SOLUTION: A boy has planned a three-hour bicycle ride. After biking at a rate of 14 mph for awhile, the bike breaks down and he rides back with his father at a rate of thirty-five mph. How f

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Question 121717: A boy has planned a three-hour bicycle ride. After biking at a rate of 14 mph for awhile, the bike breaks down and he rides back with his father at a rate of thirty-five mph. How far did the boy ride his bicycle if he returns one hour after he left?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A boy has planned a three-hour bicycle ride. After biking at a rate of 14 mph for awhile, the bike breaks down and he rides back with his father at a rate of thirty-five mph. How far did the boy ride his bicycle if he returns one hour after he left?
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Let d = distance ridden on the bike
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Write two time equations with the information we have: Time = Distance/speed
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time on the bike = d%2F14
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Time in the car = d%2F35; (distance on the bike = return distance in the car)
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Time on bike + time in car = 1 hr
d%2F14 + d%2F35 = 1
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Multiply equation by 14*35 = 490, to get rid of the denominators
490*d%2F14 + 490*d%2F35 = 490(1)
Cancel the denominators and you have:
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35d + 14d = 490
:
49d = 490
d = 490%2F10
d = 10 miles
:
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Check solution by finding the total time
10/14 + 10/35 =
.714 + .286 = 1 hr, confirms our solution
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Did this make sense to you? Any questions?