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
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