Analysis: Suppose the bus takes X hours to drop off the students and then return. At this time, the bus has traveled 20Km, and the students behind it have traveled 4Km, with a distance of 20X-4X= 16X (km). It takes (16X÷24) hours to meet. Both students have done it during this time (16x ÷). After the meeting, the distance that the bus needs to travel is100-4x-16x ÷ 24x4, and the distance that the former students need to travel is100-20x-16x ÷ 24x4, and the driving time should be equal.

The equation is (100-4x-16x24x4)/20 = (100-20x-16x24x4)/4.

The solution is X= 15/4.

Meeting time: 16X/24=5/2 (hours)

Time after the meeting: (100-4x-16x/24 * 4)/20 =15/4 (hours)

Total time:15/4+5/2+15/4 =10 (hours)

So we leave at 9 a.m.

Note: the second half of the students walk15+10 (10 km is what the bus is doing when it returns, and both students are walking at this time).

The first part of the students go 10+ 15.

Automobile: 75+50+75