Question 761849: A speed boat has a speed of 27 mph in still water. If it requires 1 1/8 hours for the boat to go 15 miles downstream and return, what is the speed of the current?
Here is how I set up the problem:
c=the current speed t= the time for downstream
27-c*(1 1/8 -t)=15
27+c*(t)=15
Do I have the problem set up right 1st of all?
I rearrange the second equation and plug it into the first to get this:
27-c(1 1/8-(15/(27+c)))=15
If this is right, how do I solve it?
Thank you so much! Abby
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A speed boat has a speed of 27 mph in still water. If it requires 1 1/8 hours for the boat to go 15 miles downstream and return, what is the speed of the current?
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upstream DATA:
dist = 15 miles ; rate = 27-c mph ; time = 15/(27-c) hrs
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downstream DATA:
dist = 15 miles ; rate = 27+c mph ; time = 15/(27+c) hrs
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Equations;
15/(27-c) + 15/(27+c) = 9/8 hrs
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Multiply thru by 8(27-c)(27+c) to get:
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8*15(27+c) + 8*15(27-c) = 9(27^2-c)^2
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120*27 + 120*27 = 9(27^2-c^2)
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2(120*27) = 9(27^2-c^2)
6*120 = 27^2-c^2
c^2 = 9
c = 3 mph (speed of the current)
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Cheers,
Stan H.
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