# NCERT Solutions For Class 8 Maths Chapter 1

## NCERT Solutions Class 8 Maths Rational Numbers

### Ncert Solutions For Class 8 Maths Chapter 1 PDF Free Download

NCERT solutions for class 8 maths chapter 1 Rational numbers are provided here. Class 8 is an important phase of a student’s life, the concepts which are taught in class 8 are essential in understanding the topics of class 9 and 10. To score good marks in class 8 mathematics examination it is advised to solve questions provided at the end of each chapter in the NCERT book. Solving NCERT questions will help you to know about the different types of questions with their difficulty level. Students can download class 8 maths chapter 1 NCERT Solutions pdf from the links given below.

A rational number can be represented as the ratio of 2 integers. All integers are also rational as they can be divided by 1, giving a ratio of 2 integers. Various floating point numbers are rational since they can be represented as fractions example, 1.25 is a rational number since it can be re-written as 125/100 or, 5/4. π is irrational since it cannot be written in fraction form. A floating point is a rational number if it matches one of the below criteria: Having a limited no. of digits after the decimal such as 15.21. If there are infinite repeating numbers after the decimal without ant pattern then the number is irrational. zero is a rational number because it can be re-written as 0/1, which is the same as 0.

Few examples of irrational and rational numbers are given below.

12 – rational

10.25 – rational

22.10 – rational

√7 – irrational

13.14 – rational

π – irrational

√8 – rational

√10 – irrational

21/7 – rational

Farey sequences present a way of orderly listing all the rational numbers. The basic algebraic procedures for merging rationals are precisely the same as for mixing fractions. It is always desirable to search rationals between any 2 members of the rationals. Therefore, the rational numbers are constant set, but at the same time, they are countable. We have provided a comprehensive study material of NCERT Solutions For Class 8 Maths Rational Numbers. The students can also download the PDFs from the link provided here.

### NCERT Solutions Class 8 Maths Chapter 1 Exercises

Exercise – 1.1

1. Using appropriate properties find the value of $\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$

Sol.   $\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$ – $\frac{-2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}$

=$\frac{-3}{5}(\frac{2}{3}+\frac{1}{6})+\frac{5}{2}$

=$\frac{-3}{5}(\frac{4+1}{6})+\frac{5}{2}$

=$\frac{-3}{5}(\frac{5}{6})+\frac{5}{2}$

=$\frac{-15}{30}+\frac{5}{2}$

=$\frac{-1}{2}+\frac{5}{2}$

=$\frac{4}{2}$

=$2$

$\ therefore\frac{-2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$ =2

1. Using appropriate properties find the value of $\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

Sol. $\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

= $\frac{-2}{5}\times \frac{3}{7}-\frac{1}{2}\times \frac{1}{2}+\frac{1}{7}\times \frac{1}{5}$

=$\frac{-6}{35}-\frac{1}{4}+\frac{1}{35}$

=$\frac{-5}{35}-\frac{1}{4}$

=$\frac{-1}{7}-\frac{1}{4}$

=$\frac{-11}{28}$

=$\frac{2}{5}\times \frac{-3}{7}-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$

=$\frac{-11}{28}$

1. Solving the additive inverse of $\frac{2}{8}$

Sol.  Additive inverse of $\frac{2}{8} is \frac{-2}{8}$

1. Solving the additive inverse of $\frac{-5}{9}$

Sol.  Additive inverse of $\frac{-5}{9} is\frac{5}{9}$

1. Solving the additive inverse of $\frac{-6}{-5}$

Sol. $\frac{-6}{-5}\frac{-6}{-5}$

Additive inverse of $\frac{6}{5} is\frac{-6}{5}$

Additive inverse of $\frac{-6}{-5}\frac{-6}{5}$

1. Solving the additive inverse of $\frac{2}{-9}$

Sol.  Additive inverse of $\frac{2}{-9} is\frac{2}{9}$

1. Verify that –(-x )=x for $x=\frac{11}{15}$

Sol. $x = \frac{11}{15}$

$- x =\frac{-11}{15}$

$-(-x)= -(\frac{-11}{15})$

=$\frac{11}{15}=x$

$\ therefore -(-x)=x$

1. Verify that –(-x )=x for $x=\frac{-13}{17}$

Sol. $x = \frac{-13}{17}$

$- x =(-\frac{-13}{17})=\frac{13}{17}$

$-(-x)= -(\frac{-13}{17})$

=$\frac{-13}{17}=x$

$\ therefore -(-x)=x$

1. Solve that multiplicative inverse of -13

Sol. Given multiplicative inverse -13 is$\frac{-1}{13}$

1. Solve that multiplicative inverse of $\frac{-13}{19}$

Sol. Given multiplicative inverse $\frac{-13}{19} is\frac{-19}{13}$

1. Solve that multiplicative inverse of $\frac{-5}{8}\times \frac{-3}{7}$

Sol. Given multiplicative inverse $\frac{-5}{8}\times \frac{-3}{7}$is $\frac{8}{5}\times \frac{7}{3}$ or $\frac{-8}{5}\times \frac{-7}{3}$

1. What is the multiplicative inverse of -1.

Sol.  The multiplicative inverse of -1 is -1.

1. Name the property under multiplicative used in each of the following.

(i) $\frac{-4}{5}\times 1=1\times \frac{-4}{5}=\frac{-4}{5}$

(ii) $\frac{-13}{17}\times \frac{-2}{7}=\frac{-2}{7}\times \frac{-13}{17}$

(iii) $\frac{-19}{29}\times \frac{29}{-19}=1$

Sol. (i) ROLE OF 1

(ii) COMMUTATIVITY

(iii) MULTIPLICATIVE  INVERSE

1. Multiply $\frac{6}{13}$ by the reciprocal of $\frac{-7}{6}$

Sol. Reciprocal of $\frac{-7}{6}$ is $\frac{6}{-7}$

$\ therefore \frac{6}{13}\times \frac{6}{-7}$

=$\frac{-36}{91}$

1. what property allows you to compute $\frac{1}{3}\times (6\times \frac{4}{3})(\frac{1}{3}\times 6)\times \frac{4}{5}$

Sol.  Associativity

1. Is $\frac{8}{9}$ the multiplicative inverse of $-1\frac{1}{8}$ ? why or why not.

Sol.

$-1\frac{1}{8} = \frac{-9}{8}$

=$\frac{8}{9}\times \frac{-9}{8}$

=$-1 \neq 1$

$\ therefore\frac{8}{9}$ is  not the multiplicative inverse of $-1\frac{1}{8}$

1. Is 0.3 the multiplicative inverse of $3\frac{1}{3}$ ? why or why not.

Sol.        $3\frac{1}{3} = \frac{10}{3} =3.3$

$3.3\times 0.3= 0.99\neq 1$

$\ therefore \, 0.3\, is\, not\, the\, multiplicative\, inverse\, of\, 3\frac{1}{3}$

Exercise 1.2:

Question 1:

Solve the following using appropriate properties:

1. ii)$\frac{2}{5}* \frac{-3}{7} – \frac{1}{6} *\frac{3}{2}+\frac{1}{14}*\frac{2}{5}$

1. ii) Use commutativity of rational numbers

$\frac{2}{5}* \frac{-3}{7} – \frac{1}{6} *\frac{3}{2}+\frac{1}{14}*\frac{2}{5}$

Using distributive property

$\frac{2}{5} * [\frac{-3}{7} + \frac{1}{14}] – \frac{1}{4}$

= $\frac{2}{5} * \frac{-3*2+1}{14} – \frac{1}{4}$

=$\frac{2}{5} * \frac{-5}{14} – \frac{1}{4}$

=$\frac{-4-7}{28} = \frac{-11}{28}$

Question 2:

(i) $\frac{-5}{-7}$

(ii) $\frac{-4}{-7}$

(iii)$\frac{2}{-5}$

(iv)$\frac{5}{-9}$

Solution:

(i) $\frac{-5}{-7}$

additive inverse = $\frac{5}{-7}$

(ii)  $\frac{-4}{-7}$ = $\frac{4}{7}$

Additive inverse = $\frac{-4}{7}$

(iii)  $\frac{2}{-5}$

Additive inverse = $\frac{2}{5}$

(iv)  $\frac{5}{-9}$

Additive inverse = $\frac{5}{9}$
Question 3:

(i)  x = $\frac{11}{15}$

(ii) x= $\frac{-13}{17}$

Solution:

x = $\frac{11}{15}$

The additive inverse of  is   x = $\frac{-11}{15}$

This equality

x = $\frac{11}{15}+frac{-11}{15}= 0$

=$-(-x) = x$

x= $\frac{-13}{17}$

The additive inverse of  x= $\frac{-13}{17}$ =  x= $\frac{13}{17}$

This equaliity

x = $\frac{-13}{17}+frac{13}{17}= 0$

=$-(-x) = x$

Question 4:

(i)  -12

(ii) $\frac{-19}{13}$

(iii) $\frac{1}{3}$

(iv)  $\frac{-3}{8} * \frac{-3}{8}$

(v)  $-1 * \frac{-3}{5}$

Solutions:

1. The multiplicative inverse of  -12 is $\frac{1}{12}$
2. The multiplicative inverse of $\frac{-19}{13}$ is$\frac{-13}{19}$
3. The multiplicative inverse of $\frac{1}{3}$  is $\frac{3}{1}$ = 3
4. The multiplicative inverse of  $-1 * \frac{-3}{5}$ is $\frac{5}{3}$

Question 5:

Find out the multiplication property that is used in the following sums:

(i) $\frac{-3}{7} * 1 = 1 * \frac{-3}{7} = \frac{-3}{7}$

(ii) $\frac{-14}{17}* \frac{-3}{4} = \frac{-3}{4} * \frac{-14}{17}$

(iii) $\frac{-17}{37} * \frac{37}{-17} = 1$

Solutions:

1. This use the multiplicative identity property and the multiplicative identity is 1
2. This uses the commutativity property
3. This uses the multiplicative inverse property

Question 6:

Find the product of  $\frac{7}{15}$ and the reciprocal of $\frac{-4}{17}$

Solution:

Reciprocal of  $\frac{-4}{17}$ =  $\frac{-17}{14}$

$\frac{7}{15}* \frac{-17}{14} = \frac{119}{60}$

Question 7:

1. Which rational number does not have a reciprocal
2. Which rational number is equal to its own reciprocal
3. Which rational number is equal to its own negative

Solution:

1. Zero is the rational number that has no reciprocal
2. One and negative one are rational numbers that are equal to their own reciprocal
3. Zero is a rational number that is also equal to its own negative.

Question 8: Answer or complete the following sentences:

1. Does zero have a reciprocal?
2. The two numbers _____ and _________ are their own reciprocals.
3. What is the reciprocal of -6?
4. What is the reciprocal of ?
5. The product of two rational numbers is always a __________.
6. The reciprocal of a positive rational number is ___________.

Solutions:

1. No it doesn’t
2. 1, -1
3. -1/6
4. X
5. Rational number
6. Positive rational number

Question 9:

Find the multiplicative inverse of the following:

1. -12
2. $\frac{-12}{18}$
3. $\frac{2}{7}$
4. -1
5. $\frac{-3}{8} * \frac{-5}{7}$
6. $-2 * \frac{-2}{7}$

Solutions:

1. $\frac{-1}{12}$
2. $\frac{-12}{18}$
3. $\frac{7}{2}$
4. -1
5. $\frac{56}{15}$
6. $\frac{7}{2}$

Question 10:

Is 0.3 the multiplicative inverse of   $3 \frac{1}{3}$? Why or why not?

$3 \frac{1}{3}= \frac{10}{3}$
$0.3*3 \frac{1}{3} = 0.3 * \frac{10}{3} = \frac{3}{10} * \frac{10}{3} = 1$
Therefore, 0.3 is the multiplicative inverse of  $3 \frac{1}{3}$