Difference between revisions of "2015 AMC 10A Problems/Problem 5"
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Mr. Patrick teaches math to <math> 15 </math> students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was <math> 80 </math>. After he graded Payton's test, the test average became <math> 81 </math>. What was Payton's score on the test? | Mr. Patrick teaches math to <math> 15 </math> students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was <math> 80 </math>. After he graded Payton's test, the test average became <math> 81 </math>. What was Payton's score on the test? | ||
− | <math> \textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D) | + | <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== | ==Solution== |
Revision as of 18:46, 1 March 2015
- The following problem is from both the 2015 AMC 12A #3 and 2015 AMC 10A #5, so both problems redirect to this page.
Contents
[hide]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?
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
Alternate Solution
The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first students each scored
. If Payton also scored an
, the average would still be
. In order to increase the overall average to
, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of
more points, so Payton needs
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 |
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.