Difference between revisions of "2013 AMC 8 Problems/Problem 24"

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==Solution==
 
==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 <math>2\pi*2/2=2\pi</math> inches, and it gains <math>2\pi</math> inches on B. So, the departure from the length of the track means that the answer is <math>\frac{200+120+160}{2}*\pi+(-2-2+2)*\pi=240\pi-2\pi=\boxed{\textbf{(A)}\ 238\pi}</math>.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2013|num-b=23|num-a=25}}
 
{{AMC8 box|year=2013|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 09:46, 27 November 2013

Problem

Solution

See Also

2013 AMC 8 (ProblemsAnswer KeyResources)
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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