2010 AMC 8 Problems/Problem 10: Difference between revisions
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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> | ||
==Solution== | |||
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>. | |||
The ratio: <math>\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}</math> | |||
==See Also== | |||
{{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 
-inch pizza when placed. If a total of 
circles of pepperoni are placed on this pizza without overlap, what fraction of the pizza is covered by pepperoni?
Solution
The pepperoni circles' diameter is 2, since 
. From that we see that the area of the circles of pepperoni is 
. The large pizza's area is 
.
The ratio: 
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| 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 | ||
