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Difference between revisions of "2011 AMC 12A Problems/Problem 10"

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== Problem ==
 
== Problem ==
 
== Solution ==
 
== Solution ==
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It is clear that only a diameter of <math>2</math> and <math>3</math> would result in the circumference being larger than the radius.
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For <math>2,</math> the radius is <math>1</math> so <math>2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi </math> The area is <math> \pi r^2 \rightarrow \pi 1^2 \right arrow \pi </math> Thus, <math> 2\pi > \pi </math> so we need snake eyes or <math>2</math> <math>1's</math> and the probability is <math> \dfrac{1}{6} \cdot  \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>
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By the same work using diameter of <math>3</math>, we find that the circumference is greater than the area. So <math>3</math> would be found by rolling a <math>1</math> and a <math>2</math> so the probability is  <math> \dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>.
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Thus, those are the only two cases however there are <math>2</math> ways to roll a <math>3</math> with <math>2</math> die so the probability is  <math>  \dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} \rightarrow \dfrac{3}{36} \rightarrow \dfrac{1}{12} \rightarrow \boxed{B} </math>
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}
 
{{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}

Revision as of 21:41, 9 February 2011

Problem

Solution

It is clear that only a diameter of $2$ and $3$ would result in the circumference being larger than the radius.

For $2,$ the radius is $1$ so $2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi$ The area is $\pi r^2 \rightarrow \pi 1^2 \right arrow \pi$ (Error compiling LaTeX. Unknown error_msg) Thus, $2\pi > \pi$ so we need snake eyes or $2$ $1's$ and the probability is $\dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36}$

By the same work using diameter of $3$, we find that the circumference is greater than the area. So $3$ would be found by rolling a $1$ and a $2$ so the probability is $\dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36}$.

Thus, those are the only two cases however there are $2$ ways to roll a $3$ with $2$ die so the probability is $\dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} \rightarrow \dfrac{3}{36} \rightarrow \dfrac{1}{12} \rightarrow \boxed{B}$

See also

2011 AMC 12A (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 AMC 12 Problems and Solutions
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