1. Divide the students except the monitor into four groups, each group 1 1 student.
2. At seven o'clock in the morning, the monitor and the first group of students 1 1 got on the school bus, and the other three groups of students set off for their destination on foot. The school bus traveled for x 1 hour, put the first group of students down and drove back, and the first group of students walked to their destination. The school bus drove back for x2 hours and met three groups of students on foot, carrying the second group of students to their destination, while the rest of the third and fourth groups of students continued to walk. The school bus traveled for x3 hours, put the second group of students down and drove back, and the second group of students walked to their destination. The school bus drove back for x4 hours, met the third and fourth groups of students, and took the third group of students to their destination, while the fourth group of students continued to walk. The school bus traveled for x5 hours, put the third group of students down and drove back, and the third group of students walked to their destination. The school bus traveled for x6 hours and met the fourth group of students. After x7 hours, it carried the fourth group of students to the destination. At this time, four groups of students arrived at the destination at the same time. The monitor was always in the car.
3. Start listing equations (if you are not interested, you can directly look at the fifth one)
For the first batch of students, school bus time * school bus speed+walking time * walking speed = distance, then it is 70 * x1+5 * (x2+x3+x4+x5+x6+x7) = 7.7;
Similarly, the second, third and fourth batches are: 70 * x3+5 * (x1+x2+x4+X5+X6+X7) = 7.7;
70 * X5+5 *(x 1+x2+x3+x4+X6+x7)= 7.7；
70 * x7+5 *(x 1+x2+x3+x4+X5+X6)= 7.7；
The school bus met the students three times on the way back. 70 * x 1-70 * x2 = 5 *(x 1+x2)；
70 * x3-70 * x4 = 5 *(x3+x4)；
70 * X5-70 * X6 = 5 *(X5+X6)；
5. These seven equations are independent of each other, and there are seven unknowns, which can be solved (seemingly troublesome, but actually very simple).
6. In fact, there is a simpler idea, that is, each group of students leave at the same time and arrive at the same time, except by car, they all walk (ignoring the time of getting on and off). So the time for each group of students to ride and walk is equal. That is to say, the time for each group of students to take the school bus is equal, that is, x1= x3 = X5 = x7; At the same time, the time for the school bus to return to meet the next batch of students is equal, that is, x2 = x4 = X6 =13/15 * x1. This greatly simplifies the calculation, that is, 70x1+5 * (3+13/15 * 3) x1= 7.7 and x1=/kloc-0.
7. To sum up, the fastest time is 3 1 minute and 7 seconds (363/700 hours).
I hope you are satisfied, please point out any mistakes!