Both cars and minibuses have to travel from A to B

The speed of a minibus is four-fifths that of a sedan.

Then V minibus: V sedan = 4:5

v = s ts t minibus: s t sedan = 4:5

T sedan: T minibus = 4:5 (S equal).

The sedan departs 11 minutes later than the minibus from place A, but arrives 7 minutes earlier than the minibus, so the sedan takes 18 minutes less than the minibus.

That is, T minibus - 18 = T sedan.

i.e. (T minibus-18): T minibus = 4:5

5t minibus - 90 = 4t minibus.

T minibus = 90

The minibus arrives in 90 minutes without stopping.

It takes 10 minutes to stop at the middle of the stop.

So it takes 90 + 10 = 100 minutes = 1 hour and 40 minutes.

10 a.m. + 1 hour 40 minutes = 11:40

Both the sedan and the minibus go from A to B, and the minibus speed is 4 5 of the speed of the sedanThe minibus stops for 10 minutes at the midpoint between the two places, and the sedan does not stop halfway. The car departs 11 minutes later than the minibus in place A and arrives in place B 7 minutes earlier, and the minibus departs at 10 o'clock, so what time is it when the car overtakes the minibus?

Solution: From the meaning of the question, it can be seen that the speed ratio of the minibus to the car is 4:5, and the actual travel time of the minibus when reaching the end point is 8 minutes longer than that of the car;

The distance is certain, and the speed is inversely proportional to the time.

It can be seen that the car takes 32 minutes to complete the journey;

The bus journey takes 40 minutes without stopping.

The minibus departs at 10 a.m. and arrives at the midpoint at 10:20 a.m. Take a 10-minute break and start walking again at 10:30;

The sedan started departing at 10:11 and arrived at the midpoint at 10:27, when the minibus was still resting, so the sedan overtook the minibus at 10:27.

12 2 (80-72) (80+72) = 456 km.

How many kilometers apart a and b are between them.

The time for driving on a normal highway is x, and the length of an ordinary highway is y.

The length of the ordinary highway travel time The driving speed, ie

y=60x ..Length of the highway = travel time Travel speed = (

The length of the highway (two-thirds of the total length) = the length of the ordinary highway (one-third of the total length) 2, there is:

2y=100( .A binary linear equation is formed by and

y=60x ..2y=100( .Solution: x = 1 hour.

y = 60 km.

Then the length of the highway = 2y = 120 km.

The driving time on ordinary roads is 1 hour and the length is 60 km;

The driving time of the highway is hourly and the length is 120 km.

What's the problem? Is it the distance from A to B?

What you are asking for, child...

I'm also asking this question, hehehehe, it's fateful.

The speed ratio of sedan to truck is 5:4, and when they meet, the sedan Shenchun Town travels to Quansen Ctrip 5 (5+4)=5 9, and the truck travels 4 9

After the speed change, the speed ratio of the car and the truck is 5* 4* 5 6

When the car arrives at place B, the car travels 4 9The wagon should be 4 9 * 6 5 = 8 15 of the full width of the roughing

10 km is the whole journey 5 9 - 8 15 = 1 45

The whole journey is 10 1 45) = 450 km.

When the road is uneven, the speed is three-quarters of the original speed, so this uneven road is more than four-thirds of the original time, which shows that the time it takes now is one-third more than the original.

And the title says that it took 12 minutes longer than planned to walk this uneven road at the original speed, which is 3=

The original speed can be calculated from the original velocity divided by the hour = 6 kmh.

Therefore, the distance between the two places is: 6 km 5:20 = 32 km.

1 3 4 4 3 4 3-1 1 3 12 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 1 3 1 3 1 3 3 3 3 3 3 3 3 3 3 3 3

5:20 320:320

Let the original speed be the brigade x, then calculate that the original speed of Chi Zhen is 6Naturally, the distance is equal to 6 times the price of the yard, and the result is.

<> such as the hail of Yuliang sail slag Qitu town.

Let the AB distance be s, so the distance traveled in 72km h is (s 2-12) and the distance traveled in 80km h is (s 2+12) according to the equation that the time of these two distances is equal: (s 2-12) 72 = (s 2 + 12) 80

Solving the above equation yields: s = 456km

The idea is:

First set half of the distance as x, and traveled x-12km at a speed of 72km; Another X+12 at a speed of 80km;The time spent on these two journeys is equal;

The column formula is: (x-12) 72=(x+12) 80