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

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==Solution==
 
==Solution==
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The pepperoni circles' diameter is 2, since <math>\dfrac{12}{6} = 2</math>. From that we see that the area of the <math>24</math> circles of pepperoni is <math>\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi</math>. The large pizza's area is <math>6^2\pi</math>. Therefore, the ratio is <math>\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}</math>
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The pepperoni circles' diameter is <math>2</math>, since <math>\dfrac{12}{6} = 2</math>. From that we see that the area of the <math>24</math> circles of pepperoni is <math>\left ( \frac{2}{2} \right )^2 (24\pi) = 24\pi</math>. The large pizza's area is <math>6^2\pi</math>. Therefore, the ratio is <math>\frac{24\pi}{36\pi} = \boxed{\textbf{(B) }\frac{2}{3}}</math>
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 +
 
 +
 
 +
 
 +
 
 +
==Video Solution by @MathTalks==
 +
https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT
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==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=9|num-a=11}}
 
{{AMC8 box|year=2010|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:55, 22 January 2024

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



Video Solution by @MathTalks

https://youtu.be/RhyRqHMXvq0?si=m1R2q8UnLRD-KksT



See Also

2010 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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