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2002 AMC 10A Problem 12

Problem 12 of 25IntermediateAlgebraArithmetic

Mr. Earl E. Bird leaves his house for work at exactly 8:008{:}00 A.M. every morning. When he averages 4040 miles per hour, he arrives at his workplace three minutes late. When he averages 6060 miles per hour, he arrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?

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Solution

Let tt hours be the on-time travel time. Since 33 minutes is 0.050.05 hours, 40(t+0.05)=60(t−0.05).40(t+0.05)=60(t-0.05). Then 40t+2=60t−3,40t+2=60t-3, so t=0.25.t=0.25. The distance is 40(0.30)=1240(0.30)=12 miles, so the required speed is 120.25=48\dfrac{12}{0.25}=48 mph. Thus, the correct answer is B.
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