2018 AMC 8 Problems/Problem 13

Revision as of 17:33, 11 November 2019 by Goldenn (talk | contribs) (Solution)

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 Lalia 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 $\boxed{\textbf{(A) }4}$.

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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