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?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
I got A. Is this right?
Sum
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It sure is.didieravoaka wrote: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?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
I got A. Is this right?
S1: We know the following equation: Average * Number Terms = Sum. Holding the sum constant, the average and the number of terms will have an inverse relationship. As the average gets smaller, the number of terms will grow larger, and vice versa. So if we know that if the average is smaller in S, and the two sums are equal, then there must be more terms in S to compensate for this smaller average. 1 alone is sufficient.
Use simple sets to test statement 2.
Say the sum for both sets is 0.
Case 1
S: {1,0,1} Median = 0
T:{1, 1, 2} Median = 1
Here S does not contain more elements than T, so the answer is NO.
Case 2:
S: {1,0,0,1} Median = 0
T:{1, 1, 2} Median = 1
Here S does contain more elements than T, so the answer is YES.
If you can get a YES and a NO, statement 2 is not sufficient.
And, as you saw, the answer is A
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Given: Sum of integers in list S = Sum of integers in list Tdidieravoaka wrote: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?
a.The average( arithmetic mean) of the integers in S is < than the average of the integers in T.
b. The median of the integers in S is greater than the median of the integers in T.
I got A. Is this right?
Required: Are the number of elements in S > number of elements in T
Statement 1: The average( arithmetic mean) of the integers in S is < than the average of the integers in T
Average = Sum/No. of elements
Since the sum is same, therefore "Number of elements in S > Number of elements in T"
SUfficient
Statement 2: The median of the integers in S is greater than the median of the integers in T
Median has not effect on the number of elements. It simply is the middle term when we write a series in ascending order.
Hence we cannot say anything about the number of elements.
Insufficient
Correct Option: A
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I don't believe that statement 1 is sufficient (unless I've completely missed something)didieravoaka wrote: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 < 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.
Target question: Does S contain more integers than T?
Given: The sum of the integers in list S is the same as the sum of the integers in list T.
Statement 1: The average (arithmetic mean) of the integers in S is < than the average of the integers in T.
There are several possible scenarios that satisfy statement 1. Here are two:
Case a: list S: {3} and list T = {1, 1, 1}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 3, and the mean of list T is 1. In this case, list S contains FEWER integers than list T does
Case b: list S: {1, 1, 1} and list T = {3}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 1, and the mean of list T is 3. In this case, list S contains MORE integers than list T does
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
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Nice catch, Brent! I checked the source on this  it's a GMATPrep question, so I assumed in the original prompt there was a stipulation that the sum is nonnegative. But there isn't. And the OA is A. So GMAC might have goofed. I'm going to send them an email. (Which, I can only assume, is going to lead to a dramatic organizational restructuring.) Exciting times![email protected] wrote:I don't believe that statement 1 is sufficient (unless I've completely missed something)didieravoaka wrote: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 < 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.
Target question: Does S contain more integers than T?
Given: The sum of the integers in list S is the same as the sum of the integers in list T.
Statement 1: The average (arithmetic mean) of the integers in S is < than the average of the integers in T.
There are several possible scenarios that satisfy statement 1. Here are two:
Case a: list S: {3} and list T = {1, 1, 1}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 3, and the mean of list T is 1. In this case, list S contains FEWER integers than list T does
Case b: list S: {1, 1, 1} and list T = {3}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 1, and the mean of list T is 3. In this case, list S contains MORE integers than list T does
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

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Nicely done. I guess I don't feel so bad for missing it, since it tripped up a couple of the gurus. When I first read it, I was thinking (A) as well. But you are absolutely correct![email protected] wrote:I don't believe that statement 1 is sufficient (unless I've completely missed something)didieravoaka wrote: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 < 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.
Target question: Does S contain more integers than T?
Given: The sum of the integers in list S is the same as the sum of the integers in list T.
Statement 1: The average (arithmetic mean) of the integers in S is < than the average of the integers in T.
There are several possible scenarios that satisfy statement 1. Here are two:
Case a: list S: {3} and list T = {1, 1, 1}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 3, and the mean of list T is 1. In this case, list S contains FEWER integers than list T does
Case b: list S: {1, 1, 1} and list T = {3}. Notice that the sum of lists S and T are equal (both equal 3). Also, the mean of list S is 1, and the mean of list T is 3. In this case, list S contains MORE integers than list T does
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
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Agreed! This could signal the demise of the GMAT, ushering in an era where MBA admissions are based on coin flips and feats of strength!![email protected] wrote:So GMAC might have goofed. I'm going to send them an email. (Which, I can only assume, is going to lead to a dramatic organizational restructuring.) Exciting times!

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They do make these mistakes from time to time, and if my experience is anything to go by, you may not get a response.[email protected] wrote:Nice catch, Brent! I checked the source on this  it's a GMATPrep question, so I assumed in the original prompt there was a stipulation that the sum is nonnegative. But there isn't. And the OA is A. So GMAC might have goofed. I'm going to send them an email. (Which, I can only assume, is going to lead to a dramatic organizational restructuring.) Exciting times!
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I'm hoping for some kind of American Gladiatorstyle obstacle course.[email protected] wrote:Agreed! This could signal the demise of the GMAT, ushering in an era where MBA admissions are based on coin flips and feats of strength!![email protected] wrote:So GMAC might have goofed. I'm going to send them an email. (Which, I can only assume, is going to lead to a dramatic organizational restructuring.) Exciting times!
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Ha! A student brought that common multiple question to class the other day and I had the exact same reaction you did.[email protected] wrote:They do make these mistakes from time to time, and if my experience is anything to go by, you may not get a response.[email protected] wrote:Nice catch, Brent! I checked the source on this  it's a GMATPrep question, so I assumed in the original prompt there was a stipulation that the sum is nonnegative. But there isn't. And the OA is A. So GMAC might have goofed. I'm going to send them an email. (Which, I can only assume, is going to lead to a dramatic organizational restructuring.) Exciting times!