NCERT Solutions for Class 7 Maths Chapter 9 - Rational Numbers

Exercise – 9.1

1. List five rational numbers between:

(i) $ - 1$ and $0$

Ans: We need to represent five rational numbers. 

so we will take number $ 5+1 = 6$

Let’s represent $ - 1$ and $ 0$ as rational numbers with a denominator 6.

So $ - 1$ can be written as 

$ \Rightarrow  - 1 = \dfrac{{ - 6}}{6}$ and $0$ with denominator 6 can be written as $0 = \dfrac{0}{6}$

now we will write numbers lying between$\dfrac{{ - 6}}{6}$ and $\dfrac{{0}}{6}$

$\therefore \dfrac{{ - 6}}{6} < \dfrac{{ - 5}}{6} < \dfrac{{ - 4}}{6} < \dfrac{{ - 3}}{6} < \dfrac{{ - 2}}{6} < \dfrac{{ - 1}}{6} < 0$

$ \Rightarrow  - 1 < \dfrac{{ - 5}}{6} < \dfrac{{ - 2}}{3} < \dfrac{{ - 1}}{2} < \dfrac{{ - 1}}{3} < \dfrac{{ - 1}}{6} < 0$

As a result, a set of five rational numbers ranging from $ - 1$ to $0$would be 

$\dfrac{{ - 5}}{6},\dfrac{{ - 2}}{3},\dfrac{{ - 1}}{2},\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{6}$

(ii) $ - 2$ and $ - 1$

Ans:  We need to represent five rational numbers. 

so we will take number $ 5+1 = 6$

Let’s represent $ - 2$ and $ - 1$ as rational numbers with a denominator 6.

$ \Rightarrow  - 2 = \dfrac{{ - 12}}{6}$and $ - 1 = \dfrac{{ - 6}}{6}$

now we will write numbers lying between $\dfrac{{ - 12}}{6}$ and $\dfrac{{ - 6}}{6}$

$\therefore \dfrac{{ - 12}}{6} < \dfrac{{ - 11}}{6} < \dfrac{{ - 10}}{6} < \dfrac{{ - 9}}{6} < \dfrac{{ - 8}}{6} < \dfrac{{ - 7}}{6} < \dfrac{{ - 6}}{6}$

$ \Rightarrow  - 2 < \dfrac{{ - 11}}{6} < \dfrac{{ - 5}}{3} < \dfrac{{ - 3}}{2} < \dfrac{{ - 4}}{3} < \dfrac{{ - 7}}{6} <  - 1$

As a result, a set of five rational numbers ranging from $ - 2$ to $ - 1$ would be

$\dfrac{{ - 11}}{6},\dfrac{{ - 5}}{3},\dfrac{{ - 3}}{2},\dfrac{{ - 4}}{3},\dfrac{{ - 7}}{6}$

(iii) $\dfrac{{ - 4}}{5}$ and $\dfrac{{ - 2}}{3}$

Ans: We need to represent five rational numbers. 

Let’s represent $\dfrac{{ - 4}}{5}$ and $\dfrac{{ - 2}}{3}$  as rational numbers

with the same denominators.

L.C.M. of both denominators $(5,3)$$=\ 15$

Let’s make denominator of both terms =$15$

$\frac{-4\times 3}{5\times 3}=\frac{-12}{15}$

Similarly, $\frac{-2\times 5}{3\times 5}=\frac{-10}{15}$

But between $-10\ \text{and}\ \text{-12}$ there is only one integer

So we will again multiply and divide our terms by a number such that the denominator will remain the same.

Let’s multiply numerator and denominator by 3

$\frac{-12\times 3}{15\times 3}=\frac{-36}{45}$ and $\frac{-10\times 3}{15\times 3}\ =\ \frac{-30}{45}$

now we will write numbers lying between $\frac{-36}{45}$ and $\frac{-30}{45}$

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

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

As a result, a set of five rational numbers ranging from $\dfrac{{ - 4}}{5}$ to $\dfrac{{ - 2}}{3}$ would be$\dfrac{{ - 7}}{9},\dfrac{{ - 34}}{{45}},\dfrac{{ - 11}}{{15}},\dfrac{{ - 32}}{{45}},\dfrac{{ - 31}}{{45}},\dfrac{{ - 2}}{3}$

(iv) $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$

Ans:  Let’s represent $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$ as rational numbers with same the denominators

L.C.M. of both denominators $(2,3)$$=\ 6$

Let’s make denominator of both terms = 

multiplying numerator and denominator by 3

$\frac{-1\times 3}{2\times 3}=\frac{-3}{6}$

Similarly multiplying numerator and denominator by 2

$\frac{2\times 2}{3\times 2}=\frac{4}{6}$

now we will write numbers lying between $\frac{-3}{6}$ and $\frac{4}{6}$

$\therefore \dfrac{{ - 3}}{6} < \dfrac{{ - 2}}{6} < \dfrac{{ - 1}}{6} < 0 < \dfrac{1}{6} < \dfrac{2}{6} < \dfrac{3}{6} < \dfrac{4}{6}$

$ \Rightarrow \dfrac{{ - 1}}{2} < \dfrac{{ - 1}}{3} < \dfrac{{ - 1}}{6} < 0 < \dfrac{1}{6} < \dfrac{1}{3} < \dfrac{1}{2} < \dfrac{2}{3}$

As a result, a set of five rational numbers ranging from $\dfrac{{ - 1}}{2}$ and $\dfrac{2}{3}$ would be

 $\dfrac{{ - 1}}{3},\dfrac{{ - 1}}{6},0,\dfrac{1}{6},\dfrac{1}{3}$

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

(i) $\dfrac{{ - 3}}{5},\dfrac{{ - 6}}{{10}},\dfrac{{ - 9}}{{15}},\dfrac{{ - 12}}{{20}},........$

Ans: $\dfrac{{ - 3 \times 1}}{{5 \times 1}},\dfrac{{ - 3 \times 2}}{{5 \times 2}},\dfrac{{ - 3 \times 3}}{{5 \times 3}},\dfrac{{ - 3 \times 4}}{{5 \times 4}},........$

As a result, a set of next four rational numbers of this pattern would be

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

(ii) $\dfrac{{ - 1}}{4},\dfrac{{ - 2}}{8},\dfrac{{ - 3}}{{12}},........$

Ans: $\dfrac{{ - 1 \times 1}}{{4 \times 1}},\dfrac{{ - 1 \times 2}}{{4 \times 2}},\dfrac{{ - 1 \times 3}}{{4 \times 3}},........$

As a result, a set of next four rational numbers of this pattern would be

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

(iii) $\dfrac{{ - 1}}{6},\dfrac{2}{{ - 12}},\dfrac{3}{{ - 18}},\dfrac{4}{{ - 24}},........$

Ans: $\dfrac{{ - 1 \times 1}}{{6 \times 1}},\dfrac{{1 \times 2}}{{ - 6 \times 2}},\dfrac{{1 \times 3}}{{ - 6 \times 3}},\dfrac{{1 \times 4}}{{ - 6 \times 4}},........$

As a result, a set of next four rational numbers of this pattern would be

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

(iv) $\dfrac{{ - 2}}{3},\dfrac{2}{{ - 3}},\dfrac{4}{{ - 6}},\dfrac{6}{{ - 9}},........$

Ans: $\dfrac{{ - 2 \times 1}}{{3 \times 1}},\dfrac{{2 \times 1}}{{ - 3 \times 1}},\dfrac{{2 \times 2}}{{ - 3 \times 2}},\dfrac{{2 \times 3}}{{ - 3 \times 3}},........$

As a result, a set of next four rational numbers of this pattern would be

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

3. Give four rational numbers equivalent to:

(i) $\dfrac{{ - 2}}{7}$

Ans: $\dfrac{{ - 2 \times 2}}{{7 \times 2}} = \dfrac{{ - 4}}{{14}},\dfrac{{ - 2 \times 3}}{{7 \times 3}} = \dfrac{{ - 6}}{{21}},\dfrac{{ - 2 \times 4}}{{7 \times 4}} = \dfrac{{ - 8}}{{28}},\dfrac{{ - 2 \times 5}}{{7 \times 5}} = \dfrac{{ - 10}}{{35}}$

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

(ii) $\dfrac{5}{{ - 3}}$

Ans: $\dfrac{{5 \times 2}}{{ - 3 \times 2}} = \dfrac{{10}}{{ - 6}},\dfrac{{5 \times 3}}{{ - 3 \times 3}} = \dfrac{{15}}{{ - 9}},\dfrac{{5 \times 4}}{{ - 3 \times 4}} = \dfrac{{20}}{{ - 12}},\dfrac{{5 \times 5}}{{ - 3 \times 5}} = \dfrac{{25}}{{ - 15}}$

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

(ii) $\dfrac{4}{9}$

Ans: $\dfrac{{4 \times 2}}{{9 \times 2}} = \dfrac{8}{{18}},\dfrac{{4 \times 3}}{{9 \times 3}} = \dfrac{{12}}{{27}},\dfrac{{4 \times 4}}{{9 \times 4}} = \dfrac{{16}}{{36}},\dfrac{{4 \times 5}}{{9 \times 5}} = \dfrac{{20}}{{35}}$

Therefore, four equivalent rational numbers are $\dfrac{8}{{18}},\dfrac{{12}}{{27}},\dfrac{{16}}{{36}},\dfrac{{20}}{{35}}$

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

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

Ans: Since,  $\dfrac{3}{4}$ lies between 0 and 1. Divide this region into 4 equal parts and then mark the part with the value equal to 3. 

(ii) $\dfrac{{ - 5}}{8}$

Ans: Since,  $\dfrac{{ - 5}}{8}$  lies between 0  and -1. Divide this region into 8 equal parts and then mark the part with the value equal to 5 away from 0 towards the negative number line. 

(iii) $\dfrac{{ - 7}}{4}$

Ans: Since, $\dfrac{{ - 7}}{4}$  lies between 0 and -2. Divide this region between 0 and -1 to 4 equal parts and -1 to -2 to 4 parts, then mark the part with the value equal to 7 away from 0 towards the negative number line. 

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

Ans: Since,  $\dfrac{7}{8}$  lies between 0 and 1. Divide this region between 0 and 1 to 8 equal parts, then mark the part with the value equal to 7 away from 0 towards the positive number line.

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.

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

$\therefore $ A$ = \dfrac{6}{3}$, P$ = \dfrac{7}{3}$, Q$ = \dfrac{8}{3}$, B $ = \dfrac{9}{3}$

Similarly, T $ = \dfrac{{ - 3}}{3}$,R $ = \dfrac{{ - 4}}{3}$,S $ = \dfrac{{ - 5}}{3}$,U $ = \dfrac{{ - 6}}{3}$

Thus, the rational numbers represented P, Q, R and S is $\dfrac{7}{3},\dfrac{8}{3},\dfrac{{ - 4}}{3},\dfrac{{ - 5}}{3}$.

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

(i) $\dfrac{{ - 7}}{{21}}$ and $\dfrac{3}{9}$

Ans: Convert it into lowest form

$\dfrac{{ - 7}}{{21}} = \dfrac{{ - 1}}{3}$ and $\dfrac{3}{9} = \dfrac{1}{3}$

Since, $\dfrac{{ - 1}}{3} \ne \dfrac{1}{3}$

Therefore, $\dfrac{{ - 7}}{{21}} \ne \dfrac{3}{9}$

(ii) $\dfrac{{ - 16}}{{20}}$ and $\dfrac{{20}}{{ - 25}}$

Ans: Convert it into lowest form

$\dfrac{{ - 16}}{{20}} = \dfrac{{ - 4}}{5}$ and $\dfrac{{20}}{{ - 25}} = \dfrac{4}{{ - 5}} = \dfrac{{ - 4}}{5}$                                                 

Since, $\dfrac{{ - 4}}{5} = \dfrac{{ - 4}}{5}$

Therefore, $\dfrac{{ - 16}}{{20}} = \dfrac{{20}}{{ - 25}}$

(iii) $\dfrac{{ - 2}}{{ - 3}}$ and $\dfrac{2}{3}$

Ans: Convert it into lowest form

$\dfrac{{ - 2}}{{ - 3}} = \dfrac{2}{3}$ and $\dfrac{2}{3} = \dfrac{2}{3}$                                                              

Since, $\dfrac{2}{3} = \dfrac{2}{3}$

Therefore, $\dfrac{{ - 2}}{{ - 3}} = \dfrac{2}{3}$

(iv) $\dfrac{{ - 3}}{5}$and $\dfrac{{ - 12}}{{20}}$

Ans: Convert it into lowest form

$\dfrac{{ - 3}}{5} = \dfrac{{ - 3}}{5}$ and $\dfrac{{ - 12}}{{20}} = \dfrac{{ - 3}}{5}$                                                        

Since, $\dfrac{{ - 3}}{5} = \dfrac{{ - 3}}{5}$

Therefore, $\dfrac{{ - 3}}{5} = \dfrac{{ - 12}}{{20}}$

(v) $\dfrac{8}{{ - 5}}$and $\dfrac{{ - 24}}{{15}}$

Ans:  Convert it into lowest form

$\dfrac{8}{{ - 5}} = \dfrac{{ - 8}}{5}$and $\dfrac{{ - 24}}{{15}} = \dfrac{{ - 8}}{5}$                                                       

Since, $\dfrac{{ - 8}}{5} = \dfrac{{ - 8}}{5}$

Therefore,$\dfrac{8}{{ - 5}} = \dfrac{{ - 24}}{{15}}$

(vi)  $\dfrac{1}{3}$and $\dfrac{{ - 1}}{9}$

Ans: Convert it into lowest form

$\dfrac{1}{3} = \dfrac{1}{3}$ and $\dfrac{{ - 1}}{9} = \dfrac{{ - 1}}{9}$

Since, $\dfrac{1}{3} \ne \dfrac{{ - 1}}{9}$

Therefore, $\dfrac{1}{3} \ne \dfrac{{ - 1}}{9}$

(vii) $\dfrac{{ - 5}}{{ - 9}}$ and  $\dfrac{5}{{ - 9}}$

Ans: Convert it into lowest form

$\dfrac{{ - 5}}{{ - 9}} = \dfrac{5}{9}$ and $\dfrac{5}{{ - 9}} = \dfrac{5}{9}$

Since, $\dfrac{5}{9} \ne \dfrac{5}{{ - 9}}$

Therefore,$\dfrac{{ - 5}}{{ - 9}} \ne \dfrac{5}{{ - 9}}$

7. Rewrite the following in the simplest form:

$\text{ }(i)\text{ }\frac{-8}{6}\text{ }(ii)\frac{25}{45}\text{ }(iii)\text{ }\frac{-44}{72}\text{ }(iv)\text{ }\frac{-8}{10}$ 

Ans: we will divide the numerator and denominator by H.C.F. of numerator and denominator 

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

(ii) $\dfrac{{25}}{{45}} = \dfrac{{25 \div 5}}{{45 \div 5}} = \dfrac{5}{9}$         [ H.C.F. of 25 and 45 is 5]

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

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

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

(i) $\dfrac{{ - 5}}{7}\boxed{}\dfrac{2}{3}$

Ans: $\dfrac{{ - 5}}{7}\boxed < \dfrac{2}{3}$  because the positive number exceeds the negative number.

(ii) $\dfrac{{ - 4}}{5}\boxed{}\dfrac{{ - 5}}{7}$

Ans: $\dfrac{{ - 4 \times 7}}{{5 \times 7}}\boxed{}\dfrac{{ - 5 \times 5}}{{7 \times 5}}$

Since, $\dfrac{{ - 28}}{{35}}\boxed < \dfrac{{ - 25}}{{35}}$

Therefore, $\dfrac{{ - 4}}{5}\boxed < \dfrac{{ - 5}}{7}$

because, if both the numbers are negative then a smaller number is considered as greater.

(iii) $\dfrac{{ - 7}}{8}\boxed{}\dfrac{{14}}{{ - 16}}$

Ans: $\dfrac{{ - 7 \times 2}}{{8 \times 2}}\boxed{}\dfrac{{14 \times ( - 1)}}{{ - 16 \times ( - 1)}}$

Since, $\dfrac{{ - 14}}{{16}}\boxed = \dfrac{{ - 14}}{{16}}$

Therefore, $\dfrac{{ - 7}}{8}\boxed = \dfrac{{14}}{{ - 16}}$

(iv) $\dfrac{{ - 8}}{5}\boxed{}\dfrac{{ - 7}}{4}$

Ans: $\dfrac{{ - 8 \times 4}}{{5 \times 4}}\boxed{}\dfrac{{ - 7 \times 5}}{{4 \times 5}}$

Since, $\dfrac{{ - 32}}{{20}}\boxed > \dfrac{{ - 35}}{{20}}$

Therefore, $\dfrac{{ - 8}}{5}\boxed > \dfrac{{ - 7}}{4}$

(v) $\dfrac{1}{{ - 3}}\boxed{}\dfrac{{ - 1}}{4}$

Ans: $\dfrac{1}{{ - 3}}\boxed < \dfrac{{ - 1}}{4}$

(vi) $\dfrac{5}{{ - 11}}\boxed{}\dfrac{{ - 5}}{{11}}$

Ans: $\dfrac{5}{{ - 11}}\boxed = \dfrac{{ - 5}}{{11}}$

(vii) $0\boxed{}\dfrac{{ - 7}}{6}$

Ans: $0\boxed > \dfrac{{ - 7}}{6}$  because zero is greater than all negative numbers.

9. Which is greater in each of the following:

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

Ans: $\dfrac{{2 \times 2}}{{3 \times 2}} = \dfrac{4}{6}$ and  $\dfrac{{5 \times 3}}{{2 \times 3}} = \dfrac{{15}}{6}$

$\because \dfrac{4}{6}\boxed < \dfrac{{15}}{6}$

$\therefore \dfrac{2}{3}\boxed < \dfrac{5}{2}$

(ii) $\dfrac{{ - 5}}{6},\dfrac{{ - 4}}{3}$

Ans: $\dfrac{{ - 5 \times 1}}{{6 \times 1}} = \dfrac{{ - 5}}{6}$ and $\dfrac{{ - 4 \times 2}}{{3 \times 2}} = \dfrac{{ - 8}}{6}$

$\because \dfrac{{ - 5}}{6}\boxed > \dfrac{{ - 8}}{6}$

$\therefore \dfrac{{ - 5}}{6}\boxed > \dfrac{{ - 4}}{3}$

(iii) $\dfrac{{ - 3}}{4},\dfrac{2}{{ - 3}}$

Ans: $\dfrac{{ - 3 \times 3}}{{4 \times 3}} = \dfrac{{ - 9}}{{12}}$ and $\dfrac{{2 \times \left( { - 4} \right)}}{{ - 3 \times \left( { - 4} \right)}} = \dfrac{{ - 8}}{{12}}$

$\because \dfrac{{ - 9}}{{12}}\boxed < \dfrac{{ - 8}}{{12}}$

$\therefore \dfrac{{ - 3}}{4}\boxed < \dfrac{2}{{ - 3}}$

(iv) $\dfrac{{ - 1}}{4},\dfrac{1}{4}$

Ans: $\dfrac{{ - 1}}{4}\boxed < \dfrac{1}{4}$ because, positive number is always greater than the negative number.

(v) $ - 3\dfrac{2}{7}, - 3\dfrac{4}{5}$

Ans: $ - 3\dfrac{2}{7} = \dfrac{{ - 23}}{7} = \dfrac{{ - 23 \times 5}}{{7 \times 5}} = \dfrac{{ - 115}}{{35}}$ and $ - 3\dfrac{4}{5} = \dfrac{{ - 19}}{5} = \dfrac{{ - 19 \times 7}}{{5 \times 7}} = \dfrac{{ - 133}}{{35}}$

$\because \dfrac{{ - 115}}{{35}}\boxed > \dfrac{{ - 133}}{{35}}$

$\therefore  - 3\dfrac{2}{7}\boxed >  - 3\dfrac{4}{5}$

10. Write the following rational numbers in ascending order:

(i) $\dfrac{{ - 3}}{5},\dfrac{{ - 2}}{5},\dfrac{{ - 1}}{5}$

Ans: $\dfrac{{ - 3}}{5} < \dfrac{{ - 2}}{5} < \dfrac{{ - 1}}{5}$

(ii) $\dfrac{1}{3},\dfrac{{ - 2}}{9},\dfrac{{ - 4}}{3}$

Ans: Firstly we will make the denominators same

$ \Rightarrow \dfrac{3}{9},\dfrac{{ - 2}}{9},\dfrac{{ - 12}}{9}$

$\because \dfrac{{ - 12}}{9} < \dfrac{{ - 2}}{9} < \dfrac{3}{9}$

$\therefore \dfrac{{ - 4}}{3} < \dfrac{{ - 2}}{9} < \dfrac{1}{3}$

(iii)  $\dfrac{{ - 3}}{7},\dfrac{{ - 3}}{2},\dfrac{{ - 3}}{4}$

Ans: $\dfrac{{ - 3}}{2} < \dfrac{{ - 3}}{4} < \dfrac{{ - 3}}{7}$

Exercise - 9.2

1. Find the Sum:

(i) $\dfrac{5}{4} + \left( {\dfrac{{ - 11}}{4}} \right)$

Ans: On opening the bracket we will get,

$\dfrac{{5 - 11}}{4} = \dfrac{{ - 6}}{4} = \dfrac{{ - 3}}{2}$

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

Ans: Firstly we will take the LCM of $3$and $5$

$\dfrac{5}{3} + \dfrac{3}{5} = \dfrac{{5 \times 5}}{{3 \times 5}} + \dfrac{{3 \times 3}}{{5 \times 3}} = \dfrac{{25}}{{15}} + \dfrac{9}{{15}}$

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

(iii) $\dfrac{{ - 9}}{{10}} + \dfrac{{22}}{{15}}$

Ans: Firstly we will take the LCM of $10$and $15$

$\dfrac{{ - 9}}{{10}} + \dfrac{{22}}{{15}} = \dfrac{{ - 9 \times 3}}{{10 \times 3}} + \dfrac{{22 \times 2}}{{15 \times 2}} = \dfrac{{ - 27}}{{30}} + \dfrac{{44}}{{30}}$

$ \Rightarrow \dfrac{{ - 27 + 44}}{{30}} = \dfrac{{17}}{{30}}$

(iv) $\dfrac{{ - 3}}{{ - 11}} + \dfrac{5}{9}$

Ans: Firstly we will take the LCM of $11$ and $9$

$\dfrac{{ - 3}}{{ - 11}} + \dfrac{5}{9} = \dfrac{{ - 3 \times 9}}{{ - 11 \times 9}} + \dfrac{{5 \times 11}}{{9 \times 11}} = \dfrac{{27}}{{99}} + \dfrac{{55}}{{99}}$

$ \Rightarrow \dfrac{{27 + 55}}{{99}} = \dfrac{{82}}{{99}}$

(v) $\dfrac{{ - 8}}{{19}} + \dfrac{{\left( { - 2} \right)}}{{57}}$

Ans: Firstly we will take the LCM of $19$and $57$

$\dfrac{{ - 8}}{{19}} + \dfrac{{\left( { - 2} \right)}}{{57}} = \dfrac{{ - 8 \times 3}}{{19 \times 3}} + \dfrac{{\left( { - 2} \right) \times 1}}{{57 \times 1}} = \dfrac{{ - 24}}{{57}} + \dfrac{{\left( { - 2} \right)}}{{57}}$

$ \Rightarrow \dfrac{{ - 24 - 2}}{{57}} = \dfrac{{ - 26}}{{57}}$

(vi) $\dfrac{{ - 2}}{3} + 0$

Ans: $\dfrac{{ - 2}}{3} + 0 = \dfrac{{ - 2}}{3}$

(vii) $ - 2\dfrac{1}{3} + 4\dfrac{3}{5}$

Ans: Firstly we will take the LCM of $3$and $5$

$ - 2\dfrac{1}{3} + 4\dfrac{3}{5} = \dfrac{{ - 7}}{3} + \dfrac{{23}}{5} = \dfrac{{ - 7 \times 5}}{{3 \times 5}} + \dfrac{{23 \times 3}}{{5 \times 3}} = \dfrac{{ - 35}}{{15}} + \dfrac{{69}}{{15}}$

$ \Rightarrow \dfrac{{ - 35 + 69}}{{15}} = \dfrac{{34}}{{15}} = 2\dfrac{4}{5}$

2. Find:

(i) $\dfrac{7}{{24}} - \dfrac{{17}}{{36}}$

Ans: Firstly we will take the LCM of $24$ and $36$

$\dfrac{7}{{24}} - \dfrac{{17}}{{36}} = \dfrac{{7 \times 3}}{{24 \times 3}} - \dfrac{{17 \times 2}}{{36 \times 2}} = \dfrac{{21}}{{72}} - \dfrac{{34}}{{72}}$

$ \Rightarrow \dfrac{{21 - 34}}{{72}} = \dfrac{{ - 13}}{{72}}$

(ii) $\dfrac{5}{{63}} - \left( {\dfrac{{ - 6}}{{21}}} \right)$

Ans: Firstly we will take the LCM of $63$and $21$

$\dfrac{5}{{63}} - \left( {\dfrac{{ - 6}}{{21}}} \right) = \dfrac{{5 \times 1}}{{63 \times 1}} - \left( {\dfrac{{ - 6 \times 3}}{{21 \times 3}}} \right) = \dfrac{5}{{63}} - \dfrac{{ - 18}}{{63}}$

$ \Rightarrow \dfrac{{5 - \left( { - 18} \right)}}{{63}} = \dfrac{{5 + 18}}{{63}} = \dfrac{{23}}{{63}}$

(iii) $\dfrac{{ - 6}}{{13}} - \left( {\dfrac{{ - 7}}{{15}}} \right)$

Ans: Firstly we will take the LCM of $13$and $15$

$\dfrac{{ - 6}}{{13}} - \left( {\dfrac{{ - 7}}{{15}}} \right) = \dfrac{{ - 6 \times 15}}{{13 \times 15}} - \left( {\dfrac{{ - 7 \times 13}}{{15 \times 13}}} \right) = \dfrac{{ - 90}}{{195}} - \left( {\dfrac{{ - 91}}{{195}}} \right)$

$ \Rightarrow \dfrac{{ - 90 - \left( { - 91} \right)}}{{195}} = \dfrac{{ - 90 + 91}}{{195}} = \dfrac{1}{{195}}$

(iv) $\dfrac{{ - 3}}{8} - \dfrac{7}{{11}}$

Ans: Firstly we will take the LCM of $8$and $11$

$\dfrac{{ - 3}}{8} - \dfrac{7}{{11}} = \dfrac{{ - 3 \times 11}}{{8 \times 11}} - \dfrac{{7 \times 8}}{{11 \times 8}} = \dfrac{{ - 33}}{{88}} - \dfrac{{56}}{{88}}$

$ \Rightarrow \dfrac{{ - 33 - 56}}{{88}} = \dfrac{{ - 89}}{{88}} =  - 1\dfrac{1}{{88}}$

(v) $ - 2\dfrac{1}{9} - 6$

Ans: $\dfrac{{ - 19}}{9} - \dfrac{6}{1} = \dfrac{{ - 19 \times 1}}{{9 \times 1}} - \dfrac{{6 \times 9}}{{1 \times 9}} = \dfrac{{ - 19}}{9} - \dfrac{{54}}{9}$

$ \Rightarrow \dfrac{{ - 19 - 54}}{9} = \dfrac{{ - 73}}{9} =  - 8\dfrac{1}{9}$

3. Find the product:

(i) $\dfrac{9}{2} \times \left( {\dfrac{{ - 7}}{4}} \right)$

Ans: $\dfrac{9}{2} \times \left( {\dfrac{{ - 7}}{4}} \right) = \dfrac{{9 \times \left( { - 7} \right)}}{{2 \times 4}}$

$ \Rightarrow \dfrac{{ - 63}}{8} =  - 7\dfrac{7}{8}$

(ii) $\dfrac{3}{{10}} \times ( - 9)$

Ans: $ \Rightarrow \dfrac{{ - 27}}{{10}} =  - 2\dfrac{7}{{10}}$

(iii) $\dfrac{{ - 6}}{5} \times \dfrac{9}{{11}}$

Ans: $\dfrac{{ - 6}}{5} \times \dfrac{9}{{11}} = \dfrac{{\left( { - 6} \right) \times 9}}{{5 \times 11}}$

$ \Rightarrow \dfrac{{ - 54}}{{55}}$

(iv) $\dfrac{3}{7} \times \left( {\dfrac{{ - 2}}{5}} \right)$

Ans: $\dfrac{3}{7} \times \left( {\dfrac{{ - 2}}{5}} \right) = \dfrac{{3 \times \left( { - 2} \right)}}{{7 \times 5}}$

$ \Rightarrow \dfrac{{ - 6}}{{35}}$

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

Ans: $\dfrac{3}{{11}} \times \dfrac{2}{5} = \dfrac{{3 \times 2}}{{11 \times 5}}$

$ \Rightarrow \dfrac{6}{{55}}$

(vi) $\dfrac{3}{{ - 5}} \times \left( {\dfrac{{ - 5}}{3}} \right)$

Ans: $\dfrac{3}{{ - 5}} \times \left( {\dfrac{{ - 5}}{3}} \right) = \dfrac{{3 \times \left( { - 5} \right)}}{{ - 5 \times 3}}$

$ \Rightarrow \dfrac{{ - 15}}{{ - 15}} = 1$

4. Find the value of:

(i) $( - 4) \div \dfrac{2}{3}$

Ans: $\left( { - 4} \right) \div \dfrac{2}{3} = \left( { - 4} \right) \times \dfrac{3}{2}$

$ \Rightarrow ( - 2) \times 3 =  - 6$

(ii) $\dfrac{{ - 3}}{5} \div 2$

Ans: $\dfrac{{ - 3}}{5} \div 2 = \dfrac{{ - 3}}{5} \times \dfrac{1}{2}$

$ \Rightarrow \dfrac{{\left( { - 3} \right) \times 1}}{{5 \times 2}} = \dfrac{{ - 3}}{{10}}$

(iii) $\dfrac{{ - 4}}{5} \div \left( { - 3} \right)$

Ans: $\dfrac{{ - 4}}{5} \div \left( { - 3} \right) = \dfrac{{ - 4}}{5} \times \dfrac{1}{{\left( { - 3} \right)}}$

$ \Rightarrow \dfrac{{( - 4) \times 1}}{{5 \times ( - 3)}} = \dfrac{{ - 4}}{{ - 15}} = \dfrac{4}{{15}}$

(iv) $\dfrac{{ - 1}}{8} \div \dfrac{3}{4}$

Ans: $\dfrac{{ - 1}}{8} \div \dfrac{3}{4} = \dfrac{{ - 1}}{8} \times \dfrac{4}{3}$

$ \Rightarrow \dfrac{{( - 1) \times 1}}{{2 \times 3}} = \dfrac{{ - 1}}{6}$

(v) $\dfrac{{ - 2}}{{13}} \div \dfrac{1}{7}$

Ans: $\dfrac{{ - 2}}{{13}} \div \dfrac{1}{7} = \dfrac{{ - 2}}{{13}} \times \dfrac{7}{1}$

$ \Rightarrow \dfrac{{( - 2) \times 7}}{{13 \times 1}} = \dfrac{{ - 14}}{{13}} =  - 1\dfrac{1}{{13}}$

(vi) $\dfrac{{ - 7}}{{12}} \div \left( {\dfrac{{ - 2}}{{13}}} \right)$

Ans: $\dfrac{{ - 7}}{{12}} \div \left( {\dfrac{{ - 2}}{{13}}} \right) = \dfrac{{ - 7}}{{12}} \times \dfrac{{13}}{{\left( { - 2} \right)}}$

$ \Rightarrow \dfrac{{\left( { - 7} \right) \times 13}}{{12 \times \left( { - 2} \right)}} = \dfrac{{ - 91}}{{ - 24}} = 3\dfrac{{19}}{{24}}$

(vii) $\dfrac{3}{{13}} \div \left( {\dfrac{{ - 4}}{{65}}} \right)$

Ans: $\dfrac{3}{{13}} \div \left( {\dfrac{{ - 4}}{{65}}} \right) = \dfrac{3}{{13}} \times \dfrac{{65}}{{\left( { - 4} \right)}}$

$ \Rightarrow \dfrac{{3 \times 65}}{{13 \times \left( { - 4} \right)}} = \dfrac{{3 \times \left( { - 5} \right)}}{{1 \times 4}} = \dfrac{{ - 15}}{4} =  - 3\dfrac{3}{4}$

NCERT Solutions for Class 7 Maths Chapter 9 – Free PDF Download

Students can download and refer to the Class 7 Maths Ch 9, which are solved by the best teachers. Some of the most difficult questions are solved in the simplest of ways. It allows students to understand the approach well. It also lets them apply their knowledge of the subject matter on other questions that are asked in the examination. Students can download and refer to solutions anytime and anywhere.

Chapter 9 – Rational Numbers

9.1 Introduction

Students are already familiar with some types of numbers. These are the natural numbers, whole numbers and integers. Natural numbers begin with 1 and go up to infinity. If you add 0 to the natural numbers, this makes whole numbers. If you add negative numbers to the above set, this is a set of integers. You were also introduced to fractions which are numbers in the form of a numerator/denominator.

This chapter extends the number system even more and introduces the concept of Rational Numbers. You will also learn about the mathematical operations of Rational Numbers.

9.2 Need for Rational Numbers

Rational numbers are a special set of numbers. This is because these can be written in a fractional form. Rational numbers can be written in the way of a ratio of p and q where p and q are both integers and q is not equal to zero.

The Rational Numbers are those numbers that are used daily. These are required because there are several measures of a quantity which cannot be adequately described by integers or natural numbers alone. There is a need for Rational Numbers to represent these numbers.

9.3 What are Rational Numbers?

This section of NCERT Class 7 Maths Chapter 9 explains what a rational number is. A rational number is a number that can be expressed as a ratio of p/q, where p and q are integers, and q does not equal to zero. The numerator and the denominator of a rational number will be integers. This means that 0 can also be a rational number as a rational number can be represented as 0/5 or 0/100 etc.

Equivalent Rational Numbers

It is possible to write Rational Numbers using different numerators and denominators. One can get the equivalent Rational Numbers by multiplying the numerator and the denominator of the rational number with the same integer. Make sure that the integer is non zero. The method is the same that is applied to obtain equivalent fractions.

So if p/q is a rational number its equivalent Rational Numbers can be (p/q) × (3/3) or (p/q) × (10/10) etc.

9.4 Positive and Negative Rational Numbers

• If both the numerator and the denominator of the rational number have the same sign, this is a positive rational number. All the positive Rational Numbers are greater than 0. This can be in the form of a (p/q) or (-p/-q).

• The negative Rational Numbers are those that have either the numerator or the denominator with an opposite sign. Negative Rational Numbers are all less than zero. The negative Rational Numbers can be in the form of (-p/q) or (p/-q).

9.5 Rational Numbers on a Number Line

The students are already aware of representing numbers on a number line. The points that lie to the right of zero are the positive integers and have a (+) sign. The points that lie to the left of zero are the negative integers, and these have a (–) sign.

When you write a natural number on the number line, these can also be thought of as Rational Numbers with a denominator 1.

This section of CBSE class 7 Maths Rational Numbers solutions explains how to represent a rational number on the number line. Draw a line and locate a point that has a coordinate of 0. This is the point of origin of the number line. If the rational number given is positive, this is located on the right of zero. If the rational number is negative, this is represented on the left of zero.

So if you have to mark (-1/2) and (½) on the number line, these are to be marked at equidistant from 0 and halfway between 0 and 1 is where you place (1/2) and halfway between 0 and -1, you need to place (-1/2).

If you need to mark 4/5 on the number line, divide the unit into four parts between 0 and 1 and represent 4/5 on it.

9.6 Rational Numbers in Standard Form

In the Rational Numbers Class 7th Maths Chapter 9, the concept of rational numbers in its standard form is explained. A rational number is in its standard form if it has a positive integer in its denominator and there are no other common factors between the numerator and the denominator instead of 1.

Like for example, 1/3 is a rational number and the only common factor between the divisor and the dividend is 1. This means that 1/3 is in its standard form.

A rational number in the standard form can also have a negative number but only in its numerator. If there is a rational number that is not in its standard form, it can be reduced to bring it to its standard form. This can be done by reducing the fraction.

9.7 Comparison of Rational Numbers

In this section of Chapter 9 Class 7 Maths, you learn how to compare two Rational Numbers. Here are the steps that need to be followed. 

• Write the two Rational Numbers in such a way that their denominators are positive. 

• Find the LCM of the positive denominators and express each of the Rational Numbers with the LCM as the common denominator.

• Then compare the two Rational Numbers obtained where the number that has a greater numerator is the higher rational number.

If you need to compare two negative Rational Numbers, compare them by ignoring their negative sign and, in the end, reverse the comparison.

If you need to compare a negative and a positive rational number, it is obvious that the negative rational number is lower than the positive rational number.

9.8 Rational Numbers Between Two Rational Numbers

If there are two Rational Numbers and m<n, the rational number between the two Rational Numbers are 1/2 (m + n).

To insert Rational Numbers between two Rational Numbers, you go back to the integers and its properties on various operations. If the two integers are non-consecutive and suppose the two integers are x and y, there can be (x - y - 1) integers between them. There can be no integers in between two consecutive integers. There can be infinite Rational Numbers that can be inserted between two Rational Numbers. This property is known as dense property.

If there are two Rational Numbers, say ‘a’ and ‘b’, and they do not have the same denominator, convert the denominator of the fractions using the LCM method to the same denominator. If there is a number that can be written between the two numbers, this is the rational number between them.

If there can be no numbers between the numerators, the numerator and the denominator should be multiplied by 10, and this will let you get the Rational Numbers between them. To get more Rational Numbers, you can multiply with multiples of 10 say 100,1000 etc.

Exercise 9.1 Solutions: 10 Questions (7 Short Questions).

9.9 Operations on Rational Numbers

This NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers section explains how you can add, subtract, multiply and divide the Rational Numbers similar to how these operations can be done on integers and fractions.

9.9.1 Addition

If there are two integers and their denominators are the same, the addition operation is done by adding the numerators and keeping the denominator the same.

Below are the steps to add two Rational Numbers with different denominators.

• If the denominator is not positive, the rational number has to be adjusted to have positive denominators.

• The next step is to find the LCM of the denominator of the given Rational Numbers.

• Now the Rational Numbers need to be converted so that they have the same denominator. This is done by multiplying the numerator and the denominator with the common multiple.

• Now add the two numbers of the converted Rational Numbers and simplify if possible. This is the final answer.

Additive Inverse

The section on Rational Numbers Class 7 NCERT Solutions explains additive inverse. Additive inverse is the opposite number. The inverse is at the same distance on the number line from 0 in the opposite direction. The additive inverse of zero is zero. The additive inverse of the Rational Numbers is the same number but with an opposite sign. When the rational number and its additive inverse are added, the result is zero.

9.9.2 Subtraction

In this section, you learn about subtraction of Rational Numbers. If there are two Rational Numbers a/b and c/d and you subtract c/d from a/b, this equals adding the additive inverse of c/d to a/b. Thus when c/d needs to be subtracted from a/b, the subtraction is written as (a/b) + (-c/d). Once the Rational Numbers are written in this form, the same method of addition of two Rational Numbers needs to be applied here.

9.9.3 Multiplication

This section explains how two Rational Numbers are multiplied. For multiplication of two Rational Numbers say a/b and c/d, all that has to be done is to multiply the numerators and the denominators of the two Rational Numbers. Thus you get a/b × c/d.

9.9.4 Division

The division is the inverse of multiplication and the same concept applies in dividing Rational Numbers as well. If there are two Rational Numbers say w/x and y/z and y/z does not equal to 0, w/x ÷ y/z = w/x × z/y.

Exercise 9.2 Solutions: 4 questions (4 Short Questions).

Key features of the NCERT solution for Class 7 Maths Chapter 9

The solutions of Rational Numbers Class 7 CBSE NCERT need to be practised and solved thoroughly to score good marks in the examination. All these solutions have been prepared very thoughtfully to ensure that students get a good grasp of the concepts.

• The Class 7 Maths Chapter 9 solutions let students have a good foundation of the topics.

• All the exercises are solved by the subject matter experts in a very meticulous manner.

• Solving and going through these solutions will help to boost the confidence of students and let them score better.

• Rational Numbers NCERT Class 7 Solutions let students get a good grasp of the topic in a very short time.