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Tuesday extra

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Tuesday extra

#1

Tuesday extra

Every day a man meets his wife at the train station after work, and she drives him home. She always arrives exactly on time. One day he catches an earlier train and arrives an hour early. He immediately begins walking home along the same route they usually drive. Eventually his wife sees him on her way to the station and drives him the rest of the way home. When they arrive home the man notices that they arrived 20 minutes earlier than usual. How much time did the man spend walking?

Re: Tuesday extra

#2

Dale Lenz

Re: Approx. 2.5 mi?

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#3

Re: Tuesday extra

(man time walking) + (man riding with wife time) = 40 minutes. Without the distance of the commute or relative speeds of the two modes of transportation I can't figure this out.

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#4

Re: Tuesday extra

The time walking equals 40 times the car velocity divided by the car velocity minus the walking velocity.

or tw = 40 x Vc /(Vc-Vw)

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#5

time, not distance ;-)

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#6

Dale Lenz

Re:

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#7

Dale Lenz

Re: less than 40 minutes, : ' >

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#8

Re: Tuesday extra

ten less than last puzzle.

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#9

;-)

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#10

sorry, not it.

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#11

Explanation

Wife leaves at same time, but gets home 20 minutes early.

Therefore her round-trip is 20 minutes shorter.

Therefore oneway (to pick up hubby) is 10 minutes shorter.

Since she leaves at the normal time, she picks up hubby 10 minutes before she usually does.

He got to the station 60 minutes before he usually does, and walks for 50 minutes before wifey picks him up.

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