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A, B, C three car each heart a certain speed from a to B, B than a 10 minutes late departure, 30 minutes after the departure to catch up with a, C than B 15 minutes late departure
Is the question like this: A, B and C start from a to B at a certain speed, B starts 10 minutes later than a, catch up with a 30 minutes later, C starts 15 minutes later than B, catch up with a 45 minutes later, how many minutes after C starts to catch up with B?
Let a speed be a meter per minute, B meter per minute be B, C meter per minute be C, and C can catch up with B in X minutes
So there are: 1. (10 + 30) a = 30b; 2. (45 + 15 + 10) a = 45C; 3. (15 + x) B = Cx
From 1.2 we get b = 4 / 3a, C = 14 / 9A. If we take B = 4 / 3a, C = 14 / 9A into 3, a will be cancelled, and the solution is x = 90
So C will catch up with B 90 minutes after departure
A, B and C drive from a to B at a certain speed. B starts 10 minutes later than C and catches up with C 40 minutes later; a starts 20 minutes later than B and leaves
Catch up with C in one hour. How long does it take for a to catch up with B? We need to use ratio instead of equation!
Let C speed be 1,
Then B speed = 1 × (10 + 40) △ 40 = 5 / 4
A speed = 1 × (10 + 20 + 100) △ 100 = 13 / 10
Time for a to catch up with B = 5 / 4 × 20 (13 / 10-5 / 4) = 500 minutes
There are three cars, a, B and C, each driving from a to B at a certain speed. B starts 10 minutes later than C, and after 40 minutes it catches up with C. A starts 20 minutes later than B, and after 50 minutes it catches up with B______ How many minutes to catch up with C?
[(20 + 50) × 50] × [(40 + 10) × 40], = [70 ／ 50] × [50 ／ 40], = 75 × 54, = 1.75 (Times); (10 + 20) × 1.75-1, = 40 (minutes). A: after 40 minutes, a catches up with C. so the answer is 40
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 after starting, a starts 10 minutes later than B, catches up with C 60 minutes after starting, and asks a how many minutes after starting to catch up with B
Suppose the speeds of a, B and C are a, B and C respectively, and a catches up with B in t minutes after starting. According to the meaning of the question, (10 + 40) C = 40B, (20 + 60) C = 60A, then there is a: B: C = 16:15:12, (10 + T) B = at, and the solution is t = 10, a catches up with B in 10 minutes
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