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Posted by Chris on January 30, 2011 – 8:42 pm

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?

This post is under “Mathemagic” and has 18 respond so far.
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18 Responds so far- Add one»

  1. 1. SP Said:

    She walked for 1 hour.

  2. 2. Wizard of Oz Said:

    I remember that someone posted a simple and very elegant solution to this the first time it was posted on ToM.

    Wish I could remember what it was!

    It has to be less than 1 hour because the chauffeur is picking her up on the way to the station, i.e. before he gets there at 5 pm, so it must be a bit earlier than that.

  3. 3. Rohan Said:

    She walked for 50 mins.

  4. 4. Edison Maxwell Said:

    Since the chauffeur drove twenty minutes less when he otherwise would have taken longer, the time she spent walking was exactly those twenty saved minutes.

  5. 5. Wizard of Oz Said:

    I remember now how we get to the boss arriving home 20 minutes early. It represents the time the chauffeur takes to get from where he picked up his boss to the station and back to the pickup point again. From this point he resumes his normal journey to the boss’s home. So he picked her up 10 minutes before 5 o’clock, which means that she walked for 50 minutes.

    Neat. I’m sure Rohan would have expressed it better if he’d taken the time to do so.

  6. 6. Edison Maxwell Said:

    Look at it this way. If the chauffeur had left an hour earlier as he should have, his boss would’ve been home an hour earlier than usual. Because she only got home 20 minutes earlier (half of the 40 minutes saved by the chauffeur), what happened to the other 40 minutes? She was walking, that’s what. Every other piece of information in the problem is irrelevant.

  7. 7. Dual Aspect Said:

    I believe Rohan and Wiz have it.

    For the car to arrive home 20 mins earlier than usual the chauffeur has saved 10 mins on each leg of the journey therefore he met the boss at 4:50 instead of 5pm and so saved ten minutes each way.

    If her met at 4:50, and she left the station at 4:00 she walked for 50 mins.

  8. 8. Chris Said:

    50 minutes it is i.e. they met at 4:50.

    Dual Aspect was the one who answered neatly the last time I posted this one.

  9. 9. Sheika Said:

    I agree with you Edison! She walked for 40 minutes.

  10. 10. Chris Said:

    Hi Edison (and Sheika). If the chauffeur had set off an hour early and hence arrived home an hour early, the driving time won’t have been reduced at all.

    However, by setting off at the usual time, the driving time was reduced by 20 minutes. That’s 10 minutes less driving each way. So the chauffeur must have met the boss 10 minutes earlier than the usual 5:00, hence they met at 4:50, hence the boss had been walking for 50 minutes.

    Jut for an extension; if the boss happened to arive home just as the chauffeur was about to leave, then she’d have got home at 4:50, and so would normally get home at 5:10.

  11. 11. Edison Maxwell Said:

    I agree. Good puzzle!

  12. 12. zaheer Said:

    ha ha ha ha

  13. 13. Serik Said:

    70 minutes

  14. 14. Eric Said:

    Chris, the first part of your reply makes sense (the 50 minutes) but I don’t agree with that last paragraph. We do not know how long it takes from her home to the station (coule be one hour, could be ten!). We cannot just make that addition because her walking speed is different from the car’s speed. The times would greatly differ. 50 minutes should be the answer anyway

  15. 15. Chris Said:

    Hi Eric, I was hypothesising. However, we can say that they could not get back before 5:10 normally.

  16. 16. Muhammad Said:


  17. 17. grimace Said:

    50mins was how long she walked

  18. 18. guest Said:

    50 minutes

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