AMC 8 · 2016 · #16
Grade 7 rate-ratioalgebraProblem
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?
Pick an answer.
AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $25$
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
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 $5 : 4$.
💡 Converting "$25\%$ more" to the fraction $\tfrac{5}{4}$ is the Grade 6 percent-as-ratio move.
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.
💡 Equal time $\Rightarrow$ distances scale exactly with speeds — that is the rate pattern (Tool #11).
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.
💡 Naming the two lap counts as variables turns the word condition into a clean equation (Tool #4).
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 $L_B$.
💡 A one-step substitution leaves a linear equation in one variable — Grade 7 algebra.
6.RP.A.3 Step 5 Use the lapping equation to read off Annie's count.
💡 The $400$ m track length never appeared — the answer depends only on the lap-count gap (Tool #13, equivalent problem).
6.RP.A.3 Turn the percentage into a speed ratio. "$25\%$ faster" means Annie's speed is $ 6.RP.A.3 Same time means distance ratio equals speed ratio. Both run for the same elapsed 6.EE.B.7 Translate "first passes" into an equation. On a closed loop, Annie first catches 7.EE.B.4 Solve the two-equation system. Substitute $L_A = \tfrac{5}{4} L_B$ from the rati 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.3Use 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.7Solve 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.4Use 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!
⭐ This AMC 8 problem only needs Grade 7 algebra — a speed ratio plus a one-step linear equation — that you already know!
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