An engineer works at a factory. A car is sent from the factory which picks up the engineer from his home and takes him to the factory. One day the engineer got ready 10 minutes before the usual pick up time and without waiting for the car, he started walking to the factory. On his way, he met the car, took it and reached the factory 4 min. before the usual time. How long (in min.) did the engineer walk before getting the car?
Answers
Usual day :
Car starts from F at t=0, reaches station at T and again reaches at the factory at time 2T.
Person reaches 'S' at T−60. Car starts at t=0 from F. Person walks for timet and reaches point P at time T−60+t. At this time car also reaches 'P'. Car comes back at 'F' at time (2T−10). That means car takes time t−5 from F→P. That means car reach at 'P' at time T−5.
Now T−5=T−60+t⇒t=55min.
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Answer:
Usual day :
Car starts from F at t=0, reaches station at T and again reaches at the factory at time 2T.
Person reaches 'S' at T-60. Car starts at t=0 from F. Person walks for timet and reaches point P at time T-60+t. At this time car also reaches 'P'. Car comes back at 'F' at time (2T-10). That means car takes time t-5 from F→P. That means car reach at 'P' at time T-5.
Now T-5-T-60+tt-55min.
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