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If x is the average (arithmetic mean) of 5 consecutive even

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If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 01 May 2012, 10:03
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If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

x= n+(n+2)+(n+4)+(n+6)+(n+8)/5

x = n+4

IMO I & II is correct
[Reveal] Spoiler: OA

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Re: If x is the average (arithmetic mean) of 5 consecutive even [#permalink] New post 01 May 2012, 14:41
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If x is the average (arithmetic mean) of 5 consecutive even integers, which of the following must be true?

I. x is an even integer.
II. x is a nonzero integer.
III. x is a multiple of 5.

(A) I only
(B) III only
(C) I and II only
(D) I and III only
(E) I, II, and III

First of all notice that we are asked which of the following MUST be true, not COULD be true.

Also notice that the average of 5 consecutive even integers equals to the median, so it's just a middle term. So, I must always be true: {even, even, even, even, even}

Next:
II. x is a nonzero integer. Not necessarily true, consider: {-4, -2, 0, 2, 4};
III. x is a multiple of 5. Not necessarily true, consider: {0, 2, 4, 6, 8}.

Answer: A.

Hope it's clear.

P.S. In your example if n=-4 then x=n+4=0, so II is not always true.
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Re: x is a nonzero integer [#permalink] New post 01 May 2012, 11:28
İİ is not correct, since nowhere mentioned that even integers MUST BE positive
I is true, since the avrg sum= (sum of even integers)/ 5 (odd integer)=even/odd=even

III is also out. u can check it , if u pick some integers
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Re: x is a nonzero integer   [#permalink] 01 May 2012, 11:28
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