fractions - class vi

Fractions

What is a Fraction?

If you divide an apple equally between two persons. each person gets half an apple, or one of the two equal parts of the apple. If you divide it equally between three persons, each person gets a third of the apple, or one of the three equal parts of the apple; if you divide it between four persons, each person gets one fourth, and so on.

1 2 1 3 1 4

1 5

Quantities such as these, which represent parts of a whole, are called fractions. 1 1 1 1 1 Fractions such as - , - . - , - . - , . . . are called unit fractions because they 2 3 4 5 6

represent one of the parts of a whole. However, a fraction need not always represent one of the parts of a whole. The following are also fractions.

3 one-fourth parts= 314

Denominator and numerator

5 one-sixth parts= 516 5 one-fourth parts= 5/4

In a fraction, the number above the line of division is called the numerator and the number below the line is called the denominator.

Examples (i) In 3 , numerator = 3 and denominator = 7. 7

(ii) In ~, numerator = 2 and denominator = 13. 13

(iii) In .!...!. , numerator = 11 and denominator = 6. 6 . N-38

Fractions N-39

Proper fractions and improper fractions If the numerator of a fraction is less than the denominator then the fraction is called a proper fraction. If the numerator is greater than or equal to the denominator then the fraction is called an improper fraction.

3 5 23 77 Examples - , - , - , -- are some proper fractions. 4 8 60 123 6 13 25 8 - , - , - , - are some improper fractions. 5 7 6 8

Mixed fractions A fraction which consists of (i) a natural number and (ii) a proper fraction is called a mixed fraction.

Examples (i) 1.!. =the natural number 1, and the proper fraction.!. together. 3 3

(ii) 2 2 = 2 + 2 (iii) 3 .!. = 3 + .!. (iv) 7 4 = 7 + 4 5 5 4 4 9 9

Since a mixed fraction is made up of a natural number and a proper fraction, it is also known as a mixed number. The natural number is the integral part of a mixed fraction, while the proper fraction is the fractional part of the mixed fraction.

Any mixed fraction can be written as an improper fraction. Conversely, any improper fraction can be written as a mixed fraction.

Examples (i) 1 3 = 4 one fourths + 3 one fo\lrths = 7 one fourths = 7 4 4

(ii) 9 = 9 one fifths = 5 one fifths + 4 one fifths = 1 + 4 = 1 4 5 . 5 5 To write a mixed fraction as an improper fraction, use:

. . _ integral part x denominator+ numerator- in the fractional part j Mixed fraction - d min t eno a or

2 3 x 5+2 17 Examples (i) 3 - = = - 5 5 5

(ii) 2 2 = 3x2+2 =8. 3 3 3

To write an improper fraction as a mixed fraction, use:

remainder _I Improper fraction = quotient in {numerator + denominator)+ d min t eno a or

- --- ------- -

Examples

Equivalent fraction

(i) 37 = 9 .!. 4 4

(ii) 37 = 6 .!. 6 6

(: in 37 + 4, quotient = 9 and remainder = 1).

(. in 37 +6, quotient = 6 and remainder = 1).

Half of each of the following circles is shaded, so the shaded portions are equal. However, in the first circle, the shaded portion represents one of two equal

N-40 ICSE Mathematics for Class 6

parts, in the second circle, two of four equal parts, in the third circle, four of eight equal parts, and in the fourth circle, three of six equal parts.

. 1 2 4 3 We can put this numerically as, - = - = - = -

2 4 8 6 Such fractions are called equivalent fractions because they are equal in value.

2 l x 2 4 l x 4 3 l x 3 Observe that -=--. - =--. -=-- 4 2 x 2 8 2 x 4 6 2 x3 1 2 + 2 4+4 3 +3 Also, - = --=-- =-- 2 4 + 2 8+4 6 +3

In general. If the numerator and the denominator of a fraction are multiplied I or divided by the same number (other than 0), we get an equivalent fraction.

Examples (i) 2 = 2 x 2 = 4 = 2 x 3 = 6 = 2 x 4 = ~, etc. 3 3 x 2 6 3x3 9 3 x 4 12

(ii) ~ = 12 + 2 = 6 = 12 +3 = 4 = 12 +6 = 2 . 18 18 + 2 9 18 +3 6 18 +6 3

Fraction in lowest terms

EXAMPLE

Solution

A fraction is said to be in lowest terms (its s implest form) if its numerator and denominator have no factor in common except 1.

1 2 5 13 18 Examples - , - , - , - , - are in lowes t terms. 5 7 9 8 7 2 15 100 22 - , - ,-- , - are not in lowest terms. 6 35 40 10

A fraction can be reduced to its lowest terms by cancelling out the common factors of the numerator and the qenominator. This can be done by (a ) prime factorisation Cb) or HCF.

135 Reduce 150

to lowest terms.

By prime factorisation 135 =5 x 27 = 5 x 3x9 = 5x3x3x3.

150 = 2 x 75 = 2 x 5 x 15 = 2 x 5 x 5 x 3.

135 .Sx 3 x 3 xZ 3 x 3 9 - - ----=--.=-

150 2 x.Sx5xZ 2 x 5 10.

ByHCF

HCFofl35and 150 = 5 x 3 = 15. 135 135 + 15 9 - - ---=- 150 150 + 15 10

Note In general, a fraction is expressed in its lowest terms.

Fractions N-41

Like fractions and unlike fractions Fractions are called like fractions if their denominators are equal. ,

1 3 6 21 . Examples - , - . - , - are like fractions. 10 10 10 10 . 6 3 9 12

- . - , - , - are unlike fractions, though they are equal fractions. 10 5 15 20

Conversion of fractions into like fractions

EXAMPLE

Solution

Steps 1. Find the LCM of the denominators of the fractions.

2. Multiply the numerato~ and denominator of each fractions by LCM ..,.. denominator.

17 5 Reduce - and - into like fractions. 24 42

2 24, 42 3 12,21

4 , 7 LCM of 24 and 42 is 2 x 3 x 4 x 7, that is, 168.

Now, 168-;-24 -:.: 7and168 -;- 42 = 4. Multiplying the numerator and denominator of the two fractions by 7 and 4 respectively,

1 7 x 7 119 5 x 4 20 24 x 7 = 168 . 42 x 4 168

the like fractions are 119 and 20 168 168

Comparison of fractions

EXAMPLE

Solution

EXAMPLE

Solution

1. If two fractions are like fractions, (a) they are equal if their numerators are equal, or (b) the fraction with the bigger numerator is bigger.

. .

2. If two fractions are unlike fractions, convert them into like fr.actions and then compare them.

13 4 Compare the fractions - and - 25 . 25

The fractions are like fractions, and 13 > 4. 13 4 - >- 25 25

5 7 Compare the fractions - and - 21 12

Let us convert the given fractions. into like fractions. 21=3x7 .. 12 = 2 x6 = 2x2x3

.. LCMof21 andl2 :::;:3x7x2 x 2 = 84. Now, 84-;-21 = 4, 84 -;- 12=7.

N-42

EXAMPLE 1

Solution

ICSE Mathematics for Class 6

Multiplying the numer~tor and denominator of the two fractions by 4 and 7 respectively,

Sx 4 20 7x 7 49 - --- - -- -

21 x 4 . 84 12 x 7 84 20 49 In the like fractions - and - , 49 > 20. 84 84

49 20 7 5 -> - , or - > - 84 84 12 21

Note If two fractions have the same numerator then the fraction with the bigger denominator is the smaller fraction.

6 6 7 7 5 5 For example: - < - , - < - . -

EXAMPLE 4

Solution

EXAMPLE 5

Solution

Fractions N-43

Now. 96+24 = 4. 96 + 16 = 6, 96+32 = 3 . Multiplying the numerator and denominator of the three given fractions by 4 , 6 and 3 respectively,

1 1 x 4 4 5 5 x 6 BO 9 9 x 3 27 24=24x4 = 96 ' 16 = 16x6 = 96 ' 32 = 32 x 3=96 .

6 54 Prove that the fractions - and - . are equivalent. 10 90

First, we change the fractions into like fractions.

10 I10. 90 LCM of 10 and 90 is 90. 1. 9 90 + 10 = 9, 90 +90 = 1

6 x 9 54 54 x 1 54 So, the like fractions are: 10

x 9

= 90

.

90x

1 = 90

As the fractions have equal numerators when their denominators .are equal, the fractions are equivalent. Alternative method Two fractions are equivalent if numerator of first fraction x denominator of second fraction = denominator of first fraction x numerator of second fraction.

a x I Symbolically: b = y if ay = bx. Here, 6 x 90 = 540, 10 x 54 = 540. So, the fractions are equivalent.

Find the numbers a and b in:

a 1 l x 2 2 Given -=-=--=-; 30 15 15 x 2 30 3 1 l x 3 3 Again - = - =--=-; b 15 15 x 3 45

1 a 3 - =- =-

15 30 b

a = 2 .

b = 45.

EXAMPLE 6 Write the following fractions in ascending order. 5 3 1 7

Solution

12' 8' 6' 10

To reduce the fractions into like fractions, we find the LCM of 12, 8, 6 and 10. 2 12, 8 , 6 , 10 2 6 , 4, 3 . 5 3 3 , 2, 3 . 5

1, 2, 1. 5 :. the LCM = 2 x 2 x 3 x 2 x 5 = 120.

Now, 120+12 = 10, 120 +8 = 15, 120 +6 = 20, 120 + 10=12. 5 5xl0 50 3 3xl5 45

12 12 x 10 - = - -= - .

120 8 8 x 15 120 7 7 x l2 84 .

=-

10 10 x l2 120 As 20 < 45 < 50 < 84.

1 3 5 7 -


Remember These

1. A fraction ts (1) proper 1f numerator< denominator, (ii) improper 1f numerator > or = denominator.

2 . (1) Mixed fractions are converted into improper fractions by using ix d fr ti integral part x denominator + numerator of fractional part m e ac on=

denominator (11) Improper fractions are changed into mixed fractions by ustng

remainder improper fraction = quotient in (numerator -:- denominator) + -----denominator

3 . If the numerator and the denominator of a fraction are multiplied or divided by the same number (other than 0), the fraction obtained is an equivalent fi:action.

4 . Fractions with equal denominators are like fractions. To change two or more fractions into Uke fractions, multiply the numerator and denominator of each fraction by (LCM of denominators -:- denominator of the fraction).

5 . To compare two fractions (proper or improper), change the fractions into like fractions. After this, the fraction with the larger numerator will be bigger.

1. Identify the proper, improper and mixed fractions.

Ci) !!_ Cit) 13 (iii) 1 ~ 13 11 11

(v) 5 .!. (vi) 120 (vii) 15 5 241 15

2 . Write the following mixed fractions as improper fractions.

(i) 2 lO (11) 3 _!_ (iii) 12 ~ 17 . 14 10

3 . Convert the following improper fractions to mixed fractions. (i) 37 (ii) 110 (iii) 413

12 43 108

4 . Fill in the blanks.

(1) ~ = ...:..:.:_ 7 42

() 11 _ .. . v - --15 45

(11) 32 =.:..:..: 80 5

(vi) 123 = ...:..:.:_ 66 22

5. Fill in the blanks with .

(i) !_ ... 21 (ii) ~ ... .!..!. 10 25 9 12

(iii) 7 2 =.:..:..: 9 9 (vii) .:.:..:= 3~

6 6

( ... ) .!..!. '!_ ill ... 12 8

(iv)_'!_!_ 100

(viii) 28 19

7 (iv) 52 -8

(. ) 1000 IV - -39

(1 ) 5 - ... v ---8 64 (viii) ~ = _:_:._:_

5 25 .

(iv)~ ... 27 7 63

Fractions

6. Write the followtng in ascending order of magnitude. (i) 18 ~ !_ (ii) ~ 28 37 (iii) ~ 17 ~

25 ' 8 ' 10 5 ' 15 ' 20 4 ' 18 ' 6

7 . Write the following fractions 1n descending order.

(i) ~' !!. . 2- (ii) ~. !_ . .!..!. ' _.! (ill) ~' ~, _!__ 32 48 40 20 30 50 15 144 84 132

ANSWERS

(iv).!..!., !_,~ . 19 36 24 48 72

N-45

1. (i) Proper (U) Improper (ill) Mixed (iv) Proper (v) Mixed (vi) Proper (V11) Improper (v1U) Improper 2 , (1) 44 (U) 43

17 14 (111) 123 (1v) 423 1 (11) 2 24 (111) 3 89 (iv) 25 25 3. (1) 3 -

10 8 12 43 108 ' 39 4. (1) 18 (U) 2 5. (l) < (U) < 6 (1) 5 7 18

. 8'10'25

7 (1) .!! .2-. ~ 48' 40' 32

Addition of like fractions

(ill) 65 (Iv) 40 (111) > (lv) =

(ii) 9 37 28 5 20' 15 (11)4 7 11 3

15'30'50'20

(v)33

(ill)~ ~ 17 4 . 6' 18

(vi) 41

(111} .!!._ .2.. 2-. 84' 132' 144

(vU) 23 (v1U) 40

i 5 19 7 11 ( v) 48' 72' 24 ' 36

Addition and Subtraction of Fractions

The sum of two (or more) like fractions is another like fraction.whose numerator is the sum of the numerators of the given fractions.

S f Uk fr ti sum of numerators um o e ac ons = - -------common denominator

Examples {i) ~ + _.! = 3 + 4 = !_ 11 11 11 11

(ii)_!_+-2_ +_3._= 4 + 7 + 2 = 13. 15 15 15 15 15

(iii) 15 +3.!=15+ 25=15 + 25 = 40 =5. 8 8 8 8 8 8

Addition of unlike fractions To find the sum of two or more unlike fractions, reduce them into like fractions and then add.

Examples (i) 15 + 5 .!. = 15 +21 = 15 +2 l x 2 8 4 8 4 8 4 x 2

= 15 + 42=15 + 42 = 57 =7.!. 8 8 8 8 8


3 20 3 3 20 31 (11) - + - + 2 - = - +-+-20 7 14 20 7 14

= 3x7 +20x20 + 3l x l0 (: LCMof20, 7, 14 = 140) 20 x 7 7 x 20 14 x 10 . 21 400 310 21 + 400 + 310

=-- +--+-- = ------140 140 140 140

= 731 =5 +~ =5~ 140 140 140

Subtraction of a fraction from a like fraction The subtraction of a fraction from a bigger like fraction gives another like fraction whose numerator is the difference of the numerators.

Examples (i) ~ - ~ = 5 - 2 = 2- 13 13 13 13

(11) 10 ~ -~ = 112 - 1 7 = 112 - 17 = 95 = 8 2. . 11 11 11 11 11 11 11

Subtraction of a fraction from an unlike fraction Convert the fractions into like fractions and then subtract as above.

Examples {t) .!_ _~ = 7 x 4 9 x 3 (: LCM of 15, 20 = 60) 15 20 15 x 4 20 x3

28 27 28-27 1 =---= = -

60 60 60 60

{ii) 4 3 _ 2 _!. = 19 _ 13 = 19x3._ 13 x 2 (: LCM of 4 , 6 = 12) 4 6 4 6 4x3 6 x2

= 5 7 - 26 = 5 7 - 26 = 31 = 2.!_ . 12 .12 12 12 12

Short cut for addition/subtraction of tWo fractions

For addition use:

EXAMPLE

Solution

numerator of the first x (LCM of denominators + denominator of the first) + numerator of the second x (LCM of denominators + denominator

Sumoftwofractlons =-------------------o_fth_ e_se_co_ nd_) LCM of denominators

For subtraction, change + into - in the formula.

7 1 Find - + - 12 18

The LCM of 12 and 18 is 36. 7 1 7 x (36 + 12} + 1x(36 + 18) 7 x 3 + 1x2 21+2 23

- + - = = =-- = - 12 18 36 36 36 36

Fractions N-47

EXAMPLE

Solution The LCM of 15 and 20 is 60. 7 9 7x(60 + 15)-9x(60+20) 7x4 - 9x3 28-27 1 --- = = = = - 15 20 60 60 60 60

Simplification

EXAMPLE

Solution

EXAMPLE

Solution

Follow these steps to simplify an expression involving the addition and s ubtraction of fractions. Steps 1. Convert the mixed fractions into improper fractions.

2. Convert all the fractions into like fractions.

3. Simplify the numerators of the fractions thus obtained.

4. Write a fraction with the number obtained in Step 3 as numerator and the common denominator in Step 2 as the denominator.

5. Reduce the fraction obtained in Step 4 to its lowest terms. If it is an improper fraction, change it into a mixed fraction.

7 8 5 Simplify - + - - - . 12 9 24

3 12,9,24 4 4, 3, 8

1, 3, 2 .. LCM of 12, 9, 24 ts3x4x3x2, that ts, 72.

Now, 72 + 12 = 6, 72 +9 =8, 72 +24 = 3.

Multiplying the numerator an1 denominator of three given fractions by 6, 8 and 3 respectively.

7 7 x 6 42 8 8x8 64 5 5x3 15 12 = 12 x 6 = 72 . 9 = 9 x 8 = 72 24 = 24 x 3 = 72 2-+~-~ = 42 + 64_15 = 42+64 - 15 = 106-15 =~ = 1~ 12 9 24 72 72 72 72 72 72 72

Short cut The LCM of 12, 9 , 24 is 72.

7 8 5 7 x (72 +12)+8x(72 +9) - 5 x (72 +24} -+- -- = ------- - ------12 9 24 72

= 7x6 +8x8-5x3 = 42 +64-15=106-15 =~ = 1 19 . 3 7 13

Simplify 8 20 - 3 30 - 40 .

72 72 72 72 72

Changing the given fractions into improper fractions, a2- = 8x20+3 = 163 . 32_ = 3x30+7 = 97

20 20 20 30 30 30

82__32__~ = 163 _ 97 -~ 20 30 40 20 30 40

The LCM of 20, 30, 40 is 120. Now, 120+20 = 6, 120 +30 = 4, 120 + 40 = 3.


Multiplying the numerator and denominator of the given fractions by 6, 4 and 3 respectively,

163x6 978 97 x 4 388 13 x 3 39 =-, -- =-. --=-

20x6 120 30x 4 120 40x3 120 8 2_ _ 32_ _ _!! = 978 - 388 - 39 = 978 - 388 - 39

20 30 40 120 120 120 120 = 978 - 427 = 551 = 42.!_.

120 120 120

Operations involving negative fractions

EXAMPLE

Solution

EXAMPLE

Solution

EXAMPLE

Solution

EXAMPLE 1

So far we have dealt with only positive fractions. Such fractions have a positive integer as the numerator. Negative fractions, on the other hand, have a negative integer as the numerator. Remember, however, that the demoninator of all fractions is a positive integer. For example, in the negative fractions:

--, - - and - 6 - that is - - , which can also be written as - 13 - 20 2 ( - 32 ) 16 11 5 5 13 20 32

- - and - - . the numerators are - 13, - 20 and -32, while 16. ii 5

the denominators are 16, 11and5. The operations for negative fractions are the same as those for positive fractions.

5 - 7 Find -+- 2 16

The LCM of 2 and 16 is 16. ....

5 - 7 5x(16 + 2) + (- 7) x (l6 + 16) - +-=---------2 16 16

= 5 x 8 + (- 7) x 1 = 40 + (- 7) = 33 = 2 _!_ .

3 - 17 Find -+ - 5 10

16 16 16 16

The LCM of 5 and 10 is 10. .. ~+ - 17 = 3 x (10 +5) - 17(10 + 10) = 3 x 2 - 17 x l = 6 - 17 = -11 =-l _l_

5 10 10 10 10 10 10

1 -4 Find - - - 6 15

1 - 4 lx(30+6)-(- 4) x (30 + 15) --=

6 15 30 l x 5-(-4) x 2 5 - (- 8) 5 + 8 13

= = = - -= - 30 30 30 30

Solved Examples

Find (i) 41 + 137 (ii) 3 _ + 2.!.!. . 100 150. 27 36

(LCM of 6 and 15 ts 30)

Fractions N-49

Solution (i) The LCM of 100 and 150 = 300. 41 137 4lx3 137 x2 123 274 ~+~ = + =~+~ 100 150 100 x 3 150 x 2 300 300

= 123 +274 =397 = 1 97 . 300 300 300

Short cut ~ + 137 = 41x(300+100) + 137 x (300 + 150) , (LCM of 100 and 150 = 300) 100 150 300

= 41 x 3+137 x 2=123 + 274 = 397 = l 97 . 300 300 300 300

(ii) 3__ +2 .!.! = 85 + 83 27 36 27 36

85 x (108 + 27) +83 x (108 + 36) =--~~-----~

108 (LCM of 27 and 36 = 108)

= 85 x 4 + 83 x 3 = 340 + 249 = 589 = 5 49 . 108 108 108 108

Alternative method 3__ + 2!! = 3 + _.!_ +2 + .!.! = 5 +_.!_ + .!.! = 5 + 4x 4+11 x 3

27 36 27 36 27 36 108 = 5 + 16 + 33 = 5 + 49 = 5 49 .

108 108 108

EXAMPLE2 Find (i) 15 -~, (ii) 4!_ - 2.!_ , (iii) 5 - 2!. . 8 12 7 14 5

Solution

EXAMPLE 3

Solution

(i) 15_~=15 x (24 + 8) - 5 x (24+12) 8 12 24

(LCM of 8 and 12 = 24)

= 15 x 3 - 5x2 = 45-10 = 35 = l.!_! 24 24 24 24

{ii) 4 _!_ _ 2~ = 29 _ 31 = 29 x (14+7)-31 x(14 + 14) 7 14 7 14 14

(LCM of 7, 14 = 14)

=29x2 - 31 x 1=58 -31 = 27 = l13 . 14 14 14 14

(111) 5-2 .!. = ~ _ !_! = 5 x(5 + 1) - 11 x (5 +5) = 5x5-11 x 1=25- 11 = 14 = 2 4 . 515 5 5 5 5 5

79 5 1 5 7 1 5 Simplify (i) - - 2 - - 1 - (il) - --+7 - - 3 - 12 6 4' 36 12 24 48

(i) 79 - 2 ~ -1.! = 79 - 17 - 5 12 6 4 12 6 4

= 79 x (12+12)-17 x (12 + 6) - 5x(12+4)_ (LCMofl2,6, 4 = l2) 12

79 x 1 - 1 7 x 2 - 5 x 3 79 - 34 - 15 = 12 12 = 79 - 49 = _30 = _30_+_6 = ~ = 2 .! .

12 12 12 +6 2 2


5 7 169 149 (11) Required value= - - - +- - - , (LCM of 36, 12, 24, 48 = 144) 36 12 24 48

= 5 x(l44 +36)-7x (144+12) + 169 x (144 + 24) - 149 x (l44 + 48)

144 5x 4-7x l 2 +169x6-149 x 3 20 - 84 + 1014 - 447

= =-------

. 144 144 = 1034 -531 = 503 = 3 2.!_ .

144 144 144 - 29 -8 - 5 2 - 7

EXAMPLE 4 Simplify (i) 4 + .12 +S , (ii) 36 + g + 27 .

Solution (i) 4 + - 29 + -8 = 4x(l2 + l) + (- 29)x(l2 + 12) +(-8)x(l2 + 3) 12 3 12

(: 4 = 4 and LCM of 1, 12, 3 = 12) 1

= 4 x l 2+(-29)xl + (-8)x4 48 - 29 - 32 48 - 61 =-13 =_ 1_!_. 12 12 12 12 12

(ii) - 5 +~+ - 7 = (- 5) x (l08 +36)+2 x (l08 + 9) +(-7) x (l08 + 27) 36 9 27 108 .

(LCM of 36, 9, 27 = 108) (-5) x 3 + 2 x 12 + (-7) x 4 -15 + 24 - 28 24 - 43 -19

= = =- 108 108 108 108

Remember These

1. To add two or more fractions, first convert all of them to like fractions.

S f l1k f ti sum of numerators of all the fractions um o e rac ons = common denominator

2. The short-cut formula for addition is: num. of the first x (LCM of denoms. + denom. of the first).

S f tw f ti +num. of the second x(LCM of denoms: + denom. of the second) um o o rac ons = LCM of denominators

S. For subtraction, replace 'sum' by 'difference' in the first formula and +' by '-' in the second formula.

4 . Negative fractions have negative integers as numerators. The operations for negative fractions are the same as these for positive fractions .

. ~;tJMHKia>-------------------~~c-1. Add the following.

(i) 2 +~ 9 9

(v) ~+~+~ 12 12 12

(11) .!. +.?. 6 6

(vi) l.?. + .!.! + ~ 8 8 8

(ill) l~ +.!_ 18 18

(vii) i +~ 9 18

3 7 9 (iv)-+-+-20 20 20 (viii) ~+ 25

. 8 24

Fractions

1 5 2 7 3 7 5 3 5 4 7 (ix) 2- + 3 - (x) -+-+- (xi) -+-+- (xii) 2 - + 1- + 3 -16 12 15 20 25 12 6 8 9 9 12 1 1 11 1 3 1 (xiii) 4- +2 - (xiv) - +5 - + 4 (xv) 3 + 4 - + 3 -6 .. 3 10 15 4 12

2. Find the following.

s.

(1) .!..!_ - .!_ (ii) 169 - 69 3 5 (ill) 1- --30 30 100 100 16 16

3 1 1 7 3 (v) 4-- - (vi) 3 --2 - (vii) 8 - 1-7 14 4 10 8

5 3 (1) What number should be added to - to get! - ? 16 8

5 (11) What number should be subtracted from 4 to get 2 - ? 13

. 5 7 (111) What number gives - when 3 - is subtracted from it? 6 12

8 3 (iv) ---15 20

(viii) 6!.!.-5 14

4. Simplify the following.

(1) 21 - .!_ + .!..!_ 25 25 25

(ii) I ~ -~-~ 7 7 7

(iii) 2~ - 1~ - 13 16 16 16

5 5 5 (iv) -+---12 8 16

(v) 3 _.!..!_ +~ 12 8

(vi) 3.!. + 4.!. - 6.!. 3 9. 6

(vii) l ~ -2~ + 4 6 8

I 3 (viii) 5 - 2 - - 1-7 5

(ix) 1 _.!:._ - 2 ~ + 5 2-15 5 10

(x) 6 - I .!_ - 2 ~ - .!_ 4 6 8

I 3 I 1 (xi) 5 - - 6-+4- - -2 4 6 12

5 . Find the value of each of the following. 2 -7 - 9 4 {1)-+ - (ii)-+-15 15 25 25

(iv) 1.i. _ .!2_ + - ll 25 40 10

5 I -3 (v) - -2 - +-6 4 5

ANSWERS

~ (iii) 13 + -7 + -5 36 36 36

(vi) 2~ + - 13 + - 10 7 14 21

N-51

1. (1) ~ (ii) 1.!. (ill) 1 ~ (iv) 19 1 (Vi) 3~ (vii) 13 (viil) l~ (Ix) 5 23 (x) 181 (v)l-9 3 20 4 18 12 48 300 (xi)l 19 (Xii) 7 !_ (xiii) 6.!. 1 5 (xiv) 10- (XV) 10-

24 12 2 6 6

2. (1)~ (ii} I (ill)~ {tv) 23 5 (Vi)!!. (vii) 6~ (Vilt) 1.!.! (V}4-15 60 14 20 14 3 . (t) 1..!..

16 (11) I~

13 (ill ) 4~

12

4 . (1) 1 (11) 0 (ill)~ 16

(iv) 35 48

(v ) 2 17 24

(Vi) l~ 18

(vii) 3 .!..!. 24 (viil) 1 :5

(iX) 4! 6

(x) 1 f9 24

5 (xi) 2 - - . 6 1 1 (ill) _!_ 73 1 (Vi) 1- 1 5 . (1) - - (ii) - - (iv) -- (v)-2-3 5 36 200 60 42


Representation of a Fraction on the Number Line

Just like integers. fractions can also be represented by points on the number line. To do so, follow the steps gtven below. Steps 1. Draw a straight line l. Mark a point 0 on it. Label this point O (zero).

2. Select a unit of length in keeping with the denominator of the fraction 3 you wish to represent. For example, if you wish to represent - , 4

choose 2 cm or 4 cm as your unit, since you can easily divide such a

unit of l~ngt:h into four segments (parts).

3. Mark points A 1 , A 2 , etc., on the right of 0 such that OA1 = 4 cm, OA 2 = 8 cm, etc. Label these points 1, 2, etc., respectively.

L O 3/4 1 2

4. Mark the points Q, R and P between 0 and A1 such that OQ = 1 cm, 3 OR = 2 cm, OP = 3 cm. Then the point P will represent - , while the

1 2( 1) 4 points Q and R will represent - and - i.e., - respectively. 4 4 2

1 5. If you wish to represent 1- on the number line, mark the point S to 4

the right of A1 such that A1S = 1 cm. Similarly, mark the points T and

U to the right of A1 , such that A1 T = 2 cm and A1 U = 3 cm, to 1 3

represent 1- and 1- 2 4

You can represent fractions on the number line in another way. Let us take the 3

example of - again. 4

Steps 1. Draw a straight line l. Mark a point 0 on it. Label the point O (zero). 2. Represent the natural numbers on the line. Let A3 represent the

number 3.

3. Divide the line segment OA3 into 4 equal parts b.y inserting points P, Q and R such that OP = PQ =QR = RA 3 . (You will learn how to do

3 this in your geometry lessons.) The point P represents the fraction - , 3 4

since - means 3 + 4. 4

L 0 3/4 1 2 3 0

Fractions N-53

Negative fr~ctions Let the point P represent the fraction 3 on the number line. If P' be a point on

4 the number line on the other side of the origin Q such that OP = OP' then

- 3 the number represented by P ' is the negative fraction - , which is also written 4

3 as --

4 p ' 0 p

-3 -2 -1 -3/4 0 3/4 1 2 3

Solved ,Examples

EXAMPLE 1 Represent the following pairs of fractions on the number line.

(i) !. and _!. (ii) ~ and !. 3 3 8 2

Solution (i) Taking note of the common denominator 3 of the fractions, let the unit of 1 length be 3 cm. Then 1 cm will represent the fraction - 3

-1 -1/3 0 1/3 P' 0 P

1 Thus, the point P represents - on the number line, where OP = 1 cm on the 3

right of 0. The point P' on the left of 0 such that OP = OP' will represent the fraction _ .!_

3

(ii) First, we write the given fractions as the like fractions ~ and i, to make it . 8 8

convenient to choose the unit of length. Now, since the common denominator is 8, let us choose the unit of length as 4 cm. Then 5 mm will represent the fraction.!..

8

0

3 4 6 6 p Q

3 Thus, the point P represents - on the number line, where OP = 3 x 5 mm on 8

4 1 the right of 0. The point Q represents - (that is,-) on the number line, where 8 2

OQ = 4x5 mm on the right of 0.

~tii4Hli------------------1. Represent each of the following fractions on the number line.

(i) .!. (ii) ~ (ill) ~ 2 3 5

(iv) 5 4

1 (v) --5


2. Represent each pair of fractions on the same number line . . 1 1 2 2 1 1 (1) - and - - (11) - and -- (111) - and -

4 4 5 5 2 3

ANSWERS

1. (1) 0 1/2 1 2 (ti} 0 213 1 0 P A1 A2 0 A1

(ill} 3/5 (iv) 0 1 5/4 2 0 p A1 0 A1 P A2

(v} -1 - 1/5 81 0 A1

2. (1) -1 -1/4 0 1/4 8 1 Q 0 p A1

(11) -1 - 215 0 215 1 81 Q 0 p A1

(ill) 0 1/3 1/2 0 A1

Multiplication and Division of Fractions

Multiplication of fractions You know that multiplication is repeated addition.

For example, 4x4=4 + 4 + 4+4 = 16. 1 1 1 1 1 l + l + l+l 4 4xl Similarly. 4 x- =-+- + - +-= - - ---3 3 3 3 3 3 3 lx3

_ product of numerators 4 and 1 product of denominators 1 and 3

4 (Remember: 4 = -} 1

Now we can write the steps for the multiplication of fractions. To multiply two or more fractions, take the following steps. Steps 1. Convert mixed fractions (if any) into improper fractions.

2. Multiply the numerators of all the fractions.

3. Multiply the denominators of all the fractions.

4. Write a fraction with the product in Step 2 as the numerator and the product in Step 3 as the denominator.

5. Reduce the fraction obtained in Step 4 to its lowest terms .

. product of numerators of all fractions Productoffractions=~~~~~~~~~~~~~

product of denominators of all fractions

Examples

Fractions

(i)3 X 7=3X7 = ~. 4 9 4x9 36

21+3 7 = =-

36 +3 12

(HCF of 21and36 is 3)

Alternatively: You can cancel out the common factors from the numerator and denominator of the product.

3 7 3 1 x 7 l x7 7 Thus, -x-= =--=- 4 9 4 x 9 3 4 x3 12

. 1 14 1 5 1 147 1 l x 7xl 7 1 (li) 2 - X-X- = - X--X- = = - = 1- 2 15 2 2 1 l-5 3 2 l x 3x2 6 6

(iii) - 24x5 = (-24)x5 = -120 =-1-204 =- 4 . 35 6 35 x 6 210 2-10 7 7

Reciprocal of an integer or a fraction

A number is called the reciprocal of another number if the product of the two numbers is 1. For every integer (:;e 0) or fraction there is a reciprocal integer or fraction. The reciprocal of a positive number is positive, while the reciprocal of a negative number is negative .

.------------.

Number Reciprocal Number Reciprocal Number Reciprocal number number number

1 5 8 - 1 7 - -8 -7 8 5 8 1 8 5 -1

7 - -8 7 5 8 8 1 3 4 4 7

5 - - -5 4 3 7 4

To write the reciprocal of a number, write the number as a proper or improper fraction and then interchange the numerator and denominator of the fraction (retaining the negative sign, if any, in the numerator). For example:

1 10 3 2 -7 -5 Reciprocal of 3 - or - = - Reciprocal of - 1- or - = - 3 3 10 5 5 7

Division of fractions 1 3 3 1 3 11 We have, 3x- =- So, -+3 =- Also, - X- =-

. 44 4 4 43 4 1 3 Observe that - is the reciprocal of 3. In other words, dividing - by 3 is 3 4 3 .

equivalent to multiplying - by the reciprocal of 3. Now we can generalise and 4

write the steps for the division of one fraction by another. Steps 1. Convert the mixed fractions (if any) into improper fractions.

2 . Multiply the dividend and the reciprocal of the divisor. 3. Reduce the product to lowest terms.


7 3 7 4 1 7 x 1 7 Examples (1) -+- = - x-=-- = - 8 4 S2 3 2 x3 6

1 5 7 21 ::f1 -8 4 1X4 4 (ii) 1-+2 - = - +-=-X- =-- = - 6 8 6 8 .6'3 ..213 3 x3 9

-4 - 5 -4-010 (W)8+-=8x- = =-10. 5 4 4'1

To find the whole when a fraction of the whole is given

Let 2 of a number = 50. So, 2 x the number = 50. 3 3

Then the number = 3 x 2 x the number 2 3 3 3 2

= - x50 =50 x- = 50 +- 2 2 3

Thus, a number = value of gtven fraction of number + fraction.

Examples (i) If 3 of a number is 12, the number = 12 + 3 = 1-24 x 5 = 20. 5 5 z

(ii) If 4 of a number is 60, the number = 60 + 4 = 0015 x 9 = 135. 9 9 ..(

Simplification of fractions An expression involving fractions is simplified by the rule of BODMAS, as in the case of integers.

EXAMPLE Simplify l~ of:+ :0 + { l ~! -(! + !)} Solution ~of 4 +!..... +{1.!.! -(2 + 3 )} =~of 4 +!..... +{23 - 2 x 4 + 3 x 3 }

16 5 10 12 3 4 16 5 10 12 12

EXAMPLE 1

Solution

EXAMPLE 2

Solution

2 6 33 Findl - x- X - 3 55 40

= ~ of 4 + !__ + { 23 _ 1 7} = ~ of 4 + !__ + 23 - 1 7 16 5 10 12 12 16 5 10 12

3 4'1 wa: 6'1 3 1 3 +7 w 5 s = - - x - x-+-=- +-=-- = - =-

16'ir2 .3'1 7 12 2 14 2 14 147 7

Solved Examples

2 6 33 5 6 33 })1 x 63 x~ 31 3 1- x - x- =-x -x - = =- 3 55 40 3 55 40 31 x 55- .51 x 46'20 20

15 l l Find 4 x - x 20- x 1 - . 22 6 3

15 1 1 4 15 121 4 42 x .l-5"5 x l%!11 x 4 2 220 1 4 x - x 20- xl - = - x - x-x - = =- = 73 - 22 6 3 1 22 6 3 -22 2 x 63 x 31 3 3

2 9 EXAMPLE3 Find 6

15+ 6

10

Fractions

Solution 6 ~ + 6 ~ = 92 + 69 = 92 x 10 = 92 4 x 162 = ~ . 15 10 15 10 15 69 183 X693

3 7 1 EXAMPLE 4 Simplify 3 4 + S x 4 6

Solution 3 7 1 15 7 25 3 -+- X4 - = -+-X -4 8 6 4 8 6

9

N-57

15 8 25 =-X-X-

4 7 6 (: division is done before multiplication)

= l-55

x 8"21

x 25 = 5 x 25 = 125 = 17 ~. -4"1 x 7 x-6 21 7 7 7

EXAMPLE 5 Simplify 3~+ ( 2i+1 ~} .3~ +(2 .!. +. 1 ~)= 17+(Z+12 )= 17 + 49 +36

5 3 7 5 . 3 r 5 21 Solution

Solution 1.!.x(4! +2.!.)= 36 x(25 + ~) 5 6 2 5 6 2 =36x25+15 =366 x 468 = 48.

5 6 .51 X 6 1

1 1 1 1 366 x255 3618 x s 1 6x5 18 7 - x4 - + 7 - x2-= + =--+-=30+ 18=48. 5 6 5 2 .51 xS1 5 1 x21 1 1

on simplification, both sides give 48. So, they are equal.

EXAMPLE 7

Solution 5 .!_ _ Lr2 .!. + {3 _.!.x(2 _ _!_ )~Jl = g .:.Lrz + {3_.!.x16-1}Jl 2 3 4 2 3 24 ~ 2 3 4 2 24

= g _rlz +{s -~}J1=g _[z+36-15] 2 3 448 2 3 48

= !!-(Z +~)= !!- 71 x481s 2 3 48 2 J'l. X2-!3 11 16 33-32 1

= --- = =-

2 3 6 6


Remember These

1 Th d t f f ti product of numerators of the fractions e pro uc o rac ons = ---------------product of denominators of the fractions

2. The reciprocal of a proper or improper fraction is obtained by interchanging the numerator and denominator of the fraction.

3. To divide a fraction by another fraction, multiply the dividend by the reciprocal of the divisor.

4. Given the size of a fraction of the whole, the whole = given size + fraction. 5. Simpllftcatlon of fractions is carried out using the rule of BODMAS.

Find the product.

1. 2 5 (ii)~ x 35 (iii) 96 x 25 (i) - x -5 18 7 8 115 54

7 9 (ill} 14 x 22 2. (i) 6 x - (ii) - x 8 8 20

- 33

3 . 2 5 (ii) 12 x 6 2 (iii) 3~ x 5! (1}2 - x -3 12 25 3

- 8 9 - 5 - 4 (iii) 18 x - 15 4. (i) 12 x - (ii)- x 30 6 75 25 27

5. Write the reciprocals of the following numbers. (i) 12 (ii) 13 (iii) _l

Find the value.

6.

7 .

(i)8 +~ 25

(i) 3! + .!. 4 8

8. Fill in the blanks. 5 (i)} X - = ... 16

(v) ~ +~ = .. . 7 7

9 . Find the following. (i) ~ x ~x..

3 16 27 (iv) 2.!. x~x 2 !

9 38 5

(vii) 1~ x 5~x .!. x 1 ~ 8 5 7 6

17 41

(ii) 14 + 7 15

(ii).!.+ l_!_ 5 10

7 (ii) -xO = ... 11

10 (vi) 0 + - = ... 17

(iii) 32 + 24 2 1 35

(iii) 11 2 + 4 3 5 4

12 5 (ili)-x-= ... 5 12

(vii) 4 3 + 1 = ... 4

(ii) 14 x 35 x 17 25 51 49

(v) 6~ x6~ x 2-8 11 10

(viii) 1.E... x l! x 12 x 12_ x l_!_ 16 7 18 10

(iv} 128 x 75 175 112

(iv} 28 x 15 56

(iv) l! x 2 2 4 5

(iv) - 35 x 5! 11 2

1 (iv) 3 -4

(iv)._ +~ 55 22

(iv} 6 2 + 13_!_ 3 13

(v) - 16 23

- 6 - 11 (iv) - x-= ... 11 6

8 (viii) 1 + - = ... 5

(iii) 42 x 39 x 24 65 56 27

(vi) 2 .!. x 2 .!. x 2 _l_ 9 5 22

Fractions

Simplify the following.

10.

11.

12.

13.

(1) 4 +2-x ~ 5 15 9

(iv) 5 .!. + ~ of.!. 4 7 2

(1) ~ + (1.?.- 3) 8 8 4

(11) 5 .!. + ~ x .!. 4 7 2

(v) .?. + 2~-..!.!x3~ 8 6 12 11

14. Prove that 8-x 2-+ - = 8 - x2-+8-,-x- 1 ( 3 5 ) 1 3 1 5 6 7 14 6 7 6 14

(ill) 4 +2- of~ 5 15 9

N-59

(vi) 1.!-2 3 of~+~+~ 5 4 11 8 10

(111) ( 4 -~)x (2 3 +3 2 ) 9 11 4 3

15. Find the number (i) ~of which is 30, (ii)~ of which is 120. 8 6

1 . (1) ~ 1 3 . (1) 1-9

5. {J) ..!.. 12

7. (i) 26

5 8 . (i) -16 l 9 . (1) -

18

10. (i) 1!.! 21

1 11. (i) -3

13. (i) 6~ 3

ANSWERS

(U) 3~ (ill) 80 (Iv) 24 2 . (1) 5! (11) 3~ 1 1 (Ut) 9- (iv) 7 -4 207 49 4 3 2

(U} 3~ (iU)l7 ! (iv) 3 4 . (1) - 10 3 2 (iv) - 18 (U) - 1- (iii} --4 5 5

(U) 17 (ill) 41 4 23 6. (i) 16~ (U)~ (ill) 2-~ 8 (iv) - (v)- - (iv) -13 13 16 15 9 45

(U)~ 11

(ill) 2~ 5

(iv) 26 51

(U) 0 (ill) l (iv) l (v) 1 (vi) 0 (VU) 4~ 4

(vitl) ~ 8

(U)~ (ill)~ (iv)!.! 3 (vi) 9.!. (VU) 2~ (vitl) 27 .!. (v)l2-15 5 18 4 2 4 2 ..

(U) 6.!_ (ill) 113 1 (v) !_?_ (vi) 13 (iv) 24 -8 14 2 24 60

(U) 2~ 4

(ill) I_!!_ 108

12. (i)~ 14

(U) ~ 8

(U) 2~ 15 . (i) 80 (U) 144 11

Applications of Fractions Understanding fractions helps us in real life. Suppose, one out of every four students in a class is a girl. We can express this as "1/4 of the students are girls", and this statement can help us determine the number of girls in the class if we know the total strength (i.e., the total number of students) of the class.

. N-60

EXAMPLE 1

Solution


Solved Examples 3 5

A housing estate has 720 residents. Of these. 8 are men, 12 are women. and the rest are children. Find the number of children in the housing estate.

Consider the total number of residents to be the whole portion, or 1. So, the 3 5 portion of children in the estate = 1 - - - - 8 12

Now, 1 _ ~ _ ~ = 24 - 9 - 10 = 24 -19 = ~ . 8 12 24 24 24

5 - of the whole are children. 24

As the total number of people in the estate is 720, 5 5 x no30

the number of children = - of 720 = = 5 x 30 = 150. 24 241 x l

Alternative method 3 3 x~90 The number of men = - of 720 = = 3 x 90 = 270. 8 .81x 1

5 5 x '72660 The number of women = - of 720 = = 5 x 60 = 300. 12 l-2'1 x l

the number of children= 720 - 270 - 300 = 720-570 = 150 . . 8

EXAMPLE 2' A man spends 15 of his salary on rent. If he is left with ~ 2940. find his salary.

Solution

EXAMPLE 3

Solution

EXAMPLE 4

. 8 15 - 8 7 The portion of the salary left with him = 1 - - = -- = - 15 15 15

7 - of his salary is ~ 2940. 15

7 . his entire salary is~ 2940 +- (: the whole =the number + given fraction)

15 nnAA420 5

= ~~v xl = ~ 420 x l5 = ~ 6300. ?1

The line segment AB is 3 of the line segment CD in length. The length of the line 5 4

segment EF is 6 of the length of AB. If CD .is 12 cm. find the length of EF. 5 The length of the line segment EF = - of the length of AB. 6 3 The length of the line segment AB= - of the length of CD. 4

EF = - of - of CD = - x - of CD = of 12 cm 5 3 (5 3 ) 5 x .81

6 4 6 4 62 x 4 5 5 5x1-2 3 . 15 1

= - of 12 cm = - x 12 cm = cm = - cm = 7 - cm. 8 8 2 2

1 1 An express train covers 871 8 km in 11 2 hours. What distance does it cover 1 (i) in 1 hour. and (ii) in 3 5 hours?

Solution

EXAMPLES

Solution

EXAMPLE 6

Solution

Fractions N-61

In 11 ! hours, the train covers 871 ! km. 2 8

(1) . . the distance covered by the train in 1 hour = 871 ! km + 11 ! 8 2

= ( 6969 + 23) km= ( 6900303 x l km= 303 km = 75 3 km. 8 2 .8'4 23"1 4 4

(11) :. in 3 ! hours, the train covers 75 3 x 3 ! km 5 4 5

= 303xl6km=303 x l-64

km = l212 km= 242 2 km. 4 5 41 x5 5 5

2 . 3 Sheela reads - of a book on the first day and - of tbe remaining portion on 5 10

the second day. If the book has 600 pages, how many more pages are left to be read?

2 3 After the first day, the portion remaining to be read = 1 - - = - 5 5

3 3 3 3 9 The portion read on the second day = - of - = - x - = - 10 5 10 5 50 2 9 50 - 20 - 9 21

the portion left to be read = 1 - - - - = = - 5 50 50 50

21 21 x.60012 the number of pages left to be read = - of 600 = = 252.

50 561

Alternative method 2 2 x 600120

The number of pages read on the first day = - of 600 = = 240. 5 .51

the number of pages left after the first day = 600 - 240 = 360. 3 3 x36036 The nwnber of pages read on the second day = - of 360 = = 108. 10 le'1

the number of pages left to be read = 360 - 108 = 252.

1 2 Mr Ali gave - of his salary to his wife. She gave - of her money to her son. The son 3 3 3

was left with Rs 500 after spending - of his money. Find Mr Ali's salary. 4

2 1 2 The portion of Mr Ali's salary that the son got = - of - = - 3 3 9

1 - 3 = ! So, ! of the son's money is Rs 500. 4 4 4

1 4 the son's money = Rs 500 + -= Rs 500 x - = Rs 2000.

4 1 2 We know that this ts - of Mr Ali's salary. 9

. 2 9 20001000 x 9 Mr Alis salary = Rs 2000 + - = Rs 2000 x - = Rs = Rs 9000. 9 2 2'1


1. In a class, ~ of the students are gtrls. If there are 72 students in all, how many girls are 8

there in the class?

2. There are 125 apples and oranges in a basket. If.!_ of the fruits in the basket are oranges. 25

how many apples does the basket have?

3. A club has 360 members, of which 80 are government officers and the rest are businessmen. What fraction of the members of the club are businessmen?

4. A man has Rs 37,500. He gtves .!_of it to his daughter and the rest to his son. How much 15

does the son get? 5 5 5. A man earns Rs 18,000 per month. He spends - of his income on household items and-9 12

of his income on other expenses. Find his monthly saving.

6. The sum of three numbers is 2 _!_ If two of them are .!_ and .!...!. find the third. 36 12 18

l 1 7 . The product of two numbers is 7 - If one of them is 2 - then find the other. 2 12

8 . The product of three numbers is 5 .!_ If two of them are 2 2 and 1 ~ Find the third. 2 7 ' 8

9. Mrs Sharma spends Rs 5250 and she is left with 2~ of her salary. Find he.r salary. 10. 39 kg of rice is taken out of a bag of rice. If~ of the rice ts left in the bag then find the

21 quantity of rice the bag contained in the beginning.

1 1 1. In a school. - of the students were absent on a certain day. If 750 students were present 16

on that day, find the total number of students in the school. 3 12. It takes 1- m of cloth to make a pair of trousers. If the cloth needed to make l 0 pairs of 4

trousers ts cut out of a 20 m length of cloth, what length of cloth will be left?

13. Mr Menon had Rs 2,25,000. He gave..!. of it to his son, ..!. of it to his daughter and the rest to 3 5

his wife. What portion of his money did his wife get and how much did she get?

14. An examination was held in two parts. Of the 630 examinees, 1-._ appeared in only the first 2 10

part and - appeared in only the second part. The rest appeared in both the parts. Find the 15

number of examinees who appeared in both the parts.

1 1 15. Kamal reads - of a book on the first day and - of the remaining portion on the second day. 3 4

What portion of the book does Kamal read in two days? If 200 pages are left to be read, find

the total number of pages in the book.

Fractions N-63

1 1 16. Reema spends- of her money in a shop . Then she spends - of what is left on the bus fare. 2 3

If she is finally left with Rs 20. how much did she have in the begtnning?

4 9 17. A container of milk is - full . When 10 L of milk is poured into it, the container becomes -

5 10 full. What is the capacity of the container?

18. Hamid deposited .!.. of his salary in a bank. Then he spent~ 3000 and found that he was left . 7 4 with - of his salary. Find is salary.

12 19. A small vessel is used to take out oil from a larger vessel full of oil. The capacity of the small

vessel is ~ of the volume of oil contained in the larger vessel. If the small vessel is filled 15 50

times to draw out oil from the larger vessel, what fraction of th'e original volume of oil is left

in the larger vessel?

2 3 20. Mr Bose gave - of his money to his wife. His wife gave - of her money to her daughter. Two 3 4

thirds of the daughter's money was stolen and she was left with~ 30,000. How much money did Mr Bose have?

ANSWERS

1. 27 2.90 s.? 9

4. Rs 20,000 5. Rs 500

5 3 3 9. Rs 6250 10. 63 kg 7.3- 8.1-6. - 5 4 6

11. 800 1 12. 2 - m 2

7 13. - : Rs 1,05,000 15. .

14. 357 1 15. - :400 2

16. Rs 60 17. 100 litres 18. ~ 18,000 19 . ..!. 20. ' 1.80,000 10

:

N-64

1. Fill in the blanks. (1) -1-32 I = " " (iv)~= 30

8 ...


Revision Exercise 2

(ii) l-131-13 = ....

()3 2_ ... v - - -3 9

2 . Use >, < or = to fill in the blanks. (1) (-5) .... (+5) {ii) 61 .... (- 16) (iii) 0 .... {- 4) (v)5 + (-8) .... (- 3) (viJl-221+1 + 221 .... o

3 9 (viii) - .... -7 21

2 3 (ix) - .... -5 7

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

3. Write the following in ascending order. (1) 10, -3, 3, -6, -4, 2, 6

(ii) 1 ~. ?... 1 3 ~ 5 8 4 7

4. Write the following in descending order. (1) The integers greater than -6 and less than 4

(ii) ?... ' ~ .!.!. ~ 9 7 13 5

5. Represent the following on the number line.

(1) The integers from (-2) to ( +2) (ii) ~ 6

2 (iv) --3

3 3 (v) - and --5 5

3 1 (vi) - and -8 4

6 . Evaluate the following. (1) 132 -(-18)

(iv) 11 - 16 - 3 (vii) 3 2.. + 1 ~

13 13

(x) 2_13 16

7. Find the following. (1) 13 +.!.!. - 1J...

24 16 12

(iv)3-1?... - 1~ 8 6

8. Find the product/quotient. (i) 15 x (+12)

(iv} {-72) + (+ 12) (vii) 56 x 1 _!_

81 35

(ii) - 68-(-25) (v) 20+15 - 40

(viii) lO +2 2 + 3 11 3

(xi) 5.!. _2 ~ 6 9

. 4 3 1 (ti) 6 - 2 --1- +3 -15 10 25

(11) (- 18) x 11 (v} 144 +(-18)

15 1 7 (viii) -x l -x-49 20 8

(iii) 1181- l-121 -1+ 61 = .... (vi) 12 =..:..:..:.

64 16

(iv) l-61 .... 161 (vii) (- 4) + (-3) ...... 4

11 8 (xi) - .... -12 9

(iii) 1.!.. 4 ..

{iii) 19 + (- 9) 1+1-91 (vi) - 40 + 17 -12 + 25 (ix) ~ + 2.. + .!.

9 12 6

(xii) 4 2- - 28 10 15

3 1 (111) 1- - 2-10 5

(iii) (- 23) x (-5) (vi) (- 175)+{-25) {ix) .!.!. + 3 _!_ x 6 ~

24 18 7

Fractions N-65

9. Fill in the blanks. (i) (-50) + .... = 5

5 . (ii) .... x (-9) = 72 (ill) .... + 16 = (-~2)

(iv) -1-x .... = 1 6

10. Simplify: (i) 65+(-13}-(-40)x(- 2)

(ill) {108 +(-9)} +(-6) (v) l~x(!_ 15 +1!)

13 10 28 8

(ii) (- 14) of(- 66) +(-11) (iv) 10 - [12-{20 -(80+ 10 -7)}) (vi) 3!-{3 3 +(2 !-~)l

2 4 6 12 ~ 11. The product of two integers is 288. If one of the integers is - 16, find the other.

4 1 1 12. A man spends - of his monthly income on house rent, - on food and - on other items. He 15 2 5

saves ~ 400 in the end. What was his monthly income? 13. The capacity of a drum is 50 L and that of a bucket is 8 3 L. If 5 buckets of water is poured

4 into the drum, what fraction of the drum remains empty?

ANSWERS 1 . (1) - 32 (11) 0 (ii1) 0 (tv) 48 (v) 33 (vi) 3

(Vil) < (Vilt) = (IX) < (x) < (xi) > 2 . (1) < (11) > (ill) > (iv) = (v) = (vi) > 7 8 3 3

3. (1) - 6, - 4, - 3. 2. 3. 6.10 CU) a'7 ' ls,14' 4. (1) 3,2,LO.-L - 2.-3.-4.-5 (11) :~~~~ 5. (1) - 2 -1 0 1. 2

63 B2 B1 0 A1 A2 A3

(U) 0 516 0 p A t

(ill) 0 514 2 0 A, p

(iv) - 1 -2/3 0 B1 a 0 A,

(v) - 1 -315 0 3/5 B1 a 0 p

(Vi) 0 114 318

6. (1) 150 (11) -43 (ill) 9 (iv) -8 (v)-5 (vi) -10 (Vil) 5 (Vill) 619 (ix)l.!...!. 3 (xi) 2!.! (x)l -33 36 16 18

(xU) 213 30

7 c11> 52!.. 9 17 7.(1) - (ill) - - (iv) --

48 150 10 24

8 . (1) + 180 c> -198 (ill) 115 (lv)-6 (v)-8 (Vi) 7 (Vii) 32 45

(Vill) ~ 32

(IX) 27 28

9 . (1) - 10 (11) -8 (ill)-192 6 10. (1)-85 (U)-84 (ill) 2 (iv) 17 (V) 19 (Vi) l~ (iv) --11 39 14

11. - 18 12.' 12.000 13 . .!. 8

)

FractionsBinder11_00351_00361_00371_00381_00391_00401_00411_00421_00431_00441_00451_0046

Binder11_0046_11_0046_2

Fractions - CopyBinder11_00471_00481_00491_0050

Binder21_00511_00521_00531_00541_00551_00561_00571_00581_00591_0060

fractions - class vi

Documents