Henry Ernest Dudeney/Modern Puzzles/40 - Pickleminster to Quickville/Solution

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Modern Puzzles by Henry Ernest Dudeney: $40$

Pickleminster to Quickville
Two trains, $A$ and $B$, leave Pickleminster for Quickville at the same time as two trains, $C$ and $D$, leave Quickville for Pickleminster.
$A$ passes $C$ $120$ miles from Pickleminster and $D$ $140$ miles from Pickleminster.
$B$ passes $C$ $126$ miles from Quickville and $D$ half-way between Pickleminster and Quickville.
Now, what is the distance from Pickleminster to Quickville?


Solution

$210$ miles.


Proof

Let $V_A$, $V_B$, $V_C$ and $V_D$ miles per hour be the speeds of $A$, $B$, $C$ and $D$ respectively.

Let $d$ be the distance between Pickleminster and Quickville.


From the given conditions:

\(\text {(1)}: \quad\) \(\ds \dfrac {120} {V_A}\) \(=\) \(\ds \dfrac {d - 120} {V_C}\) $A$ passes $C$ $120$ miles from Pickleminster
\(\text {(2)}: \quad\) \(\ds \dfrac {140} {V_A}\) \(=\) \(\ds \dfrac {d - 140} {V_D}\) $A$ passes $D$ $140$ miles from Pickleminster
\(\text {(3)}: \quad\) \(\ds \dfrac {d - 126} {V_B}\) \(=\) \(\ds \dfrac {126} {V_C}\) $B$ passes $C$ $126$ miles from Quickville
\(\text {(4)}: \quad\) \(\ds \dfrac d {2 V_B}\) \(=\) \(\ds \dfrac d {2 V_D}\) $B$ passes $D$ half-way between Pickleminster and Quickville
\(\ds \leadsto \ \ \) \(\ds V_B\) \(=\) \(\ds V_D\) simplifying $(4)$
\(\ds \leadsto \ \ \) \(\ds 140 V_B\) \(=\) \(\ds \paren {d - 140} V_A\) substituting into $(2)$ and simplifying
\(\ds \leadsto \ \ \) \(\ds 126 V_B\) \(=\) \(\ds \paren {d - 126} V_C\) simplifying $(3)$
\(\ds \leadsto \ \ \) \(\ds 120 V_C\) \(=\) \(\ds \paren {d - 120} V_A\) simplifying $(1)$
\(\ds \leadsto \ \ \) \(\ds 126 V_B\) \(=\) \(\ds \dfrac {\paren {d - 126} \paren {d - 120} V_A} {120}\) eliminating $V_C$
\(\ds \leadsto \ \ \) \(\ds 126 \dfrac {\paren {d - 140} V_A} {140}\) \(=\) \(\ds \dfrac {\paren {d - 126} \paren {d - 120} V_A} {120}\) eliminating $V_B$
\(\ds \leadsto \ \ \) \(\ds 120 \times 126 \paren {d - 140}\) \(=\) \(\ds 140 \paren {d - 126} \paren {d - 120}\) eliminating $V_A$
\(\ds \leadsto \ \ \) \(\ds 108 \paren {d - 140}\) \(=\) \(\ds \paren {d - 126} \paren {d - 120}\) simplifying
\(\ds \leadsto \ \ \) \(\ds d^2 - 354 d + 30240\) \(=\) \(\ds 0\) gathering everything together
\(\ds \leadsto \ \ \) \(\ds \paren {d - 210} \paren {d - 144}\) \(=\) \(\ds 0\) factorising

This leads to the $2$ solutions:

$d = 144$
$d = 210$

The first of these suggests that $A$ travels $140$ miles while $D$ travels $4$ miles, which would mean $A$ goes $35$ times as fast as $D$.

The second answer is more sensible: it has $A$ going twice as fast as $B$ and $D$, and $C$ going at a speed between these.

$\blacksquare$


Sources

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