Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle/Solution
Modern Puzzles by Henry Ernest Dudeney: $35$
- Sharing a Bicycle
- Anderson and Brown have to go $20$ miles and arrive at exactly the same time.
- They have only one bicycle.
- Anderson can only walk $4$ miles an hour,
- How are they to arrange the journey?
Solution
Anderson cycles $11 \tfrac 1 9$ miles, drops the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.
Thus he cycles for $1 \tfrac 1 9$ hours and walks for $2 \tfrac 2 9$ hours.
Brown thus walks $11 \tfrac 1 9$ miles, picks up the bike, and walks the rest of the way, which is $8 \tfrac 8 9$ miles.
Thus he also walks for $2 \tfrac 2 9$ hours and cycles for $1 \tfrac 1 9$ hours.
Proof
Let Anderson and Brown be denoted by $A$ and $B$ respectively.
To keep it simple, we will assume one changeover, and that $A$ starts by cycling.
Let $d$ miles be the distance from the start to where $A$ dismounts to start walking.
Let $t$ hours be the time taken to do the total journey.
Let $t_a$ hours be the time taken by $A$ to travel $d$.
Let $t_b$ hours be the time taken by $B$ to travel $d$.
We have:
\(\ds d\) | \(=\) | \(\ds 10 t_a\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5 t_b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds t_b\) | \(=\) | \(\ds \dfrac {10 t_a} 5 = 2 t_a\) | |||||||||||
\(\ds 20 - d\) | \(=\) | \(\ds 4 \paren {t - t_a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8 \paren {t - t_b}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 20 - 10 t_a\) | \(=\) | \(\ds 4 \paren {t - t_a}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 8 \paren {t - 2 t_a}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 4 t + 6 t_a\) | \(=\) | \(\ds 20\) | |||||||||||
\(\ds 8 t - 6 t_a\) | \(=\) | \(\ds 20\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 8 t + 12 t_a\) | \(=\) | \(\ds 40\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 18 t_a\) | \(=\) | \(\ds 20\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds 18 \dfrac d {10}\) | \(=\) | \(\ds 20\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds d\) | \(=\) | \(\ds \dfrac {200} {18} = 11 \tfrac 1 9\) |
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $35$. -- Sharing a Bicycle
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $63$. Sharing a Bicycle