FRACTIONS - · PDF fileWhat fractions do the shaded parts in each of the following figures...

$: FRACTIONS - · PDF fileWhat fractions do the shaded parts in each of the following figures represent? (i) ... 3 3 5 15 3 3 6 18 × ... LCM of 3, 4, 6, 8,$
1Class - VI Mathematics Question Bank

1.What fractions do the shaded parts in each of the following figures represent?

(i) (ii)

(iii) (iv)

Ans. (i)5

8(ii)

7

12(iii)

2

7(iv)

1

8

2. Write a fraction for each of the following :

(i) 4 parts out of 9 equal parts (ii) 5 parts out of 11 equal parts

(iii) two-fifths (iv) four-sevenths (v) seven-tenths

Ans. (i)4

9(ii)

5

11 (iii)

2

5(iv)

4

7 (v)

7

10

3. Point out the numerator and denominator in each of the following fractions:

(i)5

6(ii)

9

14(iii)

12

25

Ans. (i) Numerator = 5, Denominator = 6

(ii) Numerator = 9, Denominator = 14

(iii) Numerator = 12, Denominator = 25

4. Write five fractions equivalent to each of the following :

(i)2

3(ii)

7

9

Ans. (i)2 2 2 4 2 2 4 8 2 2 3 6

, , ,3 3 2 6 3 3 4 12 3 3 3 9

× × ×= = = = =

× × ×

4FRACTIONS

Class - VI Mathematics Question Bank2

2 2 5 10 2 2 6 12,

3 3 5 15 3 3 6 18

× ×= = = =

× ×

∴ Five equivalent fractions to 2

3are

4 6 8 10, , ,

6 9 12 15 and

12

18

(ii)7 7 2 14 7 3 21

,9 9 2 18 9 3 27

× ×= = = =

× ×

7 4 28 7 5 35 7 6 42

9 4 36 9 5 45 9 6 54

× × ×= = = = = =

× × ×


9are

14 21 28 35, , ,

18 27 36 45 and

42.

54

(iii)3 3 2 6 3 3 9

14 14 2 28 14 3 42

× ×= = = =

× ×

3 4 12 3 5 15 3 6 18

14 4 56 14 5 70 14 6 84

× × ×= = = = =

× × ×


14 are

6 9 12 15, , ,

28 42 56 70and

18.

84

5. Which of the following are the pairs of equivalent fractions?

(i)4

5 and

24

30(ii)

8

12 and

14

21

Ans. (i)4 24 24 6 4

,5 30 30 6 5

÷= =

÷ which are equivalent.

(ii)8 8 4 2

12 12 4 3

÷= =

÷ and

14 17 7 2.

21 21 7 3

÷= =

÷ So, both are equivalent.

6. Write an equivalent fraction of :

(i)4

5 with numerator 32 (ii)

7

9 with numerator 63

(iii)10

13 with denominator 91 (iv)

32

48with numerator 2

(v)10

13 with numerator 3 (vi)

24

56 with numerator 3

(vii)6

9 with numerator 10 (viii)

18

24 with denominator 28


Ans. (i)4 4 8 32

5 5 8 40

×= =

×(ii)

7 7 9 63

9 9 9 81

×= =

×(iii)

10 10 7 70

3 13 7 91

×= =

×

(iv)32 32 16 2

84 48 16 3

÷= =

÷(v)

21 21 7 3

28 28 7 4

÷= =

÷(vi)

24 24 8 3

56 56 8 7

÷= =

÷

(vii)6 6 3 2 2 5 10

9 9 3 3 3 5 15

÷ ×= = = =

÷ ×(viii)

18 18 6 3 3 7 21

24 24 6 4 4 7 28

÷ ×= = = =

÷ ×

7. Write the missing numerals in place holders :

(i)4

9 63= (ii)

3 18

4= (iii)

42 3

70=

Ans. (i)4 4 7 28

9 9 7 63

×= =

× (ii)

3 3 6 18

4 4 6 24

×= =

× (iii)

42 42 14 3

70 70 14 5

÷= =

×

8. Which of the following fractions are in simplest form?

(i)21

40(ii)

35

49(iii)

64

81(iv)

25

36

(v)28

39(vi)

56

65

Ans. (i)21

40 is in the simplest form (ii)

35 5

49 7= is not in the simplest form.

(iii)64 8

81 9= is not in the simplest form. (iv)

25

36 is in the simplest form.

(v)28

39is in the simplest form (vi)

56

65 is in the simplest form

9. Reduce each of the following fractions to its lowest terms

(i)72

90(ii)

140

252

Ans. (i) Now, 72 = 2 × 2 × 2 × 3 × 3 and 90 = 2 × 3 × 3 × 5

So, HCF of 72 and 90 = 2 × 3 × 3 = 18.

∴72 72 18 4

.90 90 18 5

÷= =

÷

(ii) Now , 140 = 2 × 2 × 5 × 7 and 252 = 2 × 2 × 7 × 3 × 3.

∴ H C F of 140 and 252 = 2 × 2 × 7 = 28.

∴140 140 28 5

.252 252 28 9

÷= =

÷

Hence, 140

252in its lowest terms is

5.

9


10. Point out the proper and improper fractions from the following :

(i)7

8(ii)

9

9(iii)

15

11(iv)

18

25

Ans. (i)7

,8

Proper fraction (ii)9

,9

Improper fraction

(iii)15

,11

Improper fraction (iv)18

,25

Proper fraction

11. Convert each of the following mixed numerals into an improper fraction :

(i)2

63

(ii)11

513

(iii)3

104

(iv)1

333

Ans. (i)2 (6 3) 2 18 2 20

63 3 3 3

× + += = =

(ii)11 (5 13) 11 65 11 76

513 13 13 13

× + += = =

(iii)3 (10 4) 3 40 3 43

104 4 4 4

× + += = =

(iv)1 (33 3) 1 99 1 100

333 3 3 3

× + += = =

12. Convert each of the following improper fractions into a mixed numeral :

(i)31

8(ii)

89

8(iii)

101

3(iv)

115

13(v)

208

19

Ans. (i)31

8 ∴

31 73

8 8= (ii)

89

8

111

8= . (iii)

101

3

233

3=

(iv)115

13

118

13= (v)

208

19

1810

19=

13. Convert each of the following sets of unlike fractions into like fractions :

(i)2 1 5 7 11

, , , ,3 4 6 8 12

(ii)2 1 5 7 13

, , , ,3 6 9 12 18

(iii)1 4 9 11 37

, , , ,2 7 14 21 42

Ans. (i)2 1 5 7 11

, , , ,3 4 6 8 12

LCM of 3, 4, 6, 8, 12, = 24

∴2 2 8 16 1 1 6 6

,3 3 8 24 4 4 6 24

× ×= = = =

× ×


5 5 4 20 7 7 3 21 11 11 2 22, ,

6 6 4 24 8 8 3 24 12 12 2 24

× × ×= = = = = =

× × ×

∴ Fractions become 16 6 20 21 22

, , , ,24 24 24 24 24

(ii)2 1 5 7 13

, , , ,3 6 9 12 18

. L C M of 3, 6, 9, 12, 18 = 36

∴ 2 2 12 24 1 1 6 6 5 5 4 20

, ,3 3 12 36 6 6 6 36 9 9 9 36

× × ×= = = = = =

× × ×

7 7 3 21 13 13 2 26

,12 12 3 36 18 18 2 36

× ×= = = =

× ×


, , , ,36 36 36 36 36

(iii)1 4 9 11 37

, , , ,2 7 14 21 42

L C M of 2, 7, 14, 21, 42 = 42

∴ 1 1 21 21 4 4 6 24 9 9 3 27

, , ,2 2 21 42 7 7 6 42 14 14 3 42

× × ×= = = =

× × ×

11 11 2 22 37 37 1 37

,21 21 2 42 42 42 1 42

× ×= = = =

× ×


, , , ,42 42 42 42 42

14. Fill in the place holders with > or < :

(i)7 5

9 9(ii)

9 11

13 13(iii)

3 3

5 4(iv)

7 7

9 11

Ans. (i)7 5

9 9> (ii)

9 11

13 13< (iii)

3 3

15 4< (iv)

7 7

9 11>

15. Arrange the following fractions in ascending order :

(i)2 5 7 11 13

, , , ,3 6 9 12 18

(ii)2 1 5 5 7

, , , ,3 4 6 8 12

(iii)2 4 7 11 23

, , , , .3 5 15 20 30

Ans. (i) L C M of 3, 6, 9, 12 and 18 = 36

∴2 2 12 24 5 5 6 30

, ;3 3 12 36 6 6 6 36

× ×= = = =

× ×

7 7 4 28 11 11 3 33

: ;9 9 4 36 12 18 3 36

× ×= = = =

× ×

2

2

3

3

3 – 6 – 9 – 12 – 18

3 – 3 – 9 – 6 – 9

3 – 3 – 9 – 3 – 9

1 – 1 – 3 – 1 – 3

1 – 1 – 1 – 1 – 1


13 13 2 26

18 18 2 36

×= =

× ∴ LCM = 2 × 2 × 3 × 3

Since 24 < 26 < 28 < 30 < 33, so,

24 26 28 30 33

36 36 36 36 36< < < < or

2 13 7 5 11

3 18 9 6 12< < < <

(ii) L C M of 3, 4, 6, 8 and 12 = 24

∴2 2 8 16 1 1 6 6 5 5 4 20

, ,3 3 3 24 4 4 6 24 6 6 4 24

× × ×= = = = = =

× × ×

5 5 3 15 7 7 2 14,

8 8 3 24 12 12 2 24

× ×= = = =

× ×

Since, 6 14 15 16 20

24 24 24 24 24< < < < , we have

1 7 5 2 5

4 12 8 3 6< < < <

(iii) L C M of 3, 5, 15, 20, 30, = 3 × 5 × 2 × 2 = 60.

∴ 2 2 20 40 4 4 12 48

; ;3 3 20 60 5 5 12 60

× ×= = = =

× ×

7 7 4 28 11 11 3 33

;15 15 4 60 20 20 3 60

× ×= = = =

× × and

23 23 2 46.

30 30 2 60

×= =

×

so, 28 33 40 46 48

.60 60 60 60 60

< < < < ∴ 7 11 2 23 4

.15 20 3 30 5

< < < <

16. Arrange the following fractions in descending order :

(i)4 4 4 4 4

, , , ,7 3 9 5 11

(ii) 2 5 7 9 1

, , , ,3 6 9 12 2

(iii) 17 7 19 13 9

, , , ,32 12 48 24 16

Ans. (i) Among two fractions with same numerator, the one with greater denomina-

tor is least of the two.

so, 4 4 4 4 4

.3 5 7 9 11

> > > >

(ii) L C M of 3, 6, 9, 12 and 2 = 36

∴2 2 12 24

3 3 12 36

×= =

×;

5 5 6 30 7 7 4 28; ;

6 6 6 36 9 9 4 36

× ×= = = =

× ×

9 9 3 1 1 8 18

12 12 3 2 2 18 36

× ×= = = =

× ×

Since, 30 > 28 > 27 > 24 > 18

5 7 9 2 1

6 9 12 3 2> > > >

2

2

2

3

3 – 4 – 6 – 8 – 12

3 – 2 – 3 – 4 – 6

3 – 1 – 3 – 2 – 3

3 – 1 – 3 – 1 – 3

1 – 1 – 1 – 1 – 1

2

2

3

3 – 6 – 9 – 12 – 2

3 – 3 – 9 – 6 – 1

3 – 3 – 3 – 3 – 1

1 – 1 – 3 – 1 – 1


(iii) L C M of 32, 12, 48, 24 and 16 = 96.

∴17 17 3 51 7 7 8 56

; ;32 32 3 96 12 12 8 96

× ×= = = =

× ×

19 19 2 38 13 13 4 52; ;

48 48 2 96 24 24 4 96

× ×= = = =

× ×

9 9 6 54

16 16 6 96

×= =

×

Since , 56 > 54 > 52 > 51 > 38

∴7 9 13 17 19

12 16 24 32 48> > > >

17. Find the sum:

(i)3 5

5 6+ (ii)

7 18

12 18+ (iii)

3 8 7

10 15 20+ +

(iv)8 11 13 5

9 18 27 6+ + + (v)

1 5 74 2 3

6 8 12+ +

Ans. (i) L C M of 5 and 6 = 30

3 5 18 25 43 131

5 6 30 30 30

++ = = =

(ii)7 13

12 18+ =

7 13,

12 18+ L C M of 12 and 18 = 36

= 21 26 47 11

136 36 36

+= =

(iii)3 8 7

10 15 20+ + =

3 8 7

10 15 20+ + = [L C M of 10, 15, 20 = 60]

= 18 32 21 71 11

160 60 60

+ += =

(iv)8 11 13 5

9 18 27 6+ + + [LC M of 9, 18, 27 and 6 = 54]

= 8 33 26 45

54

4 + + + =

152 76 222

54 27 27= =

(v)1 5 7

4 2 36 8 12

+ + = 25 21 43

6 8 12+ + =

100 63 86 249 310

24 24 8

+= =

L C M of 6, 8, and 12 = 24

2

2

2

2

2

3

32 – 12 – 48 – 24 – 6

16 – 6 24 – 12 – 3

8 – 3 12 – 6 – 3

4 – 3 – 6 – 3 – 3

2 – 3 – 6 – 3 – 3

1 – 3 – 3 – 3 – 3

–1 – 1 –1 – 1 – 1


18. Find the difference :

(i)2 3

5 – 33 4

(ii)1 1

5 – 48 12

(iii)1 7

6 – 36 10

Ans. (i)2 3

5 – 33 4

= 2 3 17 15

5 – 3 –3 4 3 4

= = 68 – 45 23 11

112 12 12

= = [L C M of 3 and 4 = 12]

(ii)1 1 41 49

5 – 4 –8 12 8 12

= = 123 – 98 25 1

124 24 24

= = [L C M of 8 and 12 = 24]

(iii) 1 7

6 – 36 10

37 37

–6 10

= = 185 –111 74 37 7

230 30 15 15

= = = [L C M of 6 and 10 = 30]

19. Simplify : (i)2 1 2

6 4 – 23 6 9

+ (ii)3 1 5

10 – 5 – 44 8 12

Ans. (i) 2 1 2

6 4 – 23 6 9

+ = 20 25 20

–3 6 9

+ [L C M of 3, 6, 9 = 18]

= 120 75 – 40 155 11

818 18 18

+= =

(ii)3 1 5

10 – 4 – 44 8 12

= 43 41 53

– –4 8 12

[L C M of 4, 8 and 12 = 24]

= 258 –123 –106 258 – 229

24 24= =

29 51

24 24=

20. Subtract the sum of 3

94

and 5

36

from 7

15 .12

Ans. Required number = 7 3 5

15 – 9 312 4 6

+

=

187 39 23–

12 4 6

+

= 187 39 23

– –12 4 6

[L C M of 12, 4 and 6 = 12]

= 187 –117 – 46 187 –163

12 12= =

242

12=

21. Subtract the sum of 5

212

and 3

34

from the sum of 1

73

and 1

46

Ans.1 1 5 3

7 4 – 2 33 6 12 4

+ +

=

22 25 29 15–

3 6 12 4

+ +

=

44 25 29 45–

6 12

+ +

= 69 74 138 – 74 64 16 1

– 56 12 12 12 3 3

= = = =


22. Add the difference of 3

35

and 7

210

to the difference of 7

215

and 9

320

.

Ans.3 7 9 7

3 – 2 3 – 25 10 20 15

+

=

18 27 69 37– –

5 10 20 15

+

= 36 – 27 207 –148

10 60

+

=

9 59 54 59 113 531

10 60 60 60 60

++ = = =

23. From a rope of length 1

202

m, a piece of length 5

38

m is cut off. Find the length of

the remaining rope.

Ans. Total length of rope = 1

202

m. Length cut off = 5

38

m.

Remaining rope = 1 5

20 – 32 8

= 41 29

–2 8

= 41 4 – 29 1 164 – 29

8 8

× ×= =

135 716

8 8= m.

24. The weights of three packets are 3

22

kg,1

33

kg and 2

55

kg. Find total weight of all

three packets.

Ans. ∴ Total weight of three packets = 3 1 2 11 10 27

2 3 54 3 5 4 3 5

+ + = + +

= 11 15 10 20 27 12

60

× + × + × [∵ L.C.M. of 4, 3, 5 = 60]

= 165 200 324

60

+ + =

689 2911

60 60= kg.

25. I bought fruits worth Rs 3

274

and vegetables worth Rs 1

102

. If I gave a fifty- rupee

note to the shopkeeper, how much will I get back ?

Ans. Cost of fruits Rs 3

274

.

Cost of vegetables = Rs 1

102

Total cost = 3 1

27 104 2

+ = 111 21 111 42

4 2 4

++ = = Rs

153

4


∴ Remaining money = Total – the cost of fruit and vegetables.

153 200 –153

50 –4 4

= = Rs47

4 = Rs

311

4

26. If the speed of a car is 1

505

km/ hr, find the distance covered by it in 3

35

hours.

Ans. Speed of car = 1

505

km/hours, Time = 3

35

hours

Distance = Speed × Time = 1 3

50 35 5

× = 251 18 4518

5 5 25× = =

18180

25km.

27. Find the product.

(i)11

2 1827

× Ans. 11 65 65 2 130 1

2 18 18 4327 27 3 3 3

×× = × = = =

(ii)1 1

8 121 13

× Ans. 1 1

8 121 13

× = 169 14 13 2 26 2

821 13 3 1 3 3

× = × = =

(iii)3 1 4

1 2 44 7 5

× × Ans.3 1 4 7 15 24

1 2 4 18.4 7 5 4 7 5

× × = × × =

(iv)1 3 4

3 2 25 4 11

× × Ans.16 11 26

5 4 11× × =

4 1 26

5 1 1× × =

104 420

5 5=

28. Simplify :

(i)13 2

21 7÷ Ans. (i) =

13 7 13 1 13 12

21 2 3 2 6 6× = × = =

(ii)1

3 58

÷ Ans. 1

3 58

÷ = 25 5 25 1 5

8 1 8 5 8÷ = × =

(iii)5

7 349

÷ Ans. 68 34 68 1 2

9 1 9 34 9÷ = × =

(iv)1 1

4 62 2

÷ = 9 13 9 2 9

2 2 2 13 13÷ = × =

(v)7 1

5 310 6

÷ = 57 6 3 3 9 4

110 19 5 1 5 5

× = × = =

(vi)5 11

10 17 14

÷ = 5 11 75 25 75 14

10 1 67 14 7 14 7 24

÷ = ÷ = × =

(vii)8 2

15 39 3

÷ = 8 2 143 11

15 39 3 9 3

÷ = ÷ = 143 3 13 1

49 11 3 3

× = =


29. Simplify :

(i)1 5 1

1 1 36 9 3

÷ ×

Ans.7 14 10

6 9 3÷ × =

7 9 10 1 1 5 5 12

6 14 3 2 1 1 2 2× × = × × = =

(ii)1 3

13 7

÷ of 5 1

2 18 9

+

Ans.1 3

13 7

÷ of 5 1

2 18 9

+ = 4 3

3 7÷ of

21 10 4 3 21 10

8 9 3 7 8 9+ = ÷ × +

= 4 9 10 4 8 10

3 8 9 3 9 9÷ + = × + =

32 10 32 30 62 82

27 9 27 27 27

++ = = =

(iii)1 1

6 35 10

÷ of 1 1

22 4

÷

Ans.1 1

6 35 10

÷ of 1 1

22 4

÷ = 31 31 5 1

5 10 2 4÷ × ÷ =

31 31 1 31 4 4

5 4 4 5 31 1

÷ ÷ = × ×

= 4 16 1

4 35 5 5

× = =

(iv)2 3

3 –3 11

of 3 1 2 1

2 14 4 3 3

÷ × +

Ans.2 3

3 –3 11

of 3 1 2 1

2 1 14 4 3 3

÷ × + = 11 3 11 5 5 1

–3 11 4 4 3 3

× ÷ × +

= 11 3 5 5 1

–3 4 4 3 3

÷ × + = 11 3 4 5 1

–3 4 5 3 3

× × +

= 11 1 1 11 – 3 1

–3 1 3 3

++ = =

12 – 3 93

3 3= =

(v)2

27

of 3 1 4

15 24 4 7

× ÷ of 5

28

Ans.2

27

of 3 1 4

15 24 4 7

× ÷ of 5

28

= 16 63 9 4 21

7 4 4 7 8× × ÷ ×

= 9 3 9 2

36 36 3 9 2 54.4 2 4 3

× ÷ = × × = × × =

= 4 5 1

1 – 37 4 2

÷ + ÷ = 7 5 1 1

1 –4 4 2 3

× + ×


= 7 5 1 7 1 5

– –4 4 6 4 6 4

+ = + = 21 2 –15 23 –15 8 2

12 12 12 3

+= = =

(vii)1 1 1 2 1 1

2 3 3 –13 2 3 9 8 12

+ ÷

Ans.1 1 1 2 1 1

2 3 3 –13 2 3 9 8 12

+ ÷

1 5 10 2 25 13

–3 2 3 9 8 12

+ ÷

= 1 15 20 2 75 – 26

3 6 9 24

+ ÷

=

1 35 2 49

3 6 9 24× ÷ ×

= 35 98

18 216÷ ⇒

35 216 12 30 25 4

18 98 14 7 7× = × = =

(viii)1 1 1 1 1

–4 9 2 4 3

÷ + ÷

Ans.1 1 1 1 1

–4 9 2 4 3

÷ + ÷

=

1 1 1 1 1–

4 9 2 4 3

÷ + ÷

= 9 – 4 1 3

36 2 4

÷ +

=

5 2 3 5 5 5 4 1

36 4 36 4 36 5 9

+ ÷ = ÷ = × =

30. A man earns Rs 12000 per month. He spends 7

12 on household items and

3

24 on

education of children. How much does he save ?

Ans. Total salary = Rs 12000, Amount spend on household = 7

12000 Rs 700012

× =

Amount pend on education = 3

12000 Rs 150024

× =

∴ Money saved = 12000 – 7000 – 1500 = Rs 3500

31. In a school, 25

54of the students are girls and the rest are boys. If the number of

boys is 2030, find the number of girls.

Ans. Let the total number of students be x.

∴ Fraction of girls = 25

54of x and Number of boys = 2030

As per conditions, 25

– 203054

x x = ⇒ 54 – 25

203054

x x=

⇒ 54x – 25x = 2030 × 54 ⇒ 29x = 109620


⇒ x = 109620

378029

=

Hence, number of girls = 3780 – 2030 = 1750.

32. When Ajay travelled 18km, he found that one third of his journey was still left. What

was the length of his journey ?

Ans. Let the total length of journey = x km. Distance travelled = 18km.

Journey left to covered = 1

3x

As per condition, 1

– 183

x x =

⇒ 3 – 2

183 3

x xx= = ⇒

318

2x = × ∴ x = 27

Hence, total length of his journey is 27 km.

33. A sum of Rs 56000 is shared between Mohan, Sohan and Krishna. Mohan gets

1

4of it and Sohan gets

1

5of it. How much does Krishna get ?

Ans. Total sum = Rs 56,000, Mohan get = 1

56000 Rs 140004

× =

Sohan get = 1

56000 Rs 112005

× =

Krishna get = 56000 – 14000 – 11200 = 56000 – 25200 = Rs 30800.

34. From a rope 3

124

m long, a piece of length 5

36

has been cut off. What is the length

of remaining piece ?

Ans. Length of remaining piece = 3 5

12 – 34 6

m = 51 23

–4 6

m = 153 – 46

12 m

[L C M of 4 and 6 = 12]

= 107 11

m 512 12

= m

35. On one day a labourer earned Rs 1

38 .2

Out of this money, he spent Rs 3

164

on food,

Rs 1

82

on tea and Rs 1

44

on other eatables. How much is left with him ?


Ans. One day a labourer earned= Rs 1

382

= Rs 77

2

Total expenditure of a labourer in one day = Rs 3

164

+ Rs 1

82

+Rs1

44

= Rs 67 17 17 67 34 17

4 2 4 4

+ + + + =

= Rs

118

4 = Rs

59

2

Money left with him = Rs 77 59

–2 2

= Rs 18

2= Rs 9

36. A man had Rs 15000. He gave 1

5of his money to his daughter,

1

3of the remainder

to his son and the rest to his wife. Find the share of each.

Ans. Total money a man had = Rs 15000

∴ Daughter’s share = 1

5of 15000 = Rs 3000

Son’s share = 1

3 of (15000 – 3000) =

1

3× 12000 = Rs 4000

∴ Wife’s share = Rs 15000 – (3000 + 4000) = Rs 8000

37.2

3of the students in a class are boys. If there are 17 girls in the class, how many are

boys ?

Ans. Let total students be x. ∴ Number of boys = 2

3x

As per condition, 2

– 173

xx = ⇒

3 – 217

3

x x= ⇒ 17

3

x=

⇒ x = 17 × 3 ⇒ x = 51

∴ Number of boy = 2 2 51

343 3

x×

= =

38. After reading 5

8of a book, 168 pages are left. How many pages are there in all in the

book ?

Ans. Let the number of pages in the book be x ∴ Pages read = 5

8 of x =

5

8x

As per condition, 5

– 1688

x x = ⇒ 8 – 5

1688

x x= ⇒

3168

8

x=


∴168 8

56 83

x×

= = × = 448

∴ Total pages in the book = 448.

40. A man gave 1

4 of his money to his elder son,

1

4 of the remainder to his younger son

and the rest to his daughter. What part of the money did the daughter get?

Ans. Money a man had = 1 (let) Share of elder son = 1

4 of 1 =

1

4

Share of younger son = 1

4 of

1 1 3 31

4 4 4 16

− = × =

∴ Daughter’s part = 1 3

14 16

− +

= 4 3

116

+− =

16 7 9

16 16

−=

41. In an orchard, 1

3of the trees are guava trees,

1

8are banana trees and the rest are

mango trees how many trees in all are there?

Ans. Let the total number of trees be x. Guava trees = 1

3of x

3

x=

Banana trees = 8

x. ∴ Mango tree = x – –

3 8

x x =

24 – 8 – 3 13

24 24=

x x x x

But 13

11724

x= (given) ⇒

117 24

13x

×= ∴ x = 216

Thus, total number of trees = 216.

42. The area of a rectangle is 2

465

sq. cm. If its length is 1

75

cm find, its breadth.

Ans. Breadth of the rectangle = 2 1

46 75 4

÷

cm

= 232 29

5 4

÷

cm =

232 4

5 29× =

8 4

5 1× cm =

32

5 cm =

26

5 cm

43. Mr. Sekhar had some money. He spent 1

4of it on sweets and

2

3of remainder on

fruits. The rest of money he spent on refreshment. If the amount spent on therefreshment is Rs 60, how much money did he have in the beginning ?


Ans. Let the total money in beginning be Rs x. Amount spent on sweets = 1

4of

Rs = 4

x

Remaining money = Rs –4

xx

= 4 –

4

x x = Rs

3

4

x

Amount spent on fruits = 2

3of Rs

3

4

x = Rs

2 3

3 4

x× = Rs

2

x

∴ Remaining money = Rs 3

–4 2

x x

According to statement, 604

x= ⇒ x = 60 × 4 = 240

∴ Total money in beginning = Rs 240.

44. A drum of kerosene oil is 3

4full. When 30 litres of oil are drawn from it, it is

7

12full. Find the capacity of the drum.

Ans. Let the total capacity of drum be x. Kerosene in drum = 3

4 full

After drawing 30 litres capacity = 7

12 full.

As per condition, 3 7

– 304 12

x

=

⇒ 9 – 7

3012

x

=

⇒ 2

3012

x = ⇒ 1

306

x = ⇒ x = 30 × 6 = 180 litres.

The capacity of the drum is 180 litre.

45. Two friends Albert and Robert decide to buy a car. Albert pays 3

7of the cost and

Robert pays the rest. (i) What fraction of the cost does Robert pay ?

(ii) Robert pays Rs 31540 more than Albert. Calculate the cost of the car.

Ans. (i) Let the cost of car = 1. Amount Albert pays = 3

7

∴ Robert pay = 3 7 – 3 4

1–7 7 7

= =


(ii) Excess contribution by Robert is = 4 3 1

–7 7 7

=

Let, x be the cost of the car, then 1

7 of x = 31,540 ⇒ x = 31540 × 7 = 220780.

46. I bought wheat worth Rs 3

124

rice worth Rs 3

254

and vegetables worth Rs 1

10 .4

If I gave a hundred rupee note to the shopkeeper ; how much will he return to me ?

Ans. Money given to shopkeeper = Rs 1000

Total amount of goods bought = Rs 1 3 1

10 25 102 4 4

+ +

(Wheat, Rice and Vegetable)

= 25 103 41

2 4 4+ +

= 50 103 41

4

+ += Rs

194

4

∴ Money returned by shopkeeper = Rs 194 400 –194

100 –4 4

=

= Rs 206

4 = Rs

103

2 = Rs

151

2.

47. In a school, 4

5 of the children are boys, and the number of girls is 100. Find the

number of boys.

Ans. Let the total number of boys and girls = x

Total number of boys = 4 4

of5 5

xx =

As per condition

4100

5

xx − =

⇒ 1005

x=

∴ x = 500

Thus, number of boys = total strength – number of girls

= 500 – 100

= 400.

FRACTIONS - · PDF fileWhat fractions do the shaded parts in each of the following figures...

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