# Re: trip problem

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Posted by Joel on November 05, 2002 at 18:45:40:

In Reply to: Re: trip problem posted by Denis Borris on November 05, 2002 at 18:13:00:

: : 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.
: : from beginning to end, start with one girl on bike.
: : 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 !! Very nice. Saves a lot of number-shlepping.

: 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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