Question

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

$\frac{4}{25}$

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

$\because \frac{a^m}{a^n}=(a{)}^{m-n}and{a}^m\times a^n=a^{m+n}$

$\Rightarrow x=(\frac{-2}{5}{)}^{30-28}$ and $y=(\frac{-2}{5}{)}^{2+3}$

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

=$\left(\frac{-2}{5}{)}^6\right(\because (a^m{)}^n=a^{(m\times n)})$
$x^3÷y=\left(\frac{(\frac{-2}{5}{)}^6}{(\frac{-2}{5}{)}^5}\right)$

=$(\frac{-2}{5}{)}^{6-5}$

= $\frac{-2}{5}$
$\therefore (x^3÷y{)}^2=(\frac{-2}{5}{)}^2=\frac{4}{25}$