RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

Question : 16 of 120

Marks: +1, -0

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

Solution:

Given

x = (0.08)^2

y = 1/(0.08)^2 = 156.25

z = (1 – 0.08)^2 – 1

z = 1 + (0.08)^2 – 2 × 0.08 – 1

= (0.08)^2 – 2 × 0.08

z = x – 0.16

x = z + 0.16

x > z

∴ y > x > z

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