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RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

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Question : 16 of 120

Marks: +1, -0

If

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

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

, what is the value of xyz?

1
– 1
0
$\frac{1}{2}$

Solution:

Concept:

The value of

xyz

can be found by expressing

x

and

z

in terms of

y

and simplifying.

Explanation:

From

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

, we get

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

.

From

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

, we get

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

, so

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

.

Now substitute these into

xyz

:

xyz = \left(\frac{y-1}{y}\right) \cdot y \cdot \left(\frac{1}{1-y}\right)

.

The

y

cancels out:

xyz = \frac{y-1}{1-y}

.

Since

y-1 = -(1-y)

, we have:

xyz = -1

.

Answer:

-1

, so option B is correct.

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