If the speed of a car from A to B is increased by 20%, it can arrive ahead of time. If you drive at the original speed 100 km and increase the speed by 30%, you can also arrive at the same time in advance. What is the distance between a and b?

If the vehicle speed is increased by 20%, the speed ratio before and after the speed increase is =1:(1+20%) =1:1.2 = 5: 6.

The time ratio is 6: 5.

In other words, the original time is 1, and the time taken after speeding up is 5/6.

Advance the original time by 1-5/6= 1/6.

When the vehicle speed is increased by 30%, the time ratio before and after the speed increase is =1:(1+30%) =10:13.

Time ratio = 13: 10

According to the meaning in the question, the two advanced times are the original time 1/6.

So the original time 100km after the second speed increase = (1/6)/(1-113) =18.

Then the driving time 100 km =1-13/18 = 5/18.

So the distance between Party A and Party B =100×1(5/18) = 360 kilometers.

If you drive at the original speed 100 km, it will also speed up by 30%.

Then it will exceed100km:100x30% = 30 (km).

( 1+20%):( 1+30%)= 12: 13

Full journey: 12x30=360 (km)

It takes 8 hours for the truck to go from A to B, 6 hours for the bus to go from B to A, and 2 hours for the truck to start first and then the bus to meet at a distance of 30km. What's the distance between a and b?

Think of the whole distance as 1.

Freight car speed = 1/8

Bus speed = 1/6

Then the truck runs for 2 hours 1/8×2= 1/4.

When two cars meet, go 1- 1/4=3/4.

Speed ratio of truck and bus = inverse ratio of time = 6: 8 = 3: 4.

When they met, the truck traveled 3/4×3/(3+4)=9/28.

Truck1* * has been driven 1/4+9/28=4/7 (the midpoint has been exceeded).

So the distance between Party A and Party B is 30/(4/7-1/2) = 30/(114) = 420km.