# Rational Numbers Chapter 1 Class 8 Maths NCERT Textbook With Solutions PDF

NCERT Solutions for Class 8 Maths Chapter 1‘ PDF Quick download link is given at the bottom of this article. You can see the PDF demo, size of the PDF, page numbers, and direct download Free PDF of ‘Ncert Class 8 Maths Chapter 1 Exercise Solution’ using the download button.

### Chapter 1: Rational Numbers

#### 1.1 Introduction

In Mathematics, we frequently come across simple equations to be solved. For example, the equation x + 2 = 13 (1) is solved when x = 11 because this value of x satisfies the given equation. Solution 11 is a natural number. On the other hand, for the equation x + 5 = 5 (2)

the solution gives the whole number 0 (zero). If we consider only natural numbers, equation (2) cannot be solved.

To solve equations like (2), we added the number zero to the collection of natural numbers and obtained the whole numbers. Even whole numbers will not be sufficient to solve equations of the type x + 18 = 5 (3)

Do you see ‘why’? We require the number –13 which is not a whole number. This led us to think of integers, (positive and negative). Note that the positive integers correspond to natural numbers.

One may think that we have enough numbers to solve all simple equations with the available list of integers. Consider the equations
2x = 3 (4)
5x + 7 = 0 (5)

for which we cannot find a solution from the integers. (Check this) We need the numbers 3/2 to solve equation (4) and 7/5 − to solve equation (5). This leads us to the collection of rational numbers.

We have already seen basic operations on rational numbers. We now try to explore some properties of operations on the different types of numbers seen so far.

### NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers

1. Using appropriate properties find.

(i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6

Solution:

-2/3 × 3/5 + 5/2 – 3/5 × 1/6

= -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity)

= 3/5 (-2/3 – 1/6)+ 5/2

= 3/5 ((- 4 – 1)/6)+ 5/2

= 3/5 ((–5)/6)+ 5/2 (by distributivity)

= – 15 /30 + 5/2

= – 1 /2 + 5/2

= 4/2

= 2

(ii) 2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5

Solution:

2/5 × (- 3/7) – 1/6 × 3/2 + 1/14 × 2/5

= 2/5 × (- 3/7) + 1/14 × 2/5 – (1/6 × 3/2) (by commutativity)

= 2/5 × (- 3/7 + 1/14) – 3/12

= 2/5 × ((- 6 + 1)/14) – 3/12

= 2/5 × ((- 5)/14)) – 1/4

= (-10/70) – 1/4

= – 1/7 – 1/4

= (– 4– 7)/28

= – 11/28

2. Write the additive inverse of each of the following

Solution:

(i) 2/8

Additive inverse of 2/8 is – 2/8

(ii) -5/9

Additive inverse of -5/9 is 5/9

(iii) -6/-5 = 6/5

The additive inverse of 6/5 is -6/5

(iv) 2/-9 = -2/9

The additive inverse of -2/9 is 2/9

(v) 19/-16 = -19/16

The additive inverse of -19/16 is 19/16

3. Verify that: -(-x) = x for.

(i) x = 11/15

(ii) x = -13/17

Solution:

(i) x = 11/15

We have, x = 11/15

The additive inverse of x is – x (as x + (-x) = 0)

Then, the additive inverse of 11/15 is – 11/15 (as 11/15 + (-11/15) = 0)

The same equality 11/15 + (-11/15) = 0, shows that the additive inverse of -11/15 is 11/15.

Or, – (-11/15) = 11/15

i.e., -(-x) = x

(ii) -13/17

We have, x = -13/17

The additive inverse of x is – x (as x + (-x) = 0)

Then, the additive inverse of -13/17 is 13/17 (as 13/17 + (-13/17) = 0)

The same equality (-13/17 + 13/17) = 0, shows that the additive inverse of 13/17 is -13/17.

Or, – (13/17) = -13/17,

i.e., -(-x) = x