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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, 19:47

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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.

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

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21 Feb 2012, 00:09

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Expert's post

1

This post was BOOKMARKED

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.

Answer: C.

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. _________________

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

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16 Jun 2013, 07:10

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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.

Answer: C.

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?

Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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12 Mar 2011, 20:09

Expert's post

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). _________________

Re: Which of the following CANNOT be the median of the 3 positive integers [#permalink]

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09 Apr 2011, 15:50

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. _________________

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

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21 Feb 2012, 01:38

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.

Answer: C.

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" ?

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

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21 Feb 2012, 01:41

Expert's post

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. _________________

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

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26 May 2013, 09:01

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

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

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16 Jun 2013, 07:57

Expert's post

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.

Answer: C.

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?

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

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14 Jan 2016, 00:56

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