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

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Senior Manager
Joined: 17 Aug 2005
Posts: 387

Kudos [?]: 148 [0], given: 0

Location: Boston, MA
Which of the following CANNOT be the median of the three [#permalink]

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24 May 2006, 16:43
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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

This question is from OG11. Can/should we assume that x, y and z are positive integers such that x<y<z?

Kudos [?]: 148 [0], given: 0

Senior Manager
Joined: 17 Aug 2005
Posts: 387

Kudos [?]: 148 [0], given: 0

Location: Boston, MA

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24 May 2006, 19:44
When answering the question I did not make the assumption and plugged in to get the answer (incorrectly). Is this a normal assumption to make? Pardon my ignorance, but you never know.

I guess if you really look at the question you can infer the assumption is made based on the answers. ..

Kudos [?]: 148 [0], given: 0

Senior Manager
Joined: 11 May 2006
Posts: 258

Kudos [?]: 25 [0], given: 0

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24 May 2006, 23:21
buckkitty 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

This question is from OG11. Can/should we assume that x, y and z are positive integers such that x<y<z?

i think there is no need for this assumption. The answer is C.

given x > 0, y >0, z > 0

A) x this is possible as the numbers can be y < x < z (or z < x < y)

B) z this is possible (similar reason as in A)

C) x + z
now x + z > x and x + z > z so x+z can not be the median as it is greater than 2 out of the 3 numbers

D) (x + z)/2 , y could be (x + z)/2 hence one seq could be x < y < z
so this could be a median

E) (x + z)/3 , this is also possible (same reason as D)

Kudos [?]: 25 [0], given: 0

SVP
Joined: 30 Mar 2006
Posts: 1728

Kudos [?]: 102 [0], given: 0

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25 May 2006, 00:19
iced_tea wrote:
buckkitty 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

This question is from OG11. Can/should we assume that x, y and z are positive integers such that x<y<z?

i think there is no need for this assumption. The answer is C.

given x > 0, y >0, z > 0

A) x this is possible as the numbers can be y < x < z (or z < x < y)

B) z this is possible (similar reason as in A)

C) x + z
now x + z > x and x + z > z so x+z can not be the median as it is greater than 2 out of the 3 numbers

D) (x + z)/2 , y could be (x + z)/2 hence one seq could be x < y < z
so this could be a median

E) (x + z)/3 , this is also possible (same reason as D)

True we dont have to assume here.......

Even if we do you will see that x+z is always a greater than x and z
hence for 3 numbers the greatest number can never be the median

Kudos [?]: 102 [0], given: 0

Re: PS-Integers   [#permalink] 25 May 2006, 00:19
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