A college chemistry course is divided in two sections. In section A, the average score was 92,while, in section B, the average score was 84. If the average score of all 40 students was 89, how many students are in section A?
Main principle: If the mean of X numbers is M, then the sum of all X numbers is XMklaud wrote:A college chemistry course is divided in two sections. In section A, the average score was 92,while, in section B, the average score was 84. If the average score of all 40 students was 89, how many students are in section A?
Let A = # of students in class A
Let B = # of students in class B
Sum of scores in class A = 92A (since the mean is 92)
Sum of scores in class B = 84B (since the mean is 84)
Sum of scores in classes A and B combined = (89)(40) (since the mean is 89 and there are 40 students)
. . . So, 92A + 84B = (89)(40)
We also know that A + B = 40 (40 students in total)
Now we solve the system of equations:
92A + 84B = (89)(40)
A + B = 40
Multiply both sides of blue equation by 84 to get
92A + 84B = (89)(40)
84A + 84B = (84)(40)
Subtract blue equation from green equation to get: 8A = (5)(40)
Divide both sides by 8 to get: A = (5)(5) = 25
There are 25 students in section A
Cheers,
Brent
