The sum of integers in list S is the same as the sum of the

[fahadmuhammad86]

Is there a quick/efficient way of doing this DS problem. The way i do it i.e. by putting in numbers and creating sets, is long and there is a chance of error if not done correctly. Can anyone please share there methods for this question. Greatly appreciated : )

The sum of integers in list S is the same as the sum of the integers in T. Does S contain more integers than T?
1) The mean of integers in S is less than the mean of integers in T
2) The median of integers S is greater than the median of integers T

ANSWER is: A

[nitin_prakash_khanna]

Let Set S have s integers and Set T have t integers.
lets write SUM S as sum of all integers in Set S and SUM T as Sum of integers in Set T.

We are give SUM S = SUM T

Question is asking whether s>t .

Hope its clear till this point.

Statement 1 says The mean of integers in S is less than the mean of integers in T

Which means

SUM S / s < SUM T / t (hope the ineuality is clear)
since SUM S = SUM T

1/s < 1/t (remember s and t are positive integers because they represent number of integers in a Set)

Since 1/s < 1/t implies s>t.
St.1 is SUFFICIENT

Statement 2 The median of integers S is greater than the median of integers T

Now here you need to plug in,
S could be 1,2,3, Sum = 6, Median = 2

T could be 1,1,4 , Sum = 6, Median = 1
Or T could be 3,3 Sum = 6, Median = 3

As we can see There could be multiple scenarios where two sets can have same sum, Median can be equal or different and still they can have different number of integers.

So INSUFFICIENT

Answer A

[fahadmuhammad86]

Awesome : ) this is a much efficient way of doing it. Thanks alot man i really appreciate all your help.

[RonPurewal]

does this problem state that the integers must be positive (or, at the least, non-negative)?

if not, then the actual answer is (e).

proof:
post24047.html#p24047

[harsh.singh08]

Hello, thanks for this response. I follow you up to the point where you state that Sum S / s < Sum T / t; however, how do you suddenly jump to this conclusion?

"1/s < 1/t (remember s and t are positive integers because they represent number of integers in a Set)

Since 1/s < 1/t implies s>t.
St.1 is SUFFICIENT"

Why is 1/s < 1/t and how does it imply that s > t?

Let Set S have s integers and Set T have t integers.
lets write SUM S as sum of all integers in Set S and SUM T as Sum of integers in Set T.

We are give SUM S = SUM T

Question is asking whether s>t .

Hope its clear till this point.

Statement 1 says The mean of integers in S is less than the mean of integers in T

Which means

SUM S / s < SUM T / t (hope the ineuality is clear)
since SUM S = SUM T

1/s < 1/t (remember s and t are positive integers because they represent number of integers in a Set)

Since 1/s < 1/t implies s>t.
St.1 is SUFFICIENT

Statement 2 The median of integers S is greater than the median of integers T

Now here you need to plug in,
S could be 1,2,3, Sum = 6, Median = 2

T could be 1,1,4 , Sum = 6, Median = 1
Or T could be 3,3 Sum = 6, Median = 3

As we can see There could be multiple scenarios where two sets can have same sum, Median can be equal or different and still they can have different number of integers.

So INSUFFICIENT

Answer A

[Navneet]

For option A
Supposing the condition of positive/non negative integers ( because otherwise explanation would require cases)

Considering that sum of any set of numbers = Mean * number of terms ( In fact this is very useful concept for AP) - Eq01

Option A states - The mean of integers in S is less than the mean of integers in T
Simply, If mean is less then number of terms must be more "to maintain the equivalence of the sums"

m1 *n1 = m2 *n2
If m1<m2 then n1 >n2

Option A suffice

For Option B
By testing cases - As explained in above posts

[RonPurewal]

i'm going to retire this thread, since the one i linked above is much more complete.

here's that link again:
https://www.manhattanprep.com/gmat/foru ... 24047.html

please post further discussion on that thread. thanks.