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1987 AIME Problems/Problem 10

Revision as of 23:50, 13 December 2022 by Leodoge45 (talk | contribs)

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.)

Solution 1

Let the total number of steps be $x$, the speed of the escalator be $e$ and the speed of Bob be $b$.

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 $x - 75$ 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 $\frac{e}{b} = \frac{x - 75}{75}$.

Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved $150 - x$ steps in that time. Thus $\frac{e}{3b} = \frac{150 - x}{150}$ or $\frac{e}{b} = \frac{150 - x}{50}$.

Equating the two values of $\frac{e}{b}$ we have $\frac{x - 75}{75} = \frac{150 - x}{50}$ and so $2x - 150 = 450 - 3x$ and $5x = 600$ and $x = \boxed{120}$, the answer.

Solution 2

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

Then this can be interpreted as a classic chasing problem: Bob is "behind" by $x$ steps, and since he moves at a pace of $b+e$ relative to the escalator, it will take $\frac{x}{b+e}=\frac{75}{e}$ time to get to the top.

Similarly, Al will take $\frac{x}{3b-e}=\frac{150}{e}$ time to get to the bottom.

From these two equations, we arrive at $150=\frac{ex}{3b-e}=2\cdot75=\frac{2ex}{b+e}=\frac{6ex}{3b+3e}=\frac{(ex)-(6ex)}{(3b-e)-(3b+3e)}=\frac{5x}{4}$ $\implies600=5x\implies x=\boxed{120}$, where we have used the fact that $\frac{a}{b}=\frac{c}{d}=\frac{a\pm c}{b\pm d}$ (the proportion manipulations are motivated by the desire to isolate $x$, prompting the isolation of the $150$ on one side, and the fact that if we could cancel out the $b$'s, then the $e$'s in the numerator and denominator would cancel out, resulting in an equation with $x$ by itself).

Solution 3

Let $e$ and $b$ be the speeds of the escalator and Bob, respectively.

When Al was on his way down, he took $150$ steps with a speed of $3b-e$ per step. When Bob was on his way up, he took $75$ steps with a speed of $b+e$ per step. Since Al and Bob were walking the same distance, we have \[150(3b-e)=75(b+e)\] Solving gets the ratio $\frac{e}{b}=\frac{3}{5}$.

Thus while Bob took $75$ steps to go up, the escalator contributed an extra $\frac{3}{5}\cdot75=45$ steps.

Finally, there is a total of $75+45=\boxed{120}$ 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 $x.$ Assume that Al takes $3$ steps per second and that Bob takes $1$ step per second. Since Al counts $150$ steps, it takes him \[\frac{150}{3}=50\] seconds to traverse the distance of the escalator moving downwards. Since Bob counts $75$ steps, it takes him \[\frac{75}{1}=75\] 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 $3$ steps downwards every second, but we throw hands at him immediately after each second, such that he flinches and moves himself backwards $x$ steps. This is equivalent to Al taking $3-x$ steps downwards every second. Since we discovered that it takes him $50$ 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 $50$ seconds to move from the top to the bottom of the staircase. Thus, Al takes a total of \[(3-x) \cdot 50=150-50x \qquad (\heartsuit)\] steps.

The explanation for Bob is similar except now we pick him up and place him forward $x$ steps immediately after he takes his usual step per second, and since we discovered he does this for $75$ seconds, it takes him \[(1+x) \cdot 75=75+75x \qquad (\clubsuit)\] 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 $\heartsuit$ and $\clubsuit$ to get \[150-50x=75+75x\] \[x=\frac{3}{5}.\] Therefore, substituting $x$ in the expression we discovered in $\heartsuit,$ Al takes a total of \[150-50x=150-50\left(\frac{3}{5}\right)=\boxed{120}\] steps, and we are done.

~samrocksnature

Solution 4

WLOG, let Al's speed be $15$ steps per second, so Bob's speed is $5$ steps per second. Then, Al was on the escalator for $\frac{150}{15}\ = 10$ seconds and Bob was on the escalator for $\frac{75}{5}\ = 15$ seconds. Let $r$ be the rate of the escalator, in steps per second. Then, the total amount of steps is $150 - 10r = 75 + 15r$. Al is getting $10$ seconds of resistance at rate $r$ from the escalator, while Bob is getting $15$ seconds of help at rate $r$. Solving for $r$, we hvae $r = 3$ steps per second. Then, we can plug $r$ into the previous equation or subtract/add it to Al/Bob's rate (respectively) then multiply by their respective time. Either way, we get $\boxed{120}$ and we are done.

Video Solution by OmegaLearn

https://youtu.be/4WttvHavnkM?t=456

~ pi_is_3.14

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

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