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Test Code 52, Section 6, Question 10

This topic has 1 expert reply and 1 member reply
irfan_m1973 Junior | Next Rank: 30 Posts Default Avatar
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Test Code 52, Section 6, Question 10

Post Wed Dec 02, 2009 1:30 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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

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    Post 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.

    Thanked by: irfan_m1973

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    Stuart Kovinsky GMAT Instructor
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    Post 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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