S is a finite set of numbers. Does S contain more negative numbers than positive numbers? (1) The product of all the numbers in S is -1,200. (2) There are 6 numbers in S.

Got this right but took me a while. The key for me was realizing that statement (1) alone doesn't tell us how many numbers are in the set, so we can't compare counts. Only when we combine with (2) can we figure out the exact numbers of positives and negatives.

I initially thought (1) alone would be enough because the product is negative, but then I realized we don't know the total count. Classic DS trap!

This one was medium difficulty for me. I had to consider that the product is negative, so odd number of negatives. But without knowing the total, can't compare. When combining, we have 6 numbers, so possible negative counts are 1,3,5. That means positives are 5,3,1 respectively. So in all cases, negatives are fewer than positives? Wait, if negatives=1, positives=5; if negatives=3, positives=3; if negatives=5, positives=1. So it's not always more positives? Actually, the question asks: Does S contain more negative numbers than positive numbers? That would be true only if negatives > positives. With 6 numbers, negatives > positives means negatives >=4. But odd number of negatives, so possible 5 negatives -> then positives=1, so yes, more negatives. But if negatives=3, then positives=3, not more. So it's not sufficient? Hmm, maybe I'm missing something. Let me check: product is -1200, which is negative, so odd number of negatives. With 6 numbers, negatives can be 1,3,5. For negatives=1, positives=5 -> no. For negatives=3, positives=3 -> no. For negatives=5, positives=1 -> yes. So we don't know which case. So together not sufficient? But the answer is C? Wait, maybe the product -1200 has prime factors that restrict? 1200 = 2^4 * 3 * 5^2. But numbers can be any real numbers? Probably integers? Usually in GMAT, numbers in a set are integers unless specified? Not necessarily. But even if integers, the product -1200 doesn't restrict the signs beyond parity. So it seems E? But the difficulty says 505-555, so maybe I'm overcomplicating. Let me re-read: (1) The product is -1200. (2) There are 6 numbers. So together: 6 numbers, product negative. So odd number of negatives. That gives 1,3,5 negatives. The question: Does S contain more negative numbers than positive numbers? That is, is #neg > #pos? With 6 numbers, #neg + #pos = 6 (assuming no zeros). So #neg > #pos means #neg > 3. So #neg = 5 works, but #neg = 1 or 3 does not. So we don't know. So answer should be E. But the answer choices include C, so maybe I'm wrong. Wait, could there be zeros? If there's a zero, product is 0, so no. So no zeros. So indeed, #neg can be 1,3,5. So not sufficient. So answer is E. But the problem says difficulty 505-555, so maybe it's a trick? Actually, I think the correct answer is C? Let me think: if there are 6 numbers, and product is negative, then number of negatives is odd. But also, the product is -1200, which is negative, so odd number of negatives. But does the product value matter? Not really. So together, we know odd number of negatives, but could be 1,3,5. So not sufficient. So answer is E. But many people might think it's C. So this is a good question. I'll go with E. Wait, but the choices are A,B,C,D,E. So E is possible. So my final answer: E. But the problem statement says 'Does S contain more negative numbers than positive numbers?' So if we have 5 negatives and 1 positive, yes. If 3 and 3, no. So not sufficient. So E.