Question 980736
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You have an error in the way you have written this question.  In order for the numbers you have presented to be true, one of the classes would necessarily have a negative number of students.


Let *[tex \Large a] be the number of students in Class A.  Then the total of all scores in Class A must be *[tex \Large 83a] since the total of all scores in Class A divided by the number of students in Class A is 83 by definition of "average".


Likewise, if *[tex \Large b] is the number of students in Class B, the total of all scores in Class B must be *[tex \Large 76b].


And furthermore, since *[tex \Large a\ +\ b] is the number of students in Class A and Class B and the average of these two classes is 67, the total score of both classes must be *[tex \Large 67(a\ +\ b)].


And finally, the total scores of Class A plus the total scores of Class B must equal the total scores of both classes, that is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 83a\ +\ 76b\ =\ 67(a\ +\ b)]


A little algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (83\ -\ 67)a\ +\ (76\ -\ 67)b\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16a\ +\ 11b\ =\ 0]


and the only way for this statement to be true is if one of *[tex \Large a] and *[tex \Large b] is negative.  Division by zero precludes zero students in any of the classes.


A similar result is obtained when Class B and Class C are compared.


In order for this problem to work out is if the combined A and B average is a number between 83 and 76 and the combined B and C average is a number between 76 and 75.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \