Question 325255: This is the word problem. Thanks!:
Jane and Tariq took a canoeing trip, traveling 6 mi upsteam against a 2 mi/h current. They then returned to the same point downstream. If their entire trip took 4 h, how fast can they paddle in still water? (Hint: If r is their rate, in mi/h, in still water, their rate upstream is r-2 and their rate dwonstream is r+2)
Found 3 solutions by checkley77, Alan3354, stanbon:
Answer by checkley77(12844)   (Show Source):
You can put this solution on YOUR website! D=RT
T=D/R
4=6/(R+2)+6(R-2)
4=[6(R-2)+6(R+2)]/(R+2)(R-2)
4=(6R-12+6R+12)/(R^2-4)
4=12R/(R^2-4)
4(R^2-4)=12R
4R^2-16-12R=0
4R^2-12R-16=0
4(R^2-3R-4)=0
4(R-4)(R+1)=0
R-4=0
R=4 ANS FOR THE RATE OF THE CANOE IN STILL WATER.
PROOF:
4=6/(4+2)+6/(4-2)
4=6/6+6/2
4=1+3
4=4
Answer by Alan3354(69443)  (Show Source):
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Jane and Tariq took a canoeing trip, traveling 6 mi upsteam against a 2 mi/h current.
They then returned to the same point downstream.
If their entire trip took 4 h, how fast can they paddle in still water? (Hint: If r is their rate, in mi/h, in still water, their rate upstream is r-2 and their rate downstream is r+2)
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Upstream DATA:
rate = r-2 mph ; distance = 6 miles ; time = 6/(r-2) hrs.
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Downstream DATA:
rate = r+2 mph ; distance = 6 miles ; time = 6/(r+2) hrs
-------------------------
Equations:
time + time = 4 hrs
6/(r-2) + 6/(r+2) = 4 hrs
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6(r+2) + 6(r-2) = 4(r^2-4)
12r = 4r^2 - 16
r^2 - 3r - 4 = 0
(r-4)(r+1) = 0
Positive solution:
r = 4 mph (rate in still water)
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Cheers,
Stan H.
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