Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: median of 4 numbers [#permalink]
11 Dec 2011, 13:57

Expert's post

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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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

Expert's post

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. _________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

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.

Re: Which of the following CANNOT be the median of the four [#permalink]
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....

gmatclubot

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

I am not panicking. Nope, Not at all. But I am beginning to wonder what I was thinking when I decided to work full-time and plan my cross-continent relocation...

Over the last week my Facebook wall has been flooded with most positive, almost euphoric emotions: “End of a fantastic school year”, “What a life-changing year it’s been”, “My...