Question

If $x={\left(\frac{-2}{5}\right)}^{30}÷{\left(\frac{5}{-2}\right)}^{-28}$ and $y={\left(\frac{-2}{5}\right)}^2\times {\left(\frac{-2}{5}\right)}^3$, then find the value of $(x^3÷y{)}^2$.

• A. $\frac{5}{40}$
• B. $\frac{-2}{5}$
• C. $\frac{4}{25}$
• D. $\frac{25}{4}$

Answer 1

The correct option is C $\frac{4}{25}$

Given, $x={\left(\frac{-2}{5}\right)}^{30}÷{\left(\frac{5}{-2}\right)}^{-28}$ and $y={\left(\frac{-2}{5}\right)}^2\times {\left(\frac{-2}{5}\right)}^3$

$x={\left(\frac{-2}{5}\right)}^{30}÷{\left(\frac{5}{-2}\right)}^{-28}$

$={\left(\frac{-2}{5}\right)}^{30}÷{\left(\frac{-2}{5}\right)}^{28}$$[\because (a{)}^{-m}=\frac{1}{a^m}]$

$=(\frac{-2}{5}{)}^{30-28}$ $[\because \frac{a^m}{a^n}=a^{m-n}]$

$=(\frac{-2}{5}{)}^2$

$⟹x^3=\left[\right(\frac{-2}{5}{)}^2{]}^3$

$i.e.,x^3=(\frac{-2}{5}{)}^6\cdots (i)$

$[\because (a^m{)}^n=a^{mn}]$

$y={\left(\frac{-2}{5}\right)}^2\times {\left(\frac{-2}{5}\right)}^3$

$⟹y=\left(\frac{-2}{5}{)}^{2+3}\right[\because a^m\times a^n=a^{m+n}]$

$i.e.,y=(\frac{-2}{5}{)}^5\cdots (ii)$

$\therefore \text{From (i) and (ii),}{x}^3÷y=\frac{(\frac{-2}{5}{)}^6}{(\frac{-2}{5}{)}^5}$

$=(\frac{-2}{5}{)}^{6-5}$ $[\because \frac{a^m}{a^n}=a^{m-n}]$

$=\frac{-2}{5}$

$\therefore (x^3÷y{)}^2=\frac{4}{25}$