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There were a number of excellent solutions
to this problem which revealed three basically different approaches
that I think are all worth sharing. James of Audenshaw School
offered the following solution and commentary, which reflects the
largest proportion of your answers. I like it very much because he
explains his journey very well.
Derek
adopted an approach based on a graphical solution and using similar
triangles. I have included his graph, for interest, below. I used a
graphical approach to solve this problem the first time I met it -
and I felt that my solution was quite a neat one.
The last solution, and easily the most
elegant, was presented by Ian and Charlie of the William Lovell
School. I have added a little to their solution for clarity but it
is simple and uses no algebra - quite a surprise and very, very
nice as it makes excellent use of proportionality.
What a lovely problem.
We can establish from this a couple of inequalities and I was
tempted to try this but decided it would be a pointless waste of
time though linear programming can be fun.
A bus leaving $x$ must travel to $x+y$ and back to 145 and the bus
leaving $x+y$ must travel to $x$ and then to 145. Therefore the
distances for the respective buses are: \begin{eqnarray} Bus 1: x
\rightarrow x+y \rightarrow 145\\ 2y - (145 - x)\\ Bus 2: x+y
\rightarrow x \rightarrow 145\\ y +(145- x) \end{eqnarray} At this
point it is not known which of these journeys is the longer
journey. One may expect this to be important as it is the longer
journey that has been under taken by the faster bus. However I can
assure you that this is actually irrelevant but I will assume Bus 2
to be the faster bus. This gives the following equation:
\begin{eqnarray} \frac{y+ (145-x)}{5} &=&
\frac{2y-(145-x)}{4} \\ 4(y+145-x) &= & 5(2y-145+x)\\ 1305
&=& 9x + 6y\\ 435 &=& 3x + 2y \end{eqnarray} This
equation on its own is of course unsolvable, however if the same
technique is applied to the 3rd meeting time an equation in terms
of $x$ and $y$ will be achieved again. However although it is not
relevant which bus was chosen as the faster it is very important
that you select the same bus both times.
\begin{eqnarray} Bus 1: x \rightarrow x+y \rightarrow 145
\rightarrow x \rightarrow 201 \\ 2y (201 - x)\\ Bus 2: x+y
\rightarrow x \rightarrow 145 \rightarrow x+y \rightarrow 201\\ 3y
- (201 - x) \end{eqnarray} Again we will assume that Bus 2 is the
faster bus. This will give the following equation
\begin{eqnarray} \frac{3y -(201-x)}{5} &=&
\frac{2y+(201-x)}{4} \\ 4(3y-201 +x) &= & 5(2y+201-x)\\
1809 &=& 9x + 2y \end{eqnarray} We now have two linear
equations in terms of $x$ and $y$ which we need to solve:
\begin{equation} 9x+2y= 1809 \end{equation} and \begin{equation}
3x+2y = 435 \end{equation} Solving these two equations we get
$$x=229$$ $$y=-126$$
The negative sign for $y$ has happened because
James assumed the second bus was the fastest when it was actually
the first bus.
Now all that remains is to identify which
town is $x$ and which is $x + y$.
For reasons I have already explained Bus 1
was the faster bus and we are told in the problem definition that
this bus left from Shipton and hence the slower from Veston.
In this solution Bus 1
left from $x$. Therefore $x$ is Shipton and $x + y$ is Veston. So
the milestone at Shipton is $103$ miles and the one at Veston $229$
miles.
Here is the diagram Derek
used for his solution. Although his final answer was wrong, the use
of similar triangles can result in a reasonably elegant
solution.
Finally Ian's and Charlie's solution, I have
included a diagram of the buses' journeys and made some minor
modifications to the text.
The fastest bus left from Shipton (green
lines) and the slower from Veston (blue lines). The diagram shows
the three meeting points for the buses as they travel between the
two towns.
The bus from Shipton travels $5$ units in the same time as the
bus from Veston travels $4$ units.
So first, divide the distance between the two towns into nine equal
parts.
The first meeting place is A, the second meeting place is B, and
the third meeting place is C.
The distance between B and C is : $201 - 145 = 56$
$56$ is four units and so each unit is $14$ miles.
So, to find the milestone of Veston: $201 + 28 = 229$
Milestone of Shipton: $145 - 42 = 103$