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NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers Ex 8.2

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NCERT Solutions for Maths Class 7 Chapter 8 Exercise 8.2 Rational Numbers - FREE PDF Download

NCERT Solutions for Class 7 Maths Exercise 8.2 of Chapter 8 - Rational Numbers, delves into the operations involving rational numbers. This exercise focuses on addition, subtraction, multiplication, and division of rational numbers. Mastering these operations is essential, as they are fundamental to many advanced mathematical concepts.

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Table of Content
1. NCERT Solutions for Maths Class 7 Chapter 8 Exercise 8.2 Rational Numbers - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 8 Exercise 8.2 Class 7 | Vedantu
3. Access NCERT Solutions for Maths Class 7 Chapter 8 - Rational Numbers
    3.1Exercise 8.2
4. Conclusion
5. Class 7 Maths Chapter 8: Exercises Breakdown
6. CBSE Class 7 Maths Chapter 8 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 7 Maths
8. Important Related Links for NCERT Class 7 Maths
FAQs


Vedantu's NCERT Solutions offer detailed, step-by-step explanations for each problem in Exercise 8.2. These solutions are designed by expert teachers to help students understand the concepts clearly and efficiently. Practising these problems is crucial for building a solid foundation in rational numbers, which will be immensely helpful in higher-level maths studies. Download the free PDF for comprehensive learning.


Glance on NCERT Solutions Maths Chapter 8 Exercise 8.2 Class 7 | Vedantu

  • Class 7 Maths Exercise 8.2 Solutions explains the operations of rational numbers, including addition, subtraction, multiplication, and division.

  • Rational numbers are numbers that can be expressed as a fraction of two integers, where the denominator is not zero.

  • The addition of rational numbers involves finding a common denominator and then adding the numerators.

  • Subtraction of rational numbers follows a similar process, but the numerators are subtracted instead.

  • Multiplication of rational numbers is done by multiplying the numerators and denominators separately.

  • Division of rational numbers requires multiplying by the reciprocal of the divisor, and each operation is illustrated with examples to ensure thorough understanding.

  • There are 4 fully solved questions in Class 7 Maths Ch 8 Ex 8.2.

Access NCERT Solutions for Maths Class 7 Chapter 8 - Rational Numbers

Exercise 8.2

1. Find the sum:

i. $\dfrac{5}{4} + \left( {\dfrac{{ - 11}}{4}} \right)$

Ans: As the denominator is same thus the numerators will be directly added.

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

$ = \dfrac{{5 - 11}}{4}$

$ = \dfrac{{ - 6}}{4}$

Divided both denominator and numerator by $2$.

$ = \dfrac{{ - 3}}{2}$

ii. $\dfrac{5}{3} + \dfrac{3}{5}$

Ans: To make both the numerator same, multiply the term $\dfrac{5}{3}$ by $5$ and multiply the term $\dfrac{3}{5}$ by $3$.

Thus, we will get the terms as $\dfrac{{5 \times 5}}{{3 \times 5}} = \dfrac{{25}}{{15}}$ and $\dfrac{{3 \times 3}}{{5 \times 3}} = \dfrac{9}{{15}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{25}}{{15}} + \dfrac{9}{{15}}$

$ = \dfrac{{25 + 9}}{{15}}$

$ = \dfrac{{34}}{{15}}$

$ = 2\dfrac{4}{{15}}$

iii. $\dfrac{{ - 9}}{{10}} + \dfrac{{22}}{{15}}$

Ans: To make both the numerator same, multiply the term $\dfrac{{ - 9}}{{10}}$ by $3$ and multiply the term $\dfrac{{22}}{{15}}$ by $2$.

Thus, we will get the terms as $\dfrac{{ - 9 \times 3}}{{10 \times 3}} = \dfrac{{ - 27}}{{30}}$ and $\dfrac{{22 \times 2}}{{15 \times 2}} = \dfrac{{44}}{{30}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 27}}{{30}} + \dfrac{{44}}{{30}}$

$ = \dfrac{{ - 27 + 44}}{{30}}$

$ = \dfrac{{17}}{{30}}$

iv. $\dfrac{{ - 3}}{{ - 11}} + \dfrac{5}{9}$

Ans: To make both the numerator same, multiply the term $\dfrac{{ - 3}}{{ - 11}}$ by $9$ and multiply the term $\dfrac{5}{9}$ by $11$.

Thus, we will get the terms as $\dfrac{{ - 3 \times 9}}{{ - 11 \times 9}} = \dfrac{{ - 27}}{{99}}$ and $\dfrac{{5 \times 11}}{{9 \times 11}} = \dfrac{{55}}{{99}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 27}}{{99}} + \dfrac{{55}}{{99}}$

$ = \dfrac{{27 + 55}}{{99}}$

$ = \dfrac{{82}}{{99}}$

v. $\dfrac{{ - 8}}{{19}} + \dfrac{{( - 2)}}{{57}}$

Ans: To make both the numerator same, multiply the term $\dfrac{{ - 8}}{{19}}$ by $3$ and multiply the term $\dfrac{{( - 2)}}{{57}}$ by $1$.

Thus, we will get the terms as $\dfrac{{ - 8 \times 3}}{{19 \times 3}} = \dfrac{{ - 24}}{{57}}$ and $\dfrac{{( - 2) \times 1}}{{57 \times 1}} = \dfrac{{( - 2)}}{{57}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 24}}{{57}} + \dfrac{{( - 2)}}{{57}}$

$ = \dfrac{{ - 24 - 2}}{{57}}$

$ = \dfrac{{ - 26}}{{57}}$

vi. $\dfrac{{ - 2}}{3} + 0$

Ans: Adding $0$ with any number will results the number that is $\dfrac{{ - 2}}{3} + 0 = \dfrac{{ - 2}}{3}$

vii. $ - 2\dfrac{1}{3} + 4\dfrac{3}{5}$

Ans: After simplifying the numbers will be $\dfrac{{ - 7}}{3}$and $\dfrac{{23}}{5}$.

To make both the numerator same, multiply the term  $\dfrac{{ - 7}}{3}$ by $5$ and multiply the term $\dfrac{{23}}{5}$ by $3$.

Thus, we will get the terms as $\dfrac{{ - 7 \times 5}}{{3 \times 5}} = \dfrac{{ - 35}}{{15}}$ and $\dfrac{{23 \times 3}}{{5 \times 3}} = \dfrac{{69}}{{15}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 35}}{{15}} + \dfrac{{69}}{{15}}$

$ = \dfrac{{ - 35 + 69}}{{15}}$

$ = \dfrac{{34}}{{15}}$

$ = 2\dfrac{4}{{15}}$

2. Find:

i. $\dfrac{7}{{24}} - \dfrac{{17}}{{36}}$

Ans: To make both the numerator same, multiply the term $\dfrac{7}{{24}}$ by $3$ and multiply the term $\dfrac{{17}}{{36}}$ by $2$.

Thus, we will get the terms as $\dfrac{{7 \times 3}}{{24 \times 3}} = \dfrac{{21}}{{72}}$ and $\dfrac{{17 \times 2}}{{22 \times 2}} = \dfrac{{34}}{{72}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{21}}{{72}} + \dfrac{{34}}{{72}}$

$ = \dfrac{{21 - 34}}{{72}}$

$ = \dfrac{{ - 13}}{{72}}$

ii. $\dfrac{5}{{63}} - \dfrac{{( - 6)}}{{21}}$

Ans: To make both the numerator same, multiply the term $\dfrac{5}{{63}}$ by $1$ and multiply the term $\dfrac{{( - 6)}}{{21}}$by$3$.

Thus, we will get the terms as $\dfrac{{5 \times 1}}{{63 \times 1}} = \dfrac{5}{{63}}$ and $\dfrac{{6 \times 3}}{{21 \times 3}} = \dfrac{{18}}{{63}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{5}{{63}} + \dfrac{{ - 18}}{{63}}$

$ = \dfrac{{5 - ( - 18)}}{{63}}$

$ = \dfrac{{23}}{{63}}$

iii. $\dfrac{{ - 6}}{{13}} - \dfrac{{( - 7)}}{{15}}$

Ans: To make both the numerator same, multiply the term $\dfrac{{ - 6}}{{13}}$ by $15$ and multiply the term $\dfrac{{( - 7)}}{{15}}$ by $13$.

Thus, we will get the terms as $\dfrac{{( - 6) \times 15}}{{13 \times 15}} = \dfrac{{( - 90)}}{{195}}$ and $\dfrac{{ - 7 \times 13}}{{15 \times 13}} = \dfrac{{ - 91}}{{195}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{( - 90)}}{{195}} + \dfrac{{ - 91}}{{195}}$

$ = \dfrac{{ - 90 - 91}}{{195}}$

$ = \dfrac{1}{{195}}$

iv. $\dfrac{{ - 3}}{8} - \dfrac{7}{{11}}$

Ans: To make both the numerator same, multiply the term $\dfrac{{ - 3}}{8}$by$11$ and multiply the term $\dfrac{7}{{11}}$ by $8$.

Thus, we will get the terms as $\dfrac{{ - 3 \times 11}}{{8 \times 11}} = \dfrac{{ - 33}}{{88}}$ and $\dfrac{{7 \times 8}}{{11 \times 8}} = \dfrac{{56}}{{88}}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 33}}{{88}} - \dfrac{{56}}{{88}}$

$ = \dfrac{{ - 33 - 56}}{{88}}$

$ = \dfrac{{ - 89}}{{88}}$

$ =  - 1\dfrac{1}{{88}}$

v. $ - 2\dfrac{1}{9} - 6$

Ans: Simplifying the term $ - 2\dfrac{1}{9}$ we get $\dfrac{{ - 19}}{9}$

To make both the numerator same, multiply the term  $\dfrac{{ - 19}}{9}$ by $1$ and multiply the term $\dfrac{6}{1}$ by $9$.

Thus, we will get the terms as $\dfrac{{ - 19 \times 1}}{{9 \times 1}} = \dfrac{{ - 19}}{9}$ and $\dfrac{{6 \times 9}}{{1 \times 9}} = \dfrac{{54}}{9}$

Now, As the denominator is same thus the numerators will be directly added.

$\dfrac{{ - 19}}{9} - \dfrac{{54}}{9}$

$ = \dfrac{{ - 19 - 54}}{9}$

$ = \dfrac{{ - 73}}{9}$

$ =  - 8\dfrac{1}{9}$

3. Find the product:

i. $\dfrac{9}{2} \times \dfrac{{( - 7)}}{4}$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

$\dfrac{9}{2} \times \dfrac{{( - 7)}}{4}$

$ = \dfrac{{9 \times ( - 7)}}{{2 \times 4}}$

$ = \dfrac{{ - 63}}{8}$

$ =  - 7\dfrac{7}{8}$

ii. $\dfrac{3}{{10}} \times ( - 9)$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

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

$ = \dfrac{{3 \times ( - 9)}}{{10}}$

$ = \dfrac{{ - 27}}{{10}}$

$ =  - 2\dfrac{7}{{10}}$

iii. $\dfrac{{ - 6}}{5} \times \dfrac{9}{{11}}$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

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

\[ = \dfrac{{( - 6) \times 9}}{{5 \times 11}}\]

$ = \dfrac{{ - 54}}{{55}}$

iv. $\dfrac{3}{7} \times \dfrac{{( - 2)}}{5}$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

$\dfrac{3}{7} \times \dfrac{{( - 2)}}{5}$

$ = \dfrac{{3 \times ( - 2)}}{{7 \times 5}}$

$ = \dfrac{{ - 6}}{{35}}$

v. $\dfrac{3}{{11}} \times \dfrac{2}{5}$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

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

$ = \dfrac{{3 \times ( 2)}}{{11 \times 5}}$

$ = \dfrac{{ 6}}{{55}}$

vi. $\dfrac{3}{{ - 5}} \times \dfrac{-5}{3}$

Ans: In case of multiplication, the numerator of one term is multiplied with the numerator of another term and the denominator of one term is multiplied by the denominator of another term.

Thus, 

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

$ = \dfrac{{3 \times -5}}{{( - 5) \times 3}}$

$ = \dfrac{{ - 15}}{{ - 15}}$

$ =  1$

4. Find the value of:

i. $( - 4) \div \dfrac{2}{3}$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

$ = ( - 4) \times \dfrac{3}{2}$

$ = \left( { - 2} \right) \times 3$

$ =  - 6$

ii. $\dfrac{{ - 3}}{5} \div 2$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

$\dfrac{{ - 3}}{5} \times \dfrac{1}{2}$

$ = \dfrac{{ - 3}}{{10}}$

iii. $\dfrac{{ - 4}}{5} \div ( - 3)$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

\[ = \dfrac{{ - 4}}{5} \times \dfrac{1}{{( - 3)}}\]

\[ = \dfrac{{ - 4}}{{( - 15)}}\]

$ = \dfrac{4}{{15}}$

iv. $\dfrac{{ - 1}}{8} \div \dfrac{3}{4}$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

$ = \dfrac{{ - 1}}{8} \times \dfrac{4}{3}$

$ = \dfrac{{ - 4}}{{24}}$

$ = \dfrac{{ - 1}}{6}$

v. $\dfrac{{ - 2}}{{13}} \div \dfrac{1}{7}$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

$ = \dfrac{{ - 2}}{{13}} \times 7$

$ = \dfrac{{ - 14}}{{13}}$

vi. $\dfrac{{ - 7}}{{12}} \div \dfrac{-2}{{13}}$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

$\dfrac{{ - 7}}{{12}} \div \dfrac{-2}{{13}}$

$ = \dfrac{{ - 7}}{{12}} \times \dfrac{{13}}{-2}$

$ = \dfrac{{( - 7) \times 13}}{{12 \times ( - 2)}}$

$ = \dfrac{{ - 91}}{{ - 24}}$

$ = \dfrac{{91}}{{24}}$

$ = 3\dfrac{{19}}{{24}}$

vii. $\dfrac{3}{{13}} \div \left( {\dfrac{{ - 4}}{{65}}} \right)$

Ans: In case of division the denominator of the first term will be multiplied by the numerator of the second term and vice versa.

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

$\dfrac{3}{{13}} \times \left( {\dfrac{{65}}{{ - 4}}} \right)$

$ = \dfrac{{3 \times ( - 5)}}{{1 \times 4}}$

$ = \dfrac{{ - 15}}{4}$

$ =  - 3\dfrac{3}{4}$


Conclusion

NCERT Solution for Class 7 Maths Chapter Exercise 8.2, students gain a solid understanding of performing operations with rational numbers. This exercise covers addition, subtraction, multiplication, and division of rational numbers, providing a comprehensive practice of these fundamental concepts. By working through these problems, students become proficient in handling fractions and understanding their properties. The solutions provided by Vedantu ensure clarity and accuracy, helping students build confidence in their mathematical abilities. Mastering these operations is crucial for future topics in mathematics, making this exercise an essential part of the learning journey.


Class 7 Maths Chapter 8: Exercises Breakdown

Exercises

Number of Questions

Exercise 8.1

10 Questions and Solutions



CBSE Class 7 Maths Chapter 8 Other Study Materials



Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.




Important Related Links for NCERT Class 7 Maths

Access these essential links for NCERT Class 7 Maths, offering comprehensive solutions, study guides, and additional resources to help students master language concepts and excel in their exams.


FAQs on NCERT Solutions for Class 7 Maths Chapter 8 Rational Numbers Ex 8.2

1. How many questions are there in NCERT Solutions for class 7 maths Ch 8 ex 8.2?

There are a total of four exercises in Chapter 8 of the NCERT class 7 maths Ch 8 ex 8.2 consists of 4 questions. Solutions to all these questions are being provided by Vedantu and these are available on the official website of Vedantu and the mobile application of Vedantu. The best part is that these NCERT solutions are specially designed by qualified teacher experts and these are available for students free of cost.

2. How can we solve Maths Chapter 8 Rational Numbers Class 7 Ex 8.2?

The questions that are included in the first exercise of Maths Chapter 8 Rational Numbers Class 7 Ex 8.2 are quite simple and can easily be done if the students have clear concepts about factorial notations. The NCERT contains the solved examples for the same that will help you understand the questions. For better conceptual understanding, students can look for the NCERT solutions for Maths Chapter 8 Rational Numbers Class 7 Ex 8.2 on the Vedantu website.

3. Where can I find easy ncert solutions for  class 7 maths ex 8.2?

Vedantu provides you with easy and understandable solutions for the first exercise of permutations and combinations. A highly skilled maths teacher at Vedantu prepares these solutions according to the latest pattern and in such a way that a student with no prior knowledge of permutations can also understand the topic while solving the question. The NCERT solutions are available in PDF format for download.

4. What are rational numbers in class 7 maths ex 8.2?

In class 7 maths ex 8.2 Rational numbers are numbers that can be written as a fraction where both the numerator and the denominator are integers, and the denominator is not zero. Examples include ¾ and -7/2. These numbers can be positive, negative, or zero.

5. What operations on rational numbers are covered in Exercise 8.2?

Exercise 8.2 focuses on the four main operations with rational numbers: addition, subtraction, multiplication, and division. Each operation is explained with examples to help students understand and apply the concepts effectively.

6. How do you add two rational numbers in maths class 7 Chapter 8 exercise 8.2?

In maths class 7 Chapter 8 exercise 8.2 to add two rational numbers, first find a common denominator. Adjust the numerators according to this common denominator, then add the numerators together. The result is a new fraction with the common denominator. For example, $\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$.

7. What is the process for subtracting rational numbers in class 7th exercise 8.2?

In class 7th exercise 8.2 subtracting rational numbers involves finding a common denominator, similar to addition. After adjusting the numerators, subtract the second numerator from the first. This gives you the difference as a new fraction. For example, $\frac{a}{b}-\frac{c}{d} = \frac{ad-bc}{bd}$.

8. How is the multiplication of rational numbers performed in class 7th maths Chapter 8 exercise 8.2?

Multiplication of rational numbers is straightforward. Multiply the numerators together and the denominators together in class 7th maths Chapter 8 exercise 8.2. The product of these multiplications gives you the new numerator and denominator for the result. For example, $\frac{a}{b}\times \frac{c}{d} = \frac{ac}{bd}$.

9. How do you divide one rational number by another in class 7 exercise 8.2?

To divide rational numbers, multiply the first number by the reciprocal of the second. This involves flipping the numerator and denominator of the divisor and then performing a multiplication of the resulting fractions in class 7 exercise 8.2. For example, $\frac{a}{b}\div  \frac{c}{d} = \frac{a}{b}\times   \frac{d}{c} = \frac{ad}{bc}$.