Difference between revisions of "1965 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
| − | <math>\ | + | |
| + | 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*} \\ | ||
| + | \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 <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 == | ||
{{AHSME 40p box|year=1965|num-b=27|num-a=29}} | {{AHSME 40p box|year=1965|num-b=27|num-a=29}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| − | [[Category: | + | [[Category:Introductory Algebra Problems]] |
Latest revision as of 07:28, 19 July 2024
Problem
An escalator (moving staircase) of 
uniform steps visible at all times descends at constant speed.
Two boys, 
and 
, walk down the escalator steadily as it moves, A negotiating twice as many escalator
steps per minute as . reaches the bottom after taking 
steps while reaches the bottom after taking 
steps. Then is:
Solution
If we let 's speed be 
steps/minute, then 's speed is 
steps/minute. Let 
be the time spent on the escalator, and let 
be the time spent on the escalator. Then, we know that walked down 
steps, and walked down 
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 was on the escalator for 
as long as was, only gained as many "free" steps (i.e. steps that do not have to be taken because the escalator is moving down). We know that gained 
free steps, and gained 
free steps. Thus we have the following equation: 
. Solving for gives us 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
