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

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Joined: 02 Dec 2012
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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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07 Dec 2012, 05:38
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5% (low)

Question Stats:

85% (01:01) correct 15% (01:08) wrong based on 1776 sessions

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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
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Posts: 64314
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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07 Dec 2012, 05:42
6
4
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}.

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

$$Median=middle \ term=n+2$$;

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

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

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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]

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19 Jun 2014, 12:16
1
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]

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

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30 Sep 2014, 22:11
1
$$Mean = \frac{5n+15}{3} = n+3$$

Median = n+2

Difference = 1

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

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09 Jun 2016, 13:04
1
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

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

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01 Apr 2017, 08:24
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

Mean=[5(n+3)]/5= n+3

Median is the 3rd terms (n+2)

Mean-Median= n+3-n-2=1
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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08 Aug 2018, 06:38
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

Plug in some value for $$n$$, say $$n = 1$$

Thus, the numbers in the sequence are : $$1 , 2 , 3 , 5 , 9$$

Median is 3

$$Mean = \frac{1 + 2 +3 + 5 + 9}{5}$$ = $$4$$

So, We have Mean > Mean by 1 , Answer must be (B)
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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27 Sep 2019, 06:44
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8   [#permalink] 27 Sep 2019, 06:44