Question 338239
Let y = the total number of students in Math 200.  Let {{{A = (1/5)x}}}, {{{C =(3/4)x}}}.  This is essentially a proportion problem.  Change the fractions into equivalent ones with common denominators.
{{{A = (4/20)x}}}, {{{C = (15/20)x}}}.  This leaves another {{{(1/20)x}}} unaccounted for.  Let's call that B,(it doesn't really matter). So the grades in the class are in the proportion 4:1:15, (A:B:C)
Whenever you want to solve a proportion problem, take the proportion numbers, multiply each of them by x and add them together:
{{{4x+1x+15x=20x}}} Now it's time to take into account the next piece of info:22 more students got Cs than As.  So we can change the C part of the equation (15x) into 4x+22.  Now substitute that into the equation:
{{{4x+1x+4x+22=20x}}} Solve for x.
{{{9x+22=20x}}}
{{{22=11x}}}
{{{2=x}}}
Now substitute 2 for x in the equation {{{4x+1x+15x=20x}}}.  You get {{{20*2=40}}}.
There are 40 students in Math 200.
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