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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8

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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post 07 Dec 2012, 05:38
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Question Stats:

75% (01:43) correct 24% (00:57) wrong based on 151 sessions
For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?

(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3
[Reveal] Spoiler: OA
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post 07 Dec 2012, 05:42
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Walkabout wrote:
For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8, the mean is how much greater than the median?

(A) 0
(B) 1
(C) n+l
(D) n+2
(E) n+3


Given set in ascending order is {n, n+1, n+2, n+4, n+8}.

Mean=\frac{n+(n + 1)+(n + 2)+(n + 4)+(n + 8)}{5}=n+3;

Median=middle \ term=n+2;

Difference=(n+3)-(n+2)=1.

Answer: B.
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post 21 Dec 2013, 21:29
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8   [#permalink] 21 Dec 2013, 21:29
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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8

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