Bunuel wrote:
In a class, there are 40 students. In a test, the average (arithmetic mean) score of the girls is 30, that of boys is 40, and that of the class is 32. If the score of a boy was incorrectly computed as 30 for 40, what is the correct average (arithmetic mean) score of the class?
(A) 31
(B) 31.50
(C) 32.25
(D) 41
(E) 42
Before we even make any actual computation, we note that since one of the grades was actually higher than the grade used in calculation, the correct average has to be greater than the computed average, so we can eliminate A and B. Furthermore, in a class of 40 people, a difference of 10 points in one of the grades cannot change the average grade by 9 or 10 points, so we can eliminate D and E as well. The only answer choice that makes sense here is C. Let’s verify that.
First, let’s let g = the number of girls and b = the number of boys in the class. We have:
g + b = 40
We create the weighted average equation:
(30g + 40b)/40 = 32
Multiplying both equations by 40, we have:
40g + 40b = 1600
and
30g + 40b = 1280
Subtracting these two equations, we have:
10g = 320
g = 32
So b = 8.
Since one of the boys’ scores should be 40 instead of 30, then the sum of the boys’ scores should be 8 x 40 + 10 = 330. Therefore, the correct average of the entire class should be:
(32 x 30 + 330)/40 = 1290/40 = 32.25
Answer: C