median mean again

[divya23]

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]

[cans]

sum in S = sum in T. mean(x) = mean of set X
=> mean(s)*s (s=# of integers in s) = mean(t)*t (t=#of integers in t) ---eqn1
a) mean(s)<mean(t) and for eqn1 to satisfy, s>n
Sufficient.
b) insufficient. as median alone can't determine which set has more integers.
IMO A

[smackmartine]

IMO A

let set S has x integers and set T has y integers.
Also S= T (totals are equal)

so Ms = S/x and Mt= T/y (means of S and T respectively)

1) given that Ms < Mt => S/x < T/y

cancelling S and T on both sides (because they are equal) we get , 1/x < 1/y => x>y (we know that x and y are positive)
So , Sufficient.

2) Median gives no info about #s of integers in both the sets.
So Insufficient.

[aftableo2006]

A is the correct answer

[Ian Stewart]

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.

[cans]

Nice explanation Ian.
I assumed sum to be greater than 0 and thus reached on IMO A

[krishnasty]

Nice explaination Ian..
i certainly missed to take the negative integers / sum into account