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

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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