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Difference between revisions of "2015 AMC 10A Problems/Problem 5"
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~savannahsolver
 
~savannahsolver
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== Video Solution ==
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https://youtu.be/HISL2-N5NVg?t=3425
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~ pi_is_3.14
  
 
==See also==
 
==See also==

Revision as of 22:25, 27 January 2021

The following problem is from both the 2015 AMC 12A #3 and 2015 AMC 10A #5, so both problems redirect to this page.

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?

$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$

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 $1215-1120 = \boxed{\textbf{(E) }95}$


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 $14$ students each scored $80$. If Payton also scored an $80$, the average would still be $80$. In order to increase the overall average to $81$, we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ more points, so Payton needs $80+15 = \boxed{\textbf{(E) }95}$

Video Solution

https://youtu.be/ev6z2stLJkA

~savannahsolver

Video Solution

https://youtu.be/HISL2-N5NVg?t=3425

~ pi_is_3.14

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. AMC logo.png

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