Difference between revisions of "2024 SSMO Relay Round 3 Problems/Problem 1"
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Let the speed of the escalator be <math>s</math>. We use the following distance/speed/time chart: | Let the speed of the escalator be <math>s</math>. We use the following distance/speed/time chart: | ||
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\begin{array}{c|c|c} | \begin{array}{c|c|c} | ||
\text{Distance} & \text{Speed} & \text{Time} \\\hline | \text{Distance} & \text{Speed} & \text{Time} \\\hline | ||
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225 & 200 + s & b_2 | 225 & 200 + s & b_2 | ||
\end{array} | \end{array} | ||
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Since Alice and Bob reach the bottom floor at the same time, we have <math>a_1 = b_1 + b_2</math>. Using the fact that distance equals speed times time, we write: | Since Alice and Bob reach the bottom floor at the same time, we have <math>a_1 = b_1 + b_2</math>. Using the fact that distance equals speed times time, we write: | ||
\begin{align*} | \begin{align*} | ||
| − | + | \frac{225}{200-s}&=\frac{3}{2}+\frac{225}{200+s}\implies\\ | |
| − | \frac{225}{200 - s} &= \frac{3}{2} + \frac{225}{200 + s} \\ | + | \frac{450s}{200^2-s^2} &= \frac{3}{2}\implies\\ |
| − | \frac{450s}{200^2 - s^2} &= \frac{3}{2} \ | + | s^2+300s-200^2&=0\implies\\ |
| − | + | (s+400)(s-100)&=0\implies\\ | |
| − | + | s&=\boxed{100}. | |
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| − | (s + 400)(s - 100) &= 0 \\ | ||
| − | s &= \boxed{100} | ||
\end{align*} | \end{align*} | ||
~SMO_Team | ~SMO_Team | ||
Latest revision as of 14:46, 10 September 2025
Problem
Alice and Bob on the second floor of the building right next to the escalator moving upwards. Alice and Bob each run at 200 steps per minute and the escalator is a total of 225 steps. Alice chooses to use the escalator moving upwards, while Bob chooses to run to the escalator moving downwards, 300 steps away, and then ride the escalator down. If Alice and Bob hit the bottom floor at the exact same time, find the speed of the escalators in terms of steps per minute.
Solution
Let the speed of the escalator be 
. We use the following distance/speed/time chart:
\begin{array}{c|c|c} \text{Distance} & \text{Speed} & \text{Time} \\\hline 225 & 200 - s & a_1 \\ 300 & 200 & b_1 \\ 225 & 200 + s & b_2 \end{array}
Since Alice and Bob reach the bottom floor at the same time, we have 
. Using the fact that distance equals speed times time, we write:
\begin{align*} \frac{225}{200-s}&=\frac{3}{2}+\frac{225}{200+s}\implies\\ \frac{450s}{200^2-s^2} &= \frac{3}{2}\implies\\ s^2+300s-200^2&=0\implies\\ (s+400)(s-100)&=0\implies\\ s&=\boxed{100}. \end{align*}
~SMO_Team
