A chauffeur always arrives at the train station at exactly five o'clock to pick up his boss and drive her home. One day his boss arrives an hour early, starts walking home, and is picked up by the chauffeur on the way out to the train station. They arrive at home twenty minutes earlier than usual. How long did she walk before she met her chauffeur?
For fifty minutes. She saved the chauffeur ten minutes of traveling time each way and thus was picked up at 4:50 pm rather than the usual time.
Wow, Anonymous....that was so simple....!!!
To explain the logic.......the time it takes for the chaffeur to drive to the station must be 20 minutes........to get there by 5PM, he would have to leave at 4:40PM. Then, he would get back home at 5:20PM. But...... he gets back home 20 minutes earlier than usual = 5PM....so......he leaves at 4.40PM, picks her up at 4:50PM and is back home at 5. And....at 4:50 PM....she's been walking for 50 minutes!!!!
BTW.....she must walk (1/2) the distance between her home and the station, because the chaffeur only drives for 10 minutes to pick her up instead of the customary 20.......so....he must travel (1/2) of the normal distance to the station to meet her....!!!
That's a pretty neat little logic problem that seems way harder than it really is...!!!