# Ex.9.1 Q8 Rational-Numbers Solution - NCERT Maths Class 7

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## Question

Fill in the boxes with the correct symbol out of $$>$$, $$<$$, and$$=$$.

(i) \begin{align} \frac{{ - 5}}{7} \quad\boxed{\;\;}\quad \frac{2}{3} \end{align}

(ii) \begin{align} \frac{{ - 4}}{5} \quad\boxed{\;\;}\quad \frac{{ - 5}}{7}\end{align}

(iii) \begin{align} \frac{{ - 7}}{8} \quad\boxed{\;\;}\quad \frac{{14}}{{ - 16}}\end{align}

(iv)\begin{align} \frac{{ - 8}}{5} \quad\boxed{\;\;}\quad \frac{{ - 7}}{4}\end{align}

(v)\begin{align} \frac{1}{{ - 3}} \quad\boxed{\;\;}\quad \frac{{ - 1}}{4}\end{align}

(vi)\begin{align} \frac{5}{{ - 11}} \quad\boxed{\;\;}\quad \frac{{ - 5}}{{11}}\end{align}

(vii)  \begin{align} 0 \quad\boxed{\;\;}\quad \frac{{ - 7}}{6}\end{align}

Video Solution
Rational Numbers
Ex 9.1 | Question 8

## Text Solution

What is known?

Two rational numbers.

What is unknown?

Comparison of the two rational numbers i.e., which one is smaller and which one is greater or are they both equal.

Reasoning:

In this type of questions, first find the $$LCM$$ of the denominators of both the rational numbers. Then make denominator of each rational number equal to $$LCM$$ by multiplying numerator and denominator with the same number (convert them into like fractions). Then comparison of the two numbers can be easily made.

Steps:

(i) \begin{align} \frac{{ - 5}}{7} \quad\boxed{\;\;}\quad \frac{2}{3} \end{align}

\begin{align} \frac{{ - 5 \times 3}}{{7 \times 3}} &\quad\boxed{\;\;}\quad \frac{{2 \times 7}}{{3 \times 7}}\\ \frac{{ - 15}}{{21}} &\quad\boxed{\;\;}\quad \frac{{14}}{{21}}\\ \frac{{ - 15}}{{21}} &\quad\boxed{<}\quad \frac{{14}}{{21}}\end{align}

(Positive number is greater than the negative number)

Therefore, \begin{align}\frac{{ - 5}}{7} < \frac{2}{3}\end{align}

(ii) \begin{align} \frac{{ - 4}}{5} \quad\boxed{\;\;}\quad \frac{{ - 5}}{7}\end{align}

\begin{align}\frac{{ - 4 \times 7}}{{5 \times 7}}&\quad\boxed{\;\;}\quad \frac{{ - 5 \times 5}}{{7 \times 5}}\\ \frac{{ - 28}}{{35}}&\quad\boxed{\;\;}\quad \frac{{ - 25}}{{35}}\\ \frac{{ - 28}}{{35}}&\quad\boxed{<}\quad \frac{{ - 25}}{{35}}\end{align}

Therefore, \begin{align}\frac{{ - 4}}{5}\,\, < \frac{{ - 5}}{7}\end{align}

(iii) \begin{align} \frac{{ - 7}}{8} \quad\boxed{\;\;}\quad \frac{{14}}{{ - 16}}\end{align}

\begin{align}\frac{{ - 7 \times 2}}{{8 \times 2}}&\quad\boxed{\;\;}\quad \frac{{14\times 1}}{{ - 16\times 1}}\\ \frac{{ - 14}}{{16}} & \quad\boxed{\;\;}\quad \frac{{14}}{{ - 16}}\\ \frac{{ - 14}}{{16}} & \quad\boxed{=}\quad \frac{{ - 14}}{{16}}\\\end{align}

Therefore, \begin{align}\frac{{ - 7}}{8}\,\,\,\, = \,\,\frac{{14}}{{ - 16}}\end{align}

(iv)\begin{align} \frac{{ - 8}}{5} \quad\boxed{\;\;}\quad \frac{{ - 7}}{4}\end{align}

\begin{align} \frac{{ - 8 \times 4}}{{5 \times 4}}& \quad\boxed{\;\;}\quad \frac{{ - 7 \times 5}}{{4 \times 5}}\\ \frac{{ - 32}}{{20}} &\quad\boxed{\;\;}\quad \frac{{ - 35}}{{20}}\\ \frac{{ - 8\;}}{5} &\quad\boxed{>}\quad \frac{{ - 7\;}}{4}\end{align}

Therefore, \begin{align}\frac{{ - 8}}{5}\,\,\, > \,\,\,\frac{{ - 7}}{4}\end{align}

(v)\begin{align}\frac{1}{{ - 3}} \quad\boxed{\;\;}\quad \frac{{ - 1}}{4}\end{align}

\begin{align} \frac{{1 \times 4}}{{ - 3 \times 4}}&\quad\boxed{\;\;}\quad \frac{{ - 1 \times 3}}{{4 \times 3}}\\\frac{4}{{ - 12}}&\quad\boxed{\;\;}\quad \frac{{3}}{{ - 12}}\\\frac{1}{{ - 3}}&\quad\boxed{<}\quad \frac{{ - 1}}{{4}}\end{align}

Therefore,\begin{align}\frac{1}{{ - 3}}&\quad\boxed{>}\quad \frac{{ - 1}}{{4}}\end{align}

(vi)\begin{align} \frac{5}{{ - 11}} \quad\boxed{\;\;}\quad \frac{{ - 5}}{{11}}\end{align}

\begin{align} \frac{{ - 5}}{{11}}&\quad\boxed{\;\;}\quad \frac{{ - 5}}{{11}}\\ \frac{{ - 5}}{{11}}&\quad\boxed{=}\quad \frac{{ - 5}}{{11}}\end{align}

Therefore, \begin{align}\frac{{ - 5}}{{11}}&\quad\boxed{=}\quad \frac{{ - 5}}{{11}}\end{align}

(vi)  \begin{align} 0 \quad\boxed{\;\;}\quad \frac{{ - 7}}{6}\end{align}

\begin{align}&0\quad\boxed{\;\;}\quad\frac{{ - 7}}{6} \\&0\quad\boxed{>}\quad\frac{{ - 7}}{6}\end{align}

($$0$$ is always greater than negative integer)

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