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# Which of the following CANNOT be the median of the 3

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Which of the following CANNOT be the median of the 3 [#permalink]

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15 Sep 2003, 22:51
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Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Jun 2013, 06:55, edited 2 times in total.
Edited the question and added the OA
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 13:47
Median of odd number of sequences is the middle number after arranging them in ascending order.

here as we dont know the order , it could be any of the following sequences

x, y,z ; z,y,x = > median y
x , z , y ; y ,z,x => median z
y, x,z ; z,x,y => median x

d - possible

Lets say x y z are
1 2 3 respectively

(median = 2 = (x+z)/2 = (1+3)/2)

e - possible
when x y z are
3 4 9 respectively

median = 4 = (3+9)/3 = (x+z)/3

Answer is C , median can never be sum of other two numbers

Last edited by Spidy001 on 12 Mar 2011, 17:19, edited 1 time in total.
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 15:42
IMO it's C

the median is going to be either x, y, or z

you can't add two of the number to get the median
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 18:21
Exactly !

Consider x = -1, z = -4. Here z < y < x
x + z = -5
-5 is outside the range i.e. smaller than smallest

Consider x = 4, z = 1. Here z < y < x
x + z = 5
5 is outside the range i.e. greater than greatest

khaynes3 wrote:
IMO it's C

the median is going to be either x, y, or z

you can't add two of the number to get the median
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 18:47
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gmat1220 wrote:
Exactly !

Consider x = -1, z = -4. Here z < y < x
x + z = -5
-5 is outside the range i.e. smaller than smallest

Consider x = 4, z = 1. Here z < y < x
x + z = 5
5 is outside the range i.e. greater than greatest

Only positive integers, so the first example isn't valid. That's the key to this question is the fact that all of the numbers are POSITIVE integers. If they were mixed, you could have x = -1, z = 2 and y = 1 which would be valid for x + z equals the median, but since they have to be positive, this would not work.
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 19:09
kannn wrote:
Which of the following CANNOT be the median of the three positive integers x, y, and z?

a) x
b) z
c) x + z
d) (x+z) / 2
e) (x+z) / 3

Simple yet tricky question.

The following would be my train of thought:
Median means middle value (start from the basics). Since there are three numbers, median has to be one of x, y and z. So options a and b are out.
If answer is c, d or e, then y must be the median i.e. the middle value. The middle value could be (x+z)/2 or (x+y)/3 but not (x+z) (since all the numbers are positive).
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 19:16
Yeah kudos ! @ khaynes3 I missed the point. I did not analyze this question until I saw your comment. I would have quickly marked C and moved on.
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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09 Apr 2011, 14:50
Sum of two positive integers will make the result fall between the 2nd and 3rd and so can in no way represent the 2nd or the middle number.
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Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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18 May 2011, 02:53
choice between C and E.

1,2,3 gives median = mean = 2

C remains.
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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20 Feb 2012, 23:09
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Praetorian wrote:
Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. (x+z)/2
E. (x+z)/3

The median of a set with odd number of terms is just a middle term, so it's x, y or z. Eliminate A and B right away. Now, the median can also be (x+z)/2 and (x+z)/3 (for example: {1, 2, 3} and {1, 2, 5}).

But since x, y, and z are positive integers then it no way can be x+z. Why? Because a middle term (the median) cannot possibly be greater than two terms (x and z) in a set with 3 terms.

Notice that, if we were not told that x, y, and z are positive then x+y could be the median, consider {-1, 0, 1}: -1+1=0=median.
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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21 Feb 2012, 00:38
Bunuel wrote:
Praetorian wrote:
Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. x+z/2
E. x+z/3

The median of a set with odd number of terms is just a middle term, so it's x, y or z. Eliminate A and B right away. Now, the median can also be (x+y)/2 and (x+y)/3 (for example: {1, 2, 3} and {1, 2, 5}).

But since x, y, and z are positive integers then it no way can be x+y. Why? Because a middle term (the median) cannot possibly be greater than two terms (x and y) in a set with 3 terms.

Notice that, if we were not told that x, y, and z are positive then x+y could be the median, consider {-1, 0, 1}: -1+1=0=median.

Just out of curiosity, is it always to be assumed that the variables are distinct integers? i.e. would there be cases where a gmat question names variables x, y, z without explicitly stating that theyre "distinct" ?
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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21 Feb 2012, 00:41
essarr wrote:
Just out of curiosity, is it always to be assumed that the variables are distinct integers? i.e. would there be cases where a gmat question names variables x, y, z without explicitly stating that theyre "distinct" ?

No, we should not assume that. For example here x, y, and z can be the same integer.
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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26 May 2013, 04:47
Bumping for review and further discussion.
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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26 May 2013, 08:01
If x=1, y=2, z=1 then by (C) x+z=2 which is y
But I suppose trick is to remember the median is the "middle value", when variables are arranged in ascending/descending order
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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16 Jun 2013, 06:10
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Bunuel wrote:
Praetorian wrote:
Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. x+z/2
E. x+z/3

The median of a set with odd number of terms is just a middle term, so it's x, y or z. Eliminate A and B right away. Now, the median can also be (x+y)/2 and (x+y)/3 (for example: {1, 2, 3} and {1, 2, 5}).

But since x, y, and z are positive integers then it no way can be x+y. Why? Because a middle term (the median) cannot possibly be greater than two terms (x and y) in a set with 3 terms.

Notice that, if we were not told that x, y, and z are positive then x+y could be the median, consider {-1, 0, 1}: -1+1=0=median.

You have assumed x+z/2 ==( x+z)/2.
I really see the question as unclear. Should there be brackets?

Posted from GMAT ToolKit
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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16 Jun 2013, 06:57
AbuRashid wrote:
Bunuel wrote:
Praetorian wrote:
Which of the following CANNOT be the median of the 3 positive integers x, y, and z?

A. x
B. z
C. x+z
D. x+z/2
E. x+z/3

The median of a set with odd number of terms is just a middle term, so it's x, y or z. Eliminate A and B right away. Now, the median can also be (x+y)/2 and (x+y)/3 (for example: {1, 2, 3} and {1, 2, 5}).

But since x, y, and z are positive integers then it no way can be x+y. Why? Because a middle term (the median) cannot possibly be greater than two terms (x and y) in a set with 3 terms.

Notice that, if we were not told that x, y, and z are positive then x+y could be the median, consider {-1, 0, 1}: -1+1=0=median.

You have assumed x+z/2 ==( x+z)/2.
I really see the question as unclear. Should there be brackets?

Posted from GMAT ToolKit

As brackets were missing from the options, Edited. Thank you.
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Re: Which of the following CANNOT be the median of the 3 [#permalink]

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13 Jan 2016, 23:56
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Re: Which of the following CANNOT be the median of the 3   [#permalink] 13 Jan 2016, 23:56
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