SOLUTION: The circumferences of a front and rear wheels of a carriage are, respectively, 2 and 4 feet. What distance will the carriage have passed over when the front wheel has made 8 more r

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Question 887244: The circumferences of a front and rear wheels of a carriage are, respectively, 2 and 4 feet. What distance will the carriage have passed over when the front wheel has made 8 more revolutions than the rear wheel?
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
d= distance the carriage will have passed over when the front wheel has made 8 more revolutions than the rear wheel.
d%2F2= revolutions made by the front wheel.
d%2F4= revolutions made by the rear wheel.
The problem says that when the carriage passed over the distance d
the front wheel will have made 8 ore revolutions than the rear wheel, so
d%2F2=d%2F4%2B8
Multiplying both sides of the equal sign times 4 gets rid of denominators, to get
4%2A%28d%2F2%29=4%28d%2F4%2B8%29--->4%2A%28d%2F2%29=4%28d%2F4%29%2B4%2A8--->2d=d%2B32%29
Subtracting d from both sides of the equal sign gives us
2d-d=d%2B32-d%29--->highlight%28d=32%29
The carriage will have passed over a distance of highlight%2832%29 feet.

That would take 32%2F2=16 revolutions for the front wheel and 32%2F4=8 revolutions for the rear wheel.