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Which of the following CANNOT be the median of the four

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Which of the following CANNOT be the median of the four [#permalink] New post 11 Dec 2011, 08:22
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Which of the following CANNOT be the median of the four positive integers a, b, c, and d, where a < b < c < d ?

(A) (a+c)/2
(B) (b+c)/2
(C) (a+d)/2
(D) (b+d)/2
(E) (c+d)/2
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Re: median of 4 numbers [#permalink] New post 11 Dec 2011, 13:57
dreambeliever wrote:
Which of the following CANNOT be the median of the four
positive integers a, b, c, and d, where a < b < c < d ?
(A) (a+c)/2
(B) (b+c)/2
(C) (a+d)/2
(D) (b+d)/2
(E) (c+d)/2


As written, there's something wrong with the question. Are some of the inequality signs supposed to be 'greater than or equal to' signs?

Here we know the median is the average of the two 'middle numbers', so the median is (b+c)/2. If a < b, then (a+c)/2 is less than (b+c)/2, so (a+c)/2 cannot be the median. Similarly, if c < d, then (b+c)/2 is less than (b+d)/2, so (b+d)/2 cannot be the median. Finally (c+d)/2 is certainly greater than the median as well. So as written there are three correct answers, A, D and E.
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Joined: 24 Nov 2010
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Re: median of 4 numbers [#permalink] New post 11 Dec 2011, 14:30
IanStewart wrote:
dreambeliever wrote:
Which of the following CANNOT be the median of the four
positive integers a, b, c, and d, where a < b < c < d ?
(A) (a+c)/2
(B) (b+c)/2
(C) (a+d)/2
(D) (b+d)/2
(E) (c+d)/2


As written, there's something wrong with the question. Are some of the inequality signs supposed to be 'greater than or equal to' signs?

Here we know the median is the average of the two 'middle numbers', so the median is (b+c)/2. If a < b, then (a+c)/2 is less than (b+c)/2, so (a+c)/2 cannot be the median. Similarly, if c < d, then (b+c)/2 is less than (b+d)/2, so (b+d)/2 cannot be the median. Finally (c+d)/2 is certainly greater than the median as well. So as written there are three correct answers, A, D and E.


Thanks for confirming, Ian. I thought there was something wrong with the question, too.
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Re: median of 4 numbers [#permalink] New post 11 Dec 2011, 22:00
Hi
OA is E

as given positive integers a, b, c, and d, where a < b < c < d ?

since all are positive integers arranged in increasing order. mean of these will be a no greater then b and less then c.
(A) (a+c)/2

This can be the median if resulting no is >b which is possible here
as this is less than c already

(B) (b+c)/2

This is also possible coz resulting no must be > b and <c here.
(C) (a+d)/2
This can be the median if resulting no is >b and <c which is possible here
(D) (b+d)/2
This is also possible here as resulting no is >b and may be <c which is possible here



(E) (c+d)/2

This can never be median coz this no is always greater then c which violates the 2 condition defined here for median
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Re: median of 4 numbers [#permalink] New post 12 Dec 2011, 13:23
sandeeepsharma wrote:
Hi
OA is E

as given positive integers a, b, c, and d, where a < b < c < d ?

since all are positive integers arranged in increasing order. mean of these will be a no greater then b and less then c.
(A) (a+c)/2

This can be the median if resulting no is >b which is possible here as this is less than c already


If a < b < c < d, then (a + c)/2 cannot possibly be the median of a, b, c and d. The median of those four numbers is the average of the two middle numbers, so is equal to (b+c)/2. If a < b, then (a+c)/2 is strictly less than (b+c)/2; they cannot be equal.

I googled this question, and the OA is E, but the original question is different from the one in the post above. The actual question says that a < b < c < d, in which case (c+d)/2 must be greater than the median, while all the other choices could be equal to the median.
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Re: median of 4 numbers [#permalink] New post 12 Dec 2011, 20:50
thanks for explanation.

1<2<6<7
here a+c/2 creates a no (1+6)/2 =3.5
which is greater then b so condition is true.



IanStewart wrote:
sandeeepsharma wrote:
Hi
OA is E

as given positive integers a, b, c, and d, where a < b < c < d ?

since all are positive integers arranged in increasing order. mean of these will be a no greater then b and less then c.
(A) (a+c)/2

This can be the median if resulting no is >b which is possible here as this is less than c already


Quote:
If a < b < c < d, then (a + c)/2 cannot possibly be the median of a, b, c and d. The median of those four numbers is the average of the two middle numbers, so is equal to (b+c)/2. If a < b, then (a+c)/2 is strictly less than (b+c)/2; they cannot be equal.


I googled this question, and the OA is E, but the original question is different from the one in the post above. The actual question says that a < b < c < d, in which case (c+d)/2 must be greater than the median, while all the other choices could be equal to the median.
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Re: Which of the following CANNOT be the median of the four [#permalink] New post 23 Apr 2014, 04:22
Interesting, I just got this question on a Jeff Sackman problem set that I paid $20 for.

I was looking at his explanation and thought that it HAD to be wrong, for the same reasoning as per above. His question was written exactly as above. I have sent an email but not sure if he will reply.

We will see....
Re: Which of the following CANNOT be the median of the four   [#permalink] 23 Apr 2014, 04:22
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Which of the following CANNOT be the median of the four

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