NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

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.

Latest: NCERT Class 7 Maths: Chapterwise Important Formulas with Examples and Points

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NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 - Important Formulae
NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers - Important Points
NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7
NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (Intext Questions and Exercise)
NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 Topic 9.3
NCERT rational numbers class 7 solutions Topic 9.4
NCERT Solutions for Class 7 Maths Chapter 9 Topic 9.6
NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 Topic 9.8
NCERT Solutions for Class 7 Chapter 9 Rational Numbers Maths Exercise 9.1
NCERT Solutions for Chapter 9 Maths Class 7 Rational Numbers Topic 9.9.1
NCERT Solutions for Chapter 9 Maths Class 7 Rational Numbers Topic 9.9.2
NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Topic 9.9.3
NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Topic 9.9.4
NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Exercise: 9.2
Rational Numbers Class 7 Maths Chapter 9-Topics

Once you go through NCERT Solutions for Class 7 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 NCERT . 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.

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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.

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NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7

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NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (Intext Questions and Exercise)

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

Question:1 Is the number $\frac{2}{-3}$ rational? Think about it.

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 List ten rational numbers.

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: Fill in the boxes:

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

Answer:

(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 Is 5 a positive rational number?

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.

Question:2 List five more positive rational numbers.

Answer: Five more positive rational numbers are:

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

Question:1 Is – 8 a negative rational number?

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.

Question:2 List five more negative rational numbers.

Answer: Five more negative rational numbers are:

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

Question: Which of these are negative rational numbers?

(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: Find the standard form of

(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

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

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(i) List five rational numbers between:

–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: 1(ii) List five rational numbers between:

–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: 1(iii) List five rational numbers between:

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

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: 1(iv) List five rational numbers between:

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

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

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

Question: 2(i) Write four more rational numbers in each of the following patterns:

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

Answer: 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:

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

Question: 2(ii) Write four more rational numbers in each of the following patterns:

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

Answer: 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:

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

Question: 2(iii) Write four more rational numbers in each of the following patterns:

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

Answer: 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:

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

Question: 2(iv) Write four more rational numbers in each of the following patterns:

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

Answer: 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: 3(i) Give four rational numbers equivalent to:

$\frac{-2}{7}$

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: 3(ii) Give four rational numbers equivalent to:

$\frac{5}{-3}$

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: 3(iii) Give four rational numbers equivalent to:

$\frac{4}{9}$

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

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

Question: 4(i) Draw the number line and represent the following rational numbers on it:

$\frac{3}{4}$

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

1643867890014

Question: 4(ii) Draw the number line and represent the following rational numbers on it:

$\frac{-5}{8}$

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

1643867932616

Question: 4(iii) Draw the number line and represent the following rational numbers on it:

$\frac{-7}{4}$

Answer: Representation of $\frac{-7}{4}$ on the number line, 1643867960214

Question: 4(iv) Draw the number line and represent the following rational numbers on it:

$\frac{7}{8}$

Answer: Representation of $\frac{7}{8}$ on the number line, 1643867985343

Question: 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.

1643868029468

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:

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

Question: 6 Which of the following pairs represent the same rational number?

(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.

Question: 7 Rewrite the following rational numbers in the simplest form:

(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:

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

Question: 8 Fill in the boxes with the correct symbol out of >, <, and =.

(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: 9 Which is greater in each of the following:

(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: 10(i) Write the following rational numbers in ascending order:

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

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: 10(ii) Write the following rational numbers in ascending order:

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

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: 10(iii) Write the following rational numbers in ascending order:

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

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:

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

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) Find:

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

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) Find:

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

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

Answer: 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

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

Question:2 Find $(i)\frac{7}{9}-\frac{2}{5} \ \ \ \ \ (ii) 2 \frac{1}{5}-\frac{(-1)}{3}$

Answer:

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

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

Question: What will be

(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) Find:

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

Answer: Given product:

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

Question:(ii) Find:

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

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

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

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: 3(i) Find the product:

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

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: 3(v) Find the product:

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

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(i) Find the value of:

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

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: 4(v) Find the value of:

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

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:

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

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

Benefits of NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

You will study the different types of numbers which you should know in NCERT Class 7 Mathematics book .
It will help you in your homework as all the practice questions including questions given below every topic are covered in this article.
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:

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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

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

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

NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (Intext Questions and Exercise)

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

NCERT rational numbers class 7 solutions Topic 9.4

NCERT Solutions for Class 7 Maths Chapter 9 Topic 9.6

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

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

NCERT Solutions for Chapter 9 Maths Class 7 Rational Numbers Topic 9.9.1

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

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

NCERT Solutions for Class 7th Math Chapter 9 Rational Numbers Topic 9.9.4

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

Rational Numbers Class 7 Maths Chapter 9-Topics

NCERT Solutions for Class 7 Maths Chapter Wise

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Benefits of NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers

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

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