RRB Junior Engineer Stage 1 Model Papers 7

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Question : 24 of 100

Marks: +1, -0

If

x=(0.08)^2,y=1/{(0.08)^2}

and

z=(1-0.08)^2-1

then out of the following the true relation is

$y < x \;\and x = z$
$x < y \;\and x = z$
$y < z < x$
$z < x < y$

Solution:

x=(0.08)^2

y=1/(0.08)^2=10000/64=156.25

Z=(1-0.08)^2-1

=1+(0.08)^2-2×0.08-1=(0.08)^2-2×0.08

Clearly,

z < x < y

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