Re: On a certain trip, a cyclist averaged 20 miles per hour for the 10 mil
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11 Nov 2021, 02:57
Total time take for first 10 miles with an average speed of 20 miles per hour, \(t_1 = 10/20 = 0.5 hr\)
Total time take for remaining 20 miles with an average speed of 10 miles per hour, \(t_1 = 20/16 = 1.25 hr\)
This was one way trip, which mean one-way distance = 30 miles
Total time of entire trip = 4 hrs
Time take for return trip = 4 - \((t_1 + t_2)\) = 4 - 1.75 = 2.25 hrs
Average speed for return-trip = \(\frac{30}{2.25} = \frac{120}{9} = \frac{40}{3} = 13\frac{1}{3}\)
Hence, Answer is E