# Re: trip problem

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Posted by Denis Borris on November 05, 2002 at 18:13:00:

In Reply to: Re: trip problem posted by T.Gracken on November 05, 2002 at 17:02:37:

: this problem is going to take more thought than I really want to use... so here's a little outline I believe may be the key.
: since the girls walk slow, they need to use the bike as much as possible.
: after a certain distance (or time) she will leave bike on ground and walk.
: guy will find bike before second girl and will ride it back to second girl.
: second girl will ride for certain distance and leave bike.
: now, either first girl finds bike and rides or guy finds bike and takes it back to trailing girl.
: continue/repeat/etc...
: good luck. As I said, other than this, I don't want to think that hard today.
: this problem is going to take more thought than I really want to use... so here's a little outline I believe may be the key.

Good thinking, Mr G! Just looked at it your way,
and it's really simple:

-to get to 200mile point at SAME time, the girls
HAVE to use bike same number of hours and of course
walk the SAME number of hours

-and all the guy can do with his Bike speed is
(as you say) drive the Bike back a bit, to help the girls

-so girl#1 leaves on bike @24mph, leaves it somewhere
(call it point X) and the guy (@9mph) catches up
to it and drives it back (@44mph)

-guy leaves Bike for girl#2, then heads the other
way (@9mph) to reach point 200 at same time as girl#1

-and of course girl#2 gets to point 200 on bike, at same time

-so we simply need a point X that makes that possible...

Let A=girl walk speed, B=girl bike speed
Let C=boy walk speed, D=boy bike speed
Let P= total distance

gal: bikes X miles, walks P-X miles
guy: bikes 2X-P miles, walks 2X miles
So:
X/B + (P-X)/A = 2X/C + (2X-P)/D
simplifies to:
X = [BCP(A + D)] / [2AB(C + D) + CD(B - A)]

Substitute 200/2/24/9/44 to get X = 144

So Girl#1 time is 144/24 + 56/2 = 6 + 28 = 34 hours

Guy walks 144: 144/9 = 16 hours
Bikes back 2 hours: leaves Bike at 56 mile point
Walks another 144 miles to end: 16 hours
Total: 34 hours; distance=144*2=288-44*2=200

Girl#2: walks to 56 mile point: 28 hours
Bikes remainder (144 miles) : 6 hours; total 34 hours

And I'm pretty sure we can use N (instead of 2) in
above equation, for general case:
X = [BCP(A + D)] / [NAB(C + D) + CD(B - A)]

If N=3, then you got 3 girls...and so on

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