Quantitative Aptitude Ques 1706
Question: If $ 2^{x}=3^{y}={6^{-z}}, $ then $ ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} ) $ is equal to
Options:
A) $ 0 $
B) $ 1 $
C) $ \frac{3}{2} $
D) $ -\frac{1}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
- Let $ 2^{x}=3^{y}={6^{-z}}=k $ We know that, If $ a^{x}=y, $ then $ a={y^{1/x}} $
$ \Rightarrow $ $ 2={k^{\frac{1}{x}}};3={k^{\frac{1}{y}}};6={k^{-\frac{1}{z}}} $
$ \Rightarrow $ $ {k^{\frac{1}{x}}}\times {k^{\frac{1}{y}}}={k^{-\frac{1}{z}}} $ $ [\because 2\ \times 3=6] $
$ \Rightarrow $ $ {k^{\frac{1}{x}+\frac{1}{y}}}={k^{-\frac{1}{z}}} $ $ [\because p^{m}\times p^{n}={p^{m+n}}] $
$ \Rightarrow $ $ \frac{1}{x}+\frac{1}{y}=-\frac{1}{z} $
$ \Rightarrow $ $ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 $
BETA
