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

(1998 AMC 8 #10, Solution)
 
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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>.
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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.

Revision as of 23:11, 8 November 2011

Okay, so the whole cycle, $60$ 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 $\frac{35}{60}$. This can be simplified, by factoring out the factor of $5$ common to the numerator and the denominator, to $\frac{7}{12}$ or E.

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