Ex.9.1 Q9 Rational-Numbers Solution - NCERT Maths Class 7

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Question

Which is greater in each of the following:

(i) \(\begin{align}{\frac{2}{3},\frac{5}{2}}\end{align}\)  (ii) \(\begin{align}{\frac{{ - 5}}{6},\frac{{ - 4}}{3}}\end{align}\)
(iii) \(\begin{align}\frac{{ - 3}}{4},\frac{2}{{ - 3}}\end{align}\) iv) \(\begin{align}\frac{{ - 1}}{6},\frac{2}{{ - 12}}\end{align} \)
(v) \( \begin{align}- 3\frac{2}{7}, - 3\frac{4}{5}\end{align}\)  

Text Solution

What is known?

Two rational numbers.

What is unknown?

Greater rational number of the two given rational numbers.

Reasoning:

In such type of questions take the \(L.C.M\) of denominator of both the rational numbers or convert them into like fractions. After converting them into like fractions comparison will be easy.

Steps:

(i)\(\begin{align}{\frac{2}{3},\frac{5}{2}}\end{align}\)

L.C.M of \(3\) and \( 2\) is \(6\)

\[\begin{align}{\frac{{2 \times 2}}{{3 \times 2}} = \frac{6}{9}}&\quad{{\rm{ and }}\quad\frac{{5 \times 3}}{{2 \times 3}} = \frac{{15}}{6}}\\\quad{{\text{Since}}}{\frac{4}{6} < \frac{{15}}{6}}&\quad{{\text{ so}},\frac{2}{3} < \frac{5}{2}}\end{align}\]

(ii) \(\begin{align}{\frac{{ - 5}}{6},\frac{{ - 4}}{3}}\end{align}\)

L.C.M of \(6\) and \(3\) is \(6\)

\[\begin{align}{\frac{{ - 5\times1}}{6\times1} = \frac{{ - 5}}{6}}&\quad{{\rm{ and }}\quad\frac{{ - 4 \times 2}}{{3 \times 2}} = \frac{{ - 8}}{6}}\\\rm{Since}{\frac{{ - 5}}{6} > \frac{{ - 8}}{6}}&\quad{{\rm{So, }}\frac{{ - 5}}{6} > \frac{{ - 4}}{3}}\end{align}\]

(iii) \(\begin{align}\frac{{ - 3}}{4},\frac{2}{{ - 3}}\end{align}\)

L.C.M of \(4\) and \(3\) is \(12\)

\[\begin{align}\frac{{ - 3}}{4} = \frac{{ - 3 \times 3}}{{4{\times3}}} = \frac{{ - 9}}{{12}}\;\;\;{\mkern 1mu} &\quad {\rm{and}}\qquad \frac{2}{{ - 3}} = \frac{{2{\times4}}}{{ - 3 \times 4}} = \frac{{ - 8}}{{12}}\\\rm{Since}\frac{{ - 9}}{{12}} < \frac{{ - 8}}{{12}}\;\;\;{\mkern 1mu} \;\;\;{\mkern 1mu}&\quad{\text{So,}}\;\frac{{ - 3}}{4} < \frac{2}{{ - 3}}\end{align}\]

(iv) \(\begin{align}\frac{{ - 1}}{4},\frac{1}{{ 4}}\end{align} \)

\[\begin{align}\frac{{ - 1}}{4} < \frac{1}{4} &\quad \left( {{\text{Negative number is always smaller than the positive number}}} \right)\end{align}\]

(v) \( \begin{align}- 3\frac{2}{7}, - 3\frac{4}{5}\end{align}\)

\(\begin{align} - 3\frac{2}{7} = \frac{{ - 23}}{7}\;\;{\rm{and}}\;\;\;{\mkern 1mu} 3\frac{4}{5} = \frac{{ - 19}}{5}\end{align}\)

L.C.M of \(7\) and \(5\) is \(35\)

\[\begin{align}\frac{{ - 23}}{7} &= \frac{{ - 23 \times 5}}{{7 \times 5}}\\&= \frac{{ - 115}}{{35}}\\
\frac{{ - 19}}{7} &= \frac{{ - 19\times7}}{{7 \times 5}}\\&= \frac{{ - 133}}{{35}}\end{align}\]

\(\begin{align}\rm{Since}\frac{{ - 115}}{{35}} > \frac{{ - 133}}{{35}}\end{align}\)

So,\(\begin{align}- 3\frac{2}{7}\,\,\,\,\, > \,\, - 3\frac{4}{5}\end{align}\)

  
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