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Harmonic Mean Solution?

This topic has 1 expert reply and 3 member replies
student22 Master | Next Rank: 500 Posts Default Avatar
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Harmonic Mean Solution?

Post Fri May 14, 2010 8:51 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    What is the ratio of the average (arithmetic mean) height of students in class X to the average height of students in class Y?
    (1) The average height of the students in class X is 120 centimeters.
    (2) The average height of the students in class X and class Y combined is 126 centimeters.

    OA:E

    I know that this problem has been beaten to death, but I have a question. Why can't you use a harmonic mean to solve it?

    I got C by using a harmonic mean. Since, technically an average height could be considered a rate, right? (Total height / number of students).

    Harmonic mean = (2*(120)*Y)/(120 + Y) = 126. --> Then you simply solve for Y.

    Can somebody explain to me, why the harmonic mean is not applicable in this scenario.

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    ilikaroy Junior | Next Rank: 30 Posts Default Avatar
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    Post Sat May 15, 2010 9:21 pm
    When the question is stating clearly that average height is arithmetic mean, how can you possibly take harmonic mean?

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    Stuart Kovinsky GMAT Instructor
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    Post Sun May 16, 2010 2:40 pm
    student22 wrote:
    What is the ratio of the average (arithmetic mean) height of students in class X to the average height of students in class Y?
    (1) The average height of the students in class X is 120 centimeters.
    (2) The average height of the students in class X and class Y combined is 126 centimeters.

    OA:E

    I know that this problem has been beaten to death, but I have a question. Why can't you use a harmonic mean to solve it?

    I got C by using a harmonic mean. Since, technically an average height could be considered a rate, right? (Total height / number of students).

    Harmonic mean = (2*(120)*Y)/(120 + Y) = 126. --> Then you simply solve for Y.

    Can somebody explain to me, why the harmonic mean is not applicable in this scenario.
    Hi,

    to use a weighted average formula in this kind of question, you need to know the average of each group and the weight of each group. Since we don't know the weight of each group, there's no way to solve using harmonic mean.

    Picking numbers (that are in accord with all the information given):

    1 student in x with height 120, 1 student in y with height 132. Ratio of x:y is 120:132.

    1 student in x with height 120, 2 students in y with height 129. Ratio of x:y is 120:129.

    We can get different ratios, so we don't have enough information to solve the problem.

    Thanked by: student22
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    student22 Master | Next Rank: 500 Posts Default Avatar
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    Post Mon May 17, 2010 8:24 am
    Thanks Stuart for clarifying. I mistakenly assumed that the groups here would have equal weights, when there was no real basis for me to assume that.

    frank1 Legendary Member Default Avatar
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    Post Mon May 17, 2010 7:38 pm
    isn't the core of the question
    avg of x + avg of y is not equal to avg of x and y combined?

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