AMC 8 · 2015 · #3
Easy mode Grade 6Problem
Jack and Jill are heading to a pool that is 1 mile from their house. They leave home at the same moment.
Jill rides her bike the whole way at 10 miles per hour. Jack walks the whole way at 4 miles per hour.
How many minutes before Jack does Jill arrive at the pool?
Pick an answer.
AMC 8 2015 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: 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
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)
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
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 $60$.
💡 Dividing miles by miles-per-hour cancels "miles" and leaves "hours" — Grade 6 unit-rate reasoning.
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 $60$.
💡 Same unit-rate move as Jill's step — once the method works for one rider, it works for both.
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.
💡 "Difference of two times" is the Grade 4 distance/time word-problem move — solving the third subproblem to finish the job.
6.RP.A.3 Compute Jill's travel time. She covers $1$ mile at $10$ mph, so the time in hour 6.RP.A.3 Compute Jack's travel time the same way. He covers $1$ mile at $4$ mph, so the t 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.3Use 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.1Convert 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.2Solve 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.
⭐ 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.
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