2018 AMC 8 Problems/Problem 13

Problem

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, $4x+y=410.$

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 make $x$ 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 $\boxed{\textbf{(A) }4}$ .

Solution 2

The average point is $82$ 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 is $81$ points, then the last tests should be $86$ points to keep the average point is fixed. The possible points are $86$ , $90$ , $94$ , $98$ . The answer is $\boxed{\textbf{(A) }4}$ .

Video Solution (CREATIVE ANALYSIS!!!)

https://youtu.be/LP7PQSvGLtA

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Video Solution by OmegaLearn

https://youtu.be/HISL2-N5NVg?t=3251

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Video Solutions

https://youtu.be/viShU5koGtk

https://youtu.be/Olg8fUc1KQI

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2018 AMC 8 (Problems • Answer Key • Resources)
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

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

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