# Re: trip problem

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Posted by Joel on November 05, 2002 at 18:33:55:

In Reply to: trip problem posted by RD on November 04, 2002 at 13:43:41:

: This is apparently from an old walk and bike puzzle:

: 2 girls and a guy want to travel 200 miles on a straight path.
: All 3 are together, and have one bike.
: The girls walk at 2mph, bike at 24mph.
: The guy walks at 9mph, bikes at 44mph.
: Only one can use the bike at a given time.
: What is the minimum time in which they can all
: get to the 200 mile point, and how do they do it?

: I've been trying this for a while but can't get anywhere.

I think you DO have to use calculus to prove that the minimum time is the time in which all three arrive together. However, if you assume that is the case (and I think it will turn out to be true), here's how.
Girl 1 rides distance x at 24 m/h & leaves bike. The time T1 = x/24
she then walks the rest of the 200-x miles, so T2 = (200-x)/2

Guy walks distance x at 9 m/h & finds bike. His T3 = x/9
He rides back distance y at 44 m/h. His T4 = y=44
He leaves the bike and walks to the finish. His T5 = (200-x+y)/9

Girl 2 walks distance x-y & finds bike. Her T6 = (x-y)/2
She then rides to the finish. Her T7 = (200-x+y)/24

Assuming the times are equal, T1+T2 = T3+T4+T5 = T6+T7. There are 2 unknowns so set up simultaneous equations.

x/24 + (200-x)/2 = (x-y)/2 + (200-x+y)/24

x/24 + (200-x)/2 = x/9 + y/44 + (200-x+y)/9

I'm not gonna type out all the messy arithmetic, but if you solve these you'll get x = 144, y = 88
so Girl 1's time is:
144/24 + (200-144)/2 = 34 hrs

Guy's time is
144/9 + 88/44 + (200-144+88)/9 = 34

Girl 2's time is
(144-88)/2 + (200-144+88)/24 = 34

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