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B is sufficient

Since x>|y|>0, we multiply a positive number |y| to both ends get x|y|>|y||y|>0
|y||y|=y2
So x|y|>y2
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Is x|y|>y^2 ?
or Is x|y|>|y|^2
or, Is |y|(x-|y|)>0?

Statement 1.
\(\frac{1}{y} + x^2 >3\)
There can be infinite solutions to this inequality. it is clearly not sufficient.
NOT SUFFICIENT.

Statement 2:
\(x > |y|\) or x-|y|>0
this looks very similar to question stem Is |y|(x-|y|)>0?,
so one tends to mark the answer as B.
Here is the trap, If y = 0, then Is |y|(x-|y|)>0? would be NO, if y is not equal to 0, Is |y|(x-|y|)>0? is YES.
Hence Statement 2 is also NOT SUFFICIENT.


Combining statement 1 & 2,
St 1 gives a clue that y is not equal to 0, if y had been 0, then \(\frac{1}{y}\)would be undefined.
Hence y is not equal to 0, Also, from st 2 x-|y|>0
hence |y|(x-|y|)>0? is also true, and we get a definite answer YES.

SUFFICIENT

So the answer is C

gmatbusters
Is \(x|y|>y^2\)?

1) \(\frac{1}{y} + x^2 >3\)
2) \(x > |y|\)

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