Henry Ernest Dudeney/Modern Puzzles/37 - A Side-car Problem/Solution

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Modern Puzzles by Henry Ernest Dudeney: $37$

A Side-car Problem
Atkins, Baldwin and Clarke had to go a journey of $52$ miles across country.
Atkins had a motor-bicycle with sidecar for one passenger.
How was he to take one of his companions a certain distance,
drop him on the road to walk the remainder of the way,
and return to pick up the second friend,
so that they should all arrive at their destination at exactly the same time?
The motor-bicycle could do $20$ miles per hour,
Baldwin could walk $5$ miles per hour,
and Clarke could walk $4$ miles per hour.


Solution

There are $2$ solutions:

$(1): \quad$ Atkins sets off with Clarke and drives $40$ miles, leaving Clarke to walk the remaining $12$ miles
Atkins picks up Baldwin $16$ miles from the start and continues to the destination.
$(2): \quad$ Atkins sets off with Baldwin and drives $36$ miles, leaving Baldwin to walk the remaining $16$ miles
Atkins picks up Clarke $12$ miles from the start and continues to the destination.

The whole journey takes $5$ hours.


Proof

Let Atkins, Baldwin and Clarke be denoted by $A$, $B$ and $C$ respectively.

First let us assume that:

$A$ sets off with $C$
drops him off $d_1$ miles from the start
then returns to pick up $B$ at a point $d_2$ miles from the start.

Let $t_1$ be the time $A$ and $C$ reach $d_1$.

Let $t_2$ be the time $A$ reaches $d_2$ to pick up $B$.

Let $t$ be the time they all arrive at their destination.


We have:

\(\text {(1)}: \quad\) \(\ds t_1\) \(=\) \(\ds \dfrac {d_1} {20}\) $A$ sets off with $C$
\(\text {(2)}: \quad\) \(\ds t_2 - t_1\) \(=\) \(\ds \dfrac {d_1 - d_2} {20}\) $A$ returns to pick up $B$
\(\text {(3)}: \quad\) \(\ds t - t_2\) \(=\) \(\ds \dfrac {52 - d_2} {20}\) $A$ travels to destination with $B$
\(\text {(4)}: \quad\) \(\ds t_2\) \(=\) \(\ds \dfrac {d_2} 5\) $B$ sets off walking to $d_2$
\(\text {(5)}: \quad\) \(\ds t - t_1\) \(=\) \(\ds \dfrac {52 - d_1} 4\) $C$ walks from $d_1$ to the final destination
\(\ds \leadsto \ \ \) \(\ds \dfrac {d_2} 5 - \dfrac {d_1} {20}\) \(=\) \(\ds \dfrac {d_1 - d_2} {20}\) substituting for $t_1$ and $t_2$ from $(1)$ and $(4)$ into $(2)$
\(\text {(6)}: \quad\) \(\ds \leadsto \ \ \) \(\ds d_2\) \(=\) \(\ds \dfrac {2 d_1} 5\) simplifying
\(\text {(7)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t - \dfrac {d_1} {20}\) \(=\) \(\ds \dfrac {52 - d_1} 4\) substituting for $t_1$ from $(1)$ into $(5)$
\(\text {(8)}: \quad\) \(\ds \leadsto \ \ \) \(\ds t - \dfrac {d_2} 5\) \(=\) \(\ds \dfrac {52 - d_2} {20}\) substituting for $t_2$ from $(4)$ into $(3)$
\(\text {(9)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \dfrac {52 - d_1} 4 + \dfrac {d_1} {20}\) \(=\) \(\ds \dfrac {52 - d_2} {20} + \dfrac {d_2} 5\) eliminating $t$ from between $(7)$ and $(8)$
\(\ds \leadsto \ \ \) \(\ds 5 \paren {52 - d_1} + d_1\) \(=\) \(\ds \paren {52 - d_2} + 4 d_2\) multiplying by $20$ to clear fractions
\(\text {(10)}: \quad\) \(\ds \leadsto \ \ \) \(\ds 4 \times 52 - 4 d_1\) \(=\) \(\ds 3 d_2\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {6 d_1} 5\) substituting for $d_2$ from $(6)$ into $(10)$
\(\ds \leadsto \ \ \) \(\ds d_1\) \(=\) \(\ds 40\) simplifying


Then we have:

\(\ds d_2\) \(=\) \(\ds \dfrac {2 \times 40} 5 = 16\) from $(6)$: $d_2 = \dfrac {2 d_1} 5$
\(\ds t\) \(=\) \(\ds \dfrac {52 - 40} 4 + \dfrac {40} {20} = 5\) from $(5)$ and $(1)$: $t = \dfrac {52 - d_1} 4 + \dfrac {d_1} {20}$


Running the scenario in reverse gives another solution.

$\blacksquare$


Sources

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