2010 AMC 8 Problems/Problem 9

Problem

Ryan got $80\%$ of the problems correct on a $25$-problem test, $90\%$ on a $40$-problem test, and $70\%$ on a $10$-problem test. What percent of all the problems did Ryan answer correctly?

$\textbf{(A)}\ 64 \qquad\textbf{(B)}\ 75\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 84\qquad\textbf{(E)}\ 86$

Solution

Ryan answered $(0.8)(25)=20$ problems correct on the first test, $(0.9)(40)=36$ on the second, and $(0.7)(10)=7$ on the third. This amounts to a total of $20+36+7=63$ problems correct. The total number of problems is $25+40+10=75.$ Therefore, the percentage is $\dfrac{63}{75} \rightarrow \boxed{\textbf{(D)}\ 84}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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