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

Rational numbers are numbers that can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$. This chapter introduces concepts like operations on rational numbers like addition, subtraction, multiplication and division. It also focuses on the rules and properties of the rational numbers and the way of representing the rational numbers on the number line. The NCERT Solutions in this article give the step-by-step solutions for all the exercise questions in this chapter, helping students understand the mathematical concepts accurately.

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

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1. NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers - Important Points
2. NCERT Solutions for Maths Chapter 8 Rational Numbers Class 7
3. NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers (Exercise)
4. Rational Numbers Class 7 Maths Chapter 8-Topics
5. NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers - Points to Remember
6. NCERT Solutions for Class 7 Maths Chapter Wise
7. NCERT Solutions for Class 7 Subject Wise

As these NCERT solutions for Class 7 Maths are prepared by subject matter experts from Careers360, it is one of the reliable and accurate study resources that could help the students in exam preparation. There are a total of 15 exercise questions in this chapter for which the step-by-step solutions are provided in this article.

NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers - Important Points

A rational number can be expressed in the form of $\frac{p}{q}$, Where p and q are integers and q ≠ 0.

Numerator and Denominator: In the $\frac{p}{q}$, 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: $\frac{3}{5},-\frac{5}{8},\frac{2}{7}$, etc.

Comparison of Rational Numbers:

$\begin{array}{ll}\frac{p}{q}<\frac{a}{b}& \text{ if }pb<aq\\ \frac{p}{q}>\frac{a}{b}& \text{ if }pb>aq\end{array}$

Operations on Rational Number:

• Addition of rational numbers : $\frac{p}{q}+\frac{a}{b}=\frac{(p\times b)+(a\times q)}{q\times b}$

• Subtraction of rational numbers: $\frac{p}{q}-\frac{a}{b}=\frac{(p\times b)-(a\times q)}{q\times b}$

• Multiplication of rational numbers: $\frac{p}{q}\times \frac{a}{b}=\frac{p\times a}{q\times b}$

• Division of rational numbers: $\frac{p}{q}÷\frac{a}{b}=\frac{p\times b}{q\times a}$

Reciprocal of a rational number:

Reciprocal of $\frac{p}{q}$ = $\frac{q}{p}$

The product of rational numbers with its reciprocal is always 1.

NCERT Solutions for Maths Chapter 8 Rational Numbers Class 7

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NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers (Exercise)

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

–1 and 0

Answer: To find five rational numbers between $-1and0$ we will convert each rational numbers as a denominator $5+1=6$ , we have

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

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

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

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 $5\times 3=15$ , we have $(\because LCMof5and3=15)$

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

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

$\frac{-12}{15}=\frac{-12\times 3}{15\times 3}=\frac{-36}{45}and\frac{-10}{15}=\frac{-10\times 3}{15\times 3}=\frac{-30}{45}$

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 number into its 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:

$\frac{-3}{5}=\frac{-3\times 1}{5\times 1}$ $\frac{-6}{10}=\frac{-3\times 2}{5\times 2}$ $\frac{-9}{15}=\frac{-3\times 3}{5\times 3}$ $\frac{-12}{20}=\frac{-3\times 4}{5\times 4}$

Now, following the same pattern, we have

$\frac{-3\times 5}{5\times 5}=\frac{-15}{25}$ $\frac{-3\times 6}{5\times 6}=\frac{-18}{30}$ $\frac{-3\times 7}{5\times 7}=\frac{-21}{35}$ $\frac{-3\times 8}{5\times 8}=\frac{-24}{40}$

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:

$\frac{-1}{4}=\frac{-1\times 1}{4\times 1}$ $\frac{-2}{8}=\frac{-1\times 2}{4\times 2}$ $\frac{-3}{12}=\frac{-1\times 3}{4\times 3}$

Now, following the same pattern, we have

$\frac{-1\times 4}{4\times 4}=\frac{-4}{16}$ $\frac{-1\times 5}{4\times 5}=\frac{-5}{20}$ $\frac{-1\times 6}{4\times 6}=\frac{-6}{24}$ $\frac{-1\times 7}{4\times 7}=\frac{-7}{28}$

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:

$\frac{-1}{6}=\frac{-1\times 1}{6\times 1}$ $\frac{-2}{12}=\frac{-1\times 2}{6\times 2}$ $\frac{-3}{18}=\frac{-1\times 3}{6\times 3}$ $\frac{-4}{24}=\frac{-1\times 4}{6\times 4}$

Now, following the same pattern, we have

$\frac{-1\times 5}{6\times 5}=\frac{-5}{30}$ $\frac{-1\times 6}{6\times 6}=\frac{-6}{36}$ $\frac{-1\times 7}{6\times 7}=\frac{-7}{42}$ $\frac{-1\times 8}{6\times 8}=\frac{-8}{48}$

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:

$\frac{-2}{3}=\frac{-2\times 1}{3\times 1}$ $\frac{2}{-3}=\frac{2}{-3}=\frac{2\times 1}{-3\times 1}$ $\frac{4}{-6}=\frac{-2\times 2}{3\times 2}$ $\frac{-6}{9}=\frac{-2\times 3}{3\times 3}$

Now, following the same pattern, we have

$\frac{-2\times 4}{3\times 4}=\frac{-8}{12}or\frac{8}{-12}$ $\frac{-2\times 5}{3\times 5}=\frac{-10}{15}or\frac{10}{-15}$

$\frac{-2\times 6}{3\times 6}=\frac{-12}{18}or\frac{12}{-18}$ $\frac{-2\times 7}{3\times 7}=\frac{-14}{21}or\frac{14}{-21}$

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:

$\frac{-2}{7}=\frac{-2\times 2}{7\times 2}=\frac{-4}{14}$ $\frac{-2}{7}=\frac{-2\times 3}{7\times 3}=\frac{-6}{21}$

$\frac{-2}{7}=\frac{-2\times 4}{7\times 4}=\frac{-8}{28}$ $\frac{-2}{7}=\frac{-2\times 5}{7\times 5}=\frac{-10}{35}$

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:

$\frac{5}{-3}=\frac{5\times 2}{-3\times 2}=\frac{10}{-6}$ $\frac{5}{-3}=\frac{5\times 3}{-3\times 3}=\frac{15}{-9}$

$\frac{5}{-3}=\frac{5\times 4}{-3\times 4}=\frac{20}{-12}$ $\frac{5}{-3}=\frac{5\times 5}{-3\times 5}=\frac{25}{-15}$

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:

$\frac{4}{9}=\frac{4\times 2}{9\times 2}=\frac{8}{18}$ $\frac{4}{9}=\frac{4\times 3}{9\times 3}=\frac{12}{27}$

$\frac{4}{9}=\frac{4\times 4}{9\times 4}=\frac{16}{36}$ $\frac{4}{9}=\frac{4\times 5}{9\times 5}=\frac{20}{45}$

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,

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,

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,

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,

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.

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:

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

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:

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

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 $\frac{-7}{21}and\frac{3}{9}$

$\Rightarrow \frac{-7\times 9}{21\times 9}=\frac{-63}{189}$

$\Rightarrow \frac{3\times 21}{9\times 21}=\frac{63}{189}$

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

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 $\frac{-16}{20}and\frac{20}{-25}$

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

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

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

So, they represent the same rational number.

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

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

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

$\Rightarrow \frac{-3\times 20}{5\times 20}=\frac{-60}{100}$

$\Rightarrow \frac{-12\times 5}{20\times 5}=\frac{-60}{100}$

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

So, they represent the same rational number.

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

$\Rightarrow \frac{8\times 15}{-5\times 15}=\frac{120}{-75}$

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

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

So, they represent the same rational number.

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

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

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

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

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 $-5\ne 5$.

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:

$\Rightarrow \frac{-8}{6}=\frac{-8/2}{6/2}=\frac{-4}{3}$ $\left[HCFof8and6is2\right]$

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

$\Rightarrow \frac{25}{45}=\frac{25/5}{45/5}=\frac{5}{9}$ $\left[HCFof25and45is5\right]$

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

$\Rightarrow \frac{-44}{72}=\frac{-44/4}{72/4}=\frac{-11}{18}$ $\left[HCFof44and72is4\right]$

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

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

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

(vii) $0◻\frac{-7}{6}$

Answer: (i) $\frac{-5}{7}◻\frac{2}{3}$

$\Rightarrow \frac{-5\times 3}{7\times 3}◻\frac{2\times 7}{3\times 7}$

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

Hence, $\frac{-5}{7}<\frac{2}{3}$

(ii) $\frac{-4}{5}◻\frac{-5}{7}$

$\Rightarrow \frac{-4\times 7}{5\times 7}◻\frac{-5\times 5}{7\times 5}$

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

Hence, $\frac{-4}{5}<\frac{-5}{7}$

(iii) $\frac{-7}{8}◻\frac{14}{-16}$

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

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

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

(iv) $\frac{-8}{5}◻\frac{-7}{4}$

$\Rightarrow \frac{-8\times 4}{5\times 4}◻\frac{-7\times 5}{4\times 5}$

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

Hence, $\frac{-8}{5}>\frac{-7}{4}$

(v) $\frac{1}{-3}◻\frac{-1}{4}$

$\Rightarrow \frac{1\times 4}{-3\times 4}◻\frac{-1\times -3}{4\times -3}$

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

Hence, $\frac{1}{-3}<\frac{1}{-4}$

(vi) $\frac{5}{-11}◻\frac{-5}{11}$

$\Rightarrow \frac{5\times 11}{-11\times 11}◻\frac{-5\times -11}{11\times -11}$

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

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

(vii) $0◻\frac{-7}{6}$

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) $\frac{2}{3},\frac{5}{2}$

$\Rightarrow \frac{2\times 2}{3\times 2},\frac{5\times 3}{2\times 3}$

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

Since, $\frac{15}{6}>\frac{4}{6}$

So, $\frac{5}{2}>\frac{2}{3}.$

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

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

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

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

So, $\frac{-5}{6}>\frac{-4}{3}.$

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

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

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

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

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

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

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 $3,9and3=9$.

Therefore, we have

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

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

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 $7,2and4=28$.

Therefore, we have

$\frac{-3\times 4}{7\times 4},\frac{-3\times 14}{2\times 14},\frac{-3\times 7}{4\times 7}$

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

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

Hence, the required ascending order is

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

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.

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

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:

$\Rightarrow \frac{5}{3}+\frac{3}{5}=\frac{5\times 5}{3\times 5}+\frac{3\times 3}{5\times 3}$

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

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

$\Rightarrow \frac{-9\times 3}{10\times 3}+\frac{22\times 2}{15\times 2}$

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

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,

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

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

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:

$\frac{-8\times 3}{57}+\frac{(-2)}{57}=\frac{-24-2}{57}=\frac{-26}{57}$

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

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

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,

$\Rightarrow \frac{7\times 3}{24\times 3}-\frac{17\times 2}{36\times 2}=\frac{21}{72}-\frac{34}{72}=\frac{-13}{72}$

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;

$\Rightarrow \frac{5}{63}-\left(\frac{-6\times 3}{21\times 3}\right)=\frac{5+18}{63}=\frac{23}{63}$

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,

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

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

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

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

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,

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

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

Question: 3 (i) Find the product:

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

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

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

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

Question: 3 (ii) Find the product:

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

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

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

So the value

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

Question: 3 (iii) Find the product:

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

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

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

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

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

Question: 3 (v) Find the product:

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

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

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

Question: 3 (vi) Find the product:

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

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

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

Question: 4 (i) Find the value of:

$(-4)÷\frac{2}{3}$

Answer: Given: $(-4)÷\frac{2}{3}$

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

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

Question: 4 (ii) Find the value of:

$\frac{-3}{5}÷2$

Answer: Given $\frac{-3}{5}÷2$

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

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

Question: 4 (iii) Find the value of:

$\frac{-4}{5}÷(-3)$

Answer: Given: $\frac{-4}{5}÷(-3)$

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:

$\frac{-1}{8}÷\frac{3}{4}$

Answer: Given: $\frac{-1}{8}÷\frac{3}{4}$

Simplifying it:

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

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

Question: 4 (v) Find the value of:

$\frac{-2}{13}÷\frac{1}{7}$

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

Simplifying it: we get

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

Question: 4 (vi) Find the value of:

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

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

Simplifying it: we get

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

Question: 4 (vii) Find the value of:

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

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

Simplifying it: we get

$$\begin{gathered} \Rightarrow \frac{3}{13}÷\left(\frac{-4}{65}\right)=\frac{3}{13}\times \frac{65}{-4}=\frac{3\times 65}{13\times -4} \\ since13\times 5=65 \\ \frac{3}{13}\times \frac{65}{-4}=\frac{15}{-4} \end{gathered}$$

Rational Numbers Class 7 Maths Chapter 8-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 8 Rational Numbers - Points to Remember

Rational number: A rational number is a number that can be represented as $\frac{p}{q}$, Where p and q are integers and q ≠ 0.

Numerator and Denominator: In the $\frac{p}{q}$, p is the numerator and q is the denominator.

Comparison of Rational Numbers:

$\begin{array}{ll}\frac{p}{q}<\frac{a}{b}& \text{ if }pb<aq\\ \frac{p}{q}>\frac{a}{b}& \text{ if }pb>aq\end{array}$

Operations on Rational Number:

• Addition of rational numbers : $\frac{p}{q}+\frac{a}{b}=\frac{(p\times b)+(a\times q)}{q\times b}$

• Subtraction of rational numbers: $\frac{p}{q}-\frac{a}{b}=\frac{(p\times b)-(a\times q)}{q\times b}$

• Multiplication of rational numbers: $\frac{p}{q}\times \frac{a}{b}=\frac{p\times a}{q\times b}$

• Division of rational numbers: $\frac{p}{q}÷\frac{a}{b}=\frac{p\times b}{q\times a}$