$${\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}}$$    is equal to ?

A. 1

B. 2

C. $$\frac{1}{2}$$

D. $$\frac{1}{{16}}$$

Answer: Option A

Solution(By Examveda Team)

$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$