Question 661473: Two friends living 39 mi apart,leave their homes at the same time on bicycles and travel toward each other.If one person averages 2 mi/ hr more than the other,and they meet in 11/2 hour,what is each persons average rate of cycling?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Two friends living 39 mi apart, leave their homes at the same time on bicycles and travel toward each other.If one person averages 2 mi/ hr more than the other,and they meet in 11/2 hour,what is each persons average rate of cycling?
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Faster cyclist:
time = 11/2 hrs ; distance = x miles ; rate = x/(11/2) = (2x/11) mph
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Slower cyclist:
time = 11/2 hrs ; distance = (39-x) miles ; rate = (39-x)/(11/2) = (78-2x)/11 mph
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Equation:
Faster - Slower = 2 mph
(2x/11) - (78-2x)/11 = 2 mph
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2x -(78-2x) = 22
4x = 100
x = 25 miles
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Faster rate = (2x/11) = 50/11 = 4.55 mph
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Slower cyclist = (78-2x)/11 = (78-50)/11 = 28/11 = 2.55 mph
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Cheers,
Stan H.
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