2012 AMC 10B Problem 13

1. Pick a length for the escalator that both $60$ and $24$ divide evenly. $120$ works, so pretend the escalator is $120$ steps long.
2. On the stopped escalator Clea walks $120$ steps in $60$ seconds. Her walking speed is $120 \div 60 = 2$ steps per second.
3. On the moving escalator she covers $120$ steps in $24$ seconds, which is $120 \div 24 = 5$ steps per second.
4. That $5$ is her walking speed plus the escalator's speed. So the escalator by itself moves $5 - 2 = 3$ steps per second.
5. If she just stands, only the escalator moves her: $120 \div 3 = 40$ seconds.

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

When two motions work together, their speeds add. So turn each time into a speed, do the adding or subtracting with speeds, and turn the answer back into a time at the end. This is the same as "work" problems: if Clea and the escalator together finish the job in $24$ seconds and Clea alone takes $60$, how long does the escalator take alone?

Use fractions of the escalator instead. Walking alone, Clea covers $\frac{1}{60}$ of it each second. Walking on the moving escalator, she covers $\frac{1}{24}$ each second. The escalator's share is
$$\frac{1}{24} - \frac{1}{60} = \frac{5}{120} - \frac{2}{120} = \frac{3}{120} = \frac{1}{40}$$
of the escalator per second, so it takes $40$ seconds alone.

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 $\boxed{\textbf{(B)}\ 40}$.