$${\left( {64} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}}$$ is equal to ?
A. 1
B. 2
C. $$\frac{1}{2}$$
D. $$\frac{1}{{16}}$$
Answer: Option A
Solution (By Examveda Team)
$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$This Question Belongs to Arithmetic Ability >> Surds And Indices
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