Which of the following CANNOT be the median of the four

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

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.

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.

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....