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Difference between revisions of "2019 AMC 10B Problems/Problem 18"
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==Solution==
 
==Solution==
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Let the two points that Henry walks in-between be <math>A</math> and <math>B</math>, with <math>A</math> being closer to home. In addition, let the distance that the points <math>A</math> and <math>B</math> are from his home be <math>a</math> and <math>b</math>, respectively. By symmetry, the distance from point <math>B</math> is from the gym is the same as the distance from home to point <math>A</math>. Thus, <math>a = 2 - b</math>. In addition, the distance that he walks as he repeatedly heads to home and then to the gym is <math>b - a</math>. Therefore, we are looking for the value of <math>b - a</math>. When he walks from point <math>B</math> to home, he walks <math>\frac{3}{4}</math> of the distance, ending at point <math>A</math>. Therefore, we know that <math>b - a = \frac{3}{4} \cdot b</math>. Similarily, we know <math>b - a = \frac{3}{4} \cdot (2 - a)</math>. Adding these equations, we get <math>2(b - a) = \frac{3}{4} \cdot (2 + b - a)</math>. Multiplying by <math>4</math>, we get <math>8(b - a) = 6 + 3(b - a)</math>, so <math>b - a = \frac{6}{5} = 1 \frac{1}{5}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 15:41, 14 February 2019

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 (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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