Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport/Solution

by : $75$

 * A Question of Transport

Solution

 * $14:36$

Proof
Let us refer to the start and finish points as $A$ and $B$ respectively.

Hence the object of the operation is to get all $12$ men from $A$ to $B$ as quickly as possible.

Let $d_1$ miles be the distance from $A$ to which the driver takes the first batch of soldiers.

Let $t_1$ hours past noon be the time he drops them off.

Let $d_2$ miles be the distance from $A$ to which the driver returns to pick up the second batch.

Let $t_2$ hours past noon be the time he picks them up.

Let $d_3$ miles be the distance from $A$ to which the driver takes the second batch of soldiers.

Let $t_3$ hours past noon be the time he drops them off.

Let $d_4$ miles be the distance from $A$ to which the driver returns to pick up the third batch.

Let $t_4$ hours past noon be the time he picks them up.

Let $t_5$ hours past noon be the time they all arrive at $B$.

The following diagram illustrates the situation.

The journey of the driver is shown in blue, while those of the walkers are shown in red.


 * Dudeney-Puzzles-and-Curious-Problems-75-solution.png

We have:

These can be expressed more usefully as:

This set of simultaneous linear equations can be expressed conveniently in matrix form as:


 * $\begin {pmatrix}

1 & 0 &  0 &  0 & -20 &   0 &   0 &   0 &  0 \\ 0 &  1 &  0 &  0 &   0 &  -4 &   0 &   0 &  0 \\ 1 & -1 &  0 &  0 &  20 & -20 &   0 &   0 &  0 \\ -1 &  0 &  1 &  0 &   4 &   0 &  -4 &   0 &  0 \\ 0 & -1 &  1 &  0 &   0 &  20 & -20 &   0 &  0 \\ 0 & -1 &  0 &  1 &   0 &   4 &   0 &  -4 &  0 \\ 0 &  0 &  1 & -1 &   0 &   0 &  20 & -20 &  0 \\ 0 &  0 &  1 &  0 &   0 &   0 &  -4 &   0 &  4 \\ 0 &  0 &  0 &  1 &   0 &   0 &   0 & -20 & 20 \\ \end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \\ t_5 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \end {pmatrix}$

It remains to solve this matrix equation.

In echelon form, this gives:

We are interested only in $t_5$, which we see is $\dfrac {13} 5$, that is, $2 \dfrac 3 5$ hours after noon.