A long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. Two hours later, a cyclist starts at the beginning of a trail and travels at a rate of 16 miles per hour. What is the amount of time that the cyclist travels before overtaking the runner? Do not do any rounding

Question

Mckayla Petersen

Mathematics

16 September, 01:51

A long distance runner starts at the beginning of a trail and runs at a rate of 6 miles per hour. Two hours later, a cyclist starts at the beginning of a trail and travels at a rate of 16 miles per hour. What is the amount of time that the cyclist travels before overtaking the runner? Do not do any rounding

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Answers (2)

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Valeria Barry · Answer 1

If the runner moves at 6 miles per hour for 2 hours before the cyclist starts, the runner has a 12 mile lead at the beginning. If the cyclist is going at 16 miles per hour, he is overtaking the runner at 10 miles per hour, because 16-6=10. Then, the amount of time it takes the cyclist to go 2 miles in 10 miles per hour is the answer. this is 1/5 of an hour, or 12 minutes.

Addisyn Stokes · Answer 2

Well the cyclist would catch up at 48 miles because ... 6, 12, 18, 24, 30, 36, 48, 54, 60 and 16, 32, 48, 64, 80. Now if you find the least common factor between the two that is your answer. which is48 in this case