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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink]

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21 Dec 2013, 22:29

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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink]

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12 Jan 2016, 17:35

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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

Let’s first calculate the mean (arithmetic average).

mean = sum/quantity

mean = (n + n + 1 + n + 2 + n + 4 + n + 8)/5

mean = (5n + 15)/5

mean = n + 3

Next, we determine the median. The median is the middle value when the terms are ordered from least to greatest. The terms ordered from least to greatest are as follows:

n, n + 1, n + 2, n + 4, n + 8

The median is n + 2.

Finally we are asked how much greater the mean is than the median. To determine the difference we can subtract the smaller value (the median) from the larger value (the mean) and we get:

n + 3 – (n + 2) = n + 3 – n – 2 = 1

The answer is B.
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