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Harmonic Mean Solution?

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by student22 » Fri May 14, 2010 8:51 am

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Answers

A

B

C

D

E

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Difficulty—

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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by ilikaroy » 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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by Stuart@KaplanGMAT » 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.
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by student22 » 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.
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by frank1 » 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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by [email protected] » Sun Apr 08, 2018 3:24 pm
Hi All,

We're asked for the RATIO of the average (arithmetic mean) height of students in Class X to the average height of students in Class Y. This is a great example of a 'concept question', meaning that you don't have to do much math to answer it if you recognize the concept(s) involved.

1) The average height of the students in Class X is 120 centimeters.

Fact 1 gives us the average height for Class X, but tells us nothing about the average height of Class Y.
Fact 1 is INSUFFICIENT

2) The average height of the students in Class X AND Class Y combined is 126 centimeters.

Fact 2 tells us the average height of ALL students, but we have know way to determine the average heights in just Class X or just Class Y.
Fact 2 is INSUFFICIENT

Combined, we know
-The average height of the students in Class X is 120 centimeters.
-The average height of the students in Class X AND Class Y combined is 126 centimeters.

We don't know the NUMBER of students in each class, so the information in Fact 2 will 'skew' based on the number of students in Class X vs. the number in Class Y. For example, if we have an EQUAL number of students in both classes, then the average height for Class Y would be (120+Y)/2 = 126 --> Y = 132 centimeters. However, if the number of students is NOT equal, then the average height of Class Y will change (it could be higher or lower depending on whether Class X had more or less students). Thus, there's no way to determine the exact ratio of the average heights.
Combined, INSUFFICIENT

Final Answer: E

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