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

Line 1: Line 1:
==problem==
+
==Problem==
  
A complete cycle of a tra±c light takes 60 seconds. During each cycle the light is
+
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?
 
(A)
 
1/4
 
(B)
 
1/3
 
(C)
 
5/12
 
(D)
 
1/2
 
(E)
 
7/12
 
  
==solution==
+
<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>
  
(E) 7/12:
+
==Solution==
time not green/total time
+
===Solution 1===
 +
<cmath>\frac{\text{time not green}}{\text{total time}}
 
=
 
=
(R + Y)/(R + Y + G)
+
\frac{R + Y}{R + Y + G}
 
=
 
=
(35/60)
+
\frac{35}{60}
 
=
 
=
(7/12)
+
\boxed{\text{(E)}\ \frac{7}{12}}</cmath>
:
 
  
OR
+
===Solution 2===
  
 
The probability of green is
 
The probability of green is
(25/60)
+
<math>\frac{25}{60}
 
=
 
=
(5/12)
+
\frac{5}{12}</math>,
So the probability of not green is 1-  (5/12) =
+
so the probability of not green is <math>1-  \frac{5}{12} =
(7/12)
+
\boxed{\text{(E)}\ \frac{7}{12}}</math>
 
.
 
.
  

Revision as of 12:38, 23 December 2012

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
Invalid username
Login to AoPS