Monday, June 8, 2009

Problem 34 - GMAT Statistics

Set C is a set of positive integers where 4 is least. If the average (arithmetic mean) of C equals its range, then how many positive integers does set C have?

(1) The elements in C are consecutive integers.

(2) The average (arithmetic mean) of C is 8.

8 comments:

  1. Weird question

    (1) avg of consecutive int = range = int
    So, the set size must be odd. Hence, the avg is odd if it starts with 4 (or any even). However, the range is even, given 4 is the least. This is a contradiction.

    (2) least 4, max 12, avg 8. From this, we can say the avg of remaining set, whose elements can be from 5 to 11 is 8. Insuff


    Check ur question pls

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  2. Hey Blaoism, I think you are wrong, see that the
    set can be 4 5 6 7 8 9 10 11 12, and you are right, the number of terms is odd, because there
    is a central element, but the average does not
    have to be odd, as in this set (notice that both
    average and range are eight.

    The formula to arrive at shi set is, if you call n the number of elements:
    (4 + 4 + n - 1)/2 = 4 + n - 1 -4, and n = 8.

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  4. I think the answer is A.

    (2) Avg. is 8, highest element = 4 + 8 = 12

    Set Possible = {4, 12}
    or, { 4, 8, 12 } or, {4,5,6,7,8,9,10,11,12}
    -> insufficient

    (1) Numbers are consecutive.

    Set can be represented as {4, 4+n, 4+2n.... 4+kn}
    (n=1, consecutive)
    no. of elements = k+1

    Range = 4+kn-4 = kn
    Avg. = (4 + 4+kn)/2 = (8+kn)/2
    Range = avg.
    kn = 8 => k = 8

    No. of terms = 9

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  6. Explanation of answer choice A
    F = First number
    L = Last number
    1:
    AVG: (F+L)/2
    Range = L-F
    L=3F or L=12
    4,5,6,7,8,9,10,11,12 : Avg =8, Range=8
    2: Average is 8 . Not sufficient (Eventhough it is the right average) Per this the soln could be
    4,8,12, or 4,5,6,7,8,9,10,11,12

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  7. @vikram: Pl note that avg= (F+L)/2 only if they are said to be consecutive...

    eg: 2,6,8 avg is not 5 but 5.333

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