AMC 8 · 2016 · #16

Grade 7 rate-ratioalgebra

Archetype Multiplicative Ratio with Integrality Constraints

rateratio-proportionpercentagelinear-equations-one-var convert-to-algebrapattern-recognition ↑ Prerequisites: ratio-proportionpercentage

📏 Medium solution 💡 3 insights

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Problem

Annie and Bonnie are running laps around a $400$ -meter oval track. They started together, but Annie has pulled ahead, because she runs $25\%$ faster than Bonnie. How many laps will Annie have run when she first passes Bonnie?

$\textbf{(A) }1\dfrac{1}{4}\qquad\textbf{(B) }3\dfrac{1}{3}\qquad\textbf{(C) }4\qquad\textbf{(D) }5\qquad \textbf{(E) }25$

Pick an answer.

(A)

$1\dfrac{1}{4}$

(B)

$3\dfrac{1}{3}$

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Annie and Bonnie start together on a $400$-meter oval track. Annie runs $25\%$ faster than Bonnie. Annie first passes (laps) Bonnie when she has gone exactly one full lap more than Bonnie. How many laps has Annie run at that moment?

Givens: Track length $= 400$ m (the exact length will not matter — only the lap count does); Annie's speed is $25\%$ greater than Bonnie's speed; Both runners start at the same point and at the same time; Answer choices: (A) $1\tfrac{1}{4}$, (B) $3\tfrac{1}{3}$, (C) $4$, (D) $5$, (E) $25$

Unknowns: The number of laps Annie has completed at the instant she first passes Bonnie

Understand

Plan

Primary tool: #11 Find a Pattern

Secondary: #4 Use a Variable, #13 Solve an Equivalent Problem

The key insight is a ratio pattern: when two runners go for the same time, the ratio of their distances equals the ratio of their speeds (Tool #11, Find a Pattern). Turning "$25\%$ faster" into the clean ratio $5:4$ exposes that pattern. Tool #4 (Use a Variable) lets us call Bonnie's lap count $L_B$ and Annie's $L_A$, and the lapping condition becomes the equation $L_A = L_B + 1$. Tool #13 (Solve an Equivalent Problem) reframes the geometry: the $400$ m track length never enters the calculation — "first passes" is equivalent to "the gap between them equals exactly $1$ lap." Working in lap counts instead of meters cancels the track length entirely.

Execute — Answer: D

#11 Find a Pattern 6.RP.A.3 Step 1

Turn the percentage into a speed ratio.
"$25\%$ faster" means Annie's speed is $1 + \tfrac{1}{4} = \tfrac{5}{4}$ of Bonnie's, so the ratio of their speeds is $5 : 4$.

$$\dfrac{v_A}{v_B} = 1.25 = \dfrac{5}{4}$$

💡 Converting "$25\%$ more" to the fraction $\tfrac{5}{4}$ is the Grade 6 percent-as-ratio move.

#11 Find a Pattern 6.RP.A.3 Step 2

Same time means distance ratio equals speed ratio.
Both run for the same elapsed time $t$, so $\text{distance} = \text{speed} \times t$ gives Annie's distance / Bonnie's distance $= v_A / v_B = 5/4$.
Since one lap is the same $400$ m for both, the ratio of laps equals the ratio of distances.

$$\dfrac{L_A}{L_B} = \dfrac{v_A}{v_B} = \dfrac{5}{4}$$

💡 Equal time $\Rightarrow$ distances scale exactly with speeds — that is the rate pattern (Tool #11).

#4 Use a Variable 6.EE.B.7 Step 3

Translate "first passes" into an equation.
On a closed loop, Annie first catches Bonnie from behind when she has run exactly one more lap.
Let $L_B$ be Bonnie's lap count and $L_A$ be Annie's.

$$L_A = L_B + 1$$

💡 Naming the two lap counts as variables turns the word condition into a clean equation (Tool #4).

#4 Use a Variable 7.EE.B.4 Step 4

Solve the two-equation system.
Substitute $L_A = \tfrac{5}{4} L_B$ from the ratio into $L_A = L_B + 1$ and isolate $L_B$.

$$\tfrac{5}{4} L_B = L_B + 1 \;\Rightarrow\; \tfrac{1}{4} L_B = 1 \;\Rightarrow\; L_B = 4$$

💡 A one-step substitution leaves a linear equation in one variable — Grade 7 algebra.

#13 Solve an Equivalent Problem 6.RP.A.3 Step 5

Use the lapping equation to read off Annie's count.

$$L_A = L_B + 1 = 4 + 1 = 5 \;\Rightarrow\; \textbf{(D)}$$

💡 The $400$ m track length never appeared — the answer depends only on the lap-count gap (Tool #13, equivalent problem).

[1] #11 6.RP.A.3 Turn the percentage into a speed ratio. "$25\%$ faster" means Annie's speed is $

[2] #11 6.RP.A.3 Same time means distance ratio equals speed ratio. Both run for the same elapsed

[3] #4 6.EE.B.7 Translate "first passes" into an equation. On a closed loop, Annie first catches

[4] #4 7.EE.B.4 Solve the two-equation system. Substitute $L_A = \tfrac{5}{4} L_B$ from the rati

[5] #13 6.RP.A.3 Use the lapping equation to read off Annie's count.

Review

Reasonableness: Per lap by Bonnie, Annie runs $\tfrac{5}{4}$ of a lap, so Annie's lead grows by $\tfrac{1}{4}$ lap every time Bonnie completes one lap. To open up a full $1$-lap lead, Bonnie needs $4$ laps and Annie runs $5$ laps. The numbers are small, integer, and consistent with the answer choices — and choice (D) $= 5$ matches.

Alternative: Tool #5 (Make a Table). After each Bonnie lap, list Annie's distance and the gap: Bonnie $1$ lap $\to$ Annie $1.25$ laps, gap $0.25$; Bonnie $2 \to$ Annie $2.5$, gap $0.5$; Bonnie $3 \to$ Annie $3.75$, gap $0.75$; Bonnie $4 \to$ Annie $5$, gap $1$. Annie passes Bonnie exactly when she finishes lap $5$, confirming (D).

CCSS standards used (min grade 7)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Turning "$25\%$ faster" into the speed ratio $5:4$ and transferring that ratio to lap counts because the elapsed time is the same for both runners.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Writing the lapping condition $L_A = L_B + 1$ that captures "first passes" on a closed track.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities to solve problems (Substituting $L_A = \tfrac{5}{4} L_B$ into $L_A = L_B + 1$ and solving the linear equation $\tfrac{1}{4} L_B = 1$ to get $L_B = 4$.)

⭐ This AMC 8 problem only needs Grade 7 algebra — a speed ratio plus a one-step linear equation — that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 7)

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