1987 AIME Problems/Problem 10
Contents
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 additional 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).
Solution 3
Let and be the speeds of the escalator and Bob, respectively.
When Al was on his way down, he took steps with a speed of per step. When Bob was on his way up, he took steps with a speed of per step. Since Al and Bob were walking the same distance, we have Solving gets the ratio .
Thus while Bob took steps to go up, the escalator contributed an extra steps.
Finally, there is a total of steps in the length of the escalator.
Solution 4
Please understand the machinery of an escalator before proceeding to read this solution.
Let the number of steps that disappear at the top of the escalator equal Assume that Al takes steps per second and that Bob takes step per second. Since Al counts steps, it takes him seconds to traverse the distance of the escalator moving downwards. Since Bob counts steps, it takes him seconds to traverse the distance of the escalator moving downwards.
For the sake of this solution, we activate the emergency stop button on the escalator.
Now, the escalator is not moving, or is simply a staircase. Imagine that Al is taking steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards steps. This is equivalent to Al taking steps downwards every second. Since we discovered that it takes him seconds to get from the top to the bottom of the escalator, and we are forcing Al to imitate the movement of the escalator, it also takes him seconds to move from the top to the bottom of the staircase. Thus, Al takes a total of steps.
The explanation for Bob is similar except now we pick him up and place him forward steps immediately after he takes his usual step per second, and since we discovered he does this for seconds, it takes him steps to get from the bottom to the top.
Note that because the escalator is broken and is now a staircase, Al and Bob must have had to take an equal amount of steps to get from the bottom to the top or from the top to the bottom. (Clearly, there are an equal amount of steps from the bottom to the top, and from the top to the bottom.) Therefore, we may equate and to get Therefore, substituting in the expression we discovered in Al takes a total of steps, and we are done.
~samrocksnature
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.