2010 AMC 8 Problems/Problem 10

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 $\dfrac{12}{6} = 2$. From that we see that the area of the $24$ circles of pepperoni is $\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi$. The large pizza's area is $6^2\pi$. Therefore, the ratio is $\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}$

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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