Henry Ernest Dudeney/Puzzles and Curious Problems/286 - Unlucky Breakdowns/Solution

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Puzzles and Curious Problems by Henry Ernest Dudeney: $286$

Unlucky Breakdowns
On a day of great festivities, a large crowd gathered for a day's outing and pleasure.
They all agreed to pile into a bunch of wagons, each of which was to carry the same number of people.
But ten of the wagons broke down half way, so each of the other wagons then had to carry one more person than had been planned.
As they were about to start back, it was discovered that $15$ more of these wagons had become unserviceable,
and so there were three more people in each working wagon on the way back than started out.
How many people were there in the party?


Solution

There were $900$ people, all travelling in $100$ wagons, $9$ to a wagon.

Then they were down to $90$ wagons, in which they travelled $10$ to a wagon.

Then they were down to $75$ wagons, in which they travelled $12$ to a wagon.


Proof

Let $P$ be the number of people going on the outing.

Let $W$ be the number of wagons.

Let $n$ be the number of people per wagon who started out.

We have:

\(\ds P\) \(=\) \(\ds W n\)
\(\ds \) \(=\) \(\ds \paren {W - 10} \paren {n + 1}\)
\(\ds \) \(=\) \(\ds \paren {W - 25} \paren {n + 3}\)
\(\ds \leadsto \ \ \) \(\ds W n\) \(=\) \(\ds W n - 10 n + W - 10\) multiplying out
\(\ds \) \(=\) \(\ds W n - 25 n + 3 W - 75\)
\(\ds \leadsto \ \ \) \(\ds W - 10 n\) \(=\) \(\ds 10\)
\(\ds 3 W - 25 n\) \(=\) \(\ds 75\)
\(\ds \leadsto \ \ \) \(\ds 3 W - 30 n\) \(=\) \(\ds 30\)
\(\ds \leadsto \ \ \) \(\ds 5 n\) \(=\) \(\ds 45\)
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 9\)
\(\ds \leadsto \ \ \) \(\ds W\) \(=\) \(\ds 100\)
\(\ds \leadsto \ \ \) \(\ds P\) \(=\) \(\ds 900\)

$\blacksquare$


Sources

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