It takes seven (2) students x hours to catch up with seven (1) students. In other words, when the second class leaves X hours, the total distance traveled by the two classes is the same.

4(x+ 1)=6x, and x=2 is obtained.

So the liaison went back and forth 12*2=24 (km).

2. Method without equation:

Exactly 4* 1/(6-4)=2 (hours)

2* 12=24 km

This problem seems to be of the same nature as the homing pigeon problem. You don't have to consider how the liaison actually walked, that is, the process of weakening. As long as you pay attention to the liaison who walks synchronously in two classes, he and Class 7 (Class 2) spend the same time, but they are unified anyway. As long as the speed * time, you can find the distance ~

I hope I can help you ~