30) A man arrives at the railway station nearest to his house 1½ hours before the time at which he had ordered his carriage to meet him. He sets out at once to walk at the rate of 4 km an hour, and, meeting his carriage when it had travelled 8 km, reaches home exactly 1 hour earlier than he had originally expected. How far is his home from the station, and at what rate was his carriage driven?
Answers
As they arrived 60 minutes earlier than usual time then carriage saved 60 mins on the round trip Home-Station-Home. So carriage saved 60/2=30 mins in one way Home-Meeting point.
Suppose usual arrival time of that man at the station was 16:00 (this also would be usual meeting time of the man and the carriage), he arrived 90 minutes earlier than that so he arrived at 14:30.
Now, as carriage saved 30 minutes on the way from home to station, then carriage met the man 30 minutes earlier than their usual time of meeting, which is 16:00, so they met at 15:30. Which means that the man walked for 1 hour (he arrived at 14:30 and met the carriage at 15:30) and hence covered 1 hour*4 km/hour=4 km, so the distance between the station and the home is 4+8=12 km.
As for the speed of the carriage: this 4 km (SM) would be covered by carriage in 30 minutes so the speed of the carriage is 8 km/hour.
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