AMC 8 · 2015 · #3

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition

rateunit-conversion dimensional-analysisidentify-subproblems ↑ Prerequisites: fraction-arithmeticmulti-digit-arithmetic

📏 Short solution 💡 2 insights

Problem

Jack and Jill are going swimming at a pool that is one mile from their house. They leave home simultaneously. Jill rides her bicycle to the pool at a constant speed of $10$ miles per hour. Jack walks to the pool at a constant speed of $4$ miles per hour. How many minutes before Jack does Jill arrive?

$\textbf{(A) }5\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: Jack and Jill both leave home at the same instant for a pool that is $1$ mile away. Jill bikes at a constant $10$ mph; Jack walks at a constant $4$ mph. How many minutes earlier does Jill arrive?

Givens: Distance to the pool $= 1$ mile (same for both); Jill's speed $= 10$ mph (constant); Jack's speed $= 4$ mph (constant); Both leave at the same moment; Answer choices: (A) $5$, (B) $6$, (C) $8$, (D) $9$, (E) $10$ (minutes)

Unknowns: The number of minutes between Jill's arrival and Jack's arrival

Understand

Plan

Primary tool: #8 Analyze the Units

Secondary: #7 Identify Subproblems

This is a rate problem with $\text{time} = \text{distance} / \text{speed}$. The catch is that distance is in miles and speed is in mph, so dividing gives hours — but the answer must be in minutes. Tool #8 (Analyze the Units) keeps the conversion honest: $\tfrac{\text{mile}}{\text{mile/hour}} = \text{hour}$, then $\text{hour} \times 60 = \text{minutes}$. Tool #7 (Identify Subproblems) splits the work into three clean pieces — Jill's time, Jack's time, and the difference — so each piece is a single short calculation.

Execute — Answer: D

#8 Analyze the Units 6.RP.A.3 Step 1

Compute Jill's travel time.
She covers $1$ mile at $10$ mph, so the time in hours is $1 / 10$.
Convert to minutes by multiplying by $60$.

$$t_{\text{Jill}} = \dfrac{1 \text{ mi}}{10 \text{ mph}} = \tfrac{1}{10} \text{ hr} = \tfrac{1}{10} \times 60 = 6 \text{ min}$$

💡 Dividing miles by miles-per-hour cancels "miles" and leaves "hours" — Grade 6 unit-rate reasoning.

#8 Analyze the Units 6.RP.A.3 Step 2

Compute Jack's travel time the same way.
He covers $1$ mile at $4$ mph, so the time in hours is $1 / 4$.
Convert to minutes by multiplying by $60$.

$$t_{\text{Jack}} = \dfrac{1 \text{ mi}}{4 \text{ mph}} = \tfrac{1}{4} \text{ hr} = \tfrac{1}{4} \times 60 = 15 \text{ min}$$

💡 Same unit-rate move as Jill's step — once the method works for one rider, it works for both.

#7 Identify Subproblems 4.MD.A.2 Step 3

Subtract to find how many minutes earlier Jill arrives.
Since both leave at the same instant, the gap between their arrivals equals the gap between their travel times.

$$t_{\text{Jack}} - t_{\text{Jill}} = 15 - 6 = 9 \text{ min} \;\Rightarrow\; \textbf{(D)}$$

💡 "Difference of two times" is the Grade 4 distance/time word-problem move — solving the third subproblem to finish the job.

[1] #8 6.RP.A.3 Compute Jill's travel time. She covers $1$ mile at $10$ mph, so the time in hour

[2] #8 6.RP.A.3 Compute Jack's travel time the same way. He covers $1$ mile at $4$ mph, so the t

[3] #7 4.MD.A.2 Subtract to find how many minutes earlier Jill arrives. Since both leave at the

Review

Reasonableness: Jill is $10 / 4 = 2.5$ times faster than Jack, so her travel time should be $2.5$ times shorter: $15 / 2.5 = 6$ min, matching the calculation. The gap of $9$ minutes is sensible for a $1$-mile trip at very different speeds — Jill finishes in $6$ min while Jack still has more than half the mile to go.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: the gap must equal $t_{\text{Jack}} - t_{\text{Jill}}$. From $t_{\text{Jill}} = 6$ and $t_{\text{Jack}} = 15$, the difference is $9$, so only (D) survives. The choices $5, 6, 8, 10$ would correspond to wrong arithmetic — e.g. $10$ is $t_{\text{Jack}} - 5$ (forgetting to compute Jill's time), $6$ is just Jill's own time.

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Computing each rider's travel time as $\text{distance} / \text{speed}$ — a unit-rate calculation — for both Jill ($1/10$ hr) and Jack ($1/4$ hr).)
5.MD.A.1 Convert among different-sized standard measurement units within a given system (Converting each rider's time from hours to minutes by multiplying by $60$ ($1/10 \times 60 = 6$ and $1/4 \times 60 = 15$).)
4.MD.A.2 Solve word problems involving distances, time, liquid volumes, and money (Subtracting the two arrival times ($15 - 6 = 9$) to answer the question "how many minutes before Jack does Jill arrive?")

⭐ This AMC 8 problem only needs the Grade 6 rate rule "time = distance $\div$ speed" plus a minute-to-hour conversion you learned in Grade 5.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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