Ex.9.2 Q3 Rational-Numbers Solution - NCERT Maths Class 7

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Question

Find the product:

(i)\(\begin{align}\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right]\end{align}\)

(ii)\(\begin{align}\frac{3}{{10}} \times \left( { - 9} \right)\end{align}\)

(iii)\(\begin{align}\frac{{ - 6}}{5} \times \frac{9}{{11}}\end{align}\)

(iv)\(\begin{align}\frac{3}{7} \times \left[ {\frac{{ - 2}}{5}} \right]\end{align}\)

(v)\(\begin{align}\frac{3}{{11}} \times \frac{2}{5}\end{align}\)

(vi)\(\begin{align}\frac{3}{{ - 5}} \times \frac{{ - 5}}{3}\end{align}\)

 Video Solution
Rational Numbers
Ex 9.2 | Question 3

Text Solution

What is known?

Two rational numbers

What is unknown?

Product of two rational numbers.

Reasoning:

In such type of questions find the product of numerator and denominator.

Steps:

(i)

\[\begin{align}\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right]&=\frac{9}{2} \times \left[ {\frac{{ - 7}}{4}} \right] \\&= \frac{{9 \times - 7}}{{2 \times 4}} = \frac{{ - 63}}{8}\end{align}\]

 (ii)

\[\begin{align}\frac{3}{{10}} \times \left( { - 9} \right)\,& = \frac{3}{{10}} \times \left( {\frac{{ - 9}}{1}} \right)\\ &= \;\frac{{3 \times \; - 9}}{{10}} = \frac{{ - 27}}{{10}}\end{align}\]

 (iii)

\[\begin{align}\frac{{ - 6}}{5} \times \frac{9}{{11}} = \frac{{ - 6 \times 9}}{{5 \times 11}} = \frac{{ - 54}}{{55}}\end{align}\]

 

(iv)

\[\begin{align}\frac{3}{7} \times \left[ {\frac{{ - 2}}{5}} \right] = \frac{{3 \times - 2}}{{7 \times 5}} = \frac{{ - 6}}{{35}}\end{align}\]

(v) 

\[\begin{align}\frac{3}{{11}} \times \frac{2}{5}=\frac{3}{{11}} \times \frac{2}{5}= \frac{6}{{55}}\end{align}\]

(vi) 

\[\begin{align}\frac{3}{{ - 5}} \! \times \! \frac{{ - 5}}{3} \! = \! \frac{3}{{ - 5}} \! \times  \!  \frac{{ - 5}}{3} \!  = \!  \frac{{ - 15}}{{ - 15}} \! = \! 1 \end{align}\]

  
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