Question

Al walks down an escalator that is moving up and counts $150$ steps. Bob walks up and counts $75$ steps. If Al’s walking speed is three times Bob’s, how many steps are visible at a given time? Assume this is constant.

Answer

Answer: 120

Concepts:distance rate and time system of equations

Difficulty rating: 1770

Small Hint:

Let Bob’s speed be $b,$ the escalator’s upward speed be $e,$ and the visible count be $N$

Big Hint:

Write $N$ as net speed times travel time for each person

Solution:

Bob’s time is $\frac{75}{b},$ so $N=(b+e)(\frac{75}{b}),$ or $N=75(1+\frac{e}{b}).$ Al’s time is $\frac{150}{3b}=\frac{50}{b},$ so $N=(3b-e)(\frac{50}{b}),$ or $N=150-\frac{50e}{b}.$ Equating gives $\frac{125e}{b}=75,$ hence $\frac{e}{b}=\frac{3}{5}$ and $N=75(1+\frac{3}{5})=120.$

1987 AIME Problem 10

Answer: 120

Difficulty rating: 1770

Small Hint:

Big Hint:

Solution:

Problem 10 in Other Years