Difference between revisions of "2015 AMC 12A Problems/Problem 3"
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<math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}}\ 94\qquad\textbf{(E)}\ 95</math> | <math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}}\ 94\qquad\textbf{(E)}\ 95</math> | ||
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| + | ==Solution== | ||
| + | We may assume the 14 students all scored 80%. Thus, for the average of 15 students to be 81%, each of the 14 students is one percent lower (total 14% lower), so the one other student must be 14% higher to compensate. 81+14=<math>\textbf{(E)}\ 95</math> | ||
Revision as of 19:34, 4 February 2015
Problem
Mr. Patrick teaches math to 15 students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 80. After he graded Payton's test, the class average became 81. What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}}\ 94\qquad\textbf{(E)}\ 95$ (Error compiling LaTeX. Unknown error_msg)
Solution
We may assume the 14 students all scored 80%. Thus, for the average of 15 students to be 81%, each of the 14 students is one percent lower (total 14% lower), so the one other student must be 14% higher to compensate. 81+14=
