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01 Apr 2020, 02:12
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
32% (02:25) correct
68%
(01:56)
wrong
based on 98
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[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. \(8\)
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. \(8\)
01 Apr 2020, 13:21
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base = '\(a_i\)'
height= '\(10-a_i\)'
Average area = \(\frac{1}{2n} Σ a_i * (10-a_i)\)
Average area = \(\frac{1}{n} [5Σ a_i - \frac{1}{2}Σ(a_i)^2]\)
Average area = \(5*8 -\frac{1}{2n} Σ(a_i)^2\).........(1)
Also SD = 1
SD = \( \sqrt{\frac{Σ(8-a_i)^2}{n}}\) = 1
\(\frac{Σ(8-a_i)^2}{n} = 1\)
\(\frac{1}{n}Σ64 + \frac{1}{n} Σ(a_i)^2 - \frac{16}{n}Σa_i = 1\)
\(64+ \frac{1}{n} Σ(a_i)^2 - 16*8 = 1\)
\(\frac{1}{n} Σ(a_i)^2 = 65\)........(2)
From (1) and (2)
Average area = \(5*8 -\frac{65}{2} = 7.5 \)
height= '\(10-a_i\)'
Average area = \(\frac{1}{2n} Σ a_i * (10-a_i)\)
Average area = \(\frac{1}{n} [5Σ a_i - \frac{1}{2}Σ(a_i)^2]\)
Average area = \(5*8 -\frac{1}{2n} Σ(a_i)^2\).........(1)
Also SD = 1
SD = \( \sqrt{\frac{Σ(8-a_i)^2}{n}}\) = 1
\(\frac{Σ(8-a_i)^2}{n} = 1\)
\(\frac{1}{n}Σ64 + \frac{1}{n} Σ(a_i)^2 - \frac{16}{n}Σa_i = 1\)
\(64+ \frac{1}{n} Σ(a_i)^2 - 16*8 = 1\)
\(\frac{1}{n} Σ(a_i)^2 = 65\)........(2)
From (1) and (2)
Average area = \(5*8 -\frac{65}{2} = 7.5 \)
MathRevolution
03 Apr 2020, 02:50
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=>
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
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
