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 500,000 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: Which of the following [#permalink]
27 Sep 2009, 20:02

I believe the answer to this question is A “(w+x)/2 – 1”

(w+z)/2 -> If numbers are consecutive positive integers this is median (x+y)/2 -> Since there are even numbers and they are consecutive this is median (y+z)/2 -1 -> This is possible too. Take the example of 1,2,3,4 (y+z)/2 – 1 = [(3+4)/2] -1 = 2.5 which is the median. (w+x+y+z)/4 -> for a series of consecutive numbers median = mean and this is formula for mean.

Using 1,2,3,4 as an example for all the above formulas except the first one gives you the correct median.

Re: Which of the following [#permalink]
27 Sep 2009, 21:48

1

This post received KUDOS

OA A my explanation : w,x,y,z are positive consecutive integers and w < x < y < z. Let us assume w,x,y,z are 1,2,3,4 thus median is 2.5

A] (w+x)/2 - 1....As w & x are values below median overall result of option A (avg of below median values -1) will always result in value below median..as per our nos 0.5 < 2.5...hence OA A is valid

B] (w+z)/2...w & z are extreame values hence its possible to have option B as median...2.5 = 2.5 C] (x+y)/2...as per our values 2.5 = 2.5 median...hence possible D] (y+z)/2 -1...though y and z are values above median but expression reduces value by 1 hence it is possible to have value equal to median...as per our values 2.5 = 2.5 median..hence possible E] (w+x+y+z)/4...this is a plain avg and for consecutive integers it is possible to have median = avg...hence possible

Hence oA A is valid....consider it for Kudos if u like it.. _________________

Bhushan S. If you like my post....Consider it for Kudos

Re: Which of the following [#permalink]
07 Mar 2011, 03:56

bhushan252 wrote:

OA A my explanation : w,x,y,z are positive consecutive integers and w < x < y < z. Let us assume w,x,y,z are 1,2,3,4 thus median is 2.5

A] (w+x)/2 - 1....As w & x are values below median overall result of option A (avg of below median values -1) will always result in value below median..as per our nos 0.5 < 2.5...hence OA A is valid

B] (w+z)/2...w & z are extreame values hence its possible to have option B as median...2.5 = 2.5 C] (x+y)/2...as per our values 2.5 = 2.5 median...hence possible D] (y+z)/2 -1...though y and z are values above median but expression reduces value by 1 hence it is possible to have value equal to median...as per our values 2.5 = 2.5 median..hence possible E] (w+x+y+z)/4...this is a plain avg and for consecutive integers it is possible to have median = avg...hence possible

Hence oA A is valid....consider it for Kudos if u like it..

We can use the fact that w<x<y<z and they are all positive consecutive integers, so w will be x-1 and y-2, similarly z would be x+2 and Y+1.

We know that median would be (x+y)/2

In each given choice we can replace w and z in terms of either x or y and see if it reduces to (x+y)/2.

Luckily for us, the first one itself doesn't reduce to (x+y)/2 as it becomes (w+x)/2 - 1= (y-2+x)/2-1 = (x+y)/2 -2, so answer A.

Re: Which of the following [#permalink]
07 Mar 2011, 18:26

The answer is A. To solve this, a quick way might be to take 4 positive numbers and verify the options :

1 2 3 4 -> Median = 2.5

Also, a few things are good to remember as the mean = median in an arithmetic progression and the same is equal to the mean of first and last values. This will help in shortlisting the options. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

Using the properties of consecutive positive integers we could right away eliminate B,C and E. (B) - The average of first and last terms is the median (C) - For even number of integers, the median is the average of middle terms (E) - The average is equal to median if the integers are consecutive

Remaining answer choices are A and D. For A, the average of first two consecutive numbers (w,x) will definitely be less than x making it impossible to be a median.

Low GPA MBA Acceptance Rate Analysis Many applicants worry about applying to business school if they have a low GPA. I analyzed the low GPA MBA acceptance rate at...

http://blog.davidbbaker.com/wp-content/uploads/2015/11/12249800_10153820891439090_8007573611012789132_n.jpg When you think about an MBA program, usually the last thing you think of is professional collegiate sport. (Yes American’s I’m going...