2018 AMC 8 Problems/Problem 13

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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 1

Say Laila gets a value of $x$ on her first 4 tests, and a value of $y$ on her last test. Thus, $4x+y=82 \cdot 4=410$.

Because $x$ and $y$ are different, $x$ must be less than $82$ and $y$ must be greater than $82$. When $x$ decreases by $1$, $y$ must increase by $4$ to keep the total constant.

The greatest value for $y$ is $98$ (as $y=100$ would make $x$ non-integer). In the range $83\le y\le 98$, only $4$ values for $y$ result in integer values for $x$: 86, 90, 94 and 98. Thus, the answer is $\boxed{\textbf{(A) }4}$.

Solution 2

The average score is $82$ which leads us to suppose that Laila got all $82$ points for the tests. We know that Laila got the same points in the first four tests and they are all lower than the last test. Let the first four tests be $81$ points, then the last test must be $86$ points to keep the average fixed. Continue to decrement the first four tests to identify other possible combinations. The possible points for the fifth test are $86$, $90$, $94$, $98$. The answer is $\boxed{\textbf{(A) }4}$.

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See Also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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 AJHSME/AMC 8 Problems and Solutions

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