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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:

80% (01:45) correct 20% (00:56) wrong based on 268 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] New post 19 Jun 2014, 11:07
Let's say n=2 than the set looks like this (2,3,4,6,10). The Average = 25/5=5 and the median is equal to 4 --> 5-4=1 (B)
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post 19 Jun 2014, 12:16
I did it similar to BrainLab .

plug in numbers, 1 for n.

mean = 1+2+3+5+9/5 = 20/5 = 4

median = 3

difference = 1

Plugin 2 for n

mean = 2+3+4+6+10/5 = 25/5 = 5

median = 4

difference = 1.
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post 11 Sep 2014, 09:32
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



if n=1 then 1, 2, 3, 5, 9

3 = median

mean = 20 / 5 = 4

difference =1
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8   [#permalink] 11 Sep 2014, 09:32
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