Worksheet on Fractions for Class 7

Solved Questions on Fractions

1. A rope of length 8 ⅗ metres is divided into five equal parts. What is the length of each part so obtained?

a) 1.12 metres
b) 1.32 metres
c) 1.52 metres
d) 1.72 metres

Answer: d) 1.72 metres

Explanation: Total length of rope =  8 ⅗ metres

A rope is divided into five equal parts.

Length of each part = 8 ⅗ metres ÷ 5
= 43/5 ÷ 5
= 43/5 × 1/5
= 43/25
= 1$\frac{\frac{\mathrm{18}}{}}{\frac{\mathrm{25}}{}}$ metres
= [1 + ($\frac{\frac{\mathrm{18}}{}}{\frac{\mathrm{25}}{}}$)] metres
= (1 + $\frac{\frac{\mathrm{18\; x\; 4}}{}}{\frac{\mathrm{25\; x\; 4}}{}}$) metres [Mutiplied by 4 to get denominator 100, to find decimals.]
= (1 + $\frac{\frac{\mathrm{72}}{}}{\frac{\mathrm{100}}{}}$) metres
= (1 + 0.72) metres
= 1.72 metres

2. A car runs 54 ¼ km consuming 2 litres of petrol. How much distance will it take to consume 1$\frac{\frac{2}{}}{\frac{7}{}}$ litre of petrol?

a) 34$\frac{\frac{1}{}}{\frac{8}{}}$ km
b) 34$\frac{\frac{3}{}}{\frac{8}{}}$ km
c) 34$\frac{\frac{5}{}}{\frac{8}{}}$ km
d) 34$\frac{\frac{7}{}}{\frac{8}{}}$ km

Answer: d) 34$\frac{\frac{7}{}}{\frac{8}{}}$ km

Explanation: Distance covered by consuming 2 litres of petrol = 54 ¼ km = 217/4 km

Distance covered by consuming 1 litre of petrol = 217/4 km ÷ 2
                                                                      = 217/8 km

Distance covered by consuming 1$\frac{\frac{2}{}}{\frac{7}{}}$ litre of petrol = 217/8 km × 1$\frac{\frac{2}{}}{\frac{7}{}}$ litre
                                                                         = 217/8 km × 9/7
                                                                          = 279/8 km
                                                                           = 34$\frac{\frac{7}{}}{\frac{8}{}}$ km

3. By how much does the sum of 37.786 and 29.576 exceed the sum of 38.789 and 23.548? 

a) 5.015
b) 5.025
c) 5.105
d) 5.125

Answer: b) 5.025

Explanation: Sum of 37.786 and 29.576 = 67.362

Sum of 38.789 and 23.548 = 62.337

Difference (Exceed by) =  67.362 −  62.337 = 5.025

4. What number should be added to 13$\frac{\frac{2}{}}{\frac{3}{}}$to get 12$\frac{\frac{5}{}}{\frac{7}{}}$?

a) $\frac{\frac{\mathrm{20}}{}}{\frac{\mathrm{21}}{}}$
b) $\frac{\frac{\mathrm{22}}{}}{\frac{\mathrm{21}}{}}$
c) − $\frac{\frac{\mathrm{20}}{}}{\frac{\mathrm{21}}{}}$
d) − $\frac{\frac{\mathrm{-22}}{}}{\frac{\mathrm{21}}{}}$

Answer: c) −$\frac{\frac{\mathrm{20}}{}}{\frac{\mathrm{21}}{}}$

Explanation: To find the number that should be added, subtract 13$\frac{\frac{2}{}}{\frac{3}{}}$ from 12$\frac{\frac{5}{}}{\frac{7}{}}$.

Required Number = 12$\frac{\frac{5}{}}{\frac{7}{}}$ − 13$\frac{\frac{2}{}}{\frac{3}{}}$

= $\frac{\frac{\mathrm{89}}{}}{\frac{7}{}}$ − $\frac{\frac{\mathrm{41}}{}}{\frac{3}{}}$
= $\frac{\frac{\mathrm{267\; -\; 287}}{}}{\frac{\mathrm{21}}{}}$  [LCM of 7 and 3 = 7 × 3 = 21]
= $\frac{\frac{\mathrm{-20}}{}}{\frac{\mathrm{21}}{}}$ or − $\frac{\frac{\mathrm{20}}{}}{\frac{\mathrm{21}}{}}$

5. When Katrina travelled 35 km by car, she found that (5/8)^th of her journey was still left. What was the total distance of the whole journey?

a) 93$\frac{\frac{1}{}}{\frac{3}{}}$ km
b) 93$\frac{\frac{2}{}}{\frac{3}{}}$ km
c) 93$\frac{\frac{1}{}}{\frac{8}{}}$ km
d) 93$\frac{\frac{3}{}}{\frac{8}{}}$ km

Answer: a) 93$\frac{\frac{1}{}}{\frac{3}{}}$ km

Explanation: Katrina travelled 35 km.

(5/8)^th of her journey was still left.

Fraction of her journey travelled = 1 − $\frac{\frac{5}{}}{\frac{8}{}}$

= $\frac{\frac{\mathrm{8\; -\; 5}}{}}{\frac{8}{}}$ [LCM of 1 and 8 is 8]
=  $\frac{\frac{3}{}}{\frac{8}{}}$

(3/8)^th of the total journey = Her journey travelled

⇒ (3/8)^th of the total journey = 35

⇒ Total journey = 35 × $\frac{\frac{8}{}}{\frac{3}{}}$
⇒ Total journey = $\frac{\frac{\mathrm{280}}{}}{\frac{3}{}}$
⇒ Total journey = 93$\frac{\frac{1}{}}{\frac{3}{}}$ km