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

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Modern Puzzles by Henry Ernest Dudeney: $31$

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:

\(\text {(1)}: \quad\) \(\ds d_1\) \(=\) \(\ds v_1 t\)
\(\ds \) \(=\) \(\ds v_2 \paren {t - 2}\)
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds v_1\) \(=\) \(\ds \dfrac {v_2 \paren {t - 2} } t\)
\(\text {(3)}: \quad\) \(\ds d_2\) \(=\) \(\ds v_1 \times 4 \tfrac 1 {12}\) as $16:05$ is $4 \tfrac 1 {12}$ hours after $12:00$
\(\text {(4)}: \quad\) \(\ds d_1 - d_2\) \(=\) \(\ds v_2 \paren {4 \tfrac 1 {12} - 2}\)
\(\ds \leadsto \ \ \) \(\ds v_1 t - \dfrac {49 v_1} {12}\) \(=\) \(\ds \dfrac {25 v_2} {12}\) substituting for $d_1$ and $d_2$ in $(4)$ from $(1)$ and $(3)$
\(\ds \leadsto \ \ \) \(\ds \dfrac {v_2 \paren {t - 2} } t \paren {t - \dfrac {49} {12} }\) \(=\) \(\ds \dfrac {25 v_2} {12}\) substituting for $v_1$ from $(2)$
\(\ds \leadsto \ \ \) \(\ds \paren {t - 2} \paren {12 t - 49}\) \(=\) \(\ds 25 t\) simplifying
\(\ds \leadsto \ \ \) \(\ds 12 t^2 - 24 t - 49 t + 98\) \(=\) \(\ds 25 t\) simplifying
\(\ds \leadsto \ \ \) \(\ds 6 t^2 - 49 t + 49\) \(=\) \(\ds 0\) simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {6 t - 7} \paren {t - 7}\) \(=\) \(\ds 0\) factorising
\(\ds \leadsto \ \ \) \(\ds t\) \(=\) \(\ds 7\)
\(\, \ds \text {or} \, \) \(\ds t\) \(=\) \(\ds \dfrac 7 6\)

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.

$\blacksquare$


Sources

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