The distance between Party A and Party B is 15km. From A and B, there is a bus leaving every 15 minutes at the same time. At the speed of 30km/h, Li Ming rode from A to B, and at the speed of 15km/h, after the first bus leaves for 5 minutes, Li Ming leaves from A, so he can hit the road.

According to the meaning of the question: it took Xiaoming 15÷ 15= 1 hour =60 minutes to get from a to b.

In fact, we can interpret "five minutes after the first bus leaves" as Li Minghe buses leave at the same time and arrive 65 minutes later.

So you can meet 65÷ 15≈4 cars in the trip from B to B.

You can meet 65÷ 15≈4 vehicles when you come back.

So it's 4+4=8 cars.

But this is not the final answer. We must consider whether the last bus meets Li Ming at the end of the journey.

The car that met 1 scored 15, and the second car scored 30 points, 45 points and 60 points.

There are still five minutes left, so 5 < 15.

So only three cars can meet Li Ming when he comes back.

So the answer is 4+3=7 cars.

Method 2:

It can be seen that (60+5)÷ 15+ 1≈5.3= the second place has five cars, plus 5 means that he sent a car five minutes ago.

There are (60-10-15) ÷15 ≈ 2.3 = 2 cars, negative 10 is his10 minutes later, and negative15 is his last.

* * * See 5+2=7 cars.

Method 3:

/kloc-After 0/0 minutes, you can see cars from both places at the same time, and within 25 minutes, you can also see cars from both places at the same time. After 40 minutes, he can't see the car sent by Party A, and he can't see the car sent by Party B. Similarly, after 55 minutes, you can only see the car sent by Party B, plus seven cars sent by Party B five minutes before departure.

Thank you for your adoption. I wish you progress in your studies.