SOLUTION: My problem is a word problem that goes like this: A tugboat goes 120 miles upstream in 20 hours. The return trip downstream takes 10 hours. Find the speed of the tug boat without a

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: My problem is a word problem that goes like this: A tugboat goes 120 miles upstream in 20 hours. The return trip downstream takes 10 hours. Find the speed of the tug boat without a      Log On


   



Question 632568: My problem is a word problem that goes like this: A tugboat goes 120 miles upstream in 20 hours. The return trip downstream takes 10 hours. Find the speed of the tug boat without a current, and, find the speed of the current.
So, I have two variables x= tugboat speed, and y = current speed, I believe that this is basically a rate, time, distance type of problem. I tried to solve it like this: 120x20=2400 multiply
120x10=1200 add
=3600/2=1800 divide by 2
1800-2400=-600, 1800-1200=600, I know that this is incorrect,as it is impossible for a tugboat to go -600 mph, or for a river current to move at 600 mph. I saw a similar problem on this sight, but I think the asker formatted the problem improperly, because the answer was only one part, and the problem requires a two part answer, and it just does not fit this particular problem either. Can you help me understand how to solve this problem? Thanks

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A tugboat goes 120 miles upstream in 20 hours. The return trip downstream takes 10 hours. Find the speed of the tug boat without a current, and, find the speed of the current.
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Upstream DATA:
distance = 120 miles ; time = 20 hrs ; rate = d/t = 120/20 = 6 mph
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Downstream DATA:
distance = 120 miles ; time = 10 hrs ; rate = d/t = 120/10 = 12 mph
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Equations:
b + c = 12 mph
b - c = 6 mph
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Add and solve for "b":
2b = 18 mph
boat speed = 9 mph
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Solve for "c":
b + c = 12 mph
9 + c = 12
current speed = 3 mph
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Cheers,
Stan H.
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