3 tickets for kids = 3 × Rs 2.50  Total money spent by Mike = 2 × Rs 5 + 3 × Rs 2.50 = Rs 10 + Rs 7.50 [multiplication is done] = Rs 17.50 [addition is done] The correct answer is C. Cost of 1 hamburger = Rs 2.50  Cost of 2 hamburgers = 2 × Rs 2.50 Cost of a pack of fries = Rs 0.75  Cost of 4 packs of fries = 4 × Rs 0.75 Cost of a soft drink = Rs 1.25  Cost of 2 soft drinks = 2 × Rs 1.25 Amount paid by Martha = Rs 20 Thus, amount received by Martha = Rs 20 ? 2 × Rs 2.50 ? 4 × Rs 0.75 ? 2 × Rs 1.25 = Rs 20 – Rs 5 – Rs 3 – Rs 2.50 [multiplication is done] = Rs 15 – Rs 3 – Rs 2.50 [subtraction is done] = Rs 12 – Rs 2.50 [subtraction is done] = Rs 9.50 [subtraction is done] The correct answer is B. The money Smith earned for one dog



21.

22.

23.

1  3  0.25 12 The total length of ribbon needed by Linda = 2.98 + 1.2 + 0.25 = 4.43 feet The correct answer is D.

 3 inches feet 

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22.35  $7.45 3 The correct answer is C. Cost of 100 g of coffee = Rs 2  Cost of 500 g of coffee = Rs 2 × 5 = Rs 10 [500 g contains five 100 g] Cost of 1 pack of fruit juice = $1.25 Thus, cost of 2 packs of fruit juice = $1.25 × 2 = $2.50 Total money spent in buying coffee and juice = $(10 + 2.50) = $12.50 Hence, amount of money that cashier will return to Mike = $(20 – 12.50) = $7.50 The correct answer is A. Cost of a burger = Rs 1.25  Cost of 3 burgers = $1.25 × 3 = $3.75 Cost of a glass of juice = Rs 1.20 Cost of 2 glasses of juices = $1.20 × 2 = $2.40 Hence, amount paid for the meal = $(3.75 + 2.40)= Rs 6.15 The correct answer is D. Money before purchasing cakes = Rs 5 Money spent in purchasing cakes = Rs 2.48 Money remaining after purchasing cakes = $ (5 – 2.48) = Rs 2.52 Teddy gives half the remaining money to his younger brother. $

24.

25.

26.

23  66 = Rs 15.18 100

The total money earned by Smith = Rs 15.18 × 6 = Rs 91.08 The correct answer is D. The number of hours for which the pantry boys work every day = 7 The amount earned by them in a day = 7 × 9.23 = Rs 64.61  the amount earned by them in a week = Rs 64.61 × 7 = Rs 452.27 The correct answer is A. 1 feet = 12 inches

Money spent in purchasing tickets = Rs 9.85 Money spent for food = Rs 12.50  Total money spent = $(9.85 + 12.50) = $22.25 Hence, money spent by each person

1 2

 Money left with Teddy =   $2.52 = Rs

27.

1.26 The correct answer is B. Cost of 1 kg of apples = Rs 1.45  Cost of 2 kg of apples = 2 × $1.45 = $2.90 Cost of 1 packet of strawberries = Rs 1.50  Cost of 3 packets of strawberries = 3 × $1.50 = $4.50  Total cost of the purchases = $ (2.90 + 4.50) = $7.40 Matt gives Rs 10 to the cashier. Thus, the amount of money that will be returned = $ (10 – 7.40) = $2.60 The correct answer is D.

Number System Solutions 28.

29.

30.

31.

32.

Cost of cream = Rs 3.50 Cost of strawberry = Rs 5 Number of apples purchased = 8 Cost of each apple = Rs 0.50 Therefore, total amount spent = Rs (3.50 + 5 + 8 × 0.50) = Rs (3.50 + 5 + 4) = Rs 12.50 The correct answer is A. Number of cans of grape juice = 2 Number of cans of apple juice = 4 Thus, total number of cans of fruit juice = 2 + 4 = 6 Cost of each can of fruit juice = Rs 1.50 Therefore, total amount spent on fruit juice = 6 × Rs 1.50 = Rs 9 The correct answer is C. Cost of each ticket for an adult = Rs 60 Number of tickets bought for adults = 2 Thus, amount spent on the tickets for adults = 2 × Rs 60 = Rs 120 Cost of each ticket for a child = Rs 35.5 Number of tickets bought for children = 2 Thus, amount spent on the tickets for children = 2 × Rs 35.50 = Rs 71 Thus, total amount spent on the tickets = Rs (120 + 71) = Rs 191 The correct answer is D. Cost of each ticket = Rs 3.50 Thus, cost of four tickets = 4 × Rs 3.50 = Rs 14 Cost of each packet of popcorn = Rs 0.90 Thus, cost of four packets of popcorn = 4 × Rs 0.90 = Rs 3.60 Thus, total amount spent by the friends = Rs (14 + 3.60) = Rs 17.60 The correct answer is C.

185 by Commutativity of rational numbers under addition

8  2 11      3 5 7  Thus, the expression lent

8  11  8 2      3  7 3 5 8  11 2      3  7 5 by distributivity of multiplication under addition

8  2  11       3  5  7  

of

the

given

expression

 11 8   2 8       .  3 7 3 7 33.

The correct answer is C. Same as above (repeated problem)

34.

The given rational number is 3 .

4 7 4 3 7  4 21  4 25 3     7 7 7 7

The multiplicative inverse of the rational number

3

4 25 25 or  is the reciprocal of  . 7 7 7

Therefore, the multiplicative inverse of the rational number 

25 7 is  . 7 25

The additive inverse of 



7 is the negative of 25

7 . 25

Ther efore, additive inverse of 

7 25

=

 7  7     25  25 It can be noted that

7  7   7  7     0    25  25 25  25 

 11 8   2 8   11  8   2  8             3 7   3 7   3 7   3 5  8   11 8  2   3 7 3 5

By commutativity of whole numbers under multiplication

form

8  2 11      is an equiva3 5 7 

Thus, the additive inverse of the multiplicative inverse of the number 3 35.

4 7 is . 7 25

The correct answer is C. The given expression is

 3 4 11 4  5 10  15  6  45   4 .   L.C.M (10, 15, 6, 45) = 90 The given expression can be rewritten as:  27 24 165 8  5  27  24  165  8  5  90  90  90  90   4     4 90  200 5  104 5  26       90 4   45 4  9

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7th Class Mathematics

186 Therefore, the value of the given expression is

26 . 9

Thus, the expression

6 3     4  is an equiva11  2 

lent

the

The additive inverse of a number is a number which when added to the original number gives 0.

6  6 3       4   . 11   11 2  

26 26 Now, the additive inverse of is . 9 9 26 can be written in the form of a mixed frac9 8 tion as 2 . 9

39.

The correct answer is D.

4 8  9  4   76 can be rewritten as  .  9 9 9

Thus, 

37.

38.

40.

76 9 is  . 9 76

KEY

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1

2

3

4

5

6

7

8

B

B

D

C

D

A

B

D A,B,C

10

11

12

13

14 15 16 17

C,D A,C B,D A,B D

6 3      4   11  2 

6 3     4 11  2 

1  2 11   1 2  11   exhibits     3 5 8  3 5 8

IIT JEE WORKSHEET

6  6 3       4   11   11 2   6 3 6     4  11 2 11

By distributivity of rational numbers under multiplication

4 4 and 5, which is  5  4 . 5 5

the associative property of rational numbers under addition. The correct answer is B.

  The correct answer is C. The sum of any two rational numbers is also a rational number. Therefore, rational numbers are closed under addition. The correct answer is B.



8 8  2 4   10 10  2 5

The correct answer is A. Addition is associative for rational numbers i.e., for any three rational numbers a, b, and c, a + (b + c) = (a + b) + c. Thus, the r elation 

9 4 is the multiplicative inverse of 8 . 76 9

By commutativity of rational numbers under multiplication

0.8 

product of

Now,  

 Multiplicative inverse of 

expression

Thus, the product of the additive inverse of “0.8 and the multiplicative inverse of 0.2 equals the

The multiplicative inverse of a number is a number which when multiplied to the original number gives the product as 1.

 76   9      1  9   76 

given

 Additive inverse of – 0.8 = Additive inverse of 4  4 4       5  5 5 2 22 1 0.2    10 10  2 5  Multiplicative inverse of 0.2 = Multiplicative 1 inverse of  5 5

8 2 . 9 8

of

The correct answer is D.

Thus, the additive inverse of given expression is

36.

form

V.

19

20

21

9

8

1

Matrix matching A–t B–r

C C

C–p

B

9

18 4

D–q, s

Number System Solutions

187

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 1.

264400  26.44  10000

C)

169  69 

1 1  100  2 2 5 5  10 

 26.44  10000

 5.142 



2

 10000

= 5.142 × 100 = 514.2



1 25

2 5

 

[B] 4.

45  6.708,

4.5  2.121

4500  45  100 = 6.708 × 10 = 67.08

450  4.5  100 = 2.121 × 10 = 21.21.



4500  450  67.08  21.21 = 88.29

[C] 8.

6.9  100  690 = 26.2 Itr lies between 20 to 30. [D] x 2  y2  z2  x 2  y2  y2

9.

= x×y×z



 xyz 

2

 A, B, C are correct answer

13.

9  4  13 94  9  4 9  4  3 2  A, B is correct answer s

15.

8100  81  100  81  100 = 9 × 10 = 90.

21.

22.

132  36  4  13  6  2 = 13 – 12 =1

v. A)

144  16  4  12  4  2 = 12 + 8 = 20.

B)

7456  86.34 It lies between 80 & 90  www.betoppers.com

188

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7th Class Mathematics

2. EXPONENTS AND POWERS SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

3.

4.

5.

6.

(i) 26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 (ii) 93 = 9 × 9 × 9 = 729 (iii) 112 = 11 × 11 = 121 (iv) 54 = 5 × 5 × 5 × 5 = 625 (i) 6 × 6 × 6 × 6 = 64 (ii) t × t = t2 (iii) b × b × b × b = b4 (iv) 5 × 5 × 7 × 7 × 7 = 52 × 73 (v) 2 × 2 × a × a = 22 × a2 (vi) a × a × a × c × c × c × c × d = a3 c4 d (i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 =29 (ii) 343 = 7 × 7 × 7 = 73 (iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 36 (iv) 3125 = 5 × 5 × 5 × 5 × 5 = 55 (i) 43 = 4 × 4 × 4 = 64 34 = 3 × 3 × 3 × 3 = 81 Therefore, 34 > 43 (ii) 53 = 5 × 5 × 5 =125 35 = 3 × 3 × 3 × 3 × 3 = 243 Therefore, 35 > 53 (iii) 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 82 = 8 × 8 = 64 Therefore, 28 > 82 (iv) 1002 or 2100 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 2100 = 1024 × 1024 × 1024 × 1024 ×1024 × 1024 × 1024 × 1024 ×1024 × 1024 1002 = 100 × 100 = 10000 Therefore, 2100 > 1002 (v) 210 and 102 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 102 = 10 × 10 = 100 Therefore, 210 > 102 (i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23. 34 (ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 . 5 (iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22. 33. 5 (iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 . 3 2 . 5 2 (i) 2 × 103 = 2 × 10 × 10 × 10 = 2 × 1000 = 2000 (ii) 72 × 22 = 7 × 7 × 2 × 2 = 49 × 4 = 196 (iii) 23 × 5 = 2 × 2 × 2 × 5 = 8 × 5 = 40

7.

8.

9.

(iv) 3 × 44 = 3 × 4 × 4 × 4 × 4 = 3 × 256 = 768 (v) 0 × 102 = 0 × 10 × 10 = 0 (vi) 52 × 33 = 5 × 5 × 3 × 3 × 3 = 25 × 27 = 675 (vii) 24 × 32 = 2 × 2 × 2 × 2 × 3 × 3 = 16 × 9 = 144 2 (viii) 3 × 104 = 3 × 3 × 10 × 10 × 10 × 10 = 9 × 10000 = 90000 3 (i) (–4) = (–4) × (–4) × (–4) = –64 (ii) (–3) × (–2)3 = (–3) × (–2) × (–2) × (–2) = 24 (iii) (–3)2 × (–5)2 = (–3) × (–3) × (–5) × (–5) = 9 × 25 = 225 3 3 (iv) (–2) × (–10) = (–2) × (–2) × (–2) × (–10) × (–10) × (–10) = (–8) × (–1000) = 8000 (i) 2.7 × 1012; 1.5 × 108 2.7 × 1012 > 1.5 × 108 (ii) 4 × 1014; 3 × 1017 3 × 1017 > 4 × 1014 (i) 32 × 34 × 38 = (3)2 + 4 + 8 (am × an = am+n) = 314 (ii) 615 ÷ 610 = (6)15 – 10 (am ÷ an = am–n) = 65 (iii) a3 × a2 = a(3 + 2) (am × an = am+n) = a5 (iv) 7x + 72 = 7x + 2 (am × an = am+n) (v) (52)3 ÷ 53 = 52 × 3 ÷ 53 (am)n = amn = 5 6 ÷ 53 = 5(6 – 3) (am ÷ an = am–n) = 53 (vi) 25 × 55 = (2 × 5)5 [am × bm = (a × b)m] = 105 (vii) a4 × b4 = (ab)4 [am × bm = (a × b)m] (viii) (34)3 = 34 × 3 = 312 (am)n = amn (ix) (220 ÷ 215) × 23 = (220 – 15) × 23 (am ÷ an = am–n) = 25 × 2 3 = (25 + 3) (am × an = am+n) = 28 (x) 8t ÷ 82 = 8(t – 2) (am ÷ an = am–n)

7th Class Mathematics

190 10. (i) 23  34  4 2 2  34  2  2 23  34  22   3  32 3 2 2 2 2 2 3  25 

23  2  34 3  25

(am + an = am+n)

25  34 3  25 = 25–5 × 34–1 (am  an = am+n) = 2023 = 1 × 33 =33 (ii) [(52)3 × 54] ÷ 57 = [52 × 3 × 54] ÷ 57 (am)n = amn = [56 × 54] ÷ 57 = [56 + 4] ÷ 57 (am × an = am+n) = 510 ÷ 57 = 510 – 7 (am ÷ an = am–n) = 53 (iii) 254 ÷ 53 = (5 × 5)4 ÷ 53 = (52)4 ÷ 53 = 52 × 4 ÷ 53 (am)n = amn = 5 8 ÷ 53 = 58 – 3 (am ÷ an = am–n) = 55 

3  7 2  118 3  7 2  118  (iv) 21  113 3  7  113 = 31–1 × 72–1 ×118–3 (am ÷ an = am–n) 0 1 5 = 3 × 7 × 11 = 55 = 1 × 7 × 115 = 7 × 115

(v)

37 37  43 3 3 3 3 4

(am × an = am+n)

37  37  7 (am ÷ an = am–n) 7 3 =30 = 1 (vi) 20 + 30 + 40 = 1 + 1 + 1 = 3 (vii) 20 × 30 × 40 = 1 × 1 × 1 = 1 (viii) (30 + 20) × 50 = (1 + 1) × 1 = 2

 a5  8 5 3 8 (x)  a 3   a  a  a   =a2 ×a8 = a2+8 = a10



28  a 5

2  a 23

3

 a m n  a mn   

28  a 5 26  a 3 = 28–6 × a5–3 (am ÷ an = am–n) 2 2 2 = 2 × a = (2 × a) [am × an = (a × b)m] = (2a)2 www.betoppers.com 

(am × an = am+n)

45  a 8 b3 5 5 8 5 3 2 (xi) 5 5 2  4  a  b 4 a b (am ÷ an = am–n) 0 3 1 3 4 × a × b = 1 × a × b = a 3b 2

31 (xii) (23 × 2)2 =  2  (am × an = am+n)

11. (i)

(ii)

(iii)

(iv)



28  a 5 28  a 5 28  a 5  (ix) 43  a 3  2  2 3  a 3 22 3  a 3  

(am ÷ an = am–n)

12. (i)

(ii) (iii)

(iv)

= (24)2 = 24 × 2 (am)n = amn = 28 10 × 1011 = 10011 L.H.S. = 10 × 1011 = 1011 + 1 (am × an = am+n) 12 = 10 R.H.S. = 10011 = (10 ×10)11= (102)11 = 102 × 11 = 1022 (am)n = amn As L.H.S.  R.H.S., Therefore, the given statement is false. 23 > 5 2 L.H.S. = 23 = 2 × 2 × 2 = 8 R.H.S. = 52 = 5 × 5 = 25 As 25 > 8, Therefore, the given statement is false. 23 × 3 2 = 6 5 L.H.S. = 23 × 32 = 2 × 2 × 2 × 3 × 3 = 72 R.H.S. = 65 = 7776 As L.H.S.  R.H.S., Therefore, the given statement is false. 30 = (1000)0 L.H.S. = 30 = 1 R.H.S. = (1000)0 = 1 = L.H.S. Therefore, the given statement is true. 108 × 192 = (2 × 2 × 3 × 3 × 3)×(2 × 2 × 2 × 2 × 2 × 2 × 3) = (22 × 33) × (26 × 3) = 26 + 2 × 33 + 1 (am × an = am+n) = 28 × 3 4 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2) = 36 × 2 6 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3

Exponents and Powers Solutions 5 2

13. (i)

2 

 73

3

8 7 



2 

252  73

  a m n  a mn    2  2  2   7 

210  73 3 3

191

3



7

210  73 233  7

 a m n  a mn   

210  73  210  731 (am ÷ an = am–n) 29  7 = 21 × 72 = 2 × 7 × 7 = 98 

(ii)

25  52  t 8 5  5  52  t 8  3 103  t 4  5  2  t 4 (a×b)m = (am×bm) 

511 2  t 8 53  23  t 4



54  t 8 54  3  t 8  4 m  (a ÷ an =am–n) 53  23  t 4 23

(am × an = am+n)

51  t 8 5t 4   2 2 2 8

16. (i) (ii) (iii) (iv) (v) (vi) 17. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

CONCEPTIVE WORKSHEET 1.

(i)

5

5 35  105  25 3   2  5   5  5 (iii)  57  6 5 57  25  35

35  25  55  52  57  25  35 35  25  55 2  7 5 5 5  2 3

m

m

n

m+n

50000000 = 5 × 107 7000000 = 7 × 106 3186500000 = 3.1865 × 109 390878 = 3.90878 × 105 39087.8 = 3.90878 × 104 3908.78 = 3.90878 × 103 3.84 × 108 m 3 × 108 m/s 1.2756 × 107 m 1.4 × 109 m 1 × 1011 stars 1.2 × 1010 years 3 × 1020 m 6.023 × 1022 1.353 × 109 cubic km 1.027 × 109

3125 = 5 =5 =5

=5 5 = 55

m

(a × b) = (a × b ) m

(a × a = a

)

35  25  57  7 5 5 5  2 3 = 35–5×25–5×77–7 (am ÷ an =am–n) 0 0 0 = 3 × 2 × 5 = 1 × 1 × 1 =1 14. 279404 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 0 × 101 + 4 × 100 3006194 = 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100 2806196 = 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 101 + 6 × 100 120719 = 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100 20068 = 2 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100 15. (a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100 = 86045 (b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100 = 405302 (c) 3 × 104 + 7 × 102 + 5 × 100 = 30705 (d) 9 × 105 + 2 × 102 + 3 × 101 = 900230

625 5 125 5 5 25

(ii)

5

5

5

343 1331 – 343 = –7 49 = –7 –7 –7 1331 = 11 = 11

121 11 11 1

343 7  7  7  1331 11 11 11 

73 113

Since

3

 7  ;  ;  11  am  a    bm  b 

m

(iii) a3 + b3 + 3a2b + 3ab2 = (a + b)(a2 – ab + b2) + 3ab(a + b) = (a + b)(a2 + 2ab + b2) = (a + b)(a2 + 2ab + b2) = (a + b)(a + b)2 = (a + b)3 www.betoppers.com

7th Class Mathematics

192 2.

(i)

(729)2/6 729 = 3 × 243 = 3 × 3 × 81 = 3 × 3 × 3 × 27 =3×3×3×3×9 = 36 (729)2/6 = (36)2/6 = 36 2/3 Since (am)n = amn = 32 = 9

3.

 256     625  256 = 4 =4 =4 = 44 625 = 5 =5 =5 = 54  256     625 

(iii)



4

64



1 3

1

n

a  a n

64 = 4 16 =4 4 4 = 43 1

3

64   43  3 since (am)n = amn; 1

  43 3 ; = 4

= (3125)1/5 since 64 4 4

3125 = 5 = 5 = 5 = 5 = 55

16 4 4

125 5 25 5 5 5

625 5 125 5 5 25 5 5 5 5

 44     54 

= 55 5 since (am)n = amn =5

1/ 4

;

1

3

1/ 4

4 ;   5

4x 

1 4

4 ;   5

1

1 1 3      64  4   

729   729  3 since

729 = 3 =3 =3 =3 =3 = 36

243 3 3 3 3

3

1 1  3

  64  4

a  a n

1

729   729  3   36  3

= 36 1/3 since(am)n = amn = 32 =9

(Since (am)n = amn);

1

1 12

4

 64 64 = 2 32 =2 2 =2 2 =2 2 =2 2 = 26

16 2 8 2 2 2 2

4 2

1

1

 64 12   26 12 1 12

1

n

81 3 27 3 3 9 3 3 3 3

1

6

a  a n

= (3125)1/5 = (3125)1/5

1  n  Since a  a  n 

2

1

n

1

1/ 4

  4 4  =      5  

64 = (64)1/3

Since

1/ 4

(ii)

3

since (am)n = amn;

1 2

2  2

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2

1296  1296  4

1296 = 2 =2 =2 =2 =2 =2 =2 = 24 = (2 = 64

648 2 324 2 2 162 2 2 2 81 2 2 2 3 27 2 2 2 3 3 9 2 2 2 3 3 3 3 34 3)4 since an bn = (ab)n

Exponents and Powers Solutions 1

1

4

193

1296  1296  4   6 4  4

6.



64  5 3125  3 729  4 1296 In descending order = 9 > 6 > 5 > 4 = 4

1296  3 729  5 3125  3 64 4. Express the following numbers as a product of prime factors in the exponential notation (i) 14553 (ii) 8281 (iii) 3456? Sol: (i) 14553 = 3 × 4851 = 3 × 3 × 1617 = 3 × 3 ×3 × 539 = 3 × 3 ×3 ×7 × 77 = 3 × 3 × 3 × 7 × 7 × 11 = 33 × 32 ×11 (ii) 8281 = 7 × 1183 = 7 ×7 ×169 = 7 × 7 ×13 ×13 = 33 × 72 ×11 (iii) 3456 = 2 × 1728 = 2 × 2 × 864 = 2 × 2 × 2 ×432 = 2 × 2 × 2 × 2 ×216 = 2 × 2 × 2 × 2 ×2 × 108 = 2 × 2 × 2 × 2 ×2 × 2 × 54 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 27 = 2 × 2 × 2 × 2 ×2 × 2 × 2 × 3 × 9 ×9 = 2 × 2 × 2 × 2 ×2 × 2 × 2 × 3 × 3 ×3 = 27 × 33 5.

  x a b 

a

2

  x c a  x

 ab  b

c

2

2

 ac  a 2

 a  b   a 2  ab  b2 

 xa

3

 b3



 xb

3

 c3

  x bc 

b

2

 bc  c

2





x

 b  c   b2  bc  c2 

 xc

3

x

 c  a  c2  ac  a 2 

3

 b3

 xb

3

 c3

 xc

3

a3

Since am × an = am+n = x0 =1

x x  p p x x

1



p

1

r

x x  q q x x

1



q

1

x xp  xr xr

1 1 1  q  r q r p r x x x x x x x  xq  xp xp xq xr p

xp xq xr   x p  x q  x r x p  x q  x r x p  xq  x r



x p  xq  x r 1 x p  xq  x r

3n = 729 = 3 × 243 = 3 × 3 × 81 = 3 × 3 × 3 × 27 = 3×3×3×3×9 = 3×3×3×3×3×3 n 3 = 36 n=6 (i) 3n – 3 = 36 - 3, 33 = 3 3

3 = 27

(ii) 3n + 8 = 3-6 + 8, 32 = 3 3 = 9 (iii) 3n/2 = 36/2 = 3 3 3 = 9 42

42

(iv)  3n  36   36  36  36 8.

6

42 36

 37

Since(am)n = amn 64x + 2 = 16x + 1 = 4y 3 x2

4 

  42 

x 1

 4y

43x + 6 = 42x + 2 = 4y 3x + 6 = y 3x - y = -6 ------(1) Solve (1) & (2) 3x - y = - 6 2x - y = - 2 - + +

2x + 2 = y 2x - y = -2 ------(2)

________

a3

Since (a + b)(a2 – ab + b2 ) = a3 – b3  xa

r



4

7.

q

1

1 4

mn mn  6 since (a ) = a =6 In ascensding order = 4 < 5 < 6 < 9 = 4

1



Since a0 = 1

x =-4 x = -4 in (1) -12 - y = -6 y = -12 + 6 = -6

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7th Class Mathematics

194 9.

(i) (-1)500 = (-1) (-1 ) ...... 500 times = 1(If 500 is even number) (ii) (-1)2001 = (-1) (-1) .... 2001 times = -1 (If 2001 is odd number) 0

(iii)

25 12  10  50 5

 25  =5 = 10

17.

2 5 2

4

 3 3  316

 10020   10030 

20010  20020  20030



1  1  1

n  16 3

n = 48

1 1 1

18. (73)x+1 = (72)x+21 73x + 3 = 72x + 42

5 1

2 6

4 5

6 3

3x +3 = 2x + 42

6 3

2   2   2  3   3   3 

x = 42 – 3

3/ 2 2

x = 39



25  212  218 320  318  33



25  12  18 320  18  3

19.

x ac x ba x cb   1 x bc x ca x ab

20. 5–2, 3–3, 2–5, 4–4 Since am

an = amn

2 3 (i) 3.84 108m (ii) 3 1012 m (iii) 1.353 109 cubic km (iv) 7 10-4mm (v) 1.6 10-19 columb (a – b)3 (a + b + c)2 1+1+1+1+1+1+1+1+1 = 9 Correct Answer is none.

1 1 1 1 34  33  32  3  1 14. 1   2  3  4  3 3 3 3 34 

 814

n

21. i  b, ii  c, iii  a 22. i  b, ii  a, iii  d



11. 12. 13.

n 1/ 3

3 

 3 3   34 

Since (am)n = amn

10.

5

16. 60.2× 2160.62 = 130068.12

1  3

(v)

 3     5 

n

1001 

243 3125 5

Since a0 = 1

5

0

(iv)



 3  5  5

0

1  1

 25 

15.

81  27  9  3  1 121  81 81

SUMM ATIVE WORKSHEET 1.

The solution of 21 = 2 Therefore, the point on the number line is 'F' The correct answer is B.

2.

26 = 2 × 2 × 2 × 2 × 2 × 2 = 64 34 = 3 × 3 × 3 × 3 = 81 43 = 4 × 4 × 4 = 64 62 = 6 × 6 = 36  36 < 64 < 81

 62 < 26 = 43 < 34 Thus, among the given exponents, 34 is the greatest. The correct answer is B.

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Exponents and Powers Solutions 3.

2710 – 2435 = (33)10 – (35)5 =3

3×10

–3

195 7.

5×5

2

10800

2

5400

2

2700

2

1350

= 325 (243 – 1)

3

675

= 325 × 242

3

225

= 325 × 2 × 11 × 11

3

75

= 21 × 325 × 112

5

25

5

5

= 330 – 325 = 35 + 25 – 325 = 35 × 325 – 325 = 325 (35 – 1)

Thus, the given expression can be prime factorised as 21 × 325 × 112.

1

It can be observedseen that the greatest prime factor is 11. The correct answer is C. 4.

The number 6000 can be prime factorised as: 6000 = 2 × 2 × 2 × 2 × 3 × 5 × 5 × 5 = 2 4 × 3 1 × 53 = 4 2 × 31 × 5 3 = 31 × 4 2 × 53 Comparing a1 × b2 × c3 = 31 × 42 × 53,:

2 360

 a × b × c = 3 × 4 × 5 = 60

2 180

Thus, the value of a × b × c is 60.

2 90

3x+1 × 4x = 3x × 31 × 4x

3 45

[am × an = am+n]

= 3 × 3x × 4x = 3 × (3 × 4)x

[am × bm = (ab)m]

Thus, the given expression is an equivalent form of the expression, 3 × 12x. The correct answer is A. 523 + 523 + 523 + 523 + 523 = 5x

 5 × (523) = 5x  51 × 523 = 5x  51+23 = 5x

3 15 5 5 1

= 3 × 12x

6.

8.

 10800 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 = 24 × 33 × 52 Thus, the number 10800 can be expressed in the form of exponents as 24 × 33 × 52. The correct answer is D. The number 360 can be factorised as:

a = 3, b = 4, c = 5

The correct answer is B. 5.

The number 10800 can be factorised as:

[am × an = am+n]

 524 = 5x  x = 24

 a m  a n  a  0, 1     m  n 

 360 = 2 × 2 × 2 × 3 × 3 × 5 = 23 × 32 × 5 Thus, the prime factorisation of 360 is 23 × 32 × 5. The correct answer is C. 9. The given distance can be expressed in the standard form as: 77,80,00,000 = 7.78 × 10,00,00,000 = 7.78 ×108 The correct answer is A. 10. 90252 = 90000 + 200 + 50 + 2 90252 = 9 × 10000 + 2 × 100 + 5 × 10 + 2 × 1 90252 = 9 × 104 + 2 × 102 + 5 × 101 + 2 × 100 The correct answer is A.

Thus, the value of x is 24. The correct answer is A. www.betoppers.com

7th Class Mathematics

196 11. – (–3) × (–3) × (–3) × (–3) – (–3) × (–3) × (–3) × (–3) – (–3) × (–3) × (–3) × (–3) = – (–3)4 – (–3)4 – (–3)4 = (–3)4 (– 1 – 1 – 1) = (–3)4 (–3) – (–3)5 The correct answer is A. 12. 32 + 33 + 34 + 35 = 3 2 + 3 2 × 3 + 3 2 × 3 2 + 32 × 33 = 32 (1 + 3 + 32 + 33) = 32(1 + 3 + 9 + 27) = 32(40) = 32 × 2 × 2 × 2 × 5 = 3 2 × 23 × 5 The correct answer is C. 13. 0.63 = 0.6 × 0.6 × 0.6 = 0.216 The correct answer is A. 7 0

14.

3 

 35

32

=

370  35 32

=

30  35 32

=

1  35 32

=3

5–2

[(am)n = amn]

18. 5w × 7x × 2y × 3z = 363 × 352 × 302 ×21 = (33 × 23 × 33 × 23) × (72 × 52) × (52 × 22 × 32) × (7 × 3) = 33 + 3 + 2 + 1 × 2 3 + 3 + 2 × 5 2 + 2 × 7 2 + 1 = 3 9 × 2 8 × 5 4 × 73 Since 3, 2, 5 and 7 are all prime numbers. y=8 The correct answer is D. 19. 7x × 9y × 4z = 144 × 68  7 x × 9 y × 4z = 74 × 2 4 × 38 × 28  7 x × 9 y × 4z = 7 4 × 94 × 46  z=6 The correct answer is C. 20. Result after first hop = 1 Result after second hop = 1 × 2 = 2 Result after third hop = 2 × 2 = 22 . . Result after 61st hop = 260 Result after 62nd hop = 261 Result after 63rd hop = 262  Final result result after the 63rd hop 262  60  22  4 = st result after the 61 hop 2

 a 0  1 a  0 

 am m n   n a  a 

The correct answer is B. 15. (–2)6 = (–1 × 2)6 = (–1)6 × 26 [(a × b)m = am × bm)] = 1 × 26 [(–1)even number = 1] =2×2×2×2×2×2 =4×4×4 = 43 The correct answer is A. 16. 25 + 3 × 25 + 27 = 25 + 3 × 25 + 25 ×22 = 25 (1 + 3 + 22) = 25 (1 + 3 + 4) = 25 . 8 = 25. 23 = 28 The correct answer is A. 17. 24 × (2 × 12) × (4 × 6) × (8 × 3) × (3 × 4 × 2) = 24 × 24 × 24 × 24 × 24 = 245 The correct answer is D. www.betoppers.com

The correct answer is B. 21. Number of bacteria at the end of the first day = 3.27 × 103 Number of bacteria at the end of the second day = 10 × 3.27 × 103 = 3.27 × 104 Number of bacteria at the end of the third day = 10 × 3.27 × 104 = 3.27 × 105 Number of bacteria at the end of the fourth day = 10 × 3.27 × 105 = 3.27 × 106 Number of bacteria at the end of the fifth day = 10 × 3.27 × 106 = 3.27 × 107 = 32,700,000 The correct answer is C. 22. To express the given number in standard form, the decimal point needs to be moved seven places to the right. Thus, the number 9.387 2 × 107can be expressed in standard form as 93,872,000. The correct answer is D.

Exponents and Powers Solutions

197

23. The given expression is (4x3 + 5y2)2 When x 

1 1 and y   , the expression 2 2

28.

3

becomes: 2   1 3  1   4     5       2   2

1  1   4  5  4  8

213  104 152  8  343 4



 3  7    2  5 2  3  5   23  7 3



33  73  24  54 32  5 2  23  7 3

2

[(ab)m = am × bm]

2 m n m n = 33–2 × 73–3 × 24–3 × 54–2  a  a  a 

= 31 × 70 × 21 × 52 2 5    4 4

2

= 3 × 1 × 2 × 25

= 150 Thus, the value of the given expression is 150. The correct answer is C.

2

72 7    2 4 4



49 16

The correct answer is B. 24. [(6)12 ×(35)28×(15)16]/ [(3)16×(5)43×(14)12× (21)11×(7)4] = [(3 × 2)12 (7 × 5)28 × (5 × 3)16]/ [(3)16 × (5)43 × (7 × 2)12 × (7 × 3)11 × (7)4] = [312 × 212 × 728 × 528 × 516 × 316]/ [316 × 543 × 712 × 212 × 711 × 311 × 74] = (728×328×544×212)/(327×543×727×212) = 7×3×5 Hence there are three prime factors The correct answer is D. 25. [aK + 4 – a × aK]/[a × a(K + 3)] + (1/a)3 = aK × a4 – a × aK]/aK + 4 + (1/a)3 = [aK(a 4–a)/[aK×a 4]+(1/a)3=[(54–5)/54]+(1/5)3 = [(53 – 1)/53] + (1/5)3 = (124/125) + (1/125) = 125/125 = 1 The correct answer is A. 26. 64 × 256 × 1 024 = 4x  4 3 × 4 4 × 45 = 4x

 4 3 + 4 + 5 = 4x  412 = 4x So x =12 The correct answer is C. 27.

    34 + 43  81 + 64  145 The correct answer is B.

 a 0  1 a  0 

HOTS WORKSHEET 1.

The mass of the planet Venus is approximately 4,869,000,000,000,000,000,000,000 kg. Any number can be expressed as a decimal number between 1.0 and 10.0, including 1.0, multiplied by a power of 10. The obtained form is called the standard form. The number 4,869,000,000,000,000,000,000,000 kg can be written in standard form as 4,869,000,000,000,000,000,000,000 kg = 4869 × 1021 kg = 4.869 × 103 × 1021 kg = 4.869 × 1024 kg The correct answer is C.

2. 5 5 5 5 5 5

156 25 312 5 625 125 25 5 1

(–5)6 = (–1 × 5)6 = (–1)6 × 56 [(ab)m = am × bm] = 1 × 15625 = 15625 –15625  (–5)6 Thus, alternative C is incorrectly matched. The correct answer is C. www.betoppers.com

7th Class Mathematics

198 2

3.

5   1  122  52   13     5

 144  25  

 5 

 1 5   169  169

2

  1  52 

 1



2





1 53

2 2 2 3 3 3 3 3 11

1 53

1 53

[(ab)m = am × bm]

1

 a m  a n  a m  n 

3 2

5

The correct answer is C.

4.

3

 2p    3q   62  5 2 10p   3pq   36 5

2

6.

3

 2  p    3  q   62  5  2 2 2  5  p  3  p  q    6 



[(ab)m = (a×b)m]

3

2

5

The given expression be simplified as 2



3

2 3 6 5 p q 2  32  5  62  p2 1  q 2

(am×an = am+n)



= 25–1×33–2 ×62–2×51–1×p5–3×q3–2  a m  a n  a m  n  4

0

0

2

=2 ×3×6 ×5 ×p ×q = 16 × 3 × p2q = 48p2q Thus, the required expression is 48p2q .

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2

182 a 2  5b  22 b 2 2

5 2 a 2 2   3  6   b

[(ab)m = ambm]

182  5  22 b 2 1 2

5 2 a 2 2   3  6   b

[am × an = am+n , am  an = am – n] 

182  5  22 b3 52 a 0  182  b



182 2  22  b31 521  1

= 24 × 3 × 1 × 1 × p 2 × q

The correct answer is D.

2

18a   5b   2b  2  5a   32   36b 

18a   5b   2b  2  5a   32   36b 

25  p5  33  q3  62  5 2  5  p  32  p 2  q 2  6 2 5

21384 10692 5346 2673 891 297 99 33 11 1

1  21384 = 2× 2 × 2 × 3 × 3 × 3 × 3 × 3 × 11 1+1+1 1+1+1+1+1 m n m+n =2 ×3 × 11 ( a × a = a ) = 23 × 35 × 11 Comparing 23 × 55 × 11 with x × 2y × y5, it is observed that x = 11 and y = 3 Thus, the value of x and y are 11 and 3 respectively. The correct answer is D.

1 5

5

The given number is 21384. It can be factorized as

 a m a m     m  b   b 

3

132 2

5.

3

 a 0  1 a  0  

180  22  b 2 1 4  b 2 4b 2   51 5 5 The correct answer is A. 

can

Exponents and Powers Solutions 7.

199

5020016.308 can be written as 5020016.308 = 5000000 + 20000 + 10 + 6 + 0.3 + 0.008

3 8  10 1000 6 4 1 = 5×10 +2×10 +1×10 +6×100+3×10–1+8×10–3 (100 = 1) The correct answer is A. = 5×106+2×104+1×10+6×1+

8.

243  94  8  a 5 b6 122  162  81  a 3 b 2 3

3

2 4

3

5

 b6

2

2

4 2

4

3

 b2

3 3

 33   32   23  a 5  b6

2 2

 32   24   34  a 3  b2

4

2

[(x×y)m = xm × ym] 

29  33  38  23  a 5  b6 24  32  28  34  a 3  b 2



29  3  338  a 5  b6 24 8  32  4  a 3  b 2 12



11

5

  x m  n  x mn   

[xm × xn = xm+n]

4

3

5

3  2   20  5   2  3 2     3 3 3 1004   22   32 3

1004  25  35  23 1004  223  32  33 25  35  23  6 2 3 2 3 3 25  3  35  6 23 2 3 28  35  6 5 2 3 = 28–6 × 35–5 = 22 ×30 =4×1 =4 The correct answer is B. 12. 2×(25)2 = 2 × 25×2 = 2× 22×5 = 2 × (22)5

 2  45  410 The correct answer is A.

6

2 3 a  b 212  36  a 3  b 2

= 212–12 × 311–6 × a5–3 × b6–2

9.

11.

204  54  65 1004  43  32



 2  3   3   2  a   2  3   2   3  a 2   2 

10. For negative integers x and y, the value of the expression (xn × yn) will always be positive. This is because the product of two negative integers is a positive integer. For example, (–2)4 × (–5)4 = [(–2)×(–5)]4 = 104 = 10000, which is positive. The correct answer is C.

3

18ab2   3a    4b 

 xm m n   n x  x 

= 20 × 3 5 × a 2 × b 4 [x0 = 1] = 243a2b4 The correct answer is B. The given number is 945 × 100. The numbers 945 and 100 can be prime factorised as 3 945 2 100 3 315 2 50 3 105 5 25 5 35 5 5 7 7 and 1 1 945 =3 × 3 × 3 × 5 × 7 and 100=2 × 2 × 5 × 5 945 × 100 =(3 × 3 × 3 × 5 × 7)×(2 × 2 × 5 × 5) = (2 × 2) × (3 × 3 × 3) × (5 × 5 × 5) × 7 = 2 2 × 3 3 × 53 × 7 Thus, 945 × 100 can be further expanded as 22 × 33 × 53 × 7. The correct answer is A.

13.

12a 2   ab2 

2

2

3



2  9  a  b 2   3  a    22  b  4  3  a 2   a  b2 

2

2

2



2  32  a  b 2  33  a 3   22   b 2 22  3  a 2  a 2   b 2 

2

2  32  a  b 2  33  a 3  222  b 2 22  3  a 2  a 2  b 22 2 2  3  a  b 2  33  a 3  24  b 2  22  3  a 2  a 2  b 4 21 4  32  3  a13  b 2 2  22  3  a 2  2  b 4 5 2  35  a 4  b4  2 2  3  a 4  b4 = 25–2 × 35–1 × a4–4 × b4–4 = 23 × 3 4 × a 0 × b 0 = 8 × 81 × 1 × 1 = 648 The correct answer is D. 

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7th Class Mathematics

200 4

3

3

 52   2  3  7    3  8   2   7  4  5  7 6  2 3

4

4



3

3

2 3

3

4

2

2

2 3

3

3

2

23

3

3 4

3

2 2

3

23

3

2

3

2

3 4

4

2 2

2

2

4

3

2

23  33  73  34  212   6 24  52  72  2  3

 2  49    3  7   8  3   22   2 3 710  16  3   4  3

56 23  33  73  34  212  26  36 2 4  52  7 2 6 312 3 4 5 2  3  73  6  46 2 2 3 2 5 7 6 5 215  37  73  6  10 2 2 3 2 5 7 = 56–2 × 215 – 10 × 37 – 6 × 73 – 2 = 5 4 × 2 5 × 3 1 × 71 = 5 4 × 2 4 × 21 × 2 1 × 71 = (5 × 2)4 × (2 × 3 × 7) = 104 × 421 Comparing 104 × 421 with 10x × y1, we get x = 4 and y = 42. The correct answer is C.

2  7  

56



8

15.

24   32   23  33  25

34  25  35  24 2  32 4  23  33  25  34  25  35  24 24  38  23  33  25  34  25  35  24 4+3+5–5–4 =2 × 38 + 3 – 4 –5 3 2 = 2 × 3 = 8 × 9 = 72 The correct answer is C. 984  213  242  43 17. 710  482  123

3 4

5   2  3  7   3  2  6  2  5  7 5   2  3  7   2    2  3  2   5  7   2  3  7  3  2  5    2  3 2   5  7 

3

2 5 184  63  25  2  3    2  3  2  16. 5 34  65  24 34   2  3  24

3

3 4  25  42  24  14.   1402  36 

6

11

3 2 7 3 278  96  27  811  3    3   2   2   3616  25  32  18  4  9 16  25  32   2  9 





338 326  27  2311

2

2

16

 32   25  32  2  32 324 312  27  233

2 16

2 16

 2   3 

 25  32  2  32

324 312  27  233 2216  3216  25  32  2  32 324 312  27  233  32 32 5 2 2  3  2  3  2  32 324 12  27  33  32  51 32  2 2 2 3 36 40 3 2  38 36  336 36  240 38 2 3 = 30 ×22 = 1 × 4 = 4 The correct answer is C. www.betoppers.com 

2 4

2

 33  73   23  3  223 2

3

710   24  3   22  3 4



2

24   7 2   33  73   23   32  26 2

3

710   24   32   22   33

24  7 2 4  33  73  232  32  26 710  242  32  223  33 24  78  33  73  26  32  26  710  28  32  26  33 4 6 6 2  78 3  33 2  10 8 6 2  3 7 2 3 16 2  711  35  10 14 5 7 2 3 = 216–14 × 711–10 × 35–5 = 22 × 71 ×30 = 4 × 7 ×1 = 28 The correct answer is C. 3610 6320  3  239  18. 99  164 2819  340 

10

20

 9  7   3  239   4 19  4  7   340 99   42  9  4



910  410 920  7 20  3  239  19 19 40 99  42 4 4 7 3 2 20

910  410  3   7  3  2  9 8  19 9 4  22   719  340 20

39

3

Exponents and Powers Solutions

201 `

3220  720  31  239 2219  719  340 340 1  720  239  9  42  38 19 40 2 7 3 = 9 × 16 + 341–40 × 720–19 × 239–38 = 144 + 31 ×71 × 21 = 144 + 3 × 7 × 2 = 144 + 42 = 186 The correct answer is B. 812  99 259 169   19. 417  314 1256 224  322  910  9  410 8 

3 12

2 9

2 9

14

3 6

 2   3   5    2   3 5  2 17



2312  329 2 17

2 

 314



236  318 518 236   234  314 518 224  210 236  236 34  31814  51818  24 10 2 36 2  22  34  50  34 2 = 4 × 81 + 1 + 236 –34 = 324 + 1 + 22 = 325 + 4 = 329 The correct answer is A. 20. 2 × 106 = 200 × 104 1.19 × 104 = 1.19 × 104 3.2623 × 105 = 32.623 × 104 1.41 × 106 = 141 × 104 Now, 200×104>141×104>32.623×104 >1.19× 104 Thus, the given numbers can be written in descending order as: 2 × 106 > 1.41 × 106 > 3.2623 × 105 > 1.19 × 104 The correct answer is A. 

4 9

2   2  2 

5 2

24

529 249  24 52 36 5 2 2

IIT JEE WORKSHEET KEY Q.No.

1

2

3

4

5

6

7

8

Ans

B

D

C

D

D

C

B

C

Q.No.

9

10

11

12

13

14

15

16

B, D

A, C, D

A, C

A, B, C

A, B, D

5

4

10

Ans Q.No.

17

18

Ans

1

i  c, ii  g, iii  a, iv  e, v  5/2

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 1.

2.

ab = 132 a = 13, b = 2 (a – 3)b–1 = (13 – 3)2 – 1. = 10   5 0       6  

5

5



(1)

5

 a

0

 1

  5 0        1 is true.  6    

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7th Class Mathematics

202 x 2 y2 . y 2 z 2 . x 2 z 2

5.

y2 z 2 x 2 . . z 2 y2 z 2 y2 z 2 x 2   1 z 2 y2 z2

7.

1 1  ln lm 1 m 1 n l l 



1 1  n m n l l l l m l ln m

lm ln  lm  ln ln  lm

lm ln lm  ln   1 lm  ln ln  lm lm  ln i. (–1)77 – (–1)74 = –1–(1) = –1–1 = –2 ii. ax = 1 ax =0 a=0 4 v. 1024  2 x 

18.

4

210  2 x 1

 210  4  2x 10 4

2  2x 5

2 2  2x

 

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3. ALGEBRAIC EXPRESSIONS SOLUTIONS The polynomial 5x + 6xy2 – 2x has three terms. However, the terms 5x and –2x are like terms. Since the given expression does not contain three unlike terms, it is not a trinomial. 9. 2xy, 5yx; – 3x, 6x, x; 15, 12 10. (i) (5m – 7n)+(3n – 4m + 2)+(2m – 3mn – 5) = 5m – 7n + 3n – 4m + 2 + 2m – 3mn – 5 = 5m + 2m – 4m – 7n + 3n + 2 – 5 – 3mn = 3m – 4n – 3 – 3mn Ans. (ii) 4x2y + (– 3xy2) + (– 5 xy2) + (5x2y) = 4x2y – 3xy2 – 5xy2 + 5x2y = 4x2y – 5x2y – 3xy2 – 5xy2 = 9x2y – 8xy2 Ans. (iii) (3p2q2 – 4pq + 5) + (– 10p2q2) + (15 + 9pq + 7p2q2) = 3p2q2 – 4pg + 5–10p2q2 + 15 + 9pq +7p2q2 = 3p2q2–10p2q2 +7p2q2 – 4pq +9pq+5 + 15 = 5pq + 20 Ans. (iv) (ab – 4a) + (4b – ab) + (4a – 4b) = ab – 4a + 4b – ab + 4a – 4b = ab – ab – 4a + 4a + 4b – 4b = 0 Ans. (v) (x2 – y2 – 1) + (y2 – 1 – x2) + (1 – x2 – y2) = x2 – y2 – 1 + y2 – 1 – x2 + 1 – x2 – y2 = – x2 – y2 – 1 = – (x2 + y2 + 1)Ans. 11. To get the required expression, we have to subtract (2x – 21y – 42xy) from (7x – 3y + 45xy + 7). (7x – 3y + 45xy + 7) – (2x – 21y – 42xy) = 7x – 3y + 45xy + 7 – 2x + 21y + 42xy = 7x – 2x – 3y + 21y + 45xy + 42xy + 7 = (7 – 2)x + (–3 + 21)y + (45 + 42)xy + 7 = 5x + 18y + 87xy + 7 12. (5y + 7) + (3y2 – 9y + 2) = 5y + 7 + 3y2 – 9y + 2 = 3y2 + 5y – 9y + 7 + 2 [Rearranging the terms] = 3y2 + (5y – 9y) + (7 + 2) = 3y2 + (5 – 9) y + 9 = 3y2 + (–4) y + 9 = 3y2 – 4y + 9 (4y2 – 6y) + (–2y2 + 3y – 3) = 4y2 – 6y – 2y2 + 3y – 3 = 4y2 – 2y2 – 6y + 3y–3 [Rearranging the terms] = (4y2 – 2y2) + (–6y + 3y) –3 = (4 – 2) y2 + (–6 + 3) y – 3 = 2y2 – 3y – 3 Now, subtracting the sum of (4y 2 – 6y)and (–2y2 + 3y – 3) from the sum of (5y + 7)and (3y2 – 9y + 2) is the same as subtracting 8.

FORMATIVE WORKSHEET 1.

(i) x + y (iv) 3x

2.

The terms of the expression are 

3.

4.

5.

(ii) x + 7 (v) 2x – 1

(iii) y – 7

xy ,14xy 2 , and 7

–3. –3x2yz3 = – 3 × x × x × y × z × z × z Therefore, the factors of –3x2yz3 are –3, x, x, y, z, z, and z. The tree diagram representation of the algebraic expression, 7x2y – 2xy + 5 is

Here, the like terms are 51x2y,

x2y , and –x2y.. 2

6. Terms

Coefficients of pq

2

pq

q

- 3pq

-3

15p2q2

15pq



7.

31 2 pq 5



31 p 5

An expression containing only one term is known as a monomial. Hence, among the given expressions, the monomials are (ii) 16 (iii) –21x2y2z2 An expression containing two unlike terms is known as a binomial. Hence, among the given expressions, the binomials are (i) x + 7 (iv) x2 – 3 (vi) 6xy2 – 2x2y (vii) 4x + xy An expression containing three unlike terms is known as a trinomial. Hence, among the given expressions, the trinomials are (v) 7x + 7y – 6xy (viii) 15xy2 – 7 – 2x2y

7th Class Mathematics

204 2

2

(2y – 3y– 3) from (3y – 4y + 9). This can be done as (3y2 – 4y + 9) – (2y2 – 3y – 3) = 3y2 – 4y + 9 – 2y2 + 3y + 3 = 3y2 – 2y2 – 4y + 3y + 9 + 3 [Rearranging the terms] = (3 – 2) y2 + (–4 + 3) y + (9 + 3) = y2 – y + 12 13. (i)

3 Putting x = 5 in the expression x 

3 we obtain x 

 125 

14. (i)

(ii)

x2 8, 25

x2 52  8  53  8 25 25

25  8  125  1  8  118 25

(ii) Putting k 

(iii)

1 in the expression 4k2–2k +1, 2

we obtain 4k2 – 2k + 1

(iv)

15. (i)

2

1 1  4    2   1 2 2

1 1  4   2   1  1 1  1  1 2 2 (iii) Putting x = 5, y = –2 in the expression 2x3y2 – 4x – 3xy – 5(x2y2 – x2), we obtain 2x3y2 – 4x – 3xy – 5(x2y2 – x2) = 2×53×(–2)2– 4×5–3×5×(–2)–5[52×(–2)2 –52] = 2 × 125 × 4 – 20 + 30 – 5(25 × 4 – 25) = 1000 – 20 + 30 – 5(100 – 25) = 1000 – 20 + 30 – 5 × 75 = 1000 – 20 + 30 – 375 = (1000 + 30) – (20 + 375) = 1030 – 395 = 635 (iv) Putting p = –2 in the given expression, we obtain 3p2 

(ii)

(iii)

(iv)

(v)

p  k  11 2 2

 3  2  

 2 

 k  11 2 = 3 × 4 + 1 + k – 11 = 12 + 1 + k – 11 =2 + k But the value of the given expression at p = –2 is given as 12.  2 + k = 12  k = 12 – 2 = 10 Therefore, the value of k is 10. www.betoppers.com

16. (a)

(b)

Put x = 2 in the given expression, x + 7 + 4 (x – 5) = x + 7 + 4x – 20 = 5x – 13 = 5 (2) – 13 = 10 – 13 = – 3  x + 7 + 4 (x – 5) = – 3, when x = 2 Ans. Put x = 2 in the given expression 3 (x + 2) + 5x – 7 = 3x + 6 + 5x – 7 = 8x – 1 = 8 × 2 – 1 = 16 – 1 = 15  3 (x + 2) + 5x –7 = 15, when x = 2 Ans. Put x = 2 in the given expression. 6x + 5(x – 2) = 6x + 5x – 10 = 11x – 10 = 11 (2) – 10 = 22 – 10 = 12 6x + 5 (x – 2) = 12, when x = 2 Ans. Put x = 2 in the given expression. 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11 = 11x + 7 = 11 × 2 + 7 = 22 + 7 = 29  4 (2x–1)+3x + 11= 29, when x = 2 Ans. 3x – 5 – x + 9 = 3x – x – 5 + 9 = 2x + 4 Now, put x = 3 in 2x + 4, 2x + 4 = 2 × 3 + 4 = 6 + 4=10  3x – 5 – x + 9 = 10, when x = 3 Ans. 2 – 8x + 4x + 4 = 2 + 4 – 8x + 4x = 6 – 4x Now, put x = 3 in 6 – 4x 6 – 4x = 6 – 4x3 = 6 – 12 = – 6  2 – 8x + 4x + 4 = – 6, when x = 3 3a + 5 – 8a + 1 = 3a – 8a + 5 + 1 = – 5a + 6 Now, put a = – 1 in – 5a + 6, – 5a + 6 = – 5(– 1) + 6 = 5 + 6 = 11 3a + 5 – 8a + 1 = 11, when a = – 1 Ans. 10 – 3b – 4 – 5b = 10 – 4 – 3b – 5b = 6 – 8b Now, put b = – 2 in 6 – 8b 6 – 8b = 6 – 8 × (– 2) = 6 + 16 = 22  10 – 3b – 4–5b = 22, when b = – 2 Ans. 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5 = 3a – 2b – 9 Now, put a = – 1, b = – 2 in 3a – 2b – 9, 3a – 2b – 9 = 3 (– 1) – 2 (– 2) – 9 = – 3 + 4 – 9 = – 12 + 4 = – 8  2a – 2b – 4 – 5 + a = – 8, when a = – 1, b = – 2 Ans. Put z = 10 in z3 – 3 (z – 10) z3 – 3 (z – 10) = (10)3 – 3 (10 – 10) = 1000 – 3 (0) = 1000  z3 – 3(z–10) = 1000, when z = 10 Ans. Put p = – 10 in p2 – 2p – 100 p2 – 2p – 100 = (– 10)2 – 2 (– 10) – 100 = 100 + 20 – 100 = 20 p2 – 2p – 100 = 20, when p = – 10 Ans.

Algebraic Expressions Solutions 17. Here, 2x2 + x – a = 5 Putting x = 0, we get 2 (0)2 + (0) – a = 5 –a=5 a = – 5 Ans. 18. 2 (a2 + ab2) + 3 – ab = 2a2 + 2ab + 3 – ab = 2a2 + ab + 3 Now put a = 5 and b = – 3 in 2a2 + ab + 3 2a2 + ab + 3 = 2 (5)2 + 5 x (– 3) + 3 = 2 × 25 – 15 + 3 = 50 – 15 + 3 = 38  The value of [2 (a2 + ab) + 3 – ab] is 38, when a = 5, b = – 3. 19. r 2 = x2 + y2

20. Tn = arn – 1 Tn  r n 1 a

205

15  20 24 = 50 + 12.5 = 62.5  50 

2 1 1   b c a 1 2 1   c b a ab 2a  b Here ‘c’ is the subject 24. 1 + 2 + 3 + ................. + 100 a = 1, d = 1, n = 100 c

1

 T n  n 1 r    a  21. S 

a 1 r

1 r 

r

22.

a S

S  a S

1   x1  y 2  y3   x 2  y3  y1   x 3  y1  y 2  2 Given (x1, y1) = (1,2) (x2,y2) = (4,5) (x3y3) = (6,8) 

Sn 

100  2  1  100  11 2  100  2  99 2 = 50 101 = 5050 25. mode = 3 median – 2mean 50 = 3 100 – 2 mean 2 mean = 300 – 50 2 mean = 250 mean = 125 26. A  R 2  r 2 A   R 2  r2 

  R  r  R  r 

 |1(5 – 8) + 4(8 – 2) + 6 (2 – 5)| 2

22  4  3 4  3 7

  |–3 + 24 – 18| 2

22 7 7 = 22 sqm 

   |–21 + 24|  ×3 = 1.5 Sq. units 2 2 N F 23. Median (M) = L  2 c f 80  25 50  2  20 24

 50 

n  2a   n  1 d  2

27.

b

2ac ac

b ac  2 ac 2 ac  b ac

40  25  20 24 www.betoppers.com

7th Class Mathematics

206

2 1 1   b c a

CONCEPTIVE WORKSHEET

1 2 1   a b c

1.

1 2c  b  a bc bc 2c  b Here ‘a’ is the subject. a

28. (i)

a ab multiplying with (a + b); we get c

a a  b ab c(a + b) = a ac + bc = a ac – a = bc a (c – 1)bc c a  b 

bc c 1

a (ii)

M

40 40m = N(P – R) 40m = NP – NR NR = NP – 40m

R (iii)

NP  R 

NP  40M N

1 1 1   f 4 v given that v = 15; f = 5 1 1 1   5 4 15 1 1 1   5 15 4 3 1 1  15 4 2 1  15 4 4

15  7.5 2

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(i) y – z

1 pq 4 (vii) 10 – yz (iv)

(ii)

1  x  y  (iii) z2 2

(v) x 2  y 2

(vi) 3mn + 5

(viii) ab – (a + b)

1 x  y 2 3. xy – 3 4. (a) 4x, 5 (b) 4x2, – 3xy 2 (c) 5x , 10 (d) 2x2, 5x, 6 5. (i) xy, 1 (ii) – y2, – 1 (iii) – y, 5y2; –1, 5 (iv) 4p2q, – 3pq2 ; 4, – 3 6. (i) 4 (ii) – 1 (iii) y2 (iv) – 5z 7. (i) – 3 (ii) z (iii) z2 (iv) m 8. Monomials : (i), (iii), (vi) Binomial : (iv), (v), (vii), (ix), (xii) Trinomial : (viii), (x), (xi), (xiii) 9. Like terms : (ii), (iii), (vi) Unlike terms : (i), (iv), (v). 10. (i) (3mn) + (– 5mn) + (8mn) + (– 4mn) = 3mn – 5mn + 8mn – 4mn = 3mn + 8mn – 5mn – 4mn = 11mn – 9mn = 2mn Ans. (ii) t – 8tz + 3tz – z + z – t = t – t – 8tz + 3tz – z + z = – 5tz Ans. (iii) (– 7mn + 5) + (12mn + 2) + (9mn – 8) + (– 2mn – 3) = – 7mn +5+12mn + 2 + 9mn –8– 2mn – 3 = – 7mn + 12mn +9mn –2mn +5+2 – 8 – 3 = 21mn – 9mn + 7 – 11 = 12mn – 4 Ans. (iv) (a + b – 3) + (b – a + 3) + (a – b + 3) = a + b + 3 Ans. (v) (14x + 10y – 12xy – 13) + (18 – 7x – 10y + 8xy) + 4xy = 14x + 10y – 12xy – 13 + 18 – 7x – 10y + 8xy + 4xy = (14 – 7)x + (10 – 10) y+(–12 +8+4) xy + (– 13 + 18) = 7x + 5 Ans. 11. (i) y2 – (– 5y2) = y2 + 5y2 = 6y2 Ans. (ii) – 12xy – 6xy = – 18xy Ans. (iii) (a + b) – (a – b) = a + b – a + b = 2b Ans. (iv) b(5 – a) – a (b – 5) = 5b – ab – ab + 5a = 5a + 5b – 2ab Ans. (v) (4m2 – 3mn + 8) – (– m2 + 5mn) = 4m2 – 3mn + 8 + m2 – 5mn = – 4m2 + m2 – 3mn – 5mn + 8 = 5m2 – 8mn + 8. Ans. 2.

Algebraic Expressions Solutions

207

(vi) (5x – 10) – (– x2 + 10x – 5) = 5x – 10 + x2 – 10x + 5 = x2 + 5x – 10x – 10 + 5 = x2 – 5x – 5 Ans. (vii) (3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2) = 7ab – 2a2 – 2b2 – 5a2 + 7ab – 5b2 = – 2a2 – 5a2 – 2b2 – 5b2 + 3ab + 7ab = – 7a2 – 7b2 + 10ab Ans. (viii) (5p2 + 3q2 – pq) – (4pq – 5q2 – 3p2) = 5p2 + 3q2 – pq – 4pq + 5q2 + 3p2 = 5p2 + 3p2 + 3q2 + 5q2 – pq – 4pq = 8p2 + 8q2 – 5pq. Ans. 12. (i) Put the value m = 2 in m – 2,  m–2=2–2 =0 The value of (m–2), when m = 2, is 0 Ans. (ii) Put the value m = 2 in 3m – 5,  3m – 5 = 3 (2) – 5 = 6 – 5 = 1 The value of (3m–5), when m =2, is 1 Ans. (iii) Put the value m = 2 in 9 – 5m,  9 – 5m = 9 – 5 x2 = 9 – 10 = – 1 The value of (9–5m), when m =2, is–1 Ans. (iv) Put the value m = 2 in 3m2 – 2m – 7,  3m2 – 2m – 7 =3 (2)2 – 2 (2) – 7 = 3 × 4 – 4 – 7 = 12 – 11 = 1 The value of (3m2 – 2m–7), when m = 2, is 1 Ans. (v) Put the value m = 2 in

5m  4, 2

5m 5 2 4  4  5  4 1 2 2  5m   4  , when m = 2, is The value of   2  1 Ans. 13. (i) Put p = – 2 in 4p + 7, 4p + 7 = 4x – 2 + 7 = – 8 + 7 = – l The value of (4p + 7), when p = – 2, is – 1 Ans. (ii) Put p = – 2 in – 3p2 + 4p + 7,  – 3p2 + 4p + 7 = – 3 (– 2)2 +4 (– 2) + 7 = – 3 (4) – 8 + 7 = – 12 – 8 + 7 = – 13 The value of (– 3p2 + 4p + 7), when p = – 2, is – 13 Ans. (iii) Put p = – 2 in – 2p3 – 3p2 + 4p + 7, 2p3 – 3p2 + 4p + 7 = – 2 (– 2)3 – 3 (– 2)2 + 4 (– 2) + 7 = – 2 × (– 8) – 3 x 4 + 4 x (– 2) + 7 = + 16 – 12 – 8 + 7 = 23 – 20 = 3 The value of – 2p3 – 3p2 + 4p + 7, when p = – 2, is 3 Ans.

14. (i)

Put x = – 1 in 2x – 7, 2x – 7 = 2 (–1) – 7 = – 2 – 7 = – 9  The value of (2x – 7), when x = – 1, is – 9 Ans. (ii) Put x = – 1 in – x + 2, – x + 2 = – (– 1) + 2 = 1 + 2 = 3 The value of (–x+2), when x= –1, is 3 Ans. (iii) Put x = – 1 in x2 + 2x + 1 x2 + 2x + 1 = (– 1)2 + 2 (– 1) + 1 =1–2+1=2–2=0 The value of (x2 + 2x + 1), when x = – 1, is 0 Ans. (iv) Put x = – 1, in 2x2 – x – 2. 2x2 – x – 2 = 2(– 1)2 – (– l) – 2 = 2x1 + 1 – 2 = 2 + 1 – 2 = 3 – 2 = 1 The value of (2x2 – x – 2), when x = – 1, is 1. Ans. 15. (i) Substituting a = 2, b = – 2 in a2 + b2, a2 + b2 = (2)2 + (– 2)2 = 4 + 4 = 8 a2 + b2 = 8, when a = 2, b = – 2 Ans. (ii) Substituting a = 2, b = – 2 in a2 + ab + b2 a2 + ab + b2 = (2)2 + 2 × (– 2) + (– 2)2 =4–4+4=4  a2 + ab + b2 = 4, when a=2, b= – 2 Ans. (iv) Substituting a = 2, b = – 2, in a2 – b2, a2 – b2 = (2)2 – (– 2)2 = 4 – 4 = 0 a2 – b2 = 0, when a = 2, b = – 2 Ans. 16. (i) Substituting a = 0, b = – 1, in 2a + 2b, 2a + 2b = 2 (0) + 2 (– 1) = 0 – 2 = – 2 2a + 2b = – 2, when a = 0, b = – 2 Ans. (ii) Substituting a = 0, b = – 1 in 2a2 + b2 + 1 2a2 + b2 + 1 = 2 (0)2 + (– 1)2 + 1 =2×0+1+1=0+1+1=2 2a2 + b2 + 1 = 2, when a = 0, b = – 1 Ans. (iii) Substituting a=0, b =–1 in 2a2b+2ab2 + ab, 2a2b + 2ab2 + ab = 2(0)2 (– 1) + 2 (0) (– 1)2 + (0) (– 1) = 2 × 0 × (–1) + 2 × 0 × l + 0 × (– l) =0+0+0=0  2a2b + 2ab2 + ab = 0, when a = 0, b = – 1 Ans. (iv) Substituting a = 0, b = – 1 in a2 + ab + 2, a2 + ab + 2 = (0)2 + (0) (– 1) + 2 =0+0+2=2 a2 + ab + 2 = 2, when a = 0, b = – 1 Ans.

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7th Class Mathematics

208 KEY

SUMM ATIVE WORKSHEET Q.No

17

18 19

20

21

22

Ans.

B

A

C

C

C

C

Q.No

23

24 25

26

27

28

Ans.

D

C

B

A

D

B

17. The basic unit of length is metre. 18. The basic unit of are is square metre. 19. The basic unit of volume = Basic unit of Area × Basic unit of length = m2 ×m = m3 20. Area (A) = 25 sq.units A= l2

1. 2. 3.

4.

(a) The number of cats is the unknown. (b) The sum of money is the unknown. W represents the object (model of the car) and p represents the unknown (price of the car). (a) 5xy is not an algebraic term with one unknown because it has two letters, x and y. (b) 8 r is an algebraic term with one unknown because it has only one letter, r. (c) 26 is not an algebraic term with one unknown because it has no unknowns. Terms

Coefficients

Unknown

–11x

–11

x

f 6

1 6

F

l A

 25  5 units 21. 1 metre = 100 centimetre

5.

1 22. Area of a triangle = × base × height 2

(b) 4.5 h and

1  bh 2 23. V = lbh 24. Area of a circle = r 2

1 25. Area of trapezium  d  a  b  2 26. Total angle in a area = 360°. 27. Tn = a + (n –1)d 100 = 2 + (n – 1)d  2(n – 1) = 98  n – 1 = 49 n = 48 128 28. 12.56  2  3.14 g

g

128 g

128  32 units 4

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h are like terms because they 10

have the same unknown, h. 6.

7.

The like terms with the unknown a are

3 a and 4

2a. The like terms with the unknown b are –0.2b and 5b. The like terms with the unknown c are –c and 7c. The like terms with the unknown d are 6d and 7.2d. To state like terms for a given term, change only the coefficient. (a) 3 like terms for 4p : 2p, –p, 0.5p

2 1 k : 2k, –4k, k 3 2 2u + 3u  This is an algebraic expression 5p – 7  This is an algebraic expression 0.8s – 6t + 4  This is an algebraic expression 11u + 13v – 17w  This is an algebraic expression 15a – 7b + 45  This expression has three terms 15a, –7b and 45 19t+15t  This expression has two terms: 19t, 15t. 6m – (–2m) + 4m = 6m + 2m + 4m [–(–2m) = +2m)] = (6 + 2 + 4)m = 12m

(b) 3 like terms for 8.

(a) (b) (c) (d)

9.

(a)

12.56 128  2  3.14 g

4

(a) 0.6 j and 2 g are unlike terms because they have different unknowns, j and g.

(b) 10. (a)

Algebraic Expressions Solutions (b) 12x + 7y cannot be simplified because 12x and 7y are unlike terms. (c) 5p + 9 – 3p – 6q = 5p – 3p + 9 – 6q  Separate the like terms and unlike terms = (5 – 3)p + 9 – 6q = 2p + 9 – 6q 11. Unknowns : r, s, t and u 12. 13.

14.

15.

16.

17.

18.

19.

20.

2 2 3 2 3 (a) 4xy, –6xy (b) p qs ,  0.8p qs 3 The perimeter of the triangle = (x + x + y) cm = (2x + y) cm (a) 8xy has one term, that is 8 xy. (b) ab + 2bc + 7 has three terms, that is, ab, 2bc and 7. (a) 6pq + 3pq = 9pq (b) –2x2y + 5x2y = 3x2y (c) 4ab + (–3hk) = 4ab – 3hk  (Unlike terms cannot be added). (a) 7r2s2 – (–4r2s2) – 3rs2 = 7r2s2 + (4r2s2) – 3rs2 = 11r2s2 – 3rs2 (b) 6mn + 2xy – 3mn + 5xy – 4= (6mn – 3mn) + (2xy + 5xy) – 4  (Group the like terms) = 3mn + 7xy – 4 2 2a b + xy2 – ax = 2(1)2(–2) + (–1) (3)2 – (1) (–1) = –4 – 9 + 1 = –13 + 1 = –12 Cost of a can of water = Rs. 1.50 Cost of x cans of water = Rs. 1.5x Cost of a packet of biscuits = rs. 2.70 Cost of y packets of biscuits = Rs. 2.7y  Total amount = Rs. (1.5x + 2.7y) Ans: B (5xy + 3z2) + (2xy – 4z2). = 5xy + 3z2 + 2xy – 4z2  (Remove the brackets) = 5xy + 2xy + 3z2 – 4z2  (Group the like terms and simplify) = 7xy – z2 (3p2k – 2mn) –(rs – 3mn –p2k) = 3p2k – 2mn – rs + 3mn + p 2k  (Remove brackets and change the symbol of operation for each term in the second expression) = 3p2k + p2k– 2mn + 3mn – rs  (Group the like terms and simplify). = 4p2k + mn – rs

209 21. Length of a curve in a sector 

x  2r 360

22. Let the number be x.

x  11  25 3 23. p = 2(l + b) Given:



p lb 2



p l  b 2

p  2l b 2 24. y = mx + c (or) mx + c = y  mx = y – c m=y–c =x 2 2 25. x + y = r2  y2 = r2 – x2



y  r2  x2 26. y2 = 4ax (or) 4ax = y2 (4x)(a) = y2  a

y2 4x

HOTS WORKSHEET 1. 2. 3.

(a) –21 (b) 4 Rs. (20m + 15n) 10y – 2z 4.

5.

–7x + 1

7. 9.

x+4 0.35x + 0.2y

11.

2 3

6.

3x 10 – 4x – 6y

8. 50 – 3x – 5y 10. 8x + 6y

PTR 1000 Given: I = 1000 P = 5000 R=4 R=? I

P.I  R

PTR  100

1000 I PT

PTR = 100 I

R

1000  100 5 5000  4 www.betoppers.com

7th Class Mathematics

210

9 F  C  32 5

12. Given:

 P  n 2a     v  nb  nRT 2  v 

16.

9 F   30  32 5 = 54 + 32 = 86 n 13. S   2a   n  1 d  2

P  n 2a nRT  2 v v  nb P  n 2a 

(or)

n  2a   n  1 d   S 2 2a   n  1 d  S



n 2a 

n 2a 

2S n

 n  1 d 

2S  2a n

 n  1 d 

2S  2an n



14.

1 n 2

2a   n  1 d 

d

n 2a 

a

2S  2an n  n  1

ab xb c

a  b  cb c 15. g = 9.8 m/s2 s = 40 cm t=? 1 s  gt 2 (or) 2

t

2s g

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nRtv 2  Pv  Pnb v  nb

nRtv2  Pv  Pnb  v  nb  n 2

1

2

3

4

5

B

B

D

C

A

6

7

8

9

10

C

D

C

C

D

11

12

13

14

15

C

A

A

A

C

16

17

18

19

20

AC

BC

a+3

21

22

23

24

25

a–2

2

x = 3a+1

3

3

26

27

28

29

30

1

8

2

ABCD/tspr

ABC/srp

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 20.



v  nb

ABC BCD

ab b c

x

2s g

nRtv 2  P  v  nb 

KEY

xb 1  ab c

t2 

nRtv 2 P v  nb

IIT JEE WORKSHEET

2  S  an  n  n  1

x

nRtv 2 v  nb

2  40 40  9.8 4.9



400 20  49 7

Amar get ‘a’ apples Sunil get 3 apples more than Amar  Sunil get a + 3 apples.

Algebraic Expressions Solutions 21.

22.

29.

30.

211

Prem get 5 apples less than Sunil a+3–5 = a – 2. Amar get ‘a’ apples Prem get a – 2 apples.  Prem get 2 apples less than Amar.. A) x(y – 5) – (x + y) xy – 5x – x – y xy – 6x – y B) (x – y) + (x + y) =x–y+x+y = 2x. C) x2 + xy + y2 + 3x2 + 3xy = 3x2 + y2 + 4xy D) 2x2 – 3xy – x2 + xy – 2 x2 – 2xy – 2

n 2a   n  1 d  2 A) s = 320, n = 16, a = 5 s

16  2  5  (15)d  2 40 – 10 = 5d 30 = 5d d=2 B) x is addedt any number (x + 8) and the sum multiplied by 4  4(x + 8) result is 56  4(x + 8) = 56 C) x – 8 = 3 (y – 8) 320 

 

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212

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7th Class Mathematics

4. SIMPLE EQUATIONS SOLUTIONS 6.

Here the products (4x – 7) (3x – 5) and (4x – 9) (x – 1) must be multiplied out, or written down by inspection. Forming the products, we have 5x – (12x2 – 41x + 35) = 6 – 3 (4x2 – 13x + 9); and by removing brackets. 5x – 12x2 + 41x – 35 = 6 – 12x2 + 39x – 27. The terms –12x2 may be removed from each side without altering the equality; thus 5x + 41x – 35 = 6 + 39x – 27. Transposing, 5x + 41x – 39x = 6 – 27 + 35; collecting terms, 7x = 14;  x = 2.

7.

4

FORMATIVE WORKSHEET 1.

2.

x – 5= 11 x – 5 + 5 = 11 + 5 x = 16

x =7 4 x 4= 74 4 x = 28.

3.

Add 5 to both sides.

Multiply both sides by 4.

3 8  x  2 5 3 8  x  8  2  8 5

32  x  9 x  11  8 22 22(32 – x – 9) = 8(x – 11) 704 – 22x – 198 = 8x – 88 704 – 198 + 88 = 8x +22x. 594 = 30x

3  x  10 5

3  5  5  x      10     5  3  3 x 4.

50 3

3x  2 4x  7  5 6

8.

3x – 12 +

x  5.

594  19.8 30 Multiplying throughout by 9, we have x

 3x  2   4x  7  30    30   Multiply both sides  5   6  by 30. 6 (3x + 2) = 5(4x – 7) 18x + 12 = 20x – 35 18x + 12 – 20x = – 35 –2x = –35 – 12 1 47 47  23  2 2 2 Removing brackets, 5x – 15 – 42 + 7x = 24 – 24 + 3x – 3 collecting terms, 12x – 57 = 3x – 3. Subtracting 3x from each side, we get 9x – 57 = –3. Adding 57 to each side, we have 9x = 54. Dividing by 9, x = 6.

x 9 x 1   8 22 2

9.

18x  27 9x  81 .  5x  32  35 28

transposing, 18x  27  9x  81  2x  20 . 35 28 Now clear of fractions by multiplying by 5 × 7 × 4 or 140; thus 72x – 108 + 45x + 405 = 280x – 2800;  2800 – 108 + 405 = 280x – 72x – 45x;  3097= 163x;  x = 19. Transposing, 0.375x – 0.12x = 1.185 + 1.875; collecting terms, 0.375 – 0.12x = 1.185 + 1.875; collecting terms, (0.375 – 0.12) x = 3.06, that is, 0.255x = 3.06; x  3.06 0.255 = 12

7th Class Mathematics

214 10. (i)

3x + 4 = 2x + 7 3x – 2x = 7 – 4 x=3 (ii) 5x + 8 = 3x – 2 5x – 3x = –2 – 8 2x = – 10 x = –5 (iii) 2(8x + 5) = 4(3x +1) 16x + 10 = 12x + 4 16x – 12x = 4 – 10. 4x = –6

3 2 (iv) 3(4x – 1) = 7(2x – 5) 12x – 3 = 14x – 35 12x – 14x = –35 +3 –2x = – 32 x = 16 (v) 5(x + 3) – 4(2x – 9) = 0 5x + 15 – 8x + 36 = 0 –3x + 51 = 0 3x = 51 x = 17 (vi) 3(3x –1) – 4(5 – 2x) = –10 9x – 3 – 20 + 8x = –10 17x – 23 = –10 17x = – 10 + 23 17x = 13

15x – 18 + 8 = 4 (4x) 15x – 10 = 16x. –10 = 16x – 15x. x = –10 (x)

2  8x  9  6  3x 5 2 – 8x – 45 = 5(6 – 3x) –8x – 43 = 30 – 15x –8x + 15x = 30 + 43 7x = 73

x

13 17 (vii) 9x – 2(x + 8) = 5x – 11 9x – 2x – 16 = 5x – 11 7x – 16 = 5x – 11 7x – 5x = –11 + 16 2x = 5 x

5 2 (viii) 1 – 4(2x + 3) = 5(x – 2)–3(x –1) 1 – 8x – 12 = 5x – 10 – 3x + 3 –8x –5x + 3x = –10 + 3 –1 + 12 –10x = 4 x = –4/10 x

x (ix)

2 5

3  5x  6   2  4x 4

15x  18  2  4x 4 www.betoppers.com

2 1  4x   9  3 2  x  5

x (xi)

73 . 7

7x  2 5x  3  2 3 3(7x – 2) = 2(5x –3) 21x – 6 = 10x – 6 21x – 10x = –6 + 6 11x = 0 x=0

2x  3 3x  15  3 11 11(2x +3) = (3x – 15) 22x + 33 = 9x – 45 22x – 9x = –45 – 33 13x = –78 x = –6 11. Putting u = 10, s = 15 and t = 3 into (xii)

1 (u  v)t , we have the equations: 2 1 15  (10  v)(3) v is the unknown. 2 2 15   10  v solve for v 3 10 = 10 + v v = 10 – 10 v  0 12. Let the daughter’s present age by x years. Mrs. Kumar’s present age = 3x years. In 5 years time, Mrs. Kumar’s age = (3x + 5) years. In 5 years time, her daughter’s age = (x + 5) years. Since their total age in 5 years’ time will be 62, we have: (3x + 5) + (x + 5) = 62 s

Simple Equations Solutions 3x + 5 + x + 5 = 62 4x = 62 – 10

52 4 x = 13  Hence the daughter’s present age is 13 years. 13. Let the price of a chair be Rs. x. Price of a table = Rs. (6x – 100) Price of a table and 4 charis = Rs. 1400 (6x – 100) + 4x = 1400 6x – 100 + 4x = 1400 10x = 1400 + 100

215 5.

x=

1500 = 150 10 Hence the price of a chair is Rs. 150.

6.

x

CONCEPTIVE WORKSHEET 1.

2.

x + 6= 13 x + 6 – 6 = 13 – 6 Subtract 6 from both sides x=7 –6x = 8

6x 8  6 6 Divide both sides by – 6 4 x . 3 3.

2 1 1 3 3 1 x   x 1  x  ; 3 4 9 9 4 3 clearing of fractions, 24x + 9 –4x=68–27x – 12; transposing, 24x – 4x + 27x= 68 –12 – 9, 47x = 47;  x = 1. (i) x + 8 = 9 (ii) x + 36 = –40 x=9–8 x = –40 – 36 x=1 x = –76 (iii) x – 9 = 5 (iv) x – 22 = –15 x=5+9 x = –15 + 22 x = 14 x=7 (v)

x 3 2 x=6

(vii) 4x = 24

24 4 x=6 (ix) 2x – 3 = 7 2x = 7 + 3 2x = 10

x  4 5 x = –4 × 5 x = –20 (viii) –9x = 21 (vi)

x

5(2x  9)  8  2x 3 5(2x – 9) – 3(8) = 3(2x) 10x – 45 – 24 = 6x 10x – 69 – 6x = 0 4x = 69 x

4.

7.

Removing brackets, we have 7x – 5 {x – [7 – 6x + 18]} = 3x + 1, 7x – 5 [x – 25 + 6x] = 3x + 1, 7x – 5x + 125 – 30x = 3x + 1; transposing, 7x – 5x – 30x – 3x = 1 – 125; collecting terms, –31x = – 124;  x = 4. Expressing the decimals as vulgar fractions, we have

69 4

(xi) –5x + 2 = –3 –5x = –3 – 2 –5x = –5 x=1 (ix)

1 x 9 6 4 x + 36 = 4 × 6 x + 36 = 24 x = 24 – 36 x = –12

21 9

(x) 3x + 8 = –1 3x = –1 – 8 3x = –9

9 3 x = –3 x

x=5

y y 3  7 5 2

y  y 3 10    10    10  7  5  2  2y + 5(y – 3) = 70 Multiply both sides by 10 which is 2y + 5y – 15 = 70 the LCM of 2 and 5. 7y = 70 + 15 85 y 7 1  y  12 . 7

x

(xii)

x 20 3 x 6 0 3 x–6=0 x=6 1 (x) 1  x  8 7 7 – x = –8×7 7 – x = –56 –x = –56 – 7 –x = – 63 x = 63 www.betoppers.com

7th Class Mathematics

216 8.

t 1 1 t3 4 Multiplying both sides by 4(t + 3), we have: t 5 = 4(t + 3) × t 3 4 4t = 5 (t + 3) 4t = 5t + 15 t = – 15 Here it is convenient to begin by clearing the equation of fractional coefficients. This can be done by multiplying every term on each side of the equation by the least common multiple of the denominators. Hence, multiplying throughout by 20, 16x – 6 = 4x + 5x. Subtracting 9x from each side, 7x – 6 = 0. Adding 6 to each side, 7x = 6, 4(t + 3) ×

9.

Dividing by 7, x = 6 . 7 [Verification: when x 

6 the left side 7

4 6 3 48  21 27 .      5 7 10 70 10 the right side  1  6  1  6  24  30 5 7 4 7 140 54 27   . 140 90 10. (i) Let Anne weigth = x. Then Tommy mass = x + 5 kg. (ii) 63 = x+ 5 x = 63 – 5 x = 58 11. (i) Rs (2x + 50) (ii) 208 = 2x +50 2x = 208 – 50 x = 158/2 x = 79. 12. Let the three even numbers are x, x + 2, x + 4 sum of these integers = 66.  x + x + 2 + x + 4 = 66. 3x + 6 = 66 3x = 66 – 6 3x = 60 x = 20  even integers are 20, 22, 24.

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13. Let the chocolates box be x. then 3x = 50 – 8 3x = 42 x = 14 14. A + B = 50 _________ (1) 6A + 13B = 433 _________ (2) Multiply eq (1) × 6.we get 6(A + B) = 50 × 6 6A + 6B = 30 _________ (3) subtract (3) from (2) 6A + 13B = 433 6A + 6B = 30 – – – ––––––––––––– 7B = 133 B = 19 Put the B value in eq (1) A + 19 = 50 A = 50 – 19 A = 31

SUMM ATIVE WORKSHEET KEY 1

2

3

4

5

6

7

8

C

B

C

D

D

B

C

D

9

10 11 12 13 14

15 16

D

C

B

C

D

A

B

D

17 18 19 20 C

A

A

C

HOTS WORKSHEET 5 8

1) u = – 1

2) k = –

4) p = 2 7) u = 2

5) y = 2 8) 2x+2 =10

3) d = 3 6) x = – 10 9

1 3

)

x 5 2 3 10) The equations cannot be solved for the value of x. 11) The situations do not exist. 12) Raj has certain no. of apples. Rahim has 10 apples less than Raj. Anthony has 4 times the no. of apples what Rahim has. If the sum of 5 times the apples of raj and anthony is equal to 140. Find the no. of apples each one is having.

Simple Equations Solutions 13) 4

1 2

217

14) – 4

19) x2 +2x2 – 7x + 3 = 0 24) 3

1 7

25) 8

30) 2

15) 2

16) 3

17) 3

18) 4

20) 2

21) 2

22) 5

23) 6

26) 5

27) 5

28) 7

29) 7

1 6

31) 5

IIT JEE WORKSHEET KEY Q.No.

1

2

3

4

5

6

7

8

Ans.

C

A

C

A

D

B

C

D

Q.No.

9

10

11

12

13

14

15

16

Ans.

A

D

A, B, C

B, D

A, B

A, B, D

B, C, D

7

Q.No.

17

18

19

20

21

22

23

Rs. 12

(a) 10 apples (b) Rs. 0.20

1

2

2

5

A  r; B  r, C  p, q, D  s.

Ans.

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 1.

4.

8.

9.

x + x +3 = 15 2x + 3 = 15 2x = 15 – 3 2x = 12 x=6 [A] l=y b=y–5 Perimeter = 18 2(l + b) = 18  2(y + y – 5) = 18 2(2y – 5) = 18 4y –10 = 18 4y = 18 + 10 4y = 28 y = 7. [A] Sum of all angle of quadrilateral = 360°  2y + y + 70 + 80 = 360° 3y = 360 – 150 3y = 210 y = 70 [D]

1  h  a  b   72 2 3(7x + 10) = 72 21x + 30 = 72 2x = 72 – 30 21x = 42 x = 21

12.

19.

22.

23.

2x + 1 = 5 2(2x + 1) = 2 × 5 4x + 2 = 10 3(2x + 1) = 3 × 5 6x + 3 = 15 B, D are correct answers. 15(z + 4) = 75 15z + 60 = 75 15z = 75 – 60 15z = 15 z=1 y + 3 = 4(y – 3) y + 3 = 4y – 12 3 + 12 = 4y – y 3y = 15 y = 5. d) 4 + 5(x –1) = 34 4 + 5x – 5 = 34 5x –1 = 34 5x = 35 x=7 D  s.

[B] [D]

  [A] www.betoppers.com

218

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7th Class Mathematics

5. IDENTITIES & SPECIAL PRODUCTS SOLUTIONS Multiplier :

FORMATIVE WORKSHEET

4x2 + 6

6 + 4x2

Solution: 8 – 7x – 2x2 + 3x3

1. ×

× 6

5x2 – 4x + 3 – 4x + 5

+ 4x2

48 – 42x – 12x2 + 18x3 – 20x3 +

+ 16x2 – 12x 25x2 – 20x + 15

32x2 – 28x3 – 8x4 – 12x5

+

48 – 42x + 20x2 – 10x3 – 8x4 + 12x5 – 20x3 + 41x2 – 32x + 15 So, required product : – 20x3 + 41x2 – 32x + 15. Second method : Arranging in ascending powers of x: Multiplicand : 5x2 + 3 – 4x 3 – 4x + 5x2 [in ascending powers of x] Multiplier : 5 – 4x 5 – 4x [in ascending powers of x] Solution: 3 – 4x + 5x2 × 5 – 4x

So, the required product : 48 – 42x + 20x2 – 10x3 – 8x4 + 12x5 3.

2. 3x3 – 2x2 – 7x + 8 × 4x2 + 6

3a2 – 4ax + x2 [in

Multiplier : 5x2 – 2ax + 2a2 descending powers of a]

2a2 – 2ax + 5x2 [in

3a2 – 4ax + x2 × 2a2 – 2ax + 5x2 6a4 – 8a3x + 2a2x2 +

– 6a3x + 8a2x2 – 2ax3 15a2x2 – 20ax3 + 5x4

+

15 – 20x + 25x2 + – 12x + 16x2 – 20x3 15 – 32x + 41x2 – 20x3 So, required product : 15 – 32x + 41x2 – 20x3

Multiplicand : 3a2 – 4ax + x2 descending powers of a]

6a4 – 14a3x + 25a2x2 – 22ax3 + 5x4 So, the required product : 6a4 – 14a3x + 25a2x2 – 22ax3 + 5x4 4.

Multiplicand : a3 – 3a2b + 3ab2 a3 – 3a2b + 3ab2 – b3 [in descending powers of a] Multiplier :

a2 + b2 – 2ab a2 – 2ab + b2 [in descending powers of a]

a3 – 3a2b + 3ab2 – b3 5

4

3

2

5

4

3

2

12x – 8x – 28x + 32x + 18x3 – 12x2 – 42x + 48

× a2 – 2ab + b2 a5 – 3a4b + 3a3b2 – a2b3

12x – 8x – 10x + 20x – 42x + 48 So required product : 12x5 – 8x4 – 10x3 + 20x2 – 42x + 48 Second method : Arranging in ascending powers of x. Multiplicand : 3x3 – 7x + 8 – 2x2 8 – 7x – 2x2 + 3x3

+ +

– 2a4b + 6a3b2 – 6a2 b3 + 2ab4 a3b2 – 3a2b3 + 3ab4 – b5 a5 – 5a4b + 10a3b2 – 10a2b3 + 5ab4 – b5

So, the required product : a5 – 5a4b + 10a3b2 – 10a2b3 + 5ab4 – b5

220 5.

7th Class Mathematics It can be seen that the given expressions are arranged in descending powers of x. Here, at first the product of (x2 + x + 1) and (x2 – x + 1) is to be determined and then this product is to be multiplied by (x4 – x2 + 1). First step : Second step: x2 + x + 1 x4 + x 2 + 1 2 × x –x+1 × x4 – x 2 + 1

+ +

6.

x4 + x3 + x2 – x3 – x2 – x x2 + x + 1

x8 + x6 + x4 + – x6 – x4 – x2 + x4 + x2 + 1

x4 + x2 +1 x8 + x4 +1 8 4 So, the required product : x + x + 1 In descending powers of x By detaching co-efficient. Multiplicand : 3x2 – 5 + 2x 3x2 + 2x – 5 Multiplicand : 3 + 2 – 5 Multiplier : 7 + 4x 4x + 7 Multiplier : 4 + 7 General multiplication Multiplication by detached co-efficients 3x2 + 2x – 5 3+2–5 4x + 7 4+7 12x3 + 8x2 – 20x + 21x2 + 14x – 35

12 + 8 – 20 21 + 14 – 35

12x3 + 29x2 – 6x – 35 In the above product, the descending order of powers of the variable is, x3 x2 x1 x0 and the detached co-efficients 12 + 29 – 6 – 35

7.

12 + 29 – 6 – 35 It can be seen in both the products, the successive order and the algebraic value of the co-efficients are equal and identical. Again, by index law the highest power of the variable of the product should be (2 + 1) = 3 so, the product is, 12x3 + 29x2 – 6x – 35. By arranging in descending order of powers of a and detaching the co-efficients, we get, Multiplicand : 5a3 – 4a – 3 5a3 + 0a2 – 4a – 3 ; 5 + 0 – 4 – 3 Multiplier : a2 – 4 a2 + 0a – 4 ; 1 + 0 – 4 General multiplication Multiplication by detached coefficients 5a3 + 0a2 – 4a – 3 5+0–4–3 2 × a + 0a – 4 1+0–4 5a5 + 0a4 – 4a3 – 3a2 0a4 + 0a3 – 0a2 – 0a – 20a3 – 0a2 + 16a + 12 5a5 + 0a4 – 24a3 – 3a2 + 16a + 12 By detaching the co-efficients we get, 5 + 0 – 24 – 3 + 16 + 12. By excluding the term with 0 co-efficient we get the product = 5a5 – 24a3 – 3a2 + 16a + 12

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5+0–4–3 0+0–0–0 – 20 – 0 + 16 + 12 5 + 0 – 24 – 3 + 16 + 12 It can be seen in both the products, the successive order and algebraic value of the co-efficients are identical. Now, according to index law the highest power of the variable of the product will be (3 + 2) = 5 so, the product = 5a5 + 0a4 – 24a3 – 3a2 + 16a + 12 = 5a5 – 24a3 – 3a2 + 16a + 12

Identities & Special Products 8. Both the multiplicand and the multiplier are homogeneous polynomials with two variables Multiplicand : 6x3 – y3 + 3x2y 6x3 + 3x2y – y3 (in descending powers of x) Multiplier : x2 + y2 x2 + y2 (in descending powers of x) General multiplication 6x3 + 3x2y – y3 × x2 + y2 6x5 + 3x4y +

10.

– x2y3 6x y + 3x2y3 – y5 3 2

6x5 + 3x4y + 6x3y2 + 2x2y3 – y5 It can be seen that both the multiplicand and the multiplier have absent terms. While detaching the co-efficients, those terms should be filled up. As – Multiplicand : 6x3 – y3 + 3x2y 6x3 + 3x2y + 0xy2 – y3 co-efficients : 6 + 3 + 0 – 1 Multiplier : x2 + y2 x2 + 0xy + y2 co-efficients : 1 + 0 + 1 Multiplication by detached co-efficients 6+3+0–1 1 +0+1

– 3x + 2 – 3x + 2 + – 0 Description of the method : Dividing the first term of the dividend by the first term of the dividend 6x 2 = 2x. 3x This 2x is to be placed in the quotient. Then, by multiplying the whole divisor with this term (3x – 2) 2x = 6x2 – 4x, the product is to be placed under the first two terms of the dividend and then by subtracting, the difference is – 3x + 2. by the first term of the divisor, we get

6+3+0–1 0+0+0–0 6+3+0–1

9.

6+3+6+2+0–1 Since, both the multiplicand and the multiplier are homogeneous polynomials whose multiplicand’s order is 3 and multiplier’s order is 2, so by7 the index law of multiplication the order of the product should be (3 + 2) = 5. So, the product in descending order of powers of the variable x will be, 6x5 + 3x4y + 6x3y2 + 2x2y3 + 0xy4 – y5 that is, 6x5 + 3x4y + 6x3y2 + 2x2y3 – y5 In descending order of powers of a Multiplicand : 5a3 + 0a 2b + 3ab2 – 2b3 ; coefficients : 5 + 0 + 3 – 1 – 2 Multiplier : 2a2 + 0ab – b2 ; co-efficients : 2 + 0 – 1 5+0+3–2 2+0–1 10 + 0 + 6 – 4 0+0+0–0 –5–0–3+2

221 According to index law, the order of the product should be (3 + 2) = 5 So, the product is 10a5 + 0a4b + a3b2 – 4a2b3 – 3ab4 + 2b5 that is, 10a5 + a3b2 – 4a2b3 – 3ab4 + 2b5 So, the product in descending order of powers of b is, 2b5 – 3ab4 – 4a2b3 + a3b2 + 10a5 Dividend : 6x2 + 2 – 7x 6x2 – 7x + 2 (in de scending powers of x) Divisor : 3x – 2 3x – 2) 6x2 – 7x + 2 (2x – 1 6x2 – 4x – +

3x = – 1. This –1 is to be placed in the 3x quotient on the right hand side of 2x. As before, by multiplying the whole divisor by this term (3x – 2) × (–1) = –3x + 2, the product is to be placed under the previous difference and to be subtracted from it. This time the difference becomes 0, which means, the end of division. Dividend : 17x + 4 – 15x2 4 + 17x – 15x2 (in ascending powers of x) Divisor : 5x + 1 1 + 5x (in ascending powers of x) 1 + 5x ) 4 + 17x – 15x2 (4 – 3x 4 + 20x – – Now,

11.

– 3x – 15x2 – 3x – 15x2 + + 0

10 + 0 + 1 – 4 – 3 + 2 www.betoppers.com

222 12.

7th Class Mathematics Dividend :

2a3

7a2

2a3

7a2

–a+2– – –a+2 [in descending powers of a] Divisor : a2 – 3a – 2 a2 – 3a – 2. a2 – 3a – 2 ) 2a3 – 7a2 – a + 2 ( 2a – 1 2a3 – 6a2 – 4a – + + – a2 + 3a + 2 – a2 + 3a + 2 + – –

0 13. Here, both the dividend and the divisor are arranged in the descending order of powers of the variable but some of the successive powers of the terms are absent. We know that in this case, the absent terms can be filled up by taking them with 0 co-efficient. Thus, Dividend : 81p4 – 1 81p4 + 0p3 + 0p2 + 0p – 1 Divisor : 3p – 1 3p – 1 3p – 1)81p4 + 0p3 + 0p2 + 0p – 1 ( 27p3 + 9p2 + 3p + 1 81p4 – 27p3 – + 27p3 + 0p2 + 0p – 1 27p3 – 9p2 – + 9p2 + 0p – 1 9p2 – 3p – + 3p – 1 3p – 1 – + 0 Alternative method : By keeping space between 81p4 – 1 while writing. 3p – 1 ) 81p4 – 1 ( 27p3 + 9p2 + 3p + 1 81p4 – 27p3 – + 27p3 – 1 27p3 – 9p2 – + 9p2 – 1 9p2 – 3p – + 3p – 1 3p – 1 – + www.betoppers.com

0

14.

The given expressions have more than one variable but they are not homogeneous. Dividend : 4p2 – q2 + 9 – 12p 4p2 – 12p + 9 – q2 [Arranging the Divisor : 2p + q – 3 2p – 3 + q descending powers of p] 2p – 3 + q ) 4p2 – 12p + 9 – q2 ( 2p – 3 – q 4p2 – 6p + 2pq – + –

– 6p + 9 – 2pq – q2 – 6p + 9 – 3q + – + – 2pq + 3q – q2 – 2pq + 3q – q2 + – + 15.

0 It can be seen that two terms of the dividend are absent. By filling up those two terms, the dividend becomes Dividend : 2a4 + 0a3 – 10a2 + 0a – 72 Divisor : a – 3 Detached co-efficients : 2 + 0 – 10 + 0 – 72 1–3 1 – 3) 2 + 0 – 10 + 0 – 72 (2 + 6 + 8 + 24 2–6 – + 6 – 10 + 0 – 72 6 – 18 – + 8 + 0 – 72 8 – 24 – + 24 – 72 24 – 72 – + 0 According to the index law of division, the highest power of a in the quotient is (4 – 1) = 3. Therefore, the required quotient in detached co-efficients = 2 + 6 + 8 + 24 expression = 2a3 + 6a2 + 8a + 24.

Identities & Special Products 16.

223

It can be seen that both the dividend and the divisor are arranged in descending order of powers of their variables, so 3a + 2 ) 12a2 + 14a + 8 (4a + 2 The first term of the quotient =

12a 2 = 4a 3a

17.

1 5x – 3 ) 10x2 + 15x – 12 ( 2x + 4 5 10x 2 1st term of the quotient = = 2x. 5x 2 10x – 6x 21x 21 1 4 2nd term of the quotient = 5x 5 5 – +

12a2 + 8a – – 6a + 8 6a The second term of the quotient = =2 3a 6a + 4 – – 4

18.

19.

21x – 12 63 21x – – + 5 3 5

Arranging the dividend and the divisor in descending power of a, we get, Dividend : 5a3 + 2a – 3 – 4a2 5a3 – 4a2 + 2a – 3 Divisor : a2 – 1 + 2a a2 + 2a – 1 a2 + 2a – 1 ) 5a3 – 4a2 + 2a – 3 ( 5a – 14 5a3 + 10a2 – 5a – – +

Quotient : 1st term =

– 14a2 + 7a – 3 – 14a2 – 28a + 14 + + –

2nd term =

5a 3 = 5a a2

14a 2 = – 14 a2

35a – 17 Here, both the dividend and the divisor are arranged in descending order of powers of the variable a. 4a – 5b ) 12a2 – 31ab + 22b2 ( 3a – 4b 12a2 – 15ab – +

Quotient : 1st term = 2nd term =

12a 2 = 3a 4a

16ab = – 4b 4a

– 16ab + 22b2 – 16ab + 20b2 + – 20.

2b2 Divide : x3 – x2 – 8x – 13 by x2 + 3x + 3. In the descending order of powers of x, dividend : x3 – x2 – 8x – 13 Divisor : x2 + 3x + 3 Detached co-efficients : 1 – 1 – 8 – 13 1 + 3 + 3. 1 + 3 + 3 ) 1 – 1 – 8 – 13 (1 – 4 1+3+3 According to the index law, the power of – – – x in the quotient = 3 – 2 = 1. – 4 – 11 – 13 – 4 – 12 – 12 + + + 1–1

Quotient = 1 – 4 x–4 Remainder = 1 – 1 x–1 www.betoppers.com

224 21. 22. 23.

7th Class Mathematics (0.8 m + 1.1n) (0.8 m – 1.1 n) = (0.8 m)2 – (1.1 n)2 = 0.64 m2 – 1.21 n2 7122 – 2882 = (712 + 288) (712 – 288) = 1000 × 424 = 4,24,000 (i) (a + b) = 11 Squaring both sides, (a + b)2 = 112 a2 + 2ab + b2 = 121 a2 + b2 = 121 – 2ab

(ii)

= 121 – 2 × 30 [since ab = 30]

To find (a – b) a – b)2 = a2 – 2ab + b2 = (a2 + b2) – 2ab = 61 – 2 × 30 [from (i)] = 61 – 60 = 1

(a – b) = 1

1

= 121 – 60 = 61 24.

(i)

x+

1 =5 x

x2 +

(ii)

Squaring both sides, x

1 x

x2 + 2 +

1 = 23 [from (i)] x2

Squaring both sides,

2

= 52

x4 + 2 +

1 = 25 x2

x4 +

1 = 529 x4

1 = 529 – 2 x4

1 1 = 23 x4 + 4 = 527 2 x x (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) (22)2 = a2 + b2 + c2 + 2(54) Rearranging, we get a2 + b2 + c2 = (22)2 – 2(54) = 484 – 108 = 376 x2 +

25.

a 2

26.

b 8

a = 2 a = 2 = 27. 28.

a3 8

a2 4

c 5 b 8

3

c 5 b 8

3

b3 c3 512 125

b2 64

c2 ab 25 16

×

a 2

c 5

2

3

3

b 8 a 2

bc 40 2

b 8

ca 10 c 5

2

a 2

b 8

b 8

c 5

c 5

3abc 80

Here, (a – 3b) + (3b – 5c) + (5c – a) = 0 (a – 3b)3 + (3b – 5c)3 + (5c – a)3 = 3(a – 3b) (3b – 5c) (5c – a) Removing the brackets one by one, we have a – 2b – [4a – 6b – {3a – c + (5a – 2b – 3a + c – 2b)}] = a – 2b – [4a – 6b – {3a – c + 5a – 2b – 3a + c – 2b)}] = a – 2b – [4a – 6b – 3a + c – 5a + 2b + 3a – c + 2b] = a – 2b – 4a + 6b + 3a – c + 5a – 2b – 3a + c – 2b = 2a, by collecting like terms.

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c 5

a 2

Identities & Special Products 29. The expression = – [ – 2x – {3y – 2x + 3y + 3x – 2y} + 2x] = – [ – 2x – 3y + 2x – 3y – 3x + 2y + 2x] = 2x + 3y – 2x + 3y + 3x – 2y – 2x = x + 4y. 30. The expression = 84 – 7 [ –11x – 4{ – 17x + 3(8 – 9 + 5x)}] = 84 – 7 [– 11x – 4 {– 17x + 3(5x – 1)}] = 84 – 7 [ – 11x – 4 { – 17x + 15x – 3}] = 84 – 7 [ – 11x – 4 { – 2x – 3}] = 84 – 7 [ – 11x + 8x + 12] = 84 – 7[ – 3x + 12] = 84 + 21x – 84 = 21x

225 34) 35) 37)

(i) 1 (iv) 400 9p2 – 4pq a2 + 4ab + 4b2

38)

a2

28x7y3

1) C 4) A 6) D 9A 10) A 13) D 15) A 18) D 19) (i) 0

5p4q3r2s2

1) 2) 3) 4) abcxyz 5) 28a3b3x5 6) 3a5x9y15 7) 10p3 + 6p2q 8) p5q + p3q3 – r2q 9) 15x5y + 3x4y2 – 21x5y2 10) 6p5q3r – 7p3q4r2 11) 14a2b2x2 3 4 2 12) – 5x y z 13) 14a4b3 + 28a3b4 14) – 5x2y3z2 + 3x3y2z3 – 8x2y2z2 15) p3q2r – p2q3r + p2q2r2 16) – 2p5x3 + 17) – x8y3 +

4 5 8 xy 7

18) x2 + x – 306

23) – p2

24)

25) 5a4

21 6 x 4 26) x2 – 3x + 1

10) 11) 12) 13)

27) – a + b + c

28)

2 4 a– b–c 3 15 30) 3p + 1

14)

(iii)

x2 9

2

1 a2

9 (iv) 0.25a2 + 0.2ab + 0.04b2 x2

(v) p4q2 – 2p2q2r2 + q2r4 (vi) 3x2 –

33)

7) 8)

(ii) a2 – 1 +

1 2 y 25 (i) 63.91 (ii) 38809 (iv) 6561 (v) 3481

(iii) 494209 (vi) 998001

2 3 xy + 5

b a

2

–2

2) B 5) B 7) B

3) D

11) B 14) D 16) C

12) A

8) D

17) A

(ii) 2a3

1) x4 – 4x2 + 8x + 16 2) a2 – 4b2 + 4bc – c2 3) 75a5b3 – 28a3b5 + 13a2b6 – 12ab7 4) x3 + 3xy + y3 – 1 5) x5 + y5 6) a2x3 + 27a2y6

20) 4p2q2 – 9r2 22) xy2

(i) 4x2 – 4xyz + y2z2

a 40) b

HOTS WORKSHEET

19) x2 – 42x + 441 21) 9x2 – 30xy + 25y2

29) x – 4 31) a2 + 3a + 3 32)

7 4 4 px 2

39) (a

–b)2

SUMMATIVE WORKSHEET

CONCEPTIVE WORKSHEET 8x3y6

– 4ac +

c2

(ii) 11 (iii) 1 (v) –297360 36) a2 + 2ab + b2

9 4 3 3 1 2 2 2 4 x – ax + ax – a 8 2 2 9 1 4 42 2 9 x – x + 9) y + 1 4 36 16 a2 – 3a + 1, remainder a – 6 7y2 + 5y – 3; remainder – 39y + 27 x2 – 2x + 3; remainder 31x – 15 a8 – 2a6b2 + 3a4b4 – 2a2b6 + b8

3 2 1 2 a – a– 8 4 3 15) c – y 17) – x + 2y + 6z 19) 2p 21) y 23) – 9 + q + 5r

16) x – a 18) – 25x + 2y 20) p 22) – 50z 24) – 2p + 10q – 11r

11 x – 2q 26) 2 5 27) (i) 8z(3x – y) (ii) 3x2 + 3y2 + 3z2 + 2zx – 2zy + 2xy (iii) 9(3 + 4x2) (iv) 1009027027 (v) 997002999 25)

28) 5

29)

29

30) 196 www.betoppers.com

7th Class Mathematics

226 31) 13

32) 73 33) 0

35) 52

36) 488

34)

a(a2

– 3) 37) (i) x3 + 8y3 + 64z3 – 24xyz (ii)

8 3 27 3 x – y 125 64

38) (i) 0.343x3 + 0.729y3(ii) x4 + x2 + 1

IIT JEE WORKSHEET KEY Q.No.

1

2

3

4

5

Ans.

D

D

C

C

B

Q.No.

6

7

8

9

10

Ans.

C

B

C

C

C

Q.No.

11

12

13

14

15

A, B

A, B, D

A, B, C, D

x2 + 10x + 21

16x2 + 24x + 5

18

19

20

Ans. Q.No. Ans.

16

17

2

2

2

2

16x – 24x + 5

16x – 24x + 5

16x + 16x – 5

4a4 + 18a + 45

21

22

23

24

1

6

4

A–v, B-t, C-r, D-p

Q.No. Ans.

x2y2z2

6xy2 + 8

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 1.

5 2 x 3

3 2 x 4

4 2 x 3

x2

5 3

4 3

x2

20 9 16 3 12 12

x2

2.

3.

5.

3 4

1 2 x 4

x2

9.

1 1 4 11.

36 12

=–3x2 [D] x = 1, y = 2, z = 3 5x2 = 5(1)2 = 5 –2x3z = –2(1)3(3) = –6 3y4 = 3(2)4 = 48 5x2 –2x3z + 3y4 = 5 – 6 + 48 = 47 x + y – (y – 2z) x + y – y + 2z x + 2z (a – b)(a – b) – (a + b)(a – b) (a – b)2 – (a2 – b2) a2 + b2 – 2ab – a2 + b2 2b2 – 2ab

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16.

[D]

23.

[C] 24.

18a 4 x 2 24ax 5 6ax 2 18a 4 x 2 24ax 5 6ax 2 6ax 2 = –3a3 + 4x3 [C] A) x(y – 2) + y(z – x) + z(x – y) xy – xz + yz – xy + zx – yz = 0. B) 42a2 – 8a (a – 1) – 7a(1 + 5a)+a(a–1) 42a2 – 8a2 + 8a – 7a – 35a2 + a2 –1 = 0. A, B is correct answers. (4x – 5)(4x –1) 16x2 – 4x – 20x + 5 16x2 – 24x + 5 25x2 + 70x + 49 x = –1, then 25(–1)2 + 70(–1) + 49 25(1) – 70 + 49 =4 (c) (3x2 + 5y2)(3x2 + 5y2) (3x2 + 5y2)2 = 9x4 + 25y4 + 2 × 3x2 × 5y2 = 9x4 + 25y +30x2y2

6. ARITHMETIC SOLUTIONS

FORMATIVE WORKSHEET KEY

1

2

3

4

5

6

7

8

9

10

D

A

B

B

A

C

C

B

C

C

11 12

13 14 15 16 17

18

19 20

B

D

B

C

D

21 22

23 24 25 26 27

28

29 30

C

B

C

D

C

A

B

C

D

B

B

A

C

D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. 42 is a whole number Its percent equivalent is (42 × 100)% = 4,200 % 2. Perimeter of rectangle = 2(l + w) = 120 m 2l  w 

120 m 2 2  l + w = 60 m ( l = 36m)  36 m + w = 60 m  w = 60 m – 36 m = 24 m Hence, ratio of width to 

 3.

6.

 25% 

7.

3 3  20 L   20 L  15L 3 1 4 Quantity of water in the original solution = 20 L – 15 L=5L Let x L of water be added to the solution. Total quantity of new solution = (x + 20) L Quantity of milk in the new solution = Quantity of milk in the original solution = 15 L The ratio of milk and water in the solution now is given as 1: 2. Quantity of milk in the new solution

length

24m 2  12 m 2    2:3 36 m 3  12 m 3

Cost of 2 flower pot = Rs. 1.50

Rs.1.50  Rs.0.75 2 Thus, cost of 10 flower pots = 10 × Rs.0.75 = Rs.7.50 Thus, amount spent by Diane = cost of 10 flower pots = Rs.7.50 The percent equivalent of a fraction is calculated by multiplying that fraction with 100.

5.

75 3 75%   34 100 4 0.75 4 3 0 30

1 4 Total quantity of original solution = 20 L The ratio of milk and water in the solution is given as 3:1.  Quantity of milk in the original solution 



1 1     100  %  20% 5 5 

25 25  1 1   100 25  4 4

The required fraction equivalent is

Thus, cost of each flower pot =

4.

28 20 20 0 3  4  0.75  75% = 0.75 The fraction equivalent of a percent can be calculated by dividing the percent by 100.



1   x  20  L  15L 1 2

x  20  15 3  x + 20 = 45  x = 25 Thus, the milkman should add 25 L of water to the solution. Let the number of students in the class be x. 

8.

 Number of girls in the class 

1 x of x  3 3

x 2x  3 3 It is given that 75% of the girls scored more than 75% in the last exam.  Number of boys in the class  x 

7th Class Mathematics

228  Number of girls who scored more than 75%

x 75 x 3 x x      3 100 3 4 3 4 It is also given that 50% of the boys scored less than or equal to 75% in the last exam, which implies that 100 % – 50% = 50% of the boys scored more than 75%.  Number of boys who scored more than 75%

Fraction of bottle now filled with milk =

= 75% of

2x 50 2x 1 2x x      3 100 3 2 3 3  Total number of students who scored more than

= 50% of

x x 7x   4 3 12 It is also given that 70 students scored more than 75% in the exam. 75% 



7x  70 12

9 4 9 36 9 36  9 27       16 4 64 64 64 64 64

27 of the bottle is now filled with milk. 64 10. According to the given information: x% of 700 + x = 700 Thus,

x  700  x  700 100  7x + x = 700  8x = 700  x = 87.5 Thus, the value of x is 87.5. 11. Total number of spectators = 12000 Total number of children = 2400



2400  100  20% 12000 Thus, 20% of the spectators were children. 12. Let length = l and breadth = b Let, the new breadth = x  Required percent 

12  70  x 7  x = 120  Number of boys in the class 2x 2  120   80 3 3 Occasion I: Fraction of bottle originally filled with milk = 1

25   l 1     lb  100  Then, (125% of l) × x = lb

1 1 1 Fraction drunk by the cat = of 1   1  4 4 4

 x



9.



Fraction of bottle now filled with milk  1 

1 3  4 4

Occasion II: Fraction of bottle originally filled with milk 

3 4

1 3 1 3 3 Fraction drunk by cat = of    4 4 4 4 16 Fraction of bottle now filled with milk 3 3 3 4 3 12 3 12  3 9  =        4 16 4 4 16 16 16 16 16 Occasion III: Fraction of bottle originally filled with milk  Fraction drunk by cat =

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9 9  16 64

9 16

1 9 1 9 9    of 4 16 4 16 64

100 4 b b 125 5 Decrease in breadth   4  1    b  b    100  %  20% 5 b    13. The field dimensions = 120 yards × 50 yards  The area of the field = 120 × 50 = 6,000 yards

12  6,000  6,000 100 = 720 + 6,000 = 6,720 yards. 14. Capacity of bottle A = 1 L = 1000 mL Ratio of juice concentrate to water in bottle A is 1:3. The terms in the ratio 1:3 are 1 and 3. Sum of these terms = 1 + 3 = 4  Amount of juice concentrate in bottle Material required 

1 A   1,000 mL  250 mL 4 Capacity of bottle B = 1 L = 1000 mL Ratio of juice concentrate to water in bottle B is 3:7. The terms in the ratio 3:7 are 3 and 7.

Arithmetic

229

Sum of these terms = 3 + 7 = 10  Amount of juice concentrate in bottle

3  1,000 mL  300 mL 10  Difference between the amount of juice concentrate in bottles A and B = 300 mL – 250 mL = 50 mL Thus, the amount of juice concentrate in bottle B is 50 mL more than that in bottle A. 15. It is given that out of 400 employees who lost their job, 250 were men.  Number of women employees who lost their job = 400 – 250 = 150 Therefore, percentage of women employees who lost their job B

150  150    100  %  %  37.5% 4  400  Thus, 37.5% of women employees lost their job. 16. C.P. of shirt = Rs 700 Loss% = 15% Loss = 15% of Rs 700 = Rs

15  700 = Rs 105 100

 S.P = C.P – Loss S.P. = Rs 700 – Rs 105 S.P. = Rs 595 Thus, the man sold the shirt at Rs 595. 17. Mohit, Rohit, Sumit, and Amit invested Rs 3900, Rs 3250, Rs 2925, and Rs 2925 respectively. Total investment = Rs (3900 + 3250 + 2925 + 2925) = Rs 13000 Since Sumit invested Rs 2925, percent of total

2925  100  22.5 13000 Thus, Sumit made 22.5% of the total investment. 18. It is given that Ravi bought a wrist watch for Rs 6850 and sold it to Javed at a loss of 20%. Here, C.P. = Rs 6850, loss percent = 20% S.P. = C.P. – 20% of C.P.. investment made by Sumit 

20    Rs 6850   6850  = Rs [6850 – 1370] 100   = Rs 5480 Therefore, Ravi sold the watch to Javed at Rs 5480. It is also given that Javed sold the watch that was bought at Rs 5480 to Gopal at a profit of 5%. Here, C.P. = Rs 5480, profit percent = 5% S.P. = C.P. + 5% of C.P..

5    Rs 5480   5480 = Rs[5840 + 274] 100   = Rs 5754 Thus, Gopal bought the wrist watch from Javed at Rs 5754. 19. Let the amount lent by Suman be P. Simple interest = SI = Rs 2,500 Amount paid back by Suman’s friend = A = Rs 13,750  P = A – SI = Rs 13,750 – Rs 2,500 = Rs 11,250 Time period = T = 2 years Let the rate of interest be R% p.a. SI 

PRT 100

 Rs 2, 500 

Rs 11, 250  R  2 100

Rs 2,500  100  R = Rs 11, 250  2

 R

100 9

 R  11

1 9

1 Thus, the applicable rate of interest is 11 % . 9 20. The amount paid back by Ambar includes the principal as well as the interest for one year. Let the principal borrowed by Ambar be represented as x. Thus, interest for 1 year = 12.5% of 12.5 1 x  x 100 8 Thus, amount paid back by Ambar = Rs 13,500 = Principal + Interest x

1  Rs 13, 500 = x  x 8

 1  Rs 13, 500 = 1   x  8

 Rs 13, 500 =

9 x 8

8 × Rs 13, 500 = Rs 12, 000 9 Thus, Ambar had borrowed Rs 12,000 from the bank. www.betoppers.com

 x

7th Class Mathematics

230 21. Selling price (SP) of the motorcycle = Rs 38,400 Profit percent = 20 Let the cost price of the motorcycle be denoted by CP.  Profit made in the transaction = 20% of

20 1  CP  CP 100 5 SP = CP + Profit CP 

1  Rs 38, 400 = CP  CP 5

25. Rate of interest for first 2 years = 4 × 2 = 8% Rate of interest for next 3 years = 6 × 3 = 18% Rate of interest for last 3 years = 8 × 3 = 24% Total rate of interest for 8 years = 50%

P  50 100  P = Rs. 2560 [then of 8 years is already adjusted in the total rate of interest calculation] 26. Let the usual speed and time be S and T respectively. then 1280 =

 1  Rs 38, 400 = 1   CP  5

then New speed =

6  Rs 38, 400 = CP 5

 New time =

5 × RS 38, 400 = Rs 32, 000 6 Thus, the dealer had bought the motorcycle for Rs 32,000. 22. Let C.P. = Rs. 100. Then, S.P. = Rs. 133. Let marked price be Rs. x. Then, 95% of x = 133

 CP 



95 x = 133 100

100    = 140. 95    Marked price = 40% above C.P.. 23. Let retail price = Rs. 100. Then, commission – Rs. 36.  S.P. = Rs. (100 – 36) = Rs. 64. But, profit = 8.8%.  x =  133 

1000  100  C.P. = Rs.  .  64  = Rs. 17  108.8  New commission = Rs. 12. New S.P. = Rs. (100 – 12) = Rs. 88. 

496 1000   Gain = Rs.  88   = Rs. 17 . 17  

 496 17  Gain % =    100  % = 49.6%.  17 1000  24. Let the sum be Rs. x and rate by r 

x  (r  3)  2 x  r  2  = 72 or, 2rx + 6x – 100 100 2rx = 7200  x = Rs. 1200 then

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5 S 6

6 1   T  Speed   5 Time   Given, (New time) – (Usual time) = 10 T 6 T – T  10  = 10  T = 50 min 5 5 27. Difference in the times taken at two speeds = (7 + 5) = 12 min 

=

1 hr now let the required distance be x km. 5

x x 1 –  5 6 5  6x – 5x = 6  x = 6 km 28. In both cases the student is late. Therefore, difference in timings = (30 – 5) = 25 

25 5 hr  hr. 60 12 Let the distance be x km. min =

x x 5   5 6 12



5  30 12  x = 12.5 km 29. Since the time taken in both cases is same, the ratio of the distances covered equals the ratio of the corresponding speeds. Rate at which John travels = 10 mph Rate at which Alex travels = 50 mph Thus, ratio of the distances traveled by John and Alex = ratio of their speeds 



6x  5x 

10 mph 1  50 mph 5

Arithmetic

231 4.

CONCEPTIVE WORKSHEET

75 3  8 L   8L  6 L 100 4 Quantity of water in the solution = 8L – 6L = 2L  Since 1 L of water evaporated from the solution, the solution now contains 6 L of milk and 2L – 1L = 1 L of water. It is known that Singh Sahib added 1 L of a solution containing 75% milk and 25% water. Thus, the quantity of the solution now becomes 8L. Quantity of milk in the new solution = 6 L + 75% of

KEY

8L 

1

2

3

4

5

6

7

8

9

10

B

A

B

C

C

C

C

C

C

D

11 12

13 14 15 16 17

18

19 20

B

D

C

D

A

21 22

23 24 25 26 27

28

29 30

C

D

C

B

B

B

A

B

B

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

75  1 L  6.75 L 100  Percentage composition of milk in the 1L  6L 

5 1  2  2.5 2 2 25% 

6.75 L  solution  8 L  100  84 8 %

25  0.25 100

5.

1 2  2.5 2

3.

70  x 200 It is given that the percentage score is 60%.

25 100 Multiply it with 100 to convert it into the percent equivalent



25 y 100

 x  y

y 4

 y

4x 5

70  x 3  200 5 350 + 5x = 600   5x = 250  x = 50 Thus, Appu has to score 50 marks in the third test. 1 100 33 % of 720  % of 3 3 100  720  240 3  100 240 – 200 = 40 720 

 yx

20 x 100  y = x – 20% of x Thus, y is 20% less than x.

 yx

70  x 60  200 100



6.

5y 4

70  x  60% 200



25 0.25 × 100 = ×100 = 25 100 It is given that x is 25% more than y.  x = y + 25% of y

 x  y

Sum of marks scored 40  30  x  Sum of max imum marks 80  50  70 

0.25 

 x

Let Appu’s score be x marks in the third test. Now, average percentage score



25  2.5 10 Thus, 25% is not equivalent to 2.5. 2.

Quantity of the solution = 8 L Quantity of milk in the solution = 75% of

x 5

1 Thus, 200 is 40 less than 33 % of 720. 3

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7th Class Mathematics

232 7.

Let x kg of bauxite ore contain 333 kg of aluminium.  45% of x = 333 kg



45  x  333 kg 100

333  100 kg  740 kg 45 Thus, 740 kg of bauxite ore contains 333 kg of aluminium. Let x be the number of English books in the library.  x

8.

2 3

 x  66 % of x = 600

 x

200 % of x = 600 3

200 x  600 300  5x = 600 × 3  x = 120 × 3 = 360 Thus, there are 360 English books in the library. Let the length = l cm and breadth = b cm The area = lb 17 l 10 Let the new breadth = z cm New length 170% of l 

 lb  

17 l× z 10

10 bz 17

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PRT 12000  6  3  Rs  Rs 2160 100 100 Amount = Principal + Simple interest = Rs (12000 + 2160) = Rs 14160 Thus, the amount that she has to pay at the end of the third year is Rs 14160. 13. The cost price of each sweater is Rs 1800. It is given that Aditi sold one sweater to Simran at a profit of 10%.  Pr ofit 

20  60  12 100

Pr ofit %  C.P. 100

10  1800  Rs 180 100 Therefore, the selling price of sweater is (Rs 1800 + Rs 180) i.e., Rs 1980. Aditi sold the other sweater to Sanya at a loss of 20%.

 Pr ofit 

 Loss 

 10  100  7  Decrease =   b  17 b   b  %  17  100    = 41.18% 10. Total number of students present in class VII = 60 Percentage of students who read comics = 50 Percentage of students who played indoor games = 30 It is given that the remaining students played outdoor games.  Percentage of students who played outdoor games = 100 – (50 + 30) = 20 Thus, number of students of class VII who played

outdoor games 

20  38000 100 = Rs 7600 SP = C.P – Loss = Rs 38000 – Rs 7600 = Rs 30400 Thus, Rajesh sold the scooter for Rs 30400. 12. Principal (P) = Rs 12000 Rate of interest (R) = 6% p.a. Time = 3 years Therefore, simple interest (I)  Loss = 20% of Rs 38000 = Rs 



 x

9.

11. CP of the scooter = Rs 38000 Loss percent = 20

Loss%  C.P. 100

20  1800  Rs 360 100 Therefore, selling price of the sweater = Rs 1800 – Rs 360 = Rs 1440 Thus, total money that Aditi had after selling the two sweaters = Rs 1980 + Rs 1440 = Rs 3420 14. Let the amount borrowed by Appu be P. Rate of interest = R = 8% p.a. Time = T = 4 years Let the interest paid by Appu at the end of four years be SI. Amount paid back by Appu = Rs 16,500 = P + SI

 Loss 

 Rs 16, 500 = P 

PRT 100

 Rs 16, 500 = P 

P 8 4 100

PRT    SI   100  

Arithmetic

233

 Rs 16, 500 = P 

32P 100

 Rs 16, 500 = P 

8P 25

8    Rs 16, 500 = P 1  25   

 Rs 16, 500 =

33 P 25

25    P  Rs 16,500   33    P = Rs 12, 500 Thus, Shashank had lent Rs 12,500 to Appu. 15. When Anil sold the washing machine to Dilruba: Cost price (CP) of the washing machine = Rs 8,000 Profit percent = 15  Profit = 15% of Rs 8,000 15 × Rs 8,000 = Rs 1,200 100  Selling price (SP) of the washing machine = CP + Profit = Rs 8,000 + Rs 1,200 = Rs 9,200 When Dilruba sold the washing machine to Harsh: Cost price (CP) of the washing machine = Rs 9,200 Loss percent = 15  Loss = 15% of Rs 9,200 

15 × Rs 9,200 = Rs 1,380 100  Selling price (SP) of the washing machine = CP – Loss = Rs 9,200 – Rs 1,380 = Rs 7,820 Thus, Dilruba sold the washing machine to Harsh for Rs 7,820. 16. CP = Rs. 10/kg Marked price= Rs. 12/kg (20% above CP) Since she uses a false weight of 800 g in place of 1 kg. Therefore selling price of 800 g of rice would be Rs. 12 CP of 800 g of rice is Rs. 8  Profit = (12 – 8) = Rs. 4 

Hence, profit % =

Gain 4 × 100 = × 100 = 50% CP 8

17. Let the CP of the orange be Rs. x Then, SP = 1.2x New SP = (1.2x + 1.2)

1.2x  1.2 [Since the gain is 40%] 1.4 Since the CP in both the cases is same CP =



1.2x  1.2 or, 1.4x = 1.2x + 1.2 1.4  0.2x = 1.2  x = Rs. 6 Therefore, original SP = Rs. 1.2 × 6 = Rs. 7.20 18. Let the C.P. of the horse be Rs. x. Then, C.P. of the carriage = Rs. (3000 – x).  20% of x – 10% of (3000 – x) = 2% of 3000 x=



x (3000 - x) = 60 5 10  2x – 3000 + x = 600  3x = 3600  x = 1200. Hence, C.P. of the horse = Rs. 1200. 19. Let C.P. = Rs. = 100. Then, marked price = Rs. 150. S.P. = 75% of Rs. 150 = Rs. 112.50.  Gain % = 12.50% 20. Let the marked price of each pen be Re. 1. Then, C.P. of 40 pens = Rs. 36. S.P. of 40 pens = 99% of Rs. 40 = Rs. 39.60. 

 3.60   100  % = 10%.  36  21. SI = (920 – 800) = Rs. 120  Profit % = 

120 =

800  R  3 100

 R = 5%

800  8  3 = Rs. 192 100  Amount = (800 + 192) = Rs. 992 22. Let each instalment be Rs. x, then first instalment paid after 1 year will earn a interest for 2 years at 12% and second instalment paid after 2 year will earn an interest for 1 year at the same rate as the debt has to be free off in 3 years. Interest at 8% interest =

 

 x 

x  12  1   x  12  2   x  +x =Rs.1092  100   100 

28x 31x  + x = 1092 or 28x + 31x + 25x 25 25 = 10921 × 25  x = Rs. 325 www.betoppers.com

7th Class Mathematics

234 23. Let both the trains meet x hours after 7 a.m. Then, (Distance moved by first train in x hours) + (Distance moved by other train in (x – 1) hours = 350 km 40x + 60 (x – 1) = 350  100x = 350 – 60  x= 2.9 hours Hence, both the trains will meet at 54 minutes past 9. 24. Due to stoppages the bus travels 9 km less. Hence time taken by the bus to cover 9 km is the time of the bus at stoppage.

4.



5.

6. KEY

4

5

6

7

8

C

B

D

D

D

C

C

B

9

10

11 12 13 14

15

A

D

C

A

D

A

A

7.

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. A percent can be written as a fraction by dividing it by 100

70 7  10 7   100 10  10 10 To find the percent equivalent, the fraction is multiplied with 100.  70% 

2.

i.e.

3.

8.

1 1     100  %  50% 2 2 

The percent equivalent of the fraction

3  100% = 60% 5 Thus, Meena saved 60% of her pocket money in that month. The given figure is divided into 8 equal parts, out of which, 5 parts are shaded. 5 125 1 %  62 % of this Therefore,  100%  8 2 2 figure is shaded. Let the CP of an article be Rs. x then SP = Rs. 0.9x (at a loss of 10%) New CP = Rs. 0.8x (at 20% less than original) New SP = Rs. 1.4 × 0.8x (at 40% profit) = 1.12 x Given the difference between new SP and old SP is Rs. 55.  1.12x – 0.9x = Rs. 55  0.22x = Rs. 55 or x = Rs. 250  Cost price of the article be Rs. 250. Let the retail price be Rs. 100. then SP = Rs. 80 (after a discount of 20%) Since profit = 60% SP 80 = = Rs. 50 1.6 1.6 New SP = Rs. 75 (after a commission of 25%)  Profit = Rs. (75 – 50) Rs. = 25, Hence profit %  CP =

1 is 50 2

It is given that x: y = 2: 3  x: y = 2 × 2: 3 × 2  x: y = 4: 6 ________ (1) It is also given that y: z = 2: 3  y: z = 2 × 3: 3 × 3  y: z = 6: 9 ________ (2) From (1) and (2), it can be observed that the terms of y in both the ratios are equal.  x: y: z = 4: 6: 9  x: z = 4: 9 Thus, the ratio between x and z is 4: 9.

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125  32  40 100 Thus, 125% of the required number is 40. Meena saved 3 parts out of 5 equal parts of her pocket money.  Required per cent =

SUMM ATIVE WORKSHEET 3

24  100  32 75

 125% of 32 

9  60 = 10 min. 54 Therefore, the bus stops 10 minutes per hour.

2

75x  24 100

 x

Time taken to cover 9 km =

1

Let x be the number.  75% of x = 24

25 × 100 = 50% 50 Let the original cost price be Rs. x SP at an initial profit of 15% = Rs. 1.15x New cost price = Rs. (x + 100) New SP at a profit of 10% = Rs. 1.1 (x + 100) Given that New SP = 1.1 × Original SP or 1.1 (x + 100) = 1.1 × 1.15x or 0.15x = 100  x = Rs. 666.66 =

9.

Arithmetic

235

10. Let the CP of the article be x then gain = (800 – x) and loss = (x – 275) Given (800 – x) = 20 (x – 275)  21x = 6300  x = 300  CP = Rs. 300 Now SP at a gain of 25% = 125% of 300 = Rs. 375 11. Amount (Principal + Interest) for 2 years = Rs. 1560 Amount (Principal + Interest) for 5 years = Rs. 2100 Hence, interest for 3 years = (2100 – 1560) = Rs. 540 Simple interest for 2 years = Rs. 360  Principal = (1560 – 360) = Rs. 1200

15. (a) Average speed =

Total Distance 20km +30 km 50 = = TotalTime 1hr +1hr 2 = 25 km/hr. (b) Average speed = = 24 km/hr.

HOTS WORKSHEET KEY

360  100 R= = 15% 1200  2

 1  1 12. SI for  3  2  i.e. 1 years = Rs. (1164 – 1008) 2  2  = Rs. 156 SI for 2 years = Rs. 208  Principal = Rs. (1008 – 208) = Rs. 800 Now, P = 800, T = 2 years and SI = Rs. 208 100  SI 100  208 = = 13% PT 800  2 13. Let the sum borrowed at 15% be Rs. x then sum borrowed at 18% will be Rs. (24000 – x). 

Rate =

x  15  1 (24000  x)  18  1  = 4050 100 100 or, 15x + 432000 – 18x = 405000 or x = 9000  Money borrowed at 15% = Rs. 9000 and money borrowed at 18% = Rs. (24000 – 9000) = Rs. 15000 14. Let the speed of C be x km/hr. Then speeds of B = 3x km/hr and speed of A = 6x km/hr. Therefore, the ratio of speeds of A, B and C = 6 : 3 :1 The ratio of time taken by A, B and C to cover the 1 1 : :1  1: 2 : 6 6 3

Therefore if C takes 54 minutes to cover a distance, then time taken by B to cover the same distance =

54  2 = 18 min. 6

1

2

3

4

5

6

7

8

9

10

B

B

A

D

C

B

C

B

B

C

11 12

13 14 15 16 17

18

A

A

A

D

C

B

C

D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. A percent can be written as a fraction by dividing it by 100

40 2  20 2   100 5  20 5 Let the maximum marks be x.  40% 

2.

Passing marks = 40% of x 



Therefore,

same distance =

2xy 2×20×30 1200 = = x+y 20 +30 50

40x 2x  100 5

2x  20  200 5

2x  180 5  2x = 5 × 180  x = 5 × 90 = 450 Thus, the maximum marks in the exam were 450. Let the maximum number of marks be x. Total passing marks = 82 + 83= 165  33% of x = 165



3.



33  x  165 100

165  100  500 33 Thus, the maximum number of marks in the examination is 500. x

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7th Class Mathematics

236 4.

5.

It is given that out of 400 employees who lost their job, 250 were men.  Number of women employees who lost their job = 400 – 250 = 150 Therefore, percentage of women employees who

150  150   100  %  %  37.5% lost their job   4  400  Thus, 37.5% of women employees lost their job. Let the number of students in the school be x. Percentage of students who like to play only football, only cricket, and only hockey = 45% + 35% + 15% = 95%  Percentage of students who like to play only badminton = (100 – 95) % = 5% It is given that 100 students like to play only badminton.  5% of x = 100



5  x  100 100

100  100   2000 5 Number of students who like to play only cricket =

8.

S.P. = Rs. 27.50, Profit = 10%

 100   27.50  = Rs. 25. So, C.P. = Rs.   110  When S.P. = Rs. 25.75, profit = Rs. (25.75 – 25) = Re. 0.75  0.75   100  % = 3%. 25   (S.P. of 33 m)–(C.P. of 33 m) = Gain = S.P. of 11m.  S.P. of 22 m = C.P. of 33 m. Let C.P. of each metre be Re. 1. Then, C.P. of 22 m = Rs. 22, S.P. of 22 m = Rs. 33.  Profit % = 

9.

 11   100  % = 50%. 22   10. Let the C.P. be Rs. x.  Gain % = 

1st S.P. = 125% of x =

125 5x x= ; 100 4

2nd C.P. = 80% of x =

80 4x x= . 100 5

 x

35  2000  700 100 Thus, 700 students like to play only cricket. Price of T.V. = Rs. 4000 Money spent on repairs = Rs. 500  Total C.P. = Rs. 4000 + Rs. 500 = Rs. 4500 S.P. of the T.V. = Rs. 5000 Since S.P. > C.P.  there is a gain and, Gain = S.P. – C.P. = Rs. 5000 – Rs. 4500 = Rs. 500 Gain = Rs. 500 35% of 2000 

6.

Gain % =

Gain  100 C.P.

500 1  100 = 111 % 4500 9 Let the new S.P. be Rs. x. then, (100 – loss%) : (1st S.P.) = (100 + gain %) : (2nd S.P.) =

7.

2nd S.P. = 130% of



5x 26x 21x  = 10.50  = 10.50  4 25 100

 10.50  100  x=   = 50. 21   Hence, C.P. = Rs. 50. 11. P = Rs. 25000 R = 20% per annum T = 1 year Now,

I=

PTR 100

25000  1  20 100 = Rs. 5000  His annual income is Rs. 5000 12. P = Rs. 2400 R = 6% per annum =

 100  5   100  5       1140   x 

T = 146 days =

 105  1140   x=   = 1260 95   New S.P. = Rs. 1260. 

Now,

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4x  130 4x  26x   = = . 5 25  100 5 

I =

146 years 365

PTR 2400  146  6 288 = = = 100 365  100 5

Rs. 57.60 Interest Rs. 57.60

Arithmetic

237 HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. The total volume = 15,705 L

P  R  T 7200  51  9  100 4  12  100

13. SI =

51 9    R  %;T  years  = Rs. 688.50. 4 12   14. Let P = x, then A = 2x SI = (2x – x) = x, T = 8 years R=

2.

SI  100 x  100  = 12.5% PT x 8



x 15. Let the principal be Rs. x then S.I. = Rs. 4 T = r, if R = rate per annum 

16. Total distance = (3 + 3 + 3 + 3) = 12 km. Total time = 3

3

3.

 6 + 3 + 2 +1  12×3 3 + + + =3  = hr = 10 20 30 60 60 60 5   Total Distance 12 = = 20km/hr.. 3 TotalTime 5 17. Total distance = (4 × 6 + 6 × 8) = 72 km Total time = (6 + 8) = 14 hours  Average speed =

= 18. Average speed =

24 =



y=

210  100 35  x = 600 Thus, the required number is 600. Cost of 5 L of petrol = Rs 255

4.

Total Distance 72 = TotalTime 14

Rs 42,000  Rs 3,500 12  House rent for 3 months = 3 × Rs 3,500 = Rs 10,500 Thus, the house rent for three months is Rs 10,500. 

36 1  5 km/hr 7 7

2xy xy

Rs 255  Rs 51 5  Cost of 24 L of petrol = 24 × Rs 51 = Rs 1224 Thus, it would cost Mani Rs 1224 to fill up his car’s petrol tank. It is given that the house rent for one year is Rs 42,000.  House rent for 12 months = Rs 42,000 House rent for 1 month 

 Cost of 1 L of petrol 

3

Hence, average speed =

35  x  210 100

 x

xRR x = , R2 = 25 or R = 5% 4 100

3

 225  8  Therefore, the required percentage =  15,705  ×   100% = 11.46 percent Let the required number be x. Then, 35% of x = 210

5.

2  30 y  720 + 24y = 60y 30  y

7 43 4 3 3 3 1    1 1  3 4 4 4 4 4 4 4 18 18  2 36    3.6  3.5 5 5  2 10

720 = 20 km/hr.. 36

15.2 

IIT JEE WORKSHEET

12.5% 

KEY 1

2

3

4

5

6

7

8

9 10

B

D

D

C

D

D

C

A

B

11 12 13 14 AB AC AC AC C D D D

15 AB C

152 2  76 76 78    10 25 5 5

B

6.

3.4 

12.5 125   0.125 100 1, 000

34 10

304 100 Thus, 3.4 is not equivalent to 304%. 304% 

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7th Class Mathematics

238 7.

C.P. of the cycle = Rs. 660 Profit = 25%

25 × 600 = Rs. 150 100 Now, S.P. of the cycle = C.P. + Gain = Rs. 600 + Rs. 150 = Rs. 750 Desired profit = 20% It means, if C.P. = Rs. 100, then Profit (P) = Rs. 20 =  S.P.. = C.P. + Profit = Rs. 100 + Rs. 20 Rs. 120 Now, C.P. Profit S.P. = (C.P. + Gain) (Rs) (Rs) (Rs) 100 20 120  Profit =

8.

?

3000  100 120 P = Rs. 8000 C.P. =

9.



100  100  4% 2500

Thus, the mixture contains 4% of pure juice. 21. P = Rs 80000 R = 7% per annum A = Rs 91200  P + I = Rs 91200  Rs 80000 + I = Rs 91200  I = Rs 91200 – Rs 80000 = Rs 11200 Let the time period be T. We know that,

3000 = Rs. 2500

I

1 9 years = years 2 2 R = 8% per annum T=4

 He has to pay Rs. 10880 after 4

PT R 100

 11200 

PTR 8000  9  8 Now, I = = 100 2  100 I = Rs. 2880 Now, amount = Principal + S.I. = Rs. 8000 + Rs. 2880 = Rs. 10880 1 years 2

10. P = Rs. 900

6 1  year 12 2 R = 10% per annum T = 6 months =

PTR 900  1  10 Now, I = = = Rs. 45 100 2  100 Also, A = P + S.I. = Rs. 900 + Rs. 45 = Rs. 945 Total amount paid by Neena is Rs. 945. 16. 53.32 17. 13.67% 18. 214.16% 19. 5.66% 20. Original volume of juice concentrate = 2 L = 2000 mL Amount of pure juice in the concentrate 

5 × 100

2000 mL = 100 mL 500 mL water is added to the concentrate. www.betoppers.com

Therefore, Total amount of mixture = 2000 mL + 500 mL = 2500 mL  Percentage of pure juice in the mixture

80000  T  7 100

 11200 = 5600 × T  T

11200  2 years 5600

Thus, Hardeep deposited the money for 2 years. 22. Principal deposited by Mayank = P = Rs 14,000 Amount = A = Rs 21,560  Interest earned on the principal = SI = A – P = Rs 21,560 – Rs 14,000 = Rs 7,560 Rate of interest = R = 9% p.a. Let the time period be T.

SI 

P RT 100

 Rs 7, 560 

Rs 14, 000  9  T 100

 7, 560 = 140 × 9 × T  T

7,560 6 140  9

Thus, the required time period is 6 years. 23. 5% 24. 5 Rs.

Arithmetic

239

43.5 : 24

25. (i)

(ii)

47/2 : 24

=

0.0064

2 

=

64 10000

27 : 24

=

2 72

8 100 = 0.08

23 : 1 =8:1 (iii) Speed of



0.32  0.02

the man

 18    3   km\hr 5 

54 km\hr 5

Distance covered in 26. (i)

5  54 5  hr     km  18km 3  5 3

31, 37, 41, 47

Average 

31  37  41  43  47 199  = 39 .8 5 5

(ii) (A + B)’s 1day’s work =

1 ; (B + C)’s 1day’ss 18

1 1 and (A + C)’s 1day’s work = 24 36 Therefore 2 (A + B + C)’s 1 day’s work =

work =

1 1 1   18 24 36 Therefore 2 (A + B + C)’s 1 day’s work =

1 8

1 16 Time taken by (A + B + C)’s to complete the work = 16 day’s Therefore (A + B + C)’s day’s work =

(iii)

30 60 x y 100 100 x 2  y 1



x 2 y



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240

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7th Class Mathematics

7. LINES AND ANGLES SOLUTIONS ABC  BCG (Alternate interior angles)

FORMATIVE WORKSHEET

 BCG  34

KEY

1

2

3

4

5

6

7

8

9

10

Similarly, DEC  GCE (Alternate interior angles)

C

B

C

B

C

C

C

A

C

D

 GCE  29

11 12

13 14 15 16 17

18

19 20

 BCE  BCG  GCE  34  29  63

B

A

A

D

Thus, the measure of BCE is 63°. It can be observed in the given figure that:

A

C

C

B

B

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

3.

AOD  BOC  3x – 19° = 2x + 11°

Firstly, a line FC parallel to AB and DE is

 3x – 2x = 11° + 19°  x = 30°

constructed.

B

D

F

 BOC = 2x + 11° =2×30°+11°= 60°+11° = 71°

AOC  BOC  180 (Linear pair)

 AOC = 180° – 71° = 109° A

2.

E

C

ABC  BCF (Alternate interior angles)  BCF  a  20 Similarly, CDE  DCF (Alternate interior angles)  FCD  b  50 It is given that BCD is of measure 90°.  BCF  FCD  90  a + 20° + b + 15° = 90°  a + b = 90° – 20° – 15° = 55°  a = 55 ° – b Thus, the expression given in alternative C is same as a. Firstly, a line FCG parallel to AB and DE is constructed.

4.

5.

Thus, the measure of AOC is 109°. The measure of the angle is 118°. It is known that the sum of two supplementary angles is 180°.  Supplement of 118° = 180° –118° = 62° It is also known that the sum of two complementary angles is 90°.  Complement of 62° = 28° Thus, the complement of the supplement of 118° is 28°. It is given that OE is the bisector of BOC .  BOE  COE _______ (1) It is known that sum of all the angles on a straight line is 180°.  AOC  COE  BOE  180

A

D

 92  BOE  BOE  180 [Using (1)]  2BOE  180  92  88  BOE  44

C 34°

BOD  AOC (Vertically opposite angles)

29°

 BOD  92

G B

 DOE  BOE  BOD  44  92  136 E

Thus, the measure of DOE is 136°.

7th Class Mathematics

242 6.

It is given that AB||CD and BC is a transversal.  ABC  BCD (Alternate interior angles)

 ABC  35 Similarly, AD is a transversal on the parallel lines AB and CD. It is known that the interior angles on the same side of a transversal are supplementary.  BAD  ADE  180

7.

 ADE  180  44  136 Thus, the measures of ABC and ADE are 35° and 136° respectively. It is known that the sum of all angles on a straight line is 180°.  AOC  COD  DOB  180  x + 4x + 2x + 5° = 180°  7x = 180° – 5°  7x = 175° 175  25 7  AOD  AOC  COD = x + 4x = 5x = 5 × 25° = 125° Thus, the measure of AOD is 125°. Let the measure of the angle be 7x and its complement be 11x. It is known that the sum of an angle and its complement is 90°.  7x + 11x = 90°  18x = 90°  x

8.

90  5 18  Measure of the angle = 7x = 7 × 5° = 35° Thus, the measure of the angle is 35°. It is known that the interior angles on the same side of a transversal are supplementary.  ABD  BAC  180  6y + 12° + 5y – 8° = 180°  111y = 180° – 4° = 176°

x

9.

176  16 11  ABD = 6 × 16° + 12° = 96° + 12° = 108° AB||CD and BF is a transversal.  ABD  CDF (Corresponding angles)  CDF  108 Thus, the value of x is 108°.

 y

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10. Let x and y be 17a and 19a respectively. It can be observed in the given figure that EFB and BFG form a linear pair..  EFB  BFG  180  x + y = 180°  17a + 19a = 180°  36a = 180°  a = 5°  y = 19 × 5° = 95° AB||CD and EG is a transversal.  BFG  FGC (Alternate interior angles)  y=z  z = 95° Thus, the value of z is 95°. 11. It is known that the sum of two complementary angles is 90°.  P + Q = 90°  (Q – 20°) + Q = 90°  2Q = 110°  Q = 55°  P = Q – 20° = 55° – 20° = 35° Thus, the values of P and Q are 35° and 55° respectively. 12. Let the two angles be 2x and 3x. It is known that the sum of two complementary angles is 90°.  2x + 3x = 90°  5x = 90°

90  18 5 Therefore, the two angles are 2 × 18° = 36° and 3 × 18° = 54° Thus, difference between the angles = 54°–36° = 18° 13. It can be observed in the given figure that: XDY   ADC (Vertically opposite angles)  ADC  134 It is known that interior angles on the same side of a transversal are supplementary.  ADC  DAB  180  134° + m = 180°  m = 180° – 134° = 46° Similarly, ABC  BCD  180  85° + n = 180°  n = 180° – 85° = 95° Thus, the respective values of m and n are 46° and 95°.

 x

Lines and Angles Solutions 14. It is known that two angles are supplementary if their sum is 180°. Consider the pair of angles given in alternative C: 126° and 54° It can be observed that: 126° + 54° = 180° Thus, the pair of angles 126° and 54° is supplementary. 15. Let XOZ be 2x and ZOY be x.  XOZ  ZOY  180 (Linear pair)  2x + x = 180°  3x = 180°

180  60 3  XOZ = 2x = 2 × 60° = 120° ZOY  x  60 Thus, the measures of XOZ and ZOY are 120° and 60° respectively. 16. In CDE , CD = CE [Given]  CDE is an isosceles triangle. It is known that in an isosceles triangle, base angles opposite to the equal sides are equal.  CDE  CED  30 It is also known that the measure of an exterior angle of a triangle is equal to the sum of the measures of the interior opposite angles.  BCE  CDE  CED  30  30  60 It is given that AB || CE Therefore, the pair of alternate interior angles is equal.  ABC  BCE  60 Thus, the measure of ABC is 60°. 17. It is given that line segment AY || line segment BX. We know that when two parallel lines are cut by a transversal, each pair of interior angles on the same side of the transversal is supplementary.  CAY  ACX  180 [Parallel lines AY and BX are cut by transversal AC]  40° + x = 180°  x = 180°– 40° = 140° Similarly, BAY  ABC  180 [Parallel lines AY and BX are cut by transversal AB]  (40°+ y) + 70° = 180°  110°+ y= 180°  y = 180°– 110° = 70° Thus, the values of x and y are 140° and 70° respectively.  x

243 18. It can be observed from the given figure that ONH and ONM lie on a straight line.  ONH  ONM  180  ONH = 180° – 120° = 60° ONM  LMK (Corresponding angles)  LMK  60 Similarly, EPA  EOC  EOC  65 EOC  COJ  JOF  180  65  COJ  60  180  COJ  125  180  COJ  180  125  COJ  55 LON   COJ (Vertically opposite angles)  LON  55  KLM  LON (Corresponding angles)   KLM  55 In KLM ,  KLM   LMK   MKL  180

 55  60   MKL  180  115   MKL  180  MKL  180  115  65 x   MKL (Vertically opposite angles)  x = 65° Thus, the value of x is 65°. 19. It is given that the complement of  PQR is 30°.   PQR  90  30  60  x = 60° The parallel lines AB and CD are cut by a transversal YZ.  DRZ  BPZ [Pair of corresponding angles] –––––– (1) Also, the parallel lines AB and CD are cut by a transversal WX.   BPX  x [Pair of alternate interior angles]   BPX  60   BPZ   BPX   XPZ  60  55  115 –––––– (2) Therefore, from (1) and (2),  DRZ  115  APZ and BPZ form a linear pair..   APZ   BPZ  180   APZ  115  180   APZ  180  115  65 Thus, the measure of DRZ and APZ are 115° and 65° respectively.

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7th Class Mathematics

244 20. In the given figure, PQ||RS   LXY   MYD (Corresponding angles)   LXY  85 It is given that AB||CD   ALX  LXY (Alternate interior angles)   ALX  85 Thus, the measure of  ALX is 85°. R P L

A

C

B

M

3.

Y 85° D

X

S

Q

CONCEPTIVE WORKSHEET KEY

1

2

3

4

5

6

7

8

9

10

B

D

D

B

D

B

B

A

C

B

11 12

13 14 15 16 17

18

19 20

A

B

A

B

D

A

C

B

D

2.

 CDE  88 CDE  DEF  180  88° + b = 180°  b = 180° – 88°  b = 92° Thus, the respective values of a and b are 88° and 92° respectively. Let a be 5x and b be 3x. COD is a straight line. It is known that the sum of all the angles on a straight line is 180°.  COE  EOB  BOD  180  a + 100° + b = 180°  5x + 3x = 180° – 100°  8x = 80°

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4.

C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It is known that interior angles on the same side of a transversal are supplementary.  ABC  BCD  180  92° + a = 180°  a = 180° – 92°  a = 88° BCD  CDE (Alternate interior angles)

80  10 8  a = 50° b = 30° AOC  BOD (Vertically opposite angles)  AOC  b  c = 30° Thus, the measure of angles a and c are 50° and 30° respectively. It can be observed in the given figure that AOE and EOB form a linear pair..  AOE  EOB  180  AOE = 180º – 84º = 96º AOB is a straight line.  AOE  180 It is known that the measure of a complete angle is 360º.  AOB  BOD  AOD  360  180º + 54º + AOD = 360º  AOD = 360º – 180º – 54º = 126º Thus, the measures of AOE and AOD are 96° and 126° respectively. It can be observed that AOC and BOC form a linear pair.  AOC  BOC  180  2p + 16° + 7p – 16° = 180°  9p = 180°

 x

180  20 9 Thus, the value of p is 20°. The measure of the given angle is x.  Complement of x = 90° – x According to the given condition:

 9P 

5.

x

2  90  x  3



2 2x x   90  3 3



x

2x  60 3

5x 60 3  60  x   36 3 5  Supplement of 36° = 180° – 36° = 144° Thus, the measure of supplement of angle x is 144°.



Lines and Angles Solutions 6.

7.

Angles x and y are in the ratio 8: 9, let angles x and y be 8a and 9a respectively. POQ is a straight line and it is known that the measure of a straight angle is 180°.  POR  ROQ  180  POR + x + y = 180°  95° + 8a + 9a = 180°  17a = 180° – 95° = 85°  a = 5°  x = 8 × 5° = 40° and y = 9 × 5° = 45° Thus, the values of x and y are 40° and 45° respectively. It is given that AB||CD. APE  PQC (Corresponding angles)  PQC  2x  25

PQC and PQD form a linear pair..  2x + 25° + x + 20° = 180°  3x + 45° = 180°  3x = 180° – 45°  3x = 135°

135  45 3 Thus, the value of x is 45°. It is given that PQRS is a parallelogram. We know that the opposite sides of a parallelogram are parallel.  PQ || RS and PS || QR  QSR  PQS (Alternate interior angles)  x = 25° Similarly, SQR  PSQ (Alternate interior angles)  SQR  45 It is given that TQRU is a trapezium.  QR || TU (Corresponding angles)  SQR  QTU  y = 45° Thus, the respective values of x and y are 25° and 45°. It is given that M and N are supplementary angles.  M  N  180 ________ (1) It is also given that twice the measure of M is 15° more than the measure of N .  x

8.

9.

245 Adding equations (1) and (2): 3M  180  15

 3M  195 195  65 3 Therefore, from equation (1): N  180  M  180  65  115 Thus, the measures of M and N are 65° and 115° respectively. 10. It is given that OE is the bisector of BOD .  DOE  EOB ––––––––– (1) It is known that the sum of all the angles on a straight line is 180°.  AOC  COD  DOE  EOB  180 [From (1)]  20° + 7x – 5° + 2x + 2x = 180°  111x = 180° – 15°  111x = 165°

 M 

165  15 11  EOB = 2x = 2 × 15° = 30° Thus, the measure of EOB is 30°. The correct answer is B. 11. In the given figure, AB||CD and AE is the transversal.  BAC  DCE (Pair of corresponding angles)  DCE  95 Also, CE||DF and CD is the transversal. It is known that interior angles on the same side of the transversal are supplementary.  DCE  CDF  180  CDF  180  95  CDF  8 5 Thus, the measure of CDF is 85°.

 x

12. It is given that SOQ is the complement of the angle of measure 75°.  SOQ = 90° – 75° = 15° It is known that the sum of the angles on a straight line is 180°.  POR  ROS  SOQ  180  71° + ROS + 15° = 180°  ROS = 180° – 86° = 94° Thus, the measure of ROS is 94°.

 2M  15  N

 2M  N  15 ________ (2) www.betoppers.com

7th Class Mathematics

246 13. It is given that rays OC and OD are images of each other with respect to the dotted line OX. Therefore, line OX acts as bisector of COD .  COX  DOX COD  140  COX  DOX  140

140  70 2 AOB is a straight line and OX  AB .  BOX  90  BOD  BOX  DOX  90  70  20 Thus, the measure of BOD is 20°.

 COX  DOX 

14. The angles POQ and QOR have a common  vertex O and a common arm OQ . These angles are called a pair of adjacent angles. 15. It is given that SOQ is a straight line.  SOQ  180

POS  POR  QOR  180

  POS  POR   QOR  180  ROS  QOR  180  130  QOR  180  QOR  180  130  50  POQ  POR  50  144  POR  50  POR  144  50  94 Thus, measure of POR is 94°. 16.

P A

R

B

2y + 12° C 7y – 3° S

D

Q We know that the interior angles on the same side of transversal are supplementary.  ARS  CSR  180  2y + 12° + 7y – 3° = 180°  9y + 9° = 180°  9y = 171°

171 9  y = 19°

 y

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 ARS = 2y + 12° = 2 × 19° + 12° = 50° Also, PRB  ARS (Vertically opposite angles)

 PRB  50 Thus, the measure of PRB is 50°. 17. In the given figure, it is seen that AOD and BOC are vertically opposite angles.  AOD  BOC

 BOC  84 AOD  84 



BOE  EOC  84

BOC  BOE  EOC 37  EOC  84  BOE  37

 EOC = 84° – 37° = 47° Now, COD is a straight line. Therefore, we have: COF  FOA  AOD  180 [The measure of a straight angle is 180°]  COF  79  84  180  COF  163  180  COF  180  163  17  EOC  COF  47  17  64 Thus, the complement of EOF is 90° – 64° = 26°. 18. In the given figure, DF and BG are parallel and are cut by transversal AC.  AEF  ECG [Pair of corresponding angles] [ ECG  135 ]  AEF  135 It is also seen that the angles AED and AEF form a linear pair.  AED  AEF  180  AED  135  180  AED  180  135  45 Also, BE can be regarded as the transversal for the parallel lines DF and BG.  EBC  DEB       [Pair of alternate interior angles] [ EBC  64 ]  DEB  64 N o w , AEB  AED  DEB  45  64  109 Thus, the measure of AEB is 109°. 19. In the given figure, it is seen that CEF and FED form a linear pair..  CEF  FED  180  108  FED  180  CEF  108  FED  180  108  72

Lines and Angles Solutions We know that when two parallel lines are cut by a transversal, each pair of the corresponding angles are equal. Now, AB||CD and are cut by transversal AF.  FED  FAB [Pair of corresponding angles]  FAB  72 Thus, the measure of FAB is 72°. 20. It is known that if two parallel lines are cut by a transversal, then each pair of interior angles on the same side of the transversal are supplementary. Here, AB||CD with AC as the transversal, and BAC and ACD are interior angles on the same side of the transversal.  BAC  ACD  180

3.

4.

5.

6.

 BAE  112  AEF  112 Now, BAE  BAC   CAE  112  52  CAE  CAE  112  52  60 Thus, the measure of CAE is 60°.

 BOY  YOP  YOP  44  BOY  44  44  88

SUMM ATIVE WORKSHEET KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

C

A

D

B

A

B

B C

D

B

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It can be observed that AOB is a straight line.  AOB  180  AOC  COD  BOD  180  25° + 90° + x° = 180°  x = 180º – (25º + 90º) = 180º – 115º = 65º Thus, the value of x is 65º. 2. It is known that the measure of a straight angle is 180º. 3x + 2x + x + 6 = 180°  6x = 180º – 6º  6x = 174°

174  29 6 Thus, the value of x is 29º.

 x

180  90 2 Thus, the measure of each of the angles is 90°. Let x be the required angle.  Complement = 90º – x According to the given information: x = 2 (90° – x) – 12º  x = 180° – 2x – 12º  3x = 168°  x = 56° Thus, the required angle is 34°. It can be observed that BOY and AOX are vertically opposite angles.  BOY  AOX  44 It is given that OY bisects POB . x 

 BAC  128  180  ACD  128  BAC  180  128  52 It is known that if two parallel lines are cut by a transversal, then each pair of alternate interior angles are equal. AB||EF with AE as the transversal, and AEF and BAE are a pair of alternate interior angles.  AEF  BAE

247 Let x be the angle. Since x is less than 70°, its supplement will be greater than (180º – 70º) = 110º Let one of the angles be x. It is given that both the angles are congruent. Therefore, the other angle is also x.  x + x = 180°  2x = 180°

7.

BOP and AOP forms a linear pair..  BOP  AOP  180  88º + x = 180º  x = 92° Thus, the value of x is 92º. Let the angles be 7x and 8x. It is known that the sum of the measures of complementary angles is 90°.  7x + 8x = 90°  15x = 90°

90  6 15 Thus, measures of angles are: 7 × 6º = 42º and 8 × 6º = 48º It can be observed that BOX and BOY form a linear pair.  BOX  BOY  180  BOX  50  180  BOX  130 OA is the bisector of BOX .

 x

8.

 AOX  BOA  130

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7th Class Mathematics

248

 2BOA  130  BOA  65 AOY  AOB  BOY  65  50  115  2x + 5º = 115°  2x = 110°  x = 55° Thus, the value of x is 55º. 9. In the given figure, a and b are alternate exterior angles. 10. It is known that the sum of the angles forming a linear pair is 180°.  3x – 2° + 6x + 2° = 180°  9x = 180º 180  20 9 Thus, the value of x in the given figure is 20º. 11. (i) 4624

 x

Complement of 4624 = 90° – 45°. 24  44 36 (ii)

352452 Complement

of

352452

 90  352452  54368 12. Let the required angles be (5x)° and (4x)°. Then, 5x + 4x = 90° 9x = 90° x = 10° hence the required angles measure (5 × 10)° and (4 × 10)° i.e 50° and 40° 13. (i)

1 right angle 5 1 right angle = 90° 1 1 right angle = × 90° = 18° 5 5

(ii)

2 2 striaght angle = × 180° 5 5

360 5 = 720 14. (i) Supplement of 65° = 180° – 65° = 115° (ii) Supplement of 534525 = 180 0 – 

534525 = 126 15 35

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15. Supplement of 140° = 180° – 140° = 40° 80% for angle = supplement of 140° 80% of x = 40° 80  x  40 100

2 x 1 100 100 2 x = 50° 16. we have AGH  EGB  x  (vertically opposite angles now AB||CD and EF is a transversal Therefore sum of the co-ordianate angles is 1800 Therefore AGH  GHC  180 5x = 180° x

180 5 x = 36° 17. Given that AB||CD from figure AGE  BGE  180 130° + z = 180° z = 50°  y = 50° and x = 130° 18 Given that AB||CD from figure AGE EHC are corresponding angles. AGE EHC  5x we know that CHD is a straight angle x

EHD EHC  180 5x° + 3x° = 180° 8x° = 180°

x

180 8

x

45 2

1 2 19. QOR is a straight line Therefore QOR  180 x  22

POQ  POR  180 (x + 25°) + (3x+55°) = 180° 4x° + 80° = 180°

Lines and Angles Solutions

249 3.

4x° = 180° – 80° 4x° = 100°

100 4 x° = 25°  POQ  50 and POR  130 20. Given that AOC  70 We know that AOC and BOD are vertically opposite angles Therefore AOC  BOD  70 CD is a straight line Therefore COD  180 AOC  AOD  180 x 

4.

5.

70  AOD  180 AOD  110 We know that AOD, BOC are vertiacally

opposite angles. BOC  110 6.

HOTS WORKSHEET

 AOB  180  AOP  POQ  BOQ  180  50° + 2x – 5° + x = 180°  3x + 45º = 180º  3x = 180º – 45º = 135º

KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

C

A

C

B

B B

C

D

A

B

Q.no

11 12

13

Key

C

A

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. The sum of the interior angles on the same side of transversal is 180°.  2x + 10° + 60° = 180°  2x = 180° – 70°

135  45 3 POQ = 2x – 5° = 2 × 45º – 5° = 85°

 x

7.

110  55 2 Similarly, 115° + 3y – 10° = 180°  105° + 3y = 180°  3y = 180° – 105°

 x

75  25 3  x + y = 55° + 25° = 80° Thus, the value of (x + y) in the given figure is 80º. The supplement of an acute angle is an obtuse angle. However, the complement of an obtuse angle does not exist. Therefore, the complement of supplement of an acute angle does not exist. y

2.

The measure of an obtuse angle varies from 90° to 180°. Thus, the measure of the supplement of the angle varies from (180° – 90°) to (180° – 180°) i.e., from 90° to 0°. The measure of the complement of this angle varies from (90° – 90°) to (90° – 0°) i.e., from 0° to 90°, which makes it an acute angle. It is given that PQ||RS PQR and SRQ form a pair of alternate interior angles.  SRQ  PQR  48 Let the angle be x.  Complement of the angle = 90º – x Supplement of the angle, 90º – x = 180º – (90º – x) = 180º – 90º + x = 90° + x Thus, the supplement of complement of an angle is 90° more than the angle. It can be observed that AOB is a straight angle.

8.

Thus, the measure of POQ is 85°. Let x be the required angle.  Supplement of x = 180° – x Complement of x = 90° – x According to the given information: (90° – x) + (180° – x) = 142°  270° – 2x = 142°  2x = 270° – 142°  2x = 128°  x = 64° Thus, the required angle is 64º. It can be seen from the given figure that the parallel lines x and y are cut by transversal z. A

b B

D F z

2a + 5°

3a + 45°

C

E

x

y

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7th Class Mathematics

250

9.

Now, DBC and FDE form a pair of corresponding angles.  DBC  FDE  2a + 5° = 3a + 45°  3a – 2a = 45° – 5º  a = 40°  DBC = 2a + 5° = 2 × 40° + 5° = 85° Now, ABC  DBC  180 (Linear pair)  ABC = 180° – 85° = 95° Thus, the value of b is 95°. We know that when the sum of the measures of two angles is 90°, the angles are called complementary angles.  Complement of 43° = 90° – 43° = 47° Thus, the complement of the angle of measure 43° can be represented as:

47° 10. Applying angle sum property of triangle in ABC , we obtain ABC  BCA  CAB  180  50° + BCA + 70° = 180°  BCA = 180° – (50° + 70°)  BCA = 60° It is given that BC||DE Therefore, pair of alternate interior angles is equal.  CED  BCA  60 Again by applying angle sum property of triangles in CED , we obtain

CQF and DQF form a linear pair..  CQF  DQF  180

100° + DQF = 180°

DQF = 180º – 100º = 80º Now, DQX  DQF  FQX  80  50  130 Thus, the measure of DQX is 130°. 12. In the given figure, it is seen that ABC  90 . It is given that BD is the angle bisector of ABC .

1 1  ABD  DBC  ABC   90  45 2 2 It is also seen that ABD and ABE form a linear pair.  ABE  ABD  180 [Since the angles in a linear pair are supplementary]  ABE  45  180  ABE  135 13. Two angles are said to be supplementary if the sum of their measures is 180°. In this case, each angle is said to be the supplement of the other angle. It is given that the measure of the supplement of EOC is 105°.  EOC = 180° – 105° = 75° BOC  EOC  AOE  180 [The measure of a straight angle is 180°]  40  75   AOE  180  115  AOE  180  AOE  180  115  65 Also, BOD  AOC [Vertically opposite angles]  BOD  AOE  EOC  65  75  140 Thus, the measures of AOE and BOD are 65° and 140° respectively.

CED  EDC  DCE  180  60° + CDE + 90° = 180°  CDE = 180° – (90° + 60°)  CDE = 30° 11. It is given that AB is parallel to CD. APQ and

IIT JEE WORKSHEET KEY

CQF are corresponding angles.

1

2

3

4

5

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7

8

9

10

 CQF  APQ  100

B

A

B

C

C

B

B

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A

B

13

14

BC

AD

QX is the angle bisector of CQF .  CQX  FQX 

 FQX 

CQF 2

100  50 2

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11 12 AB BC C D

Lines and Angles Solutions

251

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It is known that the measure of a complete angle is 360°.  130° + 70° + x + x + 30° = 360°  230° + 2x = 360°  2x = 130°

130  65 2 Thus, the value of x is 65º. It can be observed that AOX and XOB form a linear pair.  AOX  XOB  180  110° + XOB = 180°  XOB = 180° – 110°

1  QRS  QRS  180 3



 XOB  70 OC is the bisector of AOX .  COX  XOB  70

3.

4.

 AOC  COX  110  AOC  110  70  40 AOC and BOD are vertically opposite angles.  x = 40º Thus, the value of x is 40°. The supplement of an angle x is 180° – x. Thus, the supplement of an obtuse angle is an acute angle. It is given that OP is the bisector of AOX .

1  AOP  AOX 2 1  AOP   80  40 2 and AOX AOY form a linear pair..  AOX  AOY  180  AOY = 180° – 80° = 100° OQ is the bisector of AOY . 1  AOQ  QOY   100  50 2  POQ  AOP  AOQ  40  50  90 Thus, the measure of POQ is 90º. 5.

1 It is given that:  P RQ  QRS 3 It can be observed that  P RQ and QRS form a linear pair.   P RQ  QRS  180

3  180  135 4 Thus, the measure of QRS is 135º. Adjacent angles are a pair of angles that have a common vertex and a common arm. Here, the noncommon arms lie on either side of the common arm. In the given figure, it is seen that 4 and 6 are adjacent angles because they have a common vertex, B and a common arm, BF. Also, noncommon arms BG and BA lie on either side of BF.

 QRS 

 x

2.

4 QRS  180 3

6.

15. 16. 17. 18. 19.

20. 21. 22. 23. 24.

RAN AF

AN 

120° + 10a = 180° 10a = 180° – 120° 10a = 60° a = 60 153° + y + 20° = 180° y = 180° – 173° = 7° 1 4 8 (A)

A

D 120° x

O y

B C AB and CD are vertically opposite angles therefore AOD  COB  120 therefore y = 1200 we know that AOD  COB  120 DOB  180  COB = 1800 – 1200 = 600 therefore x = 600 therfore x + 2y = 600 + 2400 = 3000

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7th Class Mathematics

252 (B) we know that AOB is a straight angle therefore AOB  180 3x + x + 2x = 1800 6x = 1800 x = 300 (C) From figure BGH  DHC  180 BGH  30  180 BGH  150 x = 150° ( AGE  BGH ) 25. (A)

is a right angle (B)

angle is less than 900 is called an acute angle. (C)

Angle above 900 is called an obtuse angle



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8. TRIANGLES SOLUTIONS

3.

FORMATIVE WORKSHEET

C

KEY 1

2

3

4

5

6

7

8

9

10

C

D

C

C

C

B

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13 14 15 16 17

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3 km

B

C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. In  ABC, AB = 8 ft and AC = 10 ft Applying Pythagorean Theorem in right-angled  ABC: AC2 = AB2 + BC 2 (10 ft)2 = (8 ft)2 + BC2 100 ft2 = 64 ft2 BC2 (BC)2 = 100ft2 – 64 ft2 BC2 = 36 ft2  BC = 6 ft Thus, the distance between the foot of the ladder and the base of the wall is 6 ft. 2. According to the given figure, triangles ABC and DCB are right triangles right angled at B and C respectively.  By Pythagorean theorem,

4.

In  DCB, BC2 + CD2 = BD2 BC2 + (9m)2 = (15m)2 BC2 + 81 = 225 BC2 = 225 – 81 BC2 = 144 BC = 12 m Also, in  ABC: AB2 + BC 2 = AC2 (5 m)2 + (12 m)2 = AC2 25 + 144 = AC2 169 = AC2  AC = 13 m Thus, the length of the ladder AC is 13 m.

5.

A B 4 km The triangle formed by the 3 houses is a right triangle, right angled at B. Therefore, by Pythagorean Theorem, Square of hypotenuse = Sum of the squares of the remaining sides  AC2 = AB2 + BC 2 AC2 = (4 km)2 + (3 km)2 AC2 = 16 + 9 AC2 = 25  AC = 5 km Hence, the distance between the houses A and C is 5 km. In  ABD, AB = 13 m and AD = 12 m Applying Pythagorean Theorem in right-angled  ABD: AB2 = AD2 + BD2 BD2 = AB2 – AD2 BD2 = (13 m)2 – (12 m)2 BD2 = (169 – 144) m2 BD2 = 25 m2  BD = 5m CD = BC – BD = 21 m – 5m = 16 m In  ADC, AD = 12 m and CD = 16m Applying Pythagorean theorem in right angled  ADC: AC2 = AD2 + CD2 AC2 = (12 m)2 + (16 m)2 AC2 = (144 + 256) m2 AC2 = 400 m2  AC = 20m Thus, the length of side AC is 20 m. Let the angles P, Q, and R after subtraction be 4x, 5x, and 6x respectively. Therefore, the measures of angles P, Q, and R of  PQR are 4x + 15°, 5x + 10°, and 6x + 5° respectively. It is known that sum of all angles of a triangle is 180°.

7th Class Mathematics

254

6.

 (4x + 15°) + (5x + 10°) + (6x + 5°) = 180°  15x + 30° = 180°  15x = 150°  x = 10°   R = 6x + 5° = 6 × 10° + 5° = 60° + 5° = 65° Thus, the measure of  R is 65º. It is given that AB is equal to BC. Therefore,  A is equal to  C. In  ABC, applying angle sum property of triangles:  A +  B +  C = 180°   A + 90° +  A = 180°  2  A = 90°   A = 45°   A =  C = 45° As the measures of both  A and  C are less than 90º, therefore,  A and  C are not supplementary to each other.

7.

5m 12 m

8.

9.

It can be observed that the longest pole that can be placed in the cylindrical hall is the hypotenuse of a right-angled triangle of sides 12 m and 5 m.  Hypotenuse2 = (12 m)2 + (5 m)2 = 169 m2 = (13 m)2  Hypotenuse = 13 m Thus, the longest pole that can be placed inside the hall is of length 13 m. Let the measures of angles be 5x, 7x, and 8x. Applying angle sum property of triangles: 5x + 7x + 8x = 180º  20x = 180º  x = 9º  5x = 5 × 9º = 45º 7x = 7 × 9º = 63º 8x = 8 × 9º = 72º It can be observed that the measures of all the angles are acute. Thus, the triangle is an acute-angled triangle. Applying Pythagoras theorem in ABC : AB2 + BC 2 = AC2 _____ (1) Applying Pythagoras theorem in BPC : BP 2 + PC 2 = BC 2 _____ (2) 3AC2 – 4AB2 = 3AC2 – 4 [AC 2 – BC 2 ]        [Using (1)] = 4BC2 – AC2

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= 4BC2 – (2PC)2  [P is mid-point of AC] 2 2 = 4 [BC – PC ] = 4BP2 [Using (2)] 10. It can be observed that ABY and ABC form a linear pair.  ABY  ABC  180  120  ABC  180  ABC  180  120  60 Similarly, ACX and ACB form a linear pair.. ACX  ACB  180  125  ACB  180  ACB  180  125  55 In ABC : A  ABC  ACB  180  A + 60° + 55° = 180°

 A = 180° – 115° = 65° Thus, the measure of A is 65º. 11. Let the interior opposite angles be 3x and 4x.  3x + 4x = 105°  7x = 105° 105  15 7 Therefore, interior opposite angles are 3x = 3 × 15° = 45º and 4x = 4 × 15° = 60°  Third angle = 180° – (60° + 45º) = 75° Thus, the greatest interior angle of the triangle is 75°. 12. Applying angle sum property of triangles in PQR :  x

P  PQR  R  180

 35° + 90° + R = 180°  R = 180° – (90° + 35°) = 55° Again applying angle sum property of triangles in SQR : SQR  QSR  R  180

 SQR + 90° + 55° = 180°  SQR = 180° – (90° + 55°) = 35° Thus, the measure of SQR is 35º. 13. It is known that the sum of the lengths of any two sides of a triangle is greater than the third side. Also, the difference between the lengths of any two sides of a triangle is smaller than the length of the third side.  (6.8 + 8.2) cm > Third side > (8.2 – 6.8) cm  15 cm > Third side > 1.4 cm Thus, the length of third side varies from 1.4 cm to 15 cm.

Triangles

255

14. It is given that AB is parallel to CD.  OCD  OBA [Pair of alternate interior angles]  OCD  x Applying angle sum property of triangles in COD : COD  OCD  ODC  180  80° + x + 40° = 180°  x = 180° – 120° = 60° Thus, the value of x is 60°. 15. Applying exterior angle sum property of triangles in ABC : A  B  ACX  x + 3x = 100°  4x = 100°  x = 25° Thus, the value of x is 25º. 16. Shape of the pot is a trapezoid with parallel sides horizontal and longer of the parallel sides is above the shorter parallel side. Therefore, the shape congruent with the shape of the pot is figure C. 17. Only figure D can be placed over the crescent of the moon completely covering it. Figures A, B, and C are larger than the crescent of the moon. Therefore, figure D is congruent with the shape of the crescent of the moon. 18. In the given triangles, R = X = 90° (Right angle) PQ = YZ = 6.8 cm (Hypotenuse) QR = XY = 2.9 cm (Side) By RHS congruence criterion, PQR  ZYX Thus, the statement given in alternative B is correct. 19. It is given that ABCD and ABEC are parallelograms. We know that lengths of the opposite sides of a parallelogram are equal.  AD = BC, AB = DC And, AC = BE, AB = CE

A

D

B

C

E

In ADC and BCE , AD = BC AC = BE DC = CE ( AB = DC and AB = CE)  ADC  BCE (By SSS congruency criterion)

Thus, CBE = DAC = 50° (Corresponding parts of congruent triangles are equal) BCE = ADC = 70° (CPCTE)  CEB = DCA Thus, measure of CEB = 180o– CBE – BCE

= 180o – 50o – 70o = 60o 20. In AOB and DOC : AO = OD [Given] AOB = COD [Vertically opposite angles] BO = OC [Given]  AOB  DOC [By SAS congruency criterion] Congruent parts of congruent triangles are equal.  AB = CD Therefore, perimeters of AOB and DOC are equal.  Perimeter of DOC = 16 cm  OC + OD + CD = 16 cm  8 cm + 5 cm + CD = 16 cm  CD = (16 – 13) cm = 3 cm Thus, the length of CD is 3 cm. 21. It is given that: A  P C  R  A  B  C  180

 B  180   A  C   B  180   P  R   B  Q In ABC and PQR :

B  Q BC = QR [Given] C  R [By ASA congruency  ABC  PQR criterion] Thus, the given triangles are congruent by ASA congruency criterion. 22. It is given that BE = CD  BE – DE = CD – DE  BD = CE ––––– (1) In ABD and ACE : AB = AC [Given] B  C [Angles opposite to equal sides are equal] BD = CE [From (1)] Therefore, by SAS congruency criterion, ABD is congruent to ACE . www.betoppers.com

7th Class Mathematics

256 23. In AOB and DOC : OBA  OCD [Alternate interior angles] AB = CD [Given] OAB  ODC [Alternate interior angles] Therefore, by ASA congruency criterion: AOB  DOC 24. Consider two rectangles ABCD and WXYZ: A D W Z 3 cm

2 cm

6 cm Y X C B 4 cm 2 Area of ABCD = 3 cm × 4 cm = 12 cm Area of WXYZ = 2 cm × 6 cm= 12 cm2 Therefore, areas are same, but these figures are not congruent. Thus, the rectangles with equal area are not congruent. 25. ABC is congruent to QRP . Therefore, all their corresponding sides are congruent.  AB = QR BC = RP CA = PQ Thus, the relation given in alternative C is correct. 26. In ADB and ADC : AB = AC [Given] AD = AD [Common] ADB  ADC  90 Therefore, by RHS congruency criterion: ADB  ADC Corresponding parts of congruent triangles are equal.  BD = DC = 3 cm  BC = 2 × 3 cm = 6 cm According to Pythagoras theorem: AB2 = AD2 + BD2 AB2 = (4 cm)2 + (3 cm)2 AB 2 = (16 + 9) cm2  AB = 25 cm = 5 cm  Perimeter = AB + AC + BC = (5 + 5 + 6) cm = 16 cm Thus, the perimeter of ABC is 16 cm.

27. In LMN and NPQ ,

LNM  PNQ (Vertically opposite angles) MLN  NQP (Given) LN = NQ (Given)  LMN  QPN (By ASA congruency criterion) www.betoppers.com

Now, corresponding parts of congruent triangles are equal.  LM = PQ ––––––––– (1) MN = NP  (3x – 4) cm = (x + 4) cm  3x – x = 4 + 4  2x = 8 x=4 PQ = (2x + 1) cm = (2 × 4 + 1) cm = 9 cm  LM = 9 cm Thus, the length of LM is 9 cm. 28. In WXY and ZXY , XY = XY (Common) (XY bisects WXZ ) WXY  ZXY XW = XZ (Given) congruency  WXY  ZXY (By SAS criterion) Now, corresponding parts of congruent triangles are equal.  XWY  XZY –––––––– (1) XYW  XYZ Given, WXZ  60 Since, XY bisects WXZ .

1 60  30  WXY  ZXY  WXZ  2 2 Now, in WXY , WXY  30 and XWY  25 By angle sum property of triangles, we obtain WXY  XYW  XWY  180  XYW = 180° – 30° – 25° = 125° –––– (2) From (1) and (2), we obtain XYZ  XYW  125 Thus, the measure of XYZ is 125°. 29. In the given triangles, R = X = 90° (Right angle) PQ = YZ = 6.8 cm (Hypotenuse) QR = XY = 2.9 cm (Side) By RHS congruence criterion, PQR  ZYX 30. It can be seen from the given figure that, BE = 22 cm  BC + CD + DE = 22 cm  10 cm + 2 cm + DE = 22 cm  12 cm + DE = 22 cm  DE = (22 – 12) cm = 10 cm Now, BD = BC + CD = (10 + 2) cm = 12 cm CE = CD + DE = (2 + 10) cm = 12 cm  BD = CE –––– (1) In ABD and FEC ,

Triangles

257

AB = FE (Given) ABD  FEC  90 BD = CE (From (1))  ABD  FEC (SAS congruence rule)  AD = FC (CPCT)  FC = 13 cm In FEC , by Pythagoras Theorem, we get: CE2 + FE2 = FC2  (12 cm)2 + FE2 = (13 cm)2  144 cm2 + FE2 = 169 cm2  FE2 = (169 – 144) cm2 = 25 cm2  (FE)2 = (5 cm)  FE = 5 cm Perimeter of FEC = CE + FE + FC = (12 + 5 + 13) cm = 30 cm Thus, the perimeter of FEC is 30 cm.

CONCEPTIVE WORKSHEET

2.

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1  B  x 2 It is given that AB = AC. It is known that angles opposite to equal sides are equal.  B  C x  B  C  2 Applying angle sum property of triangles in ABC : A  B  C  180 x x   180 2 2  2x = 180°  x = 90°  A  90 B  C  45 Thus, ABC is a right-angled triangle.

5.

C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Let the measure of A be x.

1 × 124° = 62° 2 Thus, the value of x is 62º. It is known that the sum of the lengths of any two sides of a triangle is greater than the third side of the triangle. Consider the lengths given in alternative B: 4 cm + 6 cm = 10 cm > 8 cm 6 cm + 8 cm = 14 cm > 4 cm 4 cm + 8 cm 12 cm > 6 cm Thus, the lengths given in alternative B can be the sides of a triangle. It is known that in a triangle, the measure of exterior angle is equal to the sum of the interior opposite angles.  y + y = 3y – 68°  2y = 3y – 68°  y = 68° Thus, the value of y is 68º. Applying Pythagoras theorem: (6 cm)2 + (8 cm)2 = (Hypotenuse)2  (Hypotenuse)2 = 36 cm2 + 64 cm2  Hypotenuse2 = 100 cm2  Hypotenuse = 10 cm Thus, the length of hypotenuse is 10 cm. Applying angle sum property of triangles in PQR :

 x=

KEY 1

It is known that in a triangle, the measure of exterior angle is equal to the sum of the interior opposite angles.  2x + 1º = 55° + 70°  2x + 1º = 125°  2x = 124°

6.

P  Q  R  180

 P  2P  180 180  60 3 Thus, the measure of P is 60º. In the given figure, firstly CL is drawn parallel to XY and PQ. X Y

 P 

7.

40° A

 x

65°

L

C

x B 55° P

Q www.betoppers.com

7th Class Mathematics

258  XYC  YCL       [Alternate interior angles]  YCL  40 Similarly, LCQ  CQP  55 [Alternate interior angles]

 YCQ  YCL  LCQ = 40° + 55° = 95°

ACB  YCQ

8.

[Vertically opposite angles]

 ACB  95 Applying angle sum property of triangles in ABC : 65° + x + 95° = 180°  x = 180° – 160°  x = 20° Thus, the value of x is 20°. It is known that in an isosceles triangle, angles opposite to equal sides are equal.  A   C ––––––– (1) Applying angle sum property of triangles in ABC :

A  B  C  180  A  90  C  180  2A = 180° – 90° [Using (1)}

90  45 2 Thus, the measure of C is 45º. 9. It is given that AD is parallel to EF.  FEB  ABC [Corresponding angles]  ABC  65 Applying angle sum property of triangles in ABC : 55° + 65° + x = 180°  120° + x = 180°  x = 60° Thus, the value of x is 60º. 10. The given situation can be represented as: C

 A 

5 feet A 9 feet B

E 12 feet 9 feet

D 12 feet Here, AB and CD are trees and BD is the distance between them on the ground. Therefore, distance between their tops is AC. A line AE is drawn parallel to BD.  AE = BD = 12 feet Applying Pythagoras theorem in ACE : www.betoppers.com

AC2 = AE2 + CE2  AC2 = (12 feet)2 + (5 feet)2  AC2 = 144 feet2 + 25 feet2  AC2 = 169 feet2  AC2 = (13 feet)2  AC = 13 feet Thus, the tops of the trees are 13 feet apart. 11. It is known that the sum of the lengths of any two sides of a triangle is greater than the third side. Consider the lengths given in alternative C: 2.6 cm + 3.4 cm = 6 cm < 7 cm Therefore, the triangle is not possible in this case. Thus, the lengths given in alternative C cannot be the sides of a triangle. 12. Let the angles of the triangle be 2x, 3x, 5x. Applying angle sum property of triangles: 2x + 3x + 5x = 180°  10x = 180°  x = 18° Therefore, the angles are 2 × 18° = 36°, 3 × 18° = 54°, and 5 × 18° = 90° This means that the given triangle is a right-angled triangle. Also, all its angles are of different measures. Therefore, it is also a scalene triangle. Thus, the triangle is a scalene right triangle. 13. It is given that AB||DE. interior  ADE  BAC  90 (Alternate angles)  BAC  90 ABC is isosceles.  B  C Applying angle sum property of triangles in ABC : A  B  C  180  90° + B + B = 180°  2B = 90°

90°  45 2  BCD = A + B = 90° + 45° = 135° Thus, the measure of BCD is 135º. 14. It is known that in a triangle, the measure of an exterior angle is equal to the sum of interior opposite angles.  B + BAD = ADE  90° + (x + 15°) = 135°  105° + x = 135°  x = 135° – 105° = 30° Thus, the value of x is 30º.

 B =

Triangles

259

15. Let A and B be 3x and 5x respectively.. It is known that the measure of an exterior angle of a triangle is equal to the sum of the interior opposite angles.  ACD  A  B  160º = 3x + 5x  160° = 8x

 x=

16

17.

18.

19.

20.

160°  20 8

 B = 5 × 20º = 100° Thus, the measure of B is 100º. Figures A, B, and D can be produced by rotating the sign board but the figure C can not be produced by reflecting, rotating, or sliding the sign board. Therefore figure is not congruent to the sign board. Congruent figure have the same shape and size. Thus, the erasers drawn in figures (i) and (iii) are congruent to each other. In the given triangles, R = X = 90° (Right angle) PQ = YZ = 6.8 cm (Hypotenuse) QR = XY = 2.9 cm (Side) By RHS congruence criterion, PQR  ZYX Thus, the statement given in alternative B is correct. Consider the triangles given in alternative C AB = DE = 3 cm AC = DF = 4 cm BAC  EDF  40 By SAS congruence criterion, ABC  DEF ABC is an equilateral triangle of side 12 cm.  AB = BC = AC = 12 cm Also, ABC  ACB  BAC  60 It is given that the points D and E divide BC into three equal parts.

1 12 BC  cm  4cm 3 3 Comparing ABD and ACE , AB = AC = 12 cm ABD  ACE  60 BD = CE = 4 cm  ABD  ACE [By SAS congruence rule]  AD = AE [Corresponding parts of congruent triangles] [AD = 9 cm]  AE = 9 cm Thus, perimeter of ADE = AD + DE + AE = (9 + 4 + 9) cm = 22 cm  BD = DE = CE

21. ABDC is a parallelogram and it is known that in a parallelogram, the opposite sides are equal. AB = CD and BD = AC In ABC and DCB , AB = DC AC = DB BC = CB (Common)  ABC  DCB (By SSS congruency criterion) Thus, ABC and DCB are congruent to each other in the given figure. 22. Q

P

R

S In PQR and PSR , PR = PR (Common) PQ = PS (Given) QR = SR (Given)  PQR  PSR (SSS congruence criterion) .)  QPR  SPR and QRP  SRP (C.P.C.T.)

 PR is the bisector of QPS and QRS . 1 1   QPR  2 QPS   2  230   115   Applying angle sum property of triangles in QPR :

QPR  PQR  QRP  180

 115° + 25° + QRP = 180°  140° + QRP = 180°  QRP = 180° – 140° = 40° PR is the bisector of QRS .  QRS  2QRP = 2 × 40° = 80°

Thus, the measure of QRS is 80°. 23. PRS is a straight line.  QRS  QRP  180 ° (Linear pair)

 145° + QRP = 180°  QRP = 180° – 145° = 35°

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7th Class Mathematics

260

P Q

A

R 35° B

145° C

S

In ABC and QPR ,

ABC  QPR  90 ACB  QRP  35 BC = PR (Given)  ABC  QPR (ASA congruence criterion)  AB = QP and AC = QR (C.P.C.T.)  Perimeter of ÄABC = AB + BC + AC = QP + PR + QR  30 cm = Perimeter of PQR Thus, the perimeter of ABC is 30 cm. 24. It is given that ABC is congruent to PQR . Therefore, by CPCT, we get: AB = PQ, BC = QR and AC = PR  PQ = 7 cm and BC = 24 cm In ABC , by Pythagoras Theorem, we get: AC2 = AB2 + BC 2  AC2 = (7 cm)2 + (24 cm)2  AC2 = (49 + 576) cm2 = 625 cm2  AC2 = (25 cm)2  AC = 25 cm Now, AC = PR  PR = 25 cm Now, AC = 25 cm  AP + PC = 25 cm  AP + 20 cm = 25 cm  AP = (25 – 20) cm = 5 cm  AR = AP + PR = (5 + 25) cm = 30 cm Thus, the length of AR is 30 cm. 25. In ABC and CDE , AB = CD (Given) BC = DE (Given) ABC  CDE  60  ABC  CDE (SAS congruence rule) By CPCT, we get: AC = CE, BAC  ECD and BCA  DEC ––––– (1) In ABC , ABC  BAC  BCA  180  60  BAC  BCA  180  ECD  BCA =180°–60° = 120°(From (1)) Now, BCD is a straight line.  BCA  ACE  ECD  180 www.betoppers.com

  BCA  ECD   ACE  180  120° + ACE = 180°  ACE = 180° – 120° = 60° In ACE , AC = CE and ACE = 60°  CAE  ACE (Angles opposite to equal sides are equal) ACE  CAE  CEA  180  60  CAE  CAE  180  2CAE = 180° – 60° = 120°  CAE 

120  60 2

 CAE  60  ACE  CAE  CEA  60 Thus, ACE is an equilateral triangle. 26. It can be seen from the given figure that, BD = 23 cm.  BC + CD = 23 cm  CD = 23 cm – BC = 23 cm – 15 cm = 8 cm In ABC and CDE , AB = CD = 8 cm BC = DE = 15 cm ABC  CDE  90  ABC  CDE (SAS congruence rule) By CPCT, we get: AC = CE, BAC  DCE and BCA  DEC –––––– (1) In ABC , ABC  BAC  BCA  180

 90  DCE  BCA  180  DCE  BCA = 180° – 90° = 90° ––– (2) Now, BCD is a straight line.  BCA  ACE  DCE  180   BCA  DCE   ACE  180  90° + ACE = 180° (From (2))  ACE = 180° – 90° = 90° In ACE , AC = CE and ACE = 90° Thus, ACE is a right-angled isosceles triangle. 27. In ABD and CDB , BD = DB (Common) ABD  CDB (Alternate interior angles) ADB  CBD (Alternate interior angles)  ABD  CDB (ASA congruence rule)  AB = CD and AD = CB (CPCT) Perimeter of ABD  20 cm  AB + BD + AD = 20 cm

Triangles

261

 AB + 5 cm + AD = 20 cm  AB + AD = (20 – 5) cm = 15 cm Perimeter of quadrilateral ABCD = AB + BC + CD + AD = AB + AD + AB + AD = 2 (AB + AD) = (2 × 15) cm = 30 cm Thus, the perimeter of quadrilateral ABCD is 30 cm. 23. It can be seen from the given figure that, ABF  ABC  180 (Linear pair) DEG  DEC  180 (Linear pair)  ABF  ABC  DEG  DEC ABF  DEG (Given)  ABC  DEC ––––– (1) In ABC and DEC , AC = CD = 9 cm ABC  DEC (From (1)) ACB  DCE (Vertically opposite angles)  ABC  DEC (AAS congruence rule)  AB = DE and BC = EC (CPCT)  AB = 13 cm and BC = 11 cm Perimeter of ABC = AB + BC + CA = (13 + 11 + 9) cm = 33 cm Thus, the perimeter of ABC is 33 cm. 29. In ABD and ACE , AB = AC (Given) AD = AE (Given) BAD  CAE  30  ABD  ACE (SAS congruence rule)  BD = CE (CPCT) Now, CE = 2 cm  BD = 2 cm BC = BD + DE + CE = 2 cm + 6 cm + 2 cm = 10 cm Thus, the length of BC is 10 cm. 30. In AOC and BOD , AO = BO = 5 cm AOC  BOD (Vertically opposite angles) CAO  DBO (Alternate interior angles)  AOC  BOD (ASA congruence rule)  CO = DO (CPCT) Now, CD = 8 cm  CO + DO = 8 cm  CO + CO = 8 cm  2 CO = 8 cm

8

 CO   2  cm  4cm   Thus, the length of CO is 4 cm.

SUMM ATIVE WORKSHEET KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

B

B

A

B

C

A

C

A

B

B

Q.no

11 12

13

14

Key

D

C

A

B

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It is given that l||m. A and x are alternate angles.  A  x ––––––––– (1) It is given that AC = BC Therefore, ABC is an isosceles triangle.  A  B Applying angle sum property of triangles in ABC : A  B  ACB  180  2x + 42° = 180° {Using (1)}  2x = 180° – 42°

138  69 2 Thus, the value of x is 69º. It is known that the sum of the lengths of any two sides of a triangle is greater than the third side.  10.4 cm + 16.2 cm > x  26.6 cm > x The natural number nearest to 26.6 is 26. Thus, the greatest possible value of x is 26 cm. It can be observed in the given figure that the interior opposite angles of PRX are QPR and PQR .

 x

2.

3.

The interior opposite angles of RQZ are QPR and PRQ . Thus, QPR is common to the interior opposite 4.

angles of the exterior angles PRX and RQZ . It is known that sum of all interior angles of a triangle is 180°.  A  B  C  180  105° + 50° + C = 180°  C = 180° – 155° = 25°

B 50 2    2 :1 C 25 1 Thus, ratio of B and C is 2:1.

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7th Class Mathematics

262 5.

6.

7.

Let x be the unequal angle of the triangle. Therefore, applying angle sum property of triangles: x + 42° + 42° = 180°  x + 84° = 180°  x = 180° – 84°  x = 96° Thus, the measure of the unequal angle of the isosceles triangle is 96º. It can be seen in the given figure that ABC is a rightangled triangle. By using Pythagoras theorem in ABC , we obtain AC2 = AB2 + BC 2  AC2 = (14 m)2 + (48 m)2 = 196 m2 + 2304 m2 = 2500 m2 = (50 m)2  AC = 50 m  Distance travelled by Ishaan to reach C from A = AC = 50 m Distance travelled by Saurabh to reach C from A = AB + BC = 14 m + 48 m = 62 m  Difference = 62 m – 50 m = 12 m Thus, Saurabh travelled a distance of 12 m more than Ishaan. The screen of the flat screen television is rectangular. The length and the diagonal of the screen are given as 2 feet and 2.5 feet respectively. Let the screen be represented by a rectangle ABCD as A D

8.

Let AC be the pole and B be the point on the ground through which the rope is tied. Then, AB = 26 m, BC = 10 m A

26 m

B

9.

C 10 m The pole is perpendicular to the ground. Therefore, C  90 Applying Pythagoras Theorem in right ABC , we obtain AB2 = AC2 + BC 2  (26 m)2 = AC2 + (10 m)2  676 m2 = AC2 + 100 m2  AC2 = (676 – 100) m2  AC2 = 576 m2  AC2 = (24 m)2  AC = 24 m Thus, the height of the pole is 24 m. Let the given rectangle be ABCD and let AC be its diagonal. A D

2.5 feet

C B 2 feet It is evident that ABC is right-angled at B. [In rectangle, all angles are 90°] According to Pythagoras Theorem, it is known that in a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the legs. Therefore, we have AC2 = AB2 + BC 2  (2.5 feet)2 = AB2 + (2 feet)2  6.25 feet2 = AB2 + 4 feet2  AB2 = 6.25 feet 2 – 4 feet 2  AB2 = 2.25 feet 2 = (1.5 feet)2 Thus, the height of the screen is 1.5 feet.

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B C  AC = 25 cm and BC = 24 cm Applying Pythagoras theorem in right-angled ABC : AB2 + BC 2 = AC2  AB2 + (24 cm)2 = (25 cm)2  AB2 + 576 cm2 = 625 cm2  AB2 = 625 cm2 – 576 cm2 = 49 cm2 = (7 cm)2  AB = 7 cm  Breadth of the rectangle = AB = 7 cm Area of the rectangle = Length × Breadth = 24 cm × 7 cm = 168 cm2 Thus, the area of the rectangle is 168 cm2 . 10. AD is the altitude of ABC drawn from the vertex A. AD is perpendicular to BC and so, ABD is a right triangle. According to Pythagoras’ Theorem, it is known that in a right-angled triangle, the square of the

Triangles

263

hypotenuse is equal to the sum of the squares of the legs. Therefore, we have AB2 = AD2 + BD2  (10 cm)2 = (8 cm)2 + BD2 [AB = 10 cm, AD = 8 cm]  100 cm2 = 64 cm2 + BD2  100 cm2 – 64 cm2 = BD2  BD2 = 36 cm2  BD2 = (6 cm)2  BD = 6 cm Now, BD + DC = BC  6 cm + DC = 21 cm [BC = 21 cm]  DC = 21 cm – 6 cm = 15 cm  BD : DC =

6 cm 2   2:5 15 cm 5

Thus, the point D divides the side BC in the ratio 2:5 11. According to Pythagoras Theorem, a given triangle is a right-angled triangle, if the square of the longest side is equal to the sum of the squares of the remaining two sides. It is observed that, (11 cm)2 + (60 cm)2 = 121 cm2 + 3600 cm2 = 3721 cm2 = (61 cm)2 Thus, 11 cm, 60 cm, and 61cm can be the sides of a right-angled triangle. 12. It is given that CAB  DEF . Therefore, all the corresponding sides and angles are equal. So, we have: A  E (= 90°), BC = DF (= 13 cm) Therefore, DEF is a right triangle, where E  90 , DF = 13 cm and DE = 12 cm. According to Pythagoras’ Theorem, it is known that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.  DF2 = DE2 + EF2  (13 cm)2 = (12 cm)2 + EF2  169 cm2 = 144 cm2 + EF2  169 cm2 – 144 cm2 + EF2  EF2 = 25 cm2  EF2 = (5 cm)2  EF = 5 cm Thus, the measure of EF is 5 cm. 13. It is known that in a triangle, the sum of lengths of its any two sides is greater than the length of the third side. Consider the lengths, 1.5 cm, 3 cm, and 3.5 cm It can be observed that, 1.5 cm + 3.5 cm = 5 cm > 3 cm 3 cm + 3.5 cm = 6.5 cm > 1.5 cm 1.5 cm + 3 cm = 4.5 cm > 3.5 cm Thus, the lengths, 1.5 cm, 3 cm, and 3.5 cm, can be the sides of a triangle.

14. ΔABC is an isosceles triangle with AB = AC  ABC  ACB (Angles opposite to equal sides are equal) A 40°

B

C D Applying angle sum property of triangles in ΔABC:

ABC  ACB  BAC  180  ABC  ABC  40  180

 2 ABC  40  180  2 ABC = 180° – 40° = 140° 140 2  ABC  70 We know that the exterior angle of a triangle is equal to the sum of the interior opposite angles.  ACD  ABC  BAC  ACD = 70° + 40° = 110° Thus, the measure of ACD is 110°. The correct answer is A.

 ABC 

HOTS WORKSHEET KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

B

C

C

B

C

A

B

B B

C

Q.no

11 12

13

14 15

Key

B

C

A

C

D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. As seen in the given figure, OMN  60 (Vertically opposite angles) The measure of the exterior angle of a triangle is equal to the sum of the measures of the interior angles.  MOP  MNO  OMN

 125  MNO  60  MNO = 125° – 60° = 65° Thus, the measure of MNO is 65°. www.betoppers.com

7th Class Mathematics

264 2.

3.

4.

Let the measures of A and C be 7x and 8x respectively. Since ABC is right-angled at B, the measure of B is 90°. According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°. In ABC , A  B  C  180  7x + 90° + 8x = 180°  15x + 90° = 180°  15x = 180° – 90°  15x = 90°  x = 6° Thus, the measure of C is 8x = 8 × 6° = 48°. Since the ratio of the angles is given as 1:2:3, let the angles measure x, 2x, and 3x. According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°.  x + 2x + 3x = 180°  6x = 180°  x = 30° Thus, the angles of the triangle measure 30°, 2 × 30° = 60°, and 3 × 30° = 90°. Since all angles of the triangle are of different measures, the triangle is scalene. Also, since one of the angle measures 90°, the triangle is right-angled. Thus, the triangle is right-angled and scalene. According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°. In ABE , ABE  AEB  BAE  180  60° + AEB + 60° = 180°  120° + AEB = 180°

 AEB = 180° – 120° = 60° Since each angle of ABE measures 60°, ABE is an equilateral triangle.  AB = BE = AE Perimeter of the given figure = AB + BC + CD + DE + AE = 38 cm  AB + 10 cm + AB + 10 cm + AB = 38 cm (AB = AE = BE = CD; BC = DE = 10 cm)  3AB + 20 cm = 38 cm  3AB = 38 cm – 20 cm = 18 cm  AB = 6 cm Thus, the length of AB is 6 cm.

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5.

Since angles opposite to equal sides of a triangle are equal, BAC  BCA . According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°. In ABC , ABC  BCA  BAC  180  110  BCA  BCA  180

 BAC  BCA 

6.

 2BCA = 180° – 110° = 70°  BCA  35 BCA  DCE (Vertically opposite angles)  DCE  35 In CDE , DCE  CED  DCE  180  55° + CED + 35° = 180°  90° + CED = 180°  CED = 180° – 90° = 90° Since CED measures 90°, CDE is a rightangled triangle. Since its angles are of different measures, its sides are also of different lengths. Thus, it is also scalene. LMN  LMO  OMN  90° = 40° + OMN

7.

 OMN = 90° – 40° = 50° According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°. In OMN , OMN  ONM  MON  180  50° + ONM + 90° = 180°  ONM + 140° = 180°  ONM = 180° – 140° = 40° Thus, the measure of LNM is 40°. AXF  XZD (Alternate interior angles)  XZD  140 XZD also acts as an exterior angle to XYZ . The measure of the exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles.  XZD  XYZ  YXZ  140° = 75° + YXZ  YXZ = 140° – 75° = 65° YXZ  EXG (Vertically opposite angles)  a = 65° Thus, the value of a is 65°.

Triangles 8.

265

Let ABC be the isosceles triangle. Now, two cases arise here. The first case is when the measure of the unequal angle is 40° and the other is when the measure of each of the equal angles is 40°. Case I: When the unequal angle measures 40° ABC is isosceles with AB = AC and BAC  40 A

40°

B C In ABC , ABC  BAC   ACB  180  ABC  40  ABC  180

 ACB  ABC   2ABC = 180° – 40° = 140°  ACB  ABC  70 Thus, in this case, the difference between the measures of the unequal angle is 70° – 40° = 30°. Case II: When each of the two equal angles measure 40° ABC is isosceles with AB = AC and ABC  ACB  40

A

40°

40°

B C In ABC , ABC  BAC   ACB  180  40° + BAC + 40° = 180°

9.

 BAC = 180° – 80° = 100° Thus, in this case, the difference between the measures of the unequal angle is 100° – 40° = 60°. Thus, the possible differences between the measures of the unequal angles of the isosceles triangle are 30° and 60°. The sides of an equilateral triangle are equal in length. The perimeter of the equilateral ABC is 24 cm. 24 cm  8cm 3 Thus, width of rectangle BCFD = BC = FD = 8 cm

 AB = BC = CA =

The perimeter of the isosceles triangle DEF is 20 cm.  DE + EF + DF = 20 cm  2DE + 8 cm = 20 cm (DE = EF)  2DE = 20 cm – 8 cm = 12 cm

12 cm  6cm 2  Perimeter of the given figure = AB + BD + DE + EF + FC + CA = 8 cm + 20 cm + 6 cm + 6 cm + 20 cm + 8 cm = 68 cm 10. Since angles opposite to equal sides of a triangle are equal, ABC  ACB . According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°.  ABC  ACB  BAC  180  ABC  ABC  50  180

 DE = EF =

 ABC  ACB; BAC  50   2ABC + 50° = 180°  2ABC = 180° – 50° = 130° 130  65 2 The measure of the exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles.  ACX  ABC  BAC = 65° + 50° = 115° Thus, the magnitude of ACX is 115°. 11. The measure of the exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles.  ACY  ABC  BAC  145° = 90° + BAC  BAC = 145° – 90° = 55° Now, BAC and CAX form a linear pair..

 ABC 

 BAC   CAX  180  55° + y = 180°  y = 180° – 55° = 125° Thus, the value of y is 125°. 12. Since AB || CD, ELB  LMN (Corresponding angles)  LMN  40 According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°.  LMN  LNM  MLN  180

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7th Class Mathematics

266

 40° + 70° + x = 180°  110° + x = 180°  x = 180° – 110° = 70° Thus, the value of x is 70°. 13. According to the angle sum property of a triangle, the sum of the measures of the angles of a triangle is 180°. Two angles of the triangle measure 35° and 68°.  Measure of the third angle = 180° – (35° + 68°) = 180° – 103° = 77° Now, the measure of an exterior angle of a triangle is equal to the sum of the measures of its interior opposite angles. Thus, the measures of the three exterior angles of the triangle will be (35° + 68°), (35° + 77°), and (68° + 77°) = 103°, 112°, and 145°. Thus, 136° cannot possibly be the measure of an exterior angle of the given triangle. 14. It is known that a triangle is possible if the sum of the lengths of its any two sides is greater than the third side. Considering the dimensions 13 m, 14 m, and 9 m, it can be seen that: 13 m + 14 m = 27 m > 9 m 13 m + 9 m = 22 m > 14 m 14 m + 9 m = 23 m > 13 m Also, the perimeter is (13 m + 14 m + 9 m) = 36 m. Thus, the possible dimensions of the garden are 13 m, 14 m, and 9 m. 15. Applying the Pythagoras Theorem in a right-angled ABX : AB2 + BX2 = AX2  BX2 = AX2 – AB2  BX2 = (41 m)2 – (40 m)2  BX2 = 1681 m2 – 1600 m2  BX2 = 81 m2  BX2 = (9 m)2  BX = 9 m Applying the Pythagoras Theorem in a right-angled CDX : CD2 + DX2 = CX2  DX2 = CX2 – CD2  DX2 = (17 m)2 – (15 m)2  DX2 = 289 m2 – 225 m2  DX2 = 64 m2  DX2 = (8 m)2  DX = 8 m  Distance between the two buildings = BD = BX + DX =9m+8m = 17 m www.betoppers.com

IIT JEE WORKSHEET KEY 1

2

3

4

C

B

B

11 AB C

12

13

CD

AB

5

B

6

B

7

C

8

A

A

9

10

A

AC D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. C 60° y

B

p

100°

110°

z

E

F

D

A z = 180° – 110° = 70° y = 180° – 100° = 80° p + y + z + 60° = 360°  p = 150°

D

2.

1200

E r

3.

4.

C

500 a

b b A B a + 50° = 120° a = 70°  b = 90° – 70° = 20° Givne AE = EB,   EAB = b = 20° r + 20° + 20° = 180°  r = 180° – 40° r = 140° x + 120° + 25° = 180° (The sum of the angles in a triangle = 180°). x + 145° = 180°, x = 35° 40° + p = 165° (Exterior angle = Sum of two opposite interior angles) p = 125°

Triangles 5.

6.

7.

267

In  BDE, BD = BE.   BED =  BDE = 70°   DBE = 180° – 70° – 70° = 40° In  BCD,  CBD = 180° – 90° – 30° = 60°  z = 180° – 60° – 40° (The sum of the angles on a straight line = 180°) = 80° Let the measure of the third angle be x°. Since the sum of the three angles of a triangle is 180°, we have 52° + 72° + x = 180°  124° + x = 180°  x = 180° – 124° = 56°  The measure of the third angle is 560 . Let the measures of the given angles be 6x°, 5x° and 7x°. Since the sum of three angles of a triangle is 180°, we have 6x + 5x + 7x = 180° 18x = 180°

180  10 18  The required angles are (6  10)°, (5  10)° and (7  10)°, i.e., 60°, 50° and 70°. In  ACD y + 120° + 20° = 180° (angle sum property of triangles)  y + 140° = 180°  y = 180° – 140° = 40° For  ABC,  y is an exterior angle  x + 15° = y (exterior angle = sum of two interior opposite angles)  x + 15° = 40°  x = 40° – 15° = 25° Since, BO bisects  B and CO bisects  C, we have x

8.

9.

1 1 B – 2 2 0  C (since  A +  B +  C = 180 ; angle sum property of triangles) 

 BOC = (  A +  B +  C) –



 BOC =  A +



 BOC =

1 1 B + C 2 2

1 1 (  A +  B +  C) +  A 2 2

= 900 +

1 A 2

1  2   0 0 = (90 + 30 ) = 1200

  BOC = 900    600 

14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25.

6 9 1200 3600 7200 3 5 cm 9 cm 4 cm 8 degrees A – s; B – r; C – q; D – a A – s; B – r; C – q; D – a



1 1  B and  BCO = C 2 2 Now, in  OBC, we have  OBC +  BCO +  BOC = 180° (sum of angles of a triangle)  OBC =



1 1 B +  C +  BOC = 180° 2 2



 BOC = 180° –

1 1 B – C 2 2

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268

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7th Class Mathematics

9. QUADRILATERALS SOLUTIONS

FORMATIVE WORKSHEET KEY Q.no

1

2

3

4

5

6

Key

D

B

A

B

B

B C

Q.no

11 12

13 14

Key

A

B

9

10

B C

B

3.

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Parallelogram ABCD can be drawn as:

4.

D

7

8

C

A

B

O D C It is known that diagonals of a parallelogram divide it into two triangles of equal area.  Area of  ABD = Area of  BCD

1 × Area of ABCD 2 In a parallelogram, diagonals bisect each other.  AO is the median of  ABD. and Area of  AOB = Area of  AOD  Area of  ABD 

  Area of ΔAOB  

1 × Area of ΔABD 2

5.

1 1   × Area of ABCD 2 2

2.

Area of parallelogram ABCD 4  Area of AOB 1 Thus, the required ratio is 4: 1. Let the three angles of the quadrilateral be 2x, 5x, and 7x. Here, the greatest angle is 7x and smallest angle is 2x. It is given that the sum of the greatest and the smallest angles is 180°.  7x + 2x = 180°  9x = 180°  x = 20°  Measure of fourth angle

6.

7.

= 360° – (2x + 5x + 7x) = 360° – 14x = 360° – 14 × 20° [Putting x = 20°] = 360° – 280° = 80° Thus, the measure of the fourth angle is 80°. Since AD = BD and  ABD = 50°,  ABD is isosceles with  DAB =  ABD = 50°. In a parallelogram, adjacent angles are supplementary.   ABC = 180° –  A = 180° – 50° = 130° Thus, the measures of adjacent angles of the parallelogram are 50° and 130°. In ΔAMD,   DMA +  MDA +  DAM = 180º   DMA + 100º + 25º = 180º   DMA + 125º = 180º   DMA = 180º – 125º   DMA = 55º Since ABCD is a parallelogram, AB||CD   MAB =  DMA = 55º [Alternate interior angles] It is also given that AM = BM   MAB =  MBA = 55º In ΔMAB:  MAB +  AMB +  MBA = 180º   AMB + 55º + 55º = 180º   AMB + 110º = 180º   AMB = 180º – 110º = 70º Thus, the measure of  AMB is 70°. The sum of the interior angles of any polygon having n number of sides is (n – 2) 180°. In pentagon, n = 5  Sum of the interior angles = (5 – 2) × 180° = 540° It is known that in a rhombus, diagonals are the angle bisectors of the corresponding angles. Therefore, O is the point of intersection of the diagonals. It is also known that diagonals of a rhombus bisect each other at right angles. Thus, the measure of  AOB is 90°. It is known that the sum of the measures of the external angles of any polygon is 360°. Therefore, in particular, the sum of the measures of the external angles of an octagon is 360°.

7th Class Mathematics

270 8.

9.

Since diagonals of a rectangle bisects each other, DB = 2DO = 10 m The measure of  BCD is 90°, hence, DB2 =  DC2 +  BC2 BC  DB2  DC 2  10 2  62  8m Therefore, the area of the rectangle is 8 m × 6 m = 48 m2 It is given that WXYZ is a rhombus. It is known that in a rhombus, the diagonals bisect each other.  WO = OY and XO = OZ  2y + 1 = 5y – 11 and 3x + 2 = x + 4 2y + 1 = 5y – 11  2y – 5y = –11 –1  –3y = –12

y

12 3

y4  WO = 2y + 1 = 2 × 4 + 1 = 9 units  WY = WO + OY = 2 WO = 2 × 9 = 18 units Also, 3x + 2 = x + 4   3x – x = 4 – 2  2x = 2  x=1  XO = 3x + 2 = 3 × 1 + 2 = 5 units  XZ = XO + OZ = 2 × XO = 2 × 5 = 10 units Thus, the length of WY and XZ are 18 units and 10 units respectively. 10. Let the number of sides of the regular polygon be n. Each interior angle of the regular polygon is 150°.  Each exterior angle = 180° – 150° = 30° We know that, Exterior angle of a regular polygon

360  Number of sides  30 

360 n

360  12 30 Thus, the regular polygon has 12 sides. 11. A quadrilateral with exactly two pairs of equal consecutive sides is called a kite. 12. It can be observed that in the given quadrilateral, there are exactly two pairs of adjacent equal angles. Thus, the given quadrilateral is an isosceles trapezium. n

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13. Let PQRS be an isosceles trapezium where  S = 125° S

R

P Q Since, SR || PQ and  P =  Q. So,  R =  S= 125° Also the sum of all angles of trapezium is 360°. So,  P +  Q +  R +  S = 360°  2  P + 250° = 360°  2  P = 110°   P = 55° Thus, the measures of the other angles are 55°, 55°, and 125°. 14. Since quadrilateral PQRS is an isosceles trapezium where PQ||RS, PS = QR Therefore,  P =  Q It is known that when two parallel lines are cut by a transversal, the interior angles on the same side of the transversal are supplementary.   P +  S =  Q +  R = 180°   S =  R [  P =  Q] Thus, the relation given in alternative C is correct.

CONCEPTIVE WORKSHEET KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

B

B

D

C

B

D

A

C

D

D

Q.no

11 12

13

Key

A

D

B

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Let the length of each side of the square be a cm. It is given that the length of its diagonal is 16 2cm . Therefore, applying Pythagoras theorem in the triangle formed by a pair of adjacent sides and the diagonal:



a 2  a 2  16 2



2

2a2 = 162 × 2 a2 = 162 a = 16 Thus, perimeter of the square = 4a cm = 4 × 16 cm = 64 cm

Quadrilaterals Solutions 2.

3.

271

In a square, all the angles are 90°. s + q = 90° + 90° = 180° In a rhombus (which is also a parallelogram), adjacent angles are supplementary.  a + b = 180° Thus, s + q = a + b Hence, the relation given in alternative B is correct. The given rectangle ABCD can be drawn as: A

7.

8.

1

4.

5.

C

1

A  × 90° = 45°   DAP  2  2 In ΔADP:  DAP +  ADP +  DPA = 180°   DPA = 180° – 90° – 45° = 45° A   DAP =  DPA  ΔADP is isosceles where  A   DAP =  DPA Therefore, AD = DP = a CP = DC – DP = 3a – a = 2a Thus, CP: DP = 2a: a = 2:1 Since ABCD is a rhombus, all the sides are equal.   ABD is isosceles.   ADB =  ABD  x = y Thus, the relation given in alternative C is correct. S

R

9.

6.

P Q It is known that in a parallelogram, adjacent angles are supplementary.   Q +  R = 180°  80° +  R = 180°   R = 180° – 80° = 100° Let ABCD be the rhombus such that  A = 50°.   C = 50° [In a rhombus, opposite angles are equal]  A +  B = 180° [In a rhombus, adjacent angles are supplementary]  50° +  B = 180°   B = 180° – 50°   B = 130°

360  20 18 Thus, the regular polygon has 20 sides. We know that the diagonals of a rhombus bisect each other.   OA = OC   3x – 1 = 2x + 1  3x – 1 – 2x = 1   x – 1 = 1   x = 1 + 1 = 2 OA = (3x – 1) cm = (3 × 2 – 1) cm = 5 cm OD = (5x + 2) cm = (2 × 5 + 2) cm = 12 cm The diagonals of a rhombus bisect each other at right angles.   AOB = 90° Applying Pythagoras theorem in right triangle AOB, we obtain AB2 =  OA2 +  OB2  AB2 = (5 cm)2 + (12 cm)2  AB2 = 25 cm2 + 144 cm2  AB2 = 169 cm2  AB2 = (13 cm)2  AB = 13 cm All sides of a rhombus are equal. Therefore, perimeter of the rhombus = 4 × Side = (4 × 13) cm = 52 cm The given polygon has 6 sides.   Sum of interior angles = (n – 2) × 180° = (6 – 2) × 180° = 4 × 180° = 720° ––––––– (1)  CDE and  CDX form a linear pair..   CDE +  CDX = 180°   CDE + 35° = 180°   CDE = 145° From equation (1), we obtain A  ABC +  BCD +  CDE +  DEF +  EFA +   FAB = 720° 130° + 90° + 145° + 100° + x + 95° = 720° x + 560° = 720° x = 160° Thus, the value of x is 160º. n

B

D P Here, AD = a, DC = 3a  A = 90°

 D =  B = 130° Thus, the other angles of the rhombus are 50°, 130°, and 130°. Let the given polygon has n sides. It is known that the sum of the measures of the external angles of any polygon is 360°.  n × 18° = 360°

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7th Class Mathematics

272 10. A quadrilateral whose diagonals are perpendicular to each other and one of the diagonals bisect the other is a kite. 11. The given quadrilateral PQRS can be drawn as: S R

P Q It is known that a quadrilateral with exactly one pair of parallel sides where the non-parallel sides are of equal lengths is called an isosceles trapezium. Thus, the given quadrilateral PQRS is an isosceles trapezium. 12. It is given that l || m.  AD || BC Since lines p and q intersect  at  point  O,  they  are not parallel to each other.  AB is not parallel to CD. Thus, quadrilateral ABCD is a trapezium. 13. The diagonals of parallelograms, rectangles and squares bisect each other. In a trapezoid, they do not bisect each other.

2.

3.

SUMM ATIVE WORKSHEET KEY

1

2

3

4

5

6

7

8

9

10

D

C

C

C

D

A

B

A

A

B

11 12

13 14 15 16 17

18

19 20

A

B

A

D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It can be observed that x  ADC  180 (Linear pair)  x + 85° = 180°  x = 95° Similarly,  BAY  65  180  BAY  115 Similarly,   VCW +  WCD = 180° 95° +  WCD = 180°  WCD = 85° It is known that the sum of all the exterior angles of a polygon is 360°. Y +  CBZ +  WCD = 360°   ADX +  BAY 95° + 115° + y + 85° = 360°  y = 65°  Thus, the values of x and y are  95°  and  65° respectively. www.betoppers.com

4.

It is known that in a parallelogram, opposite sides are equal. AD = BC = 32 cm [BC = 32 cm (Given)] It is also known that the diagonals of a parallelogram bisect each other.  OA = OC and OD = OB  4x + 7 = 3x + 13  4x – 3x = 13 – 7  x = 6  OD = 2x + 7 = (2 × 6 + 7) cm = (12 + 7) cm = 19 cm Then, BD = 2 × OD = (2 × 19) cm = 38 cm Thus, perimeter of  ABD = AB + BD + AD = 35 cm + 38 cm + 32 cm = 105 cm It is given that ABCD is a parallelogram.  AB||CD [In parallelogram, opposite sides are parallel]   BDC =  ABD [Pair of alternate interior angles]   ABD = 65° [  BDC = 65°] By applying angle sum property of triangles in   OAB, we obtain  OAB +  AOB +  ABO = 180°   OAB + 95° + 65° = 180°   OAB = 180° – (95° + 65°) = 180° – 160°   OAB = 20°   BAX =  CAX –  OAB = 130° – 20° = 110° ADC =   BAX [Pair of corresponding angles]   ADC = 110° Since ABCD is a parallelogram, its opposite angles are equal.   ADC = 60° =  ABC –––––––– (1)

A

B 600

600 D C E  ADC +  ACD +  DAC = 180° (Interior angles of a triangle)   60° + 60° +   DAC = 180°   DAC = 60° –––––––– (2)   DAC =  ADC =  ACD = 60° Hence,   ADC is equilateral. [All angles are equal] Since AB || CD and AC is transversal,

Quadrilaterals Solutions

5.

6.

273

  ACD =  CAB = 60° (Alternate interior angles) Similarly,   ACB =  CAD = 60° (BC || AD, AC is transversal)   ACB =  CAB =  ABC = 60° Hence,   ACB is also equilateral. [All angles are equal] Similarly,   BCE is also equilateral. Thus, AC = AB = 5 km (Sides of same equilateral   ACB) AC = BC = 5 cm ; BC = BE = 5 cm (Sides of same equilateral  BCE)  AC = AB = BE = EC = CD = DA = 5 cm Thus, perimeter of quadrilateral ABED = AB + BE + EC + CD + DA = 5 cm + 5 cm + 5 cm + 5 cm + 5 cm = 25 cm. Sum of the given two angles = 60° + 110° = 170° the sum of the four angles of the quadrilateral is 360° therefore, the sum of the remaning two angles = 360° – 170° = 190° But they are equal . Therefore, each angle =

1  190  95 2 AB is parallel to CD, we have B A

8.

Since adjacent angles in a parallelogram have their sum equal to 180°. If these angles are in the ratio 4 : 5. 4x + 5x = 180°  9x = 180°

x

180 9

Then they are 4 

9.

i.e., 80° and 100° therefore, the angles of a parallelogram are 80°,100°,80°,100° In a rectangle; diagonals bisect each other. D

C

A therefore AC = BD given that AC = 10 BD = 10cm

10.

7.

A

D

4 cm

B

6 cm

C

Hence, the perimeter is equal to 4 + 6 + 4 + 6 = 20cm

B

C

D C D A  C  180 and B  D  180 A   40  180 and B  80  180 A  140  B  100 Since opposite sides of a parallelogram are equal, the other two sides are 4cm and 6cm.

180 180 and 5  9 9

B 400

A Given that BAD  40 opposite angles of a rhombus are equal and the sum of any two adjacent angles is 180°. therefore BAD  40 then, BCD  40 we know that sum of adjacent angles is 180° i.e., A  D  180 D  180   A = 180° – 40° D  140 we know that, opposite angles are equal B  C  180 B  40  180 B  140 Therfore anglers of a rhombus are 40°,140°,40°,140° www.betoppers.com

7th Class Mathematics

274 11.

14. Consider the following rhombus with diagonals. Consider the right triangle BOC and apply pythagoras theorem as follows

A 600

B

5cm B

D

10 O A

400 C The sides of a rhombus are all equal. In particular AD = 5cm, in triangle ABD, AB = AD. So it is isosceles. Hence ABD  ADB but BAD  60 ; ABD  ADB  120 and ABD  ADB  60 hence triangle ABD is equilateral therefore BD = AC = 5cm. that is, the diagonal BD is 5cm. 12. Given PQRS is an isosceles trapezium S R

500 P P  50 To find, Q; R; S

10 D From BOS BO2 + OC2 = BC2 BC2 = 102 + 242 = 22 52 + 22 122 = 22(52 + 122) = 22(25 + 144) BC2 = 22(169) BC2 = 22(13)2 BC = 2(13) BC = 26 Perimeter = 4 × BC = 4 × 26 = 104 meters.

Q

HOTS WORKSHEET

Q  R  50 (base angles of an isosceles trapezium are equal) P  S  180 (consecutive angles at the end of non-parallel sides are equal) 50  S  180 S  180  50 ; S  130 similarily R  130 13. Given AB//DC D

C

24

24

C

KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

C

A

B

A

B

A

C

C

C

B

Q.no

11 12

13 14 15

16 17 18

Key

D

B

A

A

C

A

C

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

It is given that AD =

1  CD, and we know that 2

AB = CD

1 AB 2 It is also given that E and F are the mid-points of CD and AB respectively. Hence, AF = FB and DE = EC Thus, AD =

400 A

B

CAB  40 CAB  ACD  40 (alternate angles) www.betoppers.com

Quadrilaterals Solutions

275

1 1 CD   AF =  AB and DE = 2 2 Since AD =

1 AB, we obtain 2

AD = AF Similarly, AD = DE Now, AB = CD –––––––– (1)

2.

  DFE = 35° –––––––– (7) From the figure, it can be seen that  DFC =  DFE +  EFC From (4) and (7), we obtain  DFC = 35° + 55° = 90° Thus,   DFC = 90° Let ABCD be the given parallelogram. A

1 1 AB = CD 2 2  AF = DE –––––––– (2) Since AF = DE and AF || DE, AFED is a parallelogram (two opposite sides are parallel and equal). Therefore, the other two opposite sides will also be parallel and equal. Thus, AD = EF –––––––– (3) Using (1), (2), and (3), we obtain AD = EF = AF = DE Thus, AFED is a rhombus as all four sides are equal and opposite sides are parallel. Similarly, FBCE is a rhombus. Now,   ECB = 110°   EFB = 110° =  ECB (Opposite angles of a rhombus are equal) We also know that diagonals of a rhombus bisect the angles at the vertices through which they pass.

D

 

1   EFC = ×110° = 55° –––––––– (4) 2 Now, EF || BC (Opposite sides of a rhombus FBCE)   DEF = 110° (Corresponding angles) Now,   DEF +  EDF +  DFE = 180° (Interior angles of triangle DEF) 110° +  EDF +  DFE = 180°   EDF +  DFE = 70° –––––––– (5) G

A

F

H

B 1100

D E C Now, DF = EF (Sides of the same rhombus)   EDF =  DFE –––––––– (6) (Angles opposite to equal sides in a Δ are equal) From (5) and (6), we obtain 2  DFE = 70°

3.

B C Let BC = AB + 4 cm It is known that in a parallelogram, opposite sides are of equal lengths.  AB = CD and BC = AD Perimeter of parallelogram ABCD = 88 cm   AB + BC + CD + DA = 88 cm  AB + (AB + 4 cm) + AB + (AB + 4 cm) = 88 cm  4AB + 8 cm = 88 cm  4AB = 80 cm  AB = 20 cm  BC = (20 + 4) cm = 24 cm Thus, the length of the longer side of the parallelogram is 24 cm. Opposite angles of a parallelogram are of equal measures.   A =  C and  B =  D It is given that the measures of  A and  B are in the ratio 7:5. Let the measure of  A be 7x and the measure of  B be 5x.   A =  C = 7x,  B =  D = 5x In a quadrilateral, the sum of the internal angles is 360°.   A +  B +  C +  D = 360°  7x + 5x + 7x + 5x = 360°  24x = 360°

360  15 24   A =  C = 7 × 15° = 105°  B =  D = 5 × 15° = 75° Thus, the measures of  A,  B,  C, and  D are respectively 105°, 75°, 105°, and 75°. x

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7th Class Mathematics

276 4.

5.

The diagonals of a rectangle are equal and bisect each other.  AC = BD; AO = OC and BO = OD It is given that AO = 2.5 cm  AC = AO + OC = 2 × AO = 2 × 2.5 cm = 5 cm Also, each internal angle of a rectangle measures 90°. Thus, ΔABC is a right triangle. By applying the Pythagoras Theorem in ΔABC: AB2 +  BC2 = AC2  AB2 + (4 cm)2 = (5 cm)2  AB2 + 16 cm2 = 25 cm2  AB2 = 25 cm2 – 16 cm2 = 9 cm2  AB2 =  (3  cm)2  AB = 3 cm Thus, perimeter of rectangle ABCD = 2 (Length + Breadth) = 2 (AB + BC) = 2(3 cm + 4 cm) = 14 cm A

B

6.

7.

8. 9. 10.

D

C It is known that all sides of a rhombus are of equal lengths.  AB = BC = CD = DA It is also given that BD = BC  AB = BC = CD = DA = BD In ΔABD, AB = BD = AD Thus, ΔABD is an equilateral triangle. Similarly, ΔBCD is also an equilateral triangle. Thus,  both  ΔABD  and ΔBCD  are  equilateral triangles. Consider a square ABCD. A D

O

B

11. 12. 13. 14. 15.

Let the diagonals AC and BD intersect at point O. All sides of a square are of the same length. Also, the two diagonals of a square are also of the same length and bisect each other at right angles.   AB = BC = CD = DA Since the diagonals bisect each other, AO = OC and BO = OD. Also, AC = BD   AO = OC = BO = OD and   AOD = 90° Thus, ΔAOD is an isosceles right triangle as AO = OD and   AOD = 90°. It is clear that ΔAOD    ΔCOD    ΔBOC    ΔAOB. Hence, the diagonals of a square divide it into four congruent isosceles right triangles. The sum of the angles of quadrilaterals = (n – 2) ×180° = (4 – 2) ×180° = 2 ×180° = 60° A quadrilaterals has 4 sides Diagonal In  ABCD A  B  C  D  360 given B  D  180  A  C  180  360  A  C  360  180 = 180° Equal Rhombus Square Square D

400 A B In a parallogram opposite angles are equal  B  D, A  C  D  40 , here sum of the angles = 360° A  B  C  D  360 A  40  C  40  360  A  C  360  80 A  C  280 2A  280

A 

C 

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C

280  140 2

A  140, C  140

Quadrilaterals Solutions

277 4. 5.

16. Trapezium 17. Kite 18.

D

C

The diagonals of a square bisect each other at 90°. Let the angles of the quadrilateral be 2x, 3x, 4x, and 6x. By angle sum property of quadrilaterals: 2x + 3x + 4x + 6x = 360°  15x = 360°

360 15   x = 24° Thus, difference between the greatest and the smallest angle is given by: 6x – 2x = 4x = 4 × 24° = 96° Consider a square ABCD. A D x

X 120 A B Straight angle = 180° x + 120° = 180° x = 180° – 120° ; x = 60°

F

6.

IIT JEE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

D

A

B

B

B

A

A

C B

D

11

12

13 BC AC AC D

O

14 15 AB AB C C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. A quadrilateral in which opposite angles are equal and diagonals are bisecting each other is always a parallelogram. 2. In square and rectangle, diagonals are always equal and they bisect each other. Thus, the given quadrilateral ABCD is necessarily a square or a rectangle. 3. A

B

D

C It is known that all sides of a rhombus are of equal lengths.  AB = BC = CD = DA It is also given that BD = BC  AB = BC = CD = DA = BD In ΔABD, AB = BD = AD Thus, ΔABD is an equilateral triangle. Similarly, ΔBCD is also an equilateral triangle. Thus,  both  ΔABD  and ΔBCD  are  equilateral triangles.

7.

B C Let the diagonals AC and BD intersect at point O. All sides of a square are of the same length. Also, the two diagonals of a square are also of the same length and bisect each other at right angles.  AB = BC = CD = DA Since the diagonals bisect each other, AO = OC and BO = OD. Also, AC = BD   AO = OC = BO = OD and   AOD = 90° Thus, ΔAOD is an isosceles right triangle as AO = OD and   AOD = 90°. It is clear that ΔAOD    ΔCOD    ΔBOC    ΔAOB. Hence, the diagonals of a square divide it into four congruent isosceles right triangles. It is given that the measure of each interior angle of the polygon is 108°.  Measure of each exterior angle of the polygon = 180° – 108° = 72° The sum of the measures of all exterior angles of any convex polygon is 360°. Thus, number of exterior angles of the

360 5 72 Since the polygon has five exterior angles, it also has five sides. Hence, the polygon is a pentagon. polygon  

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7th Class Mathematics

278 8.

A polygon which is both equiangular and equilateral is called a regular polygon. Among the given quadrilaterals, only a square has all its sides of the same length and all angles of the same measure (90°). Since a square is both equiangular and equilateral, it is a regular polygon. 9. A rhombus is a quadrilateral in which all sides and angles are of equal measure. The diagonals of a rhombus act as perpendicular bisectors, i.e., they bisect each other at right angles. 10. Since the width and length of the rectangle are in the ratio 3:4, let the width be 3x cm and the length be 4x cm.  2(3x + 4x) = 28 [Perimeter = 28 cm]  14x = 28  x = 2  Length = 3 × 2 cm = 6 cm and width = 4 × 2 cm = 8 cm Thus, area of the rectangle = 6 cm × 8 cm = 48 cm2 11. Consider the parallelogram PQRS: P

S

Q R Here, PQ||RS and PS||QR We know that for two parallel lines, the pair of interior angles on the same side of the transversal is supplementary. Thus, in parallelogram PQRS:  PQR +  QRS = 180°  QRS +  RSP = 180°  RSP +  SPQ = 180°  SPQ +  PQR = 180° Thus, in a parallelogram, each pair of adjacent angles is supplementary. Thus, statement I is correct. In a parallelogram, opposite sides are of equal lengths. However, it is not necessary that all sides are of the same length. Thus, statement III is correct. Thus, statement I and III are correct. 12. A polygon is said to be convex if all its diagonals lie inside the polygon. On the other hand, a polygon is said to be concave if at least one diagonal lies outside the polygon. Row P: DiagonalsAC and BD can be drawn for the polygon as:

A

B C Row R: The diagonals for the polygon can be drawn as:

A B

H

C

G D

13.

14.

16. 17. 18. 19.

20.

F

E Since all four diagonals lie outside the polygon, the figure represents a concave polygon. Thus, rows P and R contain correctly matched information. A rectangle is a quadrilateral in which opposite sides are of equal lengths but not all sides are equal. All interior angles measure 90° and there are two pairs of parallel lines. Its diagonals bisect each other, but not at 90°. The diagonals of parallelograms, rectangles and squares bisect each other. In a trapezoid, they do not bisect each other. AD || BC NO NO Let the regular polygon have n-sides. Thus, sum of the interior angles of the polygon = n × 120º The sum of the interior angles of a polygon of nsides is (n – 2) 180º.  n × 120º = (n – 2) × 180º 2n = 3 (n – 2) 2n = 3n – 6 n = 6 Thus, the given polygon is a regular hexagon. Let ABCD be the rhombus such that AB = BC = CD = AD = 5 cm and AC = 8 cm D C

O

A www.betoppers.com

D

B

Quadrilaterals Solutions

279

In a rhombus, diagonals are perpendicular bisectors of each other.

8 cm = 4 cm 2 In ΔAOB,   AOB = 90°  AB2 = AO2 +  OB2 [Pythagoras  theorem]  (5 cm)2 = (4 cm)2 + OB2  OB2 = (25 – 16) cm2 = 9 cm2  OB = 3 cm  BD = 2OB = 2 × 3 cm = 6 cm Thus, the length of the other diagonal is 6 cm. 21. It is given that the measure of each interior angle of the polygon is 108°.  Measure of each exterior angle of the polygon = 180° – 108° = 72° The sum of the measures of all exterior angles of any convex polygon is 360°. Thus, number of exterior angles of the  AO = OC =

360 5 72 Since the polygon has five exterior angles, it also has five sides. 22. An interior angle and its corresponding exterior angle always form a linear pair for any twodimensional figure. Measure of each exterior angle of the regular polygon = 90° Thus, measure of each interior angle of the polygon = 180° – 90° = 90° Let the polygon be of n sides. polygon  

23.

24. 25. 26.

n2  180  90 n (n – 2) × 2 = n 2n – 4 = n n=4 Thus, the given polygon is of 4 sides. The lengths of the diagonals of the rhombus are 6 cm and 8 cm It is known that the diagonals of rhombus bisect each other at an angle of 90 . So, AC and BD must bisect each other at an angle of 90. AO = 4, OB = 3 Now in right angled triangle OAB, we have AB2 = AO2 + OB2 = 42 +  32    AB2 = 25   AB =5 8 sides A – r; B – p; C – q A – r; B – p; C – s

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280

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7th Class Mathematics

10. SYMMETRY SOLUTIONS

FORMATIVE WORKSHEET

4.

The line of symmetry (or the axis of symmetry) of a figure is the line that divides the figure into two equal halves in such a way that the two halves are mirror images of each other. There is no line of symmetry in a parallelogram.

5.

A rhombus has rotational symmetry of order 2 about its centre:

KEY Q.no

1

2

3

4

5

6

7

Key

D

A

A

C

C

D

D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

2.

A scalene triangle has no rotational symmetry. Rest all the given figures are symmetric with respect to the rotational point drawn at their centers. Order of rotational symmetry of a figure is the number of different positions of the figure in which it looks exactly the same when it is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then the figure is rotated on a tracing paper about its centre. rotated by 120°

rotated by 120°

180°

A 180°

A

6.

The two lines of symmetry meet at the central point. Therefore, the central point is the centre of rotational symmetry of the parallelogram.

7.

The order of rotational symmetry of a figure is the number of different positions of the figure in which the figure looks exactly the same when the figure is rotated about the centre of the figure. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then, the figure is rotated on a tracing paper about its centre.

rotated by 120°

Here, it is seen that there are 3 different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the figure given in alternative A is 3. 3.

A

If an object looks exactly the same after a rotation, then it has a rotational symmetry. In a complete turn (of 360°), the number of times an object looks exactly the same is called the order of rotational symmetry. For an object to have a rotational symmetry of order more than 1, the angle of rotation should be an exact divisor of a complete angle i.e., 360°. It can be observed that 25º is not an exact divisor of 360° while all the other given angles are factors of 360º. Thus, for an object to have a rotational symmetry of order more than 1, it cannot have the angle of rotation as 25°.

rotated by 180°

rotated by 180°

I II III Here, it is seen that there are two different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the given figure is 2.

7th Class Mathematics

282

CONCEPTIVE WORKSHEET

6.

KEY Q.no

1

2

3

4

5

6

7

Key

D

A

D

B

B

A

B

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Only the hexagon, out of the given figures, has rotational symmetry. All the points in a hexagon are symmetrically situated around the rotational axis passing through the centre. 2. Order of rotational symmetry of a figure is the number of different positions of the figure in which it looks exactly the same when it is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then the figure is rotated on a tracing paper about its centre. rotated by 180°

N N 3.

4.

rotated by 180°

N

Here, it is seen that there are 2 different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the figure given in alternative A is 2. If a figure has rotational symmetry of order more than 1, then its angle of rotation is always a factor of one complete rotation i.e., 360°. Among the given measures, only 72° is a factor of 360° (360° = 5 × 72°). Thus, among the given measures, the angle of rotation of the figure can possibly be 72°. A scalene triangle does not have any line of symmetry. However, it has rotational symmetry of order 1 about its centre:

A

A

360° 5.

The figure drawn in alternative B has rotational symmetry of order 2: A A

180°

180°

A www.betoppers.com

The order of rotational symmetry of a figure is the number of different positions of the figure in which the figure looks exactly the same when the figure is rotated about the centre of the figure. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then, the figure is rotated on a tracing paper about its centre.

Rotated by 180°

7.

Rotated by 180°

Here, it is seen that there are two different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the given figure is 2. The line of symmetry of a figure is the line that divides the figure into equal halves in such a way that the halves are mirror images of each other. Letter Z does not have a line of symmetry. However, It, has rotational symmetry of order 2:

Q

Q

Z Z Z 180°

180°

Q

SUMM ATIVE WORKSHEET KEY Q.no

1

2

Key

D

B C

3

4

5

6

7

8

C A

B

A

C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It can be seen that there is no line of symmetry for the figure given in alternative D and the order of rotational symmetry of the figure is two.

Symmetry

283 5.

180° rotation

2.

180° rotation

It can be seen that for the figure given in alternative A, the rotational angle is 180°. Thus, its order of rotational symmetry is two.

180° rotation

180° rotation

Here, since the angle of rotation is 180°, the order of rotation is two. Thus, only in alternative D, the number of lines of symmetry is not equal to the order of rotational symmetry. The figure drawn in alternative B has rotational symmetry of order one:

6.

Thus, alternative A is correctly matched. Y

3

x(–2 , 3)

2 1 X

–3 –2 –1

1 –1

x(–2 , –3)

–2

rotated by 360°

2

3

4

–3 3.

It can be seen that in figure 2, the rotational angle is 180°. Thus, its order of rotational symmetry is two. 180° rotation

7.

180° rotation

4.

Rotated by 90°

X O 1 2 3 4 5 –1 –2 –3 –4 –5 –6 –7 The image of (5, 7); reflected in y-axis is (–5, 7).

180° rotation

Rotated by 90°

(5, 7)

–5 –4 –3 –2 –1

Thus, in figures 2 and 3, the order of rotational symmetry is two. Order of rotational symmetry of a figure is the number of different positions of the figure in which the figure looks exactly the same when the figure is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then, the figure is rotated on a tracing paper about its centre. Rotated by 90°

7 6 5 4 3 2 1

(–5, –7)

Similarly, in figure 3, the rotational angle is of 180°. Thus, its order of rotational symmetry is two. 180° rotation

The image of (–2, 3) reflected in x-axis is (–2, –3). Y

Rotated by 90°

Here, it is seen that there are four different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the figure given in alternative C is 4.

8.

Y 5 –5

Y Rotated clockwise 90°

(2, 6) 5

5

X –5

X 5 (6, –2)

When rotated through 90o clockwise p(x,y); p| (y – x). Therefore the new position of the point (2, 6) about the origin is (6, –2).

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7th Class Mathematics

284

HOTS WORKSHEET

Rotated by 72°

Rotated by 72°

Rotated by 72°

Rotated by 72°

KEY Q.no

1

2

3

4

5

Key

D

D

A

C

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

Rotational symmetry is symmetry with respect to some or all rotation axis.

Clearly, the figure 2.

3.

Rotated by 72° Here, the regular pentagon is rotated by 72° five times. The figures appear to be the same in each of these five positions. Five times 72° implies 360° or a complete rotation. Hence, a regular pentagon has rotational symmetry of the order five. The figure in alternative A can be rotated as

has no rotational symmetry..

If an object upon rotation looks exactly same as the original figure, then the figure is said to show rotational symmetry. Alphabets A and T do not show any rotational symmetry. Alphabets H and S show rotational symmetry of second order.

5.

Rotated by 90°

The order of rotational symmetry of a figure is the number of different positions of the figure in which the figure looks exactly the same when it is rotated about its centre.

Rotated by 90°

Rotated by 90°

To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. Then, the figure is rotated on a tracing paper about its centre.

These figures can now be combined to obtain the given pattern as

Z Z Z rotated by180°

rotated by180°

Here, it is seen that there are two different positions in which the figure looks exactly the same.

4.

Thus, the order of rotational symmetry of the given figure is 2. If an object, upon rotation, looks exactly same as the original figure, then the figure is said to show rotational symmetry. This order is the number of positions in which the figure appears to be the same when rotated from an angle of 0° to 360°.

IIT JEE WORKSHEET KEY 1

2

3

4

5

6

7

8

9

10

D

B

A

A

D

B

C

D

C

A

13 AB D

14 AB D

11 12 AC BC D D

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Symmetry

285

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS

Reflected Y image

6.

(–2, 7)

5

–5

1.

(2, 7) X

5

A square has 4 line symmetry. 7. Y

Y Rotated 5 clockwise 90° X (5, –1) (–1, –5)

5

A rectangle has 2 line symmetry. 2.

–5

X

8. Y

Y (8, 4)

5

A rhombus has 2 line symmetry 3.

A

5

B

5 X –5

–5

(4, –8)

A

9.

D

Rotated clockwise 90° X

B

C

A parallelogram has no line symmetry 4. 5.

S The capital letter ‘S’ has no line symmetry Y Reflected image

5

D

C

(4, 7) 10.

5

X

–5 (4, –7)

The capital letter ‘O’ has 2 line symmetry. 11. It is given that the figure when rotated at 90° looks exactly same as its original position. Therefore, this will also happen at all the multiples of 90°.

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7th Class Mathematics

286 12. The order of rotational symmetry of a figure is the number of different positions of the figure in which it looks exactly the same when it is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. 13. (A,B,D) 14. The correct answers (A,B,D) 15. The given figures can be completed as follows.

It will be a square. 16. The given figures can be completed as follows.

20. The given figures can be completed as follows.

It will be an octagon. 21. The given figure has rotational symmetry of order 3.

22. The angle of rotation of the figure is 45°. The order of rotational symmetry of the figure is

360 8 45 Thus, the figure has rotational symmetry of order 8. 23. The order of rotational symmetry of a figure is the number of different positions of the figure in which it looks exactly the same when it is rotated about its centre. To find the order of rotational symmetry of the given figure, a dot is placed at the corner of the figure to identify different positions. given by

It will be a triangle 17. The given figures can be completed as follows.

It will be a rhombus. 18. The given figures can be completed as follows.

Rotated by

Rotated by

120°

120°

Rotated by 120°

It will be a circle. 19. The given figures can be completed as follows.

Here, it can be seen that there are three different positions in which the figure looks exactly the same. Thus, the order of rotational symmetry of the given figure is three. 24. The given figure has the rotational symmetry of order 2. Rotated by 180°

25. A – q; B – r; C – p; D – w; E – v

 It will be a pentagon. www.betoppers.com

11. MENSURATION SOLUTIONS 3.

FORMATIVE WORKSHEET Key

1

2

3

4

5

6

7

8

9

10

C

*

*

D

B

D

D

*

*

*

11 12

13 14 15 16 17

18

19 20

C

C

*

*

A

B

B

B

B

Applying Pythagoras theorem in  ABC: BC2 = AB2 + AC2  BC2 = (7 cm)2 + (24 cm)2

 BC2 = 49 cm2 + 576 cm2  BC2 = 625 cm2 = (25 cm)2  BC = 25 cm Area of triangle is given by:

2.



A=

b 4a 2  b 2 4

154 4  (85)2  (154)2 m2 4

A = 2772 m2 Hence, the area of the given triangle is 2772 m2. 

2

3 2 3 p p2 a     4 4  3  12 3 2

p p2 area of square = a2 =     4  16 area of regular hexagon = 2

3 3 a2 3 3  p  p2     2 2 6 8 3 So, area of equilateral triangle : square : regular hexagon

p2

p2 p2 : = 12 3 16 8 3

1 × Base × Height 2

1 1 × AB × AC = × BC × AD 2 2 1 1 4. × 7 cm × 24 cm = × 25 cm × AD 2 2 7  24  AD   6.72cm 25 Thus, the length of AD is 6.72 cm. Assuming the sides of the triangle as a, b and c a + b + c = perimeter of triangle = 324  85 + 154 + c = 324  c = 85 Here, a = c = 85, b = 154, so, it is an isosceles triangle. Using the formula, Area =

p p each side of regular hexagon = a = 4 6 Now, area of equilateral triangle = a=

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

Let the perimeter of each polygon = p then each side of equilateral triangle a = p/3 each side of square

:

= 16 × 8 3 : 12 3 × 8 3 : 16 × 12 3 = 4:3 3 :6 In  ABD, AB2 = (26 cm)2 = 676 cm2 Now, BD2 + AD2 = (24 cm)2 + (10 cm)2 = 676 cm2  AB2 = BD2 + AD2 Therefore, by converse of Pythagoras theorem,  ABD is a right-angled triangle.  Area of  ABD =

=

1 × BD × AD 2

1 × 24 cm × 10 cm = 120 cm2 2

Now, area of  BCD =

1 × BD × CD 2

1 × 24 cm × 7 cm = 84 cm2 2  Area of quadrilateral ABCD = Area of  ABD + Area of  BCD = 120 cm2 + 84 cm2 = 204 cm2 Thus, the area of quadrilateral ABCD is 204 cm2. =

7th Class Mathematics

288 5.

3  Length of rectangle ABCD =  x  1 cm 4   x  Breadth of rectangle ABCD =   4  cm 2  Perimeter of rectangle ABCD is given by: 2 (Length + Breadth)

7.

 Area of trapezium PQRX = Area of parallelogram PQRS – Area of  PXS = 30 cm2 – 6 cm2 = 24 cm2 Thus, the area of trapezium PQRX is 24 cm2. The given information can be represented by a figure as: 5 cm 5 cm

 3  x   54 cm  2  4 x  1   2  4   cm    

50 cm

 3x x    54  2     3  4 2  

65 cm

Number of square cardboards that can be placed along the length of the rectangular cardboard

 3x  2x   54  2   3  4   54  2 

65cm  13 5cm Number of square cardboards that can be placed along the breadth of the rectangular cardboard 

5x 6 4

5x 2  x  24  60 

3



 Length of rectangle ABCD   4  24  1 cm  

= (18 + 1)cm = 19 cm

6.

50cm  10 5cm  Number of square cardboards that can be placed over the rectangular cardboard = 13 × 10 = 130 Let the length of BE be y cm. Let the height from D to the base AB be h cm. 

 24  Breadth of rectangle ABCD    4  cm  2  = (12 – 4) cm = 8 cm  Area of rectangle ABCD = Length × Breadth = 19 cm × 8 cm = 152 sq cm It is known that the area of a parallelogram is the product of its base and the corresponding height.  Area of PQRS = PS × RY = SR × PX  5 cm × 6 cm = 10 cm × PX 6cm  5cm  3cm 10cm In  PXS, applying Pythagoras theorem: PS2 = PS2 + SX2 (5 cm)2 = (3 cm)2 + SX2  SX2 = 25 cm2 – 9 cm2  SX2 = 16 cm2  SX = 4 cm

8.

Area of  ADE = 

3 × area of ABCD 4

1 3 × (6 + y) × h = ×6×h 2 4 6+y=2×

3 ×6 4

6+y=9 y=3 The length of BE is 3 cm. D

C

 PX 

 Area of  PXS =

× PX =

1 1 × Base × Height = × SX 2 2

1 × 4 cm × 3 cm = 6 cm2 2

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h

A

F

B

E

Mensuration 9.

289

Area of trapezium ABCD =

1 × (AB × DC) × DE 2

1 × (9 + 15) × 8 = 96 cm2 2 10. Let the length of CD be t cm. =

Area of ABCD =

1 × (AB + CD) × depth 2

1 × (50 + t) × 36 = 2250 2 2250 18 50 + t = 125 t = 75 The length of CD is 75 cm. 11. The distance covered by a wheel when it is rotated once is equal to the circumference of the wheel. Radius of the wheel, r = 2 m 80 cm = 2 m + 0.8 m = 2.8 m  Circumference of the wheel = 2  r 50 + t =

22 2.8  m  8.8m 7 2 Thus, distance covered by the wheel when it is rotated 12 times = 12 × 8.8 m = 105.6 m = 105 m 60 cm 12. Let O be the centre of the circle. Join PR. This can be done as:  2

A

B P

Q O

S D

R C

Let each side of square ABCD be x units long. Now, the diameter of the circle equals the length of each side of square ABCD.  Diameter of the circle = x

x 2 In square PQRS, diagonal PR = x Let the length of each side of square PQRS be a units. Applying Pythagoras theorem in right-angled  PQR: PQ2 + QR2 = PR2 x  a2 + a2 = x2  2a2 = x2  a  2  Radius of the circle =

Now, perimeter of square ABCD = 4 × x = 4x Perimeter of square PQRS x  4 a  4  2 2x 2 Perimeter of PQRS  Required ratio  Perimeter of ABCD 2 2x 1   4x 2 Thus, the ratio of the perimeters of squares PQRS and ABCD is 1: 2 . 13. Let r1 and r2 be the radius of the two circles. It is known that the circumference of a circle is given by 2ðr.  Circumference of circle with radius r1 = 2  r1 Circumference of circle with radius r2 = 2  r2 It is given that the ratio of circumference of two circles is 7: 11.



2r1 7  2r2 11



r1 7  r2 11

It is known that the area of a circle is given by  r2 . =  Ratio of areas of the circles 2

2

2r12  r1   7  49      2 2r2  r2   11  121 Thus, the ratio of the areas of the two circles is 49: 121. 14. Circumference of the garden = 88 m It is known that the circumference of a circle is given by 2  r.  2  r = 88 m

88  7 m  14m 2  22  Area of the garden, r

22  14  14m 2 = 616 m2 7 Area of the garden including path r 2 

=

22 22 6358 2 2  14  3 m 2   17  17m 2  m 7 7 7  Area of the path = 6358 2 2046 2 m  616m2  m  292.28m 2 7 7 Thus, the area of the path surrounding the garden is 292.28 m2. www.betoppers.com

7th Class Mathematics

290 15. It can be observed from the given figure that the diameter of the circle equals the side of the square i.e., 14 cm.  Diameter of the circle = 14 cm  Radius (r) of the circle = 14 cm ÷ 2 = 7 cm Area of the square = Side2 = (14 cm)2 = 196 cm2 Area of the circle = =  r2

18. The circular grass plot ABC has a ring of gravel outside it. Radius of grass plot = r = 40 m Width of gravel = W = ? It has been given that area of grass = area of gravel. el av Gr

22  7cm  7cm  154cm 2 7 Area of the shaded region = Area of square –  Area of circle = (196 – 154) cm2 = 42 cm2 A

P

C

D

Q

2 cm N

O L

B 10cm

Diameter of the semi-circular portion = 14 cm

14 cm  7cm 2 BL = BC – LC = 10 cm – 2 cm = 8 cm  NO = 8 cm It can be observed in the figure that: Area of the shaded region = Area of ABLM – Area of PNOQ + Area of semi-circle EDC = AB × BL – PN × NO 1 + ×  × (Radius)2 = 14 cm × 8 cm – 8 cm × 2 2 1 cm + ×  (7cm)2 2 Radius of the semi-circular portion =

 112cm2  16cm2 

  

22  49cm 2 = 96 cm2 + 77 14

cm2 = 173 cm2 Thus, the area of the shaded portion is 173 cm2. 17. Let the radius of the smaller circle be r. Area of smaller circle = 256  cm2   r2 = 256  cm2  r2 = 16 cm × 16 cm  r = 16 cm Radius of bigger circle = r + 2 cm = 16 cm + 2 cm = 18 cm  Area of bigger circle =  (18 cm)2 = 324  cm2 Thus, the area of the bigger circle is 324  cm2. www.betoppers.com

Gr

av

el

r2

=  W (2r + W) + 2rW – r2 = 0 2 W + 80W – 1600 = [Since r = 40] 

W2

80  802  4  1600 = 16.57 m 2 (considering the + ve value). Hence, the width of the gravel is 16.57 m. 19. Let AKBL be the circular grass plat whose circumference = C m A man walks from A to B through (i) point k (i.e., along circumference) travelling a 

C

B

r = 40 W

M E 2 cm

14cm

W

Grass plot



16.

A

W =

C metre = S1 and 2 (ii) point O (i.e., along diameter) travelling a distance = AKB =

C metre  (since circumference =  × diameter) = S2 (say). Walking speed of the person = V = 90 metre/ min. distance = AOB =

90 metre/sec. 60 Using the formula distance = speed × time, (S1 – S2) = V(t1 – t2) =

 C C  90   × 45  2   60 [Since t1 – t2 = 45 sec. (given)]

 

2 135  11 2  C = 371.25 m Hence, the circumference is 371.25 m. C×

Mensuration

291

20. Let the circumference of outer circle C1 = 62.832 m circumference of inner circle C2 = 37.6990 m The area between two circles = A1 – A2 = ? Here, using the correlation formula for circumference and area,

2.

Area (A) =



2

A=

C , 4



C2 C 2 A1 – A2 = 1  2 4 4 



A1 – A2 =



C12

Assume that the side of equilateral triangle is a metre

3 2 a rupees 4 Similarly, perimeter (P) = 3a m Cost of fencing = 25 × 3a rupees According to the equation.

Cost of paving = 10 ×

10 ×

 C 22 4

3 2 2 a m 4

3 2 a = 25 × 3a 4

a = 10 3 m a = 17.32 m Hence, the side of the equilateral triangle is 17.32 m. Let the equal side of the isosceles triangle = a the base of the isosceles triangle = b  

(C1  C2 )(C1  C2 )  A1 – A2 = 4  A1 – A2

3.

(62.832  37.6992)(62.832  37.6992) = 4  3.1416

5 b 8 Now, perimeter of isosceles triangle = 2a + b

100.5  25.13  A1 – A2 = 12.5664  A1 – A2 = 201 m2 Hence, the area between the circles is 201 m2.

CONCEPTIVE WORKSHEET

then

a=



308 = 2 ×

5 b+b 8

1

2

3

4

5

6

7

8

9

10

b = 136 5 then a= × 136 = 85 8 Using the formula

B

*

*

C

D

D

B

*

*

*

Area of isosceles triangle =

13 14 15 16 17

18

19 20



Key

11 12 A

B

D

B

B

A

*

*

*

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Length of side of triangle = 250 m Area of triangle = 1.75 hectares = 1.75 × 10,000 m2 = 17,500 m2 Area of triangle 

 17,500 m2 =

1 × Base × Altitude 2

1 × 250 m × Altitude 2

17,500  2 m  140m 250 Thus, the length of altitude of the triangle is 140 m.  Height = 



m2

4.

A=

b 4a 2  b 2 4

136 4  (85)2  (136)2 m2 = 34 × 102 4

= 3468 m2 Hence, the area of the isosceles triangle is 3468 m2. It is given that x: y: z = 2: 3: 1 Thus, let x be 2a; y be 3a, and z be a.  Area of rectangle ABCD = AB × AD = (x + y) (y + z) = (2a + 3a) (3a + a) = 5a × 4a = 20a2 Since  A =  B =  C =  D = 90°, each of the triangles MAN, MDP, OCP, and NBO is a rightangled triangle. Area of  MAN 1 1 1   AN  AM   x  y   2a  3a  3a 2 2 2 2

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7th Class Mathematics

292

 DF2 = 13.69 cm2 – 1.44 cm2  DF2 = 12.25 cm2  DF = 3.5 cm Area of the given figure = Area of BCDE + Area of  ABC + Area of  EDF

Area of  MDP 1 1 1 3a 2  MD  DP   z  y   a  3a  2 2 2 2 Area of  OCP 1 1 1   OC  CP   2z  x   2  a  2a  2a 2 2 2 2 Area of  NBO 1 1 1   NB  BO   y  x   3a  2a  3a 2 2 2 2 Now, area of non-shaded region = ar (  MAN) + ar (  MDP) + ar (  OCP) + ar (  NBO) 

3a 2 19a 2  2a 2  3a 2  2 2  Area of shaded region = Area of rectangle – Area of non-shaded region  3a 2 

1 1  BC2   BC  AB   DE  DF 2 2 1 2  1.2cm    1.2cm  0.5cm 2

7.

19a 2 21a 2  2 2 Area of shaded region  Required ratio  Area of rectangle  20a 2 

5.

6.

21a 2 21a 2 1 21  22    2 2 40 20a 20a Thus, the ratio of the area of the shaded region and that of rectangle ABCD is 21: 40. It can be observed that:  ACD =  BAC = 90° However, these are alternative interior angles. Therefore, AB is parallel to DC. It is given that AB = CD i.e., a pair of opposite sides of quadrilateral ABCD is equal and parallel. Therefore, ABCD is a parallelogram.  Area of ABCD = Base × Height = AB × AC = 4 cm × 6 cm = 24 cm2 Thus, the area of the given quadrilateral ABCD is 24 cm2. It is given that BCDE is a square of side 1.2 cm.  Area of square BCDE = (1.2 cm)2 = 1.44 cm2 Applying Pythagoras theorem in  ABC: (AB)2 = (AC)2 – (BC)2  AB2 = (1.3 cm)2 – (1.2 cm)2  AB2 = 1.69 cm2 – 1.44 cm2  AB2 = 0.25 cm2  AB = 0.5 cm Again applying Pythagoras theorem in  DEF: EF2 = ED2 + DF2 (3.7 cm)2 = (1.2 cm)2 + DF2

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1   1.2cm  3.5cm 2 = 1.44 cm2 + 0.3 cm2 + 2.1cm2 = 3.84 cm2 Thus, the area of the given figure is 3.84 cm2. Length of the side of square = 24 cm Area of square = Side × Side = 24 cm × 24 cm Length of side of parallelogram = 36 cm Area of parallelogram = Base × Altitude  36 cm × Altitude = 24 cm × 24 cm 24  24 cm = 16 cm 36 Thus, the length of altitude of parallelogram corresponding to given side is 16 cm. Area of ABCD = Area of trapezium AECD + Area of  BCE =

 Altitude = 

8.

9.

1 1   2  (4  6)  5  2  3  6  = 25 + 9 = 34 cm2 Area of the sign board = Area of trapezium ABCD – Area of  ADE

1 1  =   (6  8)  (4  3)  8  3 = 49 – 12 = 37 2 2  2 cm 10. (a) Area of the parallelogram= base × height = AB × DE =8×6 = 48 cm2 (b) If we take AD as the base, then BF is the corresponding height. ThenAD × BF = 48 6.4 × BF = 48 48 6.4 = 7.5 cm The length of BF is 7.5 cm. BF =

Mensuration

293

11. Diameter of circle = 21 cm  Circumference of circle

15. The circular garden and footpath can be drawn as:

22  21cm = 66 cm 7 The perimeter of the equilateral triangle is equal to the circumference of the circle.  3 × Side = 66 cm =  × Diameter 

66 cm = 22 cm 3 Thus, the length of the equilateral triangle is 22 cm. 12. Area of the given figure = Area of rectangle ABCD + Area of semi-circle CPD

21 m 2

1.4m

 Side 

1 = Length × Breadth + ×  × (Radius)2 2 = AB × BC +

1 1  ×  ×   AB  2 2 

2

 = AB × BC + AB2 8 Thus, the expression given in alternative B correctly represents the area of the given figure. 13. Let r be the radius of the circle.  2  r = 110 cm 22  2   r  110cm 7 110  7 r cm 22  2 35  r  cm 2  Area of the circle 22 35 35 2 =  r2   cm  cm  962.5cm 7 2 2 14. It can be observed in the given figure that the garden is in the shape of a rectangle of dimensions 10.5 m × 7 m and has a semi-circle of diameter 7 m at its one edge. 7  Radius of the semi-circular portion, r  m 2 Perimeter of garden = 10. 5 m + 7 m + 10.5  m+ r 22 7  28m   m = 28 m + 11 m 7 2 = 39 m Thus, the perimeter of the garden is 39 m.

Diameter of the garden = 21 m  Radius of the garden, r 

21 m  10.5m 2

Width of footpath = 1.4 m  Radius of garden including footpath, R = 10.5 m + 1.4 m = 11.9 m  Area of footpath = Area of garden including footpath – Area of garden  R 2  r 2



2

  11.9m   10.4m 



2



22 141.61  110.25 m2 7

22  31.36m 2 = 98.56 m2 7 Thus, the area of the footpath is 98.56 m2. 16. Let the radius of the circle be r.  Diameter = 2r  Circumference = 2  r 2r  2r  45cm 

22    r  2   2   45cm 7    r

30  45cm 7

45  7 21 cm  cm 30 2 Area of circle =  r2  r 



22 21 21 2   cm  346.5cm2 7 2 2

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7th Class Mathematics

294 17. EC = AB (opposite sides of parallelogram) = 24 cm

1  radius of semicircle CDE = EC 2 = 12 cm Perimeter of the figure = Length of arc CDE + EA + AB + BC

1  =   2 12  13  24  13  2  = (12  + 50) cm Area of the figure = Area of semicircle CDE + Area of parallelogram ABCE 1  2 =     12  24  12  2  2 = (72  + 288) cm

perpendicular = AD =

AC 2  DC 2

AD =

50 2  10 2 m AD = 48.98 m

Now, area of quadrilateral ABCD + area of  ADC =

= area of  ABC

1 1 × AB × BC + × AD . 2 2

DC

1 1 × 30 × 40 + × 48.98 × 10 m2 2 2 = 600 + 244.9 m2 = 844.9 m2. 20. Let ABC be the equilateral triangle whose each side = a. =

A

18. Using the formula Area of quadrilateral =

O r

(S  a) (S  b) (S  c) (S  d)

where

a = 36 m, b = 77 m, c = 75 m, d = 40 m

a  b  c  d 36  77  75  40  = 114 2 2 Hence, area of quadrilateral and S =

(114  36) (114  77) (114  75) (114  40) m2

78  37  39  74 m2 = 37 × 39 × 2 m2 = 2886 m2 19. In the given quadrilateral ABCD, the diagonal = AC = diameter of the circle So,  ADC =  ABC = 90o 10 90

C

o

p

40 = height of DABC 90

A

30

2 3

B

AB2  BC 2 = 30 2  40 2 m = 50 m Again, in right-angled triangle ADC,

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a



10



a = 20 3

2 3

Area of equilateral triangle =

3 2 a 4

3 × (20 3 )2 cm2 4 = 519.6 cm2

=

SUMM ATIVE WORKSHEET

o

In right-angled triangle ABC AC =

a

radius of inscribed circle =

=

D

C

B

The radius of inscribed circle = r = 10 cm using the formula for equilateral triangle

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Number of smaller squares that can be placed along

30cm 6 5cm  Number of smaller squares that can be placed over the bigger square = 6 × 6 = 36 The correct answer is C. each side of the bigger square 

Mensuration 2.

3.

4.

5.

Let the breadth of the rectangular field be x. Length of the rectangular field = 180 m It is given that the perimeter of the rectangular field is 500 m.  2 (Length + Breadth) = 500 m  2 (180 + x) = 500  180 + x = 250  x = 70  Breadth of the rectangular field = 70 m  Area of the rectangular field = Length × Breadth = 180 m × 70 m = 12600 m2 The correct answer is B. We have, l = 7m, b = 5m ; h = 4m A = 2h (l + b) = 2(4)(7 + 5) = 8(12) A = 96 square meters Hence total area of the four walls = 96 square meters. We have l = 9m, b = 7m; and h = 6m The lateral surface area = 2h (l + b) = 2(6) (9 + 7) = 12(16) = 192 Square meters. The total surface area = 2 (l b + bh + hl ) = 2[(9)(7) + (7)(6) + (6)(9)] = 2[63 + 42 + 54] = 2(159) = 318 sq meteres If ‘r’ is the radius of the circle we have, c = 4 2 r  4  r=2 Hence area of the circle = r 2    2 

6.

295 7.

A

17cm

B

15cm AB2 + BC2 = AC2 AB2 = AC2 – BC2 = 172 – 152 = 289 – 225 = 64 AB = 8cm

8.

a 2 16  b 2 36

2

a  4     b 6

2

2

x A  r 2 . 360 1 1 But it is th of the area of the circle so A  r 2 6 6

a 4 2   b 6 3 Therefore

9.

ratio

of

their

perimeter

=

4a a 2    2:3 4b b 3 Let ABCD be a parallelogram in which AB = 30m, BC = 14m, AC = 40m

D

C

x 1  r 2  r 2 360 6

360 6 x = 60 Hence the angle of the sector is 60°. x

C

1 1 Therefore Area   base  height   15  8 = 2 2 15 × 4 Area = 60 cm2 Let the sides of two squares be a and b respectively. Then

 4  Sq.units Let the radius and the angle of the sector be ‘r’ and ‘x’ respectively. If its area is A, then

i.e.,

From pythagonus theorem,

14m

A B 30m Clearly, area of parallelogram ABCDS = 2(Area of ABC ) Let a = 30, b = 14, and c = 40 www.betoppers.com

7th Class Mathematics

296 2.

1  a  b  c   42 2 Therefore Area of ABC = Then S =

s  s  a  s  b  s  c 

 42  12  28  2

Here, the area of rhombus = A = 4800 m2 side of rhombus = a = 40 m one diagonal d1 = ? second diagonal d2 = ? Using the correlation formula, (d1 + d2)2 = 4(a2 + A) and (d1 – d2)2 = 4(a2 – A)  (d1 + d2)2 = 4 [(80)2 + 4800] 

6400  4800  d1 + d2 = 212.6 _________ (i) Similarly, (d1 – d2)2 = 4[(80)2 – 4800]

 14  3  3  4  4  7  2



14 2   3 2   4 2

= 14 × 3 × 4 = 168 m2 Therefore area of parallelogram = (2 168)m2 = 336m2 10. Let the side of the square be ‘x’ then, breadth of rectangle =

3.

3 x 2

3 2 Therefore 40  x  3x 2 20x = x2; x = 20cm

HOTS WORKSHEET HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Let ABCD be a square whose each side = AB = BC = CD = DA = a = 6 cm. If r = radius of circle inscribed in the above square R = radius of circle circumscribed about the square,

R=

4.

a = 3 cm. 2

a 2 = 4.24 cm. 2

Area of inscribed circle = r 2 =

d1 + d2 = 2 1600  d1 + d2 = 80 _________ (ii) Solving (i) and (ii) d1 = 145.83 m are the length d2 = 65.83 m of two diagonals Let ABCD be a rectangle whose diagonal = d = 25 m area = A = 168 m2 length = l = ? breadth = b = ? Using the correlation formula (l + b)2 = d2 + 2A and (l – b)2 = d2 – 2A  (l + b)2 = (25)2 + 2 × 168 l + b = 31 ______ (i) Similarly, (l – b)2 = (25)2 – 2 × 168 l – b = 17 ______ (ii) From (i) and (ii) l = 24 m = length of the rectangle b = 7 m = breadth of the rectangle. Let ABCD be a rectangular grass plot whose length = l = 80 m breadth = b = 60 m Two roads of width W = 10 m (shaded part) are crossing each other at the middle of plot. Area of roads = W (l + b – W) = 10(80 + 60 – 10) m2 = 1300 m2 Cost of gravelling the roads = rate of gravelling/ m2 × area of roads = Rs. 2 × 1300 = Rs. 2600 Hence, the cost of one gravelling the roads is Rs. 2600. Let ABDC be a circular grass plot whose diameter AD = 70 m  OD = 35 m KML is a gravel walk which is 15 m from the edge (i.e., pt. D) 

 14   7  2    3  3    4  4 

then r =

d1 + d2 = 2

22 × (3)2 cm2 7

Area of circumscribed circle = R 2 22 = × (4.24) 2 7 22  (3)2 Area of inscribed circle  7 cm2 Area of circumscribed circle 22  (4.24) 2 7 = 0.5 Hence, the required ratio is 0.5. www.betoppers.com

5.

Mensuration

297 B

d = diameter of circle OD = x = required distance

d –h 2 Now, using the formula b2 = dh =r–h=

Zone A

X

Y

D

L

O O

K

15



d=

M

LD = 15 m OL = OD – LD = 35 – 15 = 20 m OY = 20 – 5 = 15 = T [Since width of gravel walk = YL = 5 m] Area of turfing = Area of section ABD – Area of gravel walk =  (35)2 –  W (2r + W) =  (35)2 –  × 5 (30 + 5) m2 = 3300 m2 Cost of turfing = Rate/m2 × Area of turfing = 2 × 3300 = Rs. 6600 Hence, the cost to turf the grass plot will be Rs. 6600. Let ABCD be the square whose side = 28 m Four equal circles are drawn about four corners A, B, C and D of the square. Have four equal circles are such that each of the which touches the other two. So, using the formula area of space enclosed by   

6 2 r 7 where 2r = 28  r = 14 four equal circles = S =

7.

S=

B b

h

A

D x O

a

8.

rl 14  28 2 = m = 196 m2 2 2 Here, radius of circle = 21 cm An arc ABC subtends an angle of 60o at the centre. Since AO = OB = 21 cm = r   OAB =  OBA = 60o =  AOB (given) Hence ABO is an equilateral  . Since the central angle 60o, so, Area of minor segment AKB (Shaded part) = 90 × r2 = 99 (21)2 = 40 cm2 Area of major segment AOB = Area of circle – Area of minor segment A=

9.

22 × 21 × 21 – 40 = 1346 cm2 7 10. Let AB and CD be two parallel chords of the circle. Then AB = radius of circle = r = 75 cm CD = diameter of circle = 150 cm If O is centre of circle, then OA = OB = AB = 75 and OAB is an equilateral  so, using the formula. Area of minor segment = 0.09 r2 (central angle = 60o) = .90 × (75)2 = 509.54 cm2 Required area between AB and CD = Area of semicircle CAD – Area of minor segment =

6 × (14)2 m2 7 = 168 m2 Hence, the area of the space between the circumference of the circles is 168 m2. Let ABC be the given arc of a circle with centre at O. BD = height of arc = h = 27 m. 

d 147 –h= – 27 = 46.5 m 2 2 Hence, the distance of the chord of the arc from the centre of the circle is 46.5 m. Here, radius = r = 14 cm length of arc = l = 28 area of sector = A = ? Using the formula So, x =

C

6.

63  63 = 147 m 27

C

E

22 75  75 r 2  – 509.54 = – 509.54 7 2 2 = 8326 cm2 Hence, the area of the required, zone = 8326 cm2. =

AB = Chord of half the arc = b = 63 m. www.betoppers.com

7th Class Mathematics

298

IIT JEE WORKSHEET Key

1

2

3

4

5

6

7

8

9

10

C

A

D

C

B B A D C

C

11

12

13

14

B,D B,C B,D A,C HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

2. 3.

Given perimeter of a square = 150cm  4a = 150 cm a = 37.5 cm Circumference

a

4.

5. 6. 15. 16. 17. 18.

(Angles opposite to equal sides are equal) In  BCD,  BCD +  CBD +  CDB = 180°  90° +  CDB +  CDB = 180°  2  CDB = 90°   CDB = 45° Also,  ADC =  ADB +  CDB  135° =  ADB + 45°   ADB = 90° Thus, ÄADB is also a right-angled triangle with  ADB = 90° and perpendicular sides as AD and BD. Applying Pythagoras theorem in right-angled  BCD: BC2 + CD2 = DB2  x2 + x2 = DB2  DB  x 2

1 1 x2 Area of  BCD   BC  CD   x  x  2 2 2

a

a Perimeter of equilateral triangle = a + a + a = 3a Area of shaded region = 2y + x cm = 2xy cm Area of remaining portion= Area of PQRS – Area of shaded region = 5y × 4x – 2xy = 20xy – 2xy = 18xy cm V = a3 = 9 × 9 × 9 cubic cm3 = 729 cubic cm3 1000 6 m by 5m 38 m2 53 cm2 Join BD as follows: B

Area

of

1 x2 2 xx 2  2 2  Area of quadrilateral ABCD  BCD + Area of  ADB 



2

2



1350

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

18  18 2  cm . x 1  2    18  18 2  cm 2 

x2 1  2

C

= Area of

2 x 2 x2 2 x 1  2    2 2 2 However, the area of the quadrilateral is given as

A

D Let BC = CD = DA = x In  BCD, BC = CD   CBD =  CDB

1  AD  DB 2

 ADB



2

  18 1  2 cm  

2

2

x2  18cm 2 2

 x 2  36cm 2  x  6cm Thus, the length of side BC is x i.e., 6 cm. The correct answer is C.

Mensuration

299

19. We have, a = 9cm, b = 6cm, Hence A 

1 a  b h 2

1 9  6 h 2 120 = 15h 60 

120 15 h = 8cm 20. Area of a square = 16cm  a2 = 16cm a = 4cm 1 Radius of a circle =  side of square 2 1  side of square 2  r  2cm 21. Area of ABCD = BC × BF = AB × DE 5 × x = 10 × 3 x = 6cm h

22. Area of rhombus = 150cm =



1 d1d 2 2

1  37.5  d 2 2

300  d2 37.5

 d 2  8cm 23. A-p, B-r, C-q, D-s 24. A-s, B-q, C-r, D-p



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300

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7th Class Mathematics

12. STATISTICS SOLUTIONS 4.

FORMATIVE WORKSHEET KEY

For calculating mean, we prepare the table as shown below: Variable Frequency f ixi (x i) (fi)

Q.no

1

2

3

4

5

6

7

8

9

10

2

5

10

Key

B

D

A

D

B

B

D

C

B

*

4

9

36

Q.no

11 12

13

6

13

78

Key

B

*

8

18

144

10

20

200

12

p

12p

C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

Range of a data set is the difference between the highest and the lowest observation.

f i = 65p

Observations of the given data can be arranged in ascending order as:

Therefore

11024, 11287, 12464, 13213, 14162, 14219, 14731, and 15613



5.

It is known that,

Sum of allobservations Number of observations

3.

2  3  5  7  11  13  17  19 77   9.625 8 8

Average duration of exercise in the week = 75 min Therefore, total duration of exercise in the week = 75 × 7 min = 525 min Total duration of exercise in the week excluding Tuesday = 120 + 90 + 75 + 45 + 100 + 60 = 490 min Thus, the duration of exercise on Tuesday was 525 min – 490 min = 35 min.

i



468  12p 65  p

468  12p 6 65  p

102  17 6 Incorrect mean of six observations = 42  Incorrect sum of six observations = 42 × 6 = 252 42 was misread as 24 while 27 was misread as 72.  Correct sum of observations = 252 – 24 –72 + 42 + 27 = 225

225  37.5 6 Thus, the correct mean of the observations is 37.5. The correct answer is B. The even numbers between 9 and 23 are 10, 12, 14, (16), 18, 20, and 22 Therefore, the median of even numbers = 16 The given data can be arranged in increasing order as: 25, 32, 35, 37, 42, 45, 90, 100, 136 Here, the number of terms is 9 i.e., odd.

 Correct mean 

Therefore, mean of the first eight prime numbers



i

6p = 102; therefore p 

The first eight prime numbers are 2, 3, 5, 7, 11, 13, 17, and 19.

Mean 

f x f

468 + 12k = 570 + 6p 6p = 570 – 468

Highest observation = 15613 and lowest observation = 11024

2.

Mean 

x

It is seen that,

Thus, the range of the number of candidates who appeared is 15613 – 11024 = 4589.

f i xi = 468+12p

6.

7.

th

 9 1  Median =   term = 5th term = 42  2  Thus, the median of the given data is 42.

7th Class Mathematics

302 8.

The given data is: 41, 60, 65, m – 2, 82, 85, 96. Here, number of observations is 7, which is odd.

(iii) A pie chart can be drawn for the above data as follows.

th

9.

 n 1  Median =   observation =  2  4thobservation = m – 2 The median of the data is given as 68.  m – 2 = 68  m = 68 + 2 = 70 Thus, the value of m is 70. It can be observed from the given bar graph that: Population of village P = 3000 Population of village Q = 2500 Population of village R = 5500 Population of village S = 4000 Population of village T = 6000

Summer 90° Winter 150°

Rainy 120°

CONCEPTIVE WORKSHEET KEY

Sum of allobservations Mean is given by: Number of observations

Q.no

1

2

3

4

5

6

7

8

9

10

Key

D

B

B

B

A

B

B

D

B

*

 Mean

population

Q.no

11 12

13

3000  2500  5500  4000  6000 21000  = 5 5 4200 10. (a) Since the bar representing cats is the tallest, cat is the most popular pet. (b) The number of children having dog as a pet are 8. 11. It can be seen from the histogram that the number of patients in the age group of 15 – 25 is 25 and in the age group 25 – 35 is 45. Thus, a total of (25 + 45) = 70 patients are below 35 years of age in the rehabilitation centre. 12. In the given circle graph, it is seen that the central angle of the sector corresponding to leather slippers is 140°. Therefore, number of girls who preferred to wear leather slippers:

Key

D

*



140   1080  420 360 Thus, 420 girls preferred to wear leather slippers. 13. (i) Winter (ii) Total number of votes = 90 + 120 + 150 = 360 No. of In Season Central angle votes fraction 90 90 Summer 90  360  90 360 360 120 120 Rainy 120  360  120 360 360 150 150 Winter 150  360  150 360 360 www.betoppers.com

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Range of a data set is the difference between the highest and the lowest observation. Observations of the given data can be arranged in ascending order as: 2.7, 2.9, 3.1, 3.2, 3.4, 3.5, 3.8, 3.9, 4, 4.1, 4.3, 4.4 It is seen that, Highest observation = 4.4 and lowest observation = 2.7 Thus, the range of distances covered by the participants is 4.4 – 2.7 = 1.7. 2. The first 8 composite numbers are 4, 6, 8, 9, 10, 12, 14, and 15. It is known that the mean is given by:,

Sum of allobservations Number of observations . 

Mean



4  6  8  9  10  12  14  15 8

78  9.75 8 Thus, the mean of the first eight composite numbers is 9.75. 

Statistics 3.

303

It is known that,

Mean 

7.

Sum of allobservations Number of observations

Therefore, mean run score =

Sum of runs scored by India in the series Number of matches

224  321  196  168  251  173  312 1645  7 7 = 235 Thus, the average run score of India in the series is 235. Temperature on Monday = 62°F Temperature on Tuesday = 72°F Temperature on Wednesday = 82°F Therefore, average temperature of the city

It is given that: x + y = 3 __________ (1) y + z = 5 __________ (2) z + x = 4 __________ (3) On adding equations (1), (2), and (3): (x + y) + (y + z) + (z + x) = 3 + 5 + 4  2x + 2y + 2z = 12

 x + y + z = 12  Mean 



4.

 5.

Thus, the mean of x, y, and z is 4. It is known that the average is given by:

Sum of allobservations Number of observations Average 

62  72  82 216  = 72°F 3 3

 81.6 

100  95  73  x  80 5

100  95  73  x  80 5

Median of a data set is the data value which lies in

 81.6 × 5 = 348 + x  408 = 348 + x  x = 408 – 348 = 60 Thus, the value of x is 60. It can be observed from the given bar graph that: Weight of Ratan = 45 kg Weight of Sanjay = 30 kg Weight of Johny = 35 kg Weight of Farhan = 40 kg  Mean weight =

the middle of the data when the data is arranged in either increasing or decreasing order, with half of the

45  30  35  40 50 kg  kg  37.5kg 4 4

Mean of 6 observations = 27  Sum of 6 observations = 27 × 6 = 162 Let x be the new number. 

162  x  25 7

 162 + x = 175  x = 175 – 162 = 13 Thus, the new number is 13. 6.

8.

x  y  z 12  4 3 3

observations above it and the other half below it. The heights (in cm) of the students can be arranged in increasing order as: 124, 124, 124, 126, 127, 128, 128, 131, 131, 132, 132, 132, 135, 135, 138, 142, 142, 143, 143,143, 144 Here, it is clearly seen that the middle observation is 132. Thus, the required median of the heights of the group of students is 132 cm.

9.

Thus, mean weight of the four boys is 37.5 kg. 10. (i) In 1989, 175 books were sold. In 1990, 475 books were sold. In 1992, 225 books were sold. (ii) From the graph, it can be concluded that 475 books were sold in the year 1990 and 225 books were sold in the year 1992. (iii) From the graph, it can be concluded that in the years 1989 and 1992, the number of books sold were less than 250.

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7th Class Mathematics

304 (iv) From the graph, it can be concluded that the number of books sold in the year 1989 is about 1 and

3 th part of 1 cm. 4

We know that the scale is taken as 1 cm = 100 books. 100 +

3 × 100 = 100 + 75 = 175 4

Therefore, about 175 books were sold in the year 1989. 11. It can be seen in the given histogram that the number of students living at a distance of 15 km or more from the school in the group of 15 – 20 is 100, in the group 20 – 25 is 25, and in the group 25 – 30 is 50.

13. The central angle for each colour can be calculated as follows. No. of In Colours Central angle people fraction 18 18 Blue 18  360  180 36 36 9 9 Green 9  360  90 36 36 6 6 Red 6  360  60 36 36 3 3 Yellow 3  360  30 36 36 The pie chart of the above data is as follows.

Yellow 30° Red 60°

Thus, a total of (100 + 25 + 50) = 175 students live within a distance of 15 km or more from the school. 12. (i)

Blue 180°

Green 90°

Number of people who like classical music = 10% This 10% represents 20 people. 100 % represents 

20  100  200 people 10

Therefore, 200 young people were surveyed. (ii) From the pie chart, it can be easily observed that the light music is represented by the maximum part of the pie chart (i.e., 40 %). Hence, most of the people like light music. (iii) Number of CD’s of classical music = 10% of 1000 

10  1000 = 100 100

Number of CD’s of semi-classical music = 20% of 1000 

20  1000 = 200 100

Number of CD’s of folk music = 30% of 1000



30  1000 = 300 100

Number of cassettes of light music = 40% of 1000 

40  1000 = 400 100

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SUMM ATIVE WORKSHEET KEY Q.no

1

2

3

4

5

6

Key

A

D

A

A

C

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It is known that

Mean 

Sum of allobservations Number of observations

It is given that mean = 25, and number of observations = 5

Sum of allobservations 5  Sum of all observations = 25 × 5 = 125 Four of the observations are given as 17, 23, 41, and 12. Let x be the fifth observation. Then, we have 17 + 23 + 41 + 12 + x = 125  93 + x = 125  x = 125 – 93 = 32 Thus, the required fifth observation is 32.  25 

Statistics 2.

The first four prime numbers are 2, 3, 5, and 7. Their cubes are (2) 3 = 8, (3) 3 = 27, (5) 3 = 125, and (7)3 = 343. It is known that the mean is given by:,

5.

It can be seen from the given histogram that, Runs scored in 0 to 5 overs = 30 Runs scored in 5 to 10 overs = 45

Sum of allobservations Number of observations .

Runs scored in 10 to 15 overs = 50

The mean of the cubes of the first four prime numbers is given by:

Therefore, total runs scored = 30 + 45 + 50 + 20 = 145

8  27  125  343 4

Thus, the team scored a total of 145 runs in the T20 cricket match.

 Required



3.

305

mean

Runs scored in 15 to 20 overs = 20



6.

503  125.75 4

Thus, the required mean of the cubes of the first four prime numbers is 125.75. Let the five observations be a, b, c, d, and e. It is given that their mean is M. 

abcd e  M ––––––––– (1) 5

If each observation is tripled, then the new observations will be 3a, 3b, 3c, 3d, and 3e. Mean = 

abcde 3a  3b  3c  3d  3e  3  5 5  

(i)

Total marks obtained by the student are 540. Hence, 540 marks represent 360º. The central angle for 105 marks has to be calculated. Central angle for 105 marks



Hindi is the subject having its central angle as 70º. Therefore, the student scored 105 marks in Hindi. (ii) Difference between the central angles of Mathematics and Hindi = 90º – 70º = 20º Marks for 20º central angle

= 3M

4.

[Using (1)] Thus, the new mean will be 3M. Let the five numbers be a, b, c, d, and e, out of which, the mean of a and b is 7.5. It is given that the mean of the five numbers is 6. 

abcde 6 5

 a + b + c + d + e = 30 –––––––– (1) Also, the mean of a and b is 7.5. 

ab  7.5 2

 a + b = 15 –––––––––– (2) Subtracting equation (2) from equation (1): a + b + c + d + e – a – b = 30 – 15  c + d + e = 15  Mean of the remaining three numbers 

c  d  e 15  5 3 3

105  360  70 540



20  540  30 360

There is a difference of 30 marks between the score obtained in Mathematics and Hindi. Therefore, 30 more marks were obtained by the student in Mathematics than in Hindi. (iii) Sum of central angles of Social Science and Mathematics = 90° + 65° = 155° Sum of central angles of Science and Hindi = 80° + 70° = 150° The sum of the central angles for Social Science and Mathematics is more than that of Science and Hindi. Therefore, the student scored more in Social Science and Mathematics than in Science and Hindi.

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7th Class Mathematics

306

HOTS WORKSHEET

24

Q.no

1

2

3

4

5

Key

B

C B

B

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. Mean of five observations = 35 Sum of five observations = 35 × 5 = 175 Mean of first two observations = 27.5 Sum of first two observations = 27.5 × 2 = 55 Mean of last two observations = 42.5  Sum of last two observations = 85 Sum of five observations = Sum of first two observations + Middle observation + Sum of last two observations  Middle observation = 175 – (55 + 85) = 35 2. The prime numbers between 20 and 40 are 23, 29, 31, and 37.

Sum of allobservations Mean is given by: Number of observations

3.

23  29  31  37 120   30  Mean = 4 4 It is known that the mean is given by:, Sum of allobservations Number of observations . It is given that mean of 12, x, 17, 12 is 11.

12  x  17  12 5  x + 41 = 55  x = 55 – 41 = 14 Therefore, the mean of x, 9, 16, 5, 7, and 3 is given 11 

14  9  16  5  7  3 54  9 6 6 Thus, the mean of the given six observations is 9. We start drawing the histogram by taking the frequency i.e, the number of teachers along the vertical axis and the class interval i.e., the age along the horizontal axis. Also, make the vertical bars of height equal to the corresponding frequncy of each class interval. Thus, the given data can be represented by the histogram as by:

4.

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Number of teachers

KEY

20 16 12 8 4 O

5.

20 25 30 35 40 45 Age (years)

Thus, the histogram given in alternative B correctly represents the given information. The central angle for each subject can be calculated as follows. Language

No. of students

Hindi

40

English

12

Marathi

9

Tamil

7

Bengali

4

In fraction

C entral angle

40 72 12 72 9 72 7 72 4 72

40 × 360º = 200º 72 12 × 360º = 60º 72 9 × 360º = 45º 72 7 × 360º = 35º 72 4 × 360º = 20º 72

A pie chart of the above data is as follows. Bangla 20°

Tamil 35° Marathi 45° English 60°

Hindi 200°

Statistics

307 4.

IIT JEE WORKSHEET

Sum of observations Number of observations

KEY 7

8

A,B, A,B, D C HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1. It is given that the arithmetic mean of w, x, and y is 10. Key

A

B D C

A

B

wxy  10 3  w + x + y = 30 ––––––––– (1) It is also given that the arithmetic mean of x, y, and z is 15. 

xyz  15 3  x + y + z = 45 ––––––––– (2) Subtracting equation (1) from equation (2): x + y + z – (w + x + y) = 45 – 30  z – w = 15  w – z = –15 Thus, the value of w – z is –15. Mean of the nine observations = 36  Sum of nine observations = 36 × 9 = 324 Mean of first six observations = 37  Sum of first six observations = 37 × 6 = 222  Sum of last three observations = 324 – 222 = 102 

2.

102  34 3 Thus, the mean of last three observations is 34. The given data can be arranged in increasing order as: 2, 3, 7, 7, 8, 9, 9, 9, 10, 11, 11, 13 It can be observed that 9 occurs maximum number of times i.e., 3. Mode = 9 Mean =  Mean of the last three observations =

3.

2  3  7  7  8  9  9  9  10  11  11  13 12 99  8.25 12  Required difference = 9 – 8.25 = 0.75 Thus, the difference between mode and mean of the given data is 0.75. 

6  9  4  2  x  15 8 6  36 + x = 48  x = 48 – 36 = 12 Thus, the value of x is 12. We will choose a scale as 1 unit = 10 children because we can represent a more clear difference between the number of students of class 7th and that of class 9th by this scale. 

9.

10.

Scale : 1cm = 10 Children

150 140 130 120 110 100 90 80 70 60 50 40 30 20 10

Tenth

6

Ninth

5

Eight

4

Seventh

3

Sixth

2

Fifth

1

Number of children

Q.no

It is known that mean is given by:,

Class 11. Since the bar representing the number of children for class fifth is the tallest, there are maximum number of children in class fifth. Similarly, since the bar representing the number of children for class tenth is the smallest, there are minimum number of children in class tenth. 12. The number of students in class sixth is 120 and the number of students in class eighth is 100. Therefore, the ratio between the number of students of class sixth and the number of students of class

120 6   6:5 100 5 13. We know that, eighth 

Sum of observations Mean = Number of observations Mean of 11 observations = 32  Sum of 11 observations = 11 × 32 = 352 Now, one new observation is included in the data set. Thus, the total number of observations now become 11 + 1 = 12. www.betoppers.com

7th Class Mathematics

308 It is given that the mean of 12 observations is 30.  Sum of 12 observations = 12 × 30 = 360 Now, the value of the 12th observation equals to the difference between the sum of 12 observations and the sum of11 observations.  Value of the 12th observation = 360 – 352 = 8 14. Median of a data set is the data value, which lies in the middle of the data with half of the observations above it and the other half below it, when the data is arranged in either increasing or decreasing order. The marks obtained by the students can be arranged in increasing order as 1, 3, 4, 5, 6, 7, 7, 7, 8, 8, 9, 9, 9, 10, 10 Here, it is clearly seen that the middle observation is 7. Thus, the required median of the marks obtained by the group of students is 7. 15. Let the six observations be a1, a2, a3, a4, a5, and a6. It is given that the mean of six observations is 9.

a1  a 2  a 3  a 4  a 5  a 6  10 6  a 1 + a 2+ a 3 + a 4+ a 5 + a 6 = 60 –––––– (1) It is also given that the mean of four observations is 12.5. 

a1  a 2  a 3  a 4  12.5 4  a 1+ a 2+ a 3 + a 4 = 50 –––––– (2) Subtracting equation (2) from equation (1): a5 + a6 = 60 – 50 = 10



a 5  a 6 10  5 2 2 Thus, the mean of the remaining two observations is 5. 16. The first 9 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, and 17. It is known that the mean is given by:, 

Sum of observations Number of observations .  Mean =

1  3  5  7  9  11  13  15  17 9

81 9 9 Thus, the mean of first nine odd natural numbers is 9. 17. A – q; B – p; C – t; D – u; E – v 

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13. FACTORIZATION SOLUTIONS

FORMATIVE WORKSHEET 1.

2. 3.

4.

5.

6. 7.

8.

9.

10.

11 12.

The terms of the expression 3a2 – 6ab have a common factor 3a  3a2 – 6ab = 3a(a – 2b). 5bx2(a2x – 3a – 4b2). Noticing that the first two terms contain a common factor x, and the last two terms a common factor b, we enclose the first two terms in one bracket, and the last two in another. Thus, x2 – ax + bx – ab = (x2 – ax) + (bx – ab) = x(x – a) + b(x – a) = (x – a) (x + b) 12a2 – 4ab – 3ax2 + bx2 = (12a2 – 4ab) – (3ax2 + bx2) = 4a(3a – b) – x2(3a – b) = (3a – b) (4a – x2). 2 25x – 16y2 = (5x)2 – (4y)2. Therefore the first factors is the sum of 5x and 4y, and the second factor is the difference of 5x and 4y.  25x2 – 16y2 = (5x + 4y) (5x – 4x).  (a + 2b)2 – 16x2 = (a + 2b)2 – (4x)2 = (a + 2b + 4x) (a + 2b – 4x). The sum of x and 2b – 3c is x + 2b – 3c, and their difference is x – (2b – 3c) = x – 2b + 3c.  x2 – (2b – 3c)2 = (x + 2b – 3c) (x – 2b + 3c). = (3x + 7y)2 – (2x – 3y)2 = {(3x + 7y) + (2x – 3y)} {(3x + 7y) – (2x – 3y)} = (3x + 7y + 2x – 3y) (3x + 7y – 2x + 3y) = (5x + 4y) (x + 10y). a2 – 2ax + x2 – 4b2 = (a2 – 2ax + x2) – 4b2 = (a – x)2 – (2b)2 = (a – x + 2b) (a – x – 2b). x4 + x2y2 + y4 = (x4 + 2x2y2 + y2) – x2y2 = (x2 + y2)2 – (xy)2 = (x2 + y2 + xy) (x2 + y2 – xy) = (x2 + xy + y2) (x2 – xy + y2). 8x3 – 27y3 = (2x)3 – (3y)3 = (2x – 3y) (4x2 + 6xy + 9y2). The expression may be written (x2 + 6x + 9) + 5 – 9; that is, x2 + 6x + 5 = (x + 3)2 – 4 = (x + 3)2 – (2)2 = (x + 3 + 2) (x + 3 – 2) = (x + 5) (x + 1).

13.

14.

15.

16a4 – 81b4 = (4a2)2 – (9b2)2 = (4a2 + 9b2) (4a2 – 9b2) = (4a2 + 9b2) (2a + 3b) (2a – 3b). The expression = p2(x3 – 8y3) – 4q2(x3 – 8y3) = (x3 – 8y3) (p2 – 4q2) = (x – 2y) (x2 + 2xy + 4y2) (p + 2q) (p – 2q). 4x2 – 25y2 + 2x + 5y = (2x + 5y) (2x – 5y) + 2x + 5y = (2x + 5y) (2x – 5y + 1).

CONCEPTIVE WORKSHEET 1) 3) 5) 7) 9) 11) 13) 15) 17) 19) 21) 23) 25)

27) 29) 31) 33)

(1 – 10x) (1 + 10x) 2) (c + 3b – 5y) (c + 3b + 5y) (a – b + x + y) (a – b – x – y) 4) (3a + 2b – c – x + 3y) (3a + 2b + c + x – 3y) (a + b – c + x – y + z) (a + b – c – x + y – z) 6) 4(x + 5) (x – 1) – (8a + 1) 8) y(2x – y) (a + b + c – d) (a + b – c + d) 10) (a – 4x – 3y) (a – 4x + 3y) y(2x + y – 18) 12) (x + y) (x + 5y) (x – 13y)2 14) (x – 18) (x – 12) (y + 3) (y – 7) 16) (2x + 3) (3x + 4) 2(9x – 5) (x + 2) 18) (2x + y) (x – 2y) (3x – 2y) (4x – 7y) 20) (x – 3) (x + 3) (x4 + 9x2 + 81) (5a – 3b) (13a2 + 21b2 – 30ab 22) (8x + 18y) (124x2 + 189y2 + 588xy) – (a + 5b) (19a2 + 43b2 + 55ab) 24) (1 + xy – x + y) (1 + xy + x – y) (x – y) (5x + y) 26) (2x + 3y) (4x + 3y) ab  a b   2  2

28) (1 – b + c) (x – 1) (x + 1) (x – 3) 30) (x + 4) (x – 1) (4 – x) (1 – x) (2 – x) (3 + 2x) 32) (x2 + 15)2 – (8x)2 (x + q) (px – 6) 34) (y – 5) (y – z)

7th Class Mathematics

310 35) 37) 39)

(a – b) (b – c) 36) (3x + 2a – y) (3x – 2a + y) 2x (8 – x) 38) (3 + x – y) (3 – x + y) x(2x – 3y) (x + 4y) 40) (2x + 5) (3x – 1)

SUMM ATIVE WORKSHEET 1)

(m – 7n) (m – 15n) 2) (x + 24y) (x + 25y) 3) (a2b2 + 25) (a2b2 + 12) 4) (x + 13y) (x + 30y) 5) (5 + x) (4 + x) 6) (a + 5) (a – 17) 7) (a + 8) (a – 19) 8) (x + 13) (x – 12) 9) (xy + 3) (xy – 8) 10) (a – 16) (a – 16) 11) (a – 19) (a – 19) 12) (ay + 24) (ay – 10) 13) (y2 + 9x2) (y2 – 3x2) 14) (a + 14bx) (a – 2bx) 15) (x2 + 25a2) (x2 – 12a2) 16) (x2 + 21a2) (x2 – 22a2) 17) (20 + x) (19 – x) 18) (14 + x) (7 – x) 19) (15 + ax) (8 – ax) 20) (17 + x) (12 – x) 21) (3x – 2y) (4x – 5y) 22) (12x + 5) (x – 3) 23) (15x – 1) (x + 15) 24) (12x – 7) (2x + 3) 25) (8x – 9) (9x – 8) 26) (1 + 7x) (5 – 3x) 27) (x2 + 4b) (x2 – 4b) 28) (ab + 2cd) (ab – 2cd) 29) (11a + 9x) (11a – 9x) 30) (7 + 10k) (7 – 10k) 31) (a + 8x3) (a – 8x3) 32) (9p2z3 + 5b) (9p2z3 – 5b) 33) (1 + 10a3b2c) (1–10a3b2c) 34) (a + b – c + x – y + z) (a + b – c – x + y – z) 35) (a – n + b + m) (a – n – b – m) 36) (3a + 2b + c + x – 2y) (3a + 2b – c – x + 2y) 37) (c + 5a – 3b) (c – 5a + 3b) 38) (7x + y + 1) (7x + y – 1) 39) (b – c + a – x) (b – c – a + x) 40) 3a(a + 2b – 2c) 41) y(2x + y – 16) 42) x(x – 14y + 2z) 43) a(4x + a – 6) 44) y(2x – y) 45) (1 + x + y) (1 – x – y)

HOTS WORKSHEET 1) 3x(2x2 – 3xy + 4y2)

2) x2 (3 + x3)

3) x(3x2 – x + 1)

4) 5(3 + 5x2)

5) 2xy2(xy – 3x + y)

6) 5x3(x2 – 2a2 – 3a3)

7) (x – y) (m – n)

8) (a – c) (a + b)

9) (x – a) (m + n)

10) (3a – b) (x – y)

11) (a + bc) (xy – z) 12) (2x + 3y) (ax – by) 13) (a – b – c) (x – y) 14) (a – y) (b – y) 15) (f2 + g2) (x2 – a) 16) (ax + by) (mx – ny) 17) (a + b) (ax + by + c) 18) (a – 2b) (a – 27b) 19) (4 – x) (3 – x) 20) (a + 27) (a + 27) 2 2 2 21) (p + 3pq + 9q ) (p – 3pq + 9q2) 22) (2m2 + 3mn + n2) (2m2 – 3mn + n2) or (4m2 – n2) (m2 – n2) 23) (6 – a) (36 + 6a + a2) 24) (ab + 8) (a2b2 – 8ab + 64) 25) (10y – 1) (100y2 + 10y + 1) 26) (pq – 3x) (p2q2 + 3pqx + 9x2) 27) (z – 4y2) (z2 + 4xy2 + 16y4) 28) (9a – 4b) (81a2 + 36ab + 16b2) 29) (x2 – 3y) (x4 + 3x2y + 9y2) 30) (2x – 9y2) (4x2 + 18xy2 + 81y4) 31) (xy – 8) (x2y2 + 8xy + 64) 32) (y2 + 13) (y2 – 12) 33) (y2 + 5) (y2 – 7) 34) (x + 13y) (x – 7y) 35) (cd + 1) (cd – 2) 36) xy(x + 9) (x – 7) 37) (3m – 4) (3m – 4) 38) (a + b + c) (a – b – c) 39) (1 + x – 3y) (1 – x + 3y) 40) (b – 29) (b + 27)

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Factorisation Solutions

311

IIT JEE WORKSHEET KEY Q.No.

1

2

3

4

5

Ans.

D

C

B

A

D

Q.No.

6

7

8

9

10

Ans.

A

B

C

D

A

Q.No.

11

12

13

14

15

Ans.

A, D

B, C, D

A, B

A, C

A, B, C, D

HINTS/SOLUTIONS FOR SECLECTED QUESTIONS 1.

5.

11.

14.

III.

x2y2 + xyz + xy + z xy(xy + z) + (xy + z) [D]  (xy + z)(xy + 1) x4 + x3 + 5x + 5 x3(x + 1) + 5 (x + 1) [D]  (x + 1)(x3 + 5) (a4 – 1) = (a2)2 – (1)2 = (a2 – 1)(a2 + 1) [ a2 – 2 b = (a + b)(a – b)] = (a2 + 1)(a – 1)(a2 + 1) [ a2 – b2 – (a + b)(a – b)] A, D are correct answers. 15x3y2 + 15x2y3 = 15(x3y2 + x2y3) = 15x2y2 (x + y)  A, C are correct answers. Paragraph Type Paragraph–I (i) 6,24, 60, 120 (ii) n(n2 + 3n + 2) (iii) Each of the term in the sequence is divisible by n. Paragraph–II 16. (8x2 + 10x + 3)m2 17. (4x + 3)m by (2x + 1)m 18.

19.

20.

20.

Temperature at 4 pm = –10°C. Temperature had risen 12°C at 11.pm.  Temperature by 11pm is –10°C + 12°C = 2°C Matrix matching A - t, B - p, C - S, Dq, E-r (C) (ax + by)2 + (bx – ay)2 = a2x2 + b2y2 + 2ax by + b2x2 + a2y2 – 2aybx = a2x2 + b2x2 + b2y2 + a2y2 = x2 (a2 + b2) + y2 (b2 + a2)  (x2 + y2) (a2 + b2) (D) (2x + 3y)2 – 16z2 = (2x + 3y)2 – (4z)2 = (2x + 3y + 4z) (2x + 3y – 4z) ( a2 – b2 = (a + b) (a – b)) (E) 16x2 – 1 = (4x)2 – (1)2 = (4x + 1) (4x – 1)

 

4x + 3 2x + 1

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312

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7th Class Mathematics

14. VISUALISING SOLID SHAPES SOLUTIONS

FORMATIVE WORKSHEET

4.

When seen from the direction indicated by the arrow, the obtained view of the solid is:

5.

The shadow of the ice-cream cone is that of a triangle.

6.

A horizontal cut can be given to the party cap as:

KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

B

D

B

C C

A

A

A

B

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

When the net is folded to form a regular hexagon based prism, it will be of the following shape:

As seen in the figure, the cross-section obtained is that of a circle.

G C

B A

D F

E

H

7.

It can be observed that G and H, A and D, B and E, C and F are the opposite faces of the prism. Thus, A and F will not be opposite faces in the prism. 2.

8.

The shape of a square pyramid is:

A hexagon can be made by using one square and two triangles as shown below: +

9. If the nets given in the alternatives A, B,and C are folded, then a square pyramid is obtained.

+

=

The given net has two triangular faces and three rectangular faces. Triangular prism has two triangular faces and three rectangular faces.

However, if the net given in alternative D is folded, then a square pyramid is not obtained. 3.

As seen in the given oblique sketch, the length of the cuboid is 7 units while its height is 4 units. The width of the cuboid has been given as 5 units.  Sum of the areas of the faces of the cuboid

= [2 × (7 × 5 + 7 × 4 + 5 × 4)] square units = [2 × (35 + 28 + 20)] square units

Therefore, triangular prism can be constructed using the given net. 10. A cube has six congruent faces which are square in shape. Thus, by combining six congruent squares, a cube can be obtained as:

= (2 × 83) square units = 166 square units

5 4

1 6

2

3

3 4 65 2 1

7th Class Mathematics

314

CONCEPTIVE WORKSHEET

6.

When seen from the direction indicated by the arrow, the obtained view of the solid is:

7.

The shadow of the can obtained is that of a circle.

8.

A tetrahedron has a triangular base and three triangular faces such that all four congruent faces are equilateral triangles.

KEY Q.no

1

2

3

4

5

6

7

8

9

10

Key

B

C C

D

B

B A

A

C

A

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

The net of the triangular prism can be

2.

It can be seen that the given arrangement is formed by two horizontal layers of cubes where the bottom layer is formed by arranging 8 cubes and the top layer is formed by arranging 3 cubes. Total number of cubes = 8 + 3 = 11 Thus, there are 11 cubes in the given arrangement.

3.

Thus, by combining four congruent equilateral triangles, a tetrahedron can be obtained as:

9.

The given cardboard piece has five triangular faces and one pentagonal face. A pentagonal pyramid has five triangular faces and one pentagonal face

The given figure can be drawn in an oblique sketch as 4 3

Sum of all edges = 4 + 4 + 4 + 4 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 40 4.

The shape of a cylinder can be drawn as shown below:

Therefore, a pentagonal pyramid can be constructed with the given cardboard piece. 10. The given net has one pentagonal and five triangular regions; similarly, a pentagonal pyramid also has a pentagonal base and five triangular faces. Thus, by folding the given net, a pentagonal pyramid can be obtained as:

Thus, the net of the cylinder can be:

5.

On being given a vertical cut, a cuboid with the same width and height will give a square crosssection.

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Visualising solid shapes

315 3.

It is given that their mean is M.

SUMM ATIVE WORKSHEET KEY Q.no

1

2

Key

A

B C

3

 4

5

C B

6

7

Mean



=

abcde 3a  3b  3c  3d  3e  3  5 5  

It is known that

Mean 

= 3M [Using (1)]

Sum of allobservations Number of observations

It is given that mean = 25, and number of observations = 5

Thus, the new mean will be 3M. 4.

Let the five numbers be a, b, c, d, and e, out of which, the mean of a and b is 7.5. It is given that the mean of the five numbers is 6.

Sum of allobservations  25  5



 Sum of all observations = 25 × 5 = 125

abcde 6 5

 a + b + c + d + e = 30 –––––––– (1)

Four of the observations are given as 17, 23, 41, and 12.

Also, the mean of a and b is 7.5.

Let x be the fifth observation.

2.

abcd e  M ––––––––– (1) 5

If each observation is tripled, then the new observations will be 3a, 3b, 3c, 3d, and 3e.

C C

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

Let the five observations be a, b, c, d, and e.

ab  7.5 2

Then, we have



17 + 23 + 41 + 12 + x = 125

 a + b = 15 –––––––––– (2)

 93 + x = 125  x = 125 – 93 = 32

Subtracting equation (2) from equation (1):

Thus, the required fifth observation is 32.

a + b + c + d + e – a – b = 30 – 15

The first four prime numbers are 2, 3, 5, and 7. Their cubes are (2) 3 = 8, (3) 3 = 27, (5) 3 = 125, and (7)3 = 343.

 c + d + e = 15  Mean of the remaining three numbers

It is known that the mean is given by:,



Sum of allobservations Number of observations .

5.

The mean of the cubes of the first four prime numbers is given by:  Required



mean

8  27  125  343  4

503  125.75 4

Thus, the required mean of the cubes of the first four prime numbers is 125.75.

c  d  e 15  5 3 3

It can be seen from the given histogram that, Runs scored in 0 to 5 overs = 30 Runs scored in 5 to 10 overs = 45 Runs scored in 10 to 15 overs = 50 Runs scored in 15 to 20 overs = 20 Therefore, total runs scored = 30 + 45 + 50 + 20 = 145 Thus, the team scored a total of 145 runs in the T20 cricket match. www.betoppers.com

7th Class Mathematics

316 6.

(i)

Total marks obtained by the student are 540. Hence, 540 marks represent 360º. The central angle for 105 marks has to be calculated. Central



angle

for

105

marks

Fourth cube = 5 Therefore the number of faces of cubes painted yellow are = 5 + 4 + 4 + 5 = 18. 2.

From the facing side of the figure; 6 cubes are there and behind that another 6 cubes are present. Therefore total number of cubes = 6 + 6 = 12

3.

In the given figure; there are 8 triangles. They are  AOB,  BOC,  COD,  DOA,  ABD,  BDC,  ABC,  ADC

105  360  70 540

Hindi is the subject having its central angle as 70º. Therefore, the student scored 105 marks in Hindi. (ii) Difference between the central angles of Mathematics and Hindi = 90º – 70º = 20º Marks



for

20º

central

4.

5.

angle

20  540  30 360

There is a difference of 30 marks between the score obtained in Mathematics and Hindi.

6.

7.

Therefore, 30 more marks were obtained by the student in Mathematics than in Hindi.

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(iii) Sum of central angles of Social Science and Mathematics = 90° + 65° = 155°

KEY

Sum of central angles of Science and Hindi = 80° + 70° = 150°

Q.no

1

2

The sum of the central angles for Social Science and Mathematics is more than that of Science and Hindi. Therefore, the student scored more in Social Science and Mathematics than in Science and Hindi.

Key

C

D C

HOTS WORKSHEET

3

4

5

6

7

8

9

C A

D

A,B, D

A, D

A,B, D

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 10. 10 11. 15 12. 7

KEY

13. Yes Q.no

1

2

Key

C

B D

3

4

5

6

B *

*

HINTS / SOLUTIONS TO THE SELECTED QUESTIONS 1.

The faces of the 4 cubes are painted yellow are visible faces of the cubes.

14. The oblique sketch of the given figure can be drawn as:

4 3

3

Those are for first cube = 5 Second cube = 4 Third cube = 4 www.betoppers.com

Therefore, the given cuboid is of dimension

Visualising solid shapes

317

3 units × 3 units × 4 units.  Required difference = 4 – 3 = 1 unit

Thus, the difference between the greatest and the smallest edge of the given cuboid is 1 unit. 15. Number of faces of the given solid, F = 8 Number of vertices of the given solid, V = 6 Number of edges of the given solid, E = 12  F + V – E = 8 + 6 – 12 = 2

Thus, the value (F + V – E) is 2. The correct answer is A. 16. 8 17. 1 18. A – s; B – r; C – q; D – a;



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