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RRB NTPC Mains or Stage 2 Model Paper 9 with solutions

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Question : 10 of 120

Marks: +1, -0

If

2^x = 3^y = 6^{-z}

\left[ \frac{1}{X} + \frac{1}{Y} + \frac{1}{Z} \right]

is equal to

0
1
$\frac{3}{2}$
$-\frac{1}{2}$

Solution:

Concept:

Let all equal powers have the same common value

k

.

Explanation:

Given

2^x=3^y=6^{-z}=k

.

So,

2=k^{1/x}

and

3=k^{1/y}

.

Also,

6^{-z}=k \Rightarrow 6=k^{-1/z}

.

Since

2 \times 3=6

,

k^{1/x} \times k^{1/y}=k^{-1/z}

.

Therefore,

\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}

.

Hence,

\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0

.

Answer:

The value is

0

, so the correct option is A.

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