divya23 wrote:The sum of integers in list S is same as sum of int in list T.Does S contains more integers than T?
1.mean of integers in S<average of intgeres in T
2.median of int in S > median of int in T
[spoiler]OA = 1[/spoiler]
Say the sum of each list is X, and we have s things in list S, and t things in list T. If Statement 1 is true, then
X/s < X/t
Now since s and t are both positive, we can multiply by st on both sides:
Xt < Xs
Now if X is positive, we can divide both sides by X to find that t < s. But if X is negative, when we divide by X on both sides, we must reverse the inequality, and we find that t > s. So Statement 1 is certainly *not* sufficient here unless you're told that the integers in each list are positive, or at the very least that their sum is positive.
You can easily generate a very simple example to see this, if the above seems abstract - any valid example using negative numbers will do. S might be {-4} and T might be {-1, -3}. Then the mean of S is smaller than the mean of T, their sums are equal, but S contains fewer elements than T.
The answer is actually E here. The sets might be:
S = {-97, -2, -1}
T = {-87, -9, -3, -1}
These sets have the same sum, the median of S (which is -2) is bigger than the median of T (which is -6), and the mean of S (which is -100/3) is smaller than the mean of T (which is -25). Here S has fewer elements than T. Using positive numbers you can easily generate an example where S has more elements than T, so the answer is E.
