Yes, I can.
And even though you haven't asked to be shown how to do it,
I'll go ahead and do that too:
Call the speed of the boat (through the water) 'B'.
Call the speed of the current (the water) 'C'.
When the boat is going 'up' the river, against the current,
his speed past the riverbank is (B - C).
When the boat is going 'down' the river, the same way as the current,
his speed past the riverbank is (B + C).
The problem says it took him 5 hours to travel 60 km against the current.
Distance = (speed) x (time)
60 km = (B - C) x (5 hours)
The problem also says it took him 3 hours to return.
The distance to return is the same 60 km.
The other direction is the same direction as the current,
so his speed on the return is (B + C).
Distance = (speed) x (time)
60 = (B + C) x (3)
Now we have two equations, so we can find 'B' and 'C'.
5B - 5C = 60
3B + 3C = 60
Multiply each side of the first equation by 3, and
multiply each side of the second equation by 5:
15B - 15C = 180
15B + 15C = 300
Add the second equation to the first one:
30B = 480
B = 480/30 = 16 km per hour.
Subtract the second equation from the first one:
-30C = -120
C = -120/-30 = 4 km per hour.
The speed of the boat through the water (B) is 16 km per hour.
The speed of the water past the riverbank is 4 km per hour.
Check:
-- When the boat is going along with the current, his speed past the riverbank
is (16 + 4) = 20 km per hour. In 3 hours, he covers (3 x 20) = 60.
-- When the boat is going against the current, his speed past the riverbank
is (16 - 4) = 12 km per hour. In 5 hours, he covers (5 x 12) = 60 km.
yay !
