## A Walking Puzzle

A man set out at noon to walk from Appleminster to Boneyham, and a friend of his started at two P.M. on the same day to walk from Boneyham to Appleminster. They met on the road at five minutes past four o'clock, and each man reached his destination at exactly the same time. Can you say at what time they both arrived? Source: Henry Ernest Dudeny, 536 PUZZLES & Curious Problems, Scribner's, 1967, #59.
[Taken from Ken's Puzzle of the Day, October 12, 1994.]

Solution: 7:00 pm.
Let the rates of the two walkers be rx and ry, the distance be d, and the meeting time be t in minutes after 2pm.
The equations for the meeting (1) and the trip (2,3) are:
(1) d = 245*rx + 125*ry
(2) d = (120+t)*rx
(3) d = t*ry.

Substituting (3) into (1):
(4) 245*rx = (t-125)*ry
(4b) rx/ry = (t-125)/245

Substituting (3) into (2):
(5) 125*ry = (t-125)*rx
(5b) rx/ry = 125/(t-125)

Equating (4b) and (5b):
(6) rx/ry = (t-125)/245 = 125/(t-125)
(6b) (t-125)^2 = 125*245
(6c) t^2 - 250t - 120*125 = 0

Solving the quadtratic equation for t:
(7) t = 125 +/- sqrt(125^2 + 120*125)
(7b) t = 125 +/- 175

Choosing positive, t = 300, so they both arrived at 7:00 pm.

Solved by: Thomas Ngo

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