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Difference between revisions of "2012 AIME I Problems/Problem 4"

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== Problem 4 ==
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== Problem ==
 
Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at <math>6,</math> <math>4,</math> and <math>2.5</math> miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are <math>n</math> miles from Dodge, and they have been traveling for <math>t</math> minutes. Find <math>n + t</math>.
 
Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at <math>6,</math> <math>4,</math> and <math>2.5</math> miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are <math>n</math> miles from Dodge, and they have been traveling for <math>t</math> minutes. Find <math>n + t</math>.
  
 
== Solution ==
 
== Solution ==
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When they meet at the milepost, Sparky has been ridden for <math>n</math> miles total. Assume Butch rides Sparky for <math>a</math> miles, and Sundance rides for <math>n-a</math> miles. Thus, we can set up an equation, given that Sparky takes <math>\frac{1}{6}</math> hours per mile, Butch takes <math>\frac{1}{4}</math>, and Sundance takes <math>\frac{2}{5} \cdot \frac{a}{6} + \frac{1}{4}(n-a) = \frac{n-a}{6} + \frac{2a}{5} \rightarrow  a = \frac{5}{19}n</math>.
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When they meet at the milepost, Sparky has been ridden for <math>n</math> miles total. Assume Butch rides Sparky for <math>a</math> miles, and Sundance rides for <math>n-a</math> miles. Thus, we can set up an equation, given that Sparky takes <math>\frac{1}{6}</math> hours per mile, Butch takes <math>\frac{1}{4}</math> hours per mile, and Sundance takes <math>\frac{2}{5}</math> hours per mile:
  
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Thus, the smallest possible value of <math>n</math> is <math>19</math>, and we plug in <math>n = 19</math> and <math>a = 5</math> to get <math>t = \frac{13}{3}</math> hours, or <math>260</math> minutes. Thus, our answer is <math>19 + 260 = \boxed{279}</math>.
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<cmath>\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow  a = \frac{5}{19}n.</cmath>
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The smallest possible integral value of <math>n</math> is <math>19</math>, so we plug in <math>n = 19</math> and <math>a = 5</math> and get <math>t = \frac{13}{3}</math> hours, or <math>260</math> minutes. So our answer is <math>19 + 260 = \boxed{279}</math>.
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Note that this solution is not rigorous because it is not guaranteed that they will switch properly to form this combination.
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== Video Solution by Richard Rusczyk ==
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https://artofproblemsolving.com/videos/amc/2012aimei/332
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~ dolphin7
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2012|n=I|num-b=3|num-a=5}}
 
{{AIME box|year=2012|n=I|num-b=3|num-a=5}}
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{{MAA Notice}}

Latest revision as of 21:28, 1 January 2023

Problem

Butch and Sundance need to get out of Dodge. To travel as quickly as possible, each alternates walking and riding their only horse, Sparky, as follows. Butch begins by walking while Sundance rides. When Sundance reaches the first of the hitching posts that are conveniently located at one-mile intervals along their route, he ties Sparky to the post and begins walking. When Butch reaches Sparky, he rides until he passes Sundance, then leaves Sparky at the next hitching post and resumes walking, and they continue in this manner. Sparky, Butch, and Sundance walk at $6,$ $4,$ and $2.5$ miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are $n$ miles from Dodge, and they have been traveling for $t$ minutes. Find $n + t$.

Solution

When they meet at the milepost, Sparky has been ridden for $n$ miles total. Assume Butch rides Sparky for $a$ miles, and Sundance rides for $n-a$ miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile:

\[\frac{a}{6} + \frac{n-a}{4} = \frac{n-a}{6} + \frac{2a}{5} \rightarrow a = \frac{5}{19}n.\]

The smallest possible integral value of $n$ is $19$, so we plug in $n = 19$ and $a = 5$ and get $t = \frac{13}{3}$ hours, or $260$ minutes. So our answer is $19 + 260 = \boxed{279}$.

Note that this solution is not rigorous because it is not guaranteed that they will switch properly to form this combination.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2012aimei/332

~ dolphin7

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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All AIME Problems and Solutions

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