2018 AMC 8 Problems/Problem 13

Revision as of 19:50, 21 November 2018 by Ben chen (talk | contribs)

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 $x$ has to be less than 82, because then she would receive a lower score on her last test. Additionaly, the greatest value for y is 100, so therefore the smallest value $x$ can be is 78. As a result, only $4$ numbers work, $78, 79,80$ and $81$. Thus the answer is $\boxed{\textbf{(A) }4}$. - song2sons

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

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