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 14:04, 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 test 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

If the average of the first $14$ peoples' scores was $80$, then the sum of all of their tests is $14*80 = 1120$. When Payton's score was added, the sum of all of the scores became $15*81 = 1215$. So, Payton's score must be $1215-1120 = \boxed{\textbf{(E) }95}$