Problem Solving (PS) | Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the

pushpitkc niks18 Hatakekakashi
amanvermagmat

How about this approach?

On seeing z, I know I can write it as\(a^2\) -\(b^2\) = (a+b) * (a-b)

or (1 - 0.08 + 1 ) * (1 - 0.08 -1 )

The second bracket gives me a negative value as final answer.

Just by looking at x and y , I know these are positive. So only C holds good.
I do not have to care about inequality between x andy.

GMATNinja Sorry to bother you on qaunt forum