Question 575063: I rowed 20 km up a river that had a steady current in 5 hours and returned downstream in 2.5 hours. How fast would I have rowed the boat in still water, and what is the speed of the current?
Let x be the speed of the boat in still water and y the speed of the current (km/h). The speed of the boat going upstream is (x –y) km/h and the speed of the boat going downstream is (x+y) km/h,.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I rowed 20 km up a river that had a steady current in 5 hours and returned downstream in 2.5 hours. How fast would I have rowed the boat in still water, and what is the speed of the current?
Let x be the speed of the boat in still water and y the speed of the current (km/h). The speed of the boat going upstream is (x –y) km/h and the speed of the boat going downstream is (x+y) km/h,.
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Upstream DATA:
distance = 20 km ; rate = (x-y)km/h ; time = 20/(x-y) hr
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Downstream DATA:
distance = 20 km ; rate = (x+y)km/h ; time = 20/(x+y) hr
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Equation:
20/(x-y) = 5 hrs
20/(x+y) = 2.5 hrs
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x-y = 4
x+y = 8
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Add and solve for "x":
2x = 12
x = 6 km/h (boat speed in calm water)
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Solve for "y":
x+y = 8
y = 2 km/h (current speed)
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Cheers,
Stan H.
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