2012 AIME I Problem 4

Below is the professionally curated solution for Problem 4 of the 2012 AIME I, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AIME I solutions, or check the answer key.

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Concepts:distance rate and timeDiophantine Equation

Difficulty rating: 2460

4.

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 6,6, 4,4, and 2.52.5 miles per hour, respectively. The first time Butch and Sundance meet at a milepost, they are nn miles from Dodge, and they have been traveling for tt minutes. Find n+t.n + t.

Solution:

Walking a mile takes Sparky 1010 minutes, Butch 15,15, and Sundance 24.24. The horse advances along the same route as the men and is ridden over each mile by exactly one of them, so if Butch walks xx of the nn miles and rides the other nx,n - x, then Sundance rides those xx miles and walks the remaining nx.n - x.

When they meet at a milepost they have been traveling for the same amount of time, so 15x+10(nx)=10x+24(nx),15x + 10(n - x) = 10x + 24(n - x), which simplifies to 19x=14n.19x = 14n. Since the handoffs happen at mileposts, xx and nn are integers, and the smallest positive solution is x=14,x = 14, n=19.n = 19.

Then t=1514+105=260t = 15 \cdot 14 + 10 \cdot 5 = 260 minutes, so n+t=19+260=279.n + t = 19 + 260 = 279.

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