 
                 
                InterviewSolution
| 1. | Simplify:(i) \((3^{2}+2^{2})\times (\frac{1}{2})^{3}\)(ii) \((3^{2}+2^{2})\times (\frac{2}{3})^{-3}\)(iii) \([(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}]\div(\frac{1}{4})^{-3}\)(iv) \((2^{2}+3^{2}-4^{2})\div (\frac{3}{2})^{2}\) | 
| Answer» (i) \((3^{2}+2^{2})\times (\frac{1}{2})^{3}\) ⇒ \((9+4)\times \frac{1}{2^{3}}\)[Using \(a^{n}=a\times a.............n\, times\)] ⇒ \(13\times \frac{1}{8}=\frac{13}{8}\) (ii) \((3^{2}+2^{2})\times (\frac{2}{3})^{-3}\) ⇒ \((9+4)\times (\frac{3}{2})^{3}\)[Using \(a^{-n}=\frac{1}{a^{n}}\) \(a^{n}=a\times a.............n\, times\)] ⇒ \(13\times \frac{27}{8}=\frac{351}{8}\) (iii) \([(\frac{1}{3})^{-3}-(\frac{1}{2})^{-3}]\div (\frac{1}{4})^{-3}\) ⇒ \([(3)^{3}-(2)^{3}]\div (4)^{3}\)[Using \(a^{-n}=\frac{1}{a^{n}}\)] ⇒ [27-8]\(\div 4^{3}\) ⇒ \(19\times \frac{1}{4^{2}}\)[Using \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)] ⇒ \(19\times \frac{1}{4^{2}}\)= \(\frac{19}{64}\)[Using \(a^{n}=a\times a.............n\, times\)] (iv) \((2^{2}+3^{2}-4^{2})\div (\frac{3}{2})^{2}\) ⇒ (4+9-16)\(\div (\frac{3}{2})^{2}\)[Using \(a^{n}=a\times a.............n\, times\)] ⇒ \((-3)\div \frac{9}{4}\) ⇒ \(-3\times \frac{4}{9}=-\frac{4}{3}\)[Using \(\frac{1}{a}\div \frac{1}{b}\)= \(\frac{1}{a}\times \frac{b}{1}\)] | |