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>. | + | 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 22:11, 8 November 2011
Okay, so the whole cycle, 
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 
. This can be simplified, by factoring out the factor of 
common to the numerator and the denominator, to 
or E.
