Mathematics Solutions for Class 6 Maths Chapter 4 - Operations On Fractions

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Page No 22:

Question 1:

Convert into improper fractions.

(i) $7 \frac{2}{5}$

(ii) $5 \frac{1}{6}$

(iii) $4 \frac{3}{4}$

(iv) $2 \frac{5}{9}$

(v) $1 \frac{5}{7}$

Answer:

(i)
$7 \frac{2}{5} = \frac{5 \times 7 + 2}{5} = \frac{35 + 2}{5} = \frac{37}{5}$

(ii)
$5 \frac{1}{6} = \frac{6 \times 5 + 1}{6} = \frac{30 + 1}{6} = \frac{31}{6}$

(iii)
$4 \frac{3}{4} = \frac{4 \times 4 + 2}{4} = \frac{16 + 2}{4} = \frac{18}{4}$

(iv)
$2 \frac{5}{9} = \frac{9 \times 2 + 5}{9} = \frac{18 + 5}{9} = \frac{23}{9}$

(v)
$1 \frac{5}{7} = \frac{7 \times 1 + 5}{7} = \frac{7 + 5}{7} = \frac{12}{7}$

Page No 22:

Question 2:

Convert into mixed numbers.

(i) $\frac{30}{7}$

(ii) $\frac{7}{4}$

(iii) $\frac{15}{12}$

(iv) $\frac{11}{8}$

(v) $\frac{21}{4}$

(vi) $\frac{20}{7}$

Answer:

(i)
$\frac{30}{7} = \frac{28 + 2}{7} = \frac{28}{7} + \frac{2}{7} = 4 + \frac{2}{7} = 4 \frac{2}{7}$

(ii)
$\frac{7}{4} = \frac{4 + 3}{4} = \frac{4}{4} + \frac{3}{4} = 1 + \frac{3}{4} = 1 \frac{3}{4}$

(iii)
$\frac{15}{12} = \frac{12 + 3}{12} = \frac{12}{12} + \frac{3}{12} = 1 + \frac{3}{12} = 1 \frac{3}{12}$

(iv)
$\frac{11}{8} = \frac{8 + 3}{8} = \frac{8}{8} + \frac{3}{8} = 1 + \frac{3}{8} = 1 \frac{3}{8}$

(v)
$\frac{21}{4} = \frac{20 + 1}{4} = \frac{20}{4} + \frac{1}{4} = 5 + \frac{1}{4} = 5 \frac{1}{4}$

(vi)
$\frac{20}{7} = \frac{14 + 6}{7} = \frac{14}{7} + \frac{6}{7} = 2 + \frac{6}{7} = 2 \frac{6}{7}$

Page No 22:

Question 3:

Write the following examples using fractions.
(i) If 9 kg rice is shared amongst 5 people, how many kilograms of rice does each person get?
( ii) To make 5 shirts of the same size, 11 metres of cloth is needed. How much cloth is needed for one shirt?

Answer:

(i) If 9 kg rice is shared amongst 5 people, then each person will get $\frac{9}{5}$ kilograms of rice.

( ii) If 11 metres of cloth is needed to make 5 shirts of the same size, then one shirt will need $\frac{11}{5}$ metres of cloth.

Page No 23:

Question 1:

Add.
(i) $6 \frac{1}{3} + 2 \frac{1}{3}$

(ii) $1 \frac{1}{4} + 3 \frac{1}{2}$

(iii) $5 \frac{1}{5} + 2 \frac{1}{7}$

(iv) $3 \frac{1}{5} + 2 \frac{1}{3}$

Answer:

(i)
$6 \frac{1}{3} + 2 \frac{1}{3} = \frac{6 \times 3 + 1}{3} + \frac{2 \times 3 + 1}{3} = \frac{18 + 1}{3} + \frac{6 + 1}{3} = \frac{19}{3} + \frac{7}{3}$
$= \frac{19 + 7}{3} = \frac{26}{3} = \frac{24 + 2}{3}$
$= \frac{24}{3} + \frac{2}{3} = 8 + \frac{2}{3} = 8 \frac{2}{3}$

(ii)
$1 \frac{1}{4} + 3 \frac{1}{2} = \frac{1 \times 4 + 1}{4} + \frac{3 \times 2 + 1}{2} = \frac{5}{4} + \frac{7}{2} = \frac{5}{4} + \frac{7 \times 2}{2 \times 2}$
$= \frac{5}{4} + \frac{14}{4} = \frac{5 + 14}{4} = \frac{19}{4} = \frac{16 + 3}{4}$
$= \frac{16}{4} + \frac{3}{4} = 4 + \frac{3}{4} = 4 \frac{3}{4}$

(iii)
$5 \frac{1}{5} + 2 \frac{1}{7} = \frac{5 \times 5 + 1}{5} + \frac{2 \times 7 + 1}{7} = \frac{26}{5} + \frac{15}{7} = \frac{26 \times 7}{5 \times 7} + \frac{15 \times 5}{7 \times 5}$
$= \frac{182}{35} + \frac{75}{35} = \frac{182 + 75}{35} = \frac{257}{35}$
$= \frac{245 + 12}{35} = \frac{245}{35} + \frac{12}{35} = 7 + \frac{12}{35} = 7 \frac{12}{35}$
(iv) $3 \frac{1}{5} + 2 \frac{1}{3}$
$3 \frac{1}{5} + 2 \frac{1}{3} = \frac{3 \times 5 + 1}{5} + \frac{2 \times 3 + 1}{3} = \frac{16}{5} + \frac{7}{3} = \frac{16 \times 3}{5 \times 3} + \frac{7 \times 5}{3 \times 5}$
$= \frac{48}{15} + \frac{35}{15} = \frac{48 + 35}{15} = \frac{83}{15}$
$= \frac{75 + 8}{15} = \frac{75}{15} + \frac{8}{15} = 5 + \frac{8}{15} = 5 \frac{8}{15}$

Page No 23:

Question 2:

Subtract.
(i) $3 \frac{1}{3} - 1 \frac{1}{4}$

(ii) $5 \frac{1}{2} - 3 \frac{1}{3}$

(iii) $7 \frac{1}{8} - 6 \frac{1}{10}$

(iv) $7 \frac{1}{2} - 3 \frac{1}{5}$

Answer:

(i)
$3 \frac{1}{3} - 1 \frac{1}{4} = \frac{10}{3} - \frac{5}{4} = \frac{10 \times 4}{3 \times 4} - \frac{5 \times 3}{4 \times 3} = \frac{40}{12} - \frac{15}{12} = \frac{40 - 15}{12}$
$= \frac{25}{12} = \frac{24 + 1}{12} = \frac{24}{12} + \frac{1}{12} = 2 + \frac{1}{12} = 2 \frac{1}{12}$

(ii)
$5 \frac{1}{2} - 3 \frac{1}{3} = \frac{11}{2} - \frac{10}{3} = \frac{11 \times 3}{2 \times 3} - \frac{10 \times 2}{3 \times 2} = \frac{33}{6} - \frac{20}{6}$
$= \frac{33 - 20}{6} = \frac{13}{6} = 2 \frac{1}{6}$

(iii)
$7 \frac{1}{8} - 6 \frac{1}{10} = \frac{57}{8} - \frac{61}{10} = \frac{57 \times 5}{8 \times 5} - \frac{61 \times 4}{10 \times 4} = \frac{285}{40} - \frac{244}{40}$
$= \frac{285 - 244}{40} = \frac{41}{40} = 1 \frac{1}{40}$

(iv)
$7 \frac{1}{2} - 3 \frac{1}{5} = \frac{15}{2} - \frac{16}{5} = \frac{15 \times 5}{2 \times 5} - \frac{16 \times 2}{5 \times 2} = \frac{75}{10} - \frac{32}{10}$
$= \frac{75 - 32}{10} = \frac{43}{10} = 4 \frac{3}{10}$

Page No 23:

Question 3:

Solve.
(1) Suyash bought $2 \frac{1}{2}$ kg of sugar and Ashish bought $3 \frac{1}{2}$ kg. How much sugar did they buy altogether? If sugar costs 32 rupees per kg, how much did they spend on the sugar they bought?
(2) Aradhana grows potatoes in $\frac{2}{5}$ part of her garden, greens in $\frac{1}{3}$ part and brinjals in the remaining part. On how much of her plot did she plant brinjals?
(3) Sandeep filled water in $\frac{4}{7}$ of an empty tank. After that, Ramakant filled $\frac{1}{4}$ part more of the same tank. Then Umesh used $\frac{3}{14}$ part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?

Answer:

(1) The amount of sugar they bought altogether = $2 \frac{1}{2} + 3 \frac{1}{2}$
$= \frac{5}{2} + \frac{7}{2} = \frac{5 + 7}{2} = \frac{12}{2}$
= 6 kg
Now, cost of 1 kg of sugar = Rs 32
Therefore, the cost of 6 kg of sugar is = 6 × 32
= Rs 192
Hence, they spend Rs 192 on the sugar they bought.

(2) The part of the garden in which Aradhana grew brinjals is given by $1 - \frac{2}{5} - \frac{1}{3}$

$= \frac{1 \times 15}{1 \times 15} - \frac{2 \times 3}{5 \times 3} - \frac{1 \times 5}{3 \times 5} = \frac{15}{15} - \frac{6}{15} - \frac{5}{15} = \frac{15 - 6 - 5}{15} = \frac{4}{15}$
Hence, Aradhana grew brinjals in $\frac{4}{15}$ part of her garden.
(3) The amount of water will be left in the tank is given by $\frac{4}{7} (560) + \frac{1}{4} (560) - \frac{3}{14} (560)$
$= 320 + 140 - 120 = 340 l$
Hence, 340 l of water will be left in the tank.

Page No 24:

Question 1:

What fractions do the points A and B show on the number lines below?
1.

2.

3.

Answer:

1. $A = \frac{5}{6} and B = \frac{10}{6}$

2. $A = \frac{3}{5} and B = \frac{7}{5}$

3. $A = \frac{3}{7} and B = \frac{10}{7}$

Page No 25:

Question 2:

Show the following fractions on the number line.

(1) $\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5}$

(2) $\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4}$

Answer:

(1) $\frac{3}{5}, \frac{6}{5}, 2 \frac{3}{5} or \frac{13}{5}$

(2) $\frac{3}{4}, \frac{5}{4}, 2 \frac{1}{4} or \frac{9}{4}$

Page No 26:

Question 1:

Multiply.

(i) $\frac{7}{5} \times \frac{1}{4}$

(ii) $\frac{6}{7} \times \frac{2}{5}$

(iii) $\frac{5}{9} \times \frac{4}{9}$

(iv) $\frac{4}{11} \times \frac{2}{7}$

(v) $\frac{1}{5} \times \frac{7}{2}$

(vi) $\frac{9}{7} \times \frac{7}{8}$

(vii) $\frac{5}{6} \times \frac{6}{5}$

(viii) $\frac{6}{17} \times \frac{3}{2}$

Answer:

(i)
$\frac{7}{5} \times \frac{1}{4} = \frac{7 \times 1}{5 \times 4} = \frac{7}{20}$

(ii)
$\frac{6}{7} \times \frac{2}{5} = \frac{6 \times 2}{7 \times 5} = \frac{12}{35}$

(iii)
$\frac{5}{9} \times \frac{4}{9} = \frac{5 \times 4}{9 \times 9} = \frac{20}{81}$

(iv)
$\frac{4}{11} \times \frac{2}{7} = \frac{4 \times 2}{11 \times 7} = \frac{8}{77}$

(v)
$\frac{1}{5} \times \frac{7}{2} = \frac{1 \times 7}{5 \times 2} = \frac{7}{10}$

(vi)
$\frac{9}{7} \times \frac{7}{8} = \frac{9}{7} \times \frac{7}{8} = \frac{9}{8}$

(vii)
$\frac{5}{6} \times \frac{6}{5} = \frac{5}{6} \times \frac{6}{5} = 1$

(viii)
$\frac{6}{17} \times \frac{3}{2} = \frac{36}{17} \times \frac{3}{2} = \frac{3 \times 3}{17} = \frac{9}{17}$

Page No 26:

Question 2:

Ashokrao planted bananas on $\frac{2}{7}$ of his field of 21 acres. What is the area of the banana plantation?

Answer:

Ashokrao planted bananas on $\frac{2}{7}$ of his field of 21 acres.
The area of the banana plantation = $\frac{2}{7} \times 21$
$= \frac{2}{7} \times 213 = 6 acres$
Hence, the area of the banana plantation is 6 acres.

Page No 26:

Question 3:

Of the total number of soldiers in our army, $\frac{4}{9}$ are posted on the northern border and one - third of them on the north - eastern border. If the number of soldiers in the north is 540000, how many are posted in the north - east?

Answer:

Let the total number of the soldiers be x.
Number of soldiers posted on the northern border = 540000
$\Rightarrow \frac{4}{9} x = 540000 \Rightarrow x = \frac{540000 \times 9}{4} = 1215000$
Now, the number of soldiers posted on the northern border = $\frac{1}{3} \times 1215000$
= 405000
Hence, 405000 soldiers are posted in the north - east.

Page No 28:

Question 1:

Write the reciprocals of the following numbers.
(i) 7

(ii) $\frac{11}{3}$

(iii) $\frac{5}{13}$

(iv) 2

(v) $\frac{6}{7}$

Answer:

(i) The reciprocal of 7 is $\frac{1}{7}$ .

(ii) The reciprocal of $\frac{11}{3}$ is $\frac{3}{11}$ .

(iii) The reciprocal of $\frac{5}{13}$ is $\frac{13}{5}$ .

(iv) The reciprocal of 2 is $\frac{1}{2}$ .

(v) The reciprocal of $\frac{6}{7}$ is $\frac{7}{6}$ .

Page No 28:

Question 2:

Carry out the following divisions.
(i) $\frac{2}{3} \div \frac{1}{4}$

(ii) $\frac{5}{9} \div \frac{3}{2}$

(iii) $\frac{3}{7} \div \frac{5}{11}$

(iv) $\frac{11}{12} \div \frac{4}{7}$

Answer:

(i)
$\frac{2}{3} \div \frac{1}{4} = \frac{2}{3} \times \frac{4}{1} = \frac{2 \times 4}{3 \times 1} = \frac{8}{3}$

(ii)
$\frac{5}{9} \div \frac{3}{2} = \frac{5}{9} \times \frac{2}{3} = \frac{5 \times 2}{9 \times 3} = \frac{10}{27}$

(iii)
$\frac{3}{7} \div \frac{5}{11} = \frac{3}{7} \times \frac{11}{5} = \frac{3 \times 11}{7 \times 5} = \frac{33}{35}$

(iv)
$\frac{11}{12} \div \frac{4}{7} = \frac{11}{12} \times \frac{7}{4} = \frac{11 \times 7}{12 \times 4} = \frac{77}{48}$

Page No 28:

Question 3:

There were 420 students participating in the Swachh Bharat campaign. They cleaned $\frac{42}{75}$ part of the town, Sevagram. What part of Sevagram did each student clean if the work was equally shared by all?

Answer:

420 students cleaned $\frac{42}{75}$ part of the Sevagram town.
1 student cleaned the part of the Sevagram town equal to $\frac{42}{75} \div 420$
$= \frac{42}{75} \times \frac{1}{420} = \frac{42}{75 \times 420} = \frac{1}{75 \times 10} = \frac{1}{750}$
Hence, each student cleaned $\frac{1}{750}$ part of the Sevagram town.

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