Question 322397: To deliver a package, a messenger must travel at a speed of 60 miles per hour on land and then use a motorboat whose speed is 20 miles per hour in still water. The messenger goes by land to a dock and then travels on a river against a current of 4 miles per hour. He reaches his destination in 4.5 hours and then returns to the starting point in 3.5 hours. How far did the messenger travel by land and how far by water?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Everything in this problem is based on the distance equals rate times time formula.
Let  represent the distance traveled on land.
Let  represent the distance traveled on the water.
Let  represent the elapsed time traveling on land.
We know the rate of speed on land because that is given as 60 mph.
The rate of speed for the outbound trip on water is the rate in still water MINUS the rate of the current (because he was traveling AGAINST the current), so 20 mph minus 4 mph is 16 mph.
Likewise, the rate of speed for the return trip on water is 24 mph.
If the total outbound trip took 4.5 hours and the land part of the trip took hours, then the water part of the outbound trip must have taken  hours.
Similarly, the water part of the return trip must have taken  hours.
Let's put some of this together. We can describe the outbound trip over water by the relationship:
)
And we can describe the return trip over water as:
)
Having two things both equal to , we can set them equal to each other:
\ =\ 24(3.5\ -\ t))
From which we can solve for  hours, the travel time on land. (Verification of this result is left as an exercise for the student).
Then, knowing that  , we can say that 
Further, says that the travel time for the outbound trip on water is  , hence the distance over water for the outbound trip is  miles.
As a check, the distance over water for the return trip better be the same as for the outbound trip:
\ =\ 48)
And our answer checks.
John
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