There are three cars, a, B and C, each driving at a certain speed from a to B. B starts 10 minutes later than C and catches up with C 40 minutes later; a starts 20 minutes later than B and catches up with C 1 hour and 40 minutes later?

The distance that C takes 130 minutes, B takes 130 × 4050 = 104 (minutes), suppose a takes X minutes, we can get: 100104 = XX + 20104x = 100 (x + 20), 104x = 100x + 2000, & nbsp; 4x = 2000, & nbsp; X = 500. A: it takes 500 minutes for a to catch up with B

There are three cars, a, B and C, each driving at a certain speed from a to B. B starts 10 minutes later than C and catches up with C 40 minutes later; a starts 20 minutes later than B and catches up with C 1 hour and 40 minutes later?

There are three cars of a, B and C each driving from a to B at a certain speed. B starts 10 minutes later than C, catches up with C 40 minutes later, and a starts 2 minutes later than B
How many minutes does it take for a to catch up with B?

After 50 minutes, a can catch up with you! The specific problem-solving process is as follows: C starts at x at zero, then it starts at y at ten, and a starts at Z at thirty. After 40 minutes, it shows that the distance is the same, so 40y = 50x, similarly 40z = 70X, -- > y = 5 / 7z. Let a catch up time be t, then ZT = 5 / 7z (T + 20) -- > t = 50
So it takes 50 points for a to catch up!