2019 AMC 10B Problems/Problem 18

Revision as of 12:22, 15 February 2019 by Xmidnightfirex (talk | contribs) (Solution)

Problem

Henry decides one morning to do a workout, and he walks $\tfrac{3}{4}$ of the way from his home to his gym. The gym is $2$ kilometers away from Henry's home. At that point, he changes his mind and walks $\tfrac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\tfrac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\tfrac{3}{4}$ of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?

$\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1\frac{1}{5} \qquad \textbf{(D) } 1\frac{1}{4} \qquad \textbf{(E) } 1\frac{1}{2}$

Solution

Let the two points that Henry walks in-between be $A$ and $B$, with $A$ being closer to home. In addition, let the distance that the points $A$ and $B$ are from his home be $a$ and $b$, respectively. By symmetry, the distance from point $B$ is from the gym is the same as the distance from home to point $A$. Thus, $a = 2 - b$. In addition, the distance that he walks as he repeatedly heads to home and then to the gym is $b - a$. Therefore, we are looking for the value of $b - a$. When he walks from point $B$ to home, he walks $\frac{3}{4}$ of the distance, ending at point $A$. Therefore, we know that $b - a = \frac{3}{4} \cdot b$. Similarily, we know $b - a = \frac{3}{4} \cdot (2 - a)$. Adding these equations, we get $2(b - a) = \frac{3}{4} \cdot (2 + b - a)$. Multiplying by $4$, we get $8(b - a) = 6 + 3(b - a)$, so $b - a = \frac{6}{5} = 1 \frac{1}{5}$.

See Also

2019 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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