| Find the square root of \[\frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04){{(0.4+0.2)}^{3}}}.\] [SSC (CGL) 2015] |
A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) 3
D) \[\frac{3}{2}\]
Correct Answer: B
Solution :
| \[\frac{(0.064-0.008)\,(0.16-0.04)}{(0.16+0.08+0.04)\,{{(0.4+0.2)}^{3}}}\] |
| \[=\ \,\frac{[{{(0.4)}^{3}}-{{(0.2)}^{3}}]\,\,[{{(0.4)}^{2}}-{{(0.2)}^{2}}]}{[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}]\,\,{{(0.4+0.2)}^{3}}}\] |
| \[[\because {{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+ab+{{b}^{2}})]\] |
| \[=\,\,\frac{(0.4-0.2)[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}]\,\,[{{(0.4)}^{2}}-{{(0.2)}^{2}}]}{[{{(0.4)}^{2}}+(0.4)(0.2)+{{(0.2)}^{2}}]\,\,{{(0.4+0.2)}^{3}}}\]\[=\,\,\frac{(0.4-0.2)(0.4-0.2)(0.4+0.2)}{{{(0.4+0.2)}^{3}}}\] |
| \[=\,\,\frac{{{(0.4-0.2)}^{2}}}{{{(0.4+0.2)}^{2}}}={{\left( \frac{0.2}{0.6} \right)}^{2}}={{\left( \frac{1}{3} \right)}^{2}}=\frac{1}{9}\] |
| But we have to find the square root of the given expression, |
| \[\therefore \] \[\sqrt{\frac{1}{9}}=\frac{1}{3}\] |
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