Difference between revisions of "2012 AIME I Problems/Problem 4"
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− | == Problem | + | == 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>. | ||
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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>. | 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>. | ||
− | + | == Video Solution by Richard Rusczyk == | |
https://artofproblemsolving.com/videos/amc/2012aimei/332 | https://artofproblemsolving.com/videos/amc/2012aimei/332 |
Latest revision as of 19:18, 24 January 2021
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 and miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are miles from Dodge, and they have been traveling for minutes. Find .
Solution
When they meet at the milepost, Sparky has been ridden for miles total. Assume Butch rides Sparky for miles, and Sundance rides for miles. Thus, we can set up an equation, given that Sparky takes hours per mile, Butch takes hours per mile, and Sundance takes hours per mile:
The smallest possible integral value of is , so we plug in and and get hours, or minutes. So our answer is .
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 | |
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.