RonPurewal Wrote:ah, evil.
to get at the heart of this problem, you must realize that "greater" is NOT the same thing as "farther away from zero".
for positive numbers, these two concepts are the same, but they are not so for negative numbers (for which the the greater number is actually closer to zero).
that realization is the crux of this problem.
see, here's the deal:
the SUM is the same for each of the two sets. therefore, since average = (sum) / (# of data points), the average will be CLOSER TO ZERO if there are more data points.
the problem is that this doesn't mean that the average is lower. if the sum is negative, then just the opposite will occur.
examples:
if S = 2, 2, 2 and T = 3, 3, then the sums are both 6, the average of S is less (2 vs. 3), and S has more integers.
if S = -3, -3 and T = -2, -2, -2, then the sums are both -6, the average of S is less (-3 vs. -2), and S has fewer integers.
so (a) is insufficient.
if T = 2, 2, 2 and S = 3, 3, then the sums are both 6, the median of S is greater (3 vs. 2), and S has fewer integers.
if T = -3, -3 and S = -2, -2, -2, then the sums are both -6, the median of S is greater (-2 vs. -3), and S has more integers.
so (b) is insufficient.
together:
this takes a little more creativity.
if S = -7, 9, 10 and T = 6, 6, then the sums are both 12, the average of S is less (4 vs. 12), the median of S is greater (9 vs. 6), and S has more integers.
if S = -6, -6 and T = -10, -9, 7, then the nums are both -12, the average of S is less (-6 vs. -4), the median of S is greater (-6 vs. -9), and S has fewer integers.
so, insufficient.
answer = (e).
incidentally, if this problem is supposed to say that the integers are positive or non-negative (and the original poster simply forgot this caveat), then the problem is much easier, and the answer is different (at a glance i think it would be (a)).
Ron, how can it be possible to be so clever? :)
Thank you for your explanations. Are real GMAT exam questions as tough as those in the GMATprep?? I wish they were similar to questions in OG.