Difference between revisions of "1987 AIME Problems/Problem 10"
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== Problem == | == 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.) | 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 == | + | == Solutions == |
− | {{ | + | === Solution 1 === |
+ | 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 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>. | ||
+ | |||
+ | 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 = \boxed{120}</math>, the answer. | ||
+ | |||
+ | === 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"). | ||
+ | |||
+ | Then this can be interpreted as a classic chasing problem: Bob is "behind" by <math>x</math> steps, and since he moves at a pace of <math>b+e</math> relative to the escalator, it will take <math>\frac{x}{b+e}=\frac{75}{e}</math> time to get to the top. | ||
+ | |||
+ | Similarly, Al will take <math>\frac{x}{3b-e}=\frac{150}{e}</math> time to get to the bottom. | ||
+ | |||
+ | From these two equations, we arrive at <math>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}</math> | ||
+ | <math>\implies600=5x\implies x=\boxed{120}</math>, where we have used the fact that <math>\frac{a}{b}=\frac{c}{d}=\frac{a\pm c}{b\pm d}</math> (the proportion manipulations are motivated by the desire to isolate <math>x</math>, prompting the isolation of the <math>150</math> on one side, and the fact that if we could cancel out the <math>b</math>'s, then the <math>e</math>'s in the numerator and denominator would cancel out, resulting in an equation with <math>x</math> by itself). | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Let <math>e</math> and <math>b</math> be the speeds of the escalator and Bob, respectively. | ||
+ | |||
+ | When Al was on his way down, he took <math>150</math> steps with a speed of <math>3b-e</math> per step. When Bob was on his way up, he took <math>75</math> steps with a speed of <math>b+e</math> per step. Since Al and Bob were walking the same distance, we have | ||
+ | <cmath>150(3b-e)=75(b+e)</cmath> | ||
+ | Solving gets the ratio <math>\frac{e}{b}=\frac{3}{5}</math>. | ||
+ | |||
+ | Thus while Bob took <math>75</math> steps to go up, the escalator contributed an extra <math>\frac{3}{5}\cdot75=45</math> steps. | ||
+ | |||
+ | Finally, there is a total of <math>75+45=\boxed{120}</math> steps in the length of the escalator. | ||
+ | |||
== See also == | == See also == | ||
− | |||
− | |||
{{AIME box|year=1987|num-b=9|num-a=11}} | {{AIME box|year=1987|num-b=9|num-a=11}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:21, 19 September 2020
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.
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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