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The sum of the integers in list S is the same as the sum of [#permalink]
18 Feb 2012, 08:09

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A

B

C

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E

Difficulty:

95% (hard)

Question Stats:

22% (02:05) correct
78% (01:06) wrong based on 756 sessions

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.

Re: The sum of the integers in list S is the same as the sum of [#permalink]
18 Feb 2012, 10:24

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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.

Re: The sum of the integers in list S is the same as the sum of [#permalink]
21 Feb 2012, 04:23

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Expert's post

prashantbacchewar wrote:

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.

Generally on DS questions when using plug-in method, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Of course algebra/math or conceptual/pure logic approach is also applicable to prove that the statement is not sufficient. It really depends on the particular problem and personal preferences to choose which approach to take.

The sum of the integers in list S is same as the sum of the integers in list T. Does S contains 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.

Any idea how to get A?

Merging similar topics. Please ask if anything remains unlcear.

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

Attachments

Crazy sets S and T.JPG [ 76.13 KiB | Viewed 15549 times ]

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.

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

Yes, there are several questions in GMAT Prep with incorrect answers. This question is one of them. A is not correct, answer should be E. _________________

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.

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

Yes, there are several questions in GMAT Prep with incorrect answers. This question is one of them. A is not correct, answer should be E.

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. _________________

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.

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

Yes, there are several questions in GMAT Prep with incorrect answers. This question is one of them. A is not correct, answer should be E.

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.

Again, answer to this question is E, not A. There are two examples in my post satisfying both statements and giving different answers. _________________

Re: sum of integers in list S [#permalink]
07 May 2013, 07:44

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.

Last edited by AbuRashid on 07 May 2013, 09:58, edited 1 time in total.

Re: sum of integers in list S [#permalink]
07 May 2013, 07:55

1

This post received KUDOS

rochak22 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 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.

Okay the Question asks if the integers in set S is more than the integers in set T ??

Statement1 :: The average (arithmetic mean) of the integers in S is less than the average of the integers in T.

Now, use smart number plugin ..... lets say if S = 2, 2, 2 and T = 3, 3, then the sums for both is 6, & the average of S is less than T and S has more integers than T. similarly, if S = -3, -3 and T = -2, -2, -2, then the sums for both is -6, the average of S is less than T and S has fewer integers than T. Therefore, Insufficient

Statement 2 :: If T = 2, 2, 2 and S = 3, 3, then the sums for both is 6 & the median of S is greater than median of T and S has fewer integers than T. & If T = -3, -3 and S = -2, -2, -2, then the sums for both -6 & the median of S is greater than median of T and S has more integers than T. Therefore, Insufficient.

1+2 .......... Lets say If S = -7, 9, 10 and T = 6, 6, then the sums for both is 12 & the average of S is less than the average pf T & the median of S is greater than the median of T and S has more integers than T. if S = -6, -6 and T = -10, -9, 7, then the sums for both is -12 & the average of S is less than the average of T & the median of S is greater than the median of T and S has fewer integers than T. Therefore, Insufficient.

Hence, E ................. _________________

If you don’t make mistakes, you’re not working hard. And Now that’s a Huge mistake.

Re: sum of integers in list S [#permalink]
08 May 2013, 01:13

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 _________________

Consider giving +1 Kudo when my post helps you. Also, Good Questions deserve Kudos..!

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.

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

Yes, there are several questions in GMAT Prep with incorrect answers. This question is one of them. A is not correct, answer should be E.

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.

but please tell me why did you consider statements 1 and 2 together..... when statement 1 alone is suffiecient : statement 1 talks about the sum as a whole so : as per the basic formula : sum/num of numbers gives mean and as given sum is equal for both sets and mean of s is less than t clearly says s has more num of integers.

but please tell me why did you consider statements 1 and 2 together..... when statement 1 alone is suffiecient : statement 1 talks about the sum as a whole so : as per the basic formula : sum/num of numbers gives mean and as given sum is equal for both sets and mean of s is less than t clearly says s has more num of integers.

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. _________________

Re: average of integers in a list [#permalink]
09 Oct 2013, 19:31

2

This post received KUDOS

pradeepss wrote:

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

1. the average of integers in S is less than average of integers in T. 2. the median of the integers in S is greater than the median of the integers in T.

Hi Pradeep

Solution :

Statement 1 : Average of S (A1) is less than average of T(A2)

A1<A2

S1 / n1 < S2 / n2

Since S1 = S2 (given)

We can surely find out whether n1 > n2 or not. Sufficient

Statement 2 :

Knowing the median of 2 sets will not let us know the number of integers in each set. (Insufficient)

Option A

Hope it helped.

Cheers Qoofi _________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Re: average of integers in a list [#permalink]
10 Oct 2013, 00:36

Expert's post

Qoofi wrote:

pradeepss wrote:

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

1. the average of integers in S is less than average of integers in T. 2. the median of the integers in S is greater than the median of the integers in T.

Hi Pradeep

Solution :

Statement 1 : Average of S (A1) is less than average of T(A2)

A1<A2

S1 / n1 < S2 / n2

Since S1 = S2 (given)

We can surely find out whether n1 > n2 or not. Sufficient

Statement 2 :

Knowing the median of 2 sets will not let us know the number of integers in each set. (Insufficient)

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