Question

If $x={\left(\frac{-2}{5}\right)}^{30}÷{\left(\frac{-2}{5}\right)}^{28}$ and $y={\left(\frac{-2}{5}\right)}^2\times {\left(\frac{-2}{5}\right)}^3$, 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 $\frac{4}{25}$

Using $a^m\times a^n=a^{m+n}$ and $a^m÷a^n=a^{m-n}$

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

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

$⟹x^3=\left(\right(-\frac{2}{5}{)}^2{)}^3=(-\frac{2}{5}{)}^6$

$⟹x^3÷y=(-\frac{2}{5}{)}^6÷(-\frac{2}{5}{)}^5=\frac{-2}{5}$

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