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There are n triangles whose summations of the length of the base and the length of the height are both 10. The average

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There are n triangles whose summations of the length of the base and the length of the height are both 10. The average

by Max@Math Revolution » Wed Apr 01, 2020 1:05 am
[GMAT math practice question]

There are n triangles whose summations of the length of the base and the length of the height are both 10. The average of the lengths of bases is 8, and the standard deviation is 1. What is the average of the areas of all the triangles?

A. 6
B. 6.5
C. 7
D. 7.5
E. 8
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Re: There are n triangles whose summations of the length of the base and the length of the height are both 10. The aver

by Max@Math Revolution » Fri Apr 03, 2020 1:38 am
=>

Assume a1, a2, …, an are the base lengths of n triangles and b1, b2, … , bn are their height lengths.
Then a1 + b1 = 10, a2 + b2 = 10, …, an + bn = 10.
We have ai + bi = 10 for i = 1, 2, … , n.

We have a1 + a2 + … + an = 8n, since the average of a1, a2, …, an is 8.
We have ((a1^2 + a2^2 + …an^2) / n) – 8^2 = 1^2 or a1^2 + a2^2 + … an^2 = 65n.

The average of the areas of the triangles is
[(1/2)a1b1 + (1/2)a2b2 + … + (1/2)anbn] / n
=[a1b1 + a2b2 + … + anbn] / (2n)
= [a1(10 - a1) + a2(10 - a2) + … + an(10 - an)] / (2n)
= [10(a1 + a2 + … + an) - (a1^2 + a2^2 + … + an^2)] / (2n)
= (10*8n) / 2n – (65n) / 2n = 40 - 65/2 = 15/2 = 7.5

Therefore, D is the answer.
Answer: D
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