Difference between revisions of "2011 AMC 10A Problems/Problem 14"

Revision as of 11:00, 4 July 2013

Problem 14

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?

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

Solution

We want the area, $\pi r^2$ , to be less than the circumference, $2 \pi r$ :

$\begin{align*} \pi r^2 &< 2 \pi r \\ r &< 2 \end{align*}$

If $r<2$ then the dice must show $(1,1),(1,2),(2,1)$ which are $3$ choices out of a total possible of $6 \times 6 =36$ , so the probability is $\frac{3}{36}=\boxed{\frac{1}{12} \ \mathbf{(B)}}$ .

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Revision as of 09:49, 8 May 2011 (view source) Kacheep (talk \| contribs) m (→‎Solution) ← Older edit		Revision as of 11:00, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 19:		Line 19:

	{{AMC10 box\|year=2011\|ab=A\|num-b=13\|num-a=15}}		{{AMC10 box\|year=2011\|ab=A\|num-b=13\|num-a=15}}
		+	{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "2011 AMC 10A Problems/Problem 14"

Revision as of 11:00, 4 July 2013

Problem 14

Solution

See Also