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

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

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