Is x < y^2 ?

Here's the answer key I wrote.


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You may be tempted to pick (E) right away, since y is nowhere to be seen in either statement.
If you WERE tempted to pick that choice — or, worse, if you actually DID pick it! — remember, answers that LOOK "too good to be true" generally ARE.
(Just as importantly, please realize that the exam makers are NOT being "tricky" when they create questions like this. There are never any "trick questions" on this exam; anything that might SEEM "tricky" will be something you can prevent with sufficient due diligence.
This is an adaptive exam, as you almost certainly know — so, many questions will have "easy-looking answers" that are wrong but that are consistently selected by LOWER-scoring test takers.)

(More generally, it would be worth your time to peruse the first few problems in the Data Sufficiency chapter of the Official Guide and/or the Official Quant Supplement — not so much to solve those problems as to develop a better intuition for what's TOO EASY for this exam. Essentially, the first few problems in the DS chapter represent the easiest things that GMAC will ever throw at you, no matter what, ever... so, if your "solution" is substantially easier than those problems, IT'S WRONG!)

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This is a "yes/no" Data Sufficiency question. As for any such question, remember the following:
• If BOTH the "yes" case AND the "no" case are possible, the statement is NOT sufficient.
• If ONLY ONE of these cases is possible—and one is impossible—the statement is SUFFICIENT.


First, TRANSLATE the "yes" and "no" cases into SPECIFICS:
• The "YES" case is \(x < y^2\).
• The "NO" case is everything else — i.e., \(x ≥ y^2\).
The GOAL — as is the case for every "yes/no" Data Sufficiency problem — is to try to make BOTH of these cases happen. If you can make them both happen, that's "NOT sufficient". If you can't — if you're stuck with just one of them, no matter what — then that's "sufficient".


Statement 1

Remember the fundamental behavior of the absolute value function:
• If the value inside the absolute-value bars is 0 or positive, then that value does not change.
• If the value inside the bars is negative, then the absolute value "flips" it to the corresponding positive value.

This statement says that, if x is put into the absolute-value function, then the result (|x|) is NOT the same as the "input" (x).
Considering the fundamental behavior just described, this is the case if and only if x is NEGATIVE.
Therefore, statement 1 is simply a very circuitous way of saying that x is negative.

Remember that squared values (such as \(y^2\)) CANNOT be negative; all squares are 0 or positive.
Since x is negative here, it follows that x must be less than any possible value of \(y^2\) — even though no information at all is provided about y itself.

Thus only the "YES" case is possible; the "NO" case is impossible. This statement is SUFFICIENT.


Statement 2

Keeping in mind the behavior of the absolute value function, as described above, let's investigate the truth of this statement for each of the three possible signs of x — positive, zero, and negative. When you consider signs, DON'T FORGET ABOUT ZERO!
• If x is positive, then |x| remains positive, but –x is negative. So this statement will be false.
• If x is zero, then both |x| and –x are equal to zero, making this statement true.
• If x is negative, then both |x| and –x will "flip the sign" to positive. So the statement will be true.
Thus statement 2 is just a very indirect way of saying that x is EITHER 0 OR NEGATIVE.

If we give x any negative value, we'll definitely get the "YES" case (as explained for statement 1).
But this time, x can be zero! By using x = 0 AND y = 0, we can make the "NO" case happen (since 0 ≥ 0).
Thus the "YES" AND "NO" cases are BOTH possible. In other words, this statement is NOT sufficient.

The correct answer is A.