# Selina Solutions Concise Maths Class 7 Chapter 3: Fraction (Including Problems) Exercise 3C

Selina Solutions Concise Maths Class 7 Chapter 3 Fraction (Including Problems) Exercise 3C help students understand the various operations on fractions like addition, subtraction, multiplication and division. It also has problems on using â€˜ofâ€™ and â€˜BODMASâ€™ in the given set of questions. Numerous solved examples are present before the exercise problems to provide a clear understanding among students. Students can self analyse their weaknesses and work on them for a better result in the annual exam. Selina Solutions Concise Maths Class 7 Chapter 3 Fraction (Including Problems) Exercise 3C PDF links are given here.

## Selina Solutions Concise Maths Class 7 Chapter 3: Fraction (Including Problems) Exercise 3C Download PDF

### Access Selina Solutions Concise Maths Class 7 Chapter 3: Fraction (Including Problems) Exercise 3C

1. Reduce to a single fraction:

(i) 1/2 + 2/3

(ii) 3/5 â€“ 1/10

(iii) 2/3 â€“ 1/6

(iv) 1 1/3 + 2 1/4

(v) 1/4 + 5/6 â€“ 1/12

(vi) 2/3 â€“ 3/5 + 3 â€“ 1/5

(vii) 2/3 â€“ 1/5 + 1/10

(viii) 2 1/2 + 2 1/3 â€“ 1 1/4

(ix) 2 5/8 â€“ 2 1/6 + 4 3/4

Solution:

(i) 1/2 + 2/3

Here the LCM of 2 and 3 is 6

= (1 Ã— 3)/ (2 Ã— 3) + (2 Ã— 2)/ (3 Ã— 2)

By further calculation

= 3/6 + 4/6

= (3 + 4)/ 6

So we get

= 7/6

= 1 1/6

(ii) 3/5 â€“ 1/10

Here the LCM of 5 and 10 is 10

= (3 Ã— 2)/ (5 Ã— 2) â€“ 1/10

By further calculation

= 6/10 â€“ 1/10

So we get

= (6 â€“ 1)/ 10

= 5/10

= 1/2

(iii) 2/3 â€“ 1/6

Here the LCM of 3 and 6 is 6

= (2 Ã— 2)/ (3 Ã— 2) â€“ 1/6

By further calculation

= 4/6 â€“ 1/6

So we get

= (4 â€“ 1)/ 6

= 3/6

= 1/2

(iv) 1 1/3 + 2 1/4

It can be written as

= 4/3 + 9/4

Here LCM of 3 and 4 is 12

= (4 Ã— 4)/ (3 Ã— 4) + (9 Ã— 3)/ (4 Ã— 3)

By further calculation

= 16/12 + 27/12

So we get

= (16 + 27)/ 12

= 43/12

= 3 7/12

(v) 1/4 + 5/6 â€“ 1/12

Here LCM of 4, 6 and 12 is 12

= (1 Ã— 3)/ (4 Ã— 3) + (5 Ã— 2)/ (6 Ã— 2) â€“ 1/12

By further calculation

= 3/12 + 10/12 â€“ 1/12

So we get

= (3 + 10 â€“ 1)/ 12

= 12/12

= 1

(vi) 2/3 â€“ 3/5 + 3 â€“ 1/5

Here LCM of 3 and 5 is 15

= (2 Ã— 5)/ (3 Ã— 5) â€“ (3 Ã— 3)/ (5 Ã— 3) + (3 Ã— 15)/ 15 â€“ (1 Ã— 3)/ (5 Ã— 3)

By further calculation

= 10/15 â€“ 9/15 + 45/15 â€“ 3/15

So we get

= (10 â€“ 9 + 45 â€“ 3)/ 15

= (55 â€“ 12)/ 15

= 43/15

= 2 13/15

(vii) 2/3 â€“ 1/5 + 1/10

Here the LCM of 3, 5 and 10 is 30

= (2 Ã— 10)/ (3 Ã— 10) â€“ (1 Ã— 6)/ (5 Ã— 6) + (1 Ã— 3)/ (10 Ã— 3)

By further calculation

= 20/30 â€“ 6/30 + 3/30

So we get

= (20 â€“ 6 + 3)/ 30

= (23 â€“ 6)/ 30

= 17/30

(viii) 2 1/2 + 2 1/3 â€“ 1 Â¼

It can be written as

= 5/2 + 7/3 â€“ 5/4

Here the LCM of 2, 3 and 4 is 12

= (5 Ã— 6)/ (2 Ã— 6) + (7 Ã— 4)/ (3 Ã— 4) â€“ (5 Ã— 3)/ (4 Ã— 3)

By further calculation

= 30/12 + 28/12 â€“ 15/12

So we get

= (30 + 28 â€“ 15)/ 12

= (58 â€“ 15)/ 12

= 43/12

= 3 7/12

(ix) 2 5/8 â€“ 2 1/6 + 4 3/4

It can be written as

= 21/8 â€“ 13/6 + 19/4

Here the LCM of 8, 6 and 4 is 24

= (21 Ã— 3)/ (8 Ã— 3) â€“ (13 Ã— 4)/ (6 Ã— 4) + (19 Ã— 6)/ (4 Ã— 6)

By further calculation

= 63/24 â€“ 52/24 + 114/24

So we get

= (63 â€“ 52 + 114)/ 24

= (177 â€“ 52)/ 24

= 125/24

= 5 5/24

2. Simplify:

(i) 3/4 Ã— 6

(ii) 2/3 Ã— 15

(iii) 3/4 Ã— 1/2

(iv) 9/12 Ã— 4/7

(v) 45 Ã— 2 1/3

(vi) 36 Ã— 3 1/4

(vii) 2 Ã· 1/3

(viii) 3 Ã· 2/5

(ix) 1 Ã· 3/5

(x) 1/3 Ã· 1/4

(xi) -5/8 Ã· 3/4

(xii) 3 3/7 Ã· 1 1/14

(xiii) 3 Â¾ Ã— 1 1/5 Ã— 20/21

Solution:

(i) 3/4 Ã— 6

It can be written as

= 3/4 Ã— 6/1

= (3 Ã— 6)/ (4 Ã— 1)

So we get

= 18/4

HCF of 18 and 4 is 2

Dividing both numerator and denominator by 2

= (18 Ã· 2)/ (4 Ã· 2)

= 9/2

= 4 1/2

(ii) 2/3 Ã— 15

It can be written as

= 2/3 Ã— 15/1

= (2 Ã— 15)/ (3 Ã— 1)

So we get

= 30/3

= 10

(iii) 3/4 Ã— 1/2

It can be written as

= (3 Ã— 1)/ (4 Ã— 2)

= 3/8

(iv) 9/12 Ã— 4/7

It can be written as

= (9 Ã— 4)/ (12 Ã— 7)

So we get

= 36/84

Here the HCF of 36 and 84 is 12

Dividing both numerator and denominator by 12

= (36 Ã· 12)/ (84 Ã· 12)

= 3/7

(v) 45 Ã— 2 1/3

It can be written as

= 45/1 Ã— 7/3

= (45 Ã— 7)/ (1 Ã— 3)

So we get

= 315/3

= 105

(vi) 36 Ã— 3 1/4

It can be written as

= 36/1 Ã— 13/4

= (36 Ã— 13)/ (4 Ã— 1)

So we get

= 468/4

= 117

(vii) 2 Ã· 1/3

It can be written as

= 2/1 Ã— 3/1

= (2 Ã— 3)/ (1 Ã— 1)

So we get

= 6/1

= 6

(viii) 3 Ã· 2/5

It can be written as

= 3/1 Ã— 5/2

= (3 Ã— 5)/ (1 Ã— 2)

So we get

= 15/2

= 7 Â½

(ix) 1 Ã· 3/5

It can be written as

= 1 Ã— 5/3

= (1 Ã— 5)/ 3

So we get

= 5/3

= 1 2/3

(x) 1/3 Ã· 1/4

It can be written as

= 1/3 Ã— 4/1

= (1 Ã— 4)/ (3 Ã— 1)

So we get

= 4/3

= 1 1/3

(xi) -5/8 Ã· 3/4

It can be written as

= -5/8 Ã— 4/3

= (-5 Ã— 4)/ (8 Ã— 3)

So we get

= – 20/24

Here the HCF of 20 and 24 is 4

So by dividing both numerator and denominator by 4

= (-20 Ã· 4)/ (24 Ã· 4)

= -5/6

(xii) 3 3/7 Ã· 1 1/14

It can be written as

= (24 Ã— 14)/ (7 Ã— 15)

So we get

= 336/105

Here the HCF of 336 and 105 is 21

So by dividing both numerator and denominator by 21

= (336 Ã· 21)/ (105 Ã· 21)

= 16/5

= 3 1/5

(xiii) 3 Â¾ Ã— 1 1/5 Ã— 20/21

It can be written as

= 15/4 Ã— 6/5 Ã— 20/21

= (15 Ã— 6 Ã— 20)/ (4 Ã— 5 Ã— 21)

So we get

= 1800/420

Here the HCF of 1800 and 420 is 60

So by dividing both numerator and denominator by 60

= (1800 Ã· 60)/ (420 Ã· 60)

= 30/7

= 4 2/7

3. Subtract:

(i) 2 from 2/3

(ii) 1/8 from 5/8

(iii) -2/5 and 2/5

(iv) â€“ 3/7 from 3/7

(v) 0 from – 4/5

(vi) 2/9 from 4/5

(vii) â€“ 4/7 from -6/11

Solution:

(i) 2 from 2/3

It can be written as

= 2/3 â€“ 2/1

LCM of 3 and 1 is 3

= 2/3 â€“ (2 Ã— 3)/ 3

= 2/3 â€“ 6/3

So we get

= (2 â€“ 6)/ 3

= -4/3

= – 1 1/3

(ii) 1/8 from 5/8

It can be written as

= 5/8 â€“ 1/8

= (5 â€“ 1)/ 8

So we get

= 4/8

= 1/2

(iii) -2/5 and 2/5

It can be written as

= 2/5 â€“ (-2/5)

= 2/5 + 2/5

So we get

= (2 + 2)/ 5

= 4/5

(iv) â€“ 3/7 from 3/7

It can be written as

= 3/7 â€“ (-3/7)

= 3/7 + 3/7

So we get

= (3 + 3)/ 7

= 6/7

(v) 0 from – 4/5

It can be written as

= -4/5 â€“ 0

= – 4/5

(vi) 2/9 from 4/5

It can be written as

= 4/5 â€“ 2/9

LCM of 5 and 9 is 45

= (4 Ã— 9)/ (5 Ã— 9) â€“ (2 Ã— 5)/ (9 Ã— 5)

By further calculation

= 36/45 â€“ 10/45

= (36 â€“ 10)/ 45

= 26/45

(vii) â€“ 4/7 from -6/11

It can be written as

= -6/11 â€“ (-4/7)

= – 6/11 + 4/7

Here LCM of 7 and 11 is 77

= (- 6 Ã— 7)/ (11 Ã— 7) + (4 Ã— 11)/ (7 Ã— 11)

By further calculation

= – 42/77 + 44/77

So we get

= (-42 + 44)/ 77

= 2/77

4. Find the value of:

(i) Â½ and 10 kg

(ii) 3/5 of 1 hour

(iii) 4/7 of 2 1/3 kg

(iv) 3 Â½ times of 2 metre

(v) 1/2 of 2 2/3

(vi) 5/11 of 4/5 of 22 kg

Solution:

(i) Â½ and 10 kg

It can be written as

= (1/2 Ã— 10) kg

= 5 kg

(ii) 3/5 of 1 hour

It can be written as

= (3/5 Ã— 60) minutes

= 3 Ã— 12

= 36 minutes

(iii) 4/7 of 2 1/3 kg

It can be written as

= (4/7 Ã— 7/3) kg

So we get

= 4/3 kg

= 1 1/3 kg

(iv) 3 Â½ times of 2 metre

It can be written as

= (7/2 Ã— 2) metres

= 7 metres

(v) 1/2 of 2 2/3

It can be written as

= 1/2 Ã— 8/3

So we get

= 4/3

= 1 1/3

(vi) 5/11 of 4/5 of 22 kg

It can be written as

= (5/11 Ã— 4/5 Ã— 22/1) kg

So we get

= 4 Ã— 2

= 8 kg

5. Simplify and reduce to a simple fraction:

(i) 3/ 3 Â¾

(ii) 3/5/ 7

(iii) 3/ 5/7

(iv) 2 1/5/ 1 1/10

(v) 2/5 of 6/11 Ã— 1 Â¼

(vi) 2 Â¼ Ã· 1/7 Ã— 1/3

(vii) 1/3 Ã— 4 2/3 Ã· 3 Â½ Ã— 1/2

(viii) 2/3 Ã— 1 Â¼ Ã· 3/7 of 2 5/8

(ix) 0 Ã· 8/11

(x) 4/5 Ã· 7/15 of 8/9

(xi) 4/5 Ã· 7/15 Ã— 8/9

(xii) 4/5 of 7/15 Ã· 8/9

(xiii) 1/2 of 3/4 Ã— 1/2 Ã· 2/3

Solution:

(i) 3/ 3 Â¾

It can be written as

= 3 / 15/4

So we get

= (3 Ã— 4)/ 15

= 4/5

(ii) 3/5/ 7

It can be written as

= 3/5 Ã— 1/7

= 3/35

(iii) 3/ 5/7

It can be written as

= 3 Ã— 7/5

So we get

= 21/5

= 4 1/5

(iv) 2 1/5/ 1 1/10

It can be written as

= 11/5 / 11/10

So we get

= 11/5 Ã— 10/11

= 2

(v) 2/5 of 6/11 Ã— 1 Â¼

It can be written as

= 2/5 of 6/11 Ã— 5/4

So we get

= 12/55 Ã— 5/4

= 3/11

(vi) 2 Â¼ Ã· 1/7 Ã— 1/3

It can be written as

= 9/4 Ã· 1/7 Ã— 1/3

So we get

= 9/4 Ã— 7/1 Ã— 1/3

= 21/4

= 5 Â¼

(vii) 1/3 Ã— 4 2/3 Ã· 3 Â½ Ã— 1/2

It can be written as

= 1/3 Ã— 14/3 Ã· 7/2 Ã— 1/2

So we get

= 1/3 Ã— 14/3 Ã— 2/7 Ã— 1/2

= 2/9

(viii) 2/3 Ã— 1 Â¼ Ã· 3/7 of 2 5/8

It can be written as

= 2/3 Ã— 5/4 Ã· 3/7 of 21/8

So we get

= 2/3 Ã— 5/4 Ã· 9/8

Here

= 2/3 Ã— 5/4 Ã— 8/9

= 20/27

(ix) 0 Ã· 8/11

It can be written as

= 0 Ã— 11/8

= 0

(x) 4/5 Ã· 7/15 of 8/9

From BODMAS rule

= 4/5 Ã· 156/135

So we get

= 4/5 Ã— 135/56

= 27/14

= 1 13/14

(xi) 4/5 Ã· 7/15 Ã— 8/9

It can be written as

= 4/5 Ã— 15/7 Ã— 8/9

So we get

= 32/21

= 1 11/21

(xii) 4/5 of 7/15 Ã· 8/9

From BODMAS rule

= 28/75 Ã· 8/9

So we get

= 28/75 Ã— 9/8

By further calculation

= (7 Ã— 3)/ (25 Ã— 2)

= 21/50

(xiii) 1/2 of 3/4 Ã— 1/2 Ã· 2/3

From BODMAS rule

= 3/8 Ã— 1/2 Ã· 2/3

So we get

= 3/8 Ã— 1/2 Ã— 3/2

= 9/32

6. A bought 3 Â¾ kg of wheat and 2 Â½ kg of rice. Find the total weight wheat and rice bought.

Solution:

It is given that

Weight of wheat = 3 Â¾ kg = 15/4 kg

Weight of rice = 2 Â½ kg = 5/2 kg

So the total weight of wheat and rice = 15/4 + 5/2

Here the LCM of 4 and 2 is 4

= (15 Ã— 1)/ (4 Ã— 1) + (5 Ã— 2)/ (2 Ã— 2)

So we get

= (15 + 10)/ 4

= 25/4 kg

= 6 Â¼ kg

7. Which is greater, 3/5 or 7/10 and by how much?

Solution:

By cross multiplying

3 Ã— 10 = 30 and 7 Ã— 5 = 35

Here 30 is smaller than 35

So 3/5 < 7/10

We know that difference between 7/10 and 3/5

= 7/10 â€“ 3/5

LCM of 10 and 5 is 10

= (7 Ã— 1)/ (10 Ã— 1) â€“ (3 Ã— 2)/ (5 Ã— 2)

So we get

= (7 â€“ 6)/ 10

= 1/10

Therefore, 7/10 is greater than 3/5 by 1/10.

8. What number should be added to 8 2/3 to get 12 5/6?

Solution:

To find the fraction we must subtract 8 2/3 from 12 5/6

So the required number = 12 5/6 â€“ 8 2/3

It can be written as

= 77/6 â€“ 26/3

Here the LCM of 3 and 6 is 6

= (77 Ã— 1)/ (6 Ã— 1) â€“ (26 Ã— 2)/ (3 Ã— 2)

By further calculation

= (77 â€“ 52)/ 6

= 25/6

= 4 1/6

9. What should be subtracted from 8 Â¾ to get 2 2/3?

Solution:

Required number = 8 Â¾ – 2 2/3

It can be written as

= 35/4 â€“ 8/3

LCM of 4 and 3 is 12

= (35 Ã— 3)/ (4 Ã— 3) â€“ (8 Ã— 4)/ (3 Ã— 4)

By further calculation

= (105 â€“ 32)/ 12

= 73/12

= 6 1/12

10. A rectangular field is 16 Â½ m long and 12 2/5 m wide. Find the perimeter of the field.

Solution:

Dimensions of rectangular field are

Length = 16 Â½ m

So the perimeter = 2 (l + b)

Substituting the values

= 2 Ã— (16 Â½ + 12 2/5)

It can be written as

= 2 Ã— (33/2 + 62/5)

LCM of 2 and 5 is 10

= 2 Ã— [(33 Ã— 5)/ (2 Ã— 5) + (62 Ã— 2)/ (5 Ã— 2)]

By further calculation

= 2 Ã— [(165 + 124)/ 10)]
So we get

= 2 Ã— 289/10

= 289/5 m

= 57 4/5 m

11. Sugar costs â‚¹ 37 Â½ per kg. Find the cost of 8 Â¾ kg sugar.

Solution:

It is given that

Cost of 1 kg sugar = â‚¹ 37 1/2

So the cost of 8 Â¾ kg sugar = 37 Â½ Ã— 8 Â¾

It can be written as

= 75/2 Ã— 35/4

= 2625/8

= â‚¹ 328 1/8

12. A motor cycle runs 31 Â¼ km consuming 1 litre of petrol. How much distance will it run consuming 1 3/5 litre of petrol?

Solution:

It is given that

Distance covered consuming 1 litre petrol = 31 Â¼ km = 125/4 km

So the distance covered consuming 1 3/5 litre petrol = 125/4 Ã— 8/5

= 1000/20

= 50 km

13. A rectangular park has length = 23 2/5 m and breadth = 16 2/3 m. Find the area of the park.

Solution:

Dimensions of rectangular park are

Length = 23 2/5 m = 117/5 m

Breadth = 16 2/3 m = 50/3 m

So the area = l Ã— b

Substituting the values

= 117/5 Ã— 50/3

= 39 Ã— 10

= 390 m2

14. Each of 40 identical boxes weighs 4 4/5 kg. Find the total weight of all the boxes.

Solution:

It is given that

Weight of one box = 4 4/5 kg = 24/5 kg

So the weight of 40 boxes = 40 Ã— 24/5

= 8 Ã— 24

= 192 kg

15. Out of 24 kg of wheat, 5/6th of wheat is consumed. Find, how much wheat is still left?

Solution:

Wheat available = 24 kg

Wheat consumed = 5/6th of 24 kg

= 5/6 Ã— 24

= 20kg

So the remaining wheat = 24 â€“ 20 kg = 4 kg

16. A rod of length 2 2/5 metre is divided into five equal parts. Find the length of each part so obtained.

Solution:

Length of rod = 2/5 m

It is given that the length of rod should be divided into 5 equal parts

So the length of each part of rod = 2 2/5 Ã· 5

It can be written as

= 12/5 Ã— 1/5

= 12/25 m

17. If A = 3 3/8 and B = 6 5/8, find: (i) A Ã· B (ii) B Ã· A.

Solution:

We know that

A = 3 3/8 = 27/8

B = 6 5/8 = 53/8

(i) A Ã· B

Substituting the values

= 27/8 Ã· 53/8

It can be written as

= 27/8 Ã— 8/53

= 27/53

(ii) B Ã· A

Substituting the values

= 53/8 Ã· 27/8

It can be written as

= 53/8 Ã— 8/27

= 53/27

= 1 26/27

18. Cost of 3 5/7 litres of oil is â‚¹ 83 Â½. Find the cost of one litre oil.

Solution:

It is given that

Cost of 3 5/7 litres of oil = â‚¹ 83 Â½

So the cost of one litre oil = â‚¹ 83 Â½ Ã· 3 5/7

It can be written as

= â‚¹ 167/2 Ã· 26/7

We get

= â‚¹ 167/2 Ã— 7/26

= â‚¹ 1169/52

= â‚¹ 22 25/52

19. The product of two numbers is 20 5/7. If one of these numbers is 6 2/3, find the other.

Solution:

It is given that

Product of two numbers = 20 5/7 = 145/7

One number = 6 2/3 = 20/3

So the other number = 145/7 Ã· 20/3

By further calculation

= 145/7 Ã— 3/20

So we get

= 87/28

= 3 3/28

20. By what number should 5 5/6 be multiplied to get 3 1/3?

Solution:

Here the required number = 3 1/3 Ã· 5 5/6

It can be written as

= 10/3 Ã· 35/6

So we get

= 10/3 Ã— 6/35

= 4/7