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

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Okay, so the whole cycle, <math>60</math> seconds, is our denominator. The light is green for 25 seconds in the cycle, so it is a color other than green for 35 seconds. So, the probability that it will not be green at a random time is <math>\frac{35}{60}</math>. This can be simplified, by factoring out the factor of <math>5</math> common to the numerator and the denominator, to <math>\frac{7}{12}</math> or E.
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==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?
 +
 
 +
<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==
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===Solution 1===
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<cmath>\frac{\text{time not green}}{\text{total time}}
 +
=
 +
\frac{R + Y}{R + Y + G}
 +
=
 +
\frac{35}{60}
 +
=
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\boxed{\text{(E)}\ \frac{7}{12}}</cmath>
 +
 
 +
===Solution 2===
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The probability of green is
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<math>\frac{25}{60}
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=
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\frac{5}{12}</math>,
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so the probability of not green is <math>1-  \frac{5}{12} =
 +
\boxed{\text{(E)}\ \frac{7}{12}}</math>
 +
.
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 +
==See Also==
 +
 
 +
{{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?

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

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 (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. AMC logo.png

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