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

Revision as of 01:33, 10 February 2011

Problem

A pair of standard $6$ -sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?

$\textbf{(A)}\ \frac{1}{36} \qquad \textbf{(B)}\ \frac{1}{12} \qquad \textbf{(C)}\ \frac{1}{6} \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ \frac{5}{18}$

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 $\pi r^2 \rightarrow \pi 1^2 \rightarrow \pi$ 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 $\dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} \rightarrow \dfrac{3}{36} \rightarrow \dfrac{1}{12} \rightarrow \boxed{B}$

@@ Line 1: / Line 1: @@
 == Problem ==
-A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
+A pair of standard <math>6</math>-sided dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
-<math> \textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{12}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{4}\qquad\textbf{(E)}\ \frac{5}{18} </math>
+<math>
+\textbf{(A)}\ \frac{1}{36} \qquad
+\textbf{(B)}\ \frac{1}{12} \qquad
+\textbf{(C)}\ \frac{1}{6} \qquad
+\textbf{(D)}\ \frac{1}{4} \qquad
+\textbf{(E)}\ \frac{5}{18} </math>
 == Solution ==

2011 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2011 AMC 12A Problems/Problem 10"

Revision as of 01:33, 10 February 2011

Problem

Solution

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