# 2015 AMC 10A Problems/Problem 5

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 $14 \cdot 80 = 1120$. When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$. 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 (CREATIVE THINKING)

https://youtu.be/TZtbmvFyGic


~Education, the Study of Everything

~savannahsolver

## Video Solution by OmegaLearn

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 