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Climbing a moving staircase

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An up escalator moves at a constant speed. Ira walks up it at $1$ step per second, one stair per step, and takes $20$ steps to reach the top. Next time she walks up at $2$ steps per second and takes $32$ steps. How many steps are visible on the escalator, that is, how many would she take if it stood still?

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Hint

Her steps plus the stairs the escalator carries in the same time make up the whole length.

Solution

Let the escalator move $v$ stairs per second and let $N$ be the number of visible steps.
The first climb lasts $\frac{20}{1} = 20$ seconds, so $N = 20 + 20 v$ .
The second climb lasts $\frac{32}{2} = 16$ seconds, so $N = 32 + 16 v$ .
Equating gives $4 v = 12$ , so $v = 3$ and $N = 20 + 20 \times 3 = 80$ .
So $80$ steps are visible.

Answer: $80$

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