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Manager  Joined: 02 Dec 2012
Posts: 172
For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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5
14 00:00

Difficulty:   5% (low)

Question Stats: 85% (01:01) correct 15% (01:08) wrong based on 1815 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
Math Expert V
Joined: 02 Sep 2009
Posts: 64947
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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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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Manager  Joined: 24 Oct 2012
Posts: 60
WE: Information Technology (Computer Software)
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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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.
Manager  Joined: 07 Apr 2014
Posts: 95
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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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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Posts: 1706
Location: India
Concentration: General Management, Technology
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Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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1
$$Mean = \frac{5n+15}{3} = n+3$$

Median = n+2

Difference = 1

Target Test Prep Representative V
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11028
Location: United States (CA)
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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1
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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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)
Director  G
Joined: 02 Sep 2016
Posts: 621
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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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

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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Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 5010
Location: India
GPA: 3.5
Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8  [#permalink]

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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

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

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