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

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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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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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

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

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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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

The median/mean of {1, 3, 5} is 3, not 4.5.
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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 .

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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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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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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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.
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