A and B leave AB at the same time and meet at 2 o'clock. After meeting, the two cars continue to move forward. When a arrives at B, B is 60km away from A. It is known that the speed of a car is a car A and B leave AB at the same time and meet at 2 o'clock. After meeting, the two vehicles continue to move forward. When a arrives at B, B is 60km away from A. It is known that the speed of B is 2 / 3 of that of a, so find the distance between AB and B! With the sixth grade knowledge, this is from the math paper

A and B leave AB at the same time and meet at 2 o'clock. After meeting, the two cars continue to move forward. When a arrives at B, B is 60km away from A. It is known that the speed of a car is a car A and B leave AB at the same time and meet at 2 o'clock. After meeting, the two vehicles continue to move forward. When a arrives at B, B is 60km away from A. It is known that the speed of B is 2 / 3 of that of a, so find the distance between AB and B! With the sixth grade knowledge, this is from the math paper

Speed ratio 3:2
The distance ratio is 3:2
60x3 △ 3-2 = 180km

Party A and Party B go from AB to ab at the same time, return to each other's departure place immediately, and meet at 60km away from B. the speed ratio of Party A and Party B is 2:3, and the distance between AB and Party B is 2:3

Can we use the equation?
The distance between the two places is x km
Then (x + 60): ((2x-60) = 2:3
3（X+60)=2(2X-60)
3X+180=4X-120
X=300

A walks 5km per hour. Two hours after starting, B goes after a by bike. If B's speed is 20km per hour, how many minutes does B catch up with a?

Let B catch up with a in X hours
5*2+5*x=20*x
15x=10
x=2/3
2 / 3 hours = 40 minutes

If the speed is 20km / h, how many minutes can I catch up with a?

5km/h*2h=10km
10/（20-5）=2/3（h）
2H / 3 = 2 / 3 * 60 = 40 points
B can catch up with a in 40 minutes