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

1965 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
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1965 AHSME Problems/Problem 28

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