The two ports A and B are 300 kilometers apart. If ship A sails from A to B on a stable water surface, and ship B sails from B to A at the same time, the two ships meet at C. If ship B sails from A to B, ship A sails from B to A at the same time, and the two ships meet at D, and C and D are 30 kilometers apart. Suppose the speed of ship A is 27 kilometers per hour, then the speed of ship B is () kilometers per hour?

Suppose the speed of ship B is X and the current speed is V, because it takes the same time for two ships to meet, so there are

Meeting for the first time: AC/(27+V) = (300-AC)/(X-V)- >

-& gt； AC(x-v)=(300-AC)(27+v)-& gt；

-& gt； AC[(x-v)+(27+v)]=300(27+v) ->

-& gt； AC(27+x)= 300(27+v)………………………………( 1)

Second meeting: ad/(x+v) = (300-ad)/(27-v)- >

-& gt； A.D. (27-V) = (300-AD) (X+V)->

-& gt； AD[(27-v)+(x+v)]= 300(x+v)-& gt；

-& gt； AD(27+x)= 300(x+v)……………………………………(2)

( 1)-(2)-& gt；

-& gt； (AC-AD)(27+x)= 300[(27+v)-(x-v)]-& gt；

-& gt； (AC-AD)(27+x)=300(27-x)

Because |AC-AD|=30, there is.

1) If AC-AD=30, then

30(27+x)=300(27-x) ->

-& gt； 27+x = 270- 10x->；

-& gt； 1 1x = 243-& gt；

-& gt； x=243/ 1 1

2) If AD-AC=30, (AC-AD=-30), then

-30(27+x)=300(27-x) ->

-27-x = 270- 10x-& gt；

-& gt； 9x = 297->；

x=33

To sum up, the speed of ship B is either 243/ 1 1 (km/h) or 33 (km/h).

Downstream velocity = still water velocity+current velocity

Backward speed = still water speed-water flow speed

Then (downstream speed-upstream speed) /2= (still water speed+water flow speed-still water speed+water flow speed) /2= water flow speed.