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# Lists S and T consist of the same number of positive

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Senior Manager
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Lists S and T consist of the same number of positive  [#permalink]

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Updated on: 11 Jul 2013, 23:58
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Difficulty:

55% (hard)

Question Stats:

65% (02:07) correct 35% (02:04) wrong based on 476 sessions

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Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

Originally posted by vaivish1723 on 17 Jan 2010, 09:45.
Last edited by Bunuel on 11 Jul 2013, 23:58, edited 1 time in total.
Edited the question and added the OA
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17 Jan 2010, 16:01
6
3
vaivish1723 wrote:
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

OA is

Pl explain

Q: is $$Median\{S\}>Mean\{T\}$$? Given: {# of terms in S}={# of terms in T}, let's say N.

(1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is $$Mean\{S\}>Mean\{T\}$$? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}.

(2) $$Sum\{S\}>Sum\{T\}$$. Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1).

(1)+(2) From (1) question became is $$Mean\{S\}>Mean\{T\}$$? --> As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is $$\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}$$ true? --> Is $$Sum\{S\}>Sum\{T\}$$? This is exactly what is said in statement (2) that $$Sum\{S\}>Sum\{T\}$$. Hence sufficient.

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24 Jan 2010, 13:24
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02 Jul 2011, 02:59
I feel there is an ambiguity in the wording of the question itself. The question says that set S and T consist of the same number of positive integers. This statement does not exclude the possibility that Set S and/or T might contain some negative numbers or zero.

Had the question said: " Set S and T consist of only the same number of positive integers then I would have agreed with the solution. What do you guys think? Also, can the original poster mention the source of this question pls?!
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03 Dec 2013, 21:13
Bunuel wrote:
vaivish1723 wrote:
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

OA is

Pl explain

Q: is $$Median\{S\}>Mean\{T\}$$? Given: {# of terms in S}={# of terms in T}, let's say N.

(1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is $$Mean\{S\}>Mean\{T\}$$? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}.

(2) $$Sum\{S\}>Sum\{T\}$$. Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1).

(1)+(2) From (1) question became is $$Mean\{S\}>Mean\{T\}$$? --> As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is $$\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}$$ true? --> Is $$Sum\{S\}>Sum\{T\}$$? This is exactly what is said in statement (2) that $$Sum\{S\}>Sum\{T\}$$. Hence sufficient.

Hi Bunuel,

In the above Question, it says there are same number of positive integers but does not explicitly rule out negative numbers or fractions etc.

After combining 2 statements and taking the case of negative numbers we can have the following as well

Let us say S= -2,0,2,4,6 and 8, Median will be 3
T= -1,1,3,5 and7, Median will be 3

Clearly there can be 2 cases and ans can be E.

I think the question needs to mention that the list contains only positive integers

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Re: Lists S and T consist of the same number of positive  [#permalink]

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04 Dec 2014, 19:07
hey Bunuel.

in your explanation for (1) and (2) combined, you derived a relationship for the Means of the two sets from statement (1). can you please explain how you got that. Statement (1) only tells us that one set has even consecutive integers and one odd. so how do you know that Mean of S > Mean of T?
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Lists S and T consist of the same number of positive  [#permalink]

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05 Dec 2014, 03:33
kritiu wrote:
hey Bunuel.

in your explanation for (1) and (2) combined, you derived a relationship for the Means of the two sets from statement (1). can you please explain how you got that. Statement (1) only tells us that one set has even consecutive integers and one odd. so how do you know that Mean of S > Mean of T?

From (1) we don't know whether $$Mean\{S\}>Mean\{T\}$$.

From (1) the question became "is $$Mean\{S\}>Mean\{T\}$$?"" Meaning that based on the information given in (1), the original question "is $$Median\{S\}>Mean\{T\}$$?" could be rephrased "is $$Mean\{S\}>Mean\{T\}$$?".

Hope it helps.
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Re: Lists S and T consist of the same number of positive  [#permalink]

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03 Apr 2015, 03:12
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.[/quote]

Hey Bunuel,

Set S=(6,8,10)...Median=8
Set T=(1,3,5)...Mean=4.5
Median of S>mean of t

If Set s=(2,4,6)...Median=4
Set T=(1,3,5)...Mean=4.5

Here,Median of S IS NOT GREATER THAN mean of T
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Lists S and T consist of the same number of positive  [#permalink]

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03 Apr 2015, 03:40
ssriva2 wrote:
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

Hey Bunuel,

Set S=(6,8,10)...Median=8
Set T=(1,3,5)...Mean=4.5
Median of S>mean of t

If Set s=(2,4,6)...Median=4
Set T=(1,3,5)...Mean=4.5

Here,Median of S IS NOT GREATER THAN mean of T

The median/mean of {1, 3, 5} is 3, not 4.5.
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Re: Lists S and T consist of the same number of positive  [#permalink]

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03 Apr 2015, 04:05
Oh mean is 3,that was indeed a silly question!
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Re: Lists S and T consist of the same number of positive  [#permalink]

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03 Apr 2015, 22:37
Hi All,
Given: List S and T have same no. of elements.
Median of S > average of T?
Statement 1 is insufficient:
Whenever it is consecutive median is always equal to mean.
Let S =2 ,4 ,6 median is 4.
Let T = 1,3,5 median(mean) is 3.
Answer to the question is YES.
Let S =2 ,4 ,6 median is 4.
Let T = 3,5,7 median(mean) is 5.
Answer to the question is NO.
So not sufficient.
Statement 2 is insufficient:
Sum of the integers in S is greater than T(here the list not necessarily be consecutive).
Let S =1 ,2 ,3 median is 2.
Let T = 0,1,2 median(mean) is 1.
Answer to the question is YES.
Let S =1 ,5 ,20 median is 5.
Let T = 5,6,7 median(mean) is 6.
Answer to the question is NO.
So not sufficient.
Together it is sufficient.
Since the S and T are consecutive even and odd and Sum of S is greater than T.
Then median of S is always greater than mean(median) of T .
So answer is together sufficient .
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Lists S and T consist of the same number of positive  [#permalink]

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Updated on: 06 Aug 2017, 05:33
Bunuel wrote:
vaivish1723 wrote:
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

OA is

Pl explain

Q: is $$Median\{S\}>Mean\{T\}$$? Given: {# of terms in S}={# of terms in T}, let's say N.

(1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is $$Mean\{S\}>Mean\{T\}$$? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}.

(2) $$Sum\{S\}>Sum\{T\}$$. Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1).

(1)+(2) From (1) question became is $$Mean\{S\}>Mean\{T\}$$? --> As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is $$\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}$$ true? --> Is $$Sum\{S\}>Sum\{T\}$$? This is exactly what is said in statement (2) that $$Sum\{S\}>Sum\{T\}$$. Hence sufficient.

I'm sorry, but I disagree. The question says that "the lists S and T consist of SAME NUMBER OF POSITIVE INTEGERS".
Now you just know that the number of positive integers are same on both lists. What if there are some negative integers as well? or infact zero?
I didn't interpret the question to say that there are same number of terms.
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Originally posted by rekhabishop on 04 Aug 2017, 17:32.
Last edited by rekhabishop on 06 Aug 2017, 05:33, edited 1 time in total.
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Re: Lists S and T consist of the same number of positive  [#permalink]

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05 Aug 2017, 11:27
We cannot assume that set A and set B do not contain negative numbers.
Either the question needs to be edited to "same number of integers" or the OA needs to be changed to E.
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Re: Lists S and T consist of the same number of positive  [#permalink]

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05 Aug 2017, 13:06
1
darn wrote:
We cannot assume that set A and set B do not contain negative numbers.
Either the question needs to be edited to "same number of integers" or the OA needs to be changed to E.

We actually can assume that! The question itself says "S and T consist of the same number of positive integers." If it said "contain the same number of positive integers", your argument would be correct. But when a GMAT problem tells you that a set 'consists of' something, you know they've told you about everything that's in the set. Since the question uses the word 'consist' and says 'positive integers', you know that both sets consist of only positive integers.
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Re: Lists S and T consist of the same number of positive  [#permalink]

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21 Jan 2019, 03:26
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Re: Lists S and T consist of the same number of positive   [#permalink] 21 Jan 2019, 03:26
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