Algebra, Arithmetic, Numbers and Numeration: A Mathematics Book for High Schools and Colleges

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Algebra, Arithmetic, Numbers and Numeration: A Mathematics Book for High Schools and Colleges

By Kingsley Augustine

Table of Contents CHAPTER 1 NUMBER BASES .................................................................................................................................... 4 CHAPTER 2 MODULAR ARITHMETIC ..................................................................................................................... 16 CHAPTER 3 STANDARD FORM AND APPROXIMATION OF NUMBERS................................................................... 23 CHAPTER 4 LAWS OF INDICES ............................................................................................................................... 28 CHAPTER 5 LOGARITHMS OF NUMBERS GREATER THAN 1 – USE OF TABLES...................................................... 35 CHAPTER 6 THEORY OF LOGARITHMS .................................................................................................................. 47 CHAPTER 7 LINEAR EQUATIONS AND CHANGE OF SUBJECT OF FORMULAE ........................................................ 54 CHAPTER 8 VARIATION.......................................................................................................................................... 62 CHAPTER 9 REVIEW OF BASIC ARITHMETIC .......................................................................................................... 77 CHAPTER 10 FRACTIONS ....................................................................................................................................... 83 CHAPTER 11 WORD PROBLEMS INVOLVING FRACTIONS ..................................................................................... 93 CHAPTER 12 DECIMALS ....................................................................................................................................... 104 CHAPTER 13 PERCENTAGE .................................................................................................................................. 119 CHAPTER 14 SIMPLE INTEREST ........................................................................................................................... 133 CHAPTER 15 COMPOUND INTEREST ................................................................................................................... 141 CHAPTER 16 RATIO ............................................................................................................................................. 149 CHAPTER 17 RATE ............................................................................................................................................... 161 CHAPTER 18 PROPORTIONAL DIVISION .............................................................................................................. 168 CHAPTER 19 AVERAGES....................................................................................................................................... 175 CHAPTER 20 MIXTURES ....................................................................................................................................... 185

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3

CHAPTER 1 NUMBER BASES For general purposes we use numbers in base ten. The place value of each digit in a base ten number such as 816ten can be expressed as follows: 816 Hundred (8 x 10 2 or 8 x 100)

Tens Unit (1 x 10 1 or 1 x 10) (6 x 10 0 or 6 x 1)

Similarly, numbers may be expressed in other bases. For example 324seven can be expressed as follows: 3 2 4seven (3 x 72)

(2 x 71)

(4 x 70)

The rule above is employed in converting numbers from one base to base ten.

Conversion of numbers from other bases to base ten Examples 1. Convert the binary number 10111two to base ten.

Solution Each of the numbers is given a power starting from 0 on the right. This power is what the base digit will be raised to, when carrying out the expansion. 1403121110two = (1x24) + (0x23) + (1x22) + (1x21) + (1x20) = 16 + 0 + 4 + 2 + 1 = 23ten Note that any number raised to power zero is equal to 1. For example, 2 0 = 1.

2. Convert 3042five to a number in base ten.

4

Solution 33024120five = (3x53) + (0x52) + (4x51) + (2x50) = (3x125) + (0x25) + 20 + (2x1) = 375 + 0 + 20 + 2 = 397ten

Conversion of numbers from base ten to other bases Here the method of repeated division is employed. The base ten numbers will be divided by the new base digit, and the remainder will be written down. ‘R’ below, denotes remainder. Examples 1. Convert 60ten to a number in base two. Solution

2 2 2 2 2 2

60 30 15 7 3 1 0

R R R R R R

0 0 1 1 1 1

Read the remainders upwards ∴

60ten = 111100two

2. Convert 587ten to a number in base eight. Solution 8 8 8 8

587 73 9 1 0

R R R R

3 1 1 1

Read the remainders upwards

5

∴

587ten = 1113eight

Bicimals Base two fractions are called bicimals. Bicimals can also be converted to decimals (base ten fractions). Similarly, fractions in other bases can be converted to base ten decimals. Examples 1. Convert the bicimal 110.011two to a decimal. Solution Powers given to the numbers after the decimal point should be negative. 1

1

1

1

121100.0-11-21-3two = (1x22) + (1x21) + (0x20)+(0x2-1)+(1x2-2) + (1x2-3) = 4 + 2 + 0 + 0 + 22 + 23 = 6 + 4 + 8 = 6 + 0.25 + 0.125 = 6.375ten

2. Convert 223.32four to base ten. Solution 3

2

3

222130.3-12-2four = (2x42) + (2x41) + (3x40) + (3x4-1) + (2x4-2) = (2x16) + 8 + (3x1) + 4 1 + 4 2 = 32 + 8 + 3 + 4 2

3

1

+ 16 = 43 + 4 + 8 = 43 + 0.75 + 0.125 = 43.875ten

Conversion of decimals to numbers in other bases Examples 1. Convert 61.875ten to base two Solution The first step is to convert 61 to base two as follows: 2 2 2 2

61 30 R 1 15 R 0 7 R 1 6

2 3 2 1 0

R 1 R 1 R 1

61ten = 111101two The decimal part is now converted as follows: 0.875:

2 x 0.875 = 1.750 2 x 0.750 = 1.500 2 x 0.500 = 1.000

Keep multiplying the decimal part by the base digit until you get to a whole number. You may stick to the original number of decimal places in the question. Finally, the answer is obtained by taking only the digits before the decimal points, i.e. 111 0.875ten = 0.111two ∴ 61.875ten = 111101.111two

2. Convert 127.75ten to base six Solution The first step is to convert 127 to base six as follows: 6 127 6 21 R 1 6 3 R 3 0 R 3 127ten = 331six The decimal part is now converted as follows: 0.75:

6 x 0.75 = 4.50 6 x 0.50 = 3.00

7

Multiply only the decimal part of each value by the base digit until you get to a whole number. Finally, the answer is obtained by taking only the digits before the decimal points, i.e. 43 0.75ten = 0.43six ∴ 127.75ten = 331.43six 3. Convert the decimal 0.5625ten to a number in base six Solution 0.5625:

6 x 0.5625 = 3.375 6 x 0.375 = 2.25 6 x 0.25 = 1.50 6 x 0.50 = 3.00

Taking only the integers of the values obtained after each multiplication gives: 0.5625ten = 0.3213six

Conversion of numbers from one base to another Examples 1. Convert 110101two to a number in base five. Solution The number 110101two will first be converted to base ten before converting the base ten value to base five. 151403120110two = (1x25) + (1x24) + (0x23) + (1x22) + (0x21) + (1x20) = 32 + 16 + 0 + 4 + 0 + 1 = 53ten Now, convert 53ten to base five as follows: 5 53 5 10 R 3 5 2 R 0 0 R 2

= 203five 8

∴ 110101two = 203five

2. Convert 317nine to a number in base six. Solution 317nine = (3x92) + (1x91) + (7x90) = (3x81) + 9 + (7x1) = 243 + 9 + 7 = 259ten The next step is to convert 259ten to base six. This gives: 6 6 6 6

259 43 7 1 0

R R R R

1 1 1 1

= 1111 ∴

317nine = 1111six

Addition and subtraction of numbers in other bases Addition and subtraction of numbers in other bases are similar to that of base ten. Numbers equal to or greater than the base digit are not written down directly. Also, a larger number cannot be subtracted from a smaller number. The following examples illustrate how numbers are added and subtracted in other bases. Note that when arranging the numbers above each other, the place value system must be maintained. This means units under units, tens under tens, and so on. Examples 1. Evaluate 11011two + 111two Solution 1 1 0 1 1 + 1 1 1 1 0 0 0 1 0two

9

Workings: 1 + 1 = 2. Since 2 should not be written down in base two, it is evaluated as 2 2 (i.e the base digit )

= 1 remainder 0. The 0 which is the remainder is written down, while 1 is added

to the next column. So, the next column becomes 1 + 1 + 1 = 3. 3 divided by 2 is 1 remainder 1. Write down 1 which is the remainder, and take the other 1 to the third column. The third column becomes 1 + 0 + 1 = 2. As before this becomes 1 remainder 0. Write 0 and take 1 to the fourth column. This column gives 1 + 1 = 2. Write 0 and take 1 to the last column, which also adds up to 2. Write 0 down and take 1. Since there is no more columns left, write down the 1 at the end.

2. Evaluate 315.46eight + 27.164eight Solution

+

3 1 5 . 4 6 2 7 . 1 6 4 3 4 4 . 6 4 4eight

Workings: From the right, 4 + 0 = 4. Next, is 6 + 6 = 12. This 12 is greater than the base digit. This is now evaluated as

12 8(i.e base digit )

= 1 remainder 4. The 4 is written down, while 1 is added to the

next column, and so on. Note that the empty space is taken to contain 0. 3. Subtract 3178nine from 6246nine Solution 6 2 4 6 - 3 1 7 8 3 0 5 7nine Workings: Since 6 - 8 will not go. Hence, 1 has to be borrowed from 4. That 1 borrowed is equal to 9 (i.e. the base digit). This 9 is added to 6 to give 15. So the first column becomes 15 – 8 = 7. Note that the 4 in the next column is now 3 since 1 has been borrowed from it. The next column becomes 3 – 7. This is impossible. 1 has to be borrowed from 2. That 1 borrowed is equal to 9 (i.e. the base digit). This 9 is now added to 3 to give 12. So it becomes 12 – 7 = 5. The 2 in the third column becomes 1. So, 1 – 1 = 0. Finally, the last column is 6 – 3 = 3.

10

4. Evaluate 100101two – 1010two Solution 1 0 0 1 0 1 1 0 1 0 1 1 0 1 1two Note that any 1 borrowed is equal to 2 (i.e. the base digit), and it is added to the number that does the borrowing. In base two subtraction, it is always “0” that does the borrowing.

Multiplication of numbers in other bases Multiplication is carried out in a similar way to addition. When numbers are multiplied and the result is greater than the base digit, the value obtained is divided by the base digit. The remainder is what is written down, while the answer is carried to the next stage. Examples 1. Evaluate 1101two x 11two

Solution 1 x 1 + 1 1 1 0 0

1 0 1 1 1 1 0 1 0 1 1 1 1two

2. What is the total age of 253seven girls whose average age is 31seven. Express your answer in base seven. Solution This is a word problem that can be expressed as follows: 𝑥 253

= 31

Where all the values are in base seven 11

By cross multiplication, x is given by: xseven = 253seven x 31seven x = 2 5 3 x 3 1 2 5 3 +1 1 2 2 1 1 5 0 3seven

∴

The total age is 11503seven

Division of numbers in other bases Division is carried out by using the usual long division method, but it should be carried out in the given base. Examples 1. Evaluate 14201five ÷ 314five Solution 24 314 1 4 2 0 1 -1133 2321 -2321 -- - ∴

14201five ÷ 314five = 24five

Workings: 1420 ÷ 314, which gives 2 with a remainder. The 2 is obtained by multiplying 314 by 1, 2, 3, etc, in base five until you obtain a value that is equal to, or just less than 1420. This 2 is written at the top of the bar and used to multiply 314 to get 1133. Then subtract 1133 from 1420 in base five to obtain 232 as the remainder. Then bring down the next digit in 14201 (i.e. 1) to it to obtain 2321. Then repeat the task

12

2. Divide 11111111two by 101two Solution The division is carried out just like the example explained above. However, since this is in base two, the only values we can obtain in the course of long division when an immediate higher number divides a number just lower than it is always 1. This makes division in base two to be very easy to solve. For example, 111two ÷ 101two = 1. There will be a remainder. Therefore, the example above is now solved as follows. 110011 101 1 1 1 1 1 1 1 1 -1 0 1 101 101 111 101 101 101 ---∴

11111111two ÷ 101two = 110011two

Workings: 111 ÷ 101 = 1. Write the 1 on top of the bar and use it to multiply 101. This gives the 101 which is written under the 111 and subtracted from it. The subtraction gives 10. Now bring down 1 from the original number to make the 10 to become 101. Repeat the process of division using this 101. When this is done, 101 is subtracted from 101 and this gives 0. Now bring down 1 from the original number. Then 1 ÷ 101 will not go, so write 0 on the bar and bring down another 1. This give 11 ÷ 101 which will also not go. So write another zero on the bar and bring down another 1. This gives 111 ÷ 101 = 1. Write the 1 on the bar and continue the division process. Note that in division in other bases, the numbers involved can be converted to base ten. Then the division is carried out in base ten and the final answer is converted back to the original base. For example, 1100two ÷ 100two = 11two, can also be solved by converting 1100 and 100 to base ten to give 12 and 4 respectively. Then, 12 ÷ 4 = 3. When 3 is converted back to base two it gives 11 which is the required answer.

More examples on number bases 1. If 32x = 122three, find the value of the base x 13

Solution The numbers have to be converted to base ten. This gives: 3120x = 122120three (3 x x1) + (2 x x 0) = (1 x 32) + (2 x 31) + (2 x 30) 3x + 2 = 9 + 6 + 2 3x + 2 = 17 3x = 17 – 2 = 15 x=

15 3

=5

2. Solve for x if 23x + 65x - 71x = 15 Solution Each number has to be converted to base ten except 15 which is already in base ten. This gives: (2 x x 1) + (3 x x 0) + (6 x x 1) + (5 x x 0) – [(7 x x 1) +(1 x x 0)] = 15 2x + 3 + 6x + 5 – (7x + 1) = 15 2x + 3 + 6x + 5 – 7x - 1 = 15 8x – 7x = 15 – 3 – 5 + 1 x=8

Exercises 1. Convert 101011two to a number in base ten. 2. Convert 436ten to base six 3. Convert 254eight to a number in base ten. 4. Convert 1011.011two to a decimal 5. Convert 3032four to base seven 14

6. Convert 6136seven to a number in base five 7. Convert 597ten to an octadecimal (base 18) number. (Hint: Take A as 10, B as 11, C as 12, and so on). 8. Evaluate 405eight – 217eight 9. Evaluate 1.101two + 11.01two 10. Find the value of P in the equation: 101Pthree + 11three = 1100three 11. Evaluate 563eight x 62eight 12. If 23n = 111two, find the value of the base n. 13. Find (101two)2 14. Evaluate 425six ÷ 11six 15. Evaluate 341five ÷ 22five 16. Express the decimal

3 4

in bicimal

17. Convert 239.68ten to base five 18. Convert 47.625ten to a number in base two 19. Convert the decimal 0.0625ten to base six 20. If 198.921875ten = mfour, find the value of m.

15

CHAPTER 2 MODULAR ARITHMETIC This is a kind of arithmetic in which remainder is of interest. For example, 9 = 1 (mod 4), since

9 4

=2

remainder 1. So, the remainder (i.e. 1) is the answer. Note that “mod” is short for modulo. The basic arithmetic operations can be carried out in modular arithmetic.

Addition In modular arithmetic, addition is represented by ⊕, while subtraction is represented by ⊝, in order to differentiate them from the usual addition and subtraction signs.

Examples Evaluate the following: 1. 5 ⊕ 8 (mod 5) 2. 67 ⊕ 38 (mod 7) Solutions 1. 5 ⊕ 8 (mod 5) = 13 (mod 5). Since 13 is greater than 5 (i.e. the modulus), divide 13 by 5 to get the remainder which is the equivalent value. This gives: 13 (mod 5) = ∴

13 5

= 2 remainder 3

5 ⊕ 8 (mod 5) = 13 (mod 5) = 3 (mod 5)

2. 67 ⊕ 38 (mod 7) = 105 (mod 7) = ∴

105 7

= 15 remainder 0

67 ⊕ 38 (mod 7) = 105 (mod 7) = 0 (mod 7)

16

Subtraction Examples 1. Find the simplest form of each of the following: a. -5 (mod 6) b. -52 (mod 11) Solutions a. -5 (mod 6) = -5 + (6x1) = -5 + 6 = 1 (mod 6). (This is obtained by adding a multiple of the modulus digit (i.e. 6) that is equal to or just greater than 5) b. -52 (mod 11) = -52 + (11x5) = -52 + 55 = 3 (mod 11). (55 is the multiple of 11 that is just greater than 52) 2. Evaluate the following: a. 21 ⊖ 6 (mod 8) b. 8 ⊖ 18 (mod 3) c. 21 ⊖ 64 (mod 9) Solution a. 21 ⊖ 6 (mod 8) = 15 (mod 8) = 7 (mod 8) b. 8 ⊖ 18 (mod 3) = -10 (mod 3) = -10 + (3x4) = -10 + 12 = 2 (mod 3) c. 21 ⊖ 64 (mod 9) = -43 (mod 9) = -43 + (9x5) = -43 + 45 = 2 (mod 9)

Multiplication In modular arithmetic, multiplication is represented by ⊗, while division is represented by ⨸ in order to differentiate them from the usual multiplication and division signs.

Examples Evaluate the following: 17

1. 5 ⊗ 7 (mod 4) 2. 21 ⊗ 65 (mod 6) 3. 3 (mod 5) ⊗ 4 (mod 5) Solution 1. 5 ⊗ 7 (mod 4) = 35 (mod 4) = 3 (mod 4) 2. 21 ⊗ 65 (mod 6). This can be done easily by simplifying 21 and 65 in modulo 6. This gives: 21 (mod 6) = 3 (mod 6). And 65 (mod 6) = 5 (mod 6) 21 ⊗ 65 (mod 6) = 3 x 5 (mod 6) = 15 (mod 6) = 3 (mod 6) 3. 3 (mod 5) ⊗ 4 (mod 5) = 12 (mod 5) = 2 (mod 5)

Division Evaluate the following: 1. 24 ⨸ 6 (mod 5) 2. 2 ⨸ 5 (mod 6) 3. 8 ⨸ 9 (mod 7)

Solutions 1. 24 ⨸ 6 (mod 5) Let 24 ⨸ 6 (mod 5) be x. This gives: 24 ÷ 6 = x 24 6

=x

x=4 2. 2 ⨸ 5 (mod 6) Let 2 ⨸ 5 (mod 6) be x 18

2 ÷ 5=x 2 5

=x

5x = 2 (mod 6) (Now, look for a multiple of 6 (i.e. the modulus) such that when it is added to 2 it gives a number that is divisible by 5). The multiple is 18 (i.e. 6 x 3) 5x = 2 + (6 x 3) 5x = 2 + 18 = 20 x=

20 5

x = 4 (mod6) 3. 8 ⨸ 9 (mod 7) Let 8 ⨸ 9 (mod 7) be x 8 9

=x

∴ 9x = 8 (mod 7) 28)

(The multiple of 7 that should be added to 8 to obtain a number divisible by 9 is

9x = 8 + (7 x 4) 9x = 8 + 28 = 36 x=

36 9

x = 4 (mod 7)

Simple equations in modular arithmetic Examples Solve the following equations: 1. 8 ⊕ x = 0 (mod 9) 2. 2x = 3 (mod 7) 19

3. 5x ⊕ 2 = 3 (mod 11) 4. 4x ⊕ 8 = 2 (mod 9) Solutions 1. 8 ⊕ x = 0 (mod 9) 8+x =0 x = - 8 (mod 9) = - 8 + (9 x 1) =-8+9 x = 1 (mod 9) 2. 2x = 3 (mod 7) 2x = 3 + (7 x 1) = 3 + 7 2x = 10 x=

10 2

x = 5 (mod 7)

3. 5x ⊕ 2 = 3 (mod 11) 5x = 3 – 2 = 1 5x = 1 + (11 x 4) 5x = 1 + 44 5x = 45 x=

45 5

x = 9 (mod 11) 4. 4x ⊕ 8 = 2 (mod 9)

20

4x = 2 – 8 4x = -6 4x = - 6 + (9 x 2) 4x = - 6 + 18 4x = 12 x=

12 4

x = 3 (mod 9)

Exercises 1. Evaluate the following: a. 28 ⊕ 62 (mod 5) b. 39 ⊕ 97 (mod 8) c. 39 ⊖ 50 (mod 7) d. 7 ⊖ 58 (mod 14)

2. Find the simplest positive form of each of the following: a. – 23 (mod 5) b. – 81 (mod 12) 3. Evaluate the following: a. 33 ⊗74 (mod 7) b. 7 (mod 8) ⊗ 13 (mod 8) c. 42 ⨸ 11 (mod 2) d. 11 ⨸ 8 (mod 5) 21

4. Solve the following equations: a. 4x = 1 (mod 7) b. 2x + 3 = 1 (mod 6)

5. Copy and complete the table below in modulo 8.

⊕

5

6

7

1 2 3 4

6. Copy and complete the table below in modulo 7.

⊗

4

5

6

2 3 4 5

22

CHAPTER 3 STANDARD FORM AND APPROXIMATION OF NUMBERS A number is in standard form if it is expressed as follows: A x 10n, where A is a number between 1 and 10 and n is either a positive or a negative whole number. Examples 1. Express 8142 in standard form. Solution At the end of a whole number, there is a decimal point. This decimal point is moved towards the left, up to the right side of the first digit. The number of times that the point is moved becomes the power of 10. 8142 = 8.142 x 103 2. Express the following numbers in standard form: a. 100000 b. 100008 c. 4562000 Solutions a. 100000 = 1 x 105 (Note that the last zero(s) after a decimal point is/are irrelevant) b. 100008 = 1.00008 x 105 c. 4562000 = 4.562 x 106

3. Express the following numbers in standard form: a. 6510.248 b. 0.04381 c. 0.00000681

23

Solutions a. 6510.248 = 6.510248 x 103 b. For numbers less than 1 and greater than 0, we move the decimal point to the right, not left like in other examples. The number of times moved is given as a negative power of 10. 0.04381 = 4.381 x 10-2 c. 0.00000681 = 6.81 x 10-6

Approximation of numbers Numbers can be approximated to the nearest hundred, ten, whole number or even to a given number decimal places or significant figures. The digits 0, 1, 2, 3 and 4 are used to round down a number, while the digits 5, 6, 7, 8 and 9 are used to round up a number. Examples 1. Approximate the number 1209849 to: a. the nearest thousand b. the nearest hundred c. the nearest ten

Solutions a. The thousand digit in the given number is 9. The number at its right side (i.e. 8) is large enough to round up 9 and make it 10. When this is done, all the numbers in front of 9 become zero. Note that when 9 becomes 10, it is written as 0, while 1 is added to the number next to it. ∴ 1209849 = 1210000 (To the nearest thousand) b. Similarly, 1209849 = 1209800 (To the nearest hundred) Note that 4 is not large enough to round up 8 (i.e. the hundred digit), so 8 remains the same. c. 1209849 = 1209850 ( To the nearest ten)

24

2. Round off 24.65 to: a. the nearest whole number b. the nearest ten Solution a. To approximate a number to the nearest whole number means to either round up or round down the digit before the decimal point. In this example, 4 is the digit before the decimal point. 24.65 = 25 (To the nearest whole number) Note that 6 (i.e. the number to the right side of 4) is large enough to round up 4 to give 5. b. 24.65 = 20 (To the nearest ten). Note that 2 is the tens digit and 4 cannot round it up.

3. Round off 381.256996 to: a. 2 decimal places b. 5 decimal places Solutions a. To approximate to two decimal places, count the first two numbers after the decimal point and see if the third number will be able to round up the second number or not. Here, 6 will round up 5 and make it 6. 381.256996 = 381.26 (To 2 decimal places)

b. 381.256996 = 381.25700 (To 5 decimal places)

Significant figures It is important to know the first significant figure in a decimal fraction. For example, the first significant figure in 0.006045 is 6. The initial zeros are not regarded as significant figures. Examples 1. Round off 4906997 to:

25

a. 1 significant figure b. 6 significant figures Solution a. 4906997 = 5000000 (To 1 significant figure). Note that 9 has rounded up the first significant figure (i.e. 4) to give 5. b. 4906997 = 4907000 (To 6 significant figures). Note that 7 has rounded up the 9 in 699 to make it 700, while the 7 itself becomes 0.

2. Approximate 0.000460794 to: a. 1 significant figure b. 4 significant figures.

Solutions a. 0.000460794 = 0.0005 (To 1 significant figure) Note that the first zeros are not counted as significant figures. b. 0.000460794 = 0.0004608 (To 4 significant figures). Note that a zero after a significant figure is counted as significant.

Exercises 1. Express 2189000 in standard form. 2. Express the following numbers in standard form: a. 2500000 b. 12050800 c. 41102000 3. Express the following numbers in standard form: 26

a. 23700.212 b. 0.2170 c. 0.000026 4. Approximate 3840196 to: a. 2 significant figures b. 3 significant figures 5. Express the following numbers in standard form: a. 814000 b. 41218004 c. 0.0001002 d. 642.42 6. Round off 149.00562 to: a. the nearest ten b. 5 significant figures c. two decimal places 7. Approximate 0.005206798 to: a. two significant figures b. 6 significant figures c. 2 decimal places

27

CHAPTER 4 LAWS OF INDICES The following are the laws of indices. They are true for all values of a, b and x ≠ 0 Law 1: x a x x b = x a+b Law 2: x a ÷ x b = x a-b Law 3: x 0 = 1 Law 4: x -a =

1 𝑥𝑎

bx -a =

or

𝑏 𝑥𝑎

or

𝑏 −𝑎 = 𝑥

𝑥 𝑎 𝑏

Examples Simplify the following: 1. 105 x 104 2. m8 ÷ m5 3.

𝑎 −8 𝑎3

4. 5x 2 x 4x 0 x 2x -6 5. y-5 ÷ b0 Solution 1. 105 x 104 = 105+4 = 109 2. m8 ÷ m5 = m8-5 = m3 3.

𝑎 −8 𝑎3

= a- 8- 3 = a-11 =

1 𝑎 11

4. 5x 2 x 4x 0 x 2x -6 = (5 x 4 x 2)x 2+0+(-6) = 40x 2-6 = 40x -4 = 5. y-5 ÷ b0 = y-5 ÷ 1 = y-5 =

1 𝑦5

28

40 𝑥4

Product of indices In applying product of indices, the following are true: (x a)b = x ab Similarly, (x ayb)c = x acybc

𝑥

𝑥𝑎

𝑦

𝑦𝑎

and ( )a =

Examples Simplify the following: 1. (h4)-5 2. (2-3)2 3. (-c3)2 4. (-4u2v)3 Solution 1. (h4)-5 = h4x(-5) = h-20 = 2. (2-3)2 = 2-3x2 = 2-6 =

1 𝑕 20

1 26

=

1 64

3. (-c3)2 = -c3x2 = -c6 = c6 (A negative number that is raised to an even number power will give a positive value). 4. (-4u2v)3 = -41x3 u2x3 v1x3 = -43u6v3 = -64u6v3

Fractional indices In applying fractional indices, the following are true: x 1/a =

𝑎

𝑥

and x a/b =

𝑏

𝑥𝑎

𝑏

or x a/b = ( 𝑥 )a

In all cases, x ≠ 0

29

Examples Simplify the following: ⅓

1. 27 2. 9

-½ ½

3. (25a2)

1

4.

9 16

16 -2/3 54

5. ( )

Solutions ⅓

3

1. 27 = 27 = 3 2. 9

-½

=

1 9½

1

=

9

=

1

(Note that

3

2

9 should be written as 9 since 2 is not usually written with the

square root sign). ½

½ 2 x ½)

3. (25a2) = 25 a(

1

4.

9 16

25

=

16

16

8

54

27

=

½

= 25 a1 = ( 25 ) x a = 5a

5 4

5. ( )-2/3 = ( )-2/3 (When the fraction is expressed in its lowest term) 8

27

27 ⅔

8

8⅔

( )-2/3 = ( )2/3 = 27

3

=

( 27 )² 3

( 8)²

=

3² 2²

=

9 4

(Note that by taking the inverse of the term in the bracket,

the negative power becomes positive)

Equations in indices Examples Solve the following equations: 30

1. 4x-1 = 64 2. n

-⅔

=9

3. 2a-3 = -16 4. 9x = 27 -½

5. 5x = 40x

Solutions 1. 4x-1 = 64 (This is solved by expressing both sides of the equation in the same base and then equating the powers. This gives: 4x-1 = 43 Equating the powers gives: x -1 = 3 x=3+1 x=4

2. n

-⅔

= 9 (In this case, the unknown letter is the base. To solve this, make the power of n to 1 by

multiplying this power by its inverse and using the same sign of the power. The other side of the equation should also be raised to the same power). This gives: -⅔ (n )-3/2 = 9-3/2

n(-2/3 n1 = n=

x -3/2)

1 ( 9)³

=

=

1 3³

1 93/2

=

1 27

1 27

31

3. 2a-3 = -16 Divide both sides by 2. a-3 =

−16 2

a-3 = -8 Now make the power of ‘a’ to be 1 by multiplying this power by its inverse. This gives: -⅓

(a-3)

a=

-⅓

= (-8) 1

(−8)⅓

a=-

=3

1 −8

=

1 −2

1 2

4. 9x = 27 Expressing both sides of the equation in the same base gives: (32)x = 33 32x = 33 Equating the powers gives 2x = 3 x=

3 2

-½

5. 5x = 40x

Divide both sides by 5. x=

40𝑥 −½ 5

-½

x = 8x

(Since 40 ÷ 5 gives 8)

32

-½

Divide both sides by x 𝑥

𝑥

−½ =

8𝑥 −½ 𝑥 −½ -½

Cancelling out x 1-(-½)

x

x÷x

= 8 (Note that x can be expressed as x1. Also, from the law of indices,

-½

∴ x

on the right hand side gives,

1-(-½)

=x

1+½

)

=8

x³/² = 8 Make the power of x to be 1 by multiplying it by its inverse. Also raise the power of 8 to the same inverse. This gives: ⅔ ⅔ (x³/²) = 8

x=8

⅔

3

x = ( 8 )2 = 22 x=4

Exercises 1. Simplify the following: a. -3(te3)4 b. (4ab3)3 c.

(–𝑎)2 𝑥 𝑎 7 (−𝑎)5

d. (-g4)5 e.

(𝑚 2 )3 𝑚 4 𝑥 (−𝑚 )

33

2. Simplify the following: a. (3a)-1 -½

b. (a2)

½

c. (49x 3)

⅔ d. (27x³/²)

3. Solve the following equations: -½

a. x

=5

b. a-2 = 9 c. 9x-2 = 27 d.

4 2𝑥 −1 16 2

= 64

34

CHAPTER 5 LOGARITHMS OF NUMBERS GREATER THAN 1 – USE OF TABLES Logarithm is another word for power. For example, 100 =10 2, and log10100 = 2. This means that the logarithm to base 10 of 100 is 2. Another example is log381 = log334 = 4. Logarithm to base 10 which is called common logarithm will be used here. Examples 1. Use the logarithm tables present in mathematical tables (commonly called four-figure tables) to find the logarithm of the following: a. 6.2 b. 29.4 c. 8 d. 438.5

Solutions a. When 6.2 is expressed in standard form it gives 6.2 x 100. The power of 10 which is 0 in this case is the “characteristic” of the logarithm of 6.2. This characteristic is the integer written down before checking the logarithm tables to get the fractional part. Log 6.2 = 0.7924 (Note that 6.2 can be expressed as 6.200 to make it up to four digits. So, 0 is the characteristic of 6.2, while 7924 is obtained by using the logarithm tables to look up 62 under 0, ‘difference’ 0. In doing this, look up the first two numbers under the third number and add the ‘difference’ of the fourth number. Since the last number is 0, it gives a ‘difference’ of 0) b. When 29.4 is expressed in standard form, it gives 2.94 x 10 1. This shows that the characteristic is 1 (i.e. the power of 10) Log 29.4 = 1.4683 (Here, look up 29 under 4). c. Log 8 = 0.9031 (8 can be expressed as 8.000. So, look up 80 under 0).

35

d. Log 438.5 = 2.6420 (Look up 43 under 8, ‘difference’ 5. The difference is obtained by using the ‘difference’ section of the logarithm tables. The ‘difference’ is added to the value obtained. Here, 43 under 8 gives a value of 6415, while 5 in the difference section gives a value of 5. Adding these values gives 6415 + 5 = 6420. So, log 438.5 = 2.6420. Note that the integer ‘2’ before the decimal point is the characteristic of 438.5). The section of the logarithm tables in which log 438.5 was obtained is as shown below. Differences X

0

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

9

43

6335

63454

6355

6365

6375

6385

6395

6405

6415

6425

1

2

3

4

5

6

7

8

9

Notice that all the examples above show that the characteristic of the logarithm of a number is one less than the number of digits before the decimal point in that number. For example, in 6.2, there is 1 number before the decimal point, so its characteristic is 0 (i.e. 1-1). Also, in 8 (or 8.000), there is also 1 number before the decimal point, so its characteristic is also 0. Similarly, in 438.5, there are 3 numbers before the decimal point, so its characteristic is 2 (i.e. 3-1). This method can be used to directly obtain the characteristic of a number.

2. Use the antilogarithm tables present in mathematical tables to find the number whose logarithm is: a. 2.142 b. 0.6165 c. 4

Solutions a. In finding the antilogarithm of a number, first ignore the integer part (i.e. the digit before the decimal point), and look up the fractional part using the antilogarithm tables. After that, you use the integer part to place the decimal point. When placing the decimal point, the number of digits before the decimal point is one greater than the integer (i.e. the opposite of what was done for logarithm). The antilog of 2.142 = 138.7 (This is obtained by first ignoring the integer part i.e. 2, and looking up 14 under 2 in the antilogarithm tables to get 1387. The 2 (i.e. the integer part), that was initially ignored is now used to place the decimal point. The digit to count is obtained by adding 1 to the 36

integer 2, to obtain 3 (i.e. 2 + 1 = 3). So, 3 digits should be counted in 1387 before placing the decimal point. This will give the final value of 138.7. b. Antilog of 0.6165 = 4.135 (Look up 61 under 6, ‘difference’ 5. The difference is obtained by using the ‘difference’ section of the antilogarithm tables. The ‘difference’ is added to the value obtained. Here, 61 under 6 gives a value of 4130, while 5 in the difference section gives a value of 5. Adding the values gives 4130 + 5 = 4135. The final step involves the use of the integer part (i.e. 0) to place the decimal point. So count 1 digit (i.e. 0 + 1 = 1) in 4135 before placing the decimal point to obtain 4.135. So, the antilog of 0.6165 = 4.135. Note that the integer ‘0’ before the decimal point was initially ignored. The section of the antilogarithm tables in which the antilog of 0.6165 was obtained is as shown below. Differences X

0

1

2

3

4

5

6

7

8

9

1

2

3

4

5

6

7

8

9

.61

4074

4083

4093

4102

4111

4121

4130

4140

4150

4159

1

2

3

4

5

6

6

8

9

c. Antilog of 4 = Antilog of 4.0000 = 10000. (Look up 00 under 0 to obtain 1000. Then count 5 digits (i.e. 4 + 1 = 5) before placing the decimal point. Notice that one zero has been added to 1000 to obtain 10000 in order to complete the 5 digits needed)

Multiplication and division of numbers by using mathematical tables Using mathematical tables to evaluate calculations is based on the laws of indices. Examples 1. Use antilogarithm tables to evaluate the following: a. 100.6112 b. 101.24 x 102.1021 c. 103.194 ÷ 100.9317

Solutions a. 100.6112 = the antilog of 0.6112 = 4.085 b. 101.24 x 102.1021 = 101.24 + 2.1021 = 103.3421 = 2198 (From the antilog of 3.3421) 37

c. 103.194 ÷ 100.9317 = 103.194 - 0.9317 = 102.2623 = 182.9 (From the antilog of 2.2623)

2. Use mathematical tables to evaluate the following: a. 715.4 x 4.31 b. 216 x 28 c. 6214 ÷ 98.76 d. e.

62.4 x 5.12 12.04 3.169 x 92.1 3.96 x 6.72

Solutions a. 715.4 x 4.31. To do this, simply add the logarithm of the numbers and then find the antilogarithm of the value obtained. No. Log 715.4 2.8545 4.31 + 0.6345 Antilog 3.4890 3083 Note that the antilog of 3.4890 gives the answer, 3083 ∴ 715.4 x 4.31 =3083

b. 216 x 28 No. 216 28 Antilog 6049

Log 2.3345 + 1.4475 3.7820

38

Note that the antilog of 3.7820 gives the answer, 6049 ∴

216 x 28 = 6049

c. 6214 ÷ 98.76 =

6214 98.76

To do this, simply subtract the logarithm of the numbers and then find the antilogarithm of the value obtained. No. Log 6214 3.7934 98.76 - 1.9946 Antilog 1.7988 62.92 Note that the antilog of 1.7988 gives the answer, 62.92 ∴ d.

6214 ÷ 98.76 = 62.92

62.4 x 5.12 12.04

In this case, add the logarithms of the numerator and subtract the logarithm of the denominator from it. The antilog of the value obtained gives the final answer No. 62.4 5.12

Log 1.7952 + 0.7093 2.5045 12.04 - 1.0806 Antilog 1.4239 26.54

∴

e.

62.4 x 5.12 12.04

= 26.54

3.169 x 92.1 3.96 x 6.72

In order to evaluate this, add the logarithms of the numerator and also add the logarithm of the denominator. Subtract the value obtained for the denominator from that obtained for the 39

numerator. This gives a value whose antilog gives the final answer. This is as evaluated below.

No. 3.169 92.1

Log 0.5009 + 1.9643 2.4652 2.4652 3.96 0.5977 6.72 +0.8274 1.4251 - 1.4251 Antilog 1.0401 10.97 Note that the antilog of 1.0401 gives 10.97 ∴

3.169 x 92.1 3.96 x 6.72

= 10.97

Calculations of powers and roots using mathematical tables In carrying out calculations when numbers are in powers and roots, find the logarithm of the number and multiply it by its power. Note that fractional powers are known as roots. Examples Use mathematical tables to evaluate the following: 1. 84.142 2.

3

403.4

3.

6.

3

21.6 1067

4. 5.

31.2

⅓

29.4 4

31.87 x 1.863

5

(6.838)3

40

7. 8.

38.32 𝑥 2.964

3

8.637 𝑥 6.285

³

(17.2)2 x 4.93 3

675000

Solutions 1. 84.142 Find the logarithm of 84.14 and multiply it by 2. Then find the antilog of the value obtained. No Log 2 84.14 1.9250 x 2 Antilog 3.8500 7079 Note that the antilog of 3.8500 gives 7079 ∴

2.

3

84.142 = 7079

31.2

Find the logarithm of 31.2 and multiply it by ⅓, and then find the antilogarithm of the value obtained. No Log 31.2 1.4942 x ⅓ Antilog 0.4981 3.148 3

3.

403.4

3

21.6

This is evaluated by subtracting the logarithm of the denominator from the logarithm of the numerator. The antilogarithm of the value obtained gives the final answer. This is as evaluated below. No. 403.4 21.6

Log 2.6057 - 1.3345 1.2712 x 3 Antilog 3.8136 6510 41

1067

4.

⅓

29.4

No 1067 21.6

Log 3.0282 - 1.4683 1.5599 x ⅓ Antilog 0.5200 3.311

5.

4

31.87 𝑥 1.863

No 31.87 21.6

Log 1.5034 +0.2702 1.7736 x ⅟4 Antilog 0.8864 7.705

6.

5

(6.838)3

This can also be expressed as: (6.838)

⅗

No. Log 6.838 0.8349 x ⅗ Antilog 0.5009 3.169

7.

3

38.32 x 2.964 8.637 x 6.285

²

42

This can also be expressed as

No. 38.32 2.964 8.637 6.285

38.32 x 2.964 8.637 x 6.285

Log 1.5834 + 0.4719 2.0553 0.9364 + 0.7983 1.7347

Antilog 1.636

8.

(17.2)2 x 4.93 3

675000

⅔

2.0553

- 1.7347 0.3206 x ⅔ 0.2137

3

Note that 675000 can also be expressed as (675000)

⅓

No. 17.22

Log 1.2355 x 2 2.4710 4.93 + 0.6928 3.1638 3.1638 3 5.8293 x ⅓ 675000 1.9431 - 1.9431 Antilog 1.2207 16.62

Note: The logarithm of a number can also be obtained directly by using calculators. In using a calculator, the antilogarithm of a number can be obtained by raising 10 to the power of that number. For example, the antilog of 0.3086 is given by 100.3086 which is equal to 2.035.

Relationship between indices and logarithms If y = ax, then logay =x. This means that x is the logarithm of y to the base a. 43

Similarly, if logay =𝑥, then y = ax

Examples 1. Write the following in index form and find the values of x: a. Log28 = x b. Log5125 = x c. Log100.001 = x Solutions a. Log28 = x.

Expressing this in index form will give:

2x = 8 2x = 23

(Since the base 2 are the same on both sides, they cancel out)

Equating the powers gives: x=3

b. Log5125 = x 5x = 125 5x = 53 The base will cancel out since they are equal. Equating the powers now gives: x=3 c. Log100.001 = x 10x = 0.001 10x = 10-3 (Note that 0.001 in standard form is 10-3) x = -3

44

2. Solve the following equations: a. Loga3 =

1 4

b. Logy0.25 = -

1 2

Solutions a. Loga3 =

1

Expressing this in index form gives:

4

¼

a = 3 (Since the base is the unknown, make its power to be 1 by multiplying this power by its inverse. Also raise the other side of the equation to the same power). ¼ 1 (a )4 = 34 (Note that the inverse of is 4) 4

¼

b. Logy0.25 = -½

y

= 0.25 -½

y

=

= a1 = a)

1 2

Express 0.25 in fraction

1

Making the power of y to be 1 gives:

4

-½ x -2)

y(

¼ x 4)

(Note that (a )4 = a(

a = 81

1

= ( )-2 4

4

y = ( )2 1

y = 42 y = 16

Exercises 1. Use mathematical tables to evaluate the following: a. 101.24 x 102.12 45

b.

10 0.25 x 10 1.214 10 0.715

c. 615 x 30.04 d. 3.254 x 38.31 x 401.5 e. f. g.

81.6 x 3.142 12.2 x 16 713.4 35 x 4.95 (314.5)2 84

28 x 5

h.

i. (10000) j.

960.5 14.02 3

k. l.

⅕

6314000 ÷ 14.2 (2 x 3.007)2

5

28.5 x 12 x 3.14 x 92

2. Solve the following equations: a. Log464 = x b. Loga1.5 =

1 3

c. Log0.25a = 4 1

2

9

3

d. Logy = -

46

CHAPTER 6 THEORY OF LOGARITHMS Logarithm can be in bases other than base 10. For example 27=33 means log327=3. For solved examples on the relationship between indices and logarithms, refer to chapter 5. Laws of logarithms The three basic laws of logarithms are: 1. Log (XY) = logX + logY 𝑋

2. Log ( ) = logX – logY 𝑌

3. Log (XY) = YlogX Examples 1. Simplify the following as far as possible: a. Log1015 + log106 b.

3 4

log81

Solutions a. Log1015 + log106 = log10(15 x 6) = log1090 b.

3 4

¾ 4 log1081 = log1081 = log10( 81)3 = log10(3)3 = log1027

2. Express the following as logarithms of single numbers: a. Log1060 – log103 b. 2 – 2log105 Solutions 60

a. Log1060 - log103 = log10( ) = log1020 3

47

100

b. 2 – 2log105 = log10100 – log1052 = log10100 – log1025 =log10(

25

) = log104

Note that 2 has been converted to log10100 as follows: let log10x = 2 x = 102 = 100 2 = log10100 (This was substituted for 2 in example 2)

3. Given that log102 = 0.3010, log103 = 0.4771 and log107 = 0.8451, evaluate the following: a. log1042 b. log1035 Solutions a. log1042 = log10(2 x 3 x 7) = log102 + log103 + log107 = 0.3010 + 0.4771 + 0.8451 = 1.6232 7 x 10

b. log1035 = log10(

2

) = log107 + log1010 - log102 = 0.8451 + 1 + 0.3010 = 1.5441 (Note that log1010

= 1, in a similar way that log10100 = 2)

1

25

2

4

4. Simplify log10

4

320

5

125

– 2log10 + log10

Solution 1 2

25

log10

4

320

5

125

– 2log10 + log10

5

320

2

125

log10( x

4

x

25 16

) = log10(

25 ½

= log10( )

320 125

4

x

125 32

)

4 320 - log10( )2 + log10 5 125

( Since 5 x 25=125, and 2 x 16=32)

Cancelling out 125 gives: 320

log10(

32

= log10(

25 320 x 125 4 42 52

)

= log1010 = 1

48

)=

5 320 x 2 log10( 16125 25

)

5. Simplify 1 + log102 Solution Not that 1 = log1010, since 101 = 10

(This is similar to what was done in example 2)

1 + log102 = log1010 + log102 (When log1010 is substituted for 1) log1010 + log102 = log10(10 x 2) = log1020

6. If log10(2x + 1) – log10(3x - 2) = 1, find x Solution log10(2x + 1) – log10(3x - 2) = 1 log10(2x + 1) – log10(3x - 2) = log1010 10 = 10. Therefore, 1 = log1010)

(Since converting 1 to logarithm in base 10 is given by

1

2𝑥+1

log10(

3𝑥−2

) = log1010

Comparing both sides of the equation above shows that: 2𝑥+1

(

3𝑥−2

) = 10

10(3x - 2) = 2x + 1 (By cross multiplication) 30x - 20 = 2x + 1 30x - 2x = 1 + 20 28x = 21 Dividing both sides by 28 gives: x= x=

21 28

3 4

(When simplified to its lowest term)

7. Without the use of tables, simplify the following: 49

a.

b. c.

log 3 16 log 3 8 log 10 8 + log 10 4 log 10 8 − log 10 4 log 8 27 log 8 9

Solutions

a.

log 3 16 log 3 8

log 3 2⁴

=

log 3 2³

=

4 log 3 2 3 log 3 2

4 log 3 2

=

3 log 3 2

Cancelling out log32 gives:

=

b.

3 4

log 10 8 + log 10 4 log 10 8 − log 10 4

=

log 10 2⁵ log 10 2

=5

c.

log 8 27 log 8 9

=

log 10 (8 x 4) 8 log 10 ( 4

5 log 10 2 log 10 2

=

)

=

log 10 32 log 10 2

5 log 10 2 log 10 2

(After cancelling out log102)

=

log 8 33/2 log 8 3²

=

=

log 8 27 1/2 log 8 3²

=

3 2

log 8 3

2log 8 3

=

= 3 2

2

log 8 (33 )1/2 log 8 3²

=

3 2

=

log 8 (33 x 1/2 ) log 8 3²

3

1

3

2

2

4

÷2= x =

8. Evaluate the following: a. Log415 b. Log721

50

Solutions a. Log415 Let log415 = 𝑥 4𝑥 = 15 Taking the logarithm to base 10 of both sides gives: Log104𝑥 = log1015 𝑥log104 = log1015 𝑥=

𝑙𝑜𝑔 10 15 𝑙𝑜𝑔 10 4

=

1.1761 0.6021

(From mathematical tables or the use of calculator)

= 1.9533 ∴ Log415 = 1.95

b. Log721 Let log721 = 𝑥 7𝑥 = 21 Taking the logarithm to base 10 of both sides gives: Log107𝑥 = log1021 𝑥log107 = log1021 𝑥=

𝑙𝑜𝑔 10 21 𝑙𝑜𝑔 10 7

=

1.3222 0.8451

(From mathematical tables or the use of calculator)

= 1.5645 ∴ Log715 = 1.56

51

Exercises 1. Simplify the following: a. 2 – lo104 b. 5 lo102 + lo105 - lo101.6 75

2. Simplify lo10

10

5

160

9

243

- 2log10 + log10

3. Given that log2 = 0.3010, log3 = 0.4771 and log5 = 0.6990, evaluate: a. log45 b. log1.2 b. log3.6 3. Solve for 𝑥 in the following equations: a. log10𝑥 – log10(2𝑥 – 1) = 2 3

b. 2log𝑥 (3 ) = 6 8

4. Simplify the following without using tables: a. b. c.

log

3

log 3 log 256 log 16 log 40 − log 5 log 16 –log 8

5. Use logarithm tables to evaluate: a. log523.69 b. log240 6. Simplify the following: a. log798 – log730 + log715 b. lo324 + log315 – 2log310 52

7. Given that log52 = 0.431 and log53 = 0.681, find the values of: a. log513.5 3

4

2

8

5

5

b. log5 + 2log5 - log5

53

CHAPTER 7 LINEAR EQUATIONS AND CHANGE OF SUBJECT OF FORMULAE Linear equations Linear equations can sometimes be expressed with brackets, with fractions or both. When solving linear equations with fractions, clear the fractions by multiplying each term of the equation by the L.C.M of the denominators of the equation. Examples 1. Solve the following equations: a. 2x + 4(3 –x) = 11 b. 6(a – 3) – 2(5a – 8) = -4 c. a – 5(2 + a) – (3a – 4) = 2(2a – 1) – 7 Solutions a. 2x + 4(3 –x) = 11 Expanding the bracket gives: 2x + 12 – 4x = 11 Collect like terms 12 - 11 = 4x – 2x 1 = 2x Divide both sides by 2 ∴ x=

1 2

b. 6(a – 3) – 2(5a – 8) = -4 Expanding the brackets gives: 6a – 18 – 10a + 16 = -4

(Note that -2 x -8 = +16)

Collect like terms: 54

6a – 10a = -4 + 18 – 16 -4a = -2 Divide both sides by -4 −4𝑎 −4

a= a=

−2

=

−4

−2 −4 1 2

c. a – 5(2 + a) – (3a – 4) = 2(2a – 1) – 7 Expanding the brackets gives: a – 10 – 5a – 3a + 4 = 4a – 2 – 7

(Note that -5 x (+a) = -5a, and - x -4 = -1 x -4 = +4)

Collect like terms: 7 + 2 + 4 – 10 = 5a – a + 3a + 4a 3 = 11a Divide both sides by 11 3 11

=

a=

11𝑎 11 3

11

2. Solve the following equations: a. b. c.

3 4

1

5

3

6

x - (x - 2) = - (2x -1)

1 6

2

(5x - 2) - (4 - x) = 1 3

5 2𝑚 −3

-

3 4

1

= +7 6

55

Solutions a.

3 4

1

5

3

6

x - (x - 2) = - (2x -1)

In order to clear the fractions, multiply each term in the equation by 12 which is the L.C.M of the denominators, i.e. 4, 3 and 6. This gives: 3𝑥

1

5

4

3

6

12( ) - 12 x (x - 2) = 12( ) - 12(2x -1) Cancelling out by using the 12 to divide the various denominators gives: 3(3x) - 4(x - 2) = 2(5) - 12(2x - 1) 9x - 4x + 8 = 10 - 24x + 12 Collect like terms: 9x - 4x + 24x = 10 + 12 – 8 29x = 14 x=

b.

14 29

1 6

2

(5x - 2) - (4 - x) = 1 3

In order to clear the fractions, multiply each term in the equation by 6 which is the L.C.M of the denominators, i.e. 6 and 3. This gives: 1

2

6

3

6 x (5x - 2) - 6 x (4 -x) = 6 x 1 Cancelling out by using the 6 to divide the various denominators gives: (5x - 2) - 2 x 2(4 -x) = 6 5x - 2 - 4(4 -x) = 6 5x - 2 - 14 + 4x = 6 5x + 4x = 6 + 2 + 14 9x = 22 56

∴ x=

c.

22 9

5 2𝑚 −3

-

3 4

1

= +7 6

Multiply each term in the equation by 12(2m - 3) which is the L.C.M of the denominators, i.e. (2m - 3), 4 and 6. This gives: 12(2m - 3)

5 2𝑚 −3

- 12(2m - 3)

3 4

1

= 12(2m - 3) + 12(2m - 3) x 7 6

Cancelling out gives: 12(5) - 3(2m - 3)3 = 2(2m - 3) + 12(2m - 3)7 60 - 9(2m - 3) = 4m - 6 + 84(2m - 3) 60 - 18m + 27 = 4m - 6 + 168m - 252 60 + 27 + 252 + 6 = 4m + 168m +18m 345 = 190m m= ∴

345 190

m=

69 38

(After equal division by 5)

Change of subject of formulae If m = b + c, then m is the subject of the formula. If its rearranged to give b = m - c, then b is now the new subject of the formula. In changing the subject of a formula, simply solve the equation for the letter which is to become the new subject. Examples 1. Make h the subject of the formula: s =

𝑤𝑑 𝑕

𝑑

(h - ) 2

57

Solution s=

𝑤𝑑 𝑕

𝑑

(h - ) 2

Expanding the bracket gives: s=

𝑤𝑑 𝑕

(h) -

𝑤𝑑 𝑑 𝑕

( ) 2

Canceling out the h gives: s = wd -

𝑤𝑑2 2𝑕

To clear fractions, multiply throughout by 2h (LCM) 𝑤𝑑2

2h(s) = 2h(wd) - 2h( 2hs = 2hwd – wd2

2𝑕

)

(Note that the 2h at the end on the right side has cancelled out).

Collect terms in h wd2 = 2hwd – 2hs Factorizing the right hand side gives: wd2 = h(2wd - 2s) Divide both sides by (2wd - 2s) 𝑤𝑑2 2𝑤𝑑 −2𝑠

=

𝑕(2𝑤𝑑 −2𝑠) 2𝑤𝑑 −2𝑠

Cancelling out the 2wd – 2s on the right hand side gives: h=

𝑤𝑑2 2𝑤𝑑 −2𝑠

2. Given that I =

𝐸 𝑅2

+ 𝑊 2 𝐿2

, express R in terms of I, E, W and L.

Solution I=

𝐸 𝑅2

+ 𝑊 2 𝐿2

58

Cross multiply I 𝑅 2 + 𝑊 2 𝐿2 = E Square both sides to remove the square root sign. (I 𝑅 2 + 𝑊 2 𝐿2 )2 = E2 I2(R2 + W2L2) = E2 I2R2 + I2W2L2 = E2 I2R2 = E2 - I2W2L2 Divide both sides by I2 R2 =

𝐸 2 − 𝑊 2 𝐼 2 𝐿2 𝐼2

This can also be simplified as follows: R2 =

𝐸 2 𝐼 2 𝑊 2 𝐿2 𝐼2

-

𝐼2

Canceling out I2 on the right side gives: R2 =

𝐸2 𝐼2

- W2L2

Take the square root of both sides in order to remove the square sign on R 2. ∴ R=

𝐸2 𝐼2

− 𝑊 2 𝐿2

3. Make x the subject of the formula R =

𝑎𝑥 − 𝑃 𝑄 + 𝑏𝑥

Solution R=

𝑎𝑥 − 𝑃 𝑄 + 𝑏𝑥

Square both sides to remove the square root sign

59

R2 =

𝑎𝑥 − 𝑃 𝑄 + 𝑏𝑥

By cross multiplication it gives: R2(Q + bx) = ax - P R2Q + R2bx = ax – P Collecting terms in x gives ax - R2bx = R2Q + P Factorizing the left hand side gives: x (a - R2b) = R2Q + P Divide both sides by (a - R2b) 𝑥(𝑎 − 𝑅 2 𝑏) (𝑎 − 𝑅 2 𝑏)

=

𝑅2𝑄 + 𝑃 𝑎 − 𝑅2𝑏

𝑅2𝑄 + 𝑃

x=

𝑎 − 𝑅2𝑏

Exercises 1. Solve the following equations: a. 5x + 2(3 –x) = 10 b. 2(2a – 3) – 5(4a – 1) = -6 c. b – 4(1 + b) – (5b – 1) = -(b – 3) – 2 2. Solve the following equations: a. b. c.

1 4

2

3

3

4

x - (x - 1) = - (5x -2)

1 6

(5x - 2) 2

2𝑛 −3

-

3 5

5 12

(3 - 2x) = 1 1

=2

2

60

𝑑

3. Make p the subject of the formula: tp = md(p - ) 3

4. Given that V =

𝑃 𝐸2

+ 𝐼2 𝐶 2

, express C in terms of I, V, E and P.

5. Make x the subject of the formula R =

𝑏𝑥 − 𝑆 𝑇 + 𝑎𝑥

61

CHAPTER 8 VARIATION Direct variation Direct variation involves the relationship between two quantities whereby an increase or decrease in one of them leads to an increase or decrease respectively in the other. The symbol  means ‘varies with’ or is ‘proportional to’. Example 1. If x varies directly as y and x =30 when y=12, find: a. the formula connecting x and y b. x when y=10 c. y when x =20

Solution a. x  y

(This means x varies directly as y)

x = Ky (Replacing the proportionality sign with the equals sign introduces a constant K) So, when x = 30 and y = 12, the equation above becomes: 30 = K x 12 30 = 12K K=

30 12

=

5 2

The formula connecting x and y is: 5

x= y 2

5

(This is obtained by substituting for K in the equation above, i.e. x = Ky) 2

b. When y = 10, x is given by:

62

5

x= y 2 5

50

2

2

x = x 10 =

= 25

∴ x = 25

c. When x = 20, y is given by: 5

x= y 2

5

20 = y 2

5y = 40 y=

40 5

y=8

2. If m varies as the square of n and m=27 when n=3, find: a. the relationship between m and n b. n when m = 48 c. m when n = 2

1 3

Solution a. m  n2

(This means m varies as n2)

m = Kn2 (Replacing the proportionality sign with the equals sign introduces a constant K) So, when m = 27 and n = 3, the equation above becomes: 27 = K x 32 27 = 9K K=

27 9

=3 63

The relationship between m and n is: m = 3n2

(This is obtained by substituting 3 for K in the equation m = Kn2)

b. When m = 48, n is given by: m = 3n2 48 = 3 x n2 48 = 3n2 n2 =

48 3

= 16

n = 16 n=4

1

c. When n = 2 , m is given by: 3

m = 3n2 7

1

7

3

3

3

m =3 x ( )2 ( Note that 2 has been converted to ) m=3x

49 9

Canceling out gives: m=

49 3

Inverse variation In inverse or indirect variation, as one quantity increases the other decreases. Examples 1. If c varies inversely as d and c=18 when y=4, find:

64

a. the formula connecting c and d b. c when d=10 c. d when c=12

Solution 1

a. c  c=

(This means c varies inversely as d)

𝑑 𝐾

(Replacing the proportionality sign with the equals sign introduces a constant K)

𝑑

So, when c = 18 and d = 4, the equation above becomes: 𝐾

18 =

4

K = 18 x 4 K = 72 The formula connecting c and d is: c=

72

𝐾

(This is obtained by substituting 72 for K in the equation above, i.e. c = )

𝑑

𝑑

b. When d = 10, c is given by: c=

72 𝑑 72

c=

10

c = 7.2 c. When c = 12, d is given by: c=

72 𝑑

12 =

72 𝑑

65

12d = 72 72

d=

12

d=6

2. If r varies inversely as the cube root of t and r=6 when t=64, find: a. the relationship between r and t b. t when r = 16 c. r when t =

8 27

Solution 1

a. r  3

𝑡

𝐾

r=3

(This means r varies inversely as the cube root of t) (Replacing the proportionality sign with the equals sign introduces a constant K)

𝑡

So, when r = 6 and t = 64, the equation above becomes: 6=3 6=

𝐾 64

𝐾 4

K = 6 x 4 = 24 The relationship between r and t is: 24

r=3

(This is obtained by substituting 24 for K in the equation r =

𝑡

b. When r = 16, t is given by: 24

r=3

𝑡

66

𝐾 3

𝑡

)

24

16 = 3 3

3

𝑡=

𝑡

24 16 3

𝑡 =

2

In order to remove the cube root, take the cube of both sides. This gives: 3

3

( 𝑡)3 = ( )3 2

∴ t=

27 8

c. When t =

8 27

, r is given by:

24

r=3 r=3

𝑡 24 8

27

24

r= 2

3

r = 24 x

3 2

After equal division by 2, it gives: r = 12 x 3 r = 36

Joint variation In joint variation, three or more quantities are related directly or inversely or both. Examples 1. If m varies directly as the square of n and inversely as p, and m=3 when n=2 and p=8, find: 67

a. the relationship between m, n and p b. m when n = 3 and p =27 1

3

2

2

c. p when m = and n = Solutions m m=

𝑛2

(This means m varies directly as the square of n and inversely as p)

𝑝

𝐾𝑛 2 𝑝

So, when m = 3, n = 2, and p = 8, the equation above becomes: 3= 3=

𝐾 𝑥 22 8 4𝐾 8

4K = 3 x 8 = 24 K=

24 4

=6

The relationship between m, n and p is: m=

6𝑛 2 𝑝

(This is obtained by substituting 6 for K in the equation m =

b. When n = 3 and p = 27, then m is given by: m= m= m= m=

6𝑛 2 𝑝 6 x 32 27 6x9 27 54 27

m=2

68

𝐾𝑛 2 𝑝

)

c. When m = m= 1 2 1 2

=

1

3

and n = , then p is given by:

2

2

6𝑛 2 𝑝 3 2

6x 𝑝 9

=

𝑝

p=9x2 P = 18

2. The weight w of a rod varies jointly as its length L and the square root of its density d. If w =12 when L = 5 and d = 9, find: a. L in terms of w and d b. w when L = 8 and d = 25 c. d when L = 20 and w = 4

Solutions a. w  L 𝑑

(This means w varies jointly as L and the square root of d)

w = KL 𝑑 So, when w = 12, L = 5, and d = 9, the equation above becomes: 12 = K x 5 x 9 12 = 15K K=

12 15

=

4 5

The formula connecting w, L and d is: 4

w= L 𝑑 5

4

(This is obtained by substituting for K in the equation w = KL 𝑑 ) 5

69

L can now be expressed in terms of w and d as follows: 4

w= L 𝑑 5

5w = 4L 𝑑 Dividing both sides of the equation by 4 𝑑 gives: 5𝑤

L=

4 𝑑

b. When L = 8 and d = 25, then w is given by: 4

w= L 𝑑 5 4

w = x 8 x 25 5 4

w= x8x5 5

Cancelling out the 5 gives: w=4x8 w = 32

c. When L = 20 and w = 4, then d is given by: 4

w= L 𝑑 5

4=

4 5

x 20 x 𝑑

4 x 5 = 4 x 20 x 𝑑 20 = 80 𝑑 𝑑 =

20 80

=

1 4

Taking the square of both sides gives: 1

( 𝑑 )2 = ( )2 4

70

d=

1 16

Partial variation The fourth type of variation is called partial variation. In partial variation, one quantity is partly constant and partly varies with the other. Two constants are involved in partial variation. Examples 1. x is partly constant and partly varies as y. When y=2, x =30, and when y=6, x =50. a. Find the formula which connects x and y. b. Find x when y=3

Solutions a. From the first sentence, we have: x = C + Ky

(Let this be equation 1) where C and K are constants.

Substituting y=2 and x =30 in this equation gives: 30 = C + 2K

(Let this be equation 2)

Similarly, when y=6 and x =50, we have: 50 = C + 6K

(Let this be equation 3)

Bringing equation 2 and 3 together gives: 30 = C + 2K

(Equation 2)

50 = C + 6K (Equation 3) Equation 3 - Equation 2: 20 = 4K Divide both sides by 4. K=

20 4

=5

Substitute 5 for K in equation 2. 71

30 = C + 2K 30 = C + (2 x 5) 30 = C + 10 30 - 10 = C C = 20 We now substitute the values of C and K into equation 1 in order to obtain the formula connecting x and y. The formula connecting x and y is now given by: x = 20 + 5y b. When y=3, x is obtained by substituting 3 for y in the formula connecting x and y. x = 20 + 5y x = 20 + (5x3) = 20 + 15 = 35 x = 35

2. m is partly constant and partly varies as n. When n=4, m=5, and when n=12, m=14. a. Find the formula which connects m and n. b. Find m when n=16 c. Find n when m=9

Solutions a. From the first sentence, we have: m = C + Kn

(Let this be equation 1) where C and K are constants.

Substituting n=4 and m=5 in equation 1 gives:

72

5 = C + 4K

(Let this be equation 2)

Similarly, when n=12 and m=14, we have: 14 = C + 12K

(Let this be equation 3)

Bringing equation 2 and 3 together gives: 5 = C + 4K

(Equation 2)

14 = C + 12K (Equation 3) Equation 3 - Equation 2: 9= 8K Divide both sides by 8. K= Substitute

9 8

9 8

for K in equation 2.

5 = C + 4K 9

5 = C + (4 x ) 8

5=C+ 5C=

9

9 2

=C

2 1 2

We now substitute the values of C and K into equation 1 in order to obtain the formula connecting m and n. The formula connecting m and n is given by: m=

1 2

9

+ n 8

b. When n=16, m is obtained by substituting 16 for n in the formula connecting m and n. m=

1 2

9

+ n 8

1

9

2

8

m = + ( x 16)

73

1

37

2

2

= + 18 = m = 18

1 2

c. When m=9, n is obtained by substituting 9 for m in the formula connecting m and n. m=

1

9

+ n

2

8

1

9𝑛

2

8

9= +( ) 917 2

1

=

2

=

9𝑛 8

9𝑛 8

17 x 8 = 9n x 2 136 = 18n n=

136 18

n=7

=

68 9

5 9

Exercises 1. If x varies directly as y and x =10 when y=8, find: a. the formula connecting x and y b. x when y=10 c. y when x =16 2. If h varies as the square root of p and h=5 when p=9, find: a. the relationship between h and p b. p when h = 20

74

c. h when p = 6

1 4

3. If p varies inversely as q and p=12 when q=3, find: a. the formula connecting p and q b. q when p=20 c. p when q=5 4. If m varies inversely as the cube root of n and m=5 when n=27, find: a. the relationship between m and n b. m when n = 8 c. n when m =

64 125

5. If a varies directly as the square of b and inversely as c, and when a=4 when b=3 and c=6, find: a. the formula connecting a, b and c b. a when b = 5 and c = 10 1

c. b when a = and c = 8 2

6. The height h of a box varies jointly as its length L and the square of its width w. If h = 20 when L = 4 and w = 3, find: a. w in terms of h and L b. w when h = 12 and w = 4 c. L when h = 8 and w = 5 7. x is partly constant and partly varies as y. When y=4, x =14, and when y=5, x =17. a. Find the relationship between x and y. b. Find x when y=8 8. E is partly constant and partly varies as F. When F=2, E=25, and when F=5, E=55. a. Find the formula which connects E and F.

75

b. Find E when F=2

1 2

c. Find F when E=40

76

CHAPTER 9 REVIEW OF BASIC ARITHMETIC In order to fully understand the content of this book, it is important to be familiar with basic arithmetic operations. The following points should remind us of these basics. 1. Numbers are usually written in the decimal place value system. Take for instance the number 2306. 418. The place value of each of the digits in the number is: 2 - Thousands 3 - Hundreds 0 - Tens 6 - Units . - Decimal point 4 - Tenths 1 - Hundredths 8 – Thousandths 2. A whole number is a number without fraction. 3 and -5 are whole numbers 3. A whole number which divides exactly into another whole number is called a factor. For example 6 is a factor of 18. 18 is a multiple of 6. A multiple is a whole number which can divide another whole number without a remainder. For example 36 is a multiple of 12. This also means that 12 is a factor of 36. 4. The numbers 16, 24 and 28 all have 4 as a factor. Therefore, 4 is a common factor of 16, 24 and 28. The highest common factor (HCF) is the largest of the common factors of a given set of numbers. For example, 2, 4 and 8 are the common factors of 16, 24 and 32. However, 8 is their HCF. The number 21 is a multiple of 3 and a multiple of 7. Therefore 21 is a common multiple of 3 and 7. The lowest common multiple (LCM) of a set of numbers is the smallest of the common multiples. For example 18 and 36 are common multiples of 6 and 9. However, 18 is their LCM 5. A prime number is a number that can divide only itself and 1, i.e. it has only two factors, 1 and itself. 2, 3, 5, 7, 11, 13, 17, 13….., are prime numbers. 1 is not a prime number. The prime factors of a number are those factors which are prime numbers. For example 2, 3 and 5 are prime factors of 60. Therefore 60 can be expressed as a product of its prime factors as follows: 60 = 2 × 2 × 3 × 5 or 60 = 22 × 3 × 5.

77

6. A fraction is a number formed when one number is placed above another with a line separating 3

3

7

7

them. The fraction means 3 ÷ 7. In the faction : 3 is the numerator ̶ is the dividing line 7 is the denominator Fractions are usually exposed in their longest terms. For example,

16 36

4

in its longest term is i.e. when 9

numerator and denominator are divided by the same number which is 4 in this case.

7. An integer is any positive or negative whole number. Integer are also referred to as directed numbers. The examples below show how to add, subtract, multiply and divide integers.

ADDITION AND SUBTRACTION OF INTEGERS (DIRECTED NUMBERS) a. -8 + 3 = -5 A traditional way of doing this is to assume that a negative number indicates an owed amount of money, while a positive number indicates a paid amount of money. So in this example, I owe someone $8 and pays $3. It means that I am still owing $5. Since I am still owing, then my answer will be negative, i.e. -5. Another way of solving this problem is to subtract the lower number from the higher number to get the answer. This is done when the two numbers have different signs. Finally, take the sign of the higher number as the sign of the answer. In this example, 3 subtracted from 8 is 5. The sign of the higher number (i.e. 8) is minus, so the answer 5 will carry a minus sign. This will finally give us -5 b. -4 + 7 = +3 Here, I owe $4 and pays $7. It means that I am going to get a balance of $3. Since I get a balance, then my answer will be positive, i.e. 3. Or, since their signs are different, subtract the lower number from the higher number to get the answer. Finally, take the sign of the higher number as the sign of the answer. In this example, 4 subtracted from 7 is 3. The sign of the higher number (i.e. 7) is plus, so the answer 3 will carry a plus sign. This will finally give us +3 or 3 c. 5 - 9 = -4 Since their signs are different, subtract 5 from 9. This gives 4. The sign of the higher number (i.e. 9) is minus, so the answer 4 will carry a minus sign. This will finally give us -4.

78

d. -2 - 5= -7 In this case, I am owing $2 and I am also owing $7. This gives means that I am owing a total sum of 7. This gives -7, since I am owing. Or, since their signs are the same, add the two numbers to get the answer. Finally, give the sign that they have, to the answer. In this example, 2 added to 5 gives 7. Their sign is minus, so the answer 7 will carry a minus sign. This will finally give us -7. e. 3 - (-4) = 3 + 4 = 7 Two negative signs that are closed to each other multiplies to become positive. f. -6 - (-2) = -6 + 2 = -4. (This is similar to (a) above) g. -14 - 17 = -31 Since their signs are the same, add the two numbers to get the answer. Finally, give the sign that they have, to the answer. In this example, 14 added to 17 gives 31. Their sign is minus, so the answer 31 will carry a minus sign. This will finally give us -31. MULTIPLICATION AND ADDITION OF INTEGERS (DIRECTED NUMBERS) a. -2 x 3 = -6 Carry out multiplication and division as the usual way of multiplying and dividing numbers. However, when one of the numbers is negative, the answer will be negative. When the two numbers are negative, the answer will be positive. b. 5 x -4 = -20 c. -2 x -6 = 12 d. -7 x -3 = 21 e. -10 ÷ 2 = f. 18 ÷ -6 =

−10 2 18

= -5

= -3

−6 −48

g. -48 ÷ -12 = h. -24 ÷ -6 =

−12 −24 −8

=4

=3

8. A rational number is any number in the form 2 1 7

3

3 5 1

3

of rational numbers are , , ,

and

21 10

𝑥 𝑦

.

79

where x and y are integers, and y ≠ 0. Examples

Non-rational numbers (irrational numbers) are numbers that cannot be expressed as exact fractions. 2 and 𝜋 (3. 14159…) are examples of irrational numbers.

9. Quantities can be obtained by a method known as simple proportion. This is used when one quantity increases as the other increases, or when one quantity decreases as the other decreases. For example: a. If 125 eggs cost $5 then the cost 50 eggs can be obtained by simple proportion as follows: 125 eggs cost to $5 Therefore,

50 eggs will cost: =

250 125

50 125

x5

= $2

A simple way of handling such a problem is to first arrange the quantities in order. Then the new quantity will be obtained by dividing the lower quantity (50 above) by the upper quantity (125 above), and then multiplying by the quantity on the upper right. Note that the quantity to be calculated a placed on the lower right (this is just a convention that can be changed). It is important to understand the arrangement process. With the example above the arrangement was done as follows: Eggs ---- $ Eggs ---- $ (value to be calculated) With this arrangement in mind, let us examine another example. b. If 72 people can eat 30kg of rice, how many people will be able to eat 40kg of rice? Solution 30kg is eaten by 72 people Therefore, 40kg will be eaten by:

40 30

x 72

= 4 x 24 (After equal division by 10 and 3) = 96 people. As explained in example (a) above, after the arrangement, the arithmetic is done by dividing the lower quantity on the left hand side (40kg) by the upper quantity (30kg), and then multiplying by the quantity on the upper right hand side. Note that the quantity to be calculated a placed on the lower right hand side.

80

c. A length of 60cm shows a temperature of 150. What temperature will be shown by 50cm? Solution 60cm corresponds to 150 Therefore, 50cm will correspond to: =

50 𝑥 5 2

50 60

x 150

(After equal division by 10 and 3)

= 125. d. A box of volume 1000cm3 can carry a total of 360 small balls. What size of box will carry exactly 270 small balls. Solution 360 corresponds to 1000cm3 Therefore, 270 will correspond to:

270 360

x 1000

= 3 x 250 (After equal division by 10, 9 and 4) = 750cm3.

Exercises 1. Evaluate the following: a. -5 + 12 = b. 12 - (- 5) = c. -18 + 7 = d. 6 - 19 = f. -4 – 6 = g. -9 - - 22 = h. -5 x -7 = i. 4 x -11 = j. -9 x 8 = 81

k. 28 ÷ -7 = l. -56 ÷ -14 = m. -42 ÷ 6 = 2. If 160 eggs cost $8 what is the cost of 60 eggs? 3. If 120 people can eat 50kg of maize, how many people will be able to eat 15kg of maize? 4. A length of 20m shows a temperature of 110. What temperature will be shown by 34m? 5. A cylinder of volume 850cm3 weighs 1.5kg when filled with water. What will be the weight of the cylinder when 500cm3 of water is in it?

82

CHAPTER 10 FRACTIONS EXAMPLES ON ADDITION AND SUBTRACTION OF FRACTIONS Simplify the following: 1. 2. 3.

3

+

4 5 6

1

+

5 1

7

2 2

+

3 1 2

4. 1 8 1

1 4

5 6

1

5 12

5. 1 - 2 + 3

6

9

10

Solutions 1.

3 5

1

+

2

The LCM of the denominators i.e. 5 and 2 is 10. Use 10 to divide each of the denominators, and 3

multiply the answer by the corresponding numerator. For the , we have 10 ÷ 5 = 2, then 2 × 3 = 6. 5

1

For the 2, we have 10 ÷ 2 = 5, then 5 × 1 = 5. So 6 and 5 becomes the numerators to be added, while 10 (LCM) becomes the new denominator. The solution is thus shown below. 3 5

+

1 2

=

= ∴

Note that

3 5 11 10

+

1 2

=

2 𝑥 3 +(5 𝑥 1) 10 6+5 10

=

11 10

11 10

can also be expressed as 1

1 10

In order to convert the improper fraction

in mixed fraction. 11 10

1

to the mixed fraction 1 , divide 11 by 10. Its gives 1 10

remainder 1. The number of times the numerator divides the denominator i.e. 1, becomes the whole number, while the remainder, i.e. 1, becomes the new numerator. The original denominator, i.e. 10 remains the same. Another example is that, if

14 3

2

is expressed as a mixed fraction it gives 4 . This is 3

obtained as follows: 83

14 ÷ 3 = 4 remainder 2. So, 4 becomes the whole number, 2 becomes the new numerator while 3 remains the denominator. ∴

2.

14 3 1

2

=4 . 3

2

+

4

5

+

3

6

The LCM of 4, 3 and 6 is 12. So divide 12 by each of the denominator and multiply by the corresponding numerator. This gives: 1 4

2

+

5

+

3

3 𝑥 1 + 4 𝑥 2 + (2 𝑥 5)

=

6

=

12 3 + 8 + 10 12

=

∴

1 4

2

+

3

5

3.

6

-

5

+

7

3

4

4

6

7

1

8

4

4. 1 -

-

(In its lowest term when you use 21 and 12 to divide 3)

(in mixed fraction)

5−3 6 2

1

6

3

= = 5

4

6

=

∴

7

1𝑥5 − 3𝑥1

=

2

12

=

= =1

6

1

21

-

1 2

=

(In its lowest term after division by 2)

1 3

5 12 7

Convert the mixed fraction 1 to an improper fraction. In order to do this, multiply the whole number by the 8

denominator and add the numerator. The answer gives the new numerator. 7

Therefore 1 = (1×8) + 7 = 8+7 = 15. So, 15 becomes the new numerator while the original 8

84

denominator remains the same. 7

15

8

8

7

1

8

4

So, 1 = ∴ 1 15 8

-

. 5

-

1 4

12

, becomes:

5

-

3 𝑥 15 − 6 𝑥 1 − (2 𝑥 5)

=

12

24 45− 6− 10

=

24 29

=

1

1

9

3

6

10

5. 1 - 2 +

24

1

1𝑥 3 +1

3

3

1 =

=

1

2𝑥 6 +1

6

6

2 = 1

1

9

3

6

10

∴ 1 - 2 + 4 3

-

13 6

+

9 10

=

5

24

3+1 3

=

12 + 1 6

4 3

=

(In improper fraction) 13 6

(In improper fraction)

becomes:

=

=

=

=1

10 𝑥 4 − 5 𝑥 13 + (3 𝑥 9) 30 40 – 65 + 27 30 2

30

=

1 15

(In its lowest term)

EXAMPLES ON MULTIPLICATION AND DIVISION OF FRACTIONS Simplify the following: 1

2

4

3

1. 2 ×

85

1

2. 1 3.

1

÷3

8

4. 1

2

1

3

4

x 4 x 2

7

3

3 4

10 1

1

9

3

5. 1 ÷ 1 6. 7.

8.

5

x 2

6 1

1 5

3

1

5

10

x 3

)

3 5

2

÷ 1

4

3

+

2

3

÷ (2 ÷

3

2

1

1

4

5

4

3

5

÷ 1 x (2 - 1 ) of

4

3 4

6 − 2 +1 1 4

1 6

1 𝑜𝑓 2

Solutions 1

2

4

3

1. 2 ×

1

Convert 2 to improper fraction and solve as follows: 4

9 4

x

2 3

=

18 12

=

3 2

(In its lowest term)

This was done by simply multiplying the numerators with each other, and then multiplying the denominators with each other. Then express the answer in its lowest term by dividing numerator and denominator by 6. 1

2. 1

7

2

1

3

4

x 4 x 2

Convert all the fractions to improper fraction. 8 7

x

14 3

x

9 4

Multiplying numerators and then denominators will give large values. Therefore a more convenient way of solving this problem is to divide the fractions to their lowest terms. Divide either across (i.e. between one numerator and another denominator) or between numerators and denominators of the 86

same fraction. Avoid dividing between one numerator and another numerator or between one denominator and another denominator. Applying this rule gives: 8

x

7

14

9

x

3

=2×2×3

4

(After equal division by 7, 4 and 3. 7 divided by 7 gives 1, while 14

divided by 7 gives 2. 8 divided by 4 gives 2, while 4 divided by 4 gives 1. Finally, 3 divided by 3 gives 1, while 9 divided by 3 gives 3. All these make the numerators to be 2, 2 and 3 respectively, while the denominators become 1 each). ∴ 2 x 2 x 3 = 12. 1

∴ 1

7

3.

1

2

1

3

4

x 4 x 2 = 12

3

1

4

8

÷3 =

8

15

÷

4

Change the division to multiplication and take the inverse of the right hand fraction. This gives: 1

15

÷

8

1

=

4

x

8

4 15

=

1 30

After equal division of 4 and 8 by 4, we will be left with

4. 1

3

3

1

5

10

÷ (2 ÷

10

1 2

x

1 15

, which multiplies out to give

)

Handling the part in the bracket gives: 3

1

5

10

2 ÷

13

=

5

=

13 5

÷ x

1 10 10 1

= 13 x 2 (After equal division of 5 and 10 by 5).

= 26 The question now simplifies to: 1

3

10

= =

13 5 13 5

÷ 26 ÷ x

26 1

(Note that a whole number can be expressed as the number divided by 1)

1 26

87

1 30

.

= =

1

1

x

5

(After dividing 13 and 26 by 13)

2

1 10

1

1

9

3

5. 1 ÷ 1

x 3

3 5

Here, the rule of BODMAS must be applied. Each letter in BODMAS stands for “Bracket”, “Of”, “Division”, “multiplication”, “addition”, “subtraction”. This means that in carrying out operation in mathematics, you must follow the order of the letters of BODMAS. Handle ‘Bracket’, before ‘of’, then ‘Division’ and so on. This shows that the division part of this example (5) should be handled first. This is done as follows: 1

1

10

9

3

9

1 ÷ 1 =

10

=

9

=

5 3

=

÷ x x

4 3 3 4 1

(After equal division of 10 and 4 by 2, and then 3 and 9 by 3)

2

5 6

The question now simplifies to: 5 6

= =

6.

5 6

3

÷ 3

5

5

÷

6

18

=

5

5 6

x

5 18

25 108

x 2

3 4

2

÷ 1

3

Convert to improper fraction. This gives: 5 6

x

11 4

÷

5 3

Apply BODMAS by handling the division part first. This gives: 88

11

5

÷

4

3

11

=

x

4

3 5

33

=

20

The question now simplifies to: 5 6

= =

7.

1 2

33

x 1

x

2 11

11

(After division of 5 and 20 by 5, and then division of 33 and 6 by 3)

4 3

=1

8

3

+

20

8

2

1

1

4

5

4

3

5

÷ 1 x (2 - 1 ) of

4

By applying the rule of BODMAS, the part in the bracket will be handled first. This gives: 1

1

9 4

4

3

4 3

2 -1 = -

3 𝑥 9 – (4 x 4)

= =

12 27−16 12

=

11 12

The question now simplifies to: 1 2

+

3 4

2

11

5

12

÷1 x

of

4 5

“Of” comes next in BODMAS. So, simplify the “Of” part as follows: 11 12

of = =

4 5 11 3

= x

11 12 1 5

x

4 5

(After division of 4 and 12 by 4)

11 15

Note that “Of” is simplified using multiplication, i.e. “x”. The question now simplifies to:

89

1

3

+

2

2

11

5

15

÷1 x

4

Next, is “Division”. So simplifying the division part gives: 3

2

3

5

4

÷1 =

4

= =

3

x

4

÷

7 5

5 7

15 28

The question now simplifies to: 1

+

2

15

x

28

11 15

The multiplication part is next. This gives: 15 28

=

x

11 15

11

(After cancelling out the 15)

28

The question now finally simplifies to: 1 2

+ = = =

8.

1 5

11 28 14 𝑥 1 + (1 𝑥 11) 28 14 + 11 28 25 28

3 4

6 − 2 +1 1 4

1 6

1 𝑜𝑓 2

Handle the numerator first. Since this involves only addition and subtraction, the entire fraction can be handled together. This gives:

90

1

3

5

4

6 - 2

1

31

6

5

+ 1 = = = =

-

11 4

+

7 6

12 𝑥 31 − 15 𝑥 11 + (10 𝑥 7) 60 372 – 165 + 70 60 277 60

The denominator is now handled as follows 1

5

2

4

4

1

1 of 2 = x = =

10 4 5 2

(In its lowest term)

The question now simplifies to: 277 60 5 2

= = =

277 60 277 60 277

=

30

÷ x x

5 2 2 5 1 5

(After dividing 2 and 60 by 2)

277 150 127

=1

150

Exercises 1. Simplify the following: a.

3 4

+

1 3

91

b. c.

2

+

5 7 8

1

1

1

+

3 3 4

6

3

7

5

10 3

d. 3 - 1 e.

2

6

5

- 1 + 4

3

6

8

2. Simplify the following: 3

5

5

9

a. 3 × 3

5

4

7

b. 1 x 6 x c.

9 10

÷5

2 5

5

5

3

8

6

11

1

2

5

9

3

6

d. 2 ÷ (1 ÷

e. 5 ÷ 7 x 2 f.

5 9

x 1

3

÷ 3

5

4

2

1

3

2

g. 2 +

h.

1

2 5

1

1 7

4

2

6 8

9

÷ 1 x (5 - ) of

1 4

2 − 3 +2 1 2

)

1 10

3 𝑜𝑓 4

92

CHAPTER 11 WORD PROBLEMS INVOLVING FRACTIONS Examples 1. Find the number of minutes in the following: a. b. c.

1

hour

5

2 15 1

of 2 hour 1

of 2 hours

3

2

Solutions a. Express the hour in minutes 1 hour = 60 minutes. ∴

1 5

= b.

1

1

60

5

5

1

hour = × 60 = ×

2 15

60 5

= 12 minutes

of 2 hours

2 hours = 2 × 60 = 120 minutes. 2 15

of 2 hours = 2

8

1

1

= x

2 15

× 120 =

2 15

x

120 1

(After dividing 15 and 120 by 15)

=2x8 = 16 minutes c.

1 3

1

of 2 hours 1

2

1

5

2

2

2 hours = 2 × 60 = 2

5

60

2

1

× 60 = ×

= 5 x 30 = 150 minutes

93

∴

1 3

1

of 2 hours 2

1

1

150

3

3

1

= x 150 = x =

150 3

= 50 minutes 2 a. Express 40 cents as a fraction of $10. b. Express 4 days as a fraction of 4 weeks c. Express 25cm as a fraction of 3m. Solutions a. The two quantities must be in the same units. So, express $10 in cents $10 = 10 × 100 (since $1= 100 cents). = 1000 cents Now express 40 cents as a fraction of 1000 cents. This gives: 40 1000

=

2 50

(After equal divisions by 10 and then by 2)

b. 7 days = 1 week so 4 weeks gives: 4 × 7 = 28 days Now express 4 days as a fraction of 28 days. This gives: 4 28

=

1 7

c. 1m = 100cm. ∴ 3m = 3 × 100 = 300cm. Now express 25cm as a fraction of 300cm. This gives: 25 300

=

1 12

(In its lowest term)

94

3. Determine the greater or greatest of the following set of fractions: a. b. c.

2 5

or

7 13

3 7 8

or

15

7 3 10

, ,

9 4 13

Solutions a.

2 5

or

3 7

Find the LCM of 5 and 7. This gives 35. Now simply as if you are adding or subtracting the fractions. This gives: 2 5

or

3 7

= =

7 𝑥 2 𝑜𝑟 (5 𝑥 3) 35 14 𝑜𝑟 15 35

Since 15 is greater than 14, its shows that the second fraction that resulted to 15 is greater. So,

3 7

is

greater.

b.

7 13

or

8 15

The LCM of 13 and 15 is 195. Use this LCM to simplify as if you are adding or subtracting the fractions. This gives: 7 13

or

8 15

=

=

15 𝑥 7 𝑜𝑟 (13 𝑥 8) 195

105 𝑜𝑟 104 195

Since 105 is greater than 104, its shows that the first fraction that resulted to 105 is greater. So, greater. Note that the LCM of 13 and 15 is 195, which can be simply obtained by multiplying 13 by 15.

95

7 13

is

c.

7 3 10

, ,

9 4 13

The LCM of 9, 4 and 13 is 468. Now simply as if you are adding or subtracting the fractions. This gives: 7 3 10

, ,

9 4 13

=

52 𝑥 7 ,

=

117 𝑥 3 , (36 𝑥 10) 468

364, 351, 360 468

The value 364 from the first fraction is the greatest. So,

7 9

is the greatest of the fractions.

4. A man earned $120 as wages. He spent $24. What fraction of his original money does he have left? Solution The amount of money he has left is: 120 - 24 = 96. Fraction of his original money left is: 96 120

=

4

(In its lowest term)

5

3

2

5

3

5. How much less than 9 is the sum of 4 and 3 ? Solution 3

2

5

3

The sum of 4 and 3 is given by: 3

2

23

5

3

5

4 +3 = = = =

+

11 3

3 𝑥 23 + (5 𝑥 11) 15 69 + 55 15 124 15

=8

4

15

96

So, 8

4

is less than 9 by:

15

9-8

4

=

15

124 15

15

= =

3

-

15 𝑥 9 − (1 𝑥 124)

=

6.

9 1

135 −124 15 11 15

of the boys in a class play football, and

4

4 7

play tennis. Every boy plays at least one of these two

games. If 36 boys play both games, how many boys are there in the class? SOLUTION Note that when a quantity has been broken into fractions, the total of the entire fraction should be equal to 1 ∴ Total fraction of boys that played the two games is given by: 3 4

4

7 𝑥 3 + (4 𝑥 4)

7

28

+ = =

Since

37 28

21+16 28

=

37 28

is greater than 1, it means that some boys have been counted twice, and these are the boys

that play both games. ∴ Fraction of boys that play both games is given by: 37 28

-1= =

=

∴

9 28

37 28

-

1 1

1 𝑥 37 +(28 𝑥 1) 28 37−28 28

=

9 28

of the boys play both games

97

But,

9 28

= 36 boys

(Those that play both games).

Therefore total boys = 1 whole number (Since total fraction should be 1). So by using the method of simple proportion the total boys is obtained as follows. If

9 28

= 36 boys

Then, 1 will be = (1 ÷ 1

9

1

28

=( ÷ 1

28

1

9

= x

9 28

) x 36

) x 36

x 36

= 28 x 4 = 112 ∴ There are 112 boys in the class

5

1

1

8

2

8

7. During a semester in a school, of the students had measles, had chickenpox and had neither. What fraction of the students in the school had both measles and chickenpox? Solution Actual fraction of students affected by the two diseases is given by: 1

1 1

8

1 8

1- = = = ∴

7 8

8 𝑥 1 − (1 𝑥 1) 8 8−1 8

=

7 8

of the student was actually affected by diseases.

But, the total fraction of student affected by the two diseases given by: 5 8

+

1 2

= =

1 𝑥 5 + (4 𝑥 1) 8 5+4 8

=

9 8

Since this is greater than

7 8

which is the actual fraction affected by the two diseases, it then means

that some students have been counted twice, and these are the student affected by both disease.

98

∴ Fraction of students who had both measles and chickenpox is given by: 9

7

-

8

8

1 𝑥 9 – (1 𝑥 7)

= = =

8 9−7 8 1

=

2 8

(In its lowest term)

4

3

8. A man shared a certain amount of money among his three sons. of the money was given to the eldest son. The second son received

5 8

7

of the remaining money.

a. What fraction of money did the third son get? b. If the first son received $3021 more than the second son, how much money was shared among them. Solution a. After giving

3 7

of the money to the first son, the fraction of the money left is 1 –

fraction must be 1. This gives: 1 1

-

3 7

=

7− 3 7

=

4 7

The second son received 5 8

of

4

8

7

5

1

2

7

= x

8

4

of this remaining fraction of . This gives: 7

7

5

= x

=

4

5

(After division of 4 and 8 by 4)

5 14

∴ The second son received

5 14

of the money.

Now, the fraction of money that the third son got would be given by:

99

3 7

, since the total

3

5

7

14

1-( +

2 𝑥 3 + (1 𝑥 5)

=1-(

14

6+5

=1-( =1=

)

14

11 14

)

) 1 11

= -

1 14

14 − 11

=

14

3 14

∴ The third son got

3 14

of the money.

b. Fractional difference between first and second son is: 3

-

5

7 14

=

2 𝑥 3 − (1 𝑥 5)

=

14

6−5 14

=

1 14

By applying simple proportion, the total money (i.e. a fraction of 1) is obtained as follows 1 14

gives $3021

Therefore 1 will give = (1 ÷ 1

14

1

1

=( x

1 14

) × 3021

) x 3021

= 14 x 3021 = 42294 ∴ The amount of money shared among them is $42294.

4

2

9

5

9. A man spent of his salary on food, and of the remainder on housing. He shared the rest equally among his four daughters. If each of the daughters received $54,000, how much was spent on food? Solution 4

After spending on food the remainder is: 9

100

4

1 4

9

1

1- = = 2 5

9

9−4 9

=

5 9

of this remainder was spent on housing.

∴ Fraction spent on housing is given by: 2

5

2

5

9

5

9

of = x

5

2

=

(After cancelling out the 5)

9

Total fraction spent on food and housing is: 4 9

2

4+2

9

9

+ =

6

2

9

3

= =

The third fraction is the fraction spent on his daughters. These three fractions must sum up to 1. So, 2

subtract from 1 in order to obtain the fraction spent on his daughters. This gives: 3

2

1 2

3

1 3

1- = = ∴

1 3

3−2 3

=

1 3

1

is the fraction spent on the daughters. Since is the fraction that was shared equally among 3

four daughters, then the fraction that each daughter received is given by: 1 3

1

4

3

1

1

1

1

3

4

12

÷4= ÷ = × =

4

By proportion method, the amount spent on food (i.e. ) can be obtained as follows: 1 12

9

(one daughter) gives $54000 4

4

1

9

9

12

So, (on food) will give ( ÷ 4

12

9

1

= x

) x 54000

x 54000

= 4 x 12 x 6000

(After dividing 54000 by 9) 101

= 288000 ∴ The Amount spent on food is $288000

Exercises 1. Find the number of minutes in the following: a. b. c.

1 12 4

of 3 hours

9 2 5

hour

2

of 1 hours 3

2 a. Express 35 cents as a fraction of $7. b. Express 9 days as a fraction of 9 weeks c. Express 120m as a fraction of 2km.

3. Determine the greater or greatest of the following set of fractions: a. b. c. d.

9 10 3 4

or

or

19 22

37 50

17 21 43

,

,

20 25 50 2 13

,

,

7

3 20 10

4. A man earned $540 as wages. He spent $90. What fraction of his original money does he have left? 1

9

2

10

5. How much less than 15 is the sum of 7 and 6 ?

102

6.

2 5

of the boys in a class play volleyball, and

7

play handball. Every boy plays at least one of these

10

two games. If 5 boys play both games, how many boys are there in the class? 7. During a dinner in a school,

5 16

5

1

8

4

of the students ate rice, ate yam and ate neither. What fraction

of the students in the school ate both rice and yam? 1

8. The money in a competition is shared among the three winners. of the money was given to the 3

2

person who came first. The person who came second received of the remaining money. 4

a. What fraction of the money did the person who came third receive? b. If the person who came first received $200 more than the second person, how much money was shared among them. 1

1

4

5

9. A boy spent of his pocket money on food, and of the remainder on book. He spent the rest equally on toy and ice cream. If the money spent on ice cream is $6, how much was spent on book? 10.

4 9

of the students in a class offer Biology, and

2 3

offer Economics. Every student offers at least

one of these two subjects. If 3 students offer both subjects, how many students are there in the class?

103

CHAPTER 12 DECIMALS Conversion of fractions to decimals Examples Convert the following fractions to decimal fractions: a. b. c. d. e. f.

3 4 1 2 1 3 1 8 12 25 5 16

Solutions a.

3 4

0.75 4 30 28 20 20 -∴

3 4

= 0.75

Just like in long division, put the numerator (i.e. 3) inside the division box and the denominator (i.e. 4) outside it. Then start by dividing 3 by 4. This is not possible since 4 is larger than 3. So put zero (0) and a decimal point on the division box, and another zero in front of 3 to make it 30. Then 30 divided by 4 will give 7. Put the 7 on the box after the decimal point and use it to multiply 4. This will give 28. Write it under 30 and subtract it from 30 to get 2. Put zero in front of this 2 to make it 20. Then start the division as before using 20. So, 20 divided by 4 will give 5. Put the 5 on the box after 7, and then use it to multiply 4. This gives 20. Write it under the first 20 and subtract. This will give zero as shown 104

by the two dashes. When you get to the point where the subtraction is zero, then value at the top of box gives the answer. 3

So, = 0.75 as shown on the box. 4

b.

1 2

0.5 2 10 10 -1

∴

2

= 0.5

1 divided by 2 will not go since 2 is larger. So put zero (0) and a decimal point on the box , and put anther zero in front of 1 make it 10. Then 10 divided by 2 will give 5. Put the 5 on the box after the decimal point and use it to multiply 2. This will give 10. Write it under the first 10 and then subtract. This gives zero, so we stop there. The value at the top of the box (i.e. 0.5) gives the answer.

c. All the other examples are done in a similar way. 1

∴

3

is converted as shown below:

0.333 3 10 9 10 9 10 9 1 ∴

1 3

= 0.333…

This is a recurring decimal. It continues and never ends. So you can stop at any number of decimal places that you want. c.

1 8

105

0.125 8 10 8 20 16 40 40 -1

∴

e.

8

= 0.125

12 25

0.48 25 120 100 200 200 --∴

f.

12 25

= 0.48

5 16

0.3125 16 50 48 20 16 40 32 80 80 -∴

5 16

= 0.3125

106

ADDITION AND SUBTRACTION OF DECIMALS Examples Simplify the following: 1. 0.2 + 0.09 2. 0.612 + 0.0134 + 0.8 3. 0.31 + 0.038 - 0.2924 4. 5.2892 - 7.3 + 3.003 5. 8.12 - 2.1422 + 0.3 - 1.197 6. 7.39 - 1.304 - 2.7 - 2.016 Solutions When adding or subtracting decimals, always arrange the numbers so that the decimal points are under each other, thereby maintaining the place value of each digit 1.

0.2 +0.09 0.29

∴ 0.2 + 0.09 = 0.29

2. 0.612 + 0.0134 + 0.8 0.612 0.0134 +0.8 _ 1.4554

3. 0.31 + 0.038 - 0.2924 Add the first two number first . 0.31 +0.038 0.348

107

Now subtract 0.2924 from 0.348 0.348 -0.2924 0.0556

(Assume there is a zero on the right side of 8)

∴ 0.31 + 0.038 - 0.2924 = 0.0556 Always assume that there are zeros at the right ends of every decimal number. They are usually not written, but it should be known that they are there. So, they can always be included when needed. This idea can help you to know that 0.348 is larger than 0.2924, since 0.348 can also be represented as 0.3480.

4. 5.2892 - 7.3 + 3.003 Add the positive values first. This gives, 5.2892+ 3.003. Note that -7.3 is a negative value. 5.2892 + 3 . 0 0 3_ 8.2922 The next step is: 8.2922 - 7.3 8.2922 - 7.3 _ 0.9922

5. 8.12 - 2.1422 + 0.3 - 1.197 Group the positive values (8.12 and 0.3) together and add them. Also group the negative values (2.1422 and 1.197) and add them. Then subtract the answer of the negative group from the answer of the positive group. This gives: (8.12 + 0.3) - (2.1422 + 1.197) Adding the positive group gives: 8.12 +0.3_ 8.42

Adding the negative group gives:

108

2.1422 +1.197_ 3.3392 Subtracting the negative group from the positive group gives: 8.42 -3.3392 5.0808 ∴

8.12 - 2.1422 + 0.3 - 1.197 = 5.0808

Note that it should be assumed that there are two zeroes at the empty space on the right side of 8.42.

6. 7.39 - 1.304 - 2.7 - 2.016 Grouping the values as explained in example 5 above gives: 7.39 - (1.304 + 2.7 + 2.016) Note that 7.39 is the only positive value in the question. So, adding the negative group gives: 1.304 2.7 +2.016 6.020 Subtracting the negative group from the positive group gives: 7.39 -6.02 1.37 ∴ 7.39 - 1.304 - 2.7 - 2.016 = 1.37 Note that 6.020 can also be expressed as 6.02 since the zero at the end is irrelevant and thus can be ignored.

MULTIPLICATION AND DIVISION 1. Without the use of table or calculator, evaluate the following: a. 1.204 × 1000 109

b. 16.092 × 100 c. 9.27 ÷ 10 d. 21.3 ÷ 1000 e. 1.25 × 10,000 f. 3.31 ÷ 10,000

Solutions To multiply a decimal or any whole number by a multiple of 10 (i.e. 10, 100, 1000 etc.), simply move the decimal point to the right by a number of times which is equal to the number of zeros present on the multiple of 10. You can also add zeros to complete the movement of the decimal point. ∴ 1.204 × 1000, means that the decimal point will move three times to the right. Three times because there are three zeros in 1000 ∴ 1.204 × 1000 = 1204 The first movement of the decimal point gives 12.04, the second movement gives 120.4, while the third movement gives 1204. There is no need of writing the decimal point at the end of the number, so the answer is 1204

b. 16. 092 × 100 =1609.2 Here, the decimal point should be moved two times, since 100 has two zeros. The first movement of the decimal point gives 160.92, while the second movement gives 1609.2 which gives the final answer.

c. To divide a decimal or any whole number by a multiple of 10, simply move the decimal point to the left by a number of times which is equal to the number of zeros. Zeros can also be added to complete the movement of the decimal point. ∴ 9.27 ÷ 10, means that the decimal point will be moved once to the left because there is one zero in 10. ∴ 9.27 ÷ 10 = 0.927

110

The movement of the decimal point to the left side gives .927. It is conventional in this case to put zero (0) just before the decimal point. So the answer becomes 0.927

d. 21.3 ÷ 1000 = 0.0213 Here the decimal point should be moved to the left three times. The first movement of the decimal point gives 2.13, the second movement gives .213 while the third movement gives .0213. Notice that one zero has been added to complete the movement of the decimal point. Also note that where there is no number before a decimal point a zero is put before the decimal point. Hence, .0213 is written as 0.0213 which gives the final answer.

e. 1.25 × 10,000 = 12500 This is multiplication, so the decimal point should be moved four times (since 10000 has four zeros) to the right. The first movement of the decimal point gives 12.5, the second movement gives 125., the third movement gives 1250. while the fourth movement gives 12500 which is the final answer. Note that when the decimal point gets to the end of a number, zeros are added to the number in order to allow more movement of the decimal point.

f. 3.31 ÷ 10000 = 0.000331 Here the decimal point should be moved to the left four times. The first movement of the decimal point gives .331, the second movement gives .0331, the third movement gives .00331 while the fourth movement gives .000331. Notice that three zeros (one at a time) have been added to complete the movement of the decimal point. Hence, .000331 is written as 0.000331 which gives the final answer.

2. Evaluate the following without using tables or calculator a. 0.34 × 6 b. 8 × 0.021 c. 0.26 × 0.0905 d. 0.0188 × 0.372 e. 0.027 ÷ 9

111

f. 0.056 ÷ 0.8 g. h. I.

6.75 𝑥 7.5 0.375 13.2 𝑥 0.051 0.2 𝑥 19.8 3.6 𝑥 4 0.5 𝑥 0.016

Solutions a. 0.34 × 6 Solve just like the usual way of carrying out multiplication, and use the total number of digits after the decimal point to put your point on the answer. In this example we have two digits after the decimal point (i.e. 3 and 4 from 0.34). So, we count two digits from the right to the left of the final answer before putting the decimal point. In some cases zeros can be used to complete the digits to be counted. The working is shown below: 0.34 × 6:

0.34 × 6 2.04

After the usual multiplication, the value obtained is 204. So count two digits from right to left and put the decimal point. This gives 2.04 ∴ 0.34 × 6 = 2.04 Note that the numbers were not multiplied by the zero before the decimal point. b. 8 × 0.021 It is easier to put the value with the lower number of digits below the one with the higher number of digits. So, put 8 below 0.021 in the multiplication, as shown below. 0.021 × 8 .168 Count three digits from left to right before putting the decimal point. This gives .168 which is written as 0.168

112

∴ 8 × 0.021 = 0.168

c. 0.26 × 0.0905 0.0905 × 0.26 5430 + 1 8 1 0_ .023530 Remember that you have to move a step to the left before starting each next stage of multiplication. For example, during the second stage, i.e. when multiplying 905 by 2 to get 1810, the multiplication was started one step to the left, which was under 3 of 5430 and not under 0 of 5430 Note that the first zero after the decimal point in 0.0905 was ignored during the multiplication. In putting the decimal point, count a total of six digits after the decimal point of each number i.e. 0, 9, 0, and 5 from 0.0905 and 2 and 6, from 0.26 making it a total of six digits. Now go to the answer and count six digits from right to left before putting the decimal point. One zero was also added to complete the sixth digits. Note that after the multiplication, 23530 was obtained. It was at the final stage of writing the decimal point that 0 and the decimal point were included to obtain .023530 ∴ 0.26 × 0.0905 = .023530, which is written as 0.023530, or 0.02353 since the last zero is irrelevant.

d. 0.0188 x 0.372 0.0188 x 0. 3 7 2 376 1316 + 564 _ .0069936

Count a total of seven digits to the left before placing the decimal point. Two zeros have been added to complete the seven digits. ∴

0.0188 × 0.372 = . 0069936 = 0.0069936 113

0.027

e. 0.027 ÷ 9

9

= 0.003

Explanation: 0 divided by 9 will not go so, write 0 and put the decimal point. Then the second 0 divided by 9 will not go, so write 0. Then 2 divided by 9 will also not go, so write 0. Finally, 27 divided by 9 is 3. So, this gives 0.003 as shown above.

f. 0.056 ÷ 0.8 =

0.056 0.8

An easier way to solve the problem is to first clear the decimal point. The number of digits after the decimal point on the numerator is three, while the number of digits after the decimal point on the denominator is one. So take the higher value which is three. Now multiply the numerator and denominator by a multiple of 10 which has three zeros. This multiple is 1000. So, multiplying numerator and denominator by 1000 gives: 0.056 x 1000 0.8 𝑥 1000

=

=

7 100

56 800

(In its lowest terms)

= 0.07 (When

7 100

is converted to decimal)

Recall that 0.056 × 1000 = 56, while 0.8 × 1000 = 800, as explained in our pervious examples. g.

6.75 𝑥 7.5 0.375

The total number of digit after the decimal point in the numerator is three, and that of the denominator is also three. So multiply both by 1000 (i.e. a multiple of 10 with three zeros) in order to clear the decimal points. This gives: 6.75 𝑥 7.5 x 1000 0.375 x 1000

=

675 x 75 375

Note that out of the three zeros used in multiplying the numerator, two zeros have cleared the decimal point from 6.75, while one zero has cleared the decimal point from 7.5. 675 x 75 375

= 27 x 5

(After equal division of numerator and denominator by 25 and then 15)

114

= 135

h.

13.2 x 0.051 0.2 x 19.8

Total digits after the decimal points on the numerator is 4, while on the denominators is 2. So take the higher value which is 4. Now multiply numerator and denominator by a multiple of 10 that has 4 zeros. This multiple is 10000. This now gives: 13.2 x 0.051 x 10000 0.2 x 19.8 x 10000

=

132 x 51 2 𝑥 198 x 100

Note that the denominator is in excess of two zeros, so write the excess as 100 as shown above. 132 x 51

=

2 𝑥 198 x 100 2 x 51 2 𝑥 3 x 100

∴

i.

17 100

=

= 0.17

3.6 x 4 0.5 x 0.016

17

=

(After equal division by 6 and then 11)

(After cancelling out the 2 and then equal division by 3)

100

(When

17 100

is converted to decimal fraction)

36 𝑥 4 x 1000

36 𝑥 4 x 1000 5 x 16

2 x 51 2 x 3 x 100

5 x 16

= 9 x 200

(After multiplying numerator and denominator by 10000) (After equal division by 4, then 4 again and finally 5)

= 1800

CONVERSION OF DECIMALS TO FRACTIONS Example Convert the following decimals to fractions 1. 0.12 2. 0.125 3. 0.064 4. 0.004 5. 0.0051

115

Solutions 1. In 0.12, the whole number that can be obtained is 12. The number of digits after the decimal point is two. By using this two, a multiple of 10 that has two zeros is 100. So divide the whole number 12 by 100, and express your answer in its lowest term. This gives: 0.12 = =

12 100 3

(When

25

12 100

is expressed in its lowest term)

2. In 0.125, the whole number that can be obtained is 125. The number of digits after the decimal point is three. By using this three, a multiple of 10 that has three zeros is 1000. So divide the whole number 125 by 1000, and express your answer in its lowest term. This gives: 0.125 = =

125 1000

1

(When

8

125 1000

is expressed in its lowest term)

3. In 0.064, the whole number that can be obtained is 64. The number of digits after the decimal point is three. So divide 64 by 1000, and express your answer in its lowest term. This gives: 0.064 = =

64 1000 8

(When

125

64 1000

is expressed in its lowest term)

4. In 0.004, the whole number that can be obtained is 4. The number of digits after the decimal point is three. So divide 4 by 1000, and express your answer in its lowest term. This gives: 0.004 =

4 1000

=

1 250

(In its lowest term)

5. In 0.0051, the whole number that can be obtained is 51. The number of digits after the decimal point is four. So divide 51 by 10000. This gives: 0.0051 =

51 10000

116

Exercises 1. Convert the following fractions to decimal fractions: a. b. c. d. e. f.

3 5 7 10 5 12 9 16 3 25 8 25

2. Simplify the following: a. 0.6 + 0.07 b. 0.333 + 0.0698 + 0.72 c. 0.0011 + 0.09 - 0.174 d. 1.237 - 3.5 + 5.01 e. 6.09 - 3 + 0.099 - 2.2 f. 4.21 - 0.2129 - 2.5 - 1.704 3. Without the use of table or calculator, evaluate the following: a. 1.005 × 100 b. 9.092 × 10000 c. 2.7 ÷ 100 d. 43.3 ÷ 10 e. 4.78 × 1000 f. 1.006 ÷ 10,000 3. Evaluate the following without using tables or calculator

117

a. 0.28 × 4 b. 5 × 0.205 c. 1.09 × 0.0243 d. 0.0495 × 0.65 e. 0.072 ÷ 12 f. 0.36 ÷ 0.04 g. h. I. 4. a. b. c. d. e.

0.45 x 5.2 0.25 4 x 0.36 0.08 x 9 1.04 x 2.5 0.13 x 0.04

Convert the following decimals to fractions 0.13 0.375 0.0125 0.00005 0.064

118

CHAPTER 13 PERCENTAGE Percentage (%) is a fraction of 100. For example 25% also means

25 100

1

or (in its lowest term) or 0.25 4

in decimal fraction. Examples 1. Express the following percentages as fractions and as decimals. a. 20% b. 45% 1

c. 33 % 3

Solutions 20

a. 20% = 1

= 0.2

5

1

=

100

5

(In its lowest term after division by 20)

(In decimal as explained in chapter 3).

1

∴ 20% = = 0.2 5

b. 45% = 9 20

45 100

9 20

20

(In its lowest term after division by 5)

= 0.45

1 3

1

33

3

100

c. 33 % =

=

9

= 0.45

∴ 45% =

=

=

100 3 100 3

=

100 3

100

1

100

3

3

(Note that 33 =

when expressed in improper fraction)

÷ 100 x

1 100

119

1

= 1

(After cancelling out the 100)

3

= 0.33 (To 2 decimal places)

3

1

1

3

3

∴ 33 % = = 0.33

2. Express the following fractions as percentage. a. b. c.

3 10 2 5 4 25

Solutions a. Simply multiply the fraction by 100. ∴

3 10

=

3 10

x 100

= 3 x 10 = 30% b.

2 5

2

= x 100 5

= 2 x 20 = 40% c.

(After dividing 100 by 10 to get 10 and then multiplying the 10 by 3)

4 25

=

4 25

(After dividing 100 by 5 to get 20 and then multiplying the 20 by 2)

x 100

= 4 x 4 = 16%

3. Express the following decimals as percentages. a. 0.21

120

b. 0.72 c . 0.625

Solutions a. Simply multiply by 100 ∴

0.21 = 0.21 x 100 = 21%

b. 0.72 = 0.72 x 100 = 72% 1

1

2

2

c. 0.625 = 0.625 x 100 = 62.5% or 62 %, since 0.5 =

4. Express: a. 2m as a percentage of 5m. b. 12.5kg as a percentage of 50kg Solutions a. 2m as a percentage of 5m is given by: 2 5

x 100 = 2 x 20 = 40%

b. 12.5kg as a percentage of 50kg is given by: 12.5 50

x 100 = 12.5 x 2 = 25%

5. Calculate the following: a. 40% of 2kg b. 5% of $16 1

c. 12 % of 40cm 2 2

d. 66 % of 24L 3

121

when converted to a fraction.

Solutions a. 40% of 2kg =

40

2

x2= x2

100

5

4

= = 0.8kg 5

b. 5% of $16 = =

4

5 100

x 16

(After equal division by 5 and 4)

5

= $0.8 = (100 x 0.8) cents

(since $1 = 100 cents)

= 80 cents 1 2

1

12

2

100

c. 12 % of 40cm =

= = =

25 2

× 40

100 25

÷ 100 × 40

2

25 2

1

x

100

= 5cm

x 40

(After equal division by 25, 4 and 2) 2 3

2

66

3

100

d. 66 % of 24L =

= = =

200 3

200

÷ 100 × 24

3

200

× 24

× 24

100

3

× 40

x

=2x8

1 100

x 24

(After equal division by 100 and 3)

122

= 16L

6. In a crate of 30 eggs, 9 are bad. What is the percentage of good eggs? Solution Number of good eggs = 30 - 9 = 21 ∴ Percentage of good eggs =

21 30

× 100

= 7 x 10 = 70%

7. A woman receives a 20% increase in her salary. If her present salary is $2010, calculate her new salary. Solution One method of solving this kind of question is to interpret it as a follows: Her present salary is 100%. Her new salary is 120% (100 + 20). So, 120% of her present salary will give her new salary. ∴ New salary =

120 100

x 2010 (This means 120% of her original/present salary)

= 12 x 201

(After cancellation of the zeros by equal division by 10 two times).

= 2412 ∴ Her new salary is $2412

8. The price of a suit is reduced by 5%. If the suit is originally marked at $1200, what is the new price of the suit? Solution This question can be interpreted as follows: the original price is 100%. The new price is 95% (100 - 5). So, 95% of the original price will give the new price. ∴

New price =

95 100

x 1200

(This means 95% the original price)

= 95 x 12 = 1140 ∴ The new price of the suit is $1140. 123

9. Increase 270kg by 30% Solution This is similar to question (7) above. 270kg corresponds to 100%. So the new value will be 130% (100 + 30) of the original value. New value =

130 100

x 270 = 13 × 27

= 351kg

10. Decrease 140m by 35% Solution This is similar to question (8) above. 140m corresponds to 100%. So, the new value will be 65% (100 - 35) of the original value. New value =

65 100

x 140 = 13 x 7

= 91m

11. 336 is the result of increasing a number by 60%. Find the number Solution This question can be interpreted as follows: 160% (100 + 60 = 160) corresponding to 336. So, the original number will correspond to 100%. 100

∴ The number is:

160

x 336 = 10 x 21 = 210

The number is 210

12. 273 is the result of decreasing a number by 25%. Find the number. Solution This question can be interpreted as follows: 75% (100 - 25) corresponds to 273. So, the number will correspond to 100% ∴ The number =

100 75

x 273

124

= 4 x 91 = 364 The number is 364.

APPLICATION OF PERCENTAGE TO PROFIT AND LOSS Usually, profit and loss are always calculated as a percentage of the cost price. Even the selling price can be calculated as a percentage of the cost price. 100% corresponds to the cost price. (100 + %profit)% corresponds to the selling price when there is a gain. (100 - %loss)% corresponds to the selling price when there is a loss.

Examples 1. A book is bought for $152 and sold at a profit of 25%. Calculate the selling price of the book. Solution Since this is a profit, the selling price will be 125% (i.e. 100 + 25 = 125) of the cost price Selling price =

125 100

x 152

= 5 x 38

(After equal division by 25 and 4).

= 190 The selling pie of the book is $190

2. A pair of shoes was bought for $120 and sold at a loss of 15%. Calculate the selling price of the shoe. Solution Since this is a loss the selling price will be 85% (100 - 15 = 85) of the cost price. Selling price =

85 100

x 120

= 17 x 6

(After equal division by 20 and 5)

= 102 The selling price of the shoe is $102 125

3. Find the percentage gain or loss when an item is bought for: a. $80 and sold for $100 b. $420 and sold for $350 Solution a. Amount gained = 100 - 80 = 20 Percentage gain = =

𝐴𝑚𝑜𝑢𝑛𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝐶𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒 20 80

x 100

x 100

=25% b. Amount lost = 420 - 350 = 70 Percentage loss = =

𝐴𝑚𝑜𝑢𝑛𝑡 𝑙𝑜𝑠𝑡 𝐶𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒 70 420

x 100

x 100

=16.7%

4. Calculate the cost price of an item which is sold for: 1

a. $940 at a profit of 17 % 2

b. $1250 at a loss of 20% Solutions 1

1

1

2

2

2

Since profit is made, then 117 % (i.e. 100 + 17 = 117 ) corresponds to the selling price of $940. So the cost price will correspond to 100% ∴

Cost price =

100 117

1 2

x 940 1

=100 ÷ 117 x 940 2

126

= 100 ÷ = 100 x

235 2 2 235

x 940 x 940

= 20 x 2 x 20

(After equal division by 5 and 47)

= $800 b. Since a loss is made, then 80% (i.e. 100 - 20 = 80) corresponds to the selling price of $1250. So, the cost price will correspond to 100% ∴

Cost price = = =

100 80 25 2

x 1250

x 125

(After equal division by 4 and 10)

3125 2

= $1562.50

5. By selling a phone for $1300, a trader gained 30%. How much money did he gain? Solution 130% (i.e. 100 + 30 = 130) corresponds to the selling price of $1300 So, the profit will correspond to 30% ∴ Amount gained (the profit) =

30 130

x 1300

= 30 x 10 = $300 Note that the cost price in this question will correspond to 100%

6. A car dealer gained $3200 on a sale. If this was an 8% profit, what was: a. the cost price of the car? b. the selling price of the car? Solutions a. 8% corresponds to the profit of $3200 127

So, the cost price will correspond to 100% ∴ Cost price =

100 8

x 3200

= 100 x 40 = $40000 b. Selling price = cost price + profit = 40,000 + 3200 =$43,200

7. By selling a dress for $3,500, a seamstress lost of 30%. For how much should she have sold it to gain 50%? Solution 70% (i.e. 100 - 30 = 70) corresponds to $3,500. So, 150% (i.e. 100 + 50 = 150) will corresponds to the new selling price. Note that the cost price corresponds to 100%. ∴ New selling price =

150 70

x 3500

= 150 x 50 = $7500

8. A man made a loss of 10% on a car he sold for $54000. If he had sold it for $57,600 what would have been his percentage loss or gain? Solution 90% (i.e. 100 -10 = 90) corresponds to $54,000. So, 100% will corresponds to the cost price ∴ Cost price =

100 90

x 54000

= 100 x 600 = $60000 If he had sold the car for $57,600, he still would have made a loss. loss = 60,000 - 57600 = 2400 ∴ Percentage loss =

2400 60000

x 100

128

= 4% He would have made a percentage loss of 4%

APPLICATION OF PERCENTAGE TO DISCOUNT AND TAX Examples 1. A worker pays 20% of his taxable income as tax. If his taxable income is $1400: a. how much tax does he pay? b. what is his take home pay if he has no allowance or benefits Solution a. 20% of his taxable income gives: 20 100

x 1400

= 20 x 14 = $280 He pays a tax of $280 b. His take home pay is = 1400 - 280 =$1120

2. A trader offers an 18% discount on all purchases during a festive season. How much would a customer pay for a table that cost $150? Solution 100% corresponds to the cost price of $150. So, 82% (i.e. 100 - 18 = 82) will correspond to the amount that the costumer would pay. ∴ Amount to be paid =

82 100

x 150

= 41 x 3 =$123 3. A man pays $630 for an item after a 10% discount has been given on the original marked price. Calculate the marked price of the item.

129

Solution 90% (100 - 10 = 90) corresponds to $630. So, 100% will corresponds to the marked priced. ∴ Marked price =

100 90

x 630

= 100 x 7 = $700

1

4. After paying 2 % of his taxable income as tax, a man takes home $1500. Calculate his taxable 2

income. Solution 1

1

1

2

2

2

97 % (i.e. 100 - 2 = 97 ) corresponds to his take home pay of $1500. So, 100% will correspond to his taxable income. ∴ His taxable income =

100 97

1 2

x 1500

1

= 100 ÷ 97 x 1500 2

= 100 ÷ = 100 x =

195 2 2 195

x 1500 x 1500

100 𝑥 2 𝑥 100 13

=

20000 13

= 1538.46 ∴ His taxable income = $1538.46

Exercises 1. Express the following percentages as fractions and as decimals. a. 40% 1

b. 12 % 2

130

2

c. 6 % 3

2. Express the following fractions as percentage. a. b. c.

21 50 7 8 9 20

3. Express the following decimals as percentages. a. 0.32 b. 0.2 c . 0.0125 4. Express: a. 5cm as a percentage of 40cm. b. 16g as a percentage of 64kg 5. Calculate the following: a. 20% of 15m b. 4% of $50 1

c. 6 % of 80kg 4

1

d. 33 % of 42days 3

6. In a class of 30 students, 6 are boys. What is the percentage of girls? 7. A woman receives a 30% increase in her salary. If her present salary is $810, calculate her new salary. 8. The price of a dress is reduced by 12%. If the dress is originally marked at $500, what is the new price of the dress? 9. Increase 120cm by 40% 10. Decrease 25L by 20% 131

11. 240 is the result of increasing a number by 20%. Find the number 12. 720 is the result of decreasing a number by 64%. Find the number. 13. A book is bought for $50 and sold at a profit of 40%. Calculate the selling price of the book. 14. A pair of shoes was bought for $200 and sold at a loss of 11%. Calculate the selling price of the shoe. 15. Find the percentage gain or loss when an item is bought for: a. $120 and sold for $90 b. $2500 and sold for $3000 16. Calculate the cost price of an item which is sold for: a. $440 at a profit of 60% b. $780 at a loss of 20% 17. By selling a phone for $132, a trader gained 10%. How much money did he gain? 18. A car dealer gained $840 on a sale. If this was a 12% profit, what was: a. the cost price of the car? b. the selling price of the car? 19. By selling a dress for $850, a seamstress lost 15%. For how much should she have sold it to gain 25%? 20. A man made a loss of 20% on a car he sold for $16000. If he had sold it for $20000 what would have been his percentage loss or gain? 21. A worker pays 10% of his taxable income as tax. If his taxable income is $820: a. how much tax does he pay? b. what is his take home pay if he has no allowance or benefits 22. A man pays $105 for an item after a 25% discount has been given on the original marked price. Calculate the marked price of the item. 1

23. After paying 12 % of his taxable income as tax, a man takes home $4200. Calculate his taxable 2

income. 132

CHAPTER 14 SIMPLE INTEREST The simple interest, I, paid on an amount of money borrowed or saved is given by: I=

𝑃𝑅𝑇 100

Where P = principal (money saved or borrowed), r = rate (given in percentage) and T = time in years. Examples 1. Calculate the simple interest on the following: a. $3000 saved for 5 years at 2% per annum 1

b. $5000 saved for 9 years at 6 % per annum 3

Solutions a. I = =

𝑃𝑅𝑇 100 3000 x 2 𝑥 5 100

= 30 x 2 x 5 = $300

b. I =

=

= =

𝑃𝑅𝑇 100 1 3

5000 x 6 x 9 100 5000 x

19 3

x9

100 5000 x 19 x 9 100 x 3

= 50 x 19 x 3 =$2850 133

2. Calculate the principal that will earn an interest of $350 in 7 years at 2% per annum. Solution I=

𝑃𝑅𝑇 100 Px2x7

350 =

100

Cross multiply 350 x 100 = P x 2 x 7 35000 = 14P Divide both side by 14 in order to make P stand alone 35000 14

=

14𝑃 14

2500 = P

(The 14 will cancel out)

P = $2500 ∴

The principal is $2500

3. Find the rate per cent per annum (p.a.) at which $210 will earn $70 in 10 years. Solution I=

𝑃𝑅𝑇 100

70 =

210 x R x 10 100

Cross multiply 210 x R x 10 = 70 x 100 2100R = 7000 Divide both sides by 2100 in order to make R stand alone 2100𝑅 2100

R=

=

10 3

7000 2100 1

=3

3

134

1

The rate is 3 % 3

4. Find the time in which $900 will earn $120 at 3% p.a. Solution I= 120=

𝑃𝑅𝑇 100 900 x 3 x T 100

Cross multiply. This gives: 900 x 3 x T = 120 x 100 2700T = 12000 Divide both sides by 2700 2700𝑇 2700

T=

=

12000 2700

40 9 4

T = 4 years 9

5. Calculate the simple interest on $15000 from 4th April 2010 to 28th may 2010 (both days inclusive) at 5% p.a. Solution From 4th of April to 28th of May gives a total of 55 days (i.e. 27 days in April + 28 days in May) By simple proportion, since 365 days correspond to 1 year, then, 55 days will correspond to: 55 365

x1=

55 365

=

11 73

years

Note that the time, T, must be expressed in years I=

=

𝑃𝑅𝑇 100

15000 x 5 x

11 73

100

135

15000 x 5 x 11

= =

100 x 73 150 x 5 x 11 73

=

8250 73

= 113.01 The simple interest is $113.01

6. How much will $8100 amount to, in 3 years at 9% per annum. Solution The simple interest should be calculated first. I= =

𝑃𝑅𝑇 100 8100 x 9 x 3 100

= 81 x 9 x 3 =$2187 The ”amount” in this question is the sum of the principle and the interest. ∴ Amount = 8100 + 2187 = $10,287

7. Calculate the rate percent p.a. at which $48540 will amount to $63102 in 6 years. Solution Interest = amount - principal I = 63102 – 48540 = 14562 I=

𝑃𝑅𝑇 100

14562 =

48540 x R x 6 100

Cross multiply

136

14562 x 100 = 48540 x R x 6 1456200 = 291240R Divide both sides by 291240 1456200 291240

=

291240 𝑅 291240

5=R R=5 The rate is 5% p.a.

8. A woman invests $2700 at 10% simple interest for 4 month. How much was her investment worth by the end of this period? Solution This question is asking for the amount i.e. principal + interest. I= But, T = ∴

I= =

4 12

𝑃𝑅𝑇 100 1

= years 3

2700 x 10 x

(Since there are 12 months in a year)

1 3

100 2700 x 10 x 1 100 x 3

= 9 x 10 = 90 The amount = principal + interest = 2700 + 90 = 2790 Her investment was worth $2790

9. A retailer has an account of $10,200 from a manufacturer who gives two months credit but offer a cash discount of 5% for prompt payment. The retailer borrows $10,200 from his bank at 9% per annum for two month and pays promptly to secure the discount. How much does he gain by transaction?

137

Solution The amount he gained due to the discount is given by: 5% of 10,200 =

5 100

x 10200

= 5 x 102 = 510 The amount he would pay the bank is the interest on the loan. I=

𝑃𝑅𝑇 100

But the time T = 2 months = ∴

I= =

10200 x 9 x

2 12

1

= years. 6

1 6

100 10200 x 9 x 1 100 x 6

= 17 x 9 = 153 ∴ Amount he gained by these two transactions = amount gained from the discount – amount (interest) paid to the bank. Amount gained = 510 - 153 = $357

10. A Nigerian visiting America changed ₦832000 to dollars at the rate of ₦320 to $2. He spent $3200 and invested the remaining amount in an American bank at 12% simple interest per annum. At the end of 9 months he transferred the capital and interest to his account in a bank in Nigeria at the rate of $3 to ₦420. What was the amount, in ₦, credited to his account? Solution ₦320 correspond to $2. So, by simple proportion, ₦832000 will correspond to an amount in dollars given by: 832000 320

=

x2

1664000 320

138

= $5200 After spending $3200, the amount remaining = 5200 - 3200 = $2000 When this $2000 is invested at 12% p.a. for 9 months, the interest is given by: I=

𝑃𝑅𝑇 100

Convert the time in months to time in years by dividing by 12. This gives, T = ∴

I= =

2000 x 12 x

9 12

3

= years 4

3 4

100 2000 x 12 x 3 100 x 4

= 5 x 12 x 3 (After equal division by 100 and 4) = $180 But, Amount = principal + interest = 2000 + 180 = $2180 ∴ $2180 is the capital and interest he transferred to Nigerian The conversion to naira is given as follows: $3 corresponds to ₦420, so, $2180 will correspond to an amount in naira given by: 2180 3

x 420

= 2180 x 140 = ₦305200

Exercises 1. Calculate the simple interest on the following: a. $5200 saved for 3 years at 6% per annum b. $1020 saved for 5 years at 10% per annum 2. Calculate the principal that will earn an interest of $121 in 4 years at 5% per annum. 3. Find the rate per cent per annum (p.a.) at which $450 will earn $160 in 8 years.

139

4. Find the time in which $5600 will earn $140 at 4% p.a. 5. Calculate the simple interest on $125000 from 10th June 2010 to 19th July 2010 (both days inclusive) at 6% p.a. 6. How much will $6000 amount to, in 5 years at 12% per annum. 7. Calculate the rate percent p.a. at which $2600 will amount to $3000 in 9 years. 8. A woman invests $45000 at 9% simple interest for 6 months. How much was her investment worth by the end of this period? 9. Calculate the time at which $8400 will amount to $8442 at 2% per annum. 10. A man invests $200000 at a rate of 5% per annum for 2 years. He withdraws his original principal invested and invests his interest for 20 years at 8% per annum. How much will his money amount to at the end of this period?

140

CHAPTER 15 COMPOUND INTEREST When money called the principal P, is invested in a bank at a rate of R% per annum compound interest for a period of T years, then the total amount that the money invested would become is given by: A = P(1 +

𝑅 T 100

)

Where A is called the amount i.e. the principal + the compound interest, C.I. This shows that compound interest, C.I. = Amount - Principal

Examples 1. Calculate the compound interest on the following: a. $1200 for 5 years at 4% p.a. b. $450 for 12 years at 10% p.a. Solutions a.

A = P(1 +

𝑅 T ) 100

= 1200(1 +

4 5 ) 100

= 1200(1 + 0.04)5 = 1200 x 1.045 = 1200 x 1.21665 = $1460 (This is the amount) ∴ Compound interest C.I = Amount - principal = 1460 - 1200 =$260 b.

A = P(1 +

𝑅 T ) 100

= 450(1 +

10 12 ) 100

141

= 450(1 + 0.1)12 = 450 x 1.112 = 450 x 3.13843 = $1412.29 (This is the amount) ∴ Compound interest C.I =1412.29 - 450 =$962.29

2. A sum of $2450 was invested in a bank at a fixed deposit rate of 6% p.a. What would the money be at the end of 3 years? Solution This is a question involving calculation of amount. A = P(1 +

𝑅 T ) 100

= 2450(1 +

6 3 ) 100

= 2450(1 + 0.06)3 = 2450 x 1.063 = 2450 x 1.191016 = $2917.99 =$2918

3. Find the principal that would amount to: 1

a. $2000 in 5 years at 4 % p.a. 2

b. $10,500 in 8 years at 2% p.a. Solutions a.

A = P(1 +

2000 = P(1 + 2000 = P(1 +

𝑅 T ) 100 4

1 2 5

100

)

4.5 5

100

)

142

2000 = P(1 + 0.045)5 2000 = P x 1.0455 2000 = P x 1.246182 2000 = 1.246182P ∴

P=

2000 1.246182

P = $1604.90 b.

A = P(1 +

𝑅 T ) 100

10500 = P(1 +

2 8 ) 100

10500 = P(1 + 0.02)8 10500 = P x 1.028 10500 = P x 1.1716594P 10500 = 1.1716594P ∴

P=

10500 1.1716594

P = $8961.65

4. Find the rate per cent to the nearest whole number at which a. $2000 will amount to $3000 in 4 years b. $8600 will amount to $12,200 in 6 years Solutions a.

A = P(1 +

𝑅 T ) 100 𝑅 4 ) 100

3000 = 2000(1 +

Divide both sides by 2000 3000 2000

=

2000 (1 +

𝑅 )4 100

2000

The 2000 will cancel out to give:

143

𝑅 4

1.5 = (1 +

100

)

Raise both sides to a power which is equal to the inverse 4 (i.e. the time) 1

1

The inverse of 4 is . Raising both sides to power gives: 4 𝑅 4 x ⅟4 1.5 = (1 + ) 100 𝑅 1 0.25

4

¼

1.5

= (1 +

1.107 - 1 = 0.107 =

1

(Note that 4 x = 1) 4

𝑅

1.107 = 1 + ∴

100

)

100 𝑅

100 𝑅

100

Cross multiply R = 0.107 x 100 R = 10.7 The rate is 11% to the nearest whole number. b.

A = P(1 +

𝑅 T ) 100 𝑅 6 ) 100

12200 = 8600(1 +

Divide both sides by 8600 12200 8600

=

8600 (1 +

𝑅 )6 100

8600

The 8600 will cancel out to give: 1.4186047 = (1 +

𝑅 6 100

)

Raise both sides to a power which is equal to the inverse 6 (i.e. the time) 1

1

The inverse of 6 is . Raising both sides to power gives: 6

(1.4186047) 1.06 = (1 + 1.06 = 1 +

⅙

𝑅 6 x ⅟6 = (1 + ) 100

𝑅 1 ) 100

6

1

(Note that 6 x = 1) 6

𝑅 100

144

∴

1.06 - 1 = 0.06 =

𝑅 100 𝑅

100

Cross multiply R = 0.06 x 100 R=6 The rate is 6%.

5. Find the time to the nearest year in which: a. $640 will amount to $960 at 10% p.a. b. $4120 will amount to $6800 at 8 % p.a. Solution a.

A = P(1 +

𝑅 T ) 100

960 = 640(1 +

10 T ) 100

Divide both sides by 640 960 640

=

640(1 + 0.1)ᵀ 640

The 640 will cancel out to give: 1.5 = 1.1T Take the logarithm of both sides log1.5 = log1.1T log1.5 = Tlog1.1 0.1761 = 0.0414T (Since log1.1 = 0.0414) Divide both sides by 0.0414 ∴

T=

0.1761 0.0414

= 4.25

The time is 4 years to the nearest year. b.

A = P(1 +

𝑅 T ) 100

145

6800 = 4120(1 +

8 T 100

)

Divide both sides by 4120 6800 4120

=

4120 (1 + 0.08)ᵀ 4120

The 4120 will cancel out to give: 1.6505 = 1.08T Take the logarithm of both sides log1.6505 = log1.08T log1.6505 = Tlog1.08 0.2176 = 0.0334T (Since log1.08 = 0.0334) Divide both sides by 0.0334 ∴

T=

0.2176 0.0334

= 6.51

The time is 7 years to the nearest year.

6. At what rate per cent p.a. compound interest will a certain amount of money be doubled when invested for 4 years? Solution Let the principal (amount invested) be P. So, when this money becomes double, then the amount, A is given by: A = 2P Recall that:

A = P(1 +

𝑅 T ) 100

Substituting 2P for A gives: 2P = P(1 +

𝑅 4 ) 100

Divide both sides by P 2𝑃 𝑃

=

𝑃(1 +

𝑅 )4 100

𝑃

The P will cancel out to give: 2 = (1 +

𝑅 4 ) 100

Raise both sides to a power which is equal to the inverse 4 (i.e. the time)

146

1

1

The inverse of 4 is . Raising both sides to power gives: 4 ¼ 𝑅 4 x ⅟4 2 = (1 + ) 100 𝑅 1

1.189 = (1 + 1.189 = 1 + ∴

1.189 - 1 =

100

4

)

𝑅 100

𝑅 100

0.189 =

𝑅 100

Cross multiply R = 0.189 x 100 = 18.9 The rate is 19% to the nearest per cent

7. How long will it take a certain amount of money invested in a bank to become four times its value when invested at 6% p.a. compound interest? Solution Let the money invested be P. When this money becomes four times its value, then it will amount to: A = 4P But,

A = P(1 +

𝑅 T ) 100

4P = P(1 +

6 T ) 100

Divide both sides by P 4𝑃 𝑃

=

𝑃(1 + 0.06)ᵀ 𝑃

4 = 1.06T Take the logarithm of both sides log4 = log1.06T log4 = Tlog1.06 0.6021 = 0.0253T Divide both sides by 0.0253 147

∴

T=

0.6021 0.0253

= 23.8

The time is 24 years to the nearest year.

Exercises 1. Calculate the e compound interest on the following: a. $2500 for 2 years at 6% p.a. b. $800 for 7 years at 5% p.a. 2. A sum of $12350 was invested in a bank at a fixed deposit rate of 10% p.a. What would the money be at the end of 4 years? 3. Find the principal that would amount to: 1

a. $5400 in 8 years at 2 % p.a. 2

b. $25500 in 3 years at 4% p.a. 4. Find the rate per cent to the nearest whole number at which a. $10000 will amount to $10500 in 3 years b. $2800 will amount to $3200 in 9 years 5. Find the time to the nearest year in which: a. $320 will amount to $380 at 5% p.a. b. $22000 will amount to $23000 at 4 % p.a. 6. At what rate per cent p.a. compound interest will a certain amount of money be tripled when invested for 12 years? 7. How long will it take a certain amount of money invested in a bank to become two times its value when invested at 10% p.a. compound interest? 8. At what rate per cent p.a. compound interest will a certain amount of money be five times its value when invested for 20 years?

148

CHAPTER 16 RATIO Ratio is a relation between two quantities showing the number of times one value is contained within the other. Examples 1. Express the following ratio in their simplest forms. a. 8 : 12 b. 40cm to 1m 1

c. 1 h : 30mins 3

1

1

2

2

d. 1 : 2 e.

3

1

2

2

3

to to

4

Solutions a. 8 : 12 = 2 : 3

(After equal division by 4)

b. 40cm to 1m (Express them in the same unit) ∴ 40cm : 1m when expressed in cm gives: 40 : 100cm (Since 1m = 100cm) 40 : 100 = 2 : 5 (After equal division by 20). 1

c. 1 h : 30min 3

1

Express 1 h in minutes. 3

1

4

3

3

1 h= h Multiply by 60 since 1h = 60mins. ∴

4 3

h=

4 3

x 60 = 4 × 20 = 80mins

149

4

∴

3

h : 30mins

= 80 : 30 =8:3

1

1

2

2

(After division by 10)

d. 1 : 2

3 5

= :

2 2

Multiply both fractions by 2, the LCM of their denominators. 3 5

∴

:

2 2

3

5

2

2

= ( x 2) : ( x 2) = 3: 5

e.

3 4

1

2

2

3

to to

The LCM of their denominations (i.e. 4, 2, and 3) is 12. So multiply each of the fractions by 12 ∴

3 1 2

: :

4 2 3

3

1

2

4

2

3

= ( x 12) : ( x 12) : ( x 12)

= (3 × 3) : (1 × 6) : (2 × 4) =9:6:8

2. Find the values of x in the ratios below. a. 25 : x = 5 : 6 b. x : 9 = 4 : 3 c. 3 : 13 = 12 : x Solutions a. If 25 : x = 5 : 6 150

25

Then

=

𝑥

5 6

Cross multiply (i.e. left hand numerator multiplies right hand denominator, while right hand numerator multiplies left hand denominator) ∴ 5x = 25 x 6 Divide both sides by 5. 5𝑥 5

=

25 𝑥 6 5

x = 5 x 6 = 30

b. If x : 9 = 4 : 3 Then,

𝑥 9

=

4 3

Cross multiply (i.e. left hand numerator multiplies right hand denominator, while right hand numerator multiplies left hand denominator) ∴ 3x = 4 x 9 Divide both sides by 3. 3𝑥 3

=

4𝑥9 3

x=4x3 = 12

c. If 3 : 13 = 12 : x Then,

3 13

=

12 𝑥

Cross multiply ∴ 3x = 13 x 12 Divide both sides by 3. 3𝑥 3

=

13 𝑥 12 3

151

x = 13 x 4 = 52

3. Increase the following quantities in the given ratios. a. $25 in the ratio 8: 5 b. 49cm in the ratio 11: 7 Solutions a. When $25 is increased in the ratio 8: 5, it means that the new increased value corresponds to the higher ratio of 8. So, let this new value be x. The ratio can now be expressed as: x : 25 = 8 : 5 𝑥

Or,

25

=

8 5

Cross multiply ∴ 5x = 25 x 8 Divide both sides by 5. 5𝑥 5

=

25 𝑥 8 5

x=5x8 = 40 The new amount is $40 b. Here the new value corresponds to the higher ratio of 11. Let this new value be x. 𝑥

∴

49

=

11 7

Cross multiply ∴ 7x = 49 x 11 Divide both sides by 7. 7𝑥 7

=

49 𝑥 11 7

x = 7 x 11 = 77 The new value is 77cm

152

4. Decrease the following quantities in the given ratios. 1

a. 5 days in the ratios 8 : 11 2

b. 1 year 8 months in the ratio 2 : 5 Solutions 1

a. When 5 days is decreased in the ratio 8 : 11, it means that the new decreased value corresponds 2

to the lower ratio of 8. Let this new value be x. The new ratio can now be expressed as: 1

x : 5 = 8 : 11 2

Or,

𝑥 1 5 2

=

8 11

Cross multiply ∴

1

11x = 5 x 8 2

11x =

11 2

x8

11x = 11 x 4 (Since 8 divided by 2 is 4) Divide both sides by 11. 11𝑥 11

=

11 𝑥 4 11

x=4 The new value is 4 days b. Express 1 year 8 months in months. This gives 12 + 8 = 20 months. (Since 1 year = 12 months) Now, the new decreased number of months corresponds to the lower ratio of 2. Let this new value be x. 𝑥

∴

20

=

2 5

Cross multiply ∴ 5x = 20 x 2 Divide both sides by 5. 5𝑥 5

=

20 𝑥 2 5

153

x=4x2 =8 The new value is 8 months

5. A gin is composed of alcohol and water. The ratio of alcohol to water in the gin is 5 : 2 by volume. Find the volume of gin which contains 35cm3 of alcohol. Solution Let the volume of water in the gin be x. The alcohol has the ratio 5, while the water has the ratio 2. 2

∴

=

5

𝑥 35

Cross multiply ∴ 5x = 2 x 35 Divide both sides by 5. 5𝑥 5

=

2 𝑥 35 5

x=2x5 = 10 The volume of water in gin is 10cm3 ∴ The volume of the gin = volume of water + volume of alcohol = 10 + 35 = 45cm3.

6. When wood is dried, its mass reduces in the ratio 7 : 11. A piece of wet wood has a mass of 5.06kg. What mass is lost when the wood dried? Solution Let the new mass of the wood be x. This new reduced mass corresponds to the lower ratio of 7. ∴

5.06 𝑥

=

11 7

Cross multiply ∴ 11x = 5.06 x 7 Divide both sides by 11. 11𝑥 11

=

5.06 𝑥 7 11

154

x = 0.46 x 7 (5.06 divided by 11 gives 0.46) = 3.22kg ∴ Mass lost = 5.06 – 3.22 = 1.84kg

7. Express the ratios below in the form n : 1 a. $2.80 : 0.70 b. 3 : 8 Solutions a. $2.80 : $0.70. In the ratio n : 1, the 1 is on the right side. Therefore, the two ratios will be divided by the ratio on the right side, i.e. 0.70. This gives: 2.80 0.70

:

0.70 0.70

=

28 7

:1

(The numerator and denominator of

2.80 0.70

have been multiplied by 10 to give

28 7

)

=4:1 The ratio is 4 : 1 b. 3 : 8 Divide the two ratios by the ratio on the right side, i.e. 8. This gives: 3 8

:

8 8

= 0.375 : 1

8. Express the ratios below in the form 1: x a. 2.0 : 1.6 b. 240g : 1.44kg Solutions a. 2.0 : 1.6 Here, divide both sides of the ratio by 2.0 in order to make the left hand side to be 1. This gives: 2.0 1.6

:

2.0 2.0

155

1.6

=1:

2.0 16

=1:

(The numerator and denominator of

20

1.6 2.0

have been multiplied by 10 to give

= 1 : 0.8 The ratio is 1 : 0.8 b. Express the two quantities in the same unit. To convert 1.44kg to g, multiply by 1000 ∴ 1.44kg = 1.44 x 1000 = 1440g So, the ratio becomes, 240 : 1440 In the form 1 : x, this gives: 240 1440 240

:

240

=1:6 ∴ The ratio is 1: 6

9. Which of the following pairs of ratios is greater? a. 17: 9, 15: 8 b. 1.5m : 40cm, 26 : 7 Solutions a. 17: 9, 15: 8 Express each ratio in fraction and compare the fractions. ∴

17: 9 =

17 9

, while 15: 8 =

15 8

The greater fraction is now obtained as follows: 17 15 9

= =

,

8

8 𝑥 17 , (9 𝑥 15) 72

(The LCM of 9 and 8 is 72)

136, 135 72

So, the first ratio which corresponds to 136 is greater ∴ The greater ratio is 17: 9 156

16 20

)

b. 1.5m : 40cm Expressing both quantities in the same unit gives: (1.5×100)cm : 40cm (Since 1m = 100cm) = 150 : 40 or 15 : 4 (In its lowest term) So, the two ratios are: 15 : 4 and 26 : 7. =

15 4

26

,

7

7 𝑥 15 , (4 𝑥 26)

=

28

=

(The LCM of 4 and 7 is 28)

105, 104 28

So, the first ratio which corresponds to 105 is greater ∴ The greater ratio is 1.5m : 40cm or 15cm : 4cm

10. The size of a farm is 15.7m by 12.3m. On a map, the length of the farm is 78.5cm. What is the scale of the map in the form 1 : n? Determine the width of the farm on the map. Solution Length of farm on the ground is 15.7m. Length of farm on the map is 78.5cm. ∴ The ratio of the length on map to length on the ground is 78.5cm : 15.7m. Expressing this length in cm gives: 78.5cm : (15.7 × 100)cm = 78.5cm : 1570cm Expressing this scale in the form 1 : n gives: 78.5 1570 78.5

:

78.5

= 1 : 20 ∴ The scale of the map is 1 : 20 Note that 20 correspond to the length of the farm on the ground. Let the width of the farm on the map be x. 1 : 20 = x : 12.3

157

∴

1 20

=

𝑥 12.3

Cross multiply 20x = 1 x 12.3 Divide both sides by 20. 20𝑥 20

=

12.3 20

x = 0.615m (Since 12.3 is in m) ∴ Width in cm = 0.615 × 100 (1m = 100cm) = 61.5cm The width of the farm on the map is 61.5cm.

11. A car covers a distance of 16km in 1 hour, and a bike covers a distance of 6.6m in 1.5 seconds. Which is faster? Solutions This is a comparison of ratios. Express the two ratios in the same units and find which is the greater ratio. 16km = (16 x 1000)m = 16000m (Since 1km = 1000m) 1 hour = 60mins = (60 x 60)sec = 3600sec (Since 1min = 60sec) ∴ 16km : 1 hour is 16000m : 3600sec = 40 : 9 (In its lowest term, after dividing by 400) The second ratio is 6.6m : 1.5sec Since the two ratios are now expressed in the same units, the greater ratio is obtained as follows: 40 : 9, 6.6 : 1.5 40 9

=

,

6.6 1.5

=

1 𝑥 40 , (6 𝑥 6.6) 9

40, 39.6 9

∴ The first ratio is greater. This ratio corresponds to the distance of 16km in 1 hour. ∴ The car is faster. Exercises 1. Express the following ratios in their simplest forms. a. 4 : 32 158

b. 50cm to 2m 4

c. 2 h : 10mins 5

1

1

2

2

d. 5 : 1 e.

3 4

to

1 12

to

5 6

2. Find the values of x in the ratios below. a. 20 : x = 4 : 3 b. x : 36 = 5 : 2 c. 7 : 9 = 21 : x

3. Increase the following quantities in the given ratios. a. $30 in the ratio 8 : 5 b. 28cm in the ratio 9 : 4 4. Decrease the following quantities in the given ratios. 1

a. 3 days in the ratios 10 : 21 3

b. 2 years 6 months in the ratio 5 : 8 5. The ratio of water to colourant in a gin is 9 : 1 by volume. Find the volume of water in the gin when the volume of the colourant is 12cm3. 6. When fabric is dried, its mass reduces in the ratio 3 : 10. A piece of wet cloth has a mass of 210g. What mass is lost when the fabric is dried? 7. Express the ratios below in the form n : 1 a. $5.4 : $0.90 b. 12 : 18 8. Express the ratios below in the form 1: x a. 2.5 : 12.5 b. 72 : 108 9. Which of the following pairs of ratios is greater? a. 21: 10, 7: 4 159

b. 2.4kg : 800g, 20 : 7 10. An amount of $1200 is increased in the ratio 4 : 3. The new value is then decreased in the ratio 5 : 8. What is the final amount obtained?

160

CHAPTER 17 RATE Quantities of different types may be compared in the form of rate. For example a man who walks a distance of 5km in 2 hours walks at a rate of 2.5km per hour. Examples 1. A woman is paid $6400 for working for a period of 20hrs. Calculate her hourly rate of pay. Solution The rate to be determined is the amount she gets per hour. This will be obtained by dividing the total amount by the total time. ∴ Hourly rate of pay =

6400 20

= $320/hour

Her hourly rate of pay is $320 per hour.

2. A worker is paid $630 for working for 8 hours. Calculate the worker’s rate of pay per day. Solution The comparison here is the amount paid divided by the time in days. 1

8

But 8 hours = (24 )days = ( )

(Since 24 hours = 1 day)

3

∴ Rate of pay per day =

630 1 3

= 630 ÷

1 3

3

= 630 x = 630 x 3 = 1890 1

The worker is paid $1890 per day.

1

3. A bus travels 160km in 1 h. calculate its average speed in km/h. 3

Solution The rate in km/h = =

distance in km time in hour 160 1 1 3

=1

160 4 3

= 160 ÷

4 3

161

3

= 160 x = 40 x 3 = 120 4

The average speed of the bus is 120km/h.

4. A village has an area of 48km2 and a population of 1440. Calculate the population density of the village per km2. Solution The rate in number of people/km2 = =

1440 48

population of village area

= 30

The population density of the village is 30people/km2

5. An iron of length 3.5m has a mass of 400.5kg. Find its mass in kg/m. Solution Mass in kg/m = =

Mass in kg length 400.5 3.5

= 114.4

Its mass in kg/m = 114.4kg/m

6. 25.2cm3 of water has a mass of 26.8g. Calculate the density of the water in g/cm3. Solution Density in g/cm3 = =

mass in g volume 26.8 25.2

= 1.06

Density of the water is 1.06g/cm3

7. A car burns petrol at the rate of 1 litre for every 3.2km travelled. How many litres will the car burn on a journey of 640km?

162

Solution The rate is 3.2km per litre, which is 3.2km/litre. This rate is constant for any distance travelled. With this constant rate, the litres of petrol burnt for a journey of 640km is obtained as follows: Total distance in km

3.2km/litre = 3.2 =

Total petrol in litres

640

(Where l is the litres of petrol)

l

3.2l = 640 ∴

640

l=

3.2

= 200

The car will burn 200 litres of petrol.

8. A container has a capacity of 20 litres. It is filled with wine whose density is 0.8kg/litre. What is the mass of the wine? Solution The rate is the density of 0.8kg/litre, which is a constant value. With this rate ke pt constant, the mass of the wine is obtained as follows: 0.8kg/litre = 0.8 =

mass in kg volume in litres

mass 20

∴ Mass = 0.8 x 20 = 16 The mass of the wine is 16kg.

1

5

2

12

9. A portion of a city has a rectangular shape which measures 1 km by 1 km. Calculate the total population of the city portion if its population density is 720people per km 2. Solution 1

5

2

12

Area of the city portion = 1 x 1 =

17 2x4

=

17 8

3

17

2

12

= x

km2

The rate is the population density of 720people/km2, which is a constant value. ∴

720people/km2 =

Total population Area in km

163

720people/km2 =

Total population 17 8

Cross multiply Total population = 720 x

17 8

= 90 x 17 = 1530 Total population of the city portion is 1530 people

10. If a man can ride a bicycle at a rate of 5m/sec, how long will it take him to ride a distance of 12km at the same rate? Solution The rate is 5m/sec. When this rate remains the same, then the time taken to ride a distance of 12km is obtained as follows: 5m/sec =

Total distance in m total time in seconds

But, 12km = (12 x 1000)m = 12000m ∴

5=

12000 t

(Where t = time in seconds)

5t = 12000 t=

12000 5

= 2400 seconds

Or, time in minutes =

2400

(Since 60 seconds = 1 minute)

60

= 40 minutes Time taken is 2400sec or 40mins.

11. I travelled at 60km/hr and took 2 hours for a certain journey. How long would it have taken me if I had travelled at 50km/h? Solution At a rate (speed) of 60km/h for 2 hours, the distance covered is obtained as follows: 60km/h = 60 =

total distance in km time in hour

total distance 2

∴ Total distance = 60 x 2 = 120km At a new rate (speed) of 50km/h across this same distance of 120km, the time taken is obtained as 164

follows: 50km/h = 50 =

Total distance in km Time in hour

120

(Where t is the time)

t

Cross multiply 50t = 120 ∴

t=

120

= 2.4 hours

50

The 0.4 hours in minutes = 0.4 x 60 = 24 minutes (Since 60 minutes make 1 hour) ∴ It would have taken me 2.4 hours or 2 hours 24 minutes.

12. A vehicle uses diesel at the rate of 2 litres for every 18km. If the price of diesel is $7 per litre, calculate the cost of diesel needed for a journey of 1080km. Solution The rate which is the distance per litre of diesel is given by: rate(km/litre) = =

18 2

distance in km diesel in litres

= 9km/litre

This rate is a constant value. So, the diesel needed for a distance of 1080km is obtained as follows: 9=

1080 l

(Where l is the litres of diesel)

9l = 1080 ∴

l=

1080 9

= 120 litres

Since the cost of diesel is $7 per litre, then the cost of 120 litres = 7 x 120 = 840 Cost of diesel needed for the journey is $840.

13. In a town with a population of 16240, an additional 812 babies were born in one year. Find the birth rate per 1000 persons. Solution The birth rate per person =

Total birth total population

165

= ∴

812 16240

= 0.05 births per person

Birth rate per 1000 persons = 0.05 x 1000 = 50 births per 1000 persons.

Exercises 1. A woman is paid $960 for working for a period of 8hrs. Calculate her hourly rate of pay. 2. A worker is paid $36144 for working for 12 hours. Calculate the worker’s rate of pay per day. 2

3. A bus travels 1248km in 2 h. calculate its average speed in km/h. 5

4. A village has an area of 56km2 and a population of 1680. Calculate the population density of the village per km2. 5. An iron of length 4.25m has a mass of 25.5kg. Find its mass in kg/m. 6. 20 litres of water has a mass of 19.5kg. Calculate the density of the water in kg/L. 7. A car burns petrol at the rate of 1 litre for every 2.4km travelled. How many litres will the car burn on a journey of 288km? 8. A container has a capacity of 800cm3. It is filled with kerosene whose density is 0.82g/cm3. What is the mass of the kerosene? 4

1

5

4

9. A portion of a city has a rectangular shape which measures 12 km by 6 km. Calculate the total population of the city portion if its population density is 140people per km2. 10. If a man can ride a bicycle at a rate of 3m/sec, how long will it take him to ride a distance of 6km at the same rate? Give your answer in hours. 11. I travelled at 40km/hr and took 5 hours for a certain journey. How long would it have taken me if I had travelled at 120km/h? 12. A vehicle uses diesel at the rate of 5 litres for every 32km. If the price of diesel is $6 per litre, calculate the cost of diesel needed for a journey of 620km. 13. In a town with a population of 124000 people, a total of 1250 people died in one year. Find the death rate per 1000 persons. 166

167

CHAPTER 18 PROPORTIONAL DIVISION Quantities can be divided or shared into two or more parts by the use of ratios. To divide a quantity into two parts in the ratio 7 : 2 means

7 7+2

7

2

9

7+2

or of the quantity, and

2

or of the quantity. 9

Examples 1. Divide $2070 between John and Kent in the ratio 4 : 5. Solution Total ratio = 4 + 5 = 9 Note that the 4 part belongs to John while the 5 part belongs to Kent, according to the order in which they were mentioned. ∴

4

John’s share = x 2070 9

= 4 x 230 = $920 5

Kent’s share = x 2070 9

∴

= 5 x 230 = $1150 John’s share is $920 while Kent’s share is $1150.

1 2

1

4 3

2

2. Ben, Jane and Dan share $1325 in the ratio 1 : : 2 . How much does each get? Solution The mixed fraction in the ratio can be expressed in improper fraction to give: 1 2

1

5 2 5

4 3

2

4 3 2

1 : :2 = : :

Convert the ratio to whole numbers by multiplying each one by the LCM of their denominators. The LCM of 4, 3 and 2 is 12. So, multiply each term in the ratio by 12. This gives: 5

2

5

4

3

2

(12 x ) : (12 x ) : (12 x ) = 15 : 8 : 30 ∴ Total ratio = 15 + 8 + 30 = 53 Ben’s share =

15 53

x 1325 = 15 x 25 (After dividing 1325 by 53 to get 25)

= $375 Jane’s share =

8 53

x 1325 = 8 x 25 = $200 168

Dan’s share =

30 53

x 1325 = 30 x 25 = $750

3. Three friends are 1.5m, 1.65m and 1.75m tall. They share $1470 in the ratio of their heights. How much does each receive? Solution The ratio is: 1.5 : 1.65 : 1.75 Total ratio = 1.5 + 1.65 + 1.75 = 4.9 1st friend’s share =

1.5 4.9

x 1470 = 1.5 x 300

(After dividing 1470 by 4.9)

= $450 2nd friend’s share = 3rd friend’s share =

1.65 4.9 1.75 4.9

x 1470 = 1.65 x 300 = $495 x 1470 = 1.75 x 300 = $525

4. A, B and C share $68 so that for every $1 that C gets B gets $2, and for every $3 that B gets, A gets $4. What is A’s share? Solution Two sets of ratios are given in the question. They are: B : C = 2 : 1 (Since when B gets $2, C gets $1) A : B = 4 : 3 (Since when B gets $3, A gets $4) B is common to the two ratios. Therefore make B equal in the two ratios. This can be obtained by multiplying the first ratio by 3 and the second ratio by 2. This gives: B : C = (3 x 2) : (3 x 1) = 6 : 3 A : B = (2 x 4) : (2 x 3) = 8 : 6 The quantity 6 is for B in the two ratios. Since B is equal in the two ratios, then the ratios can now be combined together. This gives: A:B:C=8:6:3 Total ratio = 8 + 6 + 3 = 17 ∴ A’s share =

8 17

x 68

= 8 x 4 = $32

169

5. Mary, Lopez and Alex share 425kg of rice so that for every 1kg that Mary gets Lopez gets 3kg, and for every 2kg that Lopez gets, Alex gets 3kg. Find Lopez’s share. Solution First ratio: Mary : Lopez = 1 : 3 (Since when Mary gets 1kg Lopez gets 3kg) Second ratio: Lopex : Alex = 2 : 3 (Since when Lopez gets 2kg, Alex gets 3kg) Lopez is common to the two ratios. Therefore make Lopez part equal in the two ratios. This can be obtained by multiplying the first ratio by 2 and the second ratio by 3. This gives: Mary : Lopez = (2 x 1) : (2 x 3) = 2 : 6 Lopez : Alex = (3 x 2) : (3 x 3) = 6 : 9 Since Lopez is equal in the two ratios, then the ratios can now be combined together. This gives: Mary : Lopez : Alex = 2 : 6 : 9 Total ratio = 2 + 6 + 9 = 17 ∴ Lopez’s share =

6 17

x 425

= 6 x 25 = 150kg

2

6. $8200 is shared among Cole, Lang and Jim so that Coles share is 5 of Lang’s share, and Lang’s 3

share is 4 of Jim’s share. How much did Cole receive? Solution 2

2

5

5

If Cole’s share is of Lang’s share, then Cole’s ratio is the fraction , while Lang’s ratio is the whole number 1. 2

∴ First ratio is: Cole : Lang = : 1 = 2 : 5

(After multiplying by 5)

Second ratio is:

(After multiplying by 4)

5 3

Lang : Jim = : 1 = 3 : 4 4

Lang’s part is 5 in the first ratio and 3 in the second ratio. In order to make his part equal in the two ratios, multiply the first ratio by 3 and the second ratio by 5. This gives: First ratio: Cole : Lang = (3 x 2) : (3 x 5) = 6 : 15 Second ratio: Lang : Jim = (5 x 3) : (5 x 4) = 15 : 20 Since Lang is equal in the two ratios, then the ratios can now be combined together. This gives: Cole : Lang : Jim = 6 : 15 : 20 Total ratio = 6 + 15 + 20 = 41 ∴ Cole’s share =

6 41

x 8200

= 6 x 200 = $1200

170

1

7. X, Y and Z share $1710 so that X has 2 times as much as Y, and Y has 4 times as much as Z. How 2

much does each receives? Solution 1

1

If X’s share is 2 times as much as Y’s share, then X’s ratio is 2 , while Y’s ratio is 1. 2

1

5

2

2

∴ First ratio is: X : Y = 2 : 1 = : 1 = 5 : 2

2

(After multiplying by 2)

Second ratio is: Y : Z = 4 : 1 Y’s part is 2 in the first ratio and 4 in the second ratio. In order to make his part equal in the two ratios, multiply the first ratio by 2. This gives: First ratio: X : Y = (2 x 5) : (2 x 2) = 10 : 4 Second ratio: Y : Z = 4 : 1 Since Y is equal in the two ratios, then the ratios can now be combined together. This gives: X : Y : Z = 10 : 4 : 1 Total ratio = 10 + 4 + 1 = 15 ∴ X’s share =

10 15

x 1710

= 10 x 114 = $1140 Y’s share =

4 15

x 1710

= 4 x 114 = $456 Z’s share =

1 15

x 1710

= 1 x 114 = $114

8. Kan, Steven and Alicia start a business together. Kan invests $750 for 8 months, Steven invests $600 for 5 months and Alicia invests $450 for 10 months. Find their share of a profit of $12420. Solution Kan’s investment = 750 x 8 = 6000 Steven’s investment = 600 x 5 = 3000 Alicia’s investment = 450 x 10 = 4500 ∴ The ratio of their investment is: 6000 : 3000 : 4500 = 60 : 30 : 45 (After dividing each by 100) = 12 : 6 : 9 (After dividing each by 5) Total ratio = 12 + 6 + 9 = 27 171

∴

Kan’s share =

12 27

Steven’s share = Alicia’s share =

6 27

9 27

x 12420 = 12 x 460 = $5520 x 12420 = 6 x 460 = $2760

x 12420 = 9 x 460 = $4140

9. A, B and C start a company together. A invests $24000 for 1 year, B invests $30000 for 6 months and C invests $36000 for 4 months. In any year, 20% of the profit goes to A, while 4% goes to C. The remaining profit is shared in proportion to their investment. Calculate the total amount received by each of the three partners in the company when the profit is $10500. Solution 1 year = 12 months ∴ A’s investment = 24000 x 12 = 288000 B’s investment = 30000 x 6 = 180000 C’s investment = 36000 x 4 = 144000 ∴ The ratio of their investment is given by: A : B : C = 288000 : 180000 : 144000 = 288 : 180 : 144 = 24 : 15 : 12 (After dividing each by 12) =8:5:4 (After dividing each by 3) When the profit is $10500, A’s initial share of 20% becomes 20 100

x 10500 = 20 x 105 = $2100

C’s initial share of 4% is given by: 4 100

x 10500 = 4 x 105 = $420

∴

The amount of profit left is: 10500 – 2100 – 420 = $7980 This amount is now shared based on their investment as follows: Ratio is: 8 : 5 : 4 Total ratio is = 8 + 5 + 4 = 17 ∴

A’s share =

8 17

x 7980 = $3755.29 172

B’s share = C’s share =

5 17 4 17

x 7980 = $2347.06 x 7980 = $1877.65

∴ Total amount received by A = 2100 + 3755.29 = $5855.29 B’s total amount remains $2347.06 Total amount received by C = 420 + 1877.65 = $2297.65

10. The cost of producing a machine is $225000. This cost arises from cost of materials, labour and overheads in the ratio 7 : 9 : 2 respectively. Calculate the cost of labour for producing 20 such machines. Solution Materials : Labour : Overheads = 7 : 9 : 2 Total ration is: 7 + 9 + 2 = 18 ∴ Cost of labour =

9 18

x 225000

1

= x 225000 = 112500 2

This cost of labour of $112500 is the cost for one machine. Therefore, the cost of labour for 20 machines is given by: 20 x 112500 = $225000.

Exercises 1. Divide $3060 between Larry and Locke in the ratio 2 : 7. 3 1

2. Kane, Sham and Dora share $124000 in the ratio : :

7

5 4 10

. How much does each get?

3. Three friends are 45kg, 50kg and 55kg in weight. They share $22500 in the ratio of their weights. How much does each receive? 4. A, B and C share $124 so that for every $1 that C gets B gets $3, and for every $4 that B gets, A gets $5. What is C’s share?

173

5. Agnes, Lora and Julian share 242kg of rice so that for every 1kg that Lora gets Agnes gets 4kg, and for every 2kg that Agnes gets, Julian gets 3kg. Find Lora’s share. 3

6. $14500 is shared among John, Peter and Dean so that John’s share is 4 of Dean’s share, and John’s 2

share is 5 of Peter’s share. How much did John receive? 2

7. X, Y and Z share $10800 so that X has 1 times as much as Y, and Y has 3 times as much as Z. How 3

much does each receives? 8. Jane, Hannah and Lisa start a business together. Jane invests $1200 for 6 months, Hannah invests $900 for 4 months and Lisa invests $700 for 8 months. Find their share of a profit of $5125. 9. The cost of producing a suite is $650. This cost arises from cost of materials and labour in the ratio 6 : 7 respectively. Calculate the cost of materials for producing 8 such suites. 10. An amount of $132000 was shared among Kent, Ben and Mandy in the ratio 2 : 7 : 3 respectively. Mandy decided to give 50% of his share to two non-governmental organizations in the ratio 5 : 8. What is the amount that each organization receives?

174

CHAPTER 19 AVERAGES The average of a set of numbers is a number that can be used to represent the set as a whole. It is usually near the middle of the set. Out of the many kinds of averages, the mean is the most common. It is given by: Mean =

Sum of all the numbers in the set N

Where there are N numbers in the set. Examples 1. Calculate the mean of the following set of numbers: 8, 5, 15, 12, 17, 9, 11, 3 Solution Mean = =

8 + 5 + 15 + 12 + 17 + 9 + 11 + 3

(Note that there are 8 numbers in the set)

8 80 8

= 10

The mean is 10

2. Calculate the mean of the following set of numbers: 21, 6, 14, 7, 14, 25, 20, 18, 9, 13, 7. Solution Mean =

21 + 6 + 14 + 7 + 14 + 25 + 20 + 18 + 9 + 13 + 7 11

(Note that there are 11 numbers

in the set) =

154 11

= 14

The mean is 14

3. Use a working mean (assumed mean) of 35 to calculate the mean of the data below. 17, 11, 41, 23, 51, 37. Solution Find the deviation (difference) of each value from the working mean of 35. This gives: 175

17 - 35 = -18 11 - 35 = -24 41 - 35 = 6 23 - 35 = -12 51 - 35 = 16 37 - 35 = 2 Adding these deviations gives: -18 - 24 + 6 - 12 + 16 + 2 = -30 The mean of this value gives:

−30 6

= -5

This mean value is now added to the working mean to obtain the true mean. ∴ True mean = -5 + 35 = 30 The mean of the data is 30.

4. Use a working mean (assumed mean) of 44 to calculate the mean of the data below. 45, 32, 50, 29, 61, 48, 35, 32, 54. Solution Find the deviation (difference) of each value from the working mean of 44. This gives: 45 - 44 = 1 32 - 44 = -12 50 - 44 = 6 29 - 44 = -15 61 - 44 = 17 48 - 44 = 4 35 - 44 = -9 32 - 44 = -12 54 - 44 = 10 Adding these deviations gives: 1 - 12 + 6 - 15 + 17 + 4 - 9 - 12 + 10 = -10 The mean of this value gives:

−10 9

= -1.1

This mean value is now added to the working mean to obtain the true mean. ∴ True mean = -1.1 + 44 = 42.9 The mean of the data is 42.9.

176

5. Given that the mean of 4, 8, 6, 12, 6, 8, 5, is 7, write down the mean of 42, 82, 62, 122, 62, 82, 52. Solution The mean can be determined by inspection. However, a logical way of solving this question is as explained below: Mean = 7=

Sum of all the number in a set N

Sum of the numbers 7

(Since N = 7)

∴ Sum of the numbers = 7 x 7 = 49 From the question, the second set of data was obtained from the first set by multiplying each number in the first set by 10 and adding 2. For example 4 becomes 42 as follows: (4 x 10) + 2 = 40 + 2 = 42. So, the sum of the second data can be obtained from the sum of the first data in a similar way as follows: (49 x 10) + (2 x 7) (Since there are 7 numbers multiply the 2 by 7) = 490 + 14 = 504 ∴

Mean =

504 7

= 72

This shows that by inspection, the mean of the first data which is 7 has become 72 for the second data because 2 was included at the end of each number in the first data to obtain the second data. This shows that if a specific number is put at the end of number in a first set of data to generate a second set of data, then the new mean is also obtained by putting that same number at the end of the previous mean.

6. Given that the mean of 68, 56, 48, 64, 72, 34, 36, 90, 52, and 80 is 60, write down the mean of 6807, 5607, 4807, 6407, 7207, 3407, 3607, 9007, 5207 and 8007. Solution By inspection, 07 has been written at the end of each number in the first set of data to obtain the second set of data. Similarly, 07 must be written at the end of the first mean in order to obtain the second mean. ∴ The new mean = 6007 (When 07 is written at the end of 60)

7. Calculate the value of x if the mean of 30, x, 12, 40 and 10 is x.

177

Solution Mean = x=

30 + 𝑥 + 12 + 40 + 10 5 92 + 𝑥 5

By cross multiplication it gives, it gives: 5x = 92 + x 5x - x = 92 4x = 92 ∴

x=

92 4

= 23

8. The mean age of seven students in a class is 13years 2months. If the age of one teacher is added, the mean age becomes 17years 7months. Calculate the age of the teacher. Solution Mean =

Total age N

When they are seven, then: 13years 2months =

Total age 7

∴

Total age = 7 x 13years 2months = 91years 14months = 92years 2 months (Out of the 14months, there are 1year and 2months. So, the 1 year is added to the 91 years to make it 92 years. Note that 12 months make one year). Similarly, the total age of the students and teacher, i.e. eight people is given by: Total age = 8 x 17years 7months = 136years 56months = 140years 8months (Note that in 56months, there are 4years 8months since 56 divided by 12 is equal to 4 remainder 8). Finally, the age of the teacher is given by the difference between the two total values. This give: 140years 8months – 92years 2months = 48years 6months. The age of the teacher is 48years 6months.

9. The average age of 11 players in a team is 24years 2 months. If the age of the manager is included, the average age increases to 26years. Find the age of the manager. 178

Solution Total age of 11 players = 11 x 24year 2months = 264years 22 months = 265years 10months When the age of the manager is included, they become 12 people. Since the average age of the 12 people is 26years, then their total age is: Total age = 12 x 26years = 312years ∴ The age of the manger is given by: 312year - 265years 10months = 312years 0months - 265years 10months This is evaluated as follows: 312.0 - 2 6 5 . 10 46.2 The age of the manager is 46years and 2months. Workings: 0months - 10months is not possible. So, take 1 from 2. This 1 year taken is equal to 12 months. Then add the 12 months to the 0 months. This gives 12 months. So, 12 months - 10 months = 2 months. The remaining working is carried out in the usual way of subtracting numbers. Remember that the 2 in 312 has become 1 since 1 was taken from it.

10. The average mass of nineteen students is 50.2kg. A new student of mass 44.2kg joins the students. Calculate the new average mass of the students. Solution Total mass of the nineteen students is given by: 50.2 x 19 = 953.8kg When the mass of the new student is added to it, the total mass becomes: 953.8 + 44.2 = 998kg. They are now 20 students. Therefore the average mass of the 20 students is: =

998 20

= 49.9kg

179

AVERAGE RATES Examples 1. A student lives 4km away from school. She walks 1km at 6km/h and travels the rest of the way by bus at 30km/h. a. How many minutes does the whole journey take? b. What is her average speed in km/h? Solution Distance

a. Speed =

time

The time taken to walk 1km at 6km/h is given by: 1

6= ∴ Time =

time 1 6 1

Time = hours 6

Multiply the time by 60 (since 60minutes = 1 hour) in order to convert it to minutes ∴

1 6

1

hours = x 60 6

=

60 6

= 10 minutes

Similarly, the time to travel the remaining 3km at 30km/h is given by: 30 = ∴

Time =

3 time 3 1

30

=

10

Time in minutes =

hours 1

10

x 60 = 6 minutes

∴ The total time for the whole journey = 10 minutes + 6 minutes = 16 minutes. b. Average speed =

Total distance Total time

=1 6

=

4 +

4 8 30

(Time should be in hours in order to obtain the speed in km/h)

1 10

1

1

6

10

( +

=

8 30

)

180

=4x

30 8

=

30

(After equal division by 4)

2

= 15 The average speed is 15km/h.

2. A car travels at 48km/h for the first 30km of a journey and 64km/h for the next 120km. What is the car’s average speed? Solution Distance

Speed =

time

The time taken to travel 30km at 48km/h is given by: 48 = ∴ Time =

30 time

30

=

48

5 8

5

Time = hours 8

Similarly, the time to travel 120km at 64km/h is given by: 64 = ∴

Time = =

120

time 120 64 15 8

hours

(After dividing by 8)

Total time for the journey = =

5 8 20 8

+

15 8

= 2.5 hours

Total distance = 30 + 120 = 150km Average speed = =

150 2.5

Total distance Total time

= 60

The average speed is 60km/h

181

3. For two weeks a man’s average expenses was $320 per week. For the next three weeks the man’s average expenses was $260 per week. What was his average weekly expenses for the five weeks. Solution Total expenses for the first two weeks is: 2 x 320 = $640 Total expenses for the next three weeks is: 3 x 260 = $780 ∴ His total expenses for the five weeks is: 640 + 780 = $1420 His average weekly expenses for the 5 weeks is given by: Average expenses =

1420 5

= $284

4. In a class of 30 students, 20 scored an average mark of 65% in a mathematics examination, while the remaining 10 students scored an average mark of 50%. What is the average mark of all the students in the class? Solution Total marks of the 20 students is: 20 x 65 = 1300 Total marks of the 10 students is: 10 x 50 = 500 ∴ Total marks scored by the 30 students is: 1300 + 500 = 1800 Average mark of all the students in the class given by: Average mark =

1800 30

= 60%

5. A bus travels from city A to city B, a distance of 84km. The average speed over the first 60km of the journey is 30km/h. If the average speed for the whole journey was 36km/h, calculate the average speed over the 24km part of the road. 182

Solution Average speed =

Total distance Total time

The time taken to travel 84km i.e. the whole journey at 36km/h is given by: 36 =

84 Total time

∴ Total time =

84 36

=

7 3

7

Total time = hours 3

Similarly, the time to travel 60km at 30km/h is given by: 30 = ∴

Time =

60 time

60 30

= 2hours

∴ The time spent for the 24km part of the journey = 1

1

3

3

7 3

+2

= 2 - 2 = hours Average speed over the 24km part of the journey is given by: Average speed = =

24 1 3

Total distance Total time

= 24 x

3 1

= 72 The average speed is 72km/h

Exercises 1. Calculate the mean of the following set of numbers: 4, 6, 3, 11, 6, 4, 5, 8, 7 2. Calculate the mean of the following set of numbers: 14, 7, 12, 15, 8, 18, 7, 13, 13, 8. 3. Use a working mean (assumed mean) of 60 to calculate the mean of the data below. 55, 45, 70, 68, 54, 62, 76, 50. 183

4. Use an assumed mean of 24 to calculate the mean of the data below. 18, 22, 25, 27, 24, 20, 21, 16, 19, 28. 5. Given that the mean of 5, 9, 3, 6, 2, 3, 8 and 4 is 5, write down the mean of 58, 98, 38, 68, 28, 38, 88 and 48. 6. Find the mean of 32, 36, 35, 30, 38, 35, 36, 28, 31, 27 and 35. Hence, write down the mean of 3211, 3611, 3511, 3011, 3811, 3511, 3611, 2811, 3111, 2711 and 3511. 7. Calculate the value of x if the mean of 9, x, 5, 20, 15 and 6 is x. 8. The mean age of nine students in a class is 11years 5months. If the age of the principal is added, the mean age becomes 16years 3months. Calculate the age of the principal. 9. The average age of 5 players in a team is 26years 10 months. If the age of the coach is added, the average age increases to 35years. Find the age of the coach. 10. The average weight of 25 students in a class is 62.6kg. A new student of weight 75kg joins the class. Calculate the new average weight of the students in the class. 11. A student walks 0.8km at 2km/h and travels 6.2km by riding a bicycle at 9.3km/h. a. How many minutes does the whole journey take? b. What is his average speed in km/h? 12. A car travels at 60km/h for the first 90km of a journey and 50km/h for the next 140km. What is the car’s average speed? 13. For two weeks a man’s average expenses was $420 per week. For the next 4 weeks the man’s average expenses was $680 per week. What is his average weekly expenses for the six weeks. 14. A bus travels from city A to city B, a distance of 108km. The average speed over the first 60km of the journey is 20km/h. If the average speed for the whole journey was 36km/h, calculate the average speed over the remaining 48km part of the road.

184

CHAPTER 20 MIXTURES Mixing of two or more substances can be carried out by sellers in order to have a similar or different blend of the substances mixed together. The prices of such mixtures obtained have to be determined in the process. The examples below illustrate these processes. Examples 1. 20 bags of rice costing $52 a bag is mixed with 30 bags of rice costing $40 a bag. What is the cost per bag of the mixture? Solution Total cost of the 20 bags of rice = 20 x 52 = $1040 Total cost of the 30 bags of rice = 30 x 40 = $1200 Total cost of the 50 (i.e. 20 + 30) bags of rice = 1040 + 1200 = 2240 ∴ Average cost per bag =

2240 50

= 44.8 The cost per bag of the mixture = $44.80

2. A trader mixes three sacks of sugar costing $120/sack with seven sacks of sugar costing $40/sack. If he sell the mixture at $80/sack, calculate his percentage profit. Solution Total cost of the 3 sacks = 3 x 120 = $360 Total cost of the 7 sacks = 7 x 40 = $280 Total cost of the 10 sacks = 360 + 280 = 640 ∴ The average cost per sack =

640 10

= $64

This is the cost per sack of the mixture. When the mixture is sold at $80 per sack, then the profit is: 80 - 64 = 16 The percentage profit = =

16 64

𝑃𝑟𝑜𝑓𝑖𝑡 𝐶𝑜𝑠𝑡 𝑝𝑟𝑖𝑐𝑒

x 100

x 100

185

=

1600 64

= 25%

3. Three kinds of flour at $50, $40 and $82 per drum are mixed in the ratio 2 : 3 : 5 respectively. What is the cost of the mixture per drum? Solution Each value in the ratio can be taken to be the number of drums used for the mixing. The total cost of the first flour = 50 x 2 = 100 The total cost of the second flour = 40 x 3 = 120 The total cost of the third flour = 82 x 5 = 410 Total cost of all the flour = 100 + 120 + 410 = 630 Total number of drums of the flour = 2 + 3 + 5 = 10 ∴ Average cost of the mixture =

630 10

= 63

The cost of the mixture per drum is $63.

4. A trader bought three kinds of sugar at $100 per bag, $84 per bag and $60 per bag. He mixes them in the ratio 3 : 5 : 4 respectively and sold the mixture at a profit of 20%. At what price per bag did he sell them? Solution Let us take the values in the ratio as the number of bags for each sugar. Total bags of sugar = 3 + 5 + 4 = 12 Total cost of the first sugar = 3 x 100 = 300 Total cost of the second sugar = 5 x 84 = 420 Total cost of the third sugar = 4 x 60 = 240 Total cost of the 12 bags = 300 + 420 + 240 = 960 ∴ Average cost of the mixture =

960 12

= 80

The cost of each bag of the mixture = $80 When each bag is sold at a profit of 20%, then the selling price is given by: 120 100

x 80

(Since 100% is $80, then 120% i.e. 100 + 20 = 120%, gives the selling price)

= 12 x 8 (After canceling out the zeros) = 96 He sold them at $96 per bag.

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5. In what proportion should rice at $78 and $84 per bag be mixed in order to obtain a mixture worth $82 per bag? Solution Since the mixture is worth $82 per bag, then: the profit per bag on the rice of $78 = 82 - 78 = 4 the loss per bag on the rice of $84 = 84 - 82 = 2 ∴ The ratio of profit to loss = 4 : 2 =2:1 (In its lowest term) In order for the profit to be exactly equal to the loss, the first value of the ratio should be multiplied by 1, while the second value should be multiplied by 2. This gives: (2 x 1) : (1 x 2) This means that they have to be combined in the ratio 1 : 2, which is the factors by which the original ratio is multiplied to make the ratio values equal. Hence, the proportion in which the first rice of $78 must be mixed with the second rice of $84, is 1 : 2.

6. In what ratio must honey costing $275 and $185 per gallon be mixed to produce a mixture costing $212 per gallon? Solution Since the mixture is worth $212 per gallon, then: the loss per gallon on the honey of $275 = 275 - 212 = 63 the profit per gallon on the rice of $185 = 212 - 185 = 27 ∴ The ratio of loss to profit = 63 : 27 =7:3 (In its lowest term after dividing by 9) In order for the loss to be exactly equal to the profit, the first value of the ratio should be multiplied by 3, while the second value should be multiplied by 7. This gives: (7 x 3) : (3 x 7) This means that they have to be combined in the ratio 3 : 7, which is the factors by which the original ratio is multiplied to make the ratio values equal. Hence, the proportion in which the first honey of $275 must be mixed with the second honey of $185, is 3 : 7.

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7. Apple drink which cost $26.40 per litre is diluted with water in the ratio of 5 parts of apple drink to 1 part of water. The diluted apple drink is sold at $33 per litre. Calculate the percentage profit. (Assume that the cost of the water is negligible). Solution Let us take the ratio of 5 : 1 to be the volumes in which the liquids were mixed. ∴ Cost of 5 litres of the apple drink = 5 x 26.40 = 132 The cost of 1 litre of the water = 0 ∴ The total cost of the 6 litres of the apple drink = 0 + 132 = 132 The average cost per litre =

132 6

= $22

This is the cost per litre of the diluted apple drink. When this drink is sold at $33 per litre, then the profit is: 33 - 22 = 11 ∴

Percentage profit =

11 22

x 100 =

1100 22

= 50%

8. 20 litres of orange drink which cost $42 per litre is diluted with 5 litres of water. What is the cost per litre of the diluted drink. Solution Total cost of the 20 litres of orange drink = 20 x 42 = 840 When 5 litres of water is added to the drink, the total volume of the drink becomes, 20 + 5 = 25 litres. ∴

Average cost per litre =

840 25

= 33.6

The cost per litre of the diluted drink is $33.60.

Exercises 1. 12 bags of rice costing $20 a bag is mixed with 8 bags of rice costing $15 a bag. What is the cost per bag of the mixture? 2. A trader mixes five sacks of sugar costing $42/sack with 2 sacks of sugar costing $35/sack. If he sells the mixture at $48/sack, calculate his percentage profit.

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3. Three kinds of flour at $2.80, $3.50 and $2.50 per kg are mixed in the ratio 5 : 4 : 2 respectively. What is the cost of the mixture per kg? 4. A trader bought three kinds of sugar at $60 per bag, $90 per bag and $100 per bag. He mixes them in the ratio 2 : 2 : 1 respectively and sold the mixture at a profit of 25%. At what price per bag did he sell them? 5. In what proportion should rice at $112 and $104 per bag be mixed in order to obtain a mixture worth $106 per bag? 6. In what ratio must honey costing $24.40 and $16.20 per litre be mixed to produce a mixture costing $20 per litre? 7. Apple drink which cost $18.20 per litre is diluted with water in the ratio of 3 parts of apple drink to 1 part of water. The diluted apple drink is sold at $21.5 per litre. Calculate the percentage profit. (Assume that the cost of the water is negligible). 8. 15 litres of orange drink which cost $34 per litre is diluted with 5 litres of water. What is the cost per litre of the diluted drink?

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If you have any enquiries, suggestions or information concerning this book, please contact the author through the email below.

KINGSLEY AUGUSTINE [email protected] Twitter handle: @kingzohb2

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