The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.

The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers than T?

Given: sum(S)=sum(T). Question: is t<s, where s and t are # of integers in lists S and T respectively.

(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T --> \frac{sum}{s}<\frac{sum}{t} --> cross multiply: sum*t<sum*s. Now, if sum<0 then t>s (when reducing by negative flip the sign) but if sum>0 then t<s. not sufficient.

(2) The median of the integers in S is greater than the median of the integers in T. If S={1, 1} and T={0, 0, 2} then the median of S (1) is greater than the median of T (0) and S contains less elements than T but if S={-1, -1, -1} and T={-3, 0} then the median of S (-1) is greater than the median of T (-1.5) and S contains more elements than T. Not sufficient.

(1)+(2):
If S={-1, 2, 2} and T={1, 2} then the sum is equal (3), the average of S (1) is less than the average of T (1.5), the median of S (2) is greater than the median of T (1.5) and S contains more elements than T.

If S={-2, -1} and T={-2, -2, 1} then the sum is equal (-3), the average of S (-1.5) is less than the average of T (-1), the median of S (-1.5) is greater than the median of T (-2) and S contains less elements than T.

Not sufficient.

Answer: E.
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Is the approach to solve such questions is come up with sets which satisfy and dont satisfy the conditions.

Is there any other way we can solve this type of questions.

Dear Bunuel

I ran into this question in a Gmat Prep exam and the OA is A. It works with positive integers, but it doesn't with negatives. My pick was C, but after a closer look I ended up with E as you.

My deduction is that the mean is a function of the number of items in each set. As the number of items increases, the mean decreases. In statament 1 we are told that the mean of S is smaller than that of T, but that only work for positives.

Have you ever ran into a GMAT Prep OA that is disputable?

I considered the following

S = 1, 2, 3, 4, 5
Sum is 15
Mean is 3

T = 7, 8
Sum is 15
Mean is 7.5

Therefore if mean of S < mean of T, then S must have more items than T, only if the Sum of the sets is possitive.

Considering two sets that Sum a negative number

S = -4, -3, -2, -1
Sum is -10
Mean -2.5

T = -9, -1
Sum is - 10
Mean -5

Different answer for positives and negatives, therefore not sufficient

We can not take a set that Sums 0, as the medians would be the same

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

For positives

S = 1, 2, 3, 4, 5
Sum is 15
Median is 3

T = 7, 8
Sum is 15
Median is 7.5

Answer NO

or

S= 1, 1, 1, 2, 2, 3
Sum 10
Median 1.5

T = 2, 2, 2, 4
Sum 10
Median 2

Answer No

Seems sufficient but considering two sets that Sum a negative number

S = -4, -3, -2, -1
Sum is -10
Mean -2.5
Median -2.5

T = -9, -1
Sum is - 10
Mean -5
Median -5

Answer Yes

Therefore Insufficient

I chosed C as I believed the two statements force the Sum of both sets to be positve, but I guess it was a poorly analyzed deduction.

I have never runed into a question where the OA was open to dispute, have you?

Thanks in advance

Bunuel,

I guess A sufficiently works.
Because, after reviewing your answer, I tried doing the question again with both positive and negative integers

Please see the explanation below and suggest if i went wrong anywhere.

Data Sufficiency:

1. Arithmetic mean of list S is less than Arithmetic mean of list T


Basic formula used -->
Sum = mean * Number of integers in the set

Given condition in the question:
Sum of the integers in the list S = Sum of the integers in the list T




Integers can be both positive and negative.

Let the List S = {3,-3,10,2,13}


Sum = (3-3+10+2+13) = 25
Number of integers = 5
Mean = 5 (Substitute in the formula)

Then

List T would be
Sum = 25 (Condition given in the question)
Number of integers = ?
Mean of List T Should be more than mean of List S
assume mean = 6
then number of integers in the List T = (25/6) = 4.166

Of course, Number of integers in a List is a number we can still conclude that List T will always be less in the
number of integers with Same sum and With more mean.
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Learn the Values. Knowledge comes by itself.

Thank you Bunuel.
You are a good teacher.
Attention to detail is what i have learnt from your explanations.
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S1 sufficient. Since sum S = sum T, as mean for S is less than mean for T, sum/s < sum/t => t<s where s and t are respective number of integers in S and T.

S2 insufficient.

Answer is A.

EDIT: those numbers below zero complicate things.

Let average of Set S = A1 , Sum = S1 and number of integers in the list = n1
Let average of Set T = A2, Sum = S2 and number of integers in the list = n2

(1) We know, S = A * n
A1 = S1/n1
A2 = S2/n2
Acc to statement 1, A1 < A2,
so \frac{S1}{n1} < \frac{S2}{n2}

Given, The sum of the integers in list S is the same as the sum of the integers in list T
So, S1 = S2
\frac{1}{n1} < \frac{1}{n2}
n2 < n1

(2) The median of the integers in S is greater than the median of the integers in T. - Not Sufficient


Can someone explain why, Answer is E and not A
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Consider this:

S = {-1, 0, 1}
T = {-2, -1, 0, 1, 2}
Sum of both the sets is 0. T has more integers.

or

S = {-2, -1, 0, 1, 2}
T = {-1, 0, 1}
Sum of both the sets is 0. S has more integers.

Similarly, think what happens when the sum is negative.
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