CAT4 DS absolute values question

[badpit]

Hello,

I got the question If a and b are integers, and |a| > |b|, is a · |b| < a - b?
(1) a < 0
(2) ab >= 0

The official solution greatly exceeds the possible time for solution 2-3 minutes. Is there a shortcut, which can be used for these types of questions?

[esledge]

Yes, you can save time by considering only the relevant variable properties.

For absolute value questions, the relevant number property is positive vs. negative. We also have an (integer - integer) component to the question, so the answer may also depend on the relative values of the variables. At this point, I'm glancing at the statements and thinking "They only give me sign information, nothing about relative value, so there's a good chance that the answer is E." Although this is definitely NOT standard procedure, I would probably jump ahead to the combined statements, trying to prove insufficiency.

If a = neg, and ab >= 0, then b = neg or 0. There are thus two sign cases, for which we must remember that |a| > |b|.

Case 1:
a = neg
b = 0
a*|b| = 0
a - b = neg - 0 = negative.
Is a · |b| < a - b?
Is 0 < negative?
Answer: No.

Case 2:
a = more neg
b = less neg
a*|b| = more neg * less pos (but at least 1) = more negative.
a - b = more neg - less neg = more neg + less poss = less negative.
Is a · |b| < a - b?
Is more negative < less negative?
Answer: Yes.

At this point, I would probably verify my thinking on Case 2 with a few sets of test values:
a = -2, b = -1: Is a · |b| < a - b? Is -2 < -1? Yes.
a = -100, b = -1: Is a · |b| < a - b? Is -100 < -99? Yes.
a = -100, b = -99: Is a · |b| < a - b? Is -9900 < -1? Yes.
It seems that there is a pattern, and the answer for Case 2 will always be Yes.

Since Case 1 gave a definite No answer and Case 2 gave a Yes answer (at least in the examples we tried), the combined statements are insufficient and the answer is indeed E.