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# NCERT Solutions for Class 7 Maths Exercise 9.2

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NCERT solutions for Maths Rational Numbers

## NCERT Solutions for Class 7 Maths Rational Numbers

###### Question 1.Find the sum:

(i) {tex}\frac{5}{4} + \left( {\frac{{ – 11}}{4}} \right){/tex}

(ii) {tex}\frac{5}{3} + \frac{3}{5}{/tex}

(iii) {tex}\frac{{ – 9}}{{10}} + \frac{{22}}{{15}}{/tex}

(iv) {tex}\frac{{ – 3}}{{ – 11}} + \frac{5}{9}{/tex}

(v) {tex}\frac{{ – 8}}{{19}} + \frac{{\left( { – 2} \right)}}{{57}}{/tex}

(vi) {tex}\frac{{ – 2}}{3} + 0{/tex}

(vii) {tex} – 2\frac{1}{3} + 4\frac{3}{5}{/tex}

(i) {tex}\frac{5}{4} + \left( {\frac{{ – 11}}{4}} \right){/tex} = {tex}\frac{{5 – 11}}{4}{/tex} = {tex}\frac{{ – 6}}{4} = \frac{{ – 3}}{2}{/tex}

(ii) {tex}\frac{5}{3} + \frac{3}{5}{/tex} = {tex}\frac{{5 \times 5}}{{3 \times 5}} + \frac{{3 \times 3}}{{5 \times 3}}{/tex} = {tex}\frac{{25}}{{15}} + \frac{9}{{15}}{/tex}

[L.C.M. of 3 and 5 is 15]

= {tex}\frac{{25 + 9}}{{15}} = \frac{{34}}{{15}} = 2\frac{4}{{15}}{/tex}

(iii) {tex}\frac{{ – 9}}{{10}} + \frac{{22}}{{15}}{/tex} = {tex}\frac{{ – 9 \times 3}}{{10 \times 3}} + \frac{{22 \times 2}}{{15 \times 2}}{/tex} = {tex}\frac{{ – 27}}{{30}} + \frac{{44}}{{30}}{/tex}

[L.C.M. of 10 and 15 is 30]

= {tex}\frac{{ – 27 + 44}}{{30}} = \frac{{17}}{{30}}{/tex}

(iv) {tex}\frac{{ – 3}}{{ – 11}} + \frac{5}{9}{/tex} = {tex}\frac{{ – 3 \times 9}}{{ – 11 \times 9}} + \frac{{5 \times 11}}{{9 \times 11}}{/tex} = {tex}\frac{{27}}{{99}} + \frac{{55}}{{99}}{/tex} [L.C.M. of 11 and 9 is 99]

= {tex}\frac{{27 + 55}}{{99}} = \frac{{82}}{{99}}{/tex}

(v) {tex}\frac{{ – 8}}{{19}} + \frac{{\left( { – 2} \right)}}{{57}}{/tex} = {tex}\frac{{ – 8 \times 3}}{{19 \times 3}} + \frac{{\left( { – 2} \right) \times 1}}{{57 \times 1}}{/tex} = {tex}\frac{{ – 24}}{{57}} + \frac{{\left( { – 2} \right)}}{{57}}{/tex} [L.C.M. of 19 and 57 is 57]

= {tex}\frac{{ – 24 – 2}}{{57}}{/tex} = {tex}\frac{{ – 26}}{{57}}{/tex}

(vi) {tex}\frac{{ – 2}}{3} + 0 = \frac{{ – 2}}{3}{/tex}

(vii) {tex} – 2\frac{1}{3} + 4\frac{3}{5}{/tex} = {tex}\frac{{ – 7}}{3} + \frac{{23}}{5}{/tex} = {tex}\frac{{ – 7 \times 5}}{{3 \times 5}} + \frac{{23 \times 3}}{{5 \times 3}}{/tex} = {tex}\frac{{ – 35}}{{15}} + \frac{{69}}{{15}}{/tex} [L.C.M. of 3 and 5 is 15]

= {tex}\frac{{ – 35 + 69}}{{15}}{/tex} = {tex}\frac{{34}}{{15}} = 2\frac{4}{{15}}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.2

###### Question 2.Find:

(i) {tex}\frac{7}{{24}} – \frac{{17}}{{36}}{/tex}

(ii) {tex}\frac{5}{{63}} – \left( {\frac{{ – 6}}{{21}}} \right){/tex}

(iii) {tex}\frac{{ – 6}}{{13}} – \left( {\frac{{ – 7}}{{15}}} \right){/tex}

(iv) {tex}\frac{{ – 3}}{8} – \frac{7}{{11}}{/tex}

(v) {tex} – 2\frac{1}{9} – 6{/tex}

(i) {tex}\frac{7}{{24}} – \frac{{17}}{{36}}{/tex} = {tex}\frac{{7 \times 3}}{{24 \times 3}} – \frac{{17 \times 2}}{{36 \times 2}}{/tex} = {tex}\frac{{21}}{{72}} – \frac{{34}}{{72}}{/tex}

[L.C.M. of 24 and 36 is 72]

= {tex}\frac{{21 – 34}}{{72}}{/tex} = {tex}\frac{{ – 13}}{{72}}{/tex}

(ii) {tex}\frac{5}{{63}} – \left( {\frac{{ – 6}}{{21}}} \right){/tex} = {tex}\frac{{5 \times 1}}{{63 \times 1}} – \left( {\frac{{ – 6 \times 3}}{{21 \times 3}}} \right){/tex} = {tex}\frac{5}{{63}} – \frac{{ – 18}}{{63}}{/tex} [L.C.M. of 63 and 21 is 63]

= {tex}\frac{{5 – \left( { – 18} \right)}}{{63}}{/tex} = {tex}\frac{{5 + 18}}{{63}} = \frac{{23}}{{63}}{/tex}

(iii) {tex}\frac{{ – 6}}{{13}} – \left( {\frac{{ – 7}}{{15}}} \right){/tex} = {tex}\frac{{ – 6 \times 15}}{{13 \times 15}} – \left( {\frac{{ – 7 \times 13}}{{15 \times 13}}} \right){/tex} = {tex}\frac{{ – 90}}{{195}} – \left( {\frac{{ – 91}}{{195}}} \right){/tex} [L.C.M. of 13 and 15 is 195]

= {tex}\frac{{ – 90 – \left( { – 91} \right)}}{{195}}{/tex} = {tex}\frac{{ – 90 + 91}}{{195}} = \frac{1}{{195}}{/tex}

(iv) {tex}\frac{{ – 3}}{8} – \frac{7}{{11}}{/tex} = {tex}\frac{{ – 3 \times 11}}{{8 \times 11}} – \frac{{7 \times 8}}{{11 \times 8}}{/tex} = {tex}\frac{{ – 33}}{{88}} – \frac{{56}}{{88}}{/tex}

[L.C.M. of 8 and 11 is 88]

= {tex}\frac{{ – 33 – 56}}{{88}}{/tex} = {tex}\frac{{ – 89}}{{88}} = – 1\frac{1}{{88}}{/tex}

(v) {tex} – 2\frac{1}{9} – 6{/tex} = {tex}\frac{{ – 19}}{9} – \frac{6}{1}{/tex} = {tex}\frac{{ – 19 \times 1}}{{9 \times 1}} – \frac{{6 \times 9}}{{1 \times 9}}{/tex} [L.C.M. of 9 and 1 is 9]

= {tex}\frac{{ – 19}}{9} – \frac{{54}}{9}{/tex} = {tex}\frac{{ – 19 – 54}}{9}{/tex} = {tex}\frac{{ – 73}}{9} = – 8\frac{1}{9}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.2

###### Question 3.Find the product:

(i) {tex}\frac{9}{2} \times \left( {\frac{{ – 7}}{4}} \right){/tex}

(ii) {tex}\frac{3}{{10}} \times \left( { – 9} \right){/tex}

(iii) {tex}\frac{{ – 6}}{5} \times \frac{9}{{11}}{/tex}

(iv) {tex}\frac{3}{7} \times \left( {\frac{{ – 2}}{5}} \right){/tex}

(v) {tex}\frac{3}{{11}} \times \frac{2}{5}{/tex}

(vi) {tex}\frac{3}{{ – 5}} \times \frac{5}{3}{/tex}

(i) {tex}\frac{9}{2} \times \left( {\frac{{ – 7}}{4}} \right){/tex} = {tex}\frac{{9 \times \left( { – 7} \right)}}{{2 \times 4}}{/tex} = {tex}\frac{{ – 63}}{8} = – 7\frac{7}{8}{/tex}

(ii){tex}\frac{3}{{10}} \times \left( { – 9} \right) = \frac{{3 \times \left( { – 9} \right)}}{{10}} = \frac{{ – 27}}{{10}} = – 2\frac{7}{{10}}{/tex}

(iii) {tex}\frac{{ – 6}}{5} \times \frac{9}{{11}} = \frac{{\left( { – 6} \right) \times 9}}{{5 \times 11}} = \frac{{ – 54}}{{55}}{/tex}

(iv) {tex}\frac{3}{7} \times \left( {\frac{{ – 2}}{5}} \right) = \frac{{3 \times \left( { – 2} \right)}}{{7 \times 5}} = \frac{{ – 6}}{{35}}{/tex}

(v) {tex}\frac{3}{{11}} \times \frac{2}{5} = \frac{{3 \times 2}}{{11 \times 5}} = \frac{6}{{55}}{/tex}

(vi) {tex}\frac{3}{{ – 5}} \times \left( {\frac{{ – 5}}{3}} \right) = \frac{{3 \times \left( { – 5} \right)}}{{ – 5 \times 3}} = 1{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.2

###### Question 4.Find the value of:

(i) {tex}\left( { – 4} \right) \div \frac{2}{3}{/tex}

(ii) {tex}\frac{{ – 3}}{5} \div 2{/tex}

(iii) {tex}\frac{{ – 4}}{5} \div \left( { – 3} \right){/tex}

(iv) {tex}\frac{{ – 1}}{8} \div \frac{3}{4}{/tex}

(v) {tex}\frac{{ – 2}}{{13}} \div \frac{1}{7}{/tex}

(vi) {tex}\frac{{ – 7}}{{12}} \div \left( {\frac{2}{{13}}} \right){/tex}

(vii) {tex}\frac{3}{{13}} \div \left( {\frac{{ – 4}}{{65}}} \right){/tex}

(i) {tex}\left( { – 4} \right) \div \frac{2}{3}{/tex} = {tex}\left( { – 4} \right) \times \frac{3}{2} = \left( { – 2} \right) \times 3 = – 6{/tex}

(ii) {tex}\frac{{ – 3}}{5} \div 2{/tex} = {tex}\frac{{ – 3}}{5} \times \frac{1}{2} = \frac{{\left( { – 3} \right) \times 1}}{{5 \times 2}} = \frac{{ – 3}}{{10}}{/tex}

(iii) {tex}\frac{{ – 4}}{5} \div \left( { – 3} \right){/tex} = {tex}\frac{{\left( { – 4} \right)}}{5} \times \frac{1}{{\left( { – 3} \right)}} = \frac{{\left( { – 4} \right) \times 1}}{{5 \times \left( { – 3} \right)}} = \frac{4}{{15}}{/tex}

(iv) {tex}\frac{{ – 1}}{8} \div \frac{3}{4}{/tex} = {tex}\frac{{ – 1}}{8} \times \frac{4}{3}{/tex} = {tex}\frac{{\left( { – 1} \right) \times 1}}{{2 \times 3}} = \frac{{ – 1}}{6}{/tex}

(v) {tex}\frac{{ – 2}}{{13}} \div \frac{1}{7}{/tex} = {tex}\frac{{ – 2}}{{13}} \times \frac{7}{1} = \frac{{\left( { – 2} \right) \times 7}}{{13 \times 1}} = \frac{{ – 14}}{{13}} = – 1\frac{1}{{13}}{/tex}

(vi) {tex}\frac{{ – 7}}{{12}} \div \left( {\frac{{ – 2}}{{13}}} \right){/tex} = {tex}\frac{{ – 7}}{{12}} \times \frac{{13}}{{\left( { – 2} \right)}}{/tex} = {tex}\frac{{\left( { – 7} \right) \times 13}}{{12 \times \left( { – 2} \right)}} = \frac{{ – 91}}{{24}} = 3\frac{{19}}{{24}}{/tex}

(vii) {tex}\frac{3}{{13}} \div \left( {\frac{{ – 4}}{{65}}} \right){/tex} = {tex}\frac{3}{{13}} \times \frac{{65}}{{\left( { – 4} \right)}} = \frac{{3 \times \left( { – 5} \right)}}{{1 \times 4}} = \frac{{ – 15}}{4} = – 3\frac{3}{4}{/tex}

## NCERT Solutions for Class 7 Maths Exercise 9.2

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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