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A motorboat takes 4 hours to travel 128 kilometers going upstream.
The return trip takes 2 hours going downstream. What is the rate of the boat in still water
and what is the rate of the current?
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Let u be the motorboat speed in still water (in kilometers per hour, km/h).
Let v be the rate of the current, in km/h.
Then the boat's effective speed downstream is u+v km/h,
while the boat's effective speed upstream is u-v km/h.
// It is first major point you need understand and use in this sort of problems.
Now, "speed" equation for boat floating upstream is
= u - v (1) (speed upstream = the distance divided by time upstream)
Next, "speed" equation for boat floating downstream is
= u + v (2) (speed downstream = the distance divided by time downstream)
// It is the second major point in solving such problems: you must understand and write these equation automatically !
Simplify equations (1) and (2)
u - v = 32 (3)
u + v = 64 (4)
Now add equations (3) and (4) to eliminate "v". You will get
2u = 32 + 64 = 96 ====> u = = 48.
Thus you just found the boat' speed in still water. It is 48 kilometers per hour.
Now find the current rate from equation (4) v = 64 - u = 64 - 48 = 16 km/h.
Answer. The boat' speed in still water is 48 kilometers per hour.
The current speed is 16 km/h.
Solved.
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Post-solution note
I solved the problem formally, and this formal answer and the solution both are correct.
But you should keep in your mind the following.
In this problem, the speed of the motorboat in still water is very high: 48 km/h.
Even if imagine so high speed, it will mean that the river is lowland.
But for lowland river, the current rate of 16 km/h is too high.
So, the input data in this problem are not realistic.
I wrote this post-solution note to explain this fact to you.
When a professional Math composer creates his (or her) problems, usually he (or she)
knows a subject good enough to avoid (= to exclude) such inconsistencies.
Because he (or she) does think on what he (or she) is doing.