Expert: Scott A Wilson - 9/25/2011

Question
A boat travels 10 km upstream and 10 km back.the trip upstream and back take 1.5 hrs.if the speed of the current is 5 kph,what is the speed of the boat in still water?

Answer
It is known that distance = rate * time.  It is further known that the downstream rate is 5kph faster than the regular rate and the upstream rate is 5kph slower.

Taking the distance as 10, r to be the standard rate of travel, a to be the time downstream, and b to be the time upstream, gives us the downstream equation of 10 = (r+5)a and an upstream equation of 10 = (r-5)b.

Solving each equation for time gives a = 10/(r+5) and b = 10/(r-5).
It is known the two times combine to 1.5 hours, so 1.5 = a + b = 10/(r+5) + 10(r-5).

We can then multiply 1.5 = 10/(r+5) + 10(r-5) by (r+5)(r-5) to get
1.5(r+5)(r-5) = 10(r-5) + 10(r+5).  Multiplying a bit gives 1.5(r²-25) = 10r - 50 + 10r + 50, which simplifies to 1.5(r²-25) = 20r since -50+50 is 0.

Multiplying this by 2 gives 3(r²-25) = 40r, which is 3r² - 75 = 40r.
Subtracting 40r from both sides gives 3r² - 40r - 75 = 0.
That factors into (r-15)(3r+5).  The solutions to thie are r = 15 and r = -5/3.
Since the speed needs to be positive to make sense, go with r = 15.

Looking back at the problem, this means we are going down at 20 kph, so it takes 1/2 hour,
and we are goig 10 kph upstream, so it takes 1 hour, for a total of 1.5 hours.