# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers: In earlier classes, you have studied whole numbers, natural numbers, integer numbers.

• Whole numbers:-  0, 1, 2, 3, 4.......
• Natural numbers(positive integer):-  1, 2,3, 4, 5 ......
• Integers numbers:- .....-3, -2 ,-1, 0, 1, 2, 3, 4 ........
• Fractions number:- Form $\frac{p}{q}$ where p and q are positive integers and q cannot be zero.

In this article, you will get CBSE NCERT solutions for class 7 maths chapter 9 rational numbers. What are the rational numbers?

• Rational number:- Any number which can be written in the form of p/q where $q\neq0$  and p and q are integers is known as rational numbers. So you can say that all integers are also rational numbers as they can be written as I/1.

Have you able to find the difference between the fraction number and a rational number?

Fractions number is a rational number that contains only positive integers whereas the rational number contains positive and negative integers. All fractions are rational numbers but all rational numbers are not fractions. For example, -5 is not a fraction since -5 is a negative integer but -5 is a rational number. Is zero a rational number? Think about it. Yes, zero is a rational number as 0 can be written as 0/1 or 0/2 or 0/3 .......which is of the form p/q where q is not equal to zero. Once you go through solutions of NCERT for class 7 maths chapter 9 rational numbers, you will get more clarity of the concepts. There are 14 questions in 2 exercises given in the textbook. In CBSE NCERT solutions for class 7 maths chapter 9 rational numbers, you will get all detailed explanations of all these questions including practice question given at end of the very topic. You can get NCERT Solutions by clicking on the above link. Here you will get solutions to two exercises of this chapter.

Exercise:9.1

Exercise:9.2

## The main topics of the NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers are-

9.1 Introduction

9.2 Need For Rational Numbers

9.3 What Are Rational Numbers?

9.4 Positive And Negative Rational Numbers

9.5 Rational Numbers On A Number Line

9.6 Rational Numbers In Standard Form

9.7 Comparison Of Rational Numbers

9.8 Rational Numbers Between Two Rational Numbers

9.9 Operations On Rational Numbers

## NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers are given below:

Solutions of NCERT for class 7 maths chapter 9 rational numbers topic 9.3 what are rational numbers?

Yes$\frac{2}{-3}$ is a rational number because it is written in the form: $\frac{p}{q}$ , where $p = 2\ and\ q = -3\neq 0$.

Question:2

Any ten rational numbers are:

$1, \frac{2}{3}, -\frac{1}{3}, -5, 0, 356, -\frac{39}{25}, -36, 1999, \frac{-1}{-2}$

Solutions for class 7 maths chapter 9 rational numbers topic 9.3 subtopic equivalent rational numbers

Question: Fill in the boxes:

(i)     $\frac{5}{4}=\frac{\square }{16}=\frac{25}{\square }=\frac{-15}{\square }$                     (ii)     $\frac{-3}{7}=\frac{\square }{14}=\frac{9}{\square }=\frac{-6}{\square }$

(i) $\frac{5}{4}=\frac{\square }{16}=\frac{25}{\square }=\frac{-15}{\square }$

$\frac{5}{4}$  can be written as:

$\Rightarrow \frac{5}{4} = \frac{5\times4}{4\times4} = \frac{20}{16}$

$\Rightarrow \frac{5}{4} = \frac{5\times5}{4\times5} = \frac{25}{20}$

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

Hence, we have

$\frac{5}{4}=\frac{20 }{16}=\frac{25}{20 }=\frac{-15}{-12 }$

(ii) $\frac{-3}{7}=\frac{\square }{14}=\frac{9}{\square }=\frac{-6}{\square }$

$\frac{-3}{7}$  can be written as:

$\Rightarrow \frac{-3}{7} = \frac{-3\times2}{7\times2} = \frac{-6}{14}$

$\Rightarrow \frac{-3}{7} = \frac{-3\times-3}{7\times-3} = \frac{9}{-21}$

$\Rightarrow \frac{-3}{7} = \frac{-3\times2}{7\times2} = \frac{-6}{14}$

Hence, we have

$\frac{-3}{7}=\frac{-6 }{14}=\frac{9}{-21}=\frac{-6}{14 }$

Solutions for class 7 maths chapter 9 topic 9.4 positive and negative rational numbers

Question:1

Yes, 5 can be written as a positive rational number $\frac{5}{1}$, where 5 and 1 are both positive integers and denominator not equal to zero.

Five more positive rational numbers are:

$4,\ \frac{1}{3},\ 56,\ \frac{99}{98}, 5$

Question:1

Yes$-8$  is a negative rational number because it can be written as $\frac{-8}{1}$ , where the numerator is negative integer and denominator is a positive integer.

Five more negative rational numbers are:

$-1,\ -99,\ -\frac{2}{3},\ -\frac{5}{7}, \ -27$

(i)     $\frac{-2}{3}$            (ii)     $\frac{5}{7}$            (iii)    $\frac{3}{-5}$        (iv)     0        (v)     $\frac{6}{11}$            (vi)     $\frac{-2}{-9}$

(i) $\frac{-2}{3}$ here, the numerator is -2 which is negative and the denominator is 3 which is positive.

Hence, the fraction is negative.

(ii) $\frac{5}{7}$ here, the numerator is 5 which is positive and the denominator is 7 which is also positive.

Hence, the fraction is positive.

(iii) $\frac{3}{-5}$ here, the numerator is 3 which is positive and the denominator is -5 which is negative.

Hence, the fraction is negative.

(iv) 0 zero is neither positive nor a negative number.

(v) $\frac{6}{11}$ here, the numerator is 6 which is positive and the denominator is 11 which is also positive.

Hence, the fraction is positive.

(vi) $\frac{-2}{-9}$ here, the numerator is -2 which is negative and the denominator is -9 which is also a negative integer.

Hence, the fraction is overall a positive fraction.

Solutions of NCERT for class 7 maths chapter 9 topic 9.6 rational numbers in standard form

Question: Find the standard form of

(i)     $\frac{-18}{45}$                        (ii)     $\frac{-12}{18}$

(i) Given fraction $\frac{-18}{45}$.

We can make it in the standard form :

$\frac{-18}{45} = \frac{-6\times3}{15\times3} = \frac{-2\times3\times3}{5\times3\times3} = \frac{-2}{5}$

(i) Given fraction $\frac{-12}{18}$.

We can make it in the standard form :

$\frac{-12}{18} = \frac{-6\times2}{9\times2} = \frac{-2\times3\times2}{3\times3\times2} = \frac{-2}{3}$

CBSE NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.8 rational numbers between two rational numbers

Question: Find five rational numbers between

$\frac{-5}{7} and \frac{-3}{8}$

LCM of 7 and 8 is 56.

Hence we can write given fractions as:

$\frac{-5}{7} = \frac{-5\times8}{7\times8} = \frac{-40}{56}$  and  $\frac{-3}{8} = \frac{-3\times7}{8\times7} = \frac{-21}{56}$

Therefore, we can find five rational numbers between $\frac{-5}{7} and \frac{-3}{8}$.

$\frac{-39}{56},\ \frac{-38}{56},\ \frac{-37}{56},\ \frac{-36}{56},\ and\ \frac{-35}{56}$

## Question:1(i) List five rational numbers between:

–1 and 0

To find five rational numbers between $-1\ and\ 0$ we will convert each rational numbers as a denominator $5+1 =6$, we have

$-1 = \frac{-1\times6}{6} = \frac{-6}{6}\ and\ \frac{0\times6}{6} = \frac{0}{6}$

So, we have five rational numbers between $\frac{-6}{6}\ and\ \frac{0}{6}$

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

Hence, the five rational numbers between -1 and 0 are:

$\frac{-5}{6},\frac{-4}{6},\frac{-3}{6},\frac{-2}{6}\ and\ \frac{-1}{6}.$

Question:

–2 and –1

To find five rational numbers between $-2\ and\ -1$ we will convert each rational numbers as a denominator $5+1 =6$, we have

$-2 = \frac{-2\times6}{6} = \frac{-12}{6}\ and\ \frac{-1\times6}{6} = \frac{-6}{6}$

So, we have five rational numbers between $\frac{-12}{6}\ and\ \frac{-6}{6}$

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

Hence, the required rational numbers are

$\frac{-11}{6},\frac{-5}{3},\frac{-3}{2},\frac{-4}{3}\ and\ \frac{-7}{6}.$

Question:

$\frac{-4}{5}and \frac{-2}{3}$

To find five rational numbers between $\frac{-4}{5}and \frac{-2}{3}$ we will convert each rational numbers with the denominator as $5\times3 =15$, we have  $\left ( \because LCM\ of\ 5\ and\ 3 = 15 \right )$

$\frac{-4}{5} = \frac{-4\times3}{5\times3} = \frac{-12}{15}\ and\ \frac{-2}{3} = \frac{-2\times5}{3\times5} = \frac{-10}{15}$

Since there is only one integer i.e., -11 between -12 and -10, we have to find equivalent rational numbers.

$\frac{-12}{15} = \frac{-12\times3}{15\times3} = \frac{-36}{45}\ and\ \frac{-10}{15} = \frac{-10\times3}{15\times3} = \frac{-30}{45}$

Now, we have five rational numbers possible:

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

Hence, the required rational numbers are

$\frac{-7}{9},\frac{-34}{45},\frac{-11}{15},\frac{-32}{45}\ and\ \frac{-31}{45}.$

Question:

$-\frac{1}{2} and \frac{2}{3}$

To find five rational numbers between $-\frac{1}{2} and \frac{2}{3}$ we will convert each rational numbers in their equivalent numbers, we have

Making denominator as LCM(2,3)=6

that is

$\frac{-3}{6}\ and\ \frac{4}{6}$

Now, we have five rational numbers possible:

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

Hence, the required rational numbers are

$\frac{-1}{3},\frac{-1}{6},0,\frac{1}{3}\ and\ \frac{1}{2}.$

$\frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20}$

We have the pattern:

$\frac{-3}{5} = \frac{-3\times1}{5\times1}$           $\ \frac{-6}{10} = \frac{-3\times2}{5\times2}$              $\frac{-9}{15} = \frac{-3\times3}{5\times3}$               $\frac{-12}{20} = \frac{-3\times4}{5\times4}$

Now, following the same pattern, we have

$\frac{-3\times5}{5\times5} = \frac{-15}{25}$        $\frac{-3\times6}{5\times6} = \frac{-18}{30}$            $\frac{-3\times7}{5\times7} = \frac{-21}{35}$              $\frac{-3\times8}{5\times8} = \frac{-24}{40}$

Hence, the required rational numbers are:

$\frac{-15}{25},\ \frac{-18}{30},\ \frac{-21}{35},and\ \frac{-24}{40}.$

$\frac{-1}{4},\frac{-2}{8}, \frac{-3}{12}....$

We have the pattern:

$\frac{-1}{4} = \frac{-1\times1}{4\times1}$           $\frac{-2}{8} = \frac{-1\times2}{4\times2}$              $\frac{-3}{12} = \frac{-1\times3}{4\times3}$

Now, following the same pattern, we have

$\frac{-1\times4}{4\times4} = \frac{-4}{16}$             $\frac{-1\times5}{4\times5} = \frac{-5}{20}$            $\frac{-1\times6}{4\times6} = \frac{-6}{24}$              $\frac{-1\times7}{4\times7} = \frac{-7}{28}$

Hence, the required rational numbers are:

$\frac{-4}{16},\ \frac{-5}{20},\ \frac{-6}{24},and\ \frac{-7}{28}.$

$\frac{-1}{6}, \frac{-2}{12}, \frac{-3}{18}, \frac{-4}{24}....$

We have the pattern:

$\frac{-1}{6} = \frac{-1\times1}{6\times1}$           $\frac{-2}{12} = \frac{-1\times2}{6\times2}$              $\frac{-3}{18} = \frac{-1\times3}{6\times3}$              $\frac{-4}{24} = \frac{-1\times4}{6\times4}$

Now, following the same pattern, we have

$\frac{-1\times5}{6\times5} = \frac{-5}{30}$             $\frac{-1\times6}{6\times6} = \frac{-6}{36}$            $\frac{-1\times7}{6\times7} = \frac{-7}{42}$              $\frac{-1\times8}{6\times8} = \frac{-8}{48}$

Hence, the required rational numbers are:

$\frac{-5}{30},\ \frac{-6}{36},\ \frac{-7}{42},and\ \frac{-8}{48}.$

$\frac{-2}{3}, \frac{2}{-3},\frac{4}{-6}, \frac{6}{-9}....$

We have the pattern:

$\frac{-2}{3} = \frac{-2\times1}{3\times1}$           $\frac{2}{-3} =\frac{2}{-3} = \frac{2\times1}{-3\times1}$              $\frac{4}{-6} = \frac{-2\times2}{3\times2}$              $\frac{-6}{9} = \frac{-2\times3}{3\times3}$

Now, following the same pattern, we have

$\frac{-2\times4}{3\times4} = \frac{-8}{12}\ or\ \frac{8}{-12}$                         $\frac{-2\times5}{3\times5} = \frac{-10}{15}\ or\ \frac{10}{-15}$

$\frac{-2\times6}{3\times6} = \frac{-12}{18}\ or\ \frac{12}{-18}$                     $\frac{-2\times7}{3\times7} = \frac{-14}{21}\ or\ \frac{14}{-21}$

Hence, the required rational numbers are:

$\frac{8}{-12},\ \frac{10}{-15},\ \frac{12}{-18},and\ \frac{14}{-21}.$

Question:

$\frac{-2}{7}$

$\frac{-2}{7}$ can be written as:

$\frac{-2}{7} = \frac{-2\times2}{7\times2} = \frac{-4}{14}$                 $\frac{-2}{7} = \frac{-2\times3}{7\times3} = \frac{-6}{21}$

$\frac{-2}{7} = \frac{-2\times4}{7\times4} = \frac{-8}{28}$                $\frac{-2}{7} = \frac{-2\times5}{7\times5} = \frac{-10}{35}$

Hence, the required equivalent rational numbers are

$\frac{-4}{14},\frac{-6}{21}, \frac{-8}{28},\ and\ \frac{-10}{35} .$

Question:

$\frac{5}{-3}$

$\frac{5}{-3}$ can be written as:

$\frac{5}{-3} = \frac{5\times2}{-3\times2} = \frac{10}{-6}$                 $\frac{5}{-3} = \frac{5\times3}{-3\times3} = \frac{15}{-9}$

$\frac{5}{-3} = \frac{5\times4}{-3\times4} = \frac{20}{-12}$                $\frac{5}{-3} = \frac{5\times5}{-3\times5} = \frac{25}{-15}$

Hence, the required equivalent rational numbers are

$\frac{10}{-6},\frac{15}{-9}, \frac{20}{-12},\ and\ \frac{25}{-20} .$

Question:

$\frac{4}{9}$

$\frac{4}{9}$ can be written as:

$\frac{4}{9} = \frac{4\times2}{9\times2} = \frac{8}{18}$                 $\frac{4}{9} = \frac{4\times3}{9\times3} = \frac{12}{27}$

$\frac{4}{9} = \frac{4\times4}{9\times4} = \frac{16}{36}$                $\frac{4}{9} = \frac{4\times5}{9\times5} = \frac{20}{45}$

Hence, the required equivalent rational numbers are

$\frac{8}{18},\frac{12}{27}, \frac{16}{36},\ and\ \frac{20}{45} .$

$\frac{3}{4}$

Representation of $\frac{3}{4}$ on the number line,

$\frac{-5}{8}$

Representation of $\frac{-5}{8}$ on the number line,

$\frac{-7}{4}$

Representation of $\frac{-7}{4}$ on the number line,

$\frac{7}{8}$

Representation of $\frac{7}{8}$ on the number line,

Given TR = RS = SU and AP = PQ = QB then, we have

There are two rational numbers between A and B i.e., P and Q which are at equal distances hence,

The rational numbers represented by P and Q are:

$P =2+ \frac{1}{3} = \frac{7}{3}\ and\ Q = 2+\frac{2}{3} = \frac{8}{3}$

Also, there are two rational numbers between U and T i.e., S and R which are at equal distances hence,

The rational numbers represented by S and R are:

$S =-2+ \frac{1}{3} = \frac{-5}{3}\ and\ R = -2+\frac{2}{3} = \frac{-4}{3}$

(i)     $\frac{-7}{21}\ and\ \frac{3}{9}$                 (ii)     $\frac{-16}{20} \ and\ \frac{20}{-25}$

(iii)     $\frac{-2}{-3} \ and\ \frac{2}{3}$                (iv)   $\frac{-3}{5}\ and\ \frac{-12}{20}$

(v)     $\frac{8}{-5}\ and\ \frac{-24}{15}$             (vi)  $\frac{1}{3}\ and\ \frac{-1}{9}$

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

To compare we multiply both numbers with denominators:

(i) We have $\frac{-7}{21}\ and\ \frac{3}{9}$

$\Rightarrow \frac{-7\times9}{21\times9} = \frac{-63}{189}$

$\Rightarrow \frac{3\times21}{9\times21} = \frac{63}{189}$

$\Rightarrow \frac{-63}{189} \neq \frac{63}{189}$

Here, they are equal but are in opposite signs hence, $\frac{-7}{21}\ and\ \frac{3}{9}$  do not represent the same rational numbers.

(ii)  We have $\frac{-16}{20} \ and\ \frac{20}{-25}$

$\Rightarrow \frac{-16\times-25}{20\times-25} = \frac{400}{-500}$

$\Rightarrow \frac{20\times20}{-25\times20} = \frac{400}{-500}$

$\Rightarrow \frac{400}{-500} = \frac{400}{-500}$

So, they represent the same rational number.

(iii)  We have $\frac{-2}{-3} \ and\ \frac{2}{3}$

Here, Both represents the same number as these minus signs on both numerator and denominator of $\frac{-2}{-3} = \frac{2}{3}$  will cancel out and gives the positive value.

(iv)  We have $\frac{-3}{5}\ and\ \frac{-12}{20}$

$\Rightarrow \frac{-3\times20}{5\times20} = \frac{-60}{100}$

$\Rightarrow \frac{-12\times5}{20\times5} = \frac{-60}{100}$

$\Rightarrow \frac{-60}{100} = \frac{-60}{100}$

So, they represent the same rational number.

(v)  We have $\frac{8}{-5}\ and\ \frac{-24}{15}$

$\Rightarrow \frac{8\times15}{-5\times15} = \frac{120}{-75}$

$\Rightarrow \frac{-24\times-5}{15\times-5} = \frac{120}{-75}$

$\Rightarrow \frac{120}{-75} = \frac{120}{-75}$

So, they represent the same rational number.

(vi)  We have $\frac{1}{3}\ and\ \frac{-1}{9}$

$\Rightarrow \frac{1\times9}{3\times9} = \frac{9}{27}$

$\Rightarrow \frac{-1\times3}{9\times3} = \frac{-3}{27}$

$\Rightarrow \frac{9}{27} \neq \frac{-3}{27}$

So, They do not represent the same rational number.

(vii)  We have $\frac{-5}{-9}\ and\ \frac{5}{-9}$

Here, the denominators of both are the same but $-5 \neq 5$.

So, $\frac{-5}{-9}\ and\ \frac{5}{-9}$ do not represent the same rational numbers.

(i)        $\frac{-8}{6}$                     (ii)     $\frac{22}{25}$                (iii)     $\frac{-44}{72}$            (iv)     $\frac{-8}{10}$

(i) $\frac{-8}{6}$ can be written as:

$\Rightarrow \frac{-8}{6} = \frac{-8/2}{6/2}= \frac{-4}{3}$                 $\left [ HCF\ of\ 8\ and\ 6\ is\ 2 \right ]$

(ii) $\frac{25}{45}$ can be written in the simplest form:

$\Rightarrow \frac{25}{45} = \frac{25/5}{45/5}= \frac{5}{9}$                       $\left [ HCF\ of\ 25\ and\ 45\ is\ 5 \right ]$

(iii) $\frac{-44}{72}$ can be written as in simplest form:

$\Rightarrow \frac{-44}{72} = \frac{-44/4}{72/4}= \frac{-11}{18}$            $\left [ HCF\ of\ 44\ and\ 72\ is\ 4 \right ]$

(i)     $\frac{-5}{7} \square \frac{2}{3}$                (ii)     $\frac{-4}{5} \square \frac{-5}{7}$               (iii)     $\frac{-7}{8} \square \frac{14}{-16}$

(iv)     $\frac{-8}{5} \square \frac{-7}{4}$          (v)     $\frac{1}{-3} \square \frac{-1}{4}$                (vi)     $\frac{5}{-11} \square \frac{-5}{11}$

(vii)     $0 \square \frac{-7}{6}$

(i) $\frac{-5}{7} \square \frac{2}{3}$

$\Rightarrow \frac{-5\times3}{7\times 3}\square \frac{2\times7}{3\times7}$

$\Rightarrow \frac{-15}{21} < \frac{14}{21}$

Hence, $\frac{-5}{7} < \frac{2}{3}$

(ii) $\frac{-4}{5} \square \frac{-5}{7}$

$\Rightarrow \frac{-4\times7}{5\times 7}\square \frac{-5\times5}{7\times5}$

$\Rightarrow \frac{-28}{35} < \frac{-25}{35}$

Hence, $\frac{-4}{5} < \frac{-5}{7}$

(iii) $\frac{-7}{8} \square \frac{14}{-16}$

$\Rightarrow \frac{-7\times-16}{8\times -16}\square \frac{14\times8}{-16\times8}$

$\Rightarrow \frac{112}{-128} = \frac{112}{-128}$

Hence, $\frac{-7}{8} = \frac{14}{-16}$

(iv) $\frac{-8}{5} \square \frac{-7}{4}$

$\Rightarrow \frac{-8\times4}{5\times 4}\square \frac{-7\times5}{4\times5}$

$\Rightarrow \frac{-32}{20} > \frac{-35}{20}$

Hence, $\frac{-8}{5} > \frac{-7}{4}$

(v) $\frac{1}{-3} \square \frac{-1}{4}$

$\Rightarrow \frac{1\times4}{-3\times 4}\square \frac{-1\times-3}{4\times-3}$

$\Rightarrow \frac{4}{-12} < \frac{3}{-12}$

Hence, $\frac{1}{-3} < \frac{1}{-4}$

(vi) $\frac{5}{-11} \square \frac{-5}{11}$

$\Rightarrow \frac{5\times11}{-11\times 11}\square \frac{-5\times-11}{11\times-11}$

$\Rightarrow \frac{55}{-121} = \frac{55}{-121}$

Hence, $\frac{5}{-11} = \frac{-5}{11}$

(viI)  $0 \square \frac{-7}{6}$

Zero is always greater than every negative number.

Therefore, $0 > \frac{-7}{6}$

Question:

(i)     $\frac{2}{3}, \frac{5}{2}$                           (ii)     $\frac{-5}{6}, \frac{-4}{3}$

(iii)   $\frac{-3}{4}, \frac{2}{-3}$                   (iv)     $\frac{-1}{4}, \frac{1}{4}$

(v)  $-3\frac{2}{7}, -3\frac{4}{5}$

(i)  $\frac{2}{3}, \frac{5}{2}$

$\Rightarrow \frac{2\times2}{3\times2}, \frac{5\times3}{2\times3}$

$\Rightarrow \frac{4}{6}, \frac{15}{6}$

Since, $\frac{15}{6}> \frac{4}{6}$

So, $\frac{5}{2}> \frac{2}{3}.$

(ii)  $\frac{-5}{6}, \frac{-4}{3}$

$\Rightarrow \frac{-5\times3}{6\times3}, \frac{-4\times6}{3\times6}$

$\Rightarrow \frac{-15}{18}, \frac{-24}{18}$

Since, $\frac{-15}{18}> \frac{-24}{18}$

So, $\frac{-5}{6}> \frac{-4}{3}.$

(iii)  $\frac{-3}{4}, \frac{2}{-3}$

$\Rightarrow \frac{-3\times-3}{4\times-3}, \frac{2\times4}{-3\times4}$

$\Rightarrow \frac{9}{-12}, \frac{8}{-12}$

Since, $\frac{8}{-12}> \frac{9}{-12}$

So, $\frac{2}{-3}> \frac{-3}{4}$

(iv)  $\frac{-1}{4}, \frac{1}{4}$

$\Rightarrow \frac{1}{4}> \frac{-1}{4}$

As each positive number is greater than its negative.

(v) $-3\frac{2}{7}, -3\frac{4}{5}$

$\Rightarrow \frac{-23}{7}, \frac{-19}{5} =\frac{-23\times5}{7\times5}, \frac{-19\times7}{5\times7}$

$\Rightarrow \frac{-115}{35} > \frac{-133}{35}$

So, $-3\frac{2}{7}> -3\frac{4}{5}$

Question:

$\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}$

(i) Here the denominator value is the same.

Therefore, $-3<-2<-1$

Hence, the required ascending order is

$\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}$

Question:

$\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

Given $\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

LCM of $3,9\ and\ 3 = 9$.

Therefore, we have

$\frac{1\times3}{3\times3}, \frac{-2\times1}{9\times1}, \frac{-4\times3}{3\times3}$

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

Since $\frac{-12}{9}<\frac{-2}{9}<\frac{3}{9}$

Hence, the required ascending order is

$\frac{-4}{3}<\frac{-2}{9}<\frac{1}{3}$

Question:

$\frac{-3}{7},\frac{-3}{2}, \frac{-3}{4}$

Given $\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

LCM of $7,2\ and\ 4 =28$.

Therefore, we have

$\frac{-3\times4}{7\times4}, \frac{-3\times14}{2\times14}, \frac{-3\times7}{4\times7}$

$\Rightarrow \frac{-12}{28}, \frac{-42}{28}, \frac{-21}{28}$

Since $\frac{-42}{28}<\frac{-21}{28}<\frac{-12}{28}$

Hence, the required ascending order is

$\frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}$

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.1 addition

Question: Find:

$\frac{-13}{7}+\frac{6}{7},\frac{19}{5}+\left ( \frac{-7}{5} \right )$

For the given sum:  $\frac{-13}{7}+\frac{6}{7}$

Here the denominator value is same that is 7 hence we can sum the numerator as:

$\frac{-13}{7}+\frac{6}{7}= \frac{-13+6}{7} = \frac{-7}{7} = -1$

For the given sum:  $\frac{19}{5}+\left ( \frac{-7}{5} \right )$

Here also the denominator value is the same and is equal to 5 hence we can write it as:

$\frac{19}{5}+\left ( \frac{-7}{5} \right ) = \frac{19-7}{5} = \frac{12}{5}$

Question:(i)

$\frac{-3}{7}+\frac{2}{3}$

Given sum: $\frac{-3}{7}+\frac{2}{3}$

Taking LCM of 7 and 3 we get; 21

Hence we can write the sum as:

$\Rightarrow \frac{-3\times3}{7\times3}+\frac{2\times7}{3\times7}$

$\Rightarrow \frac{-9}{21}+\frac{14}{21} = \frac{5}{21}$

Question:(ii)

$\frac{-5}{6}+\frac{-3}{11}$

Given sum: $\frac{-5}{6}+\frac{-3}{11}$

Taking LCM of 6 and 11 we get; 66

Hence we can write the sum as:

$\Rightarrow \frac{-5\times11}{6\times11}+\frac{-3\times6}{11\times6}$

$\Rightarrow \frac{-55}{66}+\frac{-18}{66} = \frac{-73}{66}$

## Solutions of NCERT for class 7 maths chapter 9 rational numbers topic 9.9.2 subtraction

The additive inverse of $\frac{-3}{9} \ is \ \frac{3}{9}$

The additive inverse of $\frac{-9}{11} \ is \ \frac{9}{11}$

The additive inverse of  $\frac{5}{7} \ is \ \frac{-5}{7}$

$(i)\frac{7}{9}-\frac{2}{5} =\frac{7 \times 5-9 \times2}{45}=\frac{17}{45} \\ (ii) 2 \frac{1}{5}-\frac{(-1)}{3}=\frac{11}{5}+\frac{1}{3}=\frac{11 \times 3+ 1 \times 5}{15}=\frac{38}{15}$

Solutions for class 7 maths chapter 9 rational numbers topic 9.9.3 multiplication

Question: What will be

(i)     $\frac{-3}{5}\times 7$                    (ii)     $\frac{-6}{5}\times (-2)$

(i) $\frac{-3}{5}\times 7$

We can write the product as:

$\frac{-3}{5}\times 7 = \frac{-3}{5}\times\frac{7}{1} = \frac{-21}{5}$

(i) $\frac{-6}{5}\times (-2)$

We can write the product as:

$\frac{-6}{5}\times (-2) = \frac{-6}{5}\times\frac{-2}{1} = \frac{12}{5}$

Question:(i)

$\frac{-3}{4}\times \frac{1}{7}$

Given product:

$\frac{-3}{4}\times \frac{1}{7} = \frac{-3\times1}{4\times7} = \frac{-3}{28}$

Question:(ii) Find:

$\frac{2}{3}\times \frac{-5}{9}$

Given product: $\frac{2}{3}\times \frac{-5}{9}$

$\Rightarrow \frac{2}{3}\times \frac{-5}{9} = \frac{2\times-5}{3\times9} = \frac{-10}{27}$

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.4 division

Question: What will be the reciprocal of

$\frac{-6}{11} and \frac{-8}{5}?$

The reciprocal of $\frac{-6}{11}$ will be:

$1\div \frac{-6}{11} = \frac{1}{\frac{-6}{11}} =\frac{11}{-6}$

The reciprocal of $\frac{-8}{5}$ will be:

$1\div \frac{-8}{5} = \frac{1}{\frac{-8}{5}} =\frac{5}{-8}$

Solutions for CBSE class 7 maths chapter 9 rational numbers-Exercise: 9.2

Question:1(i) Find the sum:

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

Given sum:  $\frac{5}{4}+\left (\frac{-11}{4} \right )$

Here the denominator is the same which is 4.

$\frac{5}{4}+\left (\frac{-11}{4} \right ) = \frac{5-11}{4} = \frac{-6}{4} = \frac{-3\times2}{2\times2} = \frac{-3}{2}$

Question:1(ii) Find the sum:

$\frac{5}{3}+\frac{3}{5}$

Given sum: $\frac{5}{3}+\frac{3}{5}$

Here the LCM of 3 and 5 is 15.

Hence, we can write the sum as:

$\Rightarrow \frac{5}{3}+\frac{3}{5} = \frac{5\times5}{3\times5}+\frac{3\times3}{5\times3}$

$\Rightarrow \frac{25}{15}+\frac{9}{15} = \frac{34}{15}$

Question:1(iii) Find the sum:

$\frac{-9}{10}+\frac{22}{15}$

Given sum:$\frac{-9}{10}+\frac{22}{15}$

Taking the LCM of 10 and 15, we have 30

$\Rightarrow \frac{-9\times3}{10\times3}+\frac{22\times2}{15\times2}$

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

Question:1(iv) Find the sum:

$\frac{-3}{-11}+\frac{5}{9}$

Given sum: $\frac{-3}{-11}+\frac{5}{9}$

Taking LCM of 11 and 9 we have,

$\Rightarrow \frac{-3\times9}{-11\times9}+\frac{5\times11}{9\times11}$

$\Rightarrow \frac{-27}{-99}+\frac{55}{99}=\frac{27+55}{99} = \frac{82}{99}$

Question:1(v) Find the sum:

$\frac{-8}{19}+\frac{(-2)}{57}$

Given sum:  $\frac{-8}{19}+\frac{(-2)}{57}$

Taking LCM of 19 and 57, we have 57

We can write the sum as:

$\frac{-8\times3}{57}+\frac{(-2)}{57} = \frac{-24-2}{57} = \frac{-26}{57}$

Question:1(vi) Find the sum:

$\frac{-2}{3}+0$

Given sum:  $\frac{-2}{3}+0$

Adding any number to zero we get, the number itself

Hence, $\frac{-2}{3}+0 = \frac{-2}{3}$

Question:1(vii) Find the sum:

$-2\frac{1}{3}+4\frac{3}{5}$

Given the sum: $-2\frac{1}{3}+4\frac{3}{5}$

Taking the LCM of 3 and 5 we have: 15

$\Rightarrow \frac{-7\times5}{3\times5}+\frac{23\times3}{5\times3} = \frac{-35}{15}+\frac{69}{15} = \frac{34}{15}$

Question:2(i) Find

$\frac{7}{24} -\frac{17}{36}$

Given sum: $\frac{7}{24} -\frac{17}{36}$

We have LCM of 24 and 36 will be, 72

Hence,

$\Rightarrow \frac{7\times3}{24\times3} -\frac{17\times2}{36\times2} = \frac{21}{72} - \frac{34}{72} = \frac{-13}{72}$

Question:2(ii) Find

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

Given   $\frac{5}{63}-\left (\frac{-6}{21} \right )$:

LCM of 63 and 21 is 63,

Then we have;

$\Rightarrow \frac{5}{63}-\left (\frac{-6\times3}{21\times3} \right ) = \frac{5+18}{63} =\frac{23}{63}$

Question:2(iii)  Find

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

Given $\frac{-6}{13}-\left (\frac{-7}{15} \right )$:

We have, LCM of 13 and 15 is 195.

Then,

$\Rightarrow \frac{-6\times15}{13\times15}-\left (\frac{-7\times13}{15\times13} \right ) = \frac{-90}{195} - \left ( \frac{-91}{195} \right ) = \frac{1}{195}$

Question:2(iv) Find

$\frac{-3}{8}-\frac{7}{11}$

Given $\frac{-3}{8}-\frac{7}{11}$:

LCM of 8 and 11 is 88, then

$\Rightarrow \frac{-3}{8}-\frac{7}{11} = \frac{-3\times11}{8\times11} - \frac{7\times8}{11\times8}$

$\Rightarrow \frac{-33}{88} - \frac{56}{88} = \frac{-33-56}{88} = \frac{-89}{88}$

Question:2(v) Find

$-2\frac{1}{9}-6$

Given: $-2\frac{1}{9}-6$

$\Rightarrow -2\frac{1}{9}-6 = \frac{-19}{9} - 6 = \frac{-19}{9} - \frac{6}{1}$

LCM of 9 and 1 will be, 9

Hence,

$\Rightarrow \frac{-19}{9} - \frac{6}{1} = \frac{-19}{9} - \frac{6\times9}{1\times9}$

$\Rightarrow \frac{-19}{9} - \frac{54}{9} = \frac{-73}{9}.$

Question:

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

Given product:  $\frac{9}{2}\times\left ( \frac{-7}{4} \right )$

$\Rightarrow \frac{9}{2}\times\left ( \frac{-7}{4} \right ) = \frac{9\times(-7)}{2\times4}$

$\Rightarrow \frac{-63}{8}$

Question:3(ii) Find the product:

$\frac{3}{10}\times (-9)$

Given $\frac{3}{10}\times (-9)$

$\Rightarrow \frac{3}{10}\times\frac{-9}{1} = \frac{3\times(-9)}{10\times1}$

So the value

$\Rightarrow \frac{-27}{10}$

Question:3(iii) Find the product:

$\frac{-6}{5}\times \frac{9}{11}$

Given product: $\frac{-6}{5}\times \frac{9}{11}$

$\Rightarrow \frac{-6}{5}\times \frac{9}{11} = \frac{-6\times9}{5\times11}$

The value of given product is

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

Question:3(iv) Find the product:

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

Given product $\frac{3}{7}\times \left (\frac{-2}{5} \right )$

$\Rightarrow \frac{3}{7}\times(\frac{-2}{5}) = \frac{3\times(-2)}{7\times5} = \frac{-6}{35}$

Question:

$\frac{3}{11}\times \frac{2}{5}$

Given product: $\frac{3}{11}\times \frac{2}{5}$

$\Rightarrow \frac{3}{11}\times \frac{2}{5} = \frac{3\times2}{11\times5} = \frac{6}{55}$

Question:3(vi) Find the product:

$\frac{3}{-5}\times \frac{-5}{3}$

Given product: $\frac{3}{-5}\times \frac{-5}{3}$

$\Rightarrow \frac{3}{-5}\times \frac{-5}{3} = \frac{3\times-5}{-5\times3} = \frac{-15}{-15} =1$

Question:

$(-4)\div \frac{2}{3}$

Given: $(-4)\div \frac{2}{3}$

Dividing $-4$ by  $\frac{2}{3}$, we get

$\Rightarrow \frac{-4}{\frac{2}{3}} =\frac{-4\times3}{2} = \frac{-12}{2} = \frac{-6}{1}$

Question:4(ii) Find the value of:

$\frac{-3}{5}\div 2$

Given $\frac{-3}{5}\div 2$

Dividing $\frac{-3}{5}$  with 2 we get,

$\Rightarrow \frac{\frac{-3}{5}}{2} = \frac{-3}{10}$

Question:4(iii) Find the value of:

$\frac{-4}{5}\div (-3)$

Given: $\frac{-4}{5}\div (-3)$

So, dividing $\frac{-4}{5}$ with -3, we get

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

Question:4(iv) Find the value of:

$\frac{-1}{8}\div \frac{3}{4}$

Given: $\frac{-1}{8}\div \frac{3}{4}$

Simplifying it:

$\Rightarrow \frac{-1}{8}\times\frac{4}{3} = \frac{-1\times4}{8\times3}$

$\Rightarrow \frac{-4}{24} = \frac{-1}{6}$

Question:

$\frac{-2}{13}\div \frac{1}{7}$

Given: $\frac{-2}{13}\div \frac{1}{7}$

Simplifying it: we get

$\Rightarrow \frac{-2}{13}\times\frac{7}{1} = \frac{-14}{13}$

Question:4(vi) Find the value of:

$\frac{-7}{12}\div \left (\frac{-2}{3} \right )$

Given: $\frac{-7}{12}\div \left (\frac{-2}{3} \right )$

Simplifying it: we get

$\Rightarrow \frac{-7}{12}\times\frac{3}{-2} = \frac{-7\times3}{12\times-2} = \frac{-21}{-24} = \frac{7\times3}{8\times3} = \frac{7}{8}$

Question:4(vii) Find the value of:

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

Given: $\frac{3}{13}\div \left (\frac{-4}{65} \right )$

Simplifying it: we get

$\\\Rightarrow \frac{3}{13}\div \left (\frac{-4}{65} \right ) = \frac{3}{13}\times\frac{65}{-4} = \frac{3\times65}{13\times-4} \\since\ 13\times5=65 \\\ \frac{3}{13}\times\frac{65}{-4} =\frac{15}{-4}$

## NCERT Solutions for Class 7 Maths - Chapter-wise

 Chapter No. Chapter Name Chapter 1 Solutions of NCERT for class 7 maths chapter 1 Integers Chapter 2 CBSE NCERT solutions for class 7 maths chapter 2 Fractions and Decimals Chapter 3 NCERT solutions for class 7 maths chapter 3 Data Handling Chapter 4 Solutions of NCERT for class 7 maths chapter 4 Simple Equations Chapter 5 CBSE NCERT solutions for class 7 maths chapter 5 Lines and Angles Chapter 6 NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties Chapter 7 Solutions of NCERT for class 7 maths chapter 7 Congruence of Triangles Chapter 8 NCERT solutions for class 7 maths chapter 8 comparing quantities Chapter 9 CBSE NCERT solutions for class 7 maths chapter 9 Rational Numbers Chapter 10 NCERT solutions for class 7 maths chapter 10 Practical Geometry Chapter 11 Solutions of NCERT for class 7 maths chapter 11 Perimeter and Area Chapter 12 CBSE NCERT solutions for class 7 maths chapter 12 Algebraic Expressions Chapter 13 NCERT solutions for class 7 maths chapter 13 Exponents and Powers Chapter 14 Solutions of NCERT for class 7 maths chapter 14 Symmetry

## Benefits of NCERT solutions for class 7 maths chapter 9 rational numbers-

• You will study the different types of numbers which you should know.

• You will also learn addition, multiplication, subtraction, and division of rational numbers.
• You will also get solutions to the practice questions given below every topic which will give you conceptual clarity.
• You should practice all the NCERT questions including examples. If you facing difficulties in doing so, you can take help from the NCERT solutions for class 7 maths chapter 9 rational numbers.

Happy learning!!!