1965 AHSME Problems/Problem 28
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 | |
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