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

by : $69$

 * The Four Cyclists

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.