Henry Ernest Dudeney/Modern Puzzles/36 - More Bicycling/Solution

by : $36$

 * More Bicycling

Solution
Anderson:
 * rides for $7 \tfrac {11} {27}$ miles, taking $44 \tfrac 4 9$ minutes
 * walks the remaining $12 \tfrac {16} {27}$ miles in $3$ hours and $8 \tfrac 8 9$ minutes.

Brown:
 * walks those $7 \tfrac {11} {27}$ miles in $1$ hour $28 \tfrac 8 9$ minutes
 * rides for $1 \tfrac {13} {27}$ miles, taking $11 \tfrac 1 9$ minutes
 * walks the remaining $11 \tfrac 1 9$ miles in $2$ hours $13 \tfrac 1 3$ minutes.

Carter:
 * walks $8 \tfrac 8 9$ miles in $2$ hour $57 \tfrac {21} {27}$ minutes
 * rides the remaining $11 \tfrac 1 9$ miles in $55 \tfrac 5 9$ minutes.

The total journey time is $3$ hours $53 \tfrac 1 3$ minutes.

Proof
Let Anderson, Brown and Carter be denoted by $A$, $B$ and $C$ respectively.

To keep it simple, we will assume:
 * $A$ starts by cycling, then leaves the bike for $B$
 * $B$ takes over cycling from $A$, then leaves the bike for $C$
 * $C$ finishes the journey on the cycle.

Let $d_1$ miles be the distance from the start to where $A$ dismounts to start walking and $B$ starts riding.

Let $d_2$ miles be the distance from $d_1$ to where $B$ dismounts to start walking.

Let $t$ hours be the time taken to do the total journey.

Let $t_{a_1}$ hours be the time taken by $A$ to travel $d_1$.

Let $t_{b_1}$ hours be the time taken by $B$ to travel $d_1$.

Let $t_{c_1}$ hours be the time taken by $C$ to travel $d_1$.

Let $t_{a_2}$ hours be the time taken by $A$ to travel $d_2$.

Let $t_{b_2}$ hours be the time taken by $B$ to travel $d_2$.

Let $t_{c_2}$ hours be the time taken by $C$ to travel $d_2$.

We have:

We set up this system of linear simultaneous equations in matrix form as:


 * $\begin {pmatrix}

1 & 0 & -10 & 0 &   0 &  0 &  0 &   0 &  0 \\ 1 & 0 &   0 & -5 &   0 &  0 &  0 &   0 &  0 \\ 1 & 0 &   0 &  0 &  -3 &  0 &  0 &   0 &  0 \\ 0 & 1 &   0 &  0 &   0 & -4 &  0 &   0 &  0 \\ 0 & 1 &   0 &  0 &   0 &  0 & -8 &   0 &  0 \\ 0 & 1 &   0 &  0 &   0 &  0 &  0 &  -3 &  0 \\ 1 & 1 &  -4 &  0 &   0 & -4 &  0 &   0 &  4 \\ 1 & 1 &   0 & -5 &   0 &  0 & -5 &   0 &  5 \\ 1 & 1 &   0 &  0 & -12 &  0 &  0 & -12 & 12 \\ \end {pmatrix} \begin {pmatrix} d_1 \\ d_2 \\ t_{a_1} \\ t_{b_1} \\ t_{c_1} \\ t_{a_2} \\ t_{b_2} \\ t_{c_2} \\ t \\ \end {pmatrix} = \begin {pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 20 \\ 20 \\ 20 \\ \end {pmatrix}$

It remains to solve this matrix equation.

In reduced echelon form, this gives:

The result can be read directly and converted into the appropriate units.