2012 AMC 10B Problem 13

Measure distance in "escalator lengths." Walking on a stopped escalator, Clea covers the whole length in $60$ seconds, so her walking speed is $\frac{1}{60}$ length per second.

Walking on the moving escalator, her speed relative to the ground is her walking speed plus the escalator's speed, and she covers the length in $24$ seconds: combined speed $\frac{1}{24}$.

Therefore the escalator's own speed is
$$\frac{1}{24} - \frac{1}{60} = \frac{5}{120} - \frac{2}{120} = \frac{3}{120} = \frac{1}{40}$$
length per second. Standing still, she needs $40$ seconds.

The answer is $\boxed{\textbf{(B)}\ 40}$.

When two motions act together, their speeds add, so convert each given time into a rate (fraction of the job per second), do the arithmetic on rates, then convert back. This is the same structure as "work" problems: two workers together take $24$ minutes, one alone takes $60$, how long for the other. Choosing the length as $1$ unit (or $120$ units to avoid fractions) keeps the numbers clean.

Let the escalator be $120$ steps long. Clea walks $120/60 = 2$ steps per second. With the escalator running she moves $120/24 = 5$ steps per second, so the escalator contributes $3$ steps per second, and $120/3 = 40$ seconds.

• Subtracting the times (60 - 24 = 36) instead of the rates; rates add, times do not.
• Averaging $60$ and $24$ to get $42$, or adding the rates $\frac{1}{60} + \frac{1}{24}$ as if the escalator's speed were the combined speed.