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sang5650
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Is x > y ?
(I) (x - y)(x - y) > 0
(II) y > 0

(1) \((x-y)^2>0\) --> the only thing we can deduce from this is that \(x\neq{y}\). Not sufficient.
(2) \(y>0\). Clearly not sufficient.

(1)+(2) \(x\neq{y}\) and \(y>0\). Not sufficient.

Answer: E.


Hi,

My questions is cant we simplify the eq as:

(x-y)^2 > 0
=> (x-y) > 0
=> x > y

Is this method wrong?

No we cannot simplify like this.

\((x-y)^2\) is never negative, as x-y is squared, hence \((x-y)^2>0\) is true in both cases when x>y and x<y.

When you are writing \(x-y>0\) after \((x-y)^2>0\), what you are actually doing is reducing (dividing) the inequality by x-y but we can not do that as we don't know whether x-y>0 or not (actually we are asked to determine this).
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just thinking , if we could change the option 1 to (x-y)(x+y) >0. what will the new answer be ?
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just thinking , if we could change the option 1 to (x-y)(x+y) >0. what will the new answer be ?

Answer would be the same: E. \(x^2-y^2>0\) and \(y>0\) does not help us to determine whether \(x>y\): consider following pairs of (x,y): (-5,3) - answer NO or (5,3) - answer YES.
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For these types of questions I find it easier to just check extreme solutions

if x = -100, y = 1 both expressions are true
if x = 4, y = 3 both expressions are true

In one example x > y in the other y > x therefore e
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