## 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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