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Re: Is x greater than y? (1) x - y^2 >0 (2) xy < 0 [#permalink]
Is x greater than y?

(1) x - y^2 >0 --> x > y^2
x = 2, y = 1 YES
x = 1/2, y = 1/2 NO

(2) xy < 0
Clearly insufficient b/c x and y are positive/negative and vice versa.

Combo pack:
x = 2, y = -1 YES
x = 1/2, y = -1/2 YES

For any pair where x is positive and y is negative x > y.
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Is x greater than y? (1) x - y^2 >0 (2) xy < 0 [#permalink]
All of the above solutions made sense, but I feel like there were some reasonings that were shorthanded or weren't stated clearly. Hope this helps for those struggling to follow.

Is x greater than y?
(1) x - y^2 >0
(2) xy < 0

Stem evaluation
Transcribes to: Is x > y ? (No simplification or reduction is available.)

1) x - y^2 > 0

Can be rewritten to state x > y^2

Test cases:
-1 > -2, but -1 !> (-2)^2 NO, x!>y^2
1 > 1/2, and 1 > (1/2)^2 YES, x>y^2
AD are out, because this given is inconclusive. So BCE
I like to always try to write down my intermediate discoveries as I go.

Intermediate discovery for 1): x is positive.

2) xy < 0
Case 1: y is neg & x is pos; x=2 & y=-2: YES
Case 2: x is neg & y is pos; x=-2 & y=2: NO

Intermediate discovery for 2): x & y signage is opposite.

Given is inconclusive again and B is out. CE are left.

1) x > y^2 provides that x is positive because y^2 will always be positive and x is greater than that.
2) Because there is a split set of cases here (one yes and the other no), we'll need to strike one to rule it a conclusive statement.
I started with Case 2 to rule out the negative case, but that should always be your approach. Proving either to be ruled out would be enough to land a second conclusive statement to hang our data sufficiency on. I clearly have a strong suspicion that Case 2 is the one to work on.

Case 2: x is neg & y is pos ; x=-2 & y=2: NO x!>y Strike Case 2 because x is negative in this case and we know x is positive from 1). This leaves us with one case and a conclusive statement.
Case 1: y is neg & x is pos; x=2 & y=-2: YES, x>y. Because x is provided to be positive in 1), then y is negative. Thus the stem x > y is true because positives > negatives.

If I simply take the intermediate discoveries for each, then I can easily determine C after AB have been ruled out.

Intermediate discoveries 1 + 2:
- x is positive.
- x & y signage is opposite.
Thus, x is positive and y is negative and x IS greater than y because positives are greater than negatives.

Choice C
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Re: Is x greater than y? (1) x - y^2 >0 (2) xy < 0 [#permalink]
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Re: Is x greater than y? (1) x - y^2 >0 (2) xy < 0 [#permalink]
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