1999 AMC 8 Problems/Problem 10
Revision as of 15:10, 4 November 2012 by Firestarly (talk | contribs)
problem
A complete cycle of a tra±c light takes 60 seconds. During each cycle the light is green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly chosen time, what is the probability that the light will NOT be green? (A) 1/4 (B) 1/3 (C) 5/12 (D) 1/2 (E) 7/12
solution
(E) 7/12: time not green/total time = (R + Y)/(R + Y + G) = (35/60) = (7/12)
OR
The probability of green is (25/60) = (5/12) So the probability of not green is 1- (5/12) = (7/12) .
See Also
| 1999 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions | ||
