# Ex.9.2 Q1 Rational-Numbers Solution - NCERT Maths Class 7

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## Question

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}

Video Solution
Rational Numbers
Ex 9.2 | Question 1

## Text 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}

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