How to solve the 3 way itinerary problem? 10

4.The sum of the speed of the two cars: 900 2 = 450 (m) The speed difference between the two cars:

900 18 = 50 (m min) Fast speed: (450 + 50) 2 = 250 (m min) Slow speed: (450-50) 2 = 200 (m min).

minutes).6.The father walked 4 kilometers at the same time from home to the first time he caught up with the son, and the father took 8 minutes less, and the father took the same time from the first time he caught up with the son to the second time, the son still walked 4 kilometers, and the father walked 4 2 8 kilometers more, from the comparison of the two situations, it can be seen that the father walked 8 kilometers in 8 minutes and 4 kilometers in 4 minutes. In this way, from home to the first time I caught up with the son, the son walked 4 kilometers, and the son shared 4 8 12 minutes.

So when the Father catches up with the Son for the second time, the Son shares 24 minutes 24 + 8 + 8 = 40 points, so when the Father catches up, it is 8:40

900/2=450km

A (450 + 50) 2 = 250

Another 200

400 200 = 2 points.

The key to this question is Dad's speed.

Dad chased him on a motorcycle and caught up with him 4 kilometers from home. Then Dad went home immediately, and when he got home, he immediately turned back to chase Xiao Ming, and when he caught up with him again, it was exactly 8 kilometers away from home, and he knew that Dad and his speed ratio was 12:4=3:1

So. 8 + 8 * 3 = 32 points.

A and B set off from school to a certain place at the same time, A is at a speed of 9 km per hour, B is at a speed of every hour, B stops for 4 hours on the way, so it is 1 hour later than A, and the distance from the school to a certain place is found.

Solution: Set the distance to xkm

x/9+1=x/

Hence x=945 17=

The school organizes a spring outing, A has something to do and didn't catch up with the tourist bus, so A takes a taxi to catch up, and the driver says "If the taxi speed is 80km per hour, you can catch up in hours, and if the taxi speed is 90km per hour, you can catch up in 40 minutes" Q: What is the estimated speed of the taxi driver?

Let the speed of the tourist car be xkm h, and the two ends of the equation are the distance that has fallen.

80-x）×

So x = 72 km h

A boat to and from A and B ports, it is known that the speed of the boat in still water is 9km per hour, the ratio of the time taken to go retrograde and forward is 2:1, one day, the day is raining, the water speed is 2 times the original, the boat round trip shares 10h, ask, how many km are the two ports of A and B?

Solution: Let the water velocity be xkm h

9+x）：(9-x)=2：1

x=3km/h

Set up a meeting YKM

y/（3×2+9）+y/（9-3×2）=10

y=25km

From A to B, first go down the mountain and walk on a flat road, someone rides a bicycle from A down the mountain at a speed of 12 km/h, and then passes through the flat road at a speed of 9 km/h, and it takes 55 minutes to get to B; When he came back, he passed through the flat land at a speed of 8 km/h, and went up the mountain at a speed of 4km/h, and it took him an hour to return to A, asking: What is the difference between A and B?

Solution: Let the difference between A and B be xkm, the mountain road AKM, the flat road BKM, A+B=X

a/12+b/9=55/60

a/4+b/8=

a=6kmb=3km

x=9km

It takes 15 minutes to go up the mountain, 10 minutes to go down, 18 minutes to go up and 12 minutes to go down.

The time from A to the summit is:

B's time at the foot of the mountain is:

Because it takes 18 1 3 6 (minutes) for B to climb the AC section, B is at the bottom of the mountain within 6 minutes when A reaches the top of the mountain, and A can see B climbing up the AC section from the top of the mountain. i.e. time of A and time of B 6.

Comparing the above data, (A65, B60), (A215, B210) are in line with the topic.

At 215 minutes, A went to the top of the mountain for the 9th time, and the second time at the top of the mountain, B was seen climbing up the AC section.

The distance traveled by train in 1 minute is 1

Each tram meets an oncoming tram every 6 minutes, indicating that the distance between the two trains is 12.

Zheng Zheng encounters an oncoming tram every 5 minutes, indicating that the distance of 5 minutes of cycling = tram travel (12-5) = 7

In the same way, Hao Hao's distance of riding for 6 minutes = (12-6) = 6, then Zheng Zheng's speed is 7 5, Xiao Wang is 6 6 = 1, and the total distance is 56

So when Zheng Zheng and Hao Hao met on the way, they had walked for 56 (7 5+1) = 70 3 minutes.

Have fun! If you don't understand, please ask!

From the inscription, it can be seen that Wu Wu's speed is the same as that of the tram. It only requires Zheng Zheng and the tram to travel from the two places at the same time, and it takes a long time to meet.

Assuming that Zheng Zheng and the tram start at the same time, the distance from the tram on the opposite side is 6*2 56=3 14

The time it takes for Zheng Zheng and the tram to meet each other from A and B at the same time is 5 (3 14) = 70 3 (minutes).

The arithmetic synthesis is: 5 (6*2 56)=70 3 (minutes) The problem can be solved in three steps.

The first sentence of the question should say that two people are ...... at the same time

The ratio of velocity is 65:70=13:14

The ratio of time is 13:14

The first time to return to the starting place at the same time, B should have exactly one more round trip than A, 2 200=400, otherwise it is impossible to meet at the starting point at the same time.

When they met for the second time, A and B traveled a total of 3 courses.

A is OK: 7 (7+11) 3 = 7 6 The distance between A and B is: 7 6-1 = 1 6 whole courses.

ab distance: 80 1 6 = 480 km.

Set: AB two places are x kilometers apart. The speed ratio of A and B is 7:11, and the ratio of the distance traveled by A and B is also 7:11.

x+80):(2x-80)=7:11

11(x+80)=7(2x-80)

11x+880=14x-560

3x=1440

x = 480 km.

A: The distance between the two places is 480 kilometers.

The ratio of A to B is 7:11

The distance ratio is also 7:11

Let AB be separated by x km.

It is easy to see that A has walked x+80 and B has walked 2x-80 (x+80) :(2x-80) = 7:11

x = 480 km between 480ab.

Let the length of the car be x and the speed be y, then.

1000+x=60y

1000-x=40y

So x=200m, y=20ms

That is, the length of the train is 200m and the speed is 20m s

It's more convenient to draw a picture and look at it)