Difference between revisions of "1999 AMC 8 Problems/Problem 10"
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− | == | + | ==Problem== |
− | A complete cycle of a | + | 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? |
− | 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? | ||
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− | + | <math>\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}</math> | |
− | + | ==Solution== | |
− | time not green | + | ===Solution 1=== |
+ | <cmath>\frac{\text{time not green}}{\text{total time}} | ||
= | = | ||
− | + | \frac{R + Y}{R + Y + G} | |
= | = | ||
− | + | \frac{35}{60} | |
= | = | ||
− | (7/ | + | \boxed{\text{(E)}\ \frac{7}{12}}</cmath> |
− | |||
− | + | ===Solution 2=== | |
The probability of green is | The probability of green is | ||
− | + | <math>\frac{25}{60} | |
= | = | ||
− | + | \frac{5}{12}</math>, | |
− | + | so the probability of not green is <math>1- \frac{5}{12} = | |
− | (7/ | + | \boxed{\text{(E)}\ \frac{7}{12}}</math> |
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{{AMC8 box|year=1999|num-b=9|num-a=11}} | {{AMC8 box|year=1999|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:34, 5 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?
Solution
Solution 1
Solution 2
The probability of green is , so the probability of not green is .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.