(2) The sum of the integers in S is greater than the sum of the integers in T.

This is insufficient, because you don't know the distribution of the numbers in each set.

For example,

S - (1, 3, 21) sum = 25
T - (1, 3, 5) sum = 9

Here, the median of S and the mean of T are equal. Notice that S has an element WAY FAR removed from the rest of the elements.

Another case...

S - (50, 51, 52)
T - (1, 2, 3)

Here, the median of S is greater than the mean of T. At this point, you know (2) is INSUFFICIENT.

Together, (1) and (2) tell you that S has consecutive evens, T has consecutive odds, and the sum of everything in S is greater than that of everything in T.

Since S has consecutive evens and T has consecutive odds, and since there's an equal number of terms in each set, we know that each term in S is either greater than or less than its corresponding term in T. (i.e. the Nth term in S is either greater than or less than the Nth term in T).

Since the sum of everything in S is greater than that of everything in T, each term in S must be greater than its corresponding term in T.

And, since the numbers are equally spaced in both lists, we know that the mean and median of each list are exactly the same.

Therefore, since mean of S is greater than mean of T, median of S must be greater than T. Final answer: C

Hey, I think the answer's gonna be C.

For B, you can have a case like:
S= 3, 8, 2, 10, 1
& T = 2, 2, 2, 2, 2

thus the median of S = Average of T

but if you take both together then the median of S will always be greater than the avg. of T