NCERT Solutions for Class 7 Maths Chapter 9: Rational Numbers - Exercise 9.2

Exercise 9.2

Refer the page 12 to 14 for the exercise 9.2 in the PDF

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}$

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Exercise 9.2

Opting for the NCERT solutions for Ex 9.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 9 Exercise 9.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

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