The downstream speed of the ship is 10+2 = 12m/s, and the upstream speed is10-2 = 8m/s.

Assuming that the walking distance of group A is x meters, the boat ride is 1800-x meters, and the total time of group A is x/4+(1800-x)/12 = (900+x)/6 seconds. Obviously, time is only related to X. If B walks y meters, the time to get B is also (900+y)/6 seconds.

To arrive at the same time, it must be x=y, that is, the walking distance of the two groups is the same. Just in a different order.

Then the retrograde distance of the ship is 1800-2x meters, and the equivalent relationship of the equation is:

The time required for the ship to go straight ahead 1800-x meters, and then go straight back 1800-2x meters is the same as the walking time of group a. ..

( 1800-x)/ 12+( 1800-2x)/8 = x/4

1800-x+2700-3x=3x

7x=4500

X = 4500/7m

Total time: (900+4500/7)/6= 1800/7 seconds =30/7 minutes =4 2/7 minutes.

You can arrive on time.