Rational Numbers Ex-9.1

NCERT solutions for Maths Rational Numbers

NCERT Solutions for Class 7 Maths Rational Numbers

Class –VII Mathematics (Ex. 9.1)

Question 1.List five rational numbers between:

(i) $– 1$ and 0

(ii) $– 2$ and $– 1$

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

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

Answer:

(i) $– 1$ and 0

Let us write $– 1$ and 0 as rational numbers with denominator 6.

$\Rightarrow$ $– 1 = \frac{{ – 6}}{6}$ and 0 = $\frac{0}{6}$

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

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

Therefore, five rational numbers between $– 1$ and 0 would be

$\frac{{ – 5}}{6},\frac{{ – 2}}{3},\frac{{ – 1}}{2},\frac{{ – 1}}{3},\frac{{ – 1}}{6}$

(ii) $– 2$ and $– 1$

Let us write $– 2$ and $– 1$ as rational numbers with denominator 6.

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

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

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

Therefore, five rational numbers between $– 2$ and $– 1$ would be

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

Question 3.Give four rational numbers equivalent to:

(i) $\frac{{ – 2}}{7}$

(ii) $\frac{5}{{ – 3}}$

(iii) $\frac{4}{9}$

Answer:

(i) $\frac{{ – 2}}{7}$

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

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

(ii) $\frac{5}{{ – 3}}$

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

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

(iii) $\frac{4}{9}$

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

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

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

(i) $\frac{3}{4}$

(ii) $\frac{{ – 5}}{8}$

(iii) $\frac{{ – 7}}{4}$

(iv) $\frac{7}{8}$

Answer:

(i) $\frac{3}{4}$

(ii) $\frac{{ – 5}}{8}$

(iii) $\frac{{ – 7}}{4}$

(iv) $\frac{7}{8}$

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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.

Answer:

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

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

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

(iii) $\frac{{ – 2}}{{ – 3}}$ and $\frac{2}{3}$

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

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

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

(vii) $\frac{{ – 5}}{{ – 9}}$ and $\frac{5}{{ – 9}}$

Answer:

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

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

$\because$ $\frac{{ – 1}}{3}$$\ne$$\frac{1}{3}$

$\therefore$ $\frac{{ – 7}}{{21}}$$\ne$$\frac{3}{9}$

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

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

[Converting into lowest term]

$\because$ $\frac{{ – 4}}{5}$ = $\frac{{ – 4}}{5}$

$\therefore$ $\frac{{ – 16}}{{20}}$ = $\frac{{20}}{{ – 25}}$

(iii) $\frac{{ – 2}}{{ – 3}}$ and $\frac{2}{3}$

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

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

$\therefore$ $\frac{{ – 2}}{{ – 3}}$ = $\frac{2}{3}$

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

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

$\because$ $\frac{{ – 3}}{5}$ = $\frac{{ – 3}}{5}$

$\therefore$ $\frac{{ – 3}}{5}$ = $\frac{{ – 12}}{{20}}$

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

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

$\because$ $\frac{{ – 8}}{5}$ = $\frac{{ – 8}}{5}$

$\therefore$ $\frac{8}{{ – 5}}$ = $\frac{{ – 24}}{{15}}$

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

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

$\because$ $\frac{1}{3}$$\ne$$\frac{{ – 1}}{9}$

$\therefore$ $\frac{1}{3}$$\ne$$\frac{{ – 1}}{9}$

(vii) $\frac{{ – 5}}{{ – 9}}$ and $\frac{5}{{ – 9}}$

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

$\because$ $\frac{5}{9}$$\ne$$\frac{5}{{ – 9}}$

$\therefore$ $\frac{{ – 5}}{{ – 9}}$$\ne$$\frac{5}{{ – 9}}$

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

(i) $\frac{{ – 8}}{6}$

(ii) $\frac{{25}}{{45}}$

(iii) $\frac{{ – 44}}{{72}}$

(iv) $\frac{{ – 8}}{{10}}$

Answer:

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

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

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

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

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

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

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

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

Answer:

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

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

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

(i) $\frac{2}{3},\frac{5}{2}$

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

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

(iv) $\frac{{ – 1}}{4},\frac{1}{4}$

(v) $– 3\frac{2}{7}, – 3\frac{4}{5}$

Answer:

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

Since $\frac{4}{6}\boxed{{\text{ < }}}\frac{{15}}{6}$

Therefore $\frac{2}{3}\boxed{{\text{ }} < {\text{ }}}\frac{5}{2}$

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

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

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

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

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

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

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

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

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

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NCERT Solutions for Class 7 Maths Exercise 9.1

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

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

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

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

Answer:

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

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

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

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

$\Rightarrow$ $\frac{{ – 3}}{2} < \frac{{ – 3}}{4} < \frac{{ – 3}}{7}$

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