A man commutes to work by train. While returning from office when he reaches the railway station, he is picked up by his chauffer (Driver). The distance between his house and the railway station is 25 km and the chauffer starts from the man's home towards the station at 5:30 PM. Take the speed of the car as 50 km/hr. One day his chauffer somehow started at 5:42 PM from the man's house towards the station, at his usual daily speed. As he did not reach on time, the man took a cab for home. After going certain distance he saw his car coming, and he covered rest of the journey by his car. He returned home exactly at the same time which he used to reach daily. What was the speed of the cab?
(Assume that, as the man comes out of the station, his car just arrives and take the time of getting the man into his car as negligible. Aslo take the time he took moving from cab to car as negligible)
50/3 km/ hr
25 km/ hr
50 km/ hr
Cannot be determined
None of the Above
Answer :50 km/ hr
Solution (Fast Approach)
As the man reaches railway station at the same time daily. It does not matter how much late the chauffer starts from home. Man covers 25 kms journey in same 30 minutes time, which he used to do daily.
hence the speed of cab should be same as the car = 50 km/ hr
Special tricky extension of the above problem
Take the normal speed of the man's car same as above (50 km/ hr). One day the chauffer starts late at 5:36 PM and drives at his usual daily speed. That day the train came early by 22 minutes. As the train reached the station early, the man started walking towards his home. After some distance he found his car coming and he covered rest of the journey with the car. What was the speed of the man if he reached home exactly at the same time as he used to reach daily. All the assumptions in the above prob holds true for this extension.
10 km/ hr
6 km/ hr
5 km/ hr
Cannot be determined
None of the Above
Answer :6 km/ hr
Solution (Fast Approach)
Car's speed is 50 km/hr and distance between house and railway station is 25 km, hence a total distance of 50 km. Car will take 60 minutes to reach the station and come back home.
Hence the man reaches home at 6:30 PM daily.
Even after starting late they reach home on time (6:30 PM). Hence total time taken by chauffer, if he starts at 5:36, is 6:30 - 5:36 = 54 min. Therefore the car travels 54/2 = 27 minutes one way towards the station.
Total Distance covered in 27 min = (50*27)/60 = 22.5 km. Hence the man has to walk 25 - 22.5 = 2.5 km
It's 5:36 + 27 = 6:03 PM when the chauffer and the man met each other.
As the train came 22 min early (5:38 PM) and the meeting time is 6:03 PM. The total time taken by the man to cover 2.5 km = 22 +3 = 25 min.
Speed of the man = (2.5*60)/25 = 6 km/ hr
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