# 2018 AMC 8 Problems/Problem 13

## Problem 13

Laila took five math tests, each worth a maximum of 100 points. Laila's score on each test was an integer between 0 and 100, inclusive. Laila received the same score on the first four tests, and she received a higher score on the last test. Her average score on the five tests was 82. How many values are possible for Laila's score on the last test? $\textbf{(A) }4\qquad\textbf{(B) }5\qquad\textbf{(C) }9\qquad\textbf{(D) }10\qquad \textbf{(E) }18$

## Solution

Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=410.$

The value $y$ has to be greater than 82, because otherwise she would receive the same score on her last test. Additionally, the greatest value for y is 98 (as y=100 would give x as a decimal), so therefore the greatest value $x$ can be is 98. As a result, only $4$ numbers work, $86, 90, 94$ and $98$. Thus the answer is $\boxed{\textbf{(A) }4}$.

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