A and B set out from a and B which are 180 kilometers apart. A rides a bicycle and B rides a motorcycle. They run at a constant speed in the same straight line. It is known that B's speed is three times that of A. if a starts one hour before B starts, and B meets a two hours later, what's the speed of a and B?

A and B set out from a and B which are 180 kilometers apart. A rides a bicycle and B rides a motorcycle. They run at a constant speed in the same straight line. It is known that B's speed is three times that of A. if a starts one hour before B starts, and B meets a two hours later, what's the speed of a and B?

Suppose the speed of a is x km / h, then the speed of B is 3x km / h. according to the meaning of the question, we get: x + 2 (x + 3x) = 180, and the solution is: x = 20. A: the speed of a and B is 20 km / h and 60 km / h respectively

A and B set out from a and B at the same time. A rode a bicycle, B drove a car, and drove along the same route at a constant speed. After three hours of departure, they met. It is known that B traveled 90 kilometers more than a when they met, and arrived at a after 1:00. What are the speeds of a and B?

As shown in the figure, let B's speed be x km / h, and the meeting point be C, then BC = x km, AC = x + 90. From: 3x = x + 90, we can get: x = 45. So: A's speed is 453 = 15 km / h, B's speed is 45 km / h. answer: A's speed is 15 km / h, B's speed is 45 km / h

Both Party A and Party B go from place a to place B. Party A rides a bicycle for 10 kilometers per hour. After 1.5 hours, Party B drives a car,
After 50 minutes, I arrived at B with A. what's the speed of B? (solve the equation with one variable and one degree)

Set B speed x km / h
10*1.5=5/6（x-10）
15=5/6x-25/3
x=32