SOLUTION: A river has a steady current. A boat travels 30 mph over still water. If a trip 85 miles downstream and back takes 10 hours, then what is the speed of the current?

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Question 106101: A river has a steady current. A boat travels 30 mph over still water. If a trip 85 miles downstream and back takes 10 hours, then what is the speed of the current?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A river has a steady current. A boat travels 30 mph over still water. If a trip 85 miles downstream and back takes 10 hours, then what is the speed of the current?
:
Let x = speed of the current
then
(30+x) speed downstream
and
(30-x) speed up-stream
:
Write a time equation: Time = Distance/speed
:
Time upstream + Time downstream = 10 hrs
85%2F%28%2830-x%29%29 + 85%2F%28%2830%2Bx%29%29 = 10
:
Multiply equation by (30-x)(30+x) to get rid of the denominators:
(30-x)(30+x)*85%2F%28%2830-x%29%29 + (30-x)(30+x)*85%2F%28%2830%2Bx%29%29 = 10(30-x)(30+x)
:
Cancel out the denominators, FOIL (30-x)(30+x) and you have;
85(30-x) + 85(30+x) = 10(900-x^2)
:
2550 - 85x + 2550 + 85x = -10x^2 + 9000
:
The 85x's cancel leaving us with:
5100 = -10x^2 + 9000
Or
10x^2 = 9000 - 5100
:
10x^2 = 3900
:
x^2 = 3900/10
:
x^2 = 390
:
x = sqrt%28390%29
:
x = 19.75 mph is the current, that seems high so let's check it in the equation
:
85%2F%28%2830-19.75%29%29 + 85%2F%28%2830%2B19.75%29%29 = 10
85%2F%28%2810.25%29%29 + 85%2F%28%2849.75%29%29 = 10
8.29 + 1.71 = 10 hrs, confirms our solution
:
Could you follow the logic here OK? Any questions?