The value of $$\frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}$$ $$\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}$$ $$\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}}$$ is = ?
A. 102
B. 105
C. 107
D. 109
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr & = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr & = \left( {{6^2} + {4^3} + {2^1}} \right) \cr & = \left( {36 + 64 + 2} \right) \cr & = 102 \cr} $$Related Questions on Surds and Indices
A. $$\frac{1}{2}$$
B. 1
C. 2
D. $$\frac{7}{2}$$
Given that 100.48 = x, 100.70 = y and xz = y2, then the value of z is close to:
A. 1.45
B. 1.88
C. 2.9
D. 3.7
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