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by gmattester » Thu Aug 21, 2008 11:02 pm
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

OA is A but according to me answer should be E
I searched the previous posts but couldn't find it
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by parallel_chase » Fri Aug 22, 2008 12:31 am
Answer should be A

if you take 2,4,6,8,10
average is 6
eliminate D and E

if you take -4,-2,0,2,4
average 0
eliminate B and C
A is left

Hence A is the answer.
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by gmattester » Fri Aug 22, 2008 5:45 am
parallel_chase wrote:Answer should be A

if you take 2,4,6,8,10
average is 6
eliminate D and E

if you take -4,-2,0,2,4
average 0
eliminate B and C
A is left

Hence A is the answer.
Thanks I was thinking in wrong way.....
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by Stuart@KaplanGMAT » Fri Aug 22, 2008 10:25 am
parallel_chase wrote:Answer should be A

if you take 2,4,6,8,10
average is 6
eliminate D and E

if you take -4,-2,0,2,4
average 0
eliminate B and C
A is left

Hence A is the answer.
I agree with your number picking, but your answer choice elimination has me confused!

From {2, 4, 6, 8, 10}, we have x = 6.

So, we can eliminate any choice that contains:

III x is a multkiple of 5

eliminate: (b), (d) and (e).

The two remaining choices are I only and I and II only. So, we automatically know that I must be true. Now we want to pick numbers to show that II need not be true:

{-4, -2, 0, 2, 4} gives us x = 0.

Now we can eliminate any choice containing:

II x is a non-zero integer.

Eliminate (c), choose (a).
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