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 22:41, 9 February 2011
Problem
Solution
It is clear that only a diameter of and would result in the circumference being larger than the radius.
For the radius is so The area is $\pi r^2 \rightarrow \pi 1^2 \right arrow \pi$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) Thus, so we need snake eyes or and the probability is
By the same work using diameter of , we find that the circumference is greater than the area. So would be found by rolling a and a so the probability is .
Thus, those are the only two cases however there are ways to roll a with die so the probability is
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 |