Problem description:

Class A and Class B started from School A and conducted military training in a military camp 75 kilometers away from the school. Class A students walk at a speed of 4km/h, while Class B students walk at a speed of 5 km/h. The school has a car with an idle speed of 40km/h and a manned speed of 20km/h, which can only carry one class at a time. Students from both classes are now required to arrive at the barracks at the same time. Ask them how long it will take at least.

Analysis:

1 Suppose you walked for x hours first and took a bus to A together.

Then X hours later, A has been separated by (20-5) x = 15x km. At this time, he came back to pick up his empty classmates.

After receiving the classmate, the classmate left 15x/(40+5)+x = 4x/3 hours.

At this time, the students in Class A walked 15x/(40+5) = x/3 hours, and walked 4*X/3 kilometers (excluding the car).

75-20x-4x/ 3 km to the left.

We still have to walk (75-20x-4x/3)/4 hours.

Just walk (75-5 * 4x/3)/20 = (15-4x/3)/4 hours.

Then (75-20x-4x/3)/4 = (75-5 * 4x/3)/20.

The solution is X=2.5.

So x+x/3+(15-4x/3)/4 =15/4+x = 6.25 hours.

Let's go for x hours first and arrive by car together.

Then after x hours, A has been separated by (20-4) x = 16x km. At this time, he returned to the next class empty.

When accepting students from Class A, Class A left16x/(40+4)+x =15x/11hour.

At this time, the students in class have already left16x/(40+4) = 4x/11hour (excluding taking the bus).

Go 5 * 4x/ 1 1km.

There are 75-20x-20x/ 1 1km left.

Still have to walk (75-20x-20x/ 1 1)/5 hours.

And a has to walk (75-4 *15x/11)/20 = (15-12x/11)/4 hours.

Then (75-20x-20x/11)/5 = (15-12x/11)/4.

X=2.8 15。

Therefore, x+4x/11+(15-12x/16438+0)/4 =15/4+12x is required.

Therefore, the minimum time for option 1 is 6.25 hours.