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AMC 10 · 2019 · #18

Grade 8 rate-ratio
Archetype Arithmetic Expression Decomposition
recursive-sequencesequences-geometriclinear-equations-two-varsystems-of-equations identify-subproblemswork-backwards ↑ Prerequisites: linear-equations-two-varsequences-geometric
📏 Medium solution 💡 3 insights

Problem

Henry decides one morning to do a workout, and he walks 34\tfrac{3}{4} of the way from his home to his gym. The gym is 22 kilometers away from Henry's home. At that point, he changes his mind and walks 34\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 34\tfrac{3}{4} of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked 34\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 AA kilometers from home and a point BB kilometers from home. What is ∣A−B∣|A-B|?

(A) 23(B) 1(C) 115(D) 114(E) 112\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}

Pick an answer.

(A)
$\frac{2}{3}$
(B)
1
(C)
$1 \frac{1}{5}$
(D)
$1 \frac{1}{4}$
(E)
$1 \frac{1}{2}$
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Toolkit + CCSS Solution

Understand

Restated: Henry's home is at $0$ km, his gym at $2$ km. He walks $3/4$ of the way to the gym, then $3/4$ of the way back home from there, then $3/4$ of the way to the gym from there, and so on, always turning around after covering $3/4$ of the remaining distance. In the long run his back-and-forth motion approaches two limit points $A$ (closer to home) and $B$ (closer to gym), measured in km from home. Find $|A - B|$.

Givens: Home at $0$ km, gym at $2$ km; Each leg covers $3/4$ of the distance from the current point to the target (home or gym); Henry alternates targets: gym, home, gym, home, $\ldots$; $A$ = limit point closer to home, $B$ = limit point closer to gym; Choices: (A) $2/3$, (B) $1$, (C) $1\tfrac{1}{5}$, (D) $1\tfrac{1}{4}$, (E) $1\tfrac{1}{2}$

Unknowns: $|A - B|$, the gap between the two limit points

Understand

Restated: Henry's home is at $0$ km, his gym at $2$ km. He walks $3/4$ of the way to the gym, then $3/4$ of the way back home from there, then $3/4$ of the way to the gym from there, and so on, always turning around after covering $3/4$ of the remaining distance. In the long run his back-and-forth motion approaches two limit points $A$ (closer to home) and $B$ (closer to gym), measured in km from home. Find $|A - B|$.

Givens: Home at $0$ km, gym at $2$ km; Each leg covers $3/4$ of the distance from the current point to the target (home or gym); Henry alternates targets: gym, home, gym, home, $\ldots$; $A$ = limit point closer to home, $B$ = limit point closer to gym; Choices: (A) $2/3$, (B) $1$, (C) $1\tfrac{1}{5}$, (D) $1\tfrac{1}{4}$, (E) $1\tfrac{1}{2}$

Plan

Primary tool: #13 Convert to Algebra

Secondary: #1 Draw a Diagram, #11 Work Backwards

Tool #1 (Diagram): a number line from $0$ to $2$ with $A$ and $B$ marked makes the geometry obvious. Tool #13 (Algebra): the limit condition gives two clean linear equations in $A$ and $B$. Tool #11 (Work Backwards): instead of simulating forward from the first walk, use the fixed-point relations $A$ and $B$ must satisfy at the limit and solve directly.

Execute — Answer: C

#1 Draw a Diagram 5.NF.B.4 Step 1
  • Draw the number line from $0$ (home) to $2$ (gym), marking $A$ between $0$ and $B$, and $B$ between $A$ and $2$.
  • From $B$, walking $3/4$ of the way toward home means moving a fraction $3/4$ of the segment from $B$ to $0$, i.e., subtracting $\tfrac{3}{4} \cdot B$ from $B$.
  • This lands at $B - \tfrac{3}{4}B = \tfrac{1}{4}B$, which must be $A$.
$$A = \dfrac{1}{4}B$$

💡 Walking $3/4$ of the way toward $0$ leaves only $1/4$ of the original distance left — so you land at $\tfrac{1}{4}$ of the starting position.

#13 Convert to Algebra 6.EE.A.2 Step 2
  • From $A$, walking $3/4$ of the way toward the gym means moving $\tfrac{3}{4}$ of the segment from $A$ to $2$.
  • New position $= A + \tfrac{3}{4}(2 - A)$.
  • This lands at $B$.
  • Simplify: $A + \tfrac{3}{2} - \tfrac{3}{4}A = \tfrac{1}{4}A + \tfrac{3}{2}$.
$$B = \dfrac{1}{4}A + \dfrac{3}{2}$$

💡 Same $1/4$ remainder trick from the home side, but the gym is at $2$ instead of $0$.

#13 Convert to Algebra 8.EE.C.7 Step 3
  • Substitute $A = \tfrac{1}{4}B$ into the second equation: $B = \tfrac{1}{4} \cdot \tfrac{1}{4}B + \tfrac{3}{2} = \tfrac{1}{16}B + \tfrac{3}{2}$.
  • Move the $B$ terms together: $B - \tfrac{1}{16}B = \tfrac{3}{2}$, so $\tfrac{15}{16}B = \tfrac{3}{2}$.
$$\dfrac{15}{16}B = \dfrac{3}{2}$$

💡 Substitute one equation into the other — a single equation in $B$.

#13 Convert to Algebra 8.EE.C.7 Step 4
  • Solve for $B$: $B = \tfrac{3}{2} \cdot \tfrac{16}{15} = \tfrac{48}{30} = \tfrac{8}{5}$.
  • Then $A = \tfrac{1}{4}B = \tfrac{1}{4} \cdot \tfrac{8}{5} = \tfrac{2}{5}$.
$$B = \dfrac{8}{5},\;\; A = \dfrac{2}{5}$$

💡 One-step division gives $B$; then $A = B/4$.

#11 Work Backwards 5.NF.A.1 Step 5
  • Compute the gap: $|A - B| = \left|\tfrac{2}{5} - \tfrac{8}{5}\right| = \tfrac{6}{5} = 1\tfrac{1}{5}$.
  • Answer: choice (C).
$$|A - B| = \dfrac{6}{5} = 1\tfrac{1}{5}$$

💡 Subtract two fractions with the same denominator — easy.

[1] #1 5.NF.B.4 Draw the number line from $0$ (home) to $2$ (gym), marking $A$ between $0$ and $
[2] #13 6.EE.A.2 From $A$, walking $3/4$ of the way toward the gym means moving $\tfrac{3}{4}$ of
[3] #13 8.EE.C.7 Substitute $A = \tfrac{1}{4}B$ into the second equation: $B = \tfrac{1}{4} \cdot
[4] #13 8.EE.C.7 Solve for $B$: $B = \tfrac{3}{2} \cdot \tfrac{16}{15} = \tfrac{48}{30} = \tfrac{
[5] #11 5.NF.A.1 Compute the gap: $|A - B| = \left|\tfrac{2}{5} - \tfrac{8}{5}\right| = \tfrac{6}

Review

Reasonableness: Sanity-check the limit points: from $B = 8/5$, walk $3/4$ toward home covers $(3/4)(8/5) = 6/5$ km, landing at $8/5 - 6/5 = 2/5 = A$. ✓. From $A = 2/5$, walk $3/4$ toward gym covers $(3/4)(2 - 2/5) = (3/4)(8/5) = 6/5$ km, landing at $2/5 + 6/5 = 8/5 = B$. ✓. The gap $6/5 = 1.2$ km is between the choices $1$ and $1.5$, matching choice (C).

Alternative: Tool #9 (Easier Problem) + Tool #5 (Pattern): simulate the first few positions. $x_0 = 0, x_1 = 3/2, x_2 = 3/2 - (3/4)(3/2) = 3/8, x_3 = 3/8 + (3/4)(2 - 3/8) = 3/8 + 39/32 = 51/32, \ldots$. The odd-indexed positions $x_1, x_3, x_5, \ldots$ converge to $B = 8/5$; the even-indexed (after start) converge to $A = 2/5$. Same gap $6/5$.

CCSS standards used (min grade 8)

  • 5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $|A - B| = 8/5 - 2/5 = 6/5$.)
  • 5.NF.B.4 Apply and extend understanding of multiplication to multiply a fraction by a fraction (Computing $\tfrac{3}{4}$ of a distance to find the landing point $A = \tfrac{1}{4}B$.)
  • 6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Writing $B = A + \tfrac{3}{4}(2 - A)$ as an algebraic expression.)
  • 8.EE.C.7 Solve linear equations in one variable (Solving $\tfrac{15}{16}B = \tfrac{3}{2}$ for $B$.)

⭐ This AMC 10 problem only needs Grade 8 linear-equation skills you already know — at the limit, Henry's two turning points $A$ and $B$ satisfy $A = B/4$ and $B = A/4 + 3/2$, giving $B = 8/5$ and $A = 2/5$, so the gap is $6/5 = 1\tfrac{1}{5}$. The answer is $\textbf{(C)}$.

⭐ This AMC 10 problem only needs Grade 8 linear-equation skills you already know — at the limit, Henry's two turning points $A$ and $B$ satisfy $A = B/4$ and $B = A/4 + 3/2$, giving $B = 8/5$ and $A = 2/5$, so the gap is $6/5 = 1\tfrac{1}{5}$. The answer is $\textbf{(C)}$.

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