Henry Ernest Dudeney/Puzzles and Curious Problems/68 - The Moving Staircase/Solution

Puzzles and Curious Problems by Henry Ernest Dudeney: $68$

"I counted $50$ steps that I made in going down the moving staircase," said Walker.

"I counted $75$ steps," said Trotman; "but I was walking down three times as quickly as you."

If the staircase were stopped, how many steps would be visible?

$100$ steps.

Let $n$ be then number of visible steps when the escalator is stopped.

Let $t$ be the unit of time taken for one step to vanish at the bottom.

If you stand still on the escalator, it takes time $n t$ to reach the bottom.

If you take $x$ steps down the escalator, it takes time $\paren {n - x} t$ to reach the bottom.

Trotman takes $75$ steps in $\paren {n - 75} t$.

That is, he takes $3$ steps in $\dfrac {n - 75} {25} t$.

Walker takes $50$ steps in $\paren {n - 50} t$.

That is, he takes $1$ step in $\dfrac {n - 50} {50} t$.

But $3$ steps taken by Trotman take as much time as $1$ step taken by Walker.

Hence:

$\dfrac {n - 75} {25} = \dfrac {n - 50} {50}$

which, after algebra, gives:

$n = 100$

$\blacksquare$