 # RS Aggarwal Solutions For Class 7 Maths Exercise 2C Chapter 2 Fractions

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## Download the PDF of RS Aggarwal Solutions For Class 7 Maths Chapter 2 Fractions – Exercise 2C      ### Access answers to Maths RS Aggarwal Solutions for Class 7 Chapter 2 – Fractions Exercise 2C

1. Write down the reciprocal of:
(i) (5/8)

Solution:-

Reciprocal of (5/8) is (8/5) [∵ ((5/8) × (8/5)) = 1]

(ii) 7

Solution:-

Reciprocal of 7 is (1/7) [∵ ((7/1) × (1/7)) = 1]

(iii) (1/12)

Solution:-

Reciprocal of (1/12) is (12/1) [∵ ((1/12) × (12/1)) = 1]

= 12

(iv) [12(3/5)]

Solution:-

Convert mixed fraction into improper fraction,

= (63/5)

Reciprocal of (63/5) is (5/63) [∵ ((63/5) × (5/63)) = 1]

2. Simplify:
(i) (4/7) ÷ (9/14)

Solution:-

We have,

= (4/7) ÷ (9/14)

= (4/7) × (14/9)

(Because reciprocal of (9/14) is (14/9)

= (4 × 14) / (7 × 9)

= (4 × 2) / (1×9)

= (8/9)

(ii) (7/10) ÷ (3/5)

Solution:-

We have,

= (7/10) ÷ (3/5)

= (7/10) × (5/3)

(Because reciprocal of (3/5) is (5/3)

= (7 × 5) / (10 × 3)

= (7 × 1) / (2 × 3)

= (7/6)

= [1(1/6)]

(iii) (8/9) ÷ (16)

Solution:-

We have,

= (8/9) ÷ (16/1)

= (8/9) × (1/16)

(Because reciprocal of (16/1) is (1/16)

= (8 × 1) / (9 × 16)

= (1 × 1) / (9 × 2)

= (1/18)

(iv) (9) ÷ (1/3)

Solution:-

We have,

= (9/1) ÷ (1/3)

= (9/1) × (3/1)

(Because reciprocal of (1/3) is (3/1)

= (9 × 3) / (1 × 1)

= 27

(v) (24) ÷ (6/7)

Solution:-

We have,

= (24/1) ÷ (6/7)

= (24/1) × (7/6)

(Because reciprocal of (6/7) is (7/6)

= (24 × 7) / (1 × 6)

= (4 × 7) / (1 × 1)

= 28

(vi) [3(3/5)] ÷ (4/5)

Solution:-

Convert mixed fraction into improper fraction,

= [3(3/5)] = (18/5)

We have,

= (18/5) ÷ (4/5)

= (18/5) × (5/4)

(Because reciprocal of (4/5) is (5/4)

= (18 × 5) / (5 × 4)

= (9 × 1) / (1 × 2)

= (9/2)

= [4(1/2)

(vii) [3(3/7)] ÷ (8/21)

Solution:-

Convert mixed fraction into improper fraction,

We have,

(Because reciprocal of (8/21) is (21/8)

= (24 × 21) / (7 × 8)

= (3 × 3) / (1 × 1)

= 9

(viii) [5(4/7)] ÷ [1(3/10)]

Solution:-

Convert mixed fraction into improper fraction,

= [5(4/7)] = (39/7)

= [1(3/10)] = (13/10)

We have,

= (39/7) ÷ (13/10)

= (39/7) × (10/13)

(Because reciprocal of (13/10) is (10/13)

= (39 × 10) / (7 × 13)

= (390) / (91)

= (30 / 7)

= [4(2/7)]

(ix) [15(3/7)] ÷ [1(23/49)]

Solution:-

Convert mixed fraction into improper fraction,

= [15(3/7)] = (108/7)

= [1(23/49)] = (72/49)

We have,

= (108/7) ÷ (72/49)

= (108/7) × (49/72)

(Because reciprocal of (72/49) is (49/72)

= (108 × 49) / (7 × 72)

= (9×7) / (1 × 6)

= (3×7) / (1×2)

= (21/2)

= [10(1 /2)

3. Divide:
(i) (11/24) by (7/8)

Solution:-

The above question can be written as,

= (11/24) ÷ (7/8)

We have,

= (11/24) × (8/7)

(Because reciprocal of (7/8) is (8/7)

= (11 × 8) / (24 × 7)

= (11 × 1) / (3 × 9)

= (11/21)

(ii) [6(7/8)] by (11/16)

Solution:-

The above question can be written as,

= [6(7/8)] ÷ (11/16)

Convert mixed fraction into improper fraction,

= [6(7/8)] = (55/8)

We have,

= (55/8) × (16/11)

(Because reciprocal of (11/16) is (16/11)

= (55 × 16) / (8 × 11)

= (5 × 2) / (1 × 1)

= 10

(iii) [5(5/9)] by [3(1/3)]

Solution:-

The above question can be written as,

= [5(5/9)] ÷ [3(1/3)]

Convert mixed fraction into improper fraction,

= [5(5/9)] = (50/9)

= [3(1/3)] = (10/3)

We have,

= (50/9) × (3/10)

(Because reciprocal of (10/3) is (3/10)

= (50 × 3) / (9 × 10)

= (5 × 1) / (3 × 1)

= (5/3)

= [1(2/3)]

(iv) 32 by [1(3/5)]

Solution:-

The above question can be written as,

= 32 ÷ [1(3/5)]

Convert mixed fraction into improper fraction,

= [1(3/5)] = (8/5)

We have,

= (32/1) × (5/8)

(Because reciprocal of (8/5) is (5/8)

= (32 × 5) / (1 × 8)

= (4 × 5) / (1 × 1)

= 20

(v) 45 by [1(4/5)]

Solution:-

The above question can be written as,

= 45 ÷ [1(4/5)]

Convert mixed fraction into improper fraction,

= [1(4/5)] = (9/5)

We have,

= (45/1) × (5/9)

(Because reciprocal of (9/5) is (5/9)

= (45 × 5) / (1 × 9)

= (5 × 5) / (1 × 1)

= 25

(vi) 63 by [2(1/4)]

Solution:-

The above question can be written as,

= 63 ÷ [2(1/4)]

Convert mixed fraction into improper fraction,

= [2(1/4)] = (9/4)

We have,

= (63/1) × (4/9)

(Because reciprocal of (9/4) is (4/9)

= (63 × 4) / (1 × 9)

= (7 × 4) / (1 × 1)

= 28

4. A rope of length [13(1/2)] m has been divided into 9 pieces of the same length. What is the length of each piece?

Solution:-

From the question,

Rope length = [13(1/2)] m = (27/2)

Number of equal pieces divided into = 9

Then we have,

= (27/2) ÷ (9/1)

= (27/2) × (1/9)

(Because reciprocal of (9/1) is (1/9)

= (27 × 1) / (2 × 9)

= (3×1) / (2×1)

= (3 / 2)

= [1(1/2)] m

Hence, the length of 9 pieces of rope is [1(1/2)] m

5. 18 boxes of nails weigh equally and their total weight is [49(1/2)] kg. How much does each box weigh?

Solution:-

From the question,

Total weight of boxes= [49(1/2)] kg = (99/2)

Number of boxes = 18

Then we have,

= (99/2) ÷ (18/1)

= (99/2) × (1/18)

(Because reciprocal of (18/1) is (1/18)

= (99 × 1) / (2 × 18)

= (11×1) / (2×2)

= (11 / 4)

= [2(3/4)] kg

Hence, the weight of each box is [2(3/4)] kg

6. By selling oranges at the rate of ₹ [6(3/4)] per orange, a man gets ₹ 378. How many oranges does he sell?

Solution:-

From the question,

Cost for 1 orange = ₹ [6(3/4)] = (27/4)

Man gets = ₹ 378

Then we have,

= (378/1) ÷ (27/4)

= (378/1) × (4/27)

(Because reciprocal of (27/4) is (4/27)

= (378 × 4) / (1 × 27)

= (42×4) / (1×3)

= (14×4) / (1×1)

= 56

Hence, the man sold 56 orange.

7. Mangos are sold at ₹ [43(1/2)] per kg. What is the weight of mangoes available for ₹ [326(1/4)]?

Solution:-

From the question,

Mangos are sold at = ₹ [43(1/2)] = (87/2)

The weight of mangos available for = ₹ [326(1/4)] = (1305/4)

Then we have,

= (1305/4) ÷ (87/2)

= (1305/4) × (2/87)

(Because reciprocal of (87/2) is (2/87)

= (1305 × 2) / (4 × 87)

= (435×1) / (2×29)

= [7(1/2)] kg

Hence, the weight of mangos available for (1305/4) is [7(1/2)] kg

8. Vikas can cover a distance of [20(2/3)] km in [7(3/4)] hours on foot. How many km per hour does he walk?

Solution:-

From the question,

Distance covered by vikas in [7(3/4)] hours on foot = [20(2/3)] km = (62/3)

Distance covered by vikas in 1 hour = (62/3) ÷ (31/4)

Then we have,

= (62/3) × (4/31)

(Because reciprocal of (31/4) is (4/31)

= (62 × 4) / (3 × 31)

= (2×4) / (3×1)

= (8) / (3)

= [2(2/3) km

Hence, Distance covered by vikas in 1 hour is [2(2/3) km

### RS Aggarwal Solutions for Class 7 Maths Exercise 2C Chapter – 2 Fractions

RS Aggarwal Solutions For Class 7 Maths Chapter 2 Fractions Exercise 2C has problems which are based on the division of fractions. This exercise contains reciprocal of fractions, there are so many sub-questions given to make students solve more along with word problems. Students are suggested to try solving the questions from RS Aggarwal Solutions book of Class 7.