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2010 AMC 8 Problems/Problem 10: Difference between revisions

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==See Also==
==See Also==
{{AMC8 box|year=2010|num-b=8|num-a=10}}
{{AMC8 box|year=2011|num-b=9|num-a=11}}

Revision as of 17:55, 4 November 2012

Problem

Six pepperoni circles will exactly fit across the diameter of a $12$-inch pizza when placed. If a total of $24$ circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?

$\textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78$

Solution

The pepperoni circles' diameter is 2, since $\frac{12}{6} = 2$. From that we see that the area of the $24$ circles of pepperoni is $(\frac{2}{2})^2*24\pi = 24\pi$. The large pizza's area is $6^2\pi$.

The ratio: $\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}$

See Also

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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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