SOLUTION: Going to parent's house in 4.5 hours. Travel by car (60 mph) to dock where I'll enter a boat (20 mph on still water with 4 mph current...against me on the way to parent's house...

Question 76842: Going to parent's house in 4.5 hours. Travel by car (60 mph) to dock where I'll enter a boat (20 mph on still water with 4 mph current...against me on the way to parent's house...with me on the way back). Takes 3.5 hours to get back home. What's the distance?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Going to parent's house in 4.5 hours. Travel by car (60 mph) to dock where I'll enter a boat (20 mph on still water with 4 mph current...against me on the way to parent's house...with me on the way back). Takes 3.5 hours to get back home. What's the distance?
:
This easiest way, is to find the time he was on the road at 60 mph and go on from there.
:
Let t = his time at 60 mph
:
Then (4.5 - t) = boat time at 16 mph
And (3.5 - t) = boat time at 24 mph
:
We know the trip there and the trip back are equal in distance; dist = speed*time:
60t + 16(4.5-t) = 60t + 24(3.5-t)
:
Subtract 60t from both sides, and multiply what's inside the brackets:
72 - 16t = 84 - 24t
-16t + 24t = 84 - 72
8t = 12
t = 12/8
t = 1.5 hrs at 60 mph
:
Total distance would be:
1.5(60) + 16(4.5-1.5) =
90 + 16(3) =
90 + 48 = 138 miles
:
Check solution using return trip:
1.5(60 + 24(3.5-1.5) =
90 + 24(2) =
90 + 48 = 138 mi also
:
Did this make sense to you?

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