1999 AMC 8 Problems/Problem 10: Difference between revisions
Firestarly (talk | contribs) No edit summary |
No edit summary |
||
| (One intermediate revision by one other user not shown) | |||
| Line 1: | Line 1: | ||
== | ==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? | |||
<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> | ||
. | . | ||
| Line 40: | Line 28: | ||
{{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 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?
Solution
Solution 1
![\[\frac{\text{time not green}}{\text{total time}} = \frac{R + Y}{R + Y + G} = \frac{35}{60} = \boxed{\text{(E)}\ \frac{7}{12}}\]](http://latex.artofproblemsolving.com/c/d/9/cd9d03f32bfb35e21abb172a8727c5ecfb98f865.png)
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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
