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Difference between revisions of "1965 AHSME Problems/Problem 28"

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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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If we let <math>Z</math>'s speed be <math>z</math> steps/minute, then <math>A</math>'s speed is <math>2z</math> steps/minute. Let <math>t_a</math> be the time <math>A</math> spent on the escalator, and let <math>t_z</math> be the time <math>Z</math> spent on the escalator. Then, we know that <math>A</math> walked down <math>2zt_a=27</math> steps, and <math>Z</math> walked down <math>zt_z=18</math> steps. Dividing the first equation by the second, we see that:
 +
\begin{align*} \\
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\frac{2zt_a}{zt_z}&=\frac{27}{18} \\
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\frac{2t_a}{t_z}&=\frac{3}{2} \\
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t_a&=\frac{3}{4}t_z \\
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\end{align*}
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Thus, because <math>A</math> was on the escalator for <math>\frac{3}{4}</math> as long as <math>Z</math> was, <math>A</math> only gained <math>\frac{3}{4}</math> as many "free" steps (i.e. steps that do not have to be taken because the escalator is moving down). We know that <math>A</math> gained <math>(n-27)</math> free steps, and <math>Z</math> gained <math>(n-18)</math> free steps. Thus we have the following equation: <math>n-27=\frac{3}{4}(n-18)</math>. Solving for <math>n</math> gives us <math>\boxed{\textbf{(B) }54}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 08:28, 19 July 2024

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

See Also

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
All AHSME Problems and Solutions

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