Difference between revisions of "2015 AMC 10A Problems/Problem 5"

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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>
 
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>
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==See also==
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{{AMC10 box|year=2015|ab=A|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 16:45, 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 $1215-1120 = \boxed{\textbf{(E) }95}$

See also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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