I have a math problem, and I don t want you to write The Journey Problem .

(6+4)=3 hours, this simple should be understandable.

The dog's run is divided into two parts:

One is the one that the dog walked before meeting B:

10*30 (10+4), i.e. the speed of the dog multiplied by the time it takes to meet;

The other is between the dog and B meeting and A and B meeting

The time difference (3-30 14) multiplied by the speed will do.

This is the most primitive algorithm, easy to understand.

In fact, the dog and A set off at the same time, although they run fast, but the time used to run from front to back is the first time to count the time when A and B meet for 3 hours, so it is very easy to calculate.

30 (6+4) = 3 hours of encounter.

No matter how the dog runs, it only runs for 3 hours, so the dog runs a total of 10*3 30km

30 divided by (4+6) = 3 hours 3 hours will meet 30 divided by (6-4) = 15 kilometers.

The two met three hours later.

When A and B met, the dog had covered a total of 30 km

Jumping out of the question, it's easy to do.

A and B walk opposite each other, A walks 6 km per hour, B walks 4 km per hour, and after a few hours, the two meet and a person takes 10 km h to cover the 30 km journey is a meaning, that is, 3 hours.

If you don't worry, you can also set up an equation to verify it, and set the two to meet after x hours, there is.

6x+4x=30 x=3

The latter problem is still the same to jump out, when A and B meet, the dog stops running, and it has taken 3 hours from the beginning to this time, and the speed of the dog has always been 10km h, that is, the distance is 10 * 3 = 30km

Don't get caught up in the topic.

Let the original speed be v km/h, then there is: 50 V (1-40%)-50 V = 1 + 1 3;

v = 25 (kmh).

It turns out to take t hours to arrive again, as is known to have:

25t=25+(t+3-1)*25*(1-40%)；

Get: t = hour, so: 25 * km.

Select: e

Question 9: Choose E for this math problem

They walked for another 2 hours and were 36 meters apart, which means that there were 36 meters, and after the two met, they walked 36 meters on their backs, and walked a total of 36 + 36 = 72 meters in these 2 hours.

Then their speed sum is 72 2 = 36 meters per hour.

At the beginning, they walked together for 2 hours and were 36 meters away from meeting, so AB was 36*2+36=108 (meters).

Hello Gou Zhibin:

Upstairs the answer is not right!

When they met, Xiao Zhang walked more than Xiao Wang:

km) more per hour (1 hour 30 minutes):

3 km) The speed of the sheet is per hour: km/h).

Don't do it, all the unknowns, what the distance, whatever, set the unknowns, it's OK, it's easy to understand, it's one-dimensional, not two-dimensional.

Set the speed of the car to x, and the speed of the car is y

x-y)*12=(x+y)*4

8x=16y

x=2yx:y=2:1

Therefore, the interval between human departures is as wide as time, which is 12 [(2-1) not opening 1]=6, or 4*[(2+1) 2]=6

The answer is 8 minutes, and the essence of this question is not the question of the encounter between people and cyclists, but the problem of chasing between buses and people, buses and cyclists. Because the departure interval is the same and the bus speed is the same, the distance between the two buses is also the same. We might as well set the distance between the two buses is S, the speed of the person is V1, the speed of the cyclist is 3V1, and the speed of the bus is V

Suppose the first bus overtakes the person, then according to the topic, the second bus exceeds the person after 10 minutes, and the distance between the person and the second car at this time is s, so the s=10(v-v1) is obtained, and the relationship between the cyclist and the bus can be obtained s=20(v-3v1), and the two equations get v=5v1, and the interval time is t=s v=40v1 5v1=8 minutes.

Solution: Let the person's walking speed be x kilometers per hour, then the question says: 4 x+10 (x+24)=14 (x+8), i.e., 64x=384, so x=6 Answer: The person's walking speed is 6 kilometers per hour

If the person walks at a speed of x km-h, the speed of cycling is (x+8) km-h, and the speed of a car is (x+24) km-h.

10/(x+24)+4/x=(4+10)/(x+8)10x(x+8)+4(x+24)(x+8)=14x(x+24)10x²+80x+4(x²+32x+192)=14x²+336x4x²+256x=4x²+128x+768128x=768

x=6 means that the person is walking at a speed of 6 km/h.

Let the walking speed be v, then there is:

4 V + 10 (V+24) = 14 (V+8)14V + 96) [V*(V+24)] = 14 (V+8)14V*(V+8) +96*(V+8) = 14*V*(V+24)14V 2 + 208V + 96*8 = 14V 2 + 336V, So, 128V = 96*8

v = 6 km/h

Let the walking time be T1, the riding time be T2, and the cycling time be T3.

t1+t2=t3

Namely. 4 V+10 (V+24)=14 (V+8) receives V=6km H.

If the walking speed is x, then the cycling speed is x+8 and the car is x+24.

The total length of land ab is 4 + 10 = 14 km.

The time to go is 4 x +10 (x+24).

The time of return is 14 (x+8).

Two equals, column equations, solve and you're done.

Walking speed is 6 km/h.