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In a street, a cyclist and a pedestrian walk in the same direction. The speed of the cyclist is three times that of the pedestrian. Every 10 minutes, a bus overtakes the pedestrian, and every 20 minutes, a bus overtakes the cyclist. If a bus leaves at the same time from the departure station, how many minutes does it leave?
Suppose the interval of each bus is 1, then according to the meaning of the question, the speed difference between the bus and the pedestrian is: 1 △ 10 = 110, the speed difference between the bus and the cyclist is: 1 △ 20 = 120, because the speed of the cyclist is three times that of the pedestrian, so the speed of the pedestrian is: (110-120) △ 2 = 140, then the bus
In a street, a cyclist and a pedestrian walk in the same direction. The speed of the cyclist is three times that of the pedestrian. Every 10 minutes, a bus overtakes the pedestrian, and every 20 minutes, a bus overtakes the cyclist. If a bus leaves at the same time from the departure station, how many minutes does it leave?
Suppose the interval of each bus is 1, then according to the meaning of the question, the speed difference between the bus and the pedestrian is: 1 △ 10 = 110, the speed difference between the bus and the cyclist is: 1 △ 20 = 120, because the speed of the cyclist is three times that of the pedestrian, so the speed of the pedestrian is: (110-120) △ 2 = 140, then the bus
On a highway, there is a cyclist and a pedestrian. The speed of the motorist is three times that of the pedestrian. Every four minutes, a bus exceeds the pedestrian, and every eight minutes, a bus exceeds the cyclist. If the bus leaves at the same time, the station leaves every few minutes
Let the distance between the front and rear buses be 1 unit length, that is to say, in 4 minutes, the distance difference between the bus and the cyclist is 4V car-4v Walker = 1, that is, V car-v Walker = &# 188;; similarly, 8V car-8v rider = 1, that is, V car-v rider = 1 / 81 ^ [&# 188; + (&# 188; - 1 / 8) ^ (3-1)] = 1 ^ [&# 188;]
RELATED INFORMATIONS
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