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2013 AMC 8 Problems/Problem 24

Revision as of 08:46, 27 November 2013 by Flyrain (talk | contribs) (→‎Problem)

Problem

Solution

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball, you see that in A and C it loses $2\pi*2/2=2\pi$ inches, and it gains $2\pi$ inches on B. So, the departure from the length of the track means that the answer is $\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}$.

See Also

2013 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 AJHSME/AMC 8 Problems and Solutions

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