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 08:28, 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.