NCERT Solutions for Class 6 Maths Chapter 7 Fractions (Ex 7.6) Exercise 7.6

Q.1

$\begin{array}{l}\left(\mathrm{a}\right)\frac{2}{3}+\frac{1}{7}\\ \left(\mathrm{b}\right)\frac{3}{10}+\frac{7}{15}\\ \left(\mathrm{c}\right)\frac{4}{9}+\frac{2}{7}\\ \left(\mathrm{d}\right)\frac{5}{7}+\frac{1}{3}\\ \left(\mathrm{e}\right)\frac{2}{5}+\frac{1}{6}\\ \left(\mathrm{f}\right)\frac{4}{5}+\frac{2}{3}\\ \left(\mathrm{g}\right)\frac{3}{4}-\frac{1}{3}\\ \left(\mathrm{h}\right)\frac{5}{6}-\frac{1}{3}\\ \left(\mathrm{i}\right)\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\\ \left(\mathrm{j}\right)\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\\ \left(\mathrm{k}\right)1\frac{1}{3}+3\frac{2}{3}\\ \left(\mathrm{l}\right)4\frac{2}{3}+3\frac{1}{4}\\ \left(\mathrm{m}\right)\frac{16}{5}-\frac{7}{5}\\ \left(\mathrm{n}\right)\frac{4}{3}-\frac{1}{2}\end{array}$

Ans.

$\begin{array}{l}\text{(a)}\frac{2}{3}+\frac{1}{7}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 7 is 21}\text{.}\\ \text{=}\frac{7\times 2+1\times 3}{21}\\ =\frac{14+3}{21}\\ =\frac{17}{21}\end{array}$

$\begin{array}{l}\text{(b)}\frac{3}{10}+\frac{7}{15}\\ \text{The L}\text{.C}\text{.M}\text{. of 10 and 15 is 30}\text{.}\\ \text{=}\frac{3\times 3+2\times 7}{30}\\ =\frac{9+14}{30}\\ =\frac{23}{30}\end{array}$

$\begin{array}{l}\text{(c)}\frac{4}{9}+\frac{2}{7}\\ \text{The L}\text{.C}\text{.M}\text{. of 9 and 7 is 63}\text{.}\\ \text{=}\frac{7\times 4+9\times 2}{63}\\ =\frac{28+18}{63}\\ =\frac{46}{63}\end{array}$

$\begin{array}{l}\text{(d)}\frac{5}{7}+\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 7 is 21}\text{.}\\ \text{=}\frac{5\times 3+1\times 7}{21}\\ =\frac{15+7}{21}\\ =\frac{22}{21}\end{array}$

$\begin{array}{l}\text{(e)}\frac{2}{5}+\frac{1}{6}\\ \text{The L}\text{.C}\text{.M}\text{. of 5 and 6 is 30}\text{.}\\ \text{=}\frac{6\times 2+5\times 1}{30}\\ =\frac{12+5}{30}\\ =\frac{17}{30}\end{array}$

$\begin{array}{l}\text{(f)}\frac{4}{5}+\frac{2}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 5 and 3 is 15}\text{.}\\ \text{=}\frac{4\times 3+5\times 2}{15}\\ =\frac{12+10}{15}\\ =\frac{22}{15}\end{array}$

$\begin{array}{l}\text{(g)}\frac{3}{4}-\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 4 and 3 is 12}\text{.}\\ \text{=}\frac{3\times 3-4\times 1}{12}\\ =\frac{9-4}{12}\\ =\frac{5}{12}\end{array}$

$\begin{array}{l}\text{(h)}\frac{5}{6}-\frac{1}{3}\\ \text{The L}\text{.C}\text{.M}\text{. of 6 and 3 is 6}\text{.}\\ \text{=}\frac{1\times 5-2\times 1}{6}\\ =\frac{5-2}{6}\\ =\frac{3}{6}=\frac{1}{2}\end{array}$

$\begin{array}{l}\text{(i)}\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\\ \text{The L}\text{.C}\text{.M}\text{. of 2, 3 and 4 is 12}\text{.}\\ \text{=}\frac{2\times 4+3\times 3+6\times 1}{12}\\ =\frac{8+9+6}{12}\\ =\frac{23}{12}\end{array}$

$\begin{array}{l}\text{(j)}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\\ \text{The L}\text{.C}\text{.M}\text{. of 2, 3 and 6 is 6}\text{.}\\ \text{=}\frac{1\times 3+1\times 2+1\times 1}{6}\\ =\frac{3+2+1}{6}\\ =\frac{6}{6}=1\end{array}$

$\begin{array}{l}\text{(k) 1}\frac{1}{3}+3\frac{2}{3}\\ \text{Converting the given fraction into improper fraction}\\ \text{=}\frac{4}{3}+\frac{11}{3}\\ =\frac{11+4}{3}\\ =\frac{15}{3}=5\end{array}$

$\begin{array}{l}\text{(l) 4}\frac{2}{3}+3\frac{1}{4}\\ \text{Converting the given fraction into improper fraction}\\ \text{=}\frac{14}{3}+\frac{13}{4}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 4 is 12}\\ =\frac{14\times 4+3\times 13}{12}\\ =\frac{56+39}{12}=\frac{95}{12}\end{array}$

$\begin{array}{l}\text{(m)}\frac{16}{5}-\frac{7}{5}\\ \text{=}\frac{16-7}{5}\\ =\frac{9}{5}\end{array}$

$\begin{array}{l}\text{(n)}\frac{4}{3}-\frac{1}{2}\\ \text{The L}\text{.C}\text{.M}\text{. of 3 and 2 is 6}\\ \text{=}\frac{4\times 2-1\times 3}{6}\\ =\frac{8-3}{6}\\ =\frac{5}{6}\end{array}$

Q.2 Sarita bought

$\frac{2}{5}$

metre of ribbon and Lalita bought

$\frac{3}{4}$

metre of ribbon. What is the total length of the ribbon they bought?

Ans.

Length of ribbon bought by Sarita =

$\frac{2}{5}$

Length of ribbon bought by Lalita =

$\frac{3}{4}$

Total length of ribbon

$\begin{array}{l}\text{=}\frac{2}{5}+\frac{3}{4}\\ =\frac{2\times 4+5\times 3}{20}\\ =\frac{8+15}{20}\\ =\frac{23}{20}\mathrm{m}\end{array}$

Q.3 Naina was given

$1\frac{1}{2}$

piece of cake and Najma was given

$1\frac{1}{3}$

piece of cake. Find the total amount of cake was given to both of them.

Ans.

Piece of cake given to Naina =

$1\frac{1}{2}=\frac{3}{2}$

Piece of cake given to Najma =

$1\frac{1}{3}=\frac{4}{3}$

Total amount of cake given

$\begin{array}{l}\text{=}\frac{3}{2}+\frac{4}{3}\\ =\frac{3\times 3+4\times 2}{6}\\ =\frac{9+8}{6}\\ =\frac{17}{6}=2\frac{5}{6}\end{array}$

Q.4

$\left(\mathrm{a}\right)\boxed{}-\frac{5}{8}=\frac{1}{4}\left(\mathrm{b}\right)\boxed{}-\frac{1}{5}=\frac{1}{2}\left(\mathrm{c}\right)\frac{1}{2}-\boxed{}=\frac{1}{6}$

Ans.

$\begin{array}{l}\text{(a)}\boxed{}-\frac{5}{8}=\frac{1}{4}\\ \Rightarrow \boxed{}=\frac{1}{4}+\frac{5}{8}\\ \Rightarrow \boxed{}=\frac{2\times 1+5\times 1}{8}\\ \Rightarrow \boxed{}=\frac{7}{8}\\ \text{(b)}\boxed{}-\frac{1}{5}=\frac{1}{2}\\ \Rightarrow \boxed{}=\frac{1}{5}+\frac{1}{2}\\ \Rightarrow \boxed{}=\frac{1\times 2+1\times 5}{10}\\ \Rightarrow \boxed{}=\frac{7}{10}\\ \text{(c)}\frac{1}{2}-\boxed{}=\frac{1}{6}\\ \Rightarrow \boxed{}=\frac{1}{2}-\frac{1}{6}\\ \Rightarrow \boxed{}=\frac{3-1}{6}=\frac{2}{6}\end{array}$

Q.5 Complete the addition-subtraction box.

Ans.

$\begin{array}{l}\text{(a)}\frac{2}{3}+\frac{4}{3}=\frac{6}{3}=2\\ \frac{1}{3}+\frac{2}{3}=\frac{3}{3}=1\\ \frac{2}{3}-\frac{1}{3}=\frac{1}{3}\\ \frac{4}{3}-\frac{2}{3}=\frac{2}{3}\\ 2-1=1\\ \end{array}$

$\begin{array}{l}\text{(b)}\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\\ \frac{1}{3}+\frac{1}{4}=\frac{7}{12}\\ \frac{1}{2}-\frac{1}{3}=\frac{1}{6}\\ \frac{1}{3}-\frac{1}{4}=\frac{1}{12}\\ \frac{5}{6}-\frac{7}{12}=\frac{3}{12}=\frac{1}{4}\\ \end{array}$

Q.6 A piece of wire

$\frac{7}{8}$

metre long broke into two pieces. One piece was

$\frac{1}{4}$

metre long. How long is the other piece?

Ans.

Length of one piece =

$\frac{1}{4}$

Length of other piece is equal to the difference of the length of the wire and the length of the first piece of wire.

So, length of other piece

$\begin{array}{l}=\frac{7}{8}-\frac{1}{4}\\ =\frac{7-2}{8}\\ =\frac{5}{8}\mathrm{m}\end{array}$

Q.7 Nandini’s house is

$\frac{9}{10}$

km from her school. She walked some distance and then took a bus for

$\frac{1}{2}$

km to reach the school. How far did she walk?

Ans.

Distance walked by Nandini
= Total distance – Distance for which she took the bus

$\begin{array}{l}\text{=}\frac{9}{10}-\frac{1}{2}\\ =\frac{9-5}{10}\\ =\frac{4}{10}\\ =\frac{2}{5}km\end{array}$

Distance walked by Nandini
= Total distance – Distance for which she took the bus

$\begin{array}{l}\text{=}\frac{9}{10}-\frac{1}{2}\\ =\frac{9-5}{10}\\ =\frac{4}{10}\\ =\frac{2}{5}km\end{array}$

Q.8 Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is

$\frac{5}{6}$

th full and Samuel’s shelf is

$\frac{2}{5}$

th full. Whose bookshelf is more full? By what fraction

Ans.

Fraction of Asha’s shelf that is full =

$\frac{5}{6}$

Fraction of Samuel’s shelf that is full =

$\frac{2}{5}$

Converting both fractions into like fractions :

$\begin{array}{l}\frac{5}{6}=\frac{5\times 5}{6\times 5}=\frac{25}{30}\\ \frac{2}{5}=\frac{2\times 6}{5\times 6}=\frac{12}{30}\\ \text{Here,}\frac{5}{6}>\frac{2}{5}\end{array}$

Hence, Asha’s bookshelf is more full.

$\begin{array}{l}\text{Difference}=\frac{25}{30}-\frac{12}{30}\\ =\frac{25-12}{30}\\ =\frac{13}{30}\end{array}$

Q.9 Jaidev takes

$2\frac{1}{5}$

minutes to walk across the school ground. Rahul takes

$\frac{7}{4}$

minutes to do the same. Who takes less time and by what fraction?

Ans.

Time taken by Jaidev =

$2\frac{1}{5}\mathrm{minutes}=\frac{11}{5}\mathrm{minutes}$

Time taken by Rahul =

$\frac{7}{4}\mathrm{minutes}$

Converting both fractions into like fractions :

$\begin{array}{l}\frac{11}{5}=\frac{11\times 4}{5\times 4}=\frac{44}{20}\\ \frac{7}{4}=\frac{7\times 5}{4\times 5}=\frac{35}{20}\\ \text{Here,}\frac{11}{5}>\frac{7}{4}\end{array}$

Hence, Rahul takes lesser time.

$\begin{array}{l}\text{Difference}=\frac{44}{20}-\frac{35}{20}\\ =\frac{44-35}{20}\\ =\frac{9}{20}\text{min}\end{array}$