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Re: median of 4 numbers [#permalink]
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.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: median of 4 numbers [#permalink]
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.

Re: median of 4 numbers [#permalink]
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

Re: median of 4 numbers [#permalink]
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.
_________________

Nov 2011: After years of development, I am now making my advanced Quant books and high-level problem sets available for sale. Contact me at ianstewartgmat at gmail.com for details.

Re: median of 4 numbers [#permalink]
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.

gmatclubot

Re: median of 4 numbers
[#permalink]
12 Dec 2011, 20:50