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== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Revision as of 16:45, 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
See also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
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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