There are two escalators, one moving up and the other moving down at the same constant speed. Two individuals start at the top of each escalator, both walking down at a pace of 1 step per second. The person walking down the downward-moving escalator reaches the bottom after taking 20 steps. The person walking down the upward-moving escalator takes 60 seconds to reach the bottom. How many steps would each escalator have if they were stationary (i.e., paused)?

Escalator Steps | General | Trading Interview

Interview Questions Math Questions Escalator Steps

There are two escalators, one moving up and the other moving down at the same constant speed. Two individuals start at the top of each escalator, both walking down at a pace of 1 step per second.

The person walking down the downward-moving escalator reaches the bottom after taking 20 steps.
The person walking down the upward-moving escalator takes 60 seconds to reach the bottom.

How many steps would each escalator have if they were stationary (i.e., paused)?

Let's define the following:

T = Total number of steps on an escalator
S = Speed of the escalator in steps per second

Next, we can work this problem out using the following equations: \begin{equation} 20 = \frac{T}{S+1} \end{equation} The +1 in this equation accounts for the extra 1 step per second that the person is walking down. The second equation is: \begin{equation} 60 = \frac{T}{-S+1} \end{equation} Here, the S is negative, because the escalator is moving up instead of down. These two equations result in the following: \begin{equation} T = 20S + 20 \end{equation} \begin{equation} T = -60S + 60 \end{equation} Subtracting these two equations from each other results in \begin{equation} 0 = 80S - 40 \end{equation} \begin{equation} S = 0.5 \end{equation} If we substitute S in the first equation, we get that \begin{equation} 20 = \frac{T}{0.5+1} \end{equation} \begin{equation} T = 20*1.5 = 30 \end{equation} So, the escalator has 30 steps.

Correct Answer: 30

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def two_sum(nums, target):
    seen = {}
    for i, n in enumerate(nums):
        if target - n in seen:
            return [seen[target - n], i]
        seen[n] = i
    return []

print(two_sum([2, 7, 11, 15], 9))  # [0, 1]

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