# Exercise 9.2 Rational-Numbers -NCERT Solutions Class 7

Go back to  'Rational Numbers'

## Chapter 9 Ex.9.2 Question 1

Find the sum:

(i) \begin{align}\frac{5}{4} + \left[ {\frac{{ - 11}}{4}} \right]{\mkern 1mu}\end{align}

(ii) \begin{align}\frac{5}{3} + \frac{3}{5}\end{align}

(iii) \begin{align}\frac{{ - 9}}{{10}} + \frac{{22}}{{15}}\end{align}

(iv) \begin{align}\frac{{ - 3}}{{ - 11}} + \,\,\frac{5}{9}\end{align}

(v) \begin{align}{\frac{{ - 8}}{{19}} + \frac{{{\rm{(}} - 2{\rm{)}}}}{{57}}}\end{align}

(vi) \begin{align}{\frac{{ - 2}}{3} + 0}\end{align}

(vii) \begin{align} - 2\frac{1}{3}{\mkern 1mu} \, + \,4\frac{3}{5}\end{align}

### Solution

What is known?

Two rational numbers

What is unknown?

Sum of two rational numbers.

Reasoning:

In such type of questions, take the $$L.C.M$$ of denominator or convert the given fractions into like fractions and then find their sum. You can also reduce the fractions to the lowest form.

Steps:

(i) \begin{align}\frac{5}{4} + \left[ {\frac{{ - 11}}{4}} \right]\end{align}

\begin{align}\frac{5}{4} + \left[ {\frac{{ - 11}}{4}} \right]{\mkern 1mu}&=\frac{5}{4} - \frac{{11}}{4}\\ &= \frac{{5 - 11}}{4}\\&= \frac{{ - 6}}{4}\\&= \frac{{ - 3}}{2}\end{align}

(ii) \begin{align}\frac{5}{3} + \frac{3}{5}\end{align}

Taking $$L.C.M$$ of $$3$$ and $$5$$, we get $$15$$

\begin{align}\frac{5}{3} + \frac{3}{5} &= \frac{{5 \times 5}}{{15}} + \frac{{3 \times 3}}{{15}}\\ &= \frac{{25}}{{15}} + \frac{9}{{15}}\\ &= \frac{{25 + 9}}{{15}}\\ &= \frac{{34}}{{15}}\end{align}

(iii) \begin{align}\frac{{ - 9}}{{10}} + \frac{{22}}{{15}}\end{align}

Taking $$L.C.M$$ of $$10$$ and $$15$$, we get $$30$$

\begin{align}\frac{{ - 9}}{{10}} + \frac{{22}}{{15}} &= \frac{{ - 9 \times 3}}{{30}} + \frac{{22 \times 2}}{{30}}\\&= \frac{{ - 27}}{{30}} + \frac{{44}}{{30}}\\&= \frac{{17}}{{30}}\end{align}

(iv) \begin{align}\frac{{ - 3}}{{ - 11}} + \,\,\frac{5}{9}\end{align}

Taking $$L.C.M$$ of $$11$$ and $$9$$, we get $$99$$

\begin{align}\frac{{ - 3}}{{ - 11}} + \frac{5}{9} &= \frac{{ - 3 \times 9}}{{ - 99}} + \frac{{5 \times 11}}{{99}}\\&= \frac{{27}}{{99}} + \frac{{55}}{{99}}\\&= \frac{{82}}{{99}}\end{align}

(v) \begin{align}{\frac{{ - 8}}{{19}} + \frac{{{\rm{(}} - 2{\rm{)}}}}{{57}}}\end{align}

Taking $$L.C.M$$ of $$19$$ and $$57$$, we get $$57$$

\begin{align}\frac{{ - 8}}{{19}} + \frac{{{\rm{(}} - {\rm{2)}}}}{{{\rm{57}}}} &= \frac{{ - 8 \times 3}}{{19\times3}} + \frac{{ - 2\times1}}{{57\times1}}\\&= \frac{{ - 24}}{{ - 57}} + \frac{{ - 2}}{{57}}\\&= \frac{{ - 24 - 2}}{{57}}\\&= \frac{{ - 26}}{{57}}\end{align}

(vi) \begin{align}{\frac{{ - 2}}{3} + 0}\end{align}

Taking $$L.C.M$$ of $$3$$ and $$1$$, we get $$3$$

\begin{align}\frac{{ - 2}}{3} + 0 &= \frac{{ - 2 \times 1}}{3} + \frac{{0 \times 3}}{3}\\&= \frac{{ - 2}}{3} + \frac{0}{3}\\&= \frac{{ - 2 + 0}}{3}\\&= \frac{{ - 2}}{3}\end{align}

(vii) \begin{align} - 2\frac{1}{3}{\mkern 1mu} \, + \,4\frac{3}{5}\end{align} = \begin{align}\frac{{ - 7}}{3} + \frac{{23}}{5}\end{align}

Taking $$L.C.M$$ of $$3$$ and $$5$$, we get $$15$$

\begin{align}&= \frac{{ - 7 \times 5}}{{15}} + \frac{{23 \times 3}}{{15}}\\&= \frac{{ - 35}}{{15}} + \frac{{69}}{{15}}\\&= \frac{{ - 35 + 69}}{{15}}\\&= \frac{{34}}{{15}}\end{align}

## Chapter 9 Ex.9.2 Question 2

Find:

(i) \begin{align}{\frac{7}{{24}} - \frac{{17}}{{36}}}\end{align}

.(ii)\begin{align}{\frac{5}{{63}} - \left[ {\frac{{ - 6}}{{21}}} \right]}\end{align}

(iii) \begin{align}{\left.{\frac{{ - 6}}{{13}} - \left[ {\frac{{ - {\rm{7}}}}{{{\rm{15}}}}} \right.} \right]} \end{align}

(iv)\begin{align}\frac{{ - 3}}{8} - \frac{7}{{11}}\end{align}

(v)\begin{align} - 2\frac{1}{9} - 6\end{align}

### Solution

What is known?

Two rational numbers

What is unknown?

Difference between the given two rational numbers.

Reasoning:

In such type of questions take the $$L.C.M$$ of denominator or convert them into like fractions, then find their difference between them. You can also reduce them to the lowest or simplest form.

Steps:

(i) \begin{align}{\frac{7}{{24}} - \frac{{17}}{{36}}}\end{align}

Taking $$L.C.M$$ of $$24$$ and $$36$$, we get $$72$$

\begin{align}\frac{7}{{24}} - \frac{{17}}{{36}} &= \frac{{7 \times 3}}{{24\times3}} - \frac{{17 \times 2}}{{36\times3}}\\&= \frac{{21}}{{72}} - \frac{{34}}{{72}}\\&= \frac{{21 - 34}}{{72}}\\&= \frac{{ - 13}}{{72}}\end{align}

(ii) \begin{align}{\frac{5}{{63}} - \left[ {\frac{{ - 6}}{{21}}} \right]}\end{align}

Taking $$L.C.M$$ of $$63$$ and $$21$$, we get $$63$$

\begin{align}\frac{5}{{63}} - \left( {\frac{{ - 6}}{{21}}} \right) &= \frac{5\times1}{{63\times1}} + \frac{{6{\times3}}}{{21\times3}}\\ &= \frac{5}{{63}} + \frac{{18}}{{63}}\\ &= \frac{{5 + 18}}{{63}}\\ &= \frac{{23}}{{63}}\end{align}

(iii) \begin{align} {\left.{\frac{{ - 6}}{{13}} - \left[ {\frac{{ - {\rm{7}}}}{{{\rm{15}}}}} \right.} \right]} \end{align}

Taking $$L.C.M$$ of $$13$$ and $$15$$, we get $$195$$

\begin{align}\frac{{ - 6}}{{13}} - \frac{{ - 7}}{{15}} &= \frac{{ - 6 \times 15}}{{13\times15}} + \frac{{7 \times 13}}{{15\times13}}\\&= \frac{{ - 90}}{{195}} + \frac{{91}}{{195}}\\&= \frac{{ - 90 + 91}}{{195}}\\&= \frac{1}{{195}}\end{align}

(iv) \begin{align}\frac{{ - 3}}{8} - \frac{7}{{11}}\end{align}

Taking $$L.C.M$$ of $$8$$ and $$11$$, we get $$88$$

\begin{align}\frac{{ - 3}}{8} - \frac{7}{{11}} &= \frac{{ - 3 \times 11}}{{8\times11}} - \frac{{7 \times 8}}{{11\times8}}\\&= \frac{{ - 33}}{{88}} - \frac{{56}}{{88}}\\&= \frac{{ - 33 - 56}}{{88}}\\&= \frac{{ - 89}}{{88}}\end{align}

(v) \begin{align} - 2\frac{1}{9} - 6\end{align}

\begin{align} - 2\frac{1}{9} - 6 = - \frac{{19}}{9} - \frac{6}{1}\end{align}

Taking $$L.C.M$$ of $$9$$ and $$1$$, we get $$9$$

\begin{align}- \frac{{19}}{9} - \frac{6}{1} &= \frac{{ - 19}}{9\times1} - \frac{{6 \times 9}}{1\times9}\\&= \frac{{ - 19}}{9} - \frac{{54}}{9}\\&= \frac{{ - 19 - 54}}{9}\\&= \frac{{ - 73}}{9}\end{align}

## Chapter 9 Ex.9.2 Question 3

Find the product:

(i)\begin{align}\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right]\end{align}

(ii)\begin{align}\frac{3}{{10}} \times \left( { - 9} \right)\end{align}

(iii)\begin{align}\frac{{ - 6}}{5} \times \frac{9}{{11}}\end{align}

(iv)\begin{align}\frac{3}{7} \times \left[ {\frac{{ - 2}}{5}} \right]\end{align}

(v)\begin{align}\frac{3}{{11}} \times \frac{2}{5}\end{align}

(vi)\begin{align}\frac{3}{{ - 5}} \times \frac{{ - 5}}{3}\end{align}

### Solution

What is known?

Two rational numbers

What is unknown?

Product of two rational numbers.

Reasoning:

In such type of questions find the product of numerator and denominator.

Steps:

(i)

\begin{align}\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right]&=\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right] \\&= \frac{{9 \times - 7}}{{2 \times 4}} = \frac{{ - 63}}{8}\end{align}

(ii)

\begin{align}\frac{3}{{10}} \times \left( { - 9} \right)\,& = \frac{3}{{10}} \times \left( {\frac{{ - 9}}{1}} \right)\\ &= \;\frac{{3 \times \; - 9}}{{10}} = \frac{{ - 27}}{{10}}\end{align}

(iii)

\begin{align}\frac{{ - 6}}{5} \times \frac{9}{{11}} = \frac{{ - 6 \times 9}}{{5 \times 11}} = \frac{{ - 54}}{{55}}\end{align}

(iv)

\begin{align}\frac{3}{7} \times \left[ {\frac{{ - 2}}{5}} \right] = \frac{{3 \times - 2}}{{7 \times 5}} = \frac{{ - 6}}{{35}}\end{align}

(v)

\begin{align}\frac{3}{{11}} \times \frac{2}{5}=\frac{3}{{11}} \times \frac{2}{5}= \frac{6}{{55}}\end{align}

(vi)

\begin{align}\frac{3}{{ - 5}} \! \times \! \frac{{ - 5}}{3} \! = \! \frac{3}{{ - 5}} \! \times \! \frac{{ - 5}}{3} \! = \! \frac{{ - 15}}{{ - 15}} \! = \! 1 \end{align}

## Chapter 9 Ex.9.2 Question 4

Find the value of:

(i) \begin{align}\,\,\left( { - 4} \right) \div \frac{2}{3}\;\;\;{\mkern 1mu}\end{align}

(ii) \begin{align}\frac{{ - 3}}{5} \div 2\end{align}

(iii) \begin{align}\frac{{ - 4}}{5} \div \left( { - 3} \right)\end{align}

(iv) \begin{align}\frac{{ - 1}}{8} \div \frac{3}{4}\;\;\end{align}

(v) \begin{align}\frac{{ - 2}}{{13}} \div \frac{1}{7}\end{align}

(vi) \begin{align}\frac{{ - 7}}{{12}} \div \left[ {\frac{{ - 2}}{{13}}} \right]\end{align}

(vii) \begin{align}\frac{3}{{13}} \div \left[ {\frac{{ - 4}}{{65}}} \right]\end{align}

### Solution

What is known?

Two numbers.

What is  unknown?

Value of given numbers.

Reasoning:

In such type of questions, convert the denominator into its reciprocal. When you convert the denominator into its reciprocal, the sign of division will also convert into multiplication and now you can find out the product of the given numbers.

Steps:

(i)

\begin{align}(- 4) \! \div \! \frac{2}{3} \! = \! - 4 \! \times \! \frac{3}{2} \! = \! -2 \! \times \! 3 \! = \! -6\end{align}

(ii)

\begin{align}\frac{{ - 3}}{5} \div 2\; = \frac{{ - 3}}{5} \times \frac{1}{2} = \frac{{ - 3}}{{10}}\end{align}

(iii)

\begin{align}\frac{{ - 4}}{5} \div \left( { - 3} \right) = \frac{{ - 4}}{5} \times \frac{{ - 1}}{3} = \frac{4}{{15}}\end{align}

(iv)

\begin{align}\frac{{ - 1}}{8} \div \frac{3}{4} = \frac{{ - 1}}{8} \times \frac{4}{3} = \frac{{ - 4}}{{24}} = \frac{{ - 1}}{6}\end{align}

(v)

\begin{align}\frac{{ - 2}}{{13}} \! \div \! \frac{1}{7} \! = \! \frac{{ - 2}}{{13}} \! \times \! \frac{7}{1} \! = \! \frac{{ - 2 \! \times \! 7}}{{13}} \! = \! \frac{{ - 14}}{{13}}\end{align}

(vi)

\begin{align}\frac{{ - 7}}{{12}} \div \left[ {\frac{{ - 2}}{{13}}} \right] &= \frac{{ - 7}}{{12}} \times \frac{{13}}{{ - 2}}\\& = \frac{{ - 7 \times 13}}{{12 \times - 2}} \\&= \frac{{ - 91}}{{ - 24}} = \frac{{91}}{{24}}\end{align}

(vii)

\begin{align}\frac{3}{{13}} \div \left[ {\frac{{ - 4}}{{65}}} \right] &= \frac{3}{{13}} \times \frac{{65}}{{ - 4}} \\&= \frac{{3 \times 65}}{{13 \times - 4}}\\& = \frac{{3 \times 5}}{{ - 4}} = \frac{{ - 15}}{4}\end{align}

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