Junior high school placement exam math problem!

2. The bus and truck depart from Party A and Party B respectively. After they met, the two cars continued to reach each other's destination, immediately returned and met again on the way. The location of the two encounters is 3000 meters apart. As we all know, the speed of freight cars is two-thirds that of passenger cars. What's the distance between a and b? (Calculation formula and problem solving process)

Solution: Take the whole journey as the unit 1.

The speed ratio of trucks and buses is 2: 3.

When we first met, the truck walked 2/5 of the whole journey and the bus walked 3/5 of the whole journey.

Because of the two encounters, the distance between the two cars is three times that of Party A and Party B, that is, 1x3=3.

The truck walked the whole distance of 3x2/5=6/5.

So the distance of the second meeting is 6/5- 1= 1/5.

The first time we met was at a distance of 3/5 from a place.

Then the distance between the two places is 3/5- 1/5=2/5.

The distance between Party A and Party B is 3000/(2/5) = 7500m.

3. Two cars, A and B, leave from two cities at the same time. Three hours later, the two cars meet at the midpoint18km. At this time, the ratio of the distance traveled by A and B is 2: 3. What are the speeds of A and B respectively?

Let the speed of A be 2a km/h and the speed of B be 3a km/h..

Total distance =(2a+3a)×3= 15a km.

Distance of line A = 15a×2/5=6a.

15a/2-6a= 18

15a- 12a=36

3a=36

a= 12

The speed of a =12x2 = 24km/h.

Speed b =12x3 = 36 km/h.

or

Think of the whole distance as 1.

Then when we met, A made 2/5.

Line B 1-2/5=3/5

Whole journey = (1/2-2/5) =110.

Whole journey =18/(110) =180 km.

The sum of the speeds of Party A and Party B =180/3 = 60 km/h.

The speed of a = 60x2/5 = 24km/h.

B =60-24 speed =36 km/h.

Two cars, A and B, start from AB at the same time and drive in opposite directions. The speed ratio of a and b is 4: 5. After the two cars met for the first time, the speed of A increased by a quarter, and the speed of B increased by a third. The two cars returned immediately after arriving in Pakistan respectively. In this way, the second meeting point is 48 kilometers away from the first meeting point, and how many kilometers away is AB?

Think of the whole journey as a unit of 1

Because time is constant, the distance ratio is the speed ratio.

So when they met, A did the whole journey of 1x4/(5+4)=4/9.

Line B 1-4/9=5/9

At this time, Party A and Party B accelerate, and the speed ratio changes from 4: 5 to 4 (1+1/4): 5 (1+1/3) = 5:10/3 = 3: 4.

The sum of the distances that A and B meet again is twice the distance of AB, which is 2.

At this time, we met for the second time, and the whole journey of Party B was 2x4/(3+4)=8/7.

The distance of the second intersection is 8/7-4/9=44/63 of the total distance.

Distance to the first meeting point is 44/63-4/9= 16/63.

AB distance = 48/(16/63) =189 km.

Xiao Ming walks home at a speed of 4 kilometers per hour along a bus route after school. Along the way, a bus overtook him from behind every 6 minutes, and another oncoming car was encountered every 4 2/7 minutes. If the bus runs at the same speed and at equal intervals, what is the release interval of the motorcycle?

This problem is two processes.

The first is the problem of tracing, and the second is the problem of meeting.

Suppose the speed of the bus is one kilometer per hour.

(a-4)x6/60=(a+4)x(30/7)/60

7a-28=5a+20

2a=48

A=24 km/h

Then the distance difference between the bus and Xiaoming is = (24-4) X6 = 120km.

So the departure time interval is 120/24=5 minutes.

Example 1, Party A and Party B set out from two places 30 kilometers apart and walked towards each other at the same time. Party A walks 6 kilometers per hour and Party B walks 4 kilometers per hour. Q: How many hours did they meet?

[Analysis] When Party A and Party B set out, they were 30 kilometers apart, and the distance between them was shortened by 6+4 = 10 (kilometers) per hour, that is, the sum of their speeds (referred to as the sum of speeds), so several of the 30 kilometers 10 kilometers met in a few hours.

Solution: 30 degrees (6+4)

=30÷ 10

= 3 hours

They met three hours later.

Example 2: Party A and Party B set out from A and B, which are100km apart, and walked towards each other at the same time. Walking, Party A's car broke down and it took 1 hour to repair it. Four hours after the departure, Party A and Party B met. It was known that Party A's speed was twice that of Party B, and Party A's car had been repaired when they met. Then, what are the network speeds of Party A and Party B respectively?

[Analysis] The speed of A is twice that of B. Therefore, B walks for 4 hours, and A only needs 2 hours to find the speed of A..

Solution: The speed of A is: 100 ÷ (4- 1+4 ÷ 2).

= 100 ÷ 5 = 20 (km/h)

The speed of B is: 20 ÷ 2 = 10 (km/h).

A: The speed of A is 20km/h, and the speed of B is10km/h. ..