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## Test Code 52, Section 6, Question 10

irfan_m1973 Junior | Next Rank: 30 Posts
Joined
13 Nov 2009
Posted:
18 messages

#### Test Code 52, Section 6, Question 10

Wed Dec 02, 2009 1:30 am
For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3

### GMAT/MBA Expert

Stuart Kovinsky GMAT Instructor
Joined
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Wed Dec 02, 2009 2:36 am
irfan_m1973 wrote:
For the positive numbers, n, n + 1, n + 2, n + 4, and n+ 8, the mean is how much greater the median?

(A) 0
(B) 1
(C) n + 1
(D) n + 2
(E) n + 3
Picking numbers is a great way to attack this question; we can also solve algebraically.

The median of a set with an odd number of terms is simply the middle term in the set, in this case n+2.

To solve for the mean, we use the average formula:

Average = sum of terms / # of terms

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

We want to know by how much the mean is greater than the median, so:

mean - median = (n + 3) - (n + 2) = 1... choose (B).

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papgust Community Manager
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Wed Dec 02, 2009 1:36 am
The mean will always be 1 greater than the median in this list of numbers. Hence, B

Try to solve this by substituting values of n.

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