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Average + Median

by carllecat » Tue Nov 17, 2009 9:09 am
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 (arithmetic mean) of the integers in T.

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

Let's say S is composed of 2,2,2,2 &2 , then the sum of these numbers is 10 and then mean is 2 and T is composed of 5 & 5 for a total sum of 10 and a mean of 5.

How can I assess 2)?
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by Kalvin » Wed Nov 18, 2009 5:59 pm
1, 1, 1, 1, 1, 100

vs

1, 26, 26, 26, 26

or

1, 1, 103

vs

1, 26, 26, 26, 26
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by gmater29 » Wed Nov 18, 2009 9:12 pm
Whats the OA ? E ?
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by 2010gmat » Thu Nov 19, 2009 3:23 am
IMO A
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by grockit_jake » Thu Nov 19, 2009 4:17 pm
The simple average formula can be expressed: Avg = SUM/# , or as SUM = Avg*#

The stimulus tells us that SUM(t) = SUM(s), therefore we know the products of their respective Avgs and #s must be equal

(1) Avg(s) < Avg(t), therefore #(t) >#(s) to make Avg*# = Avg*#. Sufficient.

(2) The stimulus does not say that the group of numbers is evenly distributed. Therefore, outliers can skew the mean in either direction while keeping the median the same. Regardless, the relationship of the median doesn't provide information about the relative averages or number of integers. Insufficient.
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by carllecat » Thu Nov 19, 2009 5:56 pm
A Statement 1 alone is sufficient.
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by akhpad » Sat Jun 05, 2010 7:31 pm
This is official problem but why OA is A.

Sum of integers in S = Sum of integers in T = SUM (Either +ve or -ve)
No of integers in S = s
No of integers in T = t

Integers may be -ve or +ve. It is not mentioned.

Statement 1:
Assume SUM = 6
6 / s < 6 / t; s > t

Assume SUM = -6
-6 / s < -6 / t; 6 / s > 6 / t; s < t

Not Sufficient


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by akhpad » Mon Jun 07, 2010 1:00 am
Anybody would like to try this.
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by kevincanspain » Tue Jun 08, 2010 12:59 am
grockit_jake wrote:The simple average formula can be expressed: Avg = SUM/# , or as SUM = Avg*#

The stimulus tells us that SUM(t) = SUM(s), therefore we know the products of their respective Avgs and #s must be equal

(1) Avg(s) < Avg(t), therefore #(t) >#(s) to make Avg*# = Avg*#. Sufficient.

(2) The stimulus does not say that the group of numbers is evenly distributed. Therefore, outliers can skew the mean in either direction while keeping the median the same. Regardless, the relationship of the median doesn't provide information about the relative averages or number of integers. Insufficient.
I'm afraid you have assumed that we are dealing with positive integers.

If x is the sum of each set, x/s and x/t are the numbers of elements in S and T, where s and t are the averages of the elements in S and T respectively.

We need to know whether x/s > x/t

(1) Given s < t , 1/s > 1/t

If x > 0 x/s > x/t YES
If x > 0, x/s < x/t NO

NOT SUFF
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by Patrick_GMATFix » Tue Jun 08, 2010 6:09 am
This is a gmatprep problem (GMATPrep question 1335) for which the official answer is A

However, like the people above, I don't see how A can be right. We know that sum = avg*#of terms, so average is sum/#of terms. Let s and t be the number of terms. Rephrase: is s>t ?

My approach to statement (1) is: This statement tells us that sum/s < sum/t. Cross multiply to get sum*t < sum*s. We then have two possibilities when we divide both sides by sum.
  • If sum>0, we get t<s. So the answer to the original Q would be YES.
  • If sum<0, we get t>s, and the answer to the original Q would be NO
Because we have conflicting data, I believe (1) is not sufficient. I simply don't know why the official answer is A.

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by MrE2All » Sat Jun 19, 2010 1:11 am
My approach to statement (1) is: This statement tells us that sum/s < sum/t. Cross multiply to get sum*t < sum*s. We then have two possibilities when we divide both sides by sum.
I don't believe you can cross multiply this inequality because, like you said, we don't know if sum is positive and don't know if we should flip the sign. I originally thought it was A but your answer made me think about it for another minute. :D
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by Patrick_GMATFix » Sat Jun 19, 2010 4:44 am
MrE2All wrote: I don't believe you can cross multiply this inequality because, like you said, we don't know if sum is positive and don't know if we should flip the sign. I originally thought it was A but your answer made me think about it for another minute. :D
I actually agree with you, but the official answer does not. If you read my full post, you'll note that at the end I wrote:
Because we have conflicting data, I believe (1) is not sufficient. I simply don't know why the official answer is A.
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by aagar2003 » Sun Jun 20, 2010 4:08 pm
Let S contain s integers from S1, S2, ......., Ss
Let T contain t integers from T1, T2, ......., Tt
Now, it is given that Sum(S) = Sum(T)

1. Sum(S)/s < Sum(T)/t

Case 1: Assume Sum(S) <0 and Sum(T)<0
Case 2: Assume Sum(S) <0 and Sum(T)>0 : Cannot Exist as Sum(S) = Sum(T)
Case 3: Assume Sum(S) >0 and Sum(T)<0 : Cannot Exist as Sum(S) = Sum(T)
Case 4: Assume Sum(S) >0 and Sum(T)>0

Case 1: Assume Sum(S) <0 and Sum(T)<0
=> |Sum(S)|/s > |Sum(T)|/t
=> s/t < |Sum(S)|/|Sum(T)| = 1
=> s < t always YES

Case 4: Assume Sum(S) >0 and Sum(T)>0
=> |Sum(S)|/s < |Sum(T)|/t
=> s/t > |Sum(S)|/|Sum(T)| = 1
=> s > t always YES

Case 1 and case 4 both gives different results depending on whether the sum of the integers is +ve or not. So (1) by itself is not sufficient by itself. Therefore, guess what OA is wrong.

Now let's see (2):
Median(S) > Median(T)
For any set of integers, Median effects neither sum nor number of integers.
Example,
S = 1,2,3 :: Median = 2 :: Sum = 6 :: #=3
T = 1,1,4 :: Median = 1 :: Sum = 6 :: #=3 (You can say there is an answer to Q that s=t always if median(S) > median (T))
We need one more to clarify
T = 1,1,1,3 :: Median = 1 :: Sum = 6 :: #=4 (=>s<t)

Hence, neither (1) nor (2) is suffcient.
Correct answer is (E)
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