A chauffeur always arrives at the train station at exactly five o'clock to
pick up his boss and drive him home. One day his boss arrives an hour early,
starts walking home, and is eventually picked up. He arrives at home twenty
minutes earlier than usual. How long did he walk before he met his
chauffeur?
-Harry
Fifty minutes. It's easier if you've seen it before.
Steve
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3 days, 4 hours, 10 minutes.
Assuming trip to and from home always take same amount of time:
Normal day:
Chauffeur leaves home at 5:00 - x
Arrives at station at 5:00
Everyone returns home at 5:00 + x
Special day
Boss starts walking at 4:00
Chauffeur leaves home at same 5:00 - x
Chauffeur meets boss at 5:00 - x + y
Everyone returns home at 5:00 - x + y + y which is 5:00 + x - 20
5:00 - x + 2y = 5:00 + x - 20
-x + 2y = x - 20
-2x + 2y = -20
-x + y = -10
Chauffeur meets boss at 5:00 - x + y = 5:00 - 10 = 4:50
Boss walked for 50 minutes.
<<Remove the del for email>>
Let L be the distance between the house and the station (turns out to be
irrelevant), V the speed of the car and v the (walking) speed of the man.
The tw be the time that the man has walked on the 'early' day.
Also, let a = v/V.
Let the time saved on the special day be ts and Te be the time by which the
boss arrives early on that day.
Clearly, the time saved on that day is the time the car has *not* travelled
as the boss is met on the way to the station. If the man has walked for a
time t1, then
ts = 2*a*tw (1)
Now, to work in the fact that the boss is early by Te. The time when the
chauffer meets his boss is the time he leaves the house (as normal) plus
the time he takes to drive to the meeting point. Let t=0 correspond to
the normal arrival time of the boss. Then, time at which they meet is
-L/V + (L-v*tw)/V = - v*tw/V = -a*tw
But, as the boss started to walk at -Te, the time they meet is also
given by
- Te + tw
So, a*tw = Te - t1
or, tw = Te/(1 + a) (2)
Eliminating a between (1) and (2)
tw = Te - ts/2 (3)
With Te = +1 and ts = 1/3 h
tw = 5/6 h
One now finds that a = v/V = 1/5.
Rather nice. Should keep this in mind.
--
Surendar Jeyadev jey...@wrc.xerox.com
Spoilt my whole day and I resolved to try and not be caught out in such a
way again.
First the problem only makes sense if you assume the chaufeur drives
directly from the house to arrive at the train station at the normally
meeting time and that the journey takes the same time in both directions.
Given this you know on this occasion the chaufeurs round trip is 20 minutes
less. eg 10 less minutes less in both directions.
Therfore he meets the boss 10 minutes early and the boss has been walking
for 50 minutes.
"Surendar Jeyadev" <jey...@wrc.xerox.bounceback.com> wrote in message
news:9vt6oh$ej6$1...@news.wrc.xerox.com...
Definitely gets the cake, IMHO.
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Awww .... don't feel so bad.
>First the problem only makes sense if you assume the chaufeur drives
>directly from the house to arrive at the train station at the normally
>meeting time and that the journey takes the same time in both directions.
>
>Given this you know on this occasion the chaufeurs round trip is 20 minutes
>less. eg 10 less minutes less in both directions.
>Therfore he meets the boss 10 minutes early and the boss has been walking
>for 50 minutes.
And, it still does not tell me the ratio of the speeds. You see, there
is a lot more that can be extracted from a problem than the answer. And,
one cannot generalise. As it often happens, the real fun begins only
after you know the answer.
--
Surendar Jeyadev jey...@wrc.xerox.com
<<snipped>>
> >Given this you know on this occasion the chaufeurs round trip is 20
minutes
> >less. eg 10 less minutes less in both directions.
> >Therfore he meets the boss 10 minutes early and the boss has been walking
> >for 50 minutes.
>
> And, it still does not tell me the ratio of the speeds. You see, there
> is a lot more that can be extracted from a problem than the answer. And,
> one cannot generalise. As it often happens, the real fun begins only
> after you know the answer.
> --
Hi SJ,
The original post did not ask for the ratio of speeds. Even if it did, IMHO
the line above your post "... 10 min early, .... walking for 50 min" etc.
gives the ratio clearly, as the distance referred to is the same, while the
times quoted are for chauffer and boss respectively.
Cheers.