Is x^2-y^2>0?

[swerve]

$$Is\ \ x^2\ y^2>0\ ?$$

$$\left(1\right)\ x^2-y^2>\left(x-y\right)$$
$$\left(2\right)\ x^2-y^2<\left(x+y\right)$$

The OA is E.

I don't have it clear. Please, can any expert explian this DS question? Thanks.

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Hi swerve,

The 'title' of this post doesn't match the question that's in it, but I'm going to assume that we're meant to answer the question "Is X^2 times Y^2 greater than 0?" This is a YES/ NO question. We can answer it by TESTing VALUES. It's worth noting that since a 'squared' term can only be 0 or positive, the ONLY time that (X^2)(Y^2) would not be greater than 0 is if one (or both) of the variables equaled 0.

Before we get to work, it's worth noting that BOTH Facts involve the same 'component' (X^2 - Y^2). In these situations, it can sometimes be useful to think about both Facts at the same time (so that you can find sets of numbers that 'fit' both Facts.

1) X^2 - Y^2 > (X - Y)

IF....
X=0 and Y= 1/2, then the answer to the question is NO
X=1 and Y= 1/2, then the answer to the question is YES
Fact 1 is INSUFFICIENT

2) X^2 - Y^2 < (X + Y)

IF...
X=0 and Y= 1/2, then the answer to the question is NO
X=1 and Y= 1/2, then the answer to the question is YES
Fact 2 is INSUFFICIENT

Combined, we already have two pairs of values that produce different answers to the given question.
Combined, INSUFFICIENT

Final Answer: E

GMAT assassins aren't born, they're made,
Rich