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

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Walkabout
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For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  07 Dec 2012, 06:38
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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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Bunuel
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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, 06:42
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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  07 Feb 2021, 20:24
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BrentGMATPrepNow
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  27 Aug 2020, 11:37
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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


To calculate the median, arrange the numbers in ascending order: {n, n + 1, n + 2, n + 4, n + 8}
Since we have an ODD number of values, the median is the middlemost term
Median = n + 2

The mean = [n + (n+1) + (n+2) + (n+4) + (n+8)]/5
= (5n + 15)/5
= n + 3

The mean is how much greater than the median?
Difference = (n + 3) - (n + 2)
= 1

Answer: B

Cheers,
Brent
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ScottTargetTestPrep
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  09 Jun 2016, 14:04
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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


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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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  19 Jun 2014, 12:07
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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, 13:16
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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, 10:32
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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



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] New post  30 Sep 2014, 23:11
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\(Mean = \frac{5n+15}{3} = n+3\)

Median = n+2

Difference = 1

Answer = B
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Shiv2016
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  01 Apr 2017, 09:24
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


Add all the terms and the answer is 5n+15
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] New post  08 Aug 2018, 07:38
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


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] New post  05 Jan 2021, 06:22
Top Contributor
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


The ascending order of the number is: \(n, n+1, n+2, n+4, n+8\)

The median \(= n+2\)

The average: \(\frac{n+ n+1+n+2+n+4+n+8}{5}=\frac{5n+15}{5}=\frac{5(n+3)}{5}=n+3\)

The difference \(=n+3-n-2=1\)

The answer is \(B\)
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  18 Oct 2021, 07:47
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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


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Answer: Option B

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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink] New post  28 May 2023, 14:06
To find the mean and median for the given sequence of positive numbers, let's calculate them:

Mean = (n + n + 1 + n + 2 + n + 4 + n + 8) / 5 = (5n + 15) / 5 = n + 3.

Now let's arrange the numbers in ascending order:

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

To find the median, we need to determine the middle value. Since there are five numbers in the sequence, the middle value will be the third number, which is n + 2.

The difference between the mean and the median is:

Mean - Median = (n + 3) - (n + 2) = n + 3 - n - 2 = 1.

Therefore, the mean is 1 greater than the median.

Hence, the correct answer is (B) 1.
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8 [#permalink]
28 May 2023, 14:06
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