Question

Evaluate: ${\left(\begin{array}{c}\frac{64}{125}\end{array}\right)}^{\frac{-2}{3}}÷\frac{1}{(\frac{256}{625}{)}^{\frac{1}{4}}}+\left(\begin{array}{c}\frac{\left(\sqrt{25}\right)}{\sqrt[3]{364}}\end{array}\right)$

• A. $\frac{9}{2}$
• B. $\frac{9}{4}$
• C. $\frac{8}{2}$
• D. $\frac{5}{2}$

Answer 1

Answer: (D)

Now,${\left(\begin{array}{c}\frac{64}{125}\end{array}\right)}^{\frac{-2}{3}}÷\frac{1}{(\frac{256}{625}{)}^{\frac{1}{4}}}+\left(\begin{array}{c}\frac{\left(\sqrt{25}\right)}{\sqrt[3]{364}}\end{array}\right)$

$={\left(\begin{array}{c}\frac{4^3}{5^3}\end{array}\right)}^{\frac{-2}{3}}÷\frac{1}{(\frac{4^4}{5^4}{)}^{\frac{1}{4}}}+\left(\begin{array}{c}\frac{5}{\sqrt[3]{34^3}}\end{array}\right)$

Using $(a^m{)}^n=a^{m\times n}$, we get,

$={\left(\begin{array}{c}\frac{4}{5}\end{array}\right)}^{3\times \frac{-2}{3}}÷\frac{1}{(\frac{4}{5}{)}^{4\times \frac{1}{4}}}+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

$={\left(\begin{array}{c}\frac{4}{5}\end{array}\right)}^{-2}÷\frac{1}{(\frac{4}{5}{)}^1}+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

$={\left(\begin{array}{c}\frac{4}{5}\end{array}\right)}^{-2}\times \frac{5}{4}+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

Using $(\frac{a}{b}{)}^{-n}=(\frac{b}{a}{)}^n$

$={\left(\begin{array}{c}\frac{5}{4}\end{array}\right)}^2\times \frac{5}{4}+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

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

$={\left(\begin{array}{c}\frac{5}{4}\end{array}\right)}^{2+1}+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

$={\left(\begin{array}{c}\frac{5}{4}\end{array}\right)}^3+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

$=\left(\begin{array}{c}\frac{125}{64}\end{array}\right)+\left(\begin{array}{c}\frac{5}{4}\end{array}\right)$

$=\left(\begin{array}{c}\frac{125}{64}\end{array}\right)+\left(\begin{array}{c}\frac{5}{4}\times \frac{16}{16}\end{array}\right)$

$=\frac{125+80}{64}$

$=\frac{205}{64}$

Hence, ${\left(\begin{array}{c}\frac{64}{125}\end{array}\right)}^{\frac{-2}{3}}÷\frac{1}{(\frac{256}{625}{)}^{\frac{1}{4}}}+\left(\begin{array}{c}\frac{\left(\sqrt{25}\right)}{\sqrt[3]{364}}\end{array}\right)=3\frac{13}{64}$