Both large and small ships can complete the freight separately, and they can all be filled up. The total cargo volume must be the common multiple of 120 and 72, assuming that it is their minimum common multiple of 360, the ship needs 360÷ 120=3 trips to complete the freight, and the ship needs 360÷72=5 trips. If the total cargo volume is 720 tons, in this case, the previous netizen's answer is wrong. They did not consider the condition that the number of trains is less than 10, and abused the conclusion that speed is inversely proportional to time.
The title itself has a disadvantage. The title doesn't say where the last ship docked. Assume that the ships have all returned to the origin.
Ship's departure speed: 33+3 = 36;
Return speed of large ship: 33* 1.2-3=36.6.
Ship speed: X+3
Ship return speed: 1.2X-3
Let the unilateral distance here be s (this s will be eliminated in the later proportion)
Round trip time of big ship: S/36+S/36.6.
Round trip time of the ship: S/(X+3)+S/( 1.2X-3)
Equation: 3* Round trip time of big ship: 5* Round trip time of small boat =5:9.
Because the data of the equation is not good, the process is omitted here, and S can be reduced, and the result is X=67.3 (km/h).
This is the speed of the ship when it is fully loaded with goods in still water.