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

(Created page with '== Problem == == Solution == == See also == {{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}')
 
(Solution)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 
== Solution ==
 
== Solution ==
 +
 +
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.
 +
 +
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>
 +
 +
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>.
 +
 +
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>
 +
 
== 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 22:41, 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. ! Missing delimiter (. inserted).) 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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS