# 1965 AHSME Problems/Problem 28

## Problem

An escalator (moving staircase) of $n$ uniform steps visible at all times descends at constant speed. Two boys, $A$ and $Z$, walk down the escalator steadily as it moves, A negotiating twice as many escalator steps per minute as $Z$. $A$ reaches the bottom after taking $27$ steps while $Z$ reaches the bottom after taking $18$ steps. Then $n$ is:

$\textbf{(A)}\ 63 \qquad \textbf{(B) }\ 54 \qquad \textbf{(C) }\ 45 \qquad \textbf{(D) }\ 36 \qquad \textbf{(E) }\ 30$

## Solution

If we let $Z$'s speed be $z$ steps/minute, then $A$'s speed is $2z$ steps/minute. Let $t_a$ be the time $A$ spent on the escalator, and let $t_z$ be the time $Z$ spent on the escalator. Then, we know that $A$ walked down $2zt_a=27$ steps, and $Z$ walked down $zt_z=18$ steps. Dividing the first equation by the second, we see that: \begin{align*} \\ \frac{2zt_a}{zt_z}&=\frac{27}{18} \\ \frac{2t_a}{t_z}&=\frac{3}{2} \\ t_a&=\frac{3}{4}t_z \\ \end{align*}

Thus, because $A$ was on the escalator for $\frac{3}{4}$ as long as $Z$ was, $A$ only gained $\frac{3}{4}$ as many "free" steps (i.e. steps that do not have to be taken because the escalator is moving down). We know that $A$ gained $(n-27)$ free steps, and $Z$ gained $(n-18)$ free steps. Thus we have the following equation: $n-27=\frac{3}{4}(n-18)$. Solving for $n$ gives us $\boxed{\textbf{(B) }54}$.