NCERT Solutions for Class 6 Maths Exercise 7.6

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Class –VI Mathematics

(Ex. 7.6)

Question 1.Solve:

(a) {tex}\frac{2}{3} + \frac{1}{7}{/tex}

(b) {tex}\frac{3}{{10}} + \frac{7}{{15}}{/tex}

(c) {tex}\frac{4}{9} + \frac{2}{7}{/tex}

(d) {tex}\frac{5}{7} + \frac{1}{3}{/tex}

(e) {tex}\frac{2}{5} + \frac{1}{6}{/tex}

(f) {tex}\frac{4}{5} + \frac{2}{3}{/tex}

(g) {tex}\frac{3}{4} – \frac{1}{3}{/tex}

(h) {tex}\frac{5}{6} – \frac{1}{3}{/tex}

(i) {tex}\frac{2}{3} + \frac{3}{4} + \frac{1}{2}{/tex}

(j) {tex}\frac{1}{2} + \frac{1}{3} + \frac{1}{6}{/tex}

(k) {tex}1\frac{1}{3} + 3\frac{2}{3}{/tex}

(l) {tex}4\frac{2}{3} + 3\frac{1}{4}{/tex}

(m) {tex}\frac{{16}}{5} – \frac{7}{5}{/tex}

(n) {tex}\frac{4}{3} – \frac{1}{2}{/tex}

Answer:

(a) L.C.M. of 3 and 7 is 21

{tex}\therefore {/tex} {tex}\frac{2}{3} + \frac{1}{7} = \frac{{2 \times 7 + 1 \times 3}}{{21}} = \frac{{14 + 3}}{{21}} = \frac{{17}}{{21}}{/tex}

(b) L.C.M. of 10 and 15 is 30

{tex}\therefore {/tex} {tex}\frac{3}{{10}} + \frac{7}{{15}} = \frac{{3 \times 3 + 7 \times 2}}{{30}} = \frac{{9 + 14}}{{30}} = \frac{{23}}{{30}}{/tex}

(c) L.C.M. of 9 and 7 is 63

{tex}\therefore {/tex} {tex}\frac{4}{9} + \frac{2}{7} = \frac{{4 \times 7 + 2 \times 9}}{{63}} = \frac{{28 + 18}}{{63}} = \frac{{46}}{{63}}{/tex}

(d)L.C.M. of 7 and 3 is 21

{tex}\therefore {/tex} {tex}\frac{5}{7} + \frac{1}{3} = \frac{{5 \times 3 + 7 \times 1}}{{21}} = \frac{{15 + 7}}{{21}} = \frac{{22}}{{21}} = 1\frac{1}{{21}}{/tex}

(e)L.C.M. of 5 and 6 is 30

{tex}\therefore {/tex} {tex}\frac{2}{5} + \frac{1}{6} = \frac{{2 \times 6 + 5 \times 1}}{{30}} = \frac{{12 + 5}}{{30}} = \frac{{17}}{{30}}{/tex}

(f)L.C.M. of 5 and 3 is 15

{tex}\therefore {/tex} {tex}\frac{4}{5} + \frac{2}{3}{/tex} = {tex}\frac{{4 \times 3 + 2 \times 5}}{{15}} = \frac{{12 + 10}}{{15}} = \frac{{22}}{{15}} = 1\frac{7}{{15}}{/tex}

(g)L.C.M. of 4 and 3 is 12

{tex}\therefore {/tex} {tex}\frac{3}{4} – \frac{1}{3} = \frac{{3 \times 3 – 4 \times 1}}{{12}} = \frac{{9 – 4}}{{12}} = \frac{5}{{12}}{/tex}

(h)L.C.M. of 6 and 3 is 6

{tex}\therefore {/tex}

{tex}\frac{5}{6}-\frac{1}{3}=\frac{5\times 1-2\times 1}{6}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}{/tex}

(i)L.C.M. of 3, 4 and 2 is 12

{tex}\therefore {/tex} {tex}\frac{2}{3} + \frac{3}{4} + \frac{1}{2} = \frac{{2 \times 4 + 3 \times 3 + 1 \times 6}}{{12}} = \frac{{6 + 9 + 6}}{{12}} = \frac{{23}}{{12}} = 1\frac{{11}}{{12}}{/tex}

(j)L.C.M. of 2, 3, and 6 is 6

{tex}\therefore {/tex}

{tex}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1\times 3+1\times 2+1\times 1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1{/tex}

(k)L.C.M. of 3 and 3 is 3

{tex}\therefore {/tex}

{tex}\frac{4}{3}+\frac{11}{3}=\frac{4+11}{3}=\frac{15}{3}=5{/tex}

(l)L.C.M. of 3 and 4 is 12

{tex}\therefore {/tex} {tex}\frac{{14}}{3} + \frac{{13}}{4} = \frac{{14 \times 4 + 13 \times 3}}{{12}} = \frac{{56 + 39}}{{12}} = \frac{{95}}{{12}} = 7\frac{{11}}{{12}}{/tex}

(m)L.C.M. of 5 and 5 is 5

{tex}\therefore {/tex} {tex}\frac{{16}}{5} – \frac{7}{5} = \frac{{16 – 7}}{5} = \frac{9}{5} = 1\frac{4}{5}{/tex}

(n)L.C.M. of 3 and 2 is 6

{tex}\therefore {/tex} {tex}\frac{4}{3} – \frac{1}{2} = \frac{{4 \times 2 – 1 \times 3}}{6} = \frac{{8 – 3}}{6} = \frac{5}{6}{/tex}

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NCERT Solutions for Class 6 Maths Exercise 7.6

Question 2.Sarika bought {tex}\frac{2}{5}{/tex} meter of ribbon and Lalita {tex}\frac{3}{4}{/tex} meter of ribbon. What is the total length of the ribbon they bought?

Answer:

Ribbon bought by Sarita = {tex}\frac{2}{5}{/tex} m and Ribbon bought by Lalita = {tex}\frac{3}{4}{/tex} m

Total length of ribbon = {tex}\frac{2}{5} + \frac{3}{4}{/tex} = {tex}\frac{{2 \times 4 + 5 \times 3}}{{20}}{/tex} [{tex}\because {/tex} L.C.M. of 5 and 4 is 20]

= {tex}\frac{{8 + 15}}{{20}} = \frac{{23}}{{20}} = 1\frac{3}{{20}}{/tex} m

Therefore, they bought {tex}1\frac{3}{{20}}{/tex} m of ribbon.

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NCERT Solutions for Class 6 Maths Exercise 7.6

Question 3.Naina was given {tex}1\frac{1}{2}{/tex} piece of cake and Najma was given {tex}1\frac{1}{3}{/tex} piece of cake. Find the total amount of cake given to both of them.

Answer:

Cake taken by Naina = {tex}1\frac{1}{2}{/tex} piece and Cake taken by Najma = {tex}1\frac{1}{3}{/tex} piece

Total cake taken = {tex}1\frac{1}{2}{/tex} + {tex}1\frac{1}{3}{/tex} = {tex}\frac{3}{2} + \frac{4}{3}{/tex} = {tex}\frac{{3 \times 3 + 4 \times 2}}{6}{/tex} [{tex}\because {/tex} L.C.M. of 2 and 3 is 6]

= {tex}\frac{{9 + 8}}{6} = \frac{{17}}{6} = 2\frac{5}{6}{/tex}

Therefore total consumption of cake is {tex}2\frac{5}{6}{/tex}.

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Question 4.Fill in the boxes:

(a) {tex}\boxed{{\text{ }}} – \frac{5}{8} = \frac{1}{4}{/tex}

(b) {tex}\boxed{{\text{ }}} – \frac{1}{5} = \frac{1}{2}{/tex}

(c) {tex}\frac{1}{2} – \boxed{{\text{ }}} = \frac{1}{6}{/tex}

Answer:

(a) {tex}\frac{1}{4} + \frac{5}{8} = \frac{{2 + 5}}{8} = \frac{7}{8}{/tex}

(b) {tex}\frac{1}{2} + \frac{1}{5} = \frac{{5 + 2}}{{10}} = \frac{7}{{10}}{/tex}

(c) {tex}\frac{1}{2} – \frac{1}{6} = \frac{{3 – 1}}{6} = \frac{2}{6}{/tex}

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NCERT Solutions for Class 6 Maths Exercise 7.6

Question 5.Complete the addition – subtraction box:

Answer:

Sol.

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Question 6.A piece of wire {tex}\frac{7}{8}{/tex} meter long broke into two pieces. One piece was {tex}\frac{1}{4}{/tex} meter long. How long is the other piece?

Answer:

Total length of wire = {tex}\frac{7}{8}{/tex} meter

Length of first part = {tex}\frac{1}{4}{/tex} meter

Remaining part = {tex}\frac{7}{8} – \frac{1}{4} = \frac{{7 \times 1 – 2 \times 1}}{8}{/tex} [{tex}\because {/tex} L.C.M. of 8 and 4 is 8]

= {tex}\frac{{7 – 2}}{8} = \frac{5}{8}{/tex} meter

Therefore, the length of remaining part is {tex}\frac{5}{8}{/tex} meter.

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NCERT Solutions for Class 6 Maths Exercise 7.6

Question 7.Nandini house is {tex}\frac{9}{{10}}{/tex} km from her school. She walked some distance and then took a bus for {tex}\frac{1}{2}{/tex} km to reach the school. How far did she walk?

Answer:

Total distance between school and house = {tex}\frac{9}{{10}}{/tex} km

Distance covered by bus = {tex}\frac{1}{2}{/tex} km

Remaining distance = {tex}\frac{9}{{10}} – \frac{1}{2} = \frac{{9 \times 1 – 1 \times 5}}{{10}}{/tex} [{tex}\because {/tex} L.C.M. of 10 and 2 is 10]

= {tex}\frac{9-5}{10}=\frac{4}{10}=\frac{2}{5}{/tex} km

Therefore, distance covered by walking us {tex}\frac{2}{5}{/tex} km.

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Question 8.Ahsa and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is {tex}\frac{5}{6}th{/tex} full and Samuel’s shelf is {tex}\frac{2}{5}th{/tex} full. Whose bookshelf is more full? By what fraction?

Answer:

{tex}\frac{5}{6}{/tex} and {tex}\frac{2}{5}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{5}{6} \times \frac{5}{5} = \frac{{25}}{{30}}{/tex} and {tex}\frac{2}{5} \times \frac{6}{6} = \frac{{12}}{{30}}{/tex} [{tex}\because {/tex} L.C.M. of 6 and 5 is 30]

{tex}\because {/tex} {tex}\frac{{25}}{{30}} > \frac{{12}}{{30}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{5}{6} > \frac{2}{5}{/tex}

{tex}\therefore {/tex} Asha’s bookshelf is more covered than Samueal.

Difference = {tex}\frac{{25}}{{30}} – \frac{{12}}{{30}} = \frac{{13}}{{30}}{/tex}

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NCERT Solutions for Class 6 Maths Exercise 7.6

Question 9.Jaidev takes {tex}2\frac{1}{5}{/tex} minutes to walk across the school ground. Rahul takes {tex}\frac{7}{4}{/tex} minutes to do same. Who takes less time and by what fraction?

Answer:

Time taken by jaidev = {tex}2\frac{1}{5}{/tex} minutes = {tex}\frac{{11}}{5}{/tex} minutes

Time taken by Rahul = {tex}\frac{7}{4}{/tex} minutes

Difference = {tex}\frac{{11}}{5} – \frac{7}{4}{/tex} = {tex}\frac{{11 \times 4 – 7 \times 5}}{{20}}{/tex} [{tex}\because {/tex} L.C.M. of 5 and 4 is 20]

= {tex}\frac{{44 – 35}}{{20}} = \frac{9}{{20}}{/tex} minutes

Thus, Rahul takes less time, which is {tex}\frac{9}{{20}}{/tex} minutes.

NCERT Solutions for Class 6 Maths Exercise 7.6

NCERT Solutions Class 6 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 6 Maths includes text book solutions from Class 6 Maths Book . NCERT Solutions for CBSE Class 6 Maths have total 14 chapters. 6 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 6 solutions PDF and Maths ncert class 6 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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