Difference between revisions of "2010 AMC 8 Problems/Problem 10"

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<math> \textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78 </math>
 
<math> \textbf{(A)}\ \frac 12 \qquad\textbf{(B)}\ \frac 23 \qquad\textbf{(C)}\ \frac 34 \qquad\textbf{(D)}\ \frac 56 \qquad\textbf{(E)}\ \frac 78 </math>
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==Solution==
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The pepperoni circles' diameter is 2, since <math>\frac{12}{6} = 2</math>. From that we see that the area of the <math>24</math> circles of pepperoni is <math>(\frac{2}{2})^2*24\pi = 24\pi</math>. The large pizza's area is <math>6^2\pi</math>.
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The ratio: <math>\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}</math>
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==See Also==
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{{AMC8 box|year=2010|num-b=8|num-a=10}}

Revision as of 23:38, 3 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

2010 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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