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Friday Puzzler -- Early Quitting time

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Friday Puzzler -- Early Quitting time

#1

Friday Puzzler -- Early Quitting time

alexy

Bob drives from home to pick up his wife, Sally, at quitting time every day, and they get home at exactly 6:00 pm. One day, Sally gets off work 1/2 hour early. Since it is a nice day, she decides not to call Bob, but just to walk toward home along the same route they drive. Bob leaves at the normal time to pick her up, sees her walking, picks her up, and they return home. They get home at 5:50. How long did Sally walk?

The route Bob takes is the same each direction, and is symmetrical, in the sense that if there is a stretch where he drives 25mph coming in to pick Sally up, they will drive that same stretch of road at 25mph going home. His car is one of those special models that make instantaneous u-turns.

Re: Friday Puzzler -- Early Quitting time

#2

Re: Friday Puzzler -- Early Quitting time

Larry Barrett

25 minutes, I think. Seems like it should depend on the distance from home to work, rate at which Bob drives, rate at which Sally walks, ect, but I don't think it does.

Re: Friday Puzzler -- Early Quitting time

#3

Correct!

alexy

When you start throwing variables in for all the things you mentioned, it gets impossibly complicated.

The simple answer is that Bob made the trip in 10 minutes less than usual, and because of the given assumption of symmetry in the trip each way, that means his trip each was was 5 minutes shorter than usual. Thus, he picked up Sally 5 minutes before her usual quitting time, after she had been walking 25 minutes,

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