Rational Numbers Ex-9.2

NCERT solutions for Maths Rational Numbers 

NCERT Solutions for Class 7 Maths Rational Numbers

Class –VII Mathematics (Ex. 9.2)

Question 1.Find the sum:

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

(ii) $\frac{5}{3} + \frac{3}{5}$

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

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

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

(vi) $\frac{{ – 2}}{3} + 0$

(vii) $– 2\frac{1}{3} + 4\frac{3}{5}$

Answer:

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

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

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

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

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

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

= $\frac{{ – 27 + 44}}{{30}} = \frac{{17}}{{30}}$

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

= $\frac{{27 + 55}}{{99}} = \frac{{82}}{{99}}$

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

= $\frac{{ – 24 – 2}}{{57}}$ = $\frac{{ – 26}}{{57}}$

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.2

Question 2.Find:

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

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

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

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

(v) $– 2\frac{1}{9} – 6$

Answer:

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

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

= $\frac{{21 – 34}}{{72}}$ = $\frac{{ – 13}}{{72}}$

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

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

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

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

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

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

= $\frac{{ – 33 – 56}}{{88}}$ = $\frac{{ – 89}}{{88}} = – 1\frac{1}{{88}}$

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.2

Question 3.Find the product:

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

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

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

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

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

(vi) $\frac{3}{{ – 5}} \times \frac{5}{3}$

Answer:

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

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.2

Question 4.Find the value of:

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

(ii) $\frac{{ – 3}}{5} \div 2$

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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