Henry Ernest Dudeney/Modern Puzzles/35 - Sharing a Bicycle/Solution

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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,
while Brown can walk $5$ miles an hour,
but Anderson can ride $10$ miles an hour to Brown's $8$ 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

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