1999 AMC 8 Problems/Problem 10

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 $1- \frac{5}{12} = \boxed{\text{(E)}\ \frac{7}{12}}$ .

1999 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AJHSME/AMC 8 Problems and Solutions

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1999 AMC 8 Problems/Problem 10

Contents

Problem

Solution

Solution 1

Solution 2

See Also