An escalator is descending at constant speed. A walks down and takes 50 steps…

Concept: In an escalator (or moving-walkway/boat-and-stream) problem, the escalator's own length — measured in steps — stays fixed no matter who is walking. For any walker whose own stepping rate is r steps per unit time, if the escalator itself moves at n steps per unit time in the same direction, then over the time the walker takes to descend, the escalator's fixed length equals the number of steps the walker personally takes plus the extra distance the escalator carries them during that same time. Writing this equation separately for two different walkers on the same escalator gives two expressions for the same fixed length, which can be equated to solve for n.

Application:

1. Let A's own stepping rate be 1 step per unit time, and let the escalator move n steps per unit time.

2. A takes 50 steps to reach the bottom, so A's own descent takes 50 units of time. In that time the escalator carries A down an additional 50n steps. So the escalator's fixed length is L = 50 + 50n.

3. B runs down and takes 90 steps in the same time that A takes 10 steps, so B's own stepping rate is 9 steps per unit time — nine times A's rate.

4. At 9 steps per unit time, B's own 90 steps take 90 ÷ 9 = 10 units of time, which matches “the same time A takes 10 steps” and confirms the rates are set up consistently.

5. In those 10 units of time the escalator carries B down an additional 10n steps. So the same fixed length is also L = 90 + 10n.

6. Equating the two expressions for the same length: 50 + 50n = 90 + 10n, so 40n = 40, giving n = 1.

7. Substituting back: L = 50 + 50(1) = 100.

Cross-check: Using B's expression instead, L = 90 + 10(1) = 100 — the same value, confirming the escalator has 100 visible steps when it is not operating.