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==
 +
 
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A complete cycle of a tra±c light takes 60 seconds. During each cycle the light is
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green for 25 seconds, yellow for 5 seconds, and red for 30 seconds. At a randomly
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chosen time, what is the probability that the light will NOT be green?
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(A)
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1/4
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(B)
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1/3
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(C)
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5/12
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(D)
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1/2
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(E)
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7/12
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==solution==
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(E) 7/12:
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time not green/total time
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=
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(R + Y)/(R + Y + G)
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=
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(35/60)
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=
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(7/12)
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:
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OR
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The probability of green is
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(25/60)
 +
=
 +
(5/12)
 +
So the probability of not green is 1-  (5/12) =
 +
(7/12)
 +
.

Revision as of 15:09, 4 November 2012

problem

A complete cycle of a tra±c 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? (A) 1/4 (B) 1/3 (C) 5/12 (D) 1/2 (E) 7/12

solution

(E) 7/12: time not green/total time = (R + Y)/(R + Y + G) = (35/60) = (7/12)

OR

The probability of green is (25/60) = (5/12) So the probability of not green is 1- (5/12) = (7/12) .

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