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Difference between revisions of "2006 AMC 8 Problems/Problem 12"

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Revision as of 00:13, 5 July 2013

Problem

Antonette gets $70 \%$ on a 10-problem test, $80 \%$ on a 20-problem test and $90 \%$ on a 30-problem test. If the three tests are combined into one 60-problem test, which percent is closest to her overall score?

$\textbf{(A)}\ 40\qquad\textbf{(B)}\ 77\qquad\textbf{(C)}\ 80\qquad\textbf{(D)}\ 83\qquad\textbf{(E)}\ 87$

Solution

$70 \% \cdot 10=7$

$80 \% \cdot 20=16$

$90 \% \cdot 30=27$

Adding them up gets $7+16+27=50$. The overall percentage correct would be $\frac{50}{60}=\frac{5}{6}=5 \cdot 16.\overline{6}=83.\overline{3} \approx \boxed{\textbf{(D)}\ 83}$.

See Also

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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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