Difference between revisions of "2015 AMC 10A Problems/Problem 5"
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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== | ||
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| + | If the average of the first <math>14</math> peoples' scores was <math>80</math>, then the sum of all of their tests is <math>14*80 = 1120</math>. When Payton's score was added, the sum of all of the scores became <math>15*81 = 1215</math>. So, Payton's score must be <math>1215-1120 = \boxed{\textbf{(E) }95}</math> | ||
Revision as of 13:04, 4 February 2015
Problem
Mr. Patrick teaches math to 
students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was 
. After he graded Payton's test, the test average became 
. 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
If the average of the first 
peoples' scores was , then the sum of all of their tests is 
. When Payton's score was added, the sum of all of the scores became 
. So, Payton's score must be 
