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# NCERT Solutions Class 7 Maths Chapter 8 Rational Numbers

Last updated date: 13th Sep 2024
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## NCERT Solutions for Maths Chapter 8 Rational Numbers Class 7 - FREE PDF Download

Class 7 Maths NCERT Solutions covers rational numbers, including both positive and negative fractions as well as whole numbers. This chapter is crucial as rational numbers are foundational for many advanced maths topics. It includes the properties of rational numbers, their representation on a number line, and operations like addition, subtraction, multiplication, and division.

Table of Content
1. NCERT Solutions for Maths Chapter 8 Rational Numbers Class 7 - FREE PDF Download
2. Glance on Maths Chapter 8 Class 7 - Rational Numbers
3. Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 7
4. Exercises Under NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers
5. Access NCERT Solutions for Class 7 Maths Chapter 8 – Rational Numbers
6. Class 7 Maths Chapter 8: Exercises Breakdown
7. Other Study Material for CBSE Class 7 Maths Chapter 8
8. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

Focus on understanding equivalent rational numbers and simplifying them. It's also important to master the rules for arithmetic operations with rational numbers, as this will make solving complex problems easier. This chapter lays the groundwork for future mathematical concepts.

## Glance on Maths Chapter 8 Class 7 - Rational Numbers

• Rational Numbers Class 7 explains rational numbers, which are numbers that can be expressed as a fraction where both the numerator and the denominator are integers, and the denominator is not zero.

• Rational numbers include both positive and negative fractions, as well as whole numbers.

• The chapter also covers the representation of rational numbers on a number line, showing their relative positions.

• Addition and subtraction of rational numbers are explained by finding a common denominator and combining the numerators.

• Multiplication of rational numbers involves multiplying the numerators together and the denominators together.

• Division of rational numbers is described as multiplying by the reciprocal of the divisor.

• The concept of equivalent rational numbers is discussed, which are different fractions that represent the same value.

• This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 8 - Rational Numbers, which you can download as PDFs.

• There are two exercises (14 fully solved questions) in maths chapter 8 rational numbers class 7 PDF.

## Access Exercise Wise NCERT Solutions for Chapter 8 Maths Class 7

 S.No. Current Syllabus Exercises of Class 7 Maths Chapter 8 1 NCERT Solutions of Class 7 Maths Rational Numbers Exercise 8.1 2 NCERT Solutions of Class 7 Maths Rational Numbers Exercise 8.2

## Exercises Under NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers

• 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

 Exercise Number of Questions Exercise 8.1 10 Questions (7 Short Questions) Exercise 8.2 4 questions (4 Short Questions)

## 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.

## 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.

## FAQs on NCERT Solutions Class 7 Maths Chapter 8 Rational Numbers

1. What are the rational numbers? How to classify them?

The word ‘rational’ is derived from the term ‘ratio’. Any number which can be expressed in the form of a ratio is a rational number. Mathematically, a rational number is defined as a number which can be expressed in the form of p/q where p and q are integers and q ≠ 0.

• A rational number can be classified into Positive and Negative rational numbers.

• Positive rational numbers are the rational numbers where both numerator and denominator of the numbers are positive integers. For example 3/8, 5/7, etc.

• Negative rational numbers are the rational numbers where the numerator is a negative integer and the denominator is a positive integer. For example -5/8, -6/7, etc.

2. How are NCERT Solutions for Class 7 Maths Chapter 8 useful for students?

NCERT Solutions for Class 7 Maths designed by experts at Vedantu is immensely helpful in exam preparation. These solutions are a one-stop destination where you can find explanations to all the exercise questions of a chapter. The solutions are provided by subject experts in a step-by-step manner. For students facing doubts in Chapter 8 of Class 7 Maths Rational Numbers, NCERT Solutions free PDF is a great way to clear all their doubts. Students can instantly understand how to solve the chapter and this will enhance their confidence as well as scores in exams.

3. What are the sub-topics included in class 7 maths chapter 8 PDF solutions?

Following are topics presented in class 7 rational numbers solutions:

• Introduction

• Need for Rational Numbers

• What are Rational Numbers

• Positive and Negative Rational Numbers

• Rational Numbers on a Number Line

• Rational Numbers in Standard Form

• Comparison of Rational Numbers

• Rational Numbers Between Two Rational Numbers

• Operations on Rational Numbers

• Addition of Rational Numbers

• Subtraction of Rational Numbers

• Multiplication of Rational Numbers

• Division of Rational Numbers

4. What are some of the questions that can be asked from Class 7 Maths Chapter 8 Rational Numbers?

Students can be asked several types of questions from the 9th chapter of Class 7 Maths in the exam. In this chapter, students are taught how to write rational numbers between two rational numbers. Students can be asked the same in the exam. Students can also be asked to draw a number line and represent rational numbers on them. Other types of questions that are likely to be asked in the exam of ncert class 7 maths chapter 8  are to find rational numbers equivalent to a particular rational number, to rewrite any rational number in its simplest form, to arrange rational numbers in ascending order, to compare between two rational numbers to find greater than or less than, etc.

5. How can we score full marks in the Class test of NCERT Solutions for rational numbers class 7 PDF?

If you want to score full marks in the class test of NCERT class 7 maths chapter 8 , you should get access to the NCERT Solutions for Class 7 Maths Chapter 8 on the Vedantu website, in which complete explanations of the chapters are given in a very easy language, and the exercises are also solved step-by-step. Practising these questions can help you in scoring full marks in your class test.

6. What are the main concepts covered in NCERT Solutions for maths class 7 chapter 8?

NCERT Solutions for maths class 7 chapter 8 on Vedantu has full explanations of Chapter 8 that are there in the NCERT Maths textbook for class 7. The main concepts covered in the NCERT Solutions in PDF are Rational Numbers, Equivalent Rational Numbers, Positive & Negative Rational Numbers, Rational Numbers On A Number Line, Rational Numbers In Standard Form, Comparison Of Rational Numbers, Rational Numbers Between Two Rational Numbers, and Arithmetic Operations On Rational Numbers.

7. Which is not a rational number in the syllabus of class 7 maths ch 8?

Any number, which cannot be expressed as a simple fraction, is not rational. For similar questions, visit Vedantu page class 7 maths ch 8 and download NCERT Solutions PDF free of cost to get access to the full explanation of the chapter along with solved exercise questions and extra related questions. Practising the class 7 maths ch 8will not only boost your confidence but also improve your scores in your tests. All the resources are also available on the Vedantu app.

8. Do I need to solve all questions provided in NCERT Solutions for Class 7 Maths?

Yes, solving all questions provided in NCERT Solutions for Class 7 Maths is immensely important because this will help you in covering all topics present in the chapter. The full completion will ensure that you can solve any questions that may come from this chapter. The NCERT Solutions Class 7 Maths has many extra questions and solved examples that will even further enhance your grip on the topics.

9. What are the chapters present in the NCERT Solutions for Class 7 Maths?

Vedantu’s NCERT Solutions Class 7 Maths has 13 chapters from the NCERT textbook;

• Chapter 1 - Integers

• Chapter 2 - Fractions and Decimals

• Chapter 3 -Data Handling

• Chapter 4- Simple Equations

• Chapter 5- Lines and Angles

• Chapter 6- The Triangle and Its Properties

• Chapter 7- Comparing Quantities

• Chapter 8- Rational Numbers

• Chapter 9 - Perimeter and Area

• Chapter 10 - Algebraic Expressions

• Chapter 11 - Exponents and Powers

• Chapter 12 - Symmetry

• Chapter 13 - Visualising Solid Shapes