Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1987 AIME Problems/Problem 10"
m (→‎Solution 1)
Line 5: Line 5:
 
Let the total number of steps be <math>x</math>, the speed of the escalator be <math>e</math> and the speed of Bob be <math>b</math>.
 
Let the total number of steps be <math>x</math>, the speed of the escalator be <math>e</math> and the speed of Bob be <math>b</math>.
  
−
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 <math>x - 75</math> 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 <math>\frac{e}{b} = \frac{x - 75}{75}</math>.   
+
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 <math>x - 75</math> 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 <math>\frac{e}{b} = \frac{x - 75}{75}</math>.   
  
 
Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved <math>150 - x</math> steps in that time.  Thus <math>\frac{e}{3b} = \frac{150 - x}{150}</math> or <math>\frac{e}{b} = \frac{150 - x}{50}</math>.
 
Similarly, in the time it took Al to walk down the escalator he saw 150 steps, so the escalator must have moved <math>150 - x</math> steps in that time.  Thus <math>\frac{e}{3b} = \frac{150 - x}{150}</math> or <math>\frac{e}{b} = \frac{150 - x}{50}</math>.
  
 
Equating the two values of <math>\frac{e}{b}</math> we have <math>\frac{x - 75}{75} = \frac{150 - x}{50}</math> and so <math>2x - 150 = 450 - 3x</math> and <math>5x = 600</math> and <math>x = 120</math>, the answer.
 
Equating the two values of <math>\frac{e}{b}</math> we have <math>\frac{x - 75}{75} = \frac{150 - x}{50}</math> and so <math>2x - 150 = 450 - 3x</math> and <math>5x = 600</math> and <math>x = 120</math>, the answer.
 +
 
=== Solution 2 ===
 
=== Solution 2 ===
 
Again, let the total number of steps be <math>x</math>, the speed of the escalator be <math>e</math> and the speed of Bob be <math>b</math> (all "per unit time").
 
Again, let the total number of steps be <math>x</math>, the speed of the escalator be <math>e</math> and the speed of Bob be <math>b</math> (all "per unit time").

Revision as of 21:23, 16 March 2017

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 $x$, 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 $x = 120$, 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).

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_10&oldid=84731"