# 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}

Video Solution
Rational Numbers
Ex 9.1 | Question 9

## 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}}\end{align}

and

\begin{align}\frac{{5 \times 3}}{{2 \times 3}} = \frac{{15}}{6}\end{align}
Since,\begin{align}{\frac{4}{6} < \frac{{15}}{6}}\end{align}
So,\begin{align}{\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}}\end{align}

and

\begin{align}\frac{{ - 4 \times 2}}{{3 \times 2}} = \frac{{ - 8}}{6}\end{align}
Since,\begin{align}{\frac{{ - 5}}{6} > \frac{{ - 8}}{6}}\end{align}
So,\begin{align}{\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}\end{align}

and

\begin{align}\frac{2}{ - 3} = \frac{2{\times4}}{ - 3 \times 4} = \frac{ - 8}{12}\end{align}
Since, \begin{align}\frac{ - 9}{12} < \frac{ - 8}{12}\end{align}
So, \begin{align}\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}\end{align}

(Negative number is always smaller than the positive number)

(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}

Since, \begin{align}\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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