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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] New post 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 [#permalink] New post 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.

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


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" ?
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Re: Which of the following CANNOT be the median of the 3 [#permalink] New post 21 Feb 2012, 00:41
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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] New post 26 May 2013, 04:47
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Re: Which of the following CANNOT be the median of the 3 [#permalink] New post 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] New post 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.

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?

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Re: Which of the following CANNOT be the median of the 3 [#permalink] New post 16 Jun 2013, 06:57
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?

Image Posted from GMAT ToolKit


As brackets were missing from the options, Edited. Thank you.
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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Which of the following CANNOT be the median of the 3   [#permalink] 16 Jun 2013, 06:57
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