Official Solution:

Even when we consider both statements together, we cannot answer the question. If \(S=\{0\}\) and \(T=\{0\}\) then the answer is NO but if \(S=\{0\}\) and \(T=\{1\}\) then the answer is YES. Not sufficient.

Answer: E
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I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

I did not understand the explanation. Can you please elaborate why statement 2 is insufficient?

Please check here: m24-q-76516.html
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Please help!

Referring to statement 2, "Neither set SS nor set TT contains negative numbers"

Does that mean the set can contain '0' which is neither negative nor positive?

Secondly, if the statement read "Neither set SS nor set TT contains negative numbers and zero" will this answer be that both statements together are sufficient (C)?

1. Yes, the sets can contain 0, because 0 is NOT negative number.
2. Yes.
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Bunuel - please help me here with below query:
If Statement 2 read "Neither set SS nor set TT contains negative numbers and zero", then we can approach as below:

Mean of Set S = Sum of elements of S (a) / Number of elements of S (b)
Mean of Set T = Sum of elements of T (c) / Number of elements of T (d)

Now, we are asked if a/b + c/d > (a+c)/(b+d) , with given that a,c > 0

\(\frac{(a+c)}{(b+d)}\) = \(\frac{a}{(b+d)}\) + \(\frac{c}{(b+d)}\)

As, b & d > 0

\(\frac{a}{(b+d)}\) < \(\frac{a}{b}\)
Similarly, \(\frac{c}{(b+d)}\) < \(\frac{c}{d}\)

Adding these 2 equations, we will have
\(\frac{a}{(b+d)}\) + \(\frac{c}{(b+d)}\) < \(\frac{a}{b}\) + \(\frac{c}{d}\)
\(\frac{(a+c)}{(b+d)}\) < \(\frac{a}{b}\) + \(\frac{c}{d}\)

Thus, B will be the answer, if we exclude possibility of 0. Please let me know error in my approach.

Thank a ton in advance!