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# NCERT Solutions for Class 7 Maths Exercise 9.1

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NCERT solutions for Maths Rational Numbers

## NCERT Solutions for Class 7 Maths Rational Numbers

###### Question 1.List five rational numbers between:

(i) {tex} – 1{/tex} and 0

(ii) {tex} – 2{/tex} and {tex} – 1{/tex}

(iii) {tex}\frac{{ – 4}}{5}{/tex} and {tex}\frac{{ – 2}}{3}{/tex}

(iv) {tex}\frac{{ – 1}}{2}{/tex} and {tex}\frac{2}{3}{/tex}

(i) {tex} – 1{/tex} and 0

Let us write {tex} – 1{/tex} and 0 as rational numbers with denominator 6.

{tex} \Rightarrow {/tex} {tex} – 1 = \frac{{ – 6}}{6}{/tex} and 0 = {tex}\frac{0}{6}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 6}}{6} < \frac{{ – 5}}{6} < \frac{{ – 4}}{6} < \frac{{ – 3}}{6} < \frac{{ – 2}}{6} < \frac{{ – 1}}{6} < 0{/tex}

{tex} \Rightarrow {/tex} {tex} – 1 < \frac{{ – 5}}{6} < \frac{{ – 2}}{3} < \frac{{ – 1}}{2} < \frac{{ – 1}}{3} < \frac{{ – 1}}{6} < 0{/tex}

Therefore, five rational numbers between {tex} – 1{/tex} and 0 would be

{tex}\frac{{ – 5}}{6},\frac{{ – 2}}{3},\frac{{ – 1}}{2},\frac{{ – 1}}{3},\frac{{ – 1}}{6}{/tex}

(ii) {tex} – 2{/tex} and {tex} – 1{/tex}

Let us write {tex} – 2{/tex} and {tex} – 1{/tex} as rational numbers with denominator 6.

{tex} \Rightarrow {/tex} {tex} – 2 = \frac{{ – 12}}{6}{/tex} and {tex} – 1 = \frac{{ – 6}}{6}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 12}}{6} < \frac{{ – 11}}{6} < \frac{{ – 10}}{6} < \frac{{ – 9}}{6} < \frac{{ – 8}}{6} < \frac{{ – 7}}{6} < \frac{{ – 6}}{6}{/tex}

{tex} \Rightarrow {/tex} {tex} – 2 < \frac{{ – 11}}{6} < \frac{{ – 5}}{3} < \frac{{ – 3}}{2} < \frac{{ – 4}}{3} < \frac{{ – 7}}{6} < – 1{/tex}

Therefore, five rational numbers between {tex} – 2{/tex} and {tex} – 1{/tex} would be

{tex}\frac{{ – 11}}{6},\frac{{ – 5}}{3},\frac{{ – 3}}{2},\frac{{ – 4}}{3},\frac{{ – 7}}{6}{/tex}

(iii) {tex}\frac{{ – 4}}{5}{/tex} and {tex}\frac{{ – 2}}{3}{/tex}

Let us write {tex}\frac{{ – 4}}{5}{/tex} and {tex}\frac{{ – 2}}{3}{/tex} as rational numbers with the same denominators.

{tex} \Rightarrow {/tex} {tex}\frac{{ – 4}}{5} = \frac{{ – 36}}{{45}}{/tex} and {tex}\frac{{ – 2}}{3} = \frac{{ – 30}}{{45}}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 36}}{{45}} < \frac{{ – 35}}{{45}} < \frac{{ – 34}}{{45}} < \frac{{ – 33}}{{45}} < \frac{{ – 32}}{{45}} < \frac{{ – 31}}{{45}} < \frac{{ – 30}}{{45}}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 4}}{5} < \frac{{ – 7}}{9} < \frac{{ – 34}}{{45}} < \frac{{ – 11}}{{15}} < \frac{{ – 32}}{{45}} < \frac{{ – 31}}{{45}} < \frac{{ – 2}}{3}{/tex}

Therefore, five rational numbers between {tex}\frac{{ – 4}}{5}{/tex} and {tex}\frac{{ – 2}}{3}{/tex} would be

{tex}\frac{{ – 7}}{9},\frac{{ – 34}}{{45}},\frac{{ – 11}}{{15}},\frac{{ – 32}}{{45}},\frac{{ – 31}}{{45}},\frac{{ – 2}}{3}{/tex}

(iv) {tex}\frac{{ – 1}}{2}{/tex} and {tex}\frac{2}{3}{/tex}

Let us write {tex}\frac{{ – 1}}{2}{/tex} and {tex}\frac{2}{3}{/tex} as rational numbers with the same denominators.

{tex} \Rightarrow {/tex} {tex}\frac{{ – 1}}{2} = \frac{{ – 3}}{6}{/tex} and {tex}\frac{2}{3} = \frac{4}{6}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 3}}{6} < \frac{{ – 2}}{6} < \frac{{ – 1}}{6} < 0 < \frac{1}{6} < \frac{2}{6} < \frac{3}{6} < \frac{4}{6}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 1}}{2} < \frac{{ – 1}}{3} < \frac{{ – 1}}{6} < 0 < \frac{1}{6} < \frac{1}{3} < \frac{1}{2} < \frac{2}{3}{/tex}

Therefore, five rational numbers between {tex}\frac{{ – 1}}{2}{/tex} and {tex}\frac{2}{3}{/tex} would be {tex}\frac{{ – 1}}{3},\frac{{ – 1}}{6},0,\frac{1}{6},\frac{1}{3}{/tex}.

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 2.Write four more rational numbers in each of the following patterns:

(i) {tex}\frac{{ – 3}}{5},\frac{{ – 6}}{{10}},\frac{{ – 9}}{{15}},\frac{{ – 12}}{{20}},………{/tex}

(ii) {tex}\frac{{ – 1}}{4},\frac{{ – 2}}{8},\frac{{ – 3}}{{12}},……….{/tex}

(iii) {tex}\frac{{ – 1}}{6},\frac{2}{{ – 12}},\frac{3}{{ – 18}},\frac{4}{{ – 24}},………{/tex}

(iv) {tex}\frac{{ – 2}}{3},\frac{2}{{ – 3}},\frac{4}{{ – 6}},\frac{6}{{ – 9}},……….{/tex}

(i) {tex}\frac{{ – 3}}{5},\frac{{ – 6}}{{10}},\frac{{ – 9}}{{15}},\frac{{ – 12}}{{20}},………{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 3 \times 1}}{{5 \times 1}},\frac{{ – 3 \times 2}}{{5 \times 2}},\frac{{ – 3 \times 3}}{{5 \times 3}},\frac{{ – 3 \times 4}}{{5 \times 4}},………{/tex}

Therefore, the next four rational numbers of this pattern would be

{tex}\frac{{ – 3 \times 5}}{{5 \times 5}},\frac{{ – 3 \times 6}}{{5 \times 6}},\frac{{ – 3 \times 7}}{{5 \times 7}},\frac{{ – 3 \times 8}}{{5 \times 8}}{/tex} = {tex}\frac{{ – 15}}{{25}},\frac{{ – 18}}{{30}},\frac{{ – 21}}{{35}},\frac{{ – 24}}{{40}}{/tex}

(ii) {tex}\frac{{ – 1}}{4},\frac{{ – 2}}{8},\frac{{ – 3}}{{12}},……….{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 1 \times 1}}{{4 \times 1}},\frac{{ – 1 \times 2}}{{4 \times 2}},\frac{{ – 1 \times 3}}{{4 \times 3}},……….{/tex}

Therefore, the next four rational numbers of this pattern would be

{tex}\frac{{ – 1 \times 4}}{{4 \times 4}},\frac{{ – 1 \times 5}}{{4 \times 5}},\frac{{ – 1 \times 6}}{{4 \times 6}},\frac{{ – 1 \times 7}}{{4 \times 7}}{/tex} = {tex}\frac{{ – 4}}{{16}},\frac{{ – 5}}{{20}},\frac{{ – 6}}{{24}},\frac{{ – 7}}{{28}}{/tex}

(iii) {tex}\frac{{ – 1}}{6},\frac{2}{{ – 12}},\frac{3}{{ – 18}},\frac{4}{{ – 24}},………{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 1 \times 1}}{{6 \times 1}},\frac{{1 \times 2}}{{ – 6 \times 2}},\frac{{1 \times 3}}{{ – 6 \times 3}},\frac{{1 \times 4}}{{ – 6 \times 4}},………{/tex}

Therefore, the next four rational numbers of this pattern would be

{tex}\frac{{1 \times 5}}{{ – 6 \times 5}},\frac{{1 \times 6}}{{ – 6 \times 6}},\frac{{1 \times 7}}{{ – 6 \times 7}},\frac{{1 \times 8}}{{ – 6 \times 8}}{/tex} = {tex}\frac{5}{{ – 30}},\frac{6}{{ – 36}},\frac{7}{{ – 42}},\frac{8}{{ – 48}}{/tex}

(iv) {tex}\frac{{ – 2}}{3},\frac{2}{{ – 3}},\frac{4}{{ – 6}},\frac{6}{{ – 9}},……….{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 2 \times 1}}{{3 \times 1}},\frac{{2 \times 1}}{{ – 3 \times 1}},\frac{{2 \times 2}}{{ – 3 \times 2}},\frac{{2 \times 3}}{{ – 3 \times 3}},……….{/tex}

Therefore, the next four rational numbers of this pattern would be

{tex}\frac{{2 \times 4}}{{ – 3 \times 4}},\frac{{2 \times 5}}{{ – 3 \times 5}},\frac{{2 \times 6}}{{ – 3 \times 6}},\frac{{2 \times 7}}{{ – 3 \times 7}}{/tex} = {tex}\frac{8}{{ – 12}},\frac{{10}}{{ – 15}},\frac{{12}}{{ – 18}},\frac{{14}}{{ – 21}}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 3.Give four rational numbers equivalent to:

(i) {tex}\frac{{ – 2}}{7}{/tex}

(ii) {tex}\frac{5}{{ – 3}}{/tex}

(iii) {tex}\frac{4}{9}{/tex}

(i) {tex}\frac{{ – 2}}{7}{/tex}

{tex}\frac{{ – 2 \times 2}}{{7 \times 2}} = \frac{{ – 4}}{{14}},{/tex}{tex}\frac{{ – 2 \times 3}}{{7 \times 3}} = \frac{{ – 6}}{{21}},{/tex}{tex}\frac{{ – 2 \times 4}}{{7 \times 4}} = \frac{{ – 8}}{{28}},{/tex}{tex}\frac{{ – 2 \times 5}}{{7 \times 5}} = \frac{{ – 10}}{{35}}{/tex}

Therefore, four equivalent rational numbers are {tex}\frac{{ – 4}}{{14}},\frac{{ – 6}}{{21}},\frac{{ – 8}}{{28}},\frac{{ – 10}}{{35}}{/tex}.

(ii) {tex}\frac{5}{{ – 3}}{/tex}

{tex}\frac{{5 \times 2}}{{ – 3 \times 2}} = \frac{{10}}{{ – 6}},{/tex}{tex}\frac{{5 \times 3}}{{ – 3 \times 3}} = \frac{{15}}{{ – 9}},{/tex}{tex}\frac{{5 \times 4}}{{ – 3 \times 4}} = \frac{{20}}{{ – 12}},{/tex}{tex}\frac{{5 \times 5}}{{ – 3 \times 5}} = \frac{{25}}{{ – 15}}{/tex}

Therefore, four equivalent rational numbers are {tex}\frac{{10}}{{ – 6}},\frac{{15}}{{ – 9}},\frac{{20}}{{ – 12}},\frac{{25}}{{ – 15}}{/tex}.

(iii) {tex}\frac{4}{9}{/tex}

{tex}\frac{{4 \times 2}}{{9 \times 2}} = \frac{8}{{18}},{/tex}{tex}\frac{{4 \times 3}}{{9 \times 3}} = \frac{{12}}{{27}},{/tex}{tex}\frac{{4 \times 4}}{{9 \times 4}} = \frac{{16}}{{36}},{/tex}{tex}\frac{{4 \times 5}}{{9 \times 5}} = \frac{{20}}{{45}}{/tex}

Therefore, four equivalent rational numbers are {tex}\frac{8}{{18}},\frac{{12}}{{27}},\frac{{16}}{{36}},\frac{{20}}{{45}}{/tex}.

###### Question 4.Draw the number line and represent the following rational numbers on it:

(i) {tex}\frac{3}{4}{/tex}

(ii) {tex}\frac{{ – 5}}{8}{/tex}

(iii) {tex}\frac{{ – 7}}{4}{/tex}

(iv) {tex}\frac{7}{8}{/tex}

(i) {tex}\frac{3}{4}{/tex}

(ii) {tex}\frac{{ – 5}}{8}{/tex}

(iii) {tex}\frac{{ – 7}}{4}{/tex}

(iv) {tex}\frac{7}{8}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 5.The points P, Q, R, S, T, U, A and B on the number line are such that, TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.

Each part which is between the two numbers is divided into 3 parts.

Therefore, A = {tex}\frac{6}{3},{/tex} P = {tex}\frac{7}{3},{/tex} Q = {tex}\frac{8}{3}{/tex} and B = {tex}\frac{9}{3}{/tex}

Similarly T = {tex}\frac{{ – 3}}{3},{/tex} R = {tex}\frac{{ – 4}}{3},{/tex} S = {tex}\frac{{ – 5}}{3}{/tex} and U = {tex}\frac{{ – 6}}{3}{/tex}

Thus, the rational numbers represented P, Q, R and S are {tex}\frac{7}{3},\frac{8}{3},\frac{{ – 4}}{3}{/tex} and {tex}\frac{{ – 5}}{3}{/tex} respectively.

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 6.Which of the following pairs represent the same rational numbers:

(i) {tex}\frac{{ – 7}}{{21}}{/tex} and {tex}\frac{3}{9}{/tex}

(ii) {tex}\frac{{ – 16}}{{20}}{/tex} and {tex}\frac{{20}}{{ – 25}}{/tex}

(iii) {tex}\frac{{ – 2}}{{ – 3}}{/tex} and {tex}\frac{2}{3}{/tex}

(iv) {tex}\frac{{ – 3}}{5}{/tex} and {tex}\frac{{ – 12}}{{20}}{/tex}

(v) {tex}\frac{8}{{ – 5}}{/tex} and {tex}\frac{{ – 24}}{{15}}{/tex}

(vi) {tex}\frac{1}{3}{/tex} and {tex}\frac{{ – 1}}{9}{/tex}

(vii) {tex}\frac{{ – 5}}{{ – 9}}{/tex} and {tex}\frac{5}{{ – 9}}{/tex}

(i) {tex}\frac{{ – 7}}{{21}}{/tex} and {tex}\frac{3}{9}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 7}}{{21}}{/tex} = {tex}\frac{{ – 1}}{3}{/tex} and {tex}\frac{3}{9}{/tex} = {tex}\frac{1}{3}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{{ – 1}}{3}{/tex}{tex} \ne {/tex}{tex}\frac{1}{3}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 7}}{{21}}{/tex}{tex} \ne {/tex}{tex}\frac{3}{9}{/tex}

(ii) {tex}\frac{{ – 16}}{{20}}{/tex} and {tex}\frac{{20}}{{ – 25}}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 16}}{{20}}{/tex} = {tex}\frac{{ – 4}}{5}{/tex} and {tex}\frac{{20}}{{ – 25}}{/tex} = {tex}\frac{4}{{ – 5}} = \frac{{ – 4}}{5}{/tex}

[Converting into lowest term]

{tex}\because {/tex} {tex}\frac{{ – 4}}{5}{/tex} = {tex}\frac{{ – 4}}{5}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 16}}{{20}}{/tex} = {tex}\frac{{20}}{{ – 25}}{/tex}

(iii) {tex}\frac{{ – 2}}{{ – 3}}{/tex} and {tex}\frac{2}{3}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 2}}{{ – 3}}{/tex} = {tex}\frac{2}{3}{/tex} and {tex}\frac{2}{3}{/tex} = {tex}\frac{2}{3}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{2}{3}{/tex} = {tex}\frac{2}{3}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 2}}{{ – 3}}{/tex} = {tex}\frac{2}{3}{/tex}

(iv) {tex}\frac{{ – 3}}{5}{/tex} and {tex}\frac{{ – 12}}{{20}}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 3}}{5}{/tex} = {tex}\frac{{ – 3}}{5}{/tex} and {tex}\frac{{ – 12}}{{20}}{/tex} = {tex}\frac{{ – 3}}{5}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{{ – 3}}{5}{/tex} = {tex}\frac{{ – 3}}{5}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 3}}{5}{/tex} = {tex}\frac{{ – 12}}{{20}}{/tex}

(v) {tex}\frac{8}{{ – 5}}{/tex} and {tex}\frac{{ – 24}}{{15}}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{8}{{ – 5}}{/tex} = {tex}\frac{{ – 8}}{5}{/tex} and {tex}\frac{{ – 24}}{{15}}{/tex} = {tex}\frac{{ – 8}}{5}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{{ – 8}}{5}{/tex} = {tex}\frac{{ – 8}}{5}{/tex}

{tex}\therefore {/tex} {tex}\frac{8}{{ – 5}}{/tex} = {tex}\frac{{ – 24}}{{15}}{/tex}

(vi) {tex}\frac{1}{3}{/tex} and {tex}\frac{{ – 1}}{9}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{1}{3}{/tex} = {tex}\frac{1}{3}{/tex} and {tex}\frac{{ – 1}}{9}{/tex} = {tex}\frac{{ – 1}}{9}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{1}{3}{/tex}{tex} \ne {/tex}{tex}\frac{{ – 1}}{9}{/tex}

{tex}\therefore {/tex} {tex}\frac{1}{3}{/tex}{tex} \ne {/tex}{tex}\frac{{ – 1}}{9}{/tex}

(vii) {tex}\frac{{ – 5}}{{ – 9}}{/tex} and {tex}\frac{5}{{ – 9}}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 5}}{{ – 9}}{/tex} = {tex}\frac{5}{9}{/tex} and {tex}\frac{5}{{ – 9}}{/tex} = {tex}\frac{5}{9}{/tex} [Converting into lowest term]

{tex}\because {/tex} {tex}\frac{5}{9}{/tex}{tex} \ne {/tex}{tex}\frac{5}{{ – 9}}{/tex}

{tex}\therefore {/tex} {tex}\frac{{ – 5}}{{ – 9}}{/tex}{tex} \ne {/tex}{tex}\frac{5}{{ – 9}}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 7.Rewrite the following rational numbers in the simplest form:

(i) {tex}\frac{{ – 8}}{6}{/tex}

(ii) {tex}\frac{{25}}{{45}}{/tex}

(iii) {tex}\frac{{ – 44}}{{72}}{/tex}

(iv) {tex}\frac{{ – 8}}{{10}}{/tex}

(i) {tex}\frac{{ – 8}}{6}{/tex} = {tex}\frac{{ – 8 \div 2}}{{6 \div 2}}{/tex} = {tex}\frac{{ – 4}}{3}{/tex} [H.C.F. of 8 and 6 is 2]

(ii){tex}\frac{{25}}{{45}}{/tex} = {tex}\frac{{25 \div 5}}{{45 \div 5}}{/tex} = {tex}\frac{5}{9}{/tex}

[H.C.F. of 25 and 45 is 5]

(iii){tex}\frac{{ – 44}}{{72}}{/tex} = {tex}\frac{{ – 44 \div 4}}{{72 \div 4}}{/tex} = {tex}\frac{{ – 11}}{{18}}{/tex} [H.C.F. of 44 and 72 is 4]

(iv) {tex}\frac{{ – 8}}{{10}}{/tex} = {tex}\frac{{ – 8 \div 2}}{{10 \div 2}}{/tex} = {tex}\frac{{ – 4}}{5}{/tex} [H.C.F. of 8 and 10 is 2]

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 8.Fill in the boxes with the correct symbol out of <, > and =:

(i) {tex}\frac{{ – 5}}{7}\boxed{{\text{ }}}\frac{2}{3}{/tex}

(ii) {tex}\frac{{ – 4}}{5}\boxed{{\text{ }}}\frac{{ – 5}}{7}{/tex}

(iii) {tex}\frac{{ – 7}}{8}\boxed{{\text{ }}}\frac{{14}}{{ – 16}}{/tex}

(iv) {tex}\frac{{ – 8}}{5}\boxed{{\text{ }}}\frac{{ – 7}}{4}{/tex}

(v) {tex}\frac{1}{{ – 3}}\boxed{{\text{ }}}\frac{{ – 1}}{4}{/tex}

(vi) {tex}\frac{5}{{ – 11}}\boxed{{\text{ }}}\frac{{ – 5}}{{11}}{/tex}

(vii) {tex}0\boxed{{\text{ }}}\frac{{ – 7}}{6}{/tex}

(i) {tex}\frac{{ – 5}}{7}\boxed{{\text{ < }}}\frac{2}{3}{/tex} Since, the positive number if greater than negative number.

(ii) {tex}\frac{{ – 4 \times 7}}{{5 \times 7}}\boxed{{\text{ }}}\frac{{ – 5 \times 5}}{{7 \times 5}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 28}}{{35}}\boxed{{\text{ < }}}\frac{{ – 25}}{{35}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 4}}{5}\boxed{{\text{ < }}}\frac{{ – 5}}{7}{/tex}

(iii) {tex}\frac{{ – 7 \times 2}}{{8 \times 2}}\boxed{{\text{ }}}\frac{{14 \times \left( { – 1} \right)}}{{ – 16 \times \left( { – 1} \right)}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 14}}{{16}}\boxed{{\text{ = }}}\frac{{ – 14}}{{16}}{/tex} {tex} \Rightarrow {/tex}{tex}\frac{{ – 7}}{8}\boxed{{\text{ = }}}\frac{{14}}{{ – 16}}{/tex}

(iv) {tex}\frac{{ – 8 \times 4}}{{5 \times 4}}\boxed{{\text{ }}}\frac{{ – 7 \times 5}}{{4 \times 5}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 32}}{{20}}\boxed{{\text{ > }}}\frac{{ – 35}}{{20}}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 8}}{5}\boxed{{\text{ > }}}\frac{{ – 7}}{4}{/tex}

(v) {tex}\frac{1}{{ – 3}}\boxed{{\text{ }}}\frac{{ – 1}}{4}{/tex} {tex} \Rightarrow {/tex}{tex}\frac{1}{{ – 3}}\boxed{{\text{ < }}}\frac{{ – 1}}{4}{/tex}

(vi) {tex}\frac{5}{{ – 11}}\boxed{{\text{ }}}\frac{{ – 5}}{{11}}{/tex} {tex} \Rightarrow {/tex}{tex}\frac{5}{{ – 11}}\boxed{{\text{ = }}}\frac{{ – 5}}{{11}}{/tex}

(vii) {tex}0\boxed{{\text{ > }}}\frac{{ – 7}}{6}{/tex} Since, 0 is greater than every negative number.

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 9.Which is greater in each of the following:

(i) {tex}\frac{2}{3},\frac{5}{2}{/tex}

(ii) {tex}\frac{{ – 5}}{6},\frac{{ – 4}}{3}{/tex}

(iii) {tex}\frac{{ – 3}}{4},\frac{2}{{ – 3}}{/tex}

(iv) {tex}\frac{{ – 1}}{4},\frac{1}{4}{/tex}

(v) {tex} – 3\frac{2}{7}, – 3\frac{4}{5}{/tex}

(i) {tex}\frac{{2 \times 2}}{{3 \times 2}} = \frac{4}{6}{/tex} and {tex}\frac{{5 \times 3}}{{2 \times 3}} = \frac{{15}}{6}{/tex}

Since {tex}\frac{4}{6}\boxed{{\text{ < }}}\frac{{15}}{6}{/tex}

Therefore {tex}\frac{2}{3}\boxed{{\text{ }} < {\text{ }}}\frac{5}{2}{/tex}

(ii) {tex}\frac{{ – 5 \times 1}}{{6 \times 1}} = \frac{{ – 5}}{6}{/tex} and {tex}\frac{{ – 4 \times 2}}{{3 \times 2}} = \frac{{ – 8}}{6}{/tex}

Since {tex}\frac{{ – 5}}{6}\boxed{{\text{ > }}}\frac{{ – 8}}{6}{/tex} Therefore {tex}\frac{{ – 5}}{6}\boxed{{\text{ > }}}\frac{{ – 4}}{3}{/tex}

(iii) {tex}\frac{{ – 3 \times 3}}{{4 \times 3}} = \frac{{ – 9}}{{12}}{/tex} and {tex}\frac{{2 \times \left( { – 4} \right)}}{{ – 3 \times \left( { – 4} \right)}} = \frac{{ – 8}}{{12}}{/tex}

Since {tex}\frac{{ – 9}}{{12}}\boxed{{\text{ < }}}\frac{{ – 8}}{{12}}{/tex}

Therefore {tex}\frac{{ – 3}}{4}\boxed{{\text{ }} < {\text{ }}}\frac{2}{{ – 3}}{/tex}

(iv) {tex}\frac{{ – 1}}{4}\boxed{{\text{ < }}}\frac{1}{4}{/tex} Since positive number is always greater than negative number.

(v) {tex} – 3\frac{2}{7} = \frac{{ – 23}}{7} = \frac{{ – 23 \times 5}}{{7 \times 5}} = \frac{{ – 115}}{{35}}{/tex} and {tex} – 3\frac{4}{5} = \frac{{ – 19}}{5} = \frac{{ – 19 \times 7}}{{5 \times 7}} = \frac{{ – 133}}{{35}}{/tex}

Since {tex}\frac{{ – 115}}{{35}}\boxed{{\text{ > }}}\frac{{ – 133}}{{35}}{/tex}

Therefore{tex} – 3\frac{2}{7}\boxed{{\text{ > }}} – 3\frac{4}{5}{/tex}

NCERT Solutions for Class 7 Maths Exercise 9.1

###### Question 10.Write the following rational numbers in ascending order:

(i) {tex}\frac{{ – 3}}{5},\frac{{ – 2}}{5},\frac{{ – 1}}{5}{/tex}

(ii) {tex}\frac{1}{3},\frac{{ – 2}}{9},\frac{{ – 4}}{3}{/tex}

(iii) {tex}\frac{{ – 3}}{7},\frac{{ – 3}}{2},\frac{{ – 3}}{4}{/tex}

(i) {tex}\frac{{ – 3}}{5},\frac{{ – 2}}{5},\frac{{ – 1}}{5}{/tex}{tex} \Rightarrow {/tex} {tex}\frac{{ – 3}}{5} < \frac{{ – 2}}{5} < \frac{{ – 1}}{5}{/tex}

(ii) {tex}\frac{1}{3},\frac{{ – 2}}{9},\frac{{ – 4}}{3}{/tex} {tex} \Rightarrow {/tex}{tex}\frac{3}{9},\frac{{ – 2}}{9},\frac{{ – 12}}{9}{/tex} [Converting into same denominator]

Now {tex}\frac{{ – 12}}{9} < \frac{{ – 2}}{9} < \frac{3}{9}{/tex} {tex} \Rightarrow {/tex} {tex}\frac{{ – 4}}{3} < \frac{{ – 2}}{9} < \frac{1}{3}{/tex}

(iii) {tex}\frac{{ – 3}}{7},\frac{{ – 3}}{2},\frac{{ – 3}}{4}{/tex}

{tex} \Rightarrow {/tex} {tex}\frac{{ – 3}}{2} < \frac{{ – 3}}{4} < \frac{{ – 3}}{7}{/tex}

## NCERT Solutions for Class 7 Maths Exercise 9.1

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