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

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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