document.write( "Question 338239: I'm sure i can figure this out if you can just tell me how to translate this into a mathematical equation:\r
\n" ); document.write( "\n" ); document.write( "three-fourths of the students in Math 200 recieved a grade of C on their test, and one-fifth recieved an A. if twenty-two more students recieved C's than recieved A's, how many students are in the class?\r
\n" ); document.write( "\n" ); document.write( "Thanks!
\n" ); document.write( "

Algebra.Com's Answer #242450 by katealdridge(100)\"\" \"About 
You can put this solution on YOUR website!
Let y = the total number of students in Math 200. Let \"A+=+%281%2F5%29x\", \"C+=%283%2F4%29x\". This is essentially a proportion problem. Change the fractions into equivalent ones with common denominators.
\n" ); document.write( "\"A+=+%284%2F20%29x\", \"C+=+%2815%2F20%29x\". This leaves another \"%281%2F20%29x\" 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)
\n" ); document.write( "Whenever you want to solve a proportion problem, take the proportion numbers, multiply each of them by x and add them together:
\n" ); document.write( "\"4x%2B1x%2B15x=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:
\n" ); document.write( "\"4x%2B1x%2B4x%2B22=20x\" Solve for x.
\n" ); document.write( "\"9x%2B22=20x\"
\n" ); document.write( "\"22=11x\"
\n" ); document.write( "\"2=x\"
\n" ); document.write( "Now substitute 2 for x in the equation \"4x%2B1x%2B15x=20x\". You get \"20%2A2=40\".
\n" ); document.write( "There are 40 students in Math 200.
\n" ); document.write( "If you have any further questions please check out my facebook page: Kate Calendrillo, tutor. I'd be happy to answer your questions. If you are satisfied with my response, please \"like\" my page on facebook.
\n" ); document.write( "
\n" );