A boatman rows to a place 30 km distant and back in 14 hours. He finds that he can row 10 km with the stream in the same time as 4 km against the stream. Find the speed (in km/h) of the stream.

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Let the speed of the boat in still water be 'b' km/h and the speed of the stream be 's' km/h.

Let us denote:
- Speed downstream = b + s
- Speed upstream = b - s

Step 1: Use the time equation for the round trip.
Time taken to go 30 km downstream = 30/(b + s)
Time taken to return 30 km upstream = 30/(b - s)
Total time = 14 hours
So,
30/(b + s) + 30/(b - s) = 14
=> [30(b - s) + 30(b + s)] / [(b + s)(b - s)] = 14
=> [30b - 30s + 30b + 30s] / (b^2 - s^2) = 14
=> (60b) / (b^2 - s^2) = 14
=> 60b = 14(b^2 - s^2)
=> 14b^2 - 14s^2 - 60b = 0
=> 7b^2 - 7s^2 - 30b = 0
=> 7b^2 - 30b - 7s^2 = 0  ...(1)

Step 2: Use the second condition.
He can row 10 km with the stream in the same time as 4 km against the stream.
So,
10/(b + s) = 4/(b - s)
Cross-multiplied:
10(b - s) = 4(b + s)
10b - 10s = 4b + 4s
10b - 4b = 10s + 4s
6b = 14s
b = (14/6)s = (7/3)s

Step 3: Substitute b = (7/3)s into equation (1):
7b^2 - 30b - 7s^2 = 0
Substitute b:
7[(7/3)s]^2 - 30*(7/3)s - 7s^2 = 0
7*(49/9)s^2 - (210/3)s - 7s^2 = 0
(343/9)s^2 - 70s - 7s^2 = 0
(343/9)s^2 - (63/9)s^2 - 70s = 0
(280/9)s^2 - 70s = 0
(280/9)s^2 = 70s
Divide both sides by s (s ≠ 0):
(280/9)s = 70
s = 70 * (9/280)
s = (9/4) km/h

Therefore, the speed of the stream is 9/4 km/h.

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