Question

a taxi leaves the station X for station y every 10 mins . simultaneously a taxi leaves station y also for station X every 10 mins .
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Answer 1

Answer:

115 mins

Explanation:

With the taxis taking the same time to travel each direction then they must be travelling the same speed in opposite directions. This doubles the closing speed and so halves the time between meeting taxis compared with the 10 minute intervals for a stationary observer. Therefore each taxi will pass another taxi every 5 minutes.

Now as each taxi is enroute for 120 minutes it will meet 24 taxis enroute as long as taxis don’t leave each station at exactly the same times. For example if the first taxi is met after 2 minutes then there will be a further 23 taxis after each 5 min period subsequently, up to and including the 2+23 × 5 = 117th minute, total 24.

However if they do leave at exactly the same times then the number passed enroute reduces to 23 taxis, assuming that you don’t count either the taxi arriving at the same time as each taxi departs or the taxi departing at the same time as each taxi arrives. In this case the first taxi enroute is met after 5 mins and the 23rd after 23 × 5 = 115 mins.

Answer 2

taxi A leaves point X and reaches point Y at time t=120 min, when taxi A parts there's already a stream of taxis towards him at the same speed and 10 minutrs apart from each other but it is obvious that they travel relatively against A at twice the speed, so A encounters a taxi every five minutes for the next 120 minutes, say he parts after the first cab arrives at A and it is taxi B then there's taxi C (10 min behind) that meets A not in 10 but in five minutes, then coming D, E, F and so on every five minutes for the next 120 minutes, that is 120/5 =24 taxis besides B, so the answer is that A meets 25 taxis enroute.