Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 02 Dec 2012
Posts: 172

For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
07 Dec 2012, 05:38
Question Stats:
85% (01:01) correct 15% (01:08) wrong based on 1815 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 64947

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
07 Dec 2012, 05:42
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.
_________________




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]
Show Tags
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.



Manager
Joined: 07 Apr 2014
Posts: 95

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



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1706
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
30 Sep 2014, 22:11
\(Mean = \frac{5n+15}{3} = n+3\)
Median = n+2
Difference = 1
Answer = B



Target Test Prep Representative
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]
Show Tags
09 Jun 2016, 13:04
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.
_________________
5star rated online GMAT quant self study course
See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews



Current Student
Joined: 10 Mar 2013
Posts: 448
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
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 > 54=1 (B)



Director
Joined: 02 Sep 2016
Posts: 621

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
01 Apr 2017, 08:24
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) MeanMedian= n+3n2=1



Board of Directors
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 5010
Location: India
GPA: 3.5
WE: Business Development (Commercial Banking)

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
08 Aug 2018, 06:38
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)
_________________



NonHuman User
Joined: 09 Sep 2013
Posts: 15370

Re: For the positive numbers, n, n + 1, n + 2, n + 4, and n + 8
[#permalink]
Show Tags
27 Sep 2019, 06:44
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




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




