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# Which of the following CANNOT be the median of the four cons

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Which of the following CANNOT be the median of the four cons [#permalink]

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27 Sep 2009, 18:35
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Which of the following CANNOT be the median of the four consecutive positive integers w, x, y, and z, where w < x < y < z ?

A. (w+x)/2 - 1
B. (w+z)/2
C. (x+y)/2
D. (y+z)/2 -1
E. (w+x+y+z)/4

Source - Mr Jeff Challenge
[Reveal] Spoiler: OA
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Re: Which of the following [#permalink]

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27 Sep 2009, 21:02
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.
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Re: Which of the following [#permalink]

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27 Sep 2009, 22:48
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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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Re: Which of the following [#permalink]

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28 Sep 2009, 05:38
goldgoldandgold wrote:
Which of the following CANNOT be the median of the four
consecutive positive integers w, x, y, and z, where
w < x < y < z ?

(w+x)/2 - 1
(w+z)/2
(x+y)/2
(y+z)/2 -1
(w+x+y+z)/4

Good luck

Source - Mr Jeff Challenge

obviously A. (w+x)/2-1 < median(w,x) <=median(w,x,y,z)
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Re: Which of the following [#permalink]

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08 Nov 2010, 05:57
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A is the answer by backsolving of the problem. take any 4 positive consecutive integers in place for W,X,Y,Z.
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Re: Which of the following [#permalink]

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07 Mar 2011, 04:56
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.
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Re: Which of the following [#permalink]

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07 Mar 2011, 10:54
you dont need to solve anything in this problem, you can notice quickly that A - is the sum of the two smallest numbers, divide by 2 and minus 1.

Means - it will always be smaller than the median and the mean...
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Re: Which of the following [#permalink]

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07 Mar 2011, 19:26
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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Re: Which of the following [#permalink]

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09 Apr 2011, 17:14
goldgoldandgold wrote:
Which of the following CANNOT be the median of the four
consecutive positive integers w, x, y, and z, where
w < x < y < z ?

(w+x)/2 - 1
(w+z)/2
(x+y)/2
(y+z)/2 -1
(w+x+y+z)/4

Good luck

Source - Mr Jeff Challenge

(A) - (w+x)/2 - 1
(B) - (w+z)/2
(C) - (x+y)/2
(D) - (y+z)/2 -1
(E) - (w+x+y+z)/4

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.

Answer (A)
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Re: Which of the following [#permalink]

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22 Apr 2011, 12:55
"A" is the answer
Avg of the bottom 2 choices will be less than mean & you will delete 1 from it which will be lower then Z thus it can't be median!!
goldgoldandgold wrote:
Which of the following CANNOT be the median of the four
consecutive positive integers w, x, y, and z, where
w < x < y < z ?

(w+x)/2 - 1
(w+z)/2
(x+y)/2
(y+z)/2 -1
(w+x+y+z)/4

Good luck

Source - Mr Jeff Challenge

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Re: Which of the following [#permalink]

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18 Nov 2011, 07:07
(A) is sum of the 2 smallest numbers divided by 2 minus 1... It's extremely low.

without any calculation, the answer is (A)
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Re: Which of the following CANNOT be the median of the four cons [#permalink]

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09 Mar 2016, 09:19
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Which of the following CANNOT be the median of the four cons [#permalink]

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23 Jul 2016, 07:33
goldgoldandgold wrote:
Which of the following CANNOT be the median of the four consecutive positive integers w, x, y, and z, where w < x < y < z ?

A. (w+x)/2 - 1
B. (w+z)/2
C. (x+y)/2
D. (y+z)/2 -1
E. (w+x+y+z)/4

Source - Mr Jeff Challenge

In sets which contains consecutive numbers MEAN=MEDIAN
so median has to be (x+y)/2

Also w+x will be smallest sum of any two numbers in the given set; dividing the smallest sum by half will give and even smaller number and then subtracting 1 from it will make the even more smaller. But mean is half of middle terms and should therefore have a value that should lie between the two middle terms of the set.
so, A is the answer
OR
use simple integer like 1,2,3,4 and see how the mean and median are 5/2 = 2.5. Now plug in the value of w=1, x=2, y=3 and z=4 one by one in each equation and find out the one that is not 2.5
That is Option A again .
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Which of the following CANNOT be the median of the four cons   [#permalink] 23 Jul 2016, 07:33
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# Which of the following CANNOT be the median of the four cons

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