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Set S consists of 5 consecutive integers, and set T consists

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VP
Joined: 22 Nov 2007
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Set S consists of 5 consecutive integers, and set T consists [#permalink]  11 Mar 2008, 11:31
Set S consists of 5 consecutive 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?

(1) The median of the numbers in set S is 0.
(2) The sum of the numbers in set S is equal to the sum of the numbers in
set T.
Director
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Re: gmatprep stats [#permalink]  11 Mar 2008, 16:13
1) The median of the numbers in set S is 0.
That means there is equal number of terms greater than and less than 0.
But this tells you nothing about Set T's relation with So question cannot be answered alone with this statement.

(2) The sum of the numbers in set S is equal to the sum of the numbers in set T.
There are infinite possibilites that sum of number is set S is equal to sum of numbers in set T, so this also cannot answer alone.

Combining these 2 has interesting effect:
Now we know that S is having median 0, so equal number of +ve and -ve numbers should be there.
S = (-2,-1,0,1,2) (This is only possibility since numbers have to be consecutive)
Note also that Sum is also 0 for set S.
So now only way we have 7 consecutive numbers resulting in 0 sum is
T = (-3,-2,-1,0,1,2,3)
So we can tell median to T is also 0.

VP
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Re: gmatprep stats [#permalink]  17 Mar 2008, 01:06
marcodonzelli wrote:
Set S consists of 5 consecutive 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?

(1) The median of the numbers in set S is 0.
(2) The sum of the numbers in set S is equal to the sum of the numbers in
set T.

the key is that the numbers are consecutive, so we have an arithmetic sequence where the mean equals the median.
1. obviously doesn't say anything alone
2. if s is the sum of the elements in S, the mean=median=s/5; and for T, the mean=median=s/7. s/5=s/7 only when s=0

c. s=0 suff
CEO
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Re: gmatprep stats [#permalink]  17 Mar 2008, 13:17
marcodonzelli wrote:
marcodonzelli wrote:
Set S consists of 5 consecutive 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?

(1) The median of the numbers in set S is 0.
(2) The sum of the numbers in set S is equal to the sum of the numbers in
set T.

the key is that the numbers are consecutive, so we have an arithmetic sequence where the mean equals the median.
1. obviously doesn't say anything alone
2. if s is the sum of the elements in S, the mean=median=s/5; and for T, the mean=median=s/7. s/5=s/7 only when s=0

c. s=0 suff

Sum of set T= sum of set S. T and S are both sets of consecutive integers. I can't find a a way for sum of set T= sum of set S when they are both consecutive sets unless the median is 0.
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Re: gmatprep stats [#permalink]  17 Mar 2008, 20:37
GMATBLACKBELT wrote:
marcodonzelli wrote:
marcodonzelli wrote:
Set S consists of 5 consecutive integers, and set T consists of seven consecutiveintegers. Is the median of the numbers in set S equal to the median of thenumbers in set T?(1) The median of the numbers in set S is 0.(2) The sum of the numbers in set S is equal to the sum of the numbers inset T.
the key is that the numbers are consecutive, so we have an arithmetic sequence where the mean equals the median. 1. obviously doesn't say anything alone2. if s is the sum of the elements in S, the mean=median=s/5; and for T, the mean=median=s/7. s/5=s/7 only when s=0c. s=0 suff
Shouldnt the answer be BSum of set T= sum of set S.  T and S are both sets of consecutive integers.  I can't find a a way for sum of set T= sum of set S when they are both consecutive sets unless the median is 0.

C.

from B:
S = 5, 6, 7 ,8, 9. median = 7 and sum = 35
T = 2, 3, 4, 5, 6, 7, 8. median = 5 and sum = 35so not sufficient.
from 1 and 2: S = -2, -1, 0, 1, 2. median = 0 and sum = 0
T = -3, -2, -1, 0, 1, 2, 3. median = 0 and sum = 0
_________________
Re: gmatprep stats   [#permalink] 17 Mar 2008, 20:37
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