Well, they don't tell us how many numbers there are in set S, so let's keep it as simple as possible. It's easier to find the median with an odd number of numbers, so we'll start with sets that have 3 elements (although we might have to try even-numbered sets, too, if we can't find good cases.)

(1)

Let's say the set is (10, 20, 30). The mean is 20. The median is also 20.

Next, try (19, 20, 21). The mean is 20. The median is also 20.

It seems like whenever we have one element larger than 20, and one element smaller than 20, the median is 20. That makes sense, because the third number in the set would have to be equal to 20. That number would be in the middle, so it would be the median.

What if we try some sets of four numbers, then? There could be two numbers smaller than 20, and two numbers bigger than 20.

(0, 0, 40, 40) - the mean is 20, and the median is 20 again (average of 0 and 20, the two middle numbers).

BUT, how about (0, 19, 30, 31)? The mean is 20 (I figured this out by making sure the sum was 80) but the median definitely isn't 20.

Insufficient.

(2) This is definitely insufficient on its own, since we could test almost any set.

(1+2) It's pretty easy to find a case where the median is equal to 20, and that fits both statements: (10,20,30) would work. What we need, then, is to find a case where the median

isn't equal to 20. That is, we want a case similar to the one that worked for statement (1), above, except where all of the numbers are even.

Let's use the same idea - a set that sums to 80 and has 4 elements, but that isn't 'evenly spaced'. One set that works here is (0, 18, 30, 32), which has a mean of 20 but a median of 24.

Insufficient.

The correct answer is E.

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