Two trains left two cities A and B at 15:00 to meet each other. The train leaving A arrived at B at 20:42

Two trains left two cities A and B at 15:00 to meet each other. The train leaving A arrived at B at 20:42, the train leaving B arrived at A 38 minutes later. Determine how many minutes after departure these trains met if the distance between A and B is 285 km.

The travel time of the 1st train will be equal to:
20: 42-15: 00 = 5 hours 42 minutes = 5 7/10 hours
a, the second:
5:42 + 0: 38 = 6 hours 20 minutes
Then the speed of the 1st train:
285: 5.7 = 50 km / h
, and the second:
285: 6 1/3 = 285 * 3/19 = 855/18 km / h
The approach speed is equal to the sum of the speeds:
50 + 855/19 = (950 + 855) / 19 = 1805/19 km / h
And the time before the meeting:
285: 1805/19 = 295 * 19/1805 = 295/95 = 59/19 hours = 3 2/19 hours.




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