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Grade 12th passMechanics

An engineer works at a factory out of town. A car is sent for him from the factory everyday and arrive at the railway station at the same time as the train. One day the engineer arrieved at the station one hour before his usual time and without waiting for the car, started walking towards factory. On his way he met the car and reached his factory 10 minutes before the usual time. For how much time (in minutes) did the engineer walk before he met the car ? The car moves with the same speed everyday.

Profile image of Dhritiman Medhi
7 Years agoGrade 12th pass
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2 Answers

Profile image of Arun
ApprovedApproved Tutor Answer7 Years ago
The time duration for which the engineer walk before he met the car is time reached before-half the time he saved as the car is travelling a two way journey. 60-5=55 mins
Car saved total time in forward and backward path since the car saved total 10 minutes then it will contain both 5 minutes of forward and 5 minutes of backward motion. But for the person the time he is save for just the backward motion of car.
Profile image of Khimraj
7 Years ago
suppose the man reaches to work in the car by travelling in one hr = 60 mins
Now the next day he reaches the place one hr before ( another 60 mins) at his place and and started walking and traveled the car and reached 10 mins before the usual time so 120 mins ( 2 hrs) - 10 mins he saved = 110 mins 
In car if the time taken to reach the place is 1 hr so 110 - 60 mins = 50 mins 
So the guy walked for 50 mins and then entered the car and it took 60 mins more to reach the place of work