Difference between revisions of "1999 AMC 8 Problems/Problem 10"

 
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Latest revision as of 23:34, 4 July 2013

Problem

A complete cycle of a traffic 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?

$\text{(A)}\ \frac{1}{4} \qquad \text{(B)}\ \frac{1}{3} \qquad \text{(C)}\ \frac{5}{12} \qquad \text{(D)}\ \frac{1}{2} \qquad \text{(E)}\ \frac{7}{12}$

Solution

Solution 1

\[\frac{\text{time not green}}{\text{total time}} = \frac{R + Y}{R + Y + G} = \frac{35}{60} = \boxed{\text{(E)}\ \frac{7}{12}}\]

Solution 2

The probability of green is $\frac{25}{60} = \frac{5}{12}$, so the probability of not green is $1-  \frac{5}{12} = \boxed{\text{(E)}\ \frac{7}{12}}$ .

See Also

1999 AMC 8 (ProblemsAnswer KeyResources)
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

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