NCERT Solutions For Class 7 Maths Chapter 8 Rational Numbers - 2025-26

• Exercise 8.1 introduces the fundamental concepts of rational numbers. It includes tasks that involve identifying rational numbers and plotting them on a number line. Additionally, students compare and order rational numbers and explore the distinctions between positive and negative rational numbers.

• Exercise 8.2 focuses on performing arithmetic operations with rational numbers, such as addition, subtraction, multiplication, and division. This exercise aims to improve students' computational skills and their ability to simplify expressions involving rational numbers.

Access NCERT Solutions for Class 7 Maths Chapter 8 – Rational Numbers

Exercise – 8.1

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 - 8.2

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}$

Class 7 Maths Chapter 8: Exercises Breakdown

Conclusion

Chapter 8 on Rational Numbers is vital for understanding the basics of rational numbers and their operations. This chapter emphasizes the properties of rational numbers, their representation on a number line, and performing arithmetic operations like addition, subtraction, multiplication, and division. Focusing on these areas is crucial as they form the foundation for more complex mathematical concepts. In the previous year's exams, about 4 to 5 questions were asked from this chapter, indicating its significance. Mastering these exercises will greatly benefit students in their academic performance.

Other Study Material for CBSE Class 7 Maths Chapter 8

Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.