Henry Ernest Dudeney/Puzzles and Curious Problems/78 - The Meeting Cars/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $78$

The Meeting Cars
The Crackhams made their first stop at Bugleminster, where they were to spend the night at a friend's house.
This friend was to leave home at the same time and ride to London to put up at the Crackhams' house.
They took the same route, and each car went at its own uniform speed.
They kept a look-out for one another, and met forty miles from Bugleminster.
George that evening worked out the following little puzzle:
"I find that if, on our respective arrivals, we had each at once proceeded on the return journey at the same speeds
we should meet $48$ miles from London."
If this were so, what is the distance from London to Bugleminster?


Solution

$72$ miles.


Proof

let $D$ be the distance from Bugleminster to London.

Let $v_1$ and $v_2$ miles per hour be the speed of Crackham and his friend respectively.

Let $t_1$ hours be the time after setting out that they met.

Let $t_2$ hours be the time after originally setting out that they would have met for the second time.

We have:

\(\ds t_1\) \(=\) \(\ds \dfrac {40} {v_1}\) They kept a look-out for one another, and met forty miles from Bugleminster.
\(\ds \) \(=\) \(\ds \dfrac {D - 40} {v_2}\)
\(\ds t_2\) \(=\) \(\ds \dfrac {D + 48} {v_1}\) if ... we had each at once proceeded on the return journey at the same speeds we should meet $48$ miles from London.
\(\ds \) \(=\) \(\ds \dfrac {D + \paren {D - 48} } {v_2}\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {v_1} {v_2}\) \(=\) \(\ds \dfrac {40} {D - 40}\) eliminating $t_1$ and $t_2$
\(\ds \) \(=\) \(\ds \dfrac {D + 48} {2 D - 48}\)
\(\ds \leadsto \ \ \) \(\ds 40 \paren {2 D - 48}\) \(=\) \(\ds \paren {D - 40} \paren {D + 48}\) eliminating $v_1$ and $v_2$
\(\ds \leadsto \ \ \) \(\ds D^2\) \(=\) \(\ds 72 D\) after simplification
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds 72\) as $D$ is not zero

$\blacksquare$


Sources

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