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Set S consistes of 5 consecutive integers, Set T consists of

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Set S consistes of 5 consecutive integers, Set T consists of [#permalink] New post 16 May 2006, 23:23
Set S consistes of 5 consecutive integers, Set T consists of 7 consecutive integers.

is the Median of S equal to the median of T?

1) Median of S = 0

2) Sum of numbers in S = Sum of numbers in T

OA later[color=darkblue][/color]
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 [#permalink] New post 16 May 2006, 23:59
B it is

The median of a odd-sized set of consecutive integers is equal to its average

1) obviously insufficient
2) If S1 is equal to S2 , then their averages will be different due to different denominator, hence sufficient

This one is wrong, see the correct explanation below.

Last edited by deowl on 17 May 2006, 07:40, edited 2 times in total.
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 [#permalink] New post 17 May 2006, 07:00
I think it's B.
Sum of S is x+(x+1)+..(x+4)=5x+10
Sum of T is 5y+10
B says: 5y+10=5y+10 => y=x => median is the same
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 [#permalink] New post 17 May 2006, 07:02
B aswell.

In this case if the sums equal, then the median should be same.
s: -2, -1, 0, 1, 2
t:-3, -1, -1, 0, 1, 2, 3
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 [#permalink] New post 17 May 2006, 07:04
B.

St1: Clearly INSUFF
St2: This can only be true if both the sets are as follows

S = {-2,-1,0,1,2}
T = {-3,-2,-1,0,1,2,3}
Median of both is same.: SUFF
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 [#permalink] New post 17 May 2006, 07:32
C it is

1) insuff

2) at least two options exist
{2,3,4,5,6,7,8} and {5,6,7,8,9} - medians are different
actually much more sets can be found where medians are different
see my WRONG explanation above
{-3, -2, -1, 0, 1, 2, 3 } and {-2, -1, 0, 1, 2} - medians are equal

(1) + (2) together leave only the possibility in which the average (and the median ) is zero so it is sufficient.
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 [#permalink] New post 17 May 2006, 07:43
deowl wrote:
C it is

1) insuff

2) at least two options exist
{2,3,4,5,6,7,8} and {5,6,7,8,9} - medians are different
actually much more sets can be found where medians are different
see my WRONG explanation above
{-3, -2, -1, 0, 1, 2, 3 } and {-2, -1, 0, 1, 2} - medians are equal

(1) + (2) together leave only the possibility in which the average (and the median ) is zero so it is sufficient.


Oops I missed this :wall
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 [#permalink] New post 17 May 2006, 07:52
Thats a nice catch deowl, this was a tricky one
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 [#permalink] New post 17 May 2006, 07:57
Yeah, guys. Cross our fingers and hope this will not happen on G day.
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 [#permalink] New post 17 May 2006, 09:11
deowl wrote:
Yeah, guys. Cross our fingers and hope this will not happen on G day.


thanks deowl, nice work.
I tried to produce the same result but couldn't in a short time. Def a tricky q.
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 [#permalink] New post 19 May 2006, 04:08
I think it's B.
Sum of S is x+(x+1)+..(x+4)=5x+10
Sum of T is 5y+10
B says: 5y+10=5y+10 => y=x => median is the same
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 [#permalink] New post 20 May 2006, 11:21
deowl wrote:
C it is

1) insuff

2) at least two options exist
{2,3,4,5,6,7,8} and {5,6,7,8,9} - medians are different
actually much more sets can be found where medians are different
see my WRONG explanation above
{-3, -2, -1, 0, 1, 2, 3 } and {-2, -1, 0, 1, 2} - medians are equal

(1) + (2) together leave only the possibility in which the average (and the median ) is zero so it is sufficient.


Can you provide the 2 examples you have in mind that show b is not suff?

consecutive is the key here

thank you
  [#permalink] 20 May 2006, 11:21
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