1987 AIME Problems/Problem 10
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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
| 1987 AIME (Problems • Resources) | ||
| Preceded by Problem 9 | Followed by Problem 11 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
