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 17:45, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
