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Senior Manager  Joined: 12 Mar 2009
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Lists S and T consist of the same number of positive integer  [#permalink]

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Question Stats: 62% (02:05) correct 38% (01:59) wrong based on 350 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 10 May 2009, 12:12.
Last edited by Bunuel on 04 Jun 2013, 04:47, edited 1 time in total.
Edited the question and added the OA
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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.

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vaivish1723 wrote:
Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T?

1. S is consecutive even integers and T is consecutive odd integers.
2. The sum of S is greater than sum of T.

We'll use simple sets of 5 numbers (once we find out both conditions):

S: [E, E+2, E+4, E+6, E+8] - we want median from Set S
T: [O, O+2, O+4, O+6, O+8] - we want mean from Set T

(1) Clearly not enough; 1 set could start with very small numbers & the other very big (& vice versa);
(2) If we don't know statement 1, we can pick any numbers for both sets, and adjust each's mean & median to whatever we want, also not enough.

(combined): We can make a few sets using both conditions combined:

i.e. S: [2, 4, 6, 8, 10]; T: [1, 3, 5, 7, 9] Median of S = 6, mean of T = 5; in this case, answer - yes.
S: [2, 4, 6, 8]; T: [1, 3, 5, 7] Median of S = 5, mean of T = 4; yes again.
S: [2, 4, 6]; T: [1, 3, 5] Median of S = 4, mean of T = 3; yes

So, now we can pretty much be sure we can answer this question reliably; so answer (C).

---------------
My opinion:
It can be
S: [2, 4, 6, 8]; T: [1, 3, 5, 7]
or

S: [ 6, 8,9,10]; T: [1, 3, 5, 7] too
As the question states only about the same number of positive integers and the Stat(1) says S is consecutive even integers and T is consecutive odd integers.does that mean both S and T should be consecutive?---hence I would go with 'E'.

Correct me ,if I am wrong.
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Quote:
---------------
My opinion:
It can be
S: [2, 4, 6, 8]; T: [1, 3, 5, 7]
or

S: [2, 4, 6, 8]; T: [7,9, 11, 13] too as the question states only about the same number of positive integers---hence I would go with 'E'.

Correct me ,if I am wrong.

Under the second condition, the sum of S must be greater than the sum of T, therefore the sets above do not qualify.

S: [ 6, 8,10,12]; T: [1, 3, 5, 7]

If this is the case,what would be the answer?
Will the answer be E?
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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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This can be solved the following ( Hope it is right) without pluging numbers.

First: it is given that the two sets have same number of elements

Stmt 1: Since each set has consecutive numbers then for each set the median will equal the mean.
That's for odd and even consecutive sets.
Ex: S { 1,2,3,4} mean= 2,5. Median = 2,5
T{ 1,2,3,4,5} mean = 3 and Median= 3
however it is Ins. since it gives us nothing else to help in determining the requested

Stmt 2: since the sum of the first set is bigger than the second, it means the first has bigger mean however nothing is given about the median..

Together, yes
the set with bigger mean has bigger median and that's applicable on set 2.. please correct me if there's anything wrong
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vaivish1723 wrote:
Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T?

1. S is consecutive even integers and T is consecutive odd integers.
2. The sum of S is greater than sum of T.

ST. (1) two possible scenarios

S{2,4,6} and T{1,3,5}

or S{2,4,6} and T{3,5,7} both satsify the conditions under statement (1). So, st. 1 is insufficient.

ST. (2) (insufficient)

Look at the examples

1. S{1,3,5} and t{0,0,8)
and S(10,40,60), while T{ 0, 45, 54}. St. 2 is also insufficient.

Combining ST 1 and St. 2, we can conclude that the median of S is greater thean the mean of T.
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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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vannu wrote:
vaivish1723 wrote:
Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T?

1. S is consecutive even integers and T is consecutive odd integers.
2. The sum of S is greater than sum of T.

We'll use simple sets of 5 numbers (once we find out both conditions):

S: [E, E+2, E+4, E+6, E+8] - we want median from Set S
T: [O, O+2, O+4, O+6, O+8] - we want mean from Set T

(1) Clearly not enough; 1 set could start with very small numbers & the other very big (& vice versa);
(2) If we don't know statement 1, we can pick any numbers for both sets, and adjust each's mean & median to whatever we want, also not enough.

(combined): We can make a few sets using both conditions combined:

i.e. S: [2, 4, 6, 8, 10]; T: [1, 3, 5, 7, 9] Median of S = 6, mean of T = 5; in this case, answer - yes.
S: [2, 4, 6, 8]; T: [1, 3, 5, 7] Median of S = 5, mean of T = 4; yes again.
S: [2, 4, 6]; T: [1, 3, 5] Median of S = 4, mean of T = 3; yes

So, now we can pretty much be sure we can answer this question reliably; so answer (C).

---------------
My opinion:
It can be
S: [2, 4, 6, 8]; T: [1, 3, 5, 7]
or

S: [ 6, 8,9,10]; T: [1, 3, 5, 7] too
As the question states only about the same number of positive integers and the Stat(1) says S is consecutive even integers and T is consecutive odd integers.does that mean both S and T should be consecutive?---hence I would go with 'E'.

Correct me ,if I am wrong.

I think if u opted for E then u r write.
My opinion:
It can be
S: [2, 4, 6, 8]; T: [1, 3, 5, 7] - Median can be 4 or 6 which is still insufficient
or

S: [ 6, 8,9,10]; T: [1, 3, 5, 7] Yes Median (S) > Avg (T)
We dont know the number of elements in two set. If the number of elements is odd then yes we can find the relationship by combining both 1 and 2 statements while if it is even then we cant. So, E must be OA.

In either case the Median of set S will be more than avg of set T.
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ykaiim wrote:
I think if u opted for E then u r write.
My opinion:
It can be
S: [2, 4, 6, 8]; T: [1, 3, 5, 7] - Median can be 4 or 6 which is still insufficient
or

S: [ 6, 8,9,10]; T: [1, 3, 5, 7] Yes Median (S) > Avg (T)
We dont know the number of elements in two set. If the number of elements is odd then yes we can find the relationship by combining both 1 and 2 statements while if it is even then we cant. So, E must be OA.

In either case the Median of set S will be more than avg of set T.

S: [2, 4, 6, 8] - Median = (4+6)/2 = 5
T: [1, 3, 5, 7] - Mean = (1+7)/2 = 4

Hence, here also Median of S > Mean of T .

So, Answer should be C only.
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From 1 we can two sets as :
-6 -4 -2 ---- 2 4 6 - {S}

-5 -3 -1 ---- 1 3 5 - {T}

From 2 we can have two sets as :

1 1 1 22 25 --------- 5 5 5 5 5 - {S}

1 1 3 4 5 -------- 1 1 1 1 1 - {T}

So insufficient

From 1 and 2, sufficient, so C.
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Re: Lists S and T consist of the same number of positive integer  [#permalink]

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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.

So as per the translation of this statement ( in red) it means that the set could be Random example irrespective of the statements would be T(-1,-2,-3,-4, 5,6,7) S( -2,-3,-4,1,2,3)

It doesn't mean all of them are positive right?
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Re: Lists S and T consist of the same number of positive integer  [#permalink]

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_________________ Re: Lists S and T consist of the same number of positive integer   [#permalink] 10 Feb 2019, 19:09
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