Question 1: When can the liaison catch up with Class 5, Grade 1?

Solution: Suppose it takes X hours to catch up with Class 5, Grade 1. According to the meaning of the question:

12X=4(X+ 1)

X=0.5

A: It takes 0.5 hours for the liaison to catch up with Class 5, Grade 1.

Question 2: When will Class 6 of Grade 1 catch up with Class 5 of Grade 1?

Solution: Suppose it takes X hours to catch up with Class 5, Grade 1. According to the meaning of the question:

6X=4(X+ 1)

X=2

Answer: It takes 2 hours for Class 6 of Grade 1 to catch up with Class 5 of Grade 1.

It takes four times to transport all teachers and students.

If the car transports 25 people for the first time and walks X kilometers, it will return to pick it up.

After the meeting, 25 people were transported for the second time and returned after walking X kilometers.

In this way, 25 people were transported for the third time to walk X kilometers and then came back to pick them up.

The fourth time, if the rest of the people walk X kilometers and just reach the finish line, then this method takes the shortest time (at this time, all the teachers and students take the bus and walk at the same time, so they reach the finish line at the same time).

It takes X/55 hours for the first car to walk x kilometers, and then pedestrians walk (X/55)*5=X/ 1 1 (kilometers).

At this time, the distance between cars and pedestrians is x-x/11= (10x)/1(km).

The time when the car returns to meet the pedestrian is [(10x)/1]/(5+55) = x/66 (hours), and the pedestrian has already walked (x/66) * 5km during this time.

So we left as soon as we met, x/1+(x/66) * 5 = x/6 (km).

Similarly, the car walked back X/6 kilometers for the second time, and X/6 kilometers for the third time. At this time, the remaining teachers and students who walked * * * walked (X/6)*3 kilometers, just x kilometers from the finish line. So:

(X/6)*3+X=33，

The solution is X=22.

So the shortest time to reach the destination is 22/55+(33-22)/5=2.6 hours.