1987 AIME Problem 10

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

Let Bob’s speed be

b,

the escalator’s upward speed be

e,

and the visible count be

N

Write

N

as net speed times travel time for each person

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.

Problem 10 in other years