Question 975660
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Let *[tex \Large x] represent the number of students who pass and let *[tex \Large y] represent the number of students who fail.  If the average score of the students who pass is 65, then the total of all scores of students who pass must be *[tex \Large 65x].  Likewise the total of all scores of students who fail is *[tex \Large 35y].  Then there are two ways to represent the total of all of the students' scores, either *[tex \Large 65x\ +\ 35y] or *[tex \Large 53(x\ +\ y)].  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 65x\ +\ 35y\ =\ 53(x\ +\ y)]


A little algebra music, Sammy,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12x\ =\ 18y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x\ =\ 3y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x}{y}\ =\ \frac{3}{2}]


Hence there are 3 passing scores for every 2 failure scores, hence there are a multiple of 5 scores and the same multiple of 3 passing scores.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{5}\ =\ 60%]


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

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