Henry Ernest Dudeney/Modern Puzzles/31 - A Walking Puzzle

by : $31$

 * A Walking Puzzle
 * A man set out at noon to walk from Appleminster to Boneyham,
 * and a friend of his started at $2$ p.m. on the same day to walk from Boneyham to Appleminster.
 * They met on the road at $5$ minutes past $4$ o'clock
 * and each man reached his destination at exactly the same time.


 * Can you say what time they both arrived?

Solution

 * $19:00$, that is, $7$ p.m.

Proof
Let Appleminster and Boneyham be referred to as $A$ and $B$.

Let $t$ be the number of hours after $12:00$ at which both men arrived.

Let $d_1$ be the distance from $A$ to $B$.

Let $d_2$ be the distance from $A$ at which the two met.

Let $v_1$ be the speed of the first man.

Let $v_2$ be the speed of the second man.

We have:

The solution $t = \dfrac 7 6$ leads to a time of $13:10$ which is invalid.

Hence we have $t = 7$, that is, $19:00$ or $7$ p.m.