simple for you maybe...

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: : : 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 : : 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 : : thank you for doing the hard part (and the complement). I just didn't want to think any further. Very nice setup! you actually make it seem easier than I thought it would be!!

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