Quantitative Aptitude Ques 2302

Question: Find the square root of $ \frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04){{(0.4+0.2)}^{3}}}. $

Options:

A) $ \frac{2}{3} $

B) $ \frac{1}{3} $

C) 3

D) $ \frac{3}{2} $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ \frac{(0.064-0.008),(0.16-0.04)}{(0.16+0.08+0.04),{{(0.4+0.2)}^{3}}} $ $ =\ ,\frac{[{{(0.4)}^{3}}-{{(0.2)}^{3}}][{{(0.4)}^{2}}-{{(0.2)}^{2}}]}{[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}]{{(0.4+0.2)}^{3}}} $ $ [\because a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})] $ $ =\frac{(0.4-0.2)[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}][{{(0.4)}^{2}}-{{(0.2)}^{2}}]}{[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}]{{(0.4+0.2)}^{3}}} $ $ =\frac{(0.4-0.2)(0.4-0.2)(0.4+0.2)}{{{(0.4+0.2)}^{3}}} $ $ =\frac{{{(0.4-0.2)}^{2}}}{{{(0.4+0.2)}^{2}}}={{( \frac{0.2}{0.6} )}^{2}}={{( \frac{1}{3} )}^{2}}=\frac{1}{9} $ But we have to find the square root of the given expression,

$ \therefore $ $ \sqrt{\frac{1}{9}}=\frac{1}{3} $

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