1987 AIME Problems/Problem 10

Problem

Al walks down to the bottom of an escalator that is moving up and he counts 150 steps. His friend, Bob, walks up to the top of the escalator and counts 75 steps. If Al's speed of walking (in steps per unit time) is three times Bob's walking speed, how many steps are visible on the escalator at a given time? (Assume that this value is constant.)

Solutions

Solution 1

Let the total number of steps be , the speed of the escalator be and the speed of Bob be .

In the time it took Bob to climb up the escalator he saw 75 steps and also climbed the entire escalator. Thus the contribution of the escalator must have been an addition steps. Since Bob and the escalator were both moving at a constant speed over the time it took Bob to climb, the ratio of their distances covered is the same as the ratio of their speeds, so .

Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved steps in that time. Thus or .

Equating the two values of we have and so and and , the answer.

Solution 2

Again, let the total number of steps be , the speed of the escalator be and the speed of Bob be (all "per unit time").

Then this can be interpreted as a classic chasing problem: Bob is "behind" by steps, and since he moves at a pace of relative to the escalator, it will take time to get to the top.

Similarly, Al will take time to get to the bottom.

From these two equations, we arrive at , where we have used the fact that (the proportion manipulations are motivated by the desire to isolate , prompting the isolation of the on one side, and the fact that if we could cancel out the 's, then the 's in the numerator and denominator would cancel out, resulting in an equation with by itself).

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.