Henry Ernest Dudeney/Puzzles and Curious Problems/69 - The Four Cyclists/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $69$

The Four Cyclists
The four circles represent cinder paths.
Dudeney-Puzzles-and-Curious-Problems-69.png
The four cyclists started at noon.
Each person rode round a different circle,
one at the rate of $6$ miles an hour,
another at the rate of $9$ miles an hour,
another at the rate of $12$ miles an hour,
and the fourth at the rate of $15$ miles an hour.
They agreed to ride until all met at the centre, from which they started, for the fourth time.
The distances around each circle was exactly one-third of a mile.
When did they finish their ride?


Solution

$26$ minutes and $40$ seconds past noon.


Proof

Let $t$ be the time in hours at which they all arrive together.

The times taken to complete one circuit are respectively:

$\dfrac 1 3 \div 6 = \dfrac 1 {18}$ hours
$\dfrac 1 3 \div 9 = \dfrac 1 {27}$ hours
$\dfrac 1 3 \div 12 = \dfrac 1 {36}$ hours
$\dfrac 1 3 \div 15 = \dfrac 1 {45}$ hours

We have that $\lcm {18, 27, 36, 45} = 540$

Hence let us clear the fractions by letting time $h$ be the time period that is $h = 1 {540}$ of an hour.

Let $t_1$, $t_2$, $t_3$ and $t_4$ be the times taken by each cyclist in units of $h$.

We have:

$t_1 = 30$
$t_2 = 20$
$t_3 = 15$
$t_4 = 12$

Now we have that:

$\lcm {30, 20, 15, 12} = 60$

which is the smallest number of time units $h$ at which they are all at the middle together.

Thus:

$t = \dfrac {60} {540} = \dfrac 1 9$

or $6 \tfrac 2 3$ minutes.

Four times this is $26 \tfrac 2 3$ minutes, or $26$ minutes and $40$ seconds.

$\blacksquare$


Sources

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