4 Answers

The average time of the trip, which cancels out the current, is 4 hours, so the boat’s motor is pushing it at 7.5 km/h. You can figure out the rest.

Here’s what we know:
distance = rate X time. Let s = the speed of the river, r = the speed of the boat relative to the river. Then
30 Km = (r + s) X 3 [The speed downriver is the boat’s speed plus the river’s speed]
and
30 Km = (r – s) X 5 [The speed upriver is the boat’s speed minus the river’s speed]
Since 30 Km = 30 Km, (r + s) X 3 = (r – s) X 5
Solving for s:
3r + 3s = 5r – 5s
8s = 2r
s = r/4
Putting this back in the first equation,
30 Km = (r + r/4) X 3.
Solving for r:
15r/4 = 30 Km
So r = 8 Km/hr (speed of the boat) and
s = 2 Km/hr (speed of the river)
Source(s): high school algebra 
3×30 = 90km.
90/5=18km speed while returning.
3018=12
River helps boat while going and slows it down while returning. Therefore, 12/2=6km change is the on boat’s speed. Therefore, river’s flowing speed is 6km.

Fast enough to slow it by 2 hrs