Physics, asked by average66, 11 months ago

A taxi leaves the station X for station Y every 10 minutes. Simultaneously, a taxi leaves the station Y also for station X every 10 minutes. The taxis move at the same constant speed and go from X to Y or vice-versa in 2 hours. How many taxis coming from the other side will each taxi meet enroute from Y to X ?​

Answers

Answered by neerajmehta449p61pbp
39

Answer:

There will be 11 taxis at a time in between stations coz  after every 10 minutes lefts one .hence each taxi meets 22 taxis during too & fro . so including herself there will  (11+11+1)23 taxis  on the way to meet each other en-route .

Answered by archishman37
8

Answer:

The taxi will meet 12 other taxis.

Explanation:

  • Let two taxi start at the same time from X & Y.
  • As their speed is same, the taxi will not meet any one of the taxi starting from Y.
  • So it can meet the taxis starting from X only.
  • So in the two hours in which the taxi reaches X, 12 [2*60/10](time*time interval of starting of a taxi) taxis will start off.
  • But due to same speed, it will meet the first taxi after 1 hour, second at 1 hour & 5min, third at 1 hour and 10 min, and so on.
  • The last one(the 13th taxi) will meet it at X.
  • So it will meet 12 taxis enroute from Y to X.
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