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
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $31$. -- A Walking Puzzle
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $59$. A Walking Puzzle