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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..
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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.
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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.