A and B vehicles leave from ab station at the same time. The speed of B vehicle is 9 / 10 of that of a vehicle. After the two vehicles meet at 5km from the midpoint, the two vehicles return to the original speed

A and B vehicles leave from ab station at the same time. The speed of B vehicle is 9 / 10 of that of a vehicle. After the two vehicles meet at 5km from the midpoint, the two vehicles return to the original speed

The title is not complete
When car a arrives at the destination, how far is car B from the destination?
Analysis: when meeting, a walks one more share than B, 10km away
The total distance of the two cars is 19, that is, the whole journey = 10 × 19 = 190km
When a is finished, B is 9 / 10
Therefore, there is still 1 / 10, that is, 19 km
5 × 2 ÷ (1 / 19) × (1-9 / 10) = 19 km

The walking speed ratio of a and B is 7:5, they walk from AB to ab at the same time, and meet half an hour later. If they start from AB in the same direction at the same time, how many hours will a catch up with B?
Come on!
Pay attention to go in opposite directions first. The problem is to start in the same direction
I want the process

7 + 5 = 12: Party A and Party B took 12 shares in total
When 12 / 2 = 6, 12 copies is 1, when walking 0.5, so AB has 6 copies
6 / (7-5) = 3A chasing B is to walk 6km more than B at a certain time, 7-5 more per hour = 2 formulas: distance / speed difference = time