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Set S consistes of 5 consecutive integers, Set T consists of [#permalink]
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17 May 2006, 00:23
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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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Joined: 09 Mar 2006
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B it is
The median of a oddsized 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, 08:40, edited 2 times in total.



Intern
Joined: 23 Feb 2006
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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



Manager
Joined: 14 Mar 2006
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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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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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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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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
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Director
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Thats a nice catch deowl, this was a tricky one



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Yeah, guys. Cross our fingers and hope this will not happen on G day.



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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.



Intern
Joined: 23 Feb 2006
Posts: 39

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



Senior Manager
Joined: 09 Aug 2005
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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










