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

by : $40$

 * 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:

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.