Henry Ernest Dudeney/Puzzles and Curious Problems/75 - A Question of Transport/Solution
Puzzles and Curious Problems by Henry Ernest Dudeney: $75$
- A Question of Transport
- Twelve soldiers had to get to a place twenty miles distant with the quickest possible dispatch,
- and all had to arrive at the same time.
- They requisitioned the services of a man with a small motor-car.
- "I can do twenty miles an hour," he said, "but I cannot carry more than four men at a time.
- At what rate do you walk?"
- "All of us can do a steady four miles an hour," they replied.
- "Very well," exclaimed the driver, "then I will go ahead with four men,
- drop them somewhere on the road to walk,
- then return and pick up four more (who will then be somewhere on the road),
- drop them off also, and return for the last four.
- So all you have to do is to keep walking while you are on your feet, and I will do the rest."
- As they started at noon, what was the exact time that they all arrived together?
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.
We have:
\(\ds d_1\) | \(=\) | \(\ds 20 t_1\) | "I can do twenty miles an hour," he said | |||||||||||
\(\ds d_2\) | \(=\) | \(\ds 4 t_2\) | "All of us can do a steady four miles an hour," they replied. | |||||||||||
\(\ds d_1 - d_2\) | \(=\) | \(\ds 20 \paren {t_2 - t_1}\) | ||||||||||||
\(\ds d_3 - d_1\) | \(=\) | \(\ds 4 \paren {t_3 - t_1}\) | ||||||||||||
\(\ds d_3 - d_2\) | \(=\) | \(\ds 20 \paren {t_3 - t_2}\) | ||||||||||||
\(\ds d_4 - d_2\) | \(=\) | \(\ds 4 \paren {t_4 - t_2}\) | ||||||||||||
\(\ds d_3 - d_4\) | \(=\) | \(\ds 20 \paren {t_4 - t_3}\) | ||||||||||||
\(\ds 20 - d_3\) | \(=\) | \(\ds 4 \paren {t_5 - t_3}\) | ||||||||||||
\(\ds 20 - d_4\) | \(=\) | \(\ds 20 \paren {t_5 - t_4}\) |
These can be expressed more usefully as:
\(\ds d_1 - 20 t_1\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds d_2 - 4 t_2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds d_1 - d_2 + 20 t_1 - 20 t_2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds -d_1 + d_3 + 4 t_1 - 4 t_3\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds -d_2 + d_3 + 20 t_2 - 20 t_3\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds -d_2 + d_4 + 4 t_2 - 4 t_4\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds d_3 - d_4 + 20 t_3 - 20 t_4\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds d_3 - 4 t_3 + 4 t_5\) | \(=\) | \(\ds 20\) | ||||||||||||
\(\ds d_4 - 20 t_4 + 20 t_5\) | \(=\) | \(\ds 20\) |
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:
- $\begin {pmatrix} 1 & 0 & 0 & 0 & -20 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & -4 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & -16 & 0 & -4 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -4 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & -5/2 & 3/2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & -1 & -1/6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5/6 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ d_3 \\ d_4 \\ t_1 \\ t_2 \\ t_3 \\ t_4 \\ t_5 \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ -5/6 \\ 25/6 \\ 13/5 \end {pmatrix}$
We are interested only in $t_5$, which we see is $\dfrac {13} 5$, that is, $2 \dfrac 3 5$ hours after noon.
$\blacksquare$
Sources
- 1932: Henry Ernest Dudeney: Puzzles and Curious Problems ... (previous) ... (next): Solutions: $75$. -- A Question of Transport
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $89$. A Question of Transport