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: 2ztaztz=27182tatz=32ta=34tz

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 $\boxed{\textbf{(B) }54}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 27
Followed by
Problem 29
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