AMC 8 · 2016 · #16

Easy mode Grade 7

Problem

Annie and Bonnie are running around a $400$ -meter oval track. They start side by side at the same spot.

Annie runs $25\%$ faster than Bonnie. That means whenever Bonnie runs $4$ laps, Annie runs $5$ laps in the same time.

After a while, Annie pulls so far ahead that she catches up to Bonnie from behind and passes her on the track for the first time.

How many laps has Annie run by the moment she first passes Bonnie?

Pick an answer.

(A)

$1\dfrac{1}{4}$

(B)

$3\dfrac{1}{3}$

(C)

(D)

(E)

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.

View mode:

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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