Every day, Bill takes the train to travel from his work back to Llareggub, his place of residence. Usually, he arrives at the station of Llareggub at six o'clock, and exactly at that moment he is picked up by his wife by car. Yesterday evening, Bill took an earlier train, without informing his wife, and therefore he already was at the station of Llareggub at five o'clock. He decided to walk part of the way to meet his wife. When he met the car with his wife, he drove home with her. In this way, they were home ten minutes earlier than normal.
For how long did Bill walk yesterday evening?
Originally posted by CoconutI think this is right, right?
55 mins
The wife usually leaves at some specific time, travels to the station in time t(normal), picks up the husband, and travels home in the same amount of time t(normal).
Yesterday, the wife left at the same specific time but met her husband on the road having traveled only time t(early), and drove home in the same amount of time t(early). When they arrived home, it was 10 minutes earlier than usual.
So, we have:
2*t(normal) - 2*t(early) = 10 minutes
t(normal) - t(early) = 5 minutes
Since we also know that when the wife drives for time t(normal) she arrives at the station at 6:00pm, she must have picked up her husband 5 minutes early at 5:55pm. Since the husband left the station at 5:00pm, he must have been walking for 55 minutes.
Originally posted by PBE6YEP! Correct. Too easy, I guess.
I think this is right, right?
The wife usually leaves at some specific time, travels to the station in time t(normal), picks up the husband, and travels home in the same amount of time t(normal).
Yesterday, the wife left at the same specific time but met her husband on the road having traveled only time t(early), and drove home in the same amount of time ...[text shortened]... 5:55pm. Since the husband left the station at 5:00pm, he must have been walking for 55 minutes.
BTW - Elegant solution, you...