NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers - Download PDF

# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers - Download PDF

Edited By Ravindra Pindel | Updated on Feb 07, 2024 06:06 PM IST

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers: In earlier classes, you have studied whole numbers, natural numbers, integer numbers. In this article, you will get CBSE NCERT rational numbers class 7 solutions. 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.

Once you go through you will get more clarity of the concepts. You must refer to the NCERT Syllabus for Class 7 Maths for better understanding. There are 14 questions in 2 exercises given in . 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.

## NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 - Important Formulae

Rational number = p/q, Where p and q are integers and q ≠ 0.

Numerator and Denominator: In the p/q , p is the numerator and q is the denominator.

Comparison of Rational Numbers:

p/q < a/b If pb < aq

p/q > a/b If pb > aq

Operations on Rational Number:

• Addition of rational numbers : (p/q) + (a/b) = ((p×b) + (a×q))/(q×b)

• Subtraction of rational numbers: (p/q) - (a/b) = ((p×b) - (a×q))/(q×b)

• Multiplication of rational numbers: p/q) × (a/b) = (p×a)/(q×b)

• Division of rational numbers: (p/q) ÷ (a/b) = (p×b)/(q×a)

Reciprocal of p/q = q/p

(Rational number)(Reciprocal) = 1

## NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers - Important Points

A rational can be expressed in the form of p/q , Where p and q are integers and q ≠ 0.

Numerator and Denominator: In the p/q , p is the numerator and q is the denominator.

We obtain another equivalent rational number by multiplying the numerator and denominator with the same nonzero integer.

+ sign and positive integer: a position to the right of 0.

- sign and negative integer: a position to the left of 0.

Rational Numbers in Standard Form:

Its denominator is a positive integer.

The numerator and denominator have no common factor other than 1.

Examples: 3/5 , -5/8 , 2/7, etc.

Comparison of Rational Numbers:

p/q < a/b If pb < aq

p/q > a/b If pb > aq

Operations on Rational Number:

• Addition of rational numbers : (p/q) + (a/b) = ((p×b) + (a×q))/(q×b)

• Subtraction of rational numbers: (p/q) - (a/b) = ((p×b) - (a×q))/(q×b)

• Multiplication of rational numbers: p/q) × (a/b) = (p×a)/(q×b)

• Division of rational numbers: (p/q) ÷ (a/b) = (p×b)/(q×a)

Reciprocal of a rational number:

Reciprocal of p/q = q/p

Product of rational numbers with its reciprocal is always 1.

Free download NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers PDF for CBSE Exam.

## NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 Topic 9.3

Answer: 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

Answer: Any ten rational numbers are:

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

Question:

(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

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

## NCERT rational numbers class 7 solutions Topic 9.4

Question:1

Answer: 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.

Answer: Five more positive rational numbers are:

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

Question:1

Answer: 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.

Answer: 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) $\frac{-2}{-9}$

Answer: (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 a 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.

## NCERT Solutions for Class 7 Maths Chapter 9 Topic 9.6

Question:

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

Answer: (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 :

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

## NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 Topic 9.8

Question: Find five rational numbers between

Answer: 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 $\frac{-5}{7} and \frac{-3}{8}$ .

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

## NCERT Solutions for Class 7 Chapter 9 Rational Numbers Maths Exercise 9.1

Question:

–1 and 0

Answer: 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

Answer: 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:

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

Answer: 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:

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

Answer: 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

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

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

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

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:

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

Question:

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

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

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:

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

Question:

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

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

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:

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

Question:

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

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

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:

$\frac{-2}{7}$

Answer: $\frac{-2}{7}$ 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:

$\frac{5}{-3}$

Answer: $\frac{5}{-3}$ 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:

$\frac{4}{9}$

Answer: $\frac{4}{9}$ 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

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

$\frac{3}{4}$

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

$\frac{-5}{8}$

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

Question:

$\frac{-7}{4}$

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

$\frac{7}{8}$

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

Answer: 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) $\frac{-5}{-9}\ and\ \frac{5}{-9}$

Answer: 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) $\frac{-8}{10}$

Answer: (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) $0 \square \frac{-7}{6}$

Answer: (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) $-3\frac{2}{7}, -3\frac{4}{5}$

Answer: (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:

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

Answer: (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:

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

Answer: 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:

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

Answer: 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 Chapter 9 Maths Class 7 Rational Numbers Topic 9.9.1

Question: Find:

Answer: 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)

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

Answer: 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)

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

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

Taking LCM of 6 and 11 we get; 66

Hence we can write the sum as:

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

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

## NCERT Solutions for Chapter 9 Maths Class 7 Rational Numbers Topic 9.9.2

Question:1 What will be the additive inverse of

-3/9, -9/11, 5/7

$\frac{-3}{9} \ is \ \frac{3}{9}$

$\frac{-9}{11} \ is \ \frac{9}{11}$

$\frac{5}{7} \ is \ \frac{-5}{7}$

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

## NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Topic 9.9.3

Question:

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

Answer: (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)

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

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

Question:(ii)

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

Answer: 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 7th Math Chapter 9 Rational Numbers Topic 9.9.4

Question: What will be the reciprocal of

Answer: 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:

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

## NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Exercise: 9.2

Question: 1(i) Find the sum:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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 :

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

Answer: 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:

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

Answer: 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:

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

Answer: 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

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

Answer: 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

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

Answer: 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

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

Answer: 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

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

Answer: 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

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Answer: 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:

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

Simplifying it: we get

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

## Rational Numbers Class 7 Maths Chapter 9-Topics

• Need For Rational Numbers
• What Are The Rational Numbers?
• Positive And Negative Rational Numbers
• Rational Numbers On A Number Line
• Rational Numbers In A Standard Form
• Comparison Of Rational Numbers
• Rational Numbers Between Two Rational Numbers
• Operations On Rational Numbers

### NCERT Solutions for Class 7 Maths Chapter Wise

 Chapter No. Chapter Name 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 Congruence of Triangles Chapter 8 Comparing quantities Chapter 9 Rational Numbers Chapter 10 Practical Geometry Chapter 11 Perimeter and Area Chapter 12 Algebraic Expressions Chapter 13 Exponents and Powers Chapter 14 Symmetry Chapter 15 Visualising Solid Shapes

### NCERT Solutions for Class 7 Subject Wise

 NCERT Solutions for Class 7 Maths NCERT Solutions for Class 7 Science

• ### You will study the different types of numbers which you should know in NCERT Class 7 Mathematics book .

• 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!!!

Also Check NCERT Books and NCERT Syllabus here:

1. What is rational number?

A number can be expressed in the form of p/q, where p and q are integers and q is not equal to 0 is called a rational number. Students can deeper understanding of rational number in class 7 math chapter 9. You can download rational numbers class 7 pdf to study both offline and online mode.

2. How do you perform operations with rational numbers?
• For addition and subtraction, find a common denominator, perform the operation on the numerators, and keep the denominator the same.
• For multiplication, multiply the numerators and denominators separately.
• For division, multiply by the reciprocal of the second number (divisor) and follow the multiplication rule.
3. How many exercises in NCERT solution Class 7 Maths chapter 9?

There are 2 exercises in NCERT solution Class 7 Maths chapter 9.

Rational Numbers Exercise 9.1- 10 questions

Rational Numbers Exercise 9.2- 4 questions

4. How do you convert a recurring decimal into a fraction?
• To convert a recurring decimal into a fraction, represent the recurring part as 'x' and subtract it from the whole decimal.
• Solve for 'x' to get an equation. Multiply both sides by the appropriate power of 10 to eliminate the recurring part.
• This will give you an equation that you can solve to find the fraction equivalent of the recurring decimal.

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