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Set S consists of five consectuve integers, and set T [#permalink]
 17 Apr 2008, 17:43
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Set S consists of five consectuve integers, and set T consists of seven consecutive integers. Is the median of the numbers in set S equal to the median of the numbers in set T?
C1. The median of the numbers in set S is 0.
C2. The sum of the numbers in set S is equal to the sum of the numbers in set T.
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Re: DS median of sets [#permalink]
17 Apr 2008, 19:20
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Given set S = {a-2d,a-d,a,a+d,a+2d}, where d = 1, and a is any integer.
So median of set S = a, and Average = 5a/5 = a
Given set T = {b-3d,b-2d,b-d,b,b+d,b+2d,b+3d}, where d = 1, and b is any integer.
So median of set T = b, and Average = 7b/7 = b
Statement 1:
Tells us median of set S = 0 => a = 0, But this statement tells us nothing about T, so insufficient.
Statement 2:
Tells us 5a = 7b => a = 7b/5, which means a (median of set S) is not equal to b(median of set T) provided b is not equal to 0, otherwise a will become 0, so this statement alone is not sufficient.
Combining both statements:
From statement one we know a=0, so statement two tells us b is also zero. So both sets have same median.
Answer C.
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Re: DS median of sets [#permalink]
17 Apr 2008, 20:34
abhijit_sen wrote: Given set S = {a-2d,a-d,a,a+d,a+2d}, where d = 1, and a is any integer.
So median of set S = a, and Average = 5a/5 = a
Given set T = {b-3d,b-2d,b-d,b,b+d,b+2d,b+3d}, where d = 1, and b is any integer.
So median of set T = b, and Average = 7b/7 = b
Statement 1:
Tells us median of set S = 0 => a = 0, But this statement tells us nothing about T, so insufficient.
Statement 2:
Tells us 5a = 7b => a = 7b/5, which means a (median of set S) is not equal to b(median of set T) provided b is not equal to 0, otherwise a will become 0, so this statement alone is not sufficient.
Combining both statements:
From statement one we know a=0, so statement two tells us b is also zero. So both sets have same median.
Answer C. I like your reasoning! 
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Re: DS median of sets [#permalink]
18 Apr 2008, 23:57
abhijit_sen wrote: Given set S = {a-2d,a-d,a,a+d,a+2d}, where d = 1, and a is any integer.
So median of set S = a, and Average = 5a/5 = a
Given set T = {b-3d,b-2d,b-d,b,b+d,b+2d,b+3d}, where d = 1, and b is any integer.
So median of set T = b, and Average = 7b/7 = b
Statement 1:
Tells us median of set S = 0 => a = 0, But this statement tells us nothing about T, so insufficient.
Statement 2:
Tells us 5a = 7b => a = 7b/5, which means a (median of set S) is not equal to b(median of set T) provided b is not equal to 0, otherwise a will become 0, so this statement alone is not sufficient.
Combining both statements:
From statement one we know a=0, so statement two tells us b is also zero. So both sets have same median.
Answer C. Abhijit, My hunch is, Answer could be B here I think , for 2 sets of "consecutive" number, 5a=7b is possible only when a=b=0 Which makes B SUFFICIENT. Can you think of example where B is not sufficient ?
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Re: DS median of sets [#permalink]
19 Apr 2008, 00:39
kyatin wrote: Abhijit,
My hunch is, Answer could be B here
I think , for 2 sets of "consecutive" number, 5a=7b is possible only when a=b=0
Which makes B SUFFICIENT.
Can you think of example where B is not sufficient ? s={5,6,7,8,9} t={2,3,4,5,6,7,8} sum=35 but medians are different.
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Re: DS median of sets [#permalink]
19 Apr 2008, 08:11
walker wrote: kyatin wrote: Abhijit,
My hunch is, Answer could be B here
I think , for 2 sets of "consecutive" number, 5a=7b is possible only when a=b=0
Which makes B SUFFICIENT.
Can you think of example where B is not sufficient ? s={5,6,7,8,9} t={2,3,4,5,6,7,8} sum=35 but medians are different. heheh. that was easy. I did use his equation to just plug in numbers...but created two sequences by incorrect calculations.. 
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Re: DS median of sets
[#permalink]
19 Apr 2008, 08:11
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