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A motorboat that travels with a speed of 20 km/hour in still water has traveled 36 km against the current and 22 km with the current, having spent 3 hours on the entire trip. Find the speed of the current of the river.

Mar 8, 2020

#1
+1247
+2

Explanation:

Let s be the speed of the current.

Let x be the time it has traveled against the current and let y be the time it traveled with the current.

Obviously, d=rt.

The rate at going against is 20-s km/hr and with is 20+s km/hr.

d=rt, or, solving for t, t = d/r.

We set t to equal x, so x = d/r.

The distance is 36, as stated, so x = 36/r

And the rate is 20-s, so x = 36/20-s.

Similarly, for y, y=22/20+s

Add those two equations to get x+y=(36/20-s)+(22/20+s).

We know that x+y=3 since the total time traveled was 3 hours.

System of equations: $$x+y=\frac{36}{20-s}+\frac{22}{20+s}, x+y=3$$.

We can now solve for s. I believe that you can solve this yourself, so I won't go over the hassle.

You should eventually get s = -20/3 or 2. Obviously, only the positive number works.

Therefore, the answer is 2 km/h.

(Can somebody check this? I'm not 100% sure about it. Thanks!)

You are very welcome!

:P

Mar 8, 2020
#2
+28025
+2

Similarly:

rate x time = distance           or      distance/rate = time

36/(20-c) + 22/(20+c)  =3

36 (20+c)  + 22(20-c) = 3 (20+c)(20-c)

720 +36c +440-22c = 3 (400-c^2)

1160 +14c= 1200-3c^2

3c^2 +14c -40 =0          Using quadratic formula    c = 2 km/hr

Mar 8, 2020