Let's assume the speed of the boat in still water is 'b' km/h, and the speed of the stream is 's' km/h.
When the boat rows with the stream, its effective speed is increased by the speed of the stream. Therefore, the speed of the boat with the stream is (b + s) km/h. We are given that the boat can row 1 km with the stream in 10 minutes, which is equivalent to 1/6 hours.
Using the formula: Speed = Distance / Time, we can write the equation:
(b + s) = 1 / (1/6) (b + s) = 6 km/h
Similarly, when the boat rows against the stream, its effective speed is decreased by the speed of the stream. Therefore, the speed of the boat against the stream is (b - s) km/h. We are given that the boat can row 1 km against the stream in 20 minutes, which is equivalent to 1/3 hours.
Using the same formula, we can write the equation:
(b - s) = 1 / (1/3) (b - s) = 3 km/h
Now we have a system of two equations:
(b + s) = 6 (b - s) = 3
We can solve this system of equations to find the values of 'b' and 's'.
Adding the two equations together:
(b + s) + (b - s) = 6 + 3 2b = 9 b = 9 / 2 b = 4.5 km/h
So, the speed of the boat in still water is 4.5 km/h.