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Lists S and T consist of the same number of positive integers. Is the

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Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   Updated on: 22 May 2021, 13:27
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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.
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C

Originally posted by vaivish1723 on 10 May 2009, 12:12.
Last edited by Bunuel on 22 May 2021, 13:27, edited 2 times in total.
Edited the question and added the OA
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Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   17 Jan 2010, 17:01
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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
Show Spoiler
c


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.

Answer: C.
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   17 Feb 2011, 13:22
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Original question is:

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?

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

(1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers.

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

\(Sum\{S\}>Sum\{T\}\). 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.

Answer: C.
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   18 May 2009, 00:35
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S and T have the same number of integers say n . Is Median of S > Median of T?

1. S is consecutive even integers and T is consecutive odd integers.
So S could be 2,4,6 or 111112,1111114.
So Insufficient.
2. The sum of S is greater than sum of T.
If the sets are equally spaced then, Mean = Median. sum of S > sum of T
then (sum of S)/n > (sum of T)/n
If the set are not, Then S = (1,1,1,1,1) T=(-1,-2,1,2,3) here sum of S > sum of T but Median of S = Median of T

Together

Yes we know S and T are equally sapced. So as seen in Statement two yes. So Ans . C
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   05 Aug 2017, 14:06
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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 integers. Is the [#permalink] New post   14 Oct 2009, 12:50
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C

Median = Mean for consec odd/even #s

If sum of S is greater than sum of T, then median and mean of S are greater than median and mean of T
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   05 Aug 2017, 12:27
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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 integers. Is the [#permalink] New post   11 Jun 2020, 01:58
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GDT wrote:
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
Show Spoiler
c


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.

Answer: C.


VeritasKarishma

The ques says that set S and T consist of same no. of positive integers but it doesn't say that these are the only elements in the set. the set may also contain 0 or negative integers. In that case wouldn't E be the answer


consist of = made up of

So we are given "...lists S and T are made up of same no. of positive integers"
So this is how the lists are made. There are no non-positive integers in them.

This is an official GMAT question so why fret over it. It gives you an important takeaway!
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   04 Dec 2014, 20:07
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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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   05 Dec 2014, 04:33
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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 \(Mean\{S\}>Mean\{T\}\)?".

Please re-read the solution.

Hope it helps.
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   03 Apr 2015, 04:12
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,
PLease explain this.When I take below sets I get answer E.

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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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   03 Apr 2015, 04:40
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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,
PLease explain this.When I take below sets I get answer E.

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 integers. Is the [#permalink] New post   03 Apr 2015, 23:37
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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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   09 Jun 2020, 07:27
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
Show Spoiler
c


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.

Answer: C.


VeritasKarishma

The ques says that set S and T consist of same no. of positive integers but it doesn't say that these are the only elements in the set. the set may also contain 0 or negative integers. In that case wouldn't E be the answer
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   24 Aug 2023, 08:40
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
Show Spoiler
c


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.

Answer: C.


hi, could someone please explain why my working is incorrect?
combining statement 1 and 2:

let S={100,102,103}; median=102
T={1,3,5}; mean = 3
YES, median of the integers in S greater than the average (arithmetic mean) of the integers in T.

OR

let S={0,2,4,6,8}; median=4
T= {1,3,5,7}; mean=4
(S and T consist of the same number of positive integers)
NO
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   24 Aug 2023, 09:01
Expert Reply
TA396 wrote:
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
Show Spoiler
c


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.

Answer: C.


hi, could someone please explain why my working is incorrect?
combining statement 1 and 2:

let S={100,102,103}; median=102
T={1,3,5}; mean = 3
YES, median of the integers in S greater than the average (arithmetic mean) of the integers in T.

OR

let S={0,2,4,6,8}; median=4
T= {1,3,5,7}; mean=4
(S and T consist of the same number of positive integers)
NO


Your interpretation of "Lists S and T consist of the same number of positive integers" is not correct. While you interpret it to mean that the count of positive integers in lists S and T are identical, it actually means that both lists are composed solely of positive integers, and the quantity in each is equal.
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   25 Aug 2023, 04:23
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
Show Spoiler
c


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.

Answer: C.



If we consider a set
S - 2,4,6
T - 5
dont we get a NO as an answer?

Bunuel
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Re: Lists S and T consist of the same number of positive integers. Is the [#permalink] New post   25 Aug 2023, 04:43
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Its_me_aka_ak wrote:
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
Show Spoiler
c


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.

Answer: C.



If we consider a set
S - 2,4,6
T - 5
dont we get a NO as an answer?

Bunuel


Check the highlighted part:

Lists S and T consist of the same number of positive integers.
GMAT Club Bot
gmatclubot
Re: Lists S and T consist of the same number of positive integers. Is the [#permalink]
25 Aug 2023, 04:43
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Bunuel
Math Expert
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