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A certain assistant professor in the Art History department at University X is being evaluated for promotion. One of the requirements is a performance score of at least 4.0. The performance score is the weighted average of 3 component scores: one for research, one for teaching, and one for service, with the scores for research and teaching each weighted 40% and the score for service weighted 20%. Each component score is between 0.0 and 5.0, inclusive.

Consistent with the given information, select for Minimum service score the least possible score the professor can receive for service and still achieve a performance score of at least 4.0, and select for Minimum research score the least possible score the professor can receive for research and still achieve a performance score of at least 4.0. Make only two selections, one in each column.


shvm_sin7
Hi, I did not understand as to why did we maximize teaching and research score to find minimum service score? Should'nt we minimize the both so as to gete the bare minimum of service score required which will guarantee us a weighted score of atleast 4? Please help

We are given that 0.4R + 0.4T + 0.2S must be greater than or equal to 4:

0.4R + 0.4T + 0.2S ≥ 4

General rule for such problems:

  • To maximize one quantity, minimize the others.
  • To minimize one quantity, maximize the others.

Minimizing S:

To find the minimum value of S for which 0.4R + 0.4T + 0.2S ≥ 4 holds, we need to maximize R and T. Since the maximum value for both R and T is 5, we get 0.4(5) + 0.4(5) + 0.2S ≥ 4, which simplifies to S ≥ 0. Thus, the minimum value of S is 0.

Minimizing R:

To find the minimum value of R for which 0.4R + 0.4T + 0.2S ≥ 4 holds, we maximize T and S. Again, the maximum values for T and S are both 5: 0.4R + 0.4(5) + 0.2(5) ≥ 4. This simplifies to R ≥ 2.5. Thus, the minimum value of R is 2.5.
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I didnt understand that why didnt we divide the equation by 3 i.e (.4x + .4y + .2z)/3 since it is weighted average
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I didnt understand that why didnt we divide the equation by 3 i.e (.4x + .4y + .2z)/3 since it is weighted average

No, what you are doing is normal average.

Weighted average is due to amount/weight of each. Here it is 40%, 40% and 20%, and that is why we are multiplying R, S and T by 0.4, 0.4 and 0.2.
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I didnt understand that why didnt we divide the equation by 3 i.e (.4x + .4y + .2z)/3 since it is weighted average
­In weighted averages you have to sum the weights and not the number of items
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Hi, I did not understand as to why did we maximize teaching and research score to find minimum service score? Should'nt we minimize the both so as to gete the bare minimum of service score required which will guarantee us a weighted score of atleast 4? Please help
GMATCoachBen
The key here is if we want to minimize one variable, we need to maximize the other variables:

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But arents we searching for the weighted average - so wouldnt we need to divide the weighted sum by 3?
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Weighted Average = ((Wi*Xi), i=1...n)/((Wi), i=1...n).

Here we don't divide by total observations, but sum of the weights.

In this case here the sum = 0.4+0.4+0.2 = 1.0

This makes the equation for weighted average = ((Wi*Xi), i=1...n) only.
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