AMC 8 · 2001 · #15
Grade 6 rate-ratioalgebraProblem
Homer began peeling a pile of 44 potatoes at the rate of 3 potatoes per minute. Four minutes later Christen joined him and peeled at the rate of 5 potatoes per minute. When they finished, how many potatoes had Christen peeled?
Pick an answer.
AMC 8 2001 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: Homer starts peeling a pile of $44$ potatoes at $3$ per minute. After $4$ minutes Christen joins and peels at $5$ per minute. They keep peeling until the pile is gone. How many potatoes did Christen peel?
Givens: Total pile: $44$ potatoes; Homer's rate: $3$ potatoes/minute, starting at time $0$; Christen's rate: $5$ potatoes/minute, starting at time $4$ minutes; Both stop the moment the pile is finished; Answer choices: (A) $20$, (B) $24$, (C) $32$, (D) $33$, (E) $40$
Unknowns: The number of potatoes Christen peeled before the pile ran out
Understand
Restated: Homer starts peeling a pile of $44$ potatoes at $3$ per minute. After $4$ minutes Christen joins and peels at $5$ per minute. They keep peeling until the pile is gone. How many potatoes did Christen peel?
Givens: Total pile: $44$ potatoes; Homer's rate: $3$ potatoes/minute, starting at time $0$; Christen's rate: $5$ potatoes/minute, starting at time $4$ minutes; Both stop the moment the pile is finished; Answer choices: (A) $20$, (B) $24$, (C) $32$, (D) $33$, (E) $40$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #8 Analyze the Units
The story has two distinct phases — Homer alone for the first $4$ minutes, then Homer and Christen together. Tool #7 (Identify Subproblems) splits the work along that natural seam: (a) count what Homer peeled alone and subtract from $44$ to get what is left, then (b) work out how long the two of them need to finish that remainder, and finally read off Christen's share. Tool #8 (Analyze the Units) keeps every step honest: every multiplication is $\text{potatoes/minute} \times \text{minutes} = \text{potatoes}$, and the rate-addition step pools $3 + 5$ correctly because both rates share the same unit.
Execute — Answer: A
5.NBT.B.5 Step 1 - Subproblem 1: count what Homer peeled in the first $4$ minutes alone.
- Multiply his rate by the time he worked solo.
💡 Rate $\times$ time $=$ amount. The minute units cancel, leaving potatoes — exactly what we want to count.
4.OA.A.3 Step 2 - Find how many potatoes are still in the pile when Christen joins.
- Subtract Homer's $12$ from the starting $44$.
💡 Total minus what is done leaves what is left — the bridge between phase 1 and phase 2.
5.NBT.B.5 Step 3 - Subproblem 2: combine the two rates.
- Once Christen joins, they peel from the same pile at the same time, so add the rates.
💡 Rates with the same units add directly. Two peelers acting together is one combined rate of $8$ potatoes per minute.
6.RP.A.3 Step 4 - Find how long phase 2 lasts.
- Divide the remaining $32$ potatoes by the combined rate $8$ potatoes/minute.
💡 Dividing potatoes by potatoes-per-minute leaves minutes — the time to finish at the joint pace.
5.NBT.B.5 Step 5 - Count Christen's potatoes.
- Multiply her rate by the time she actually peeled, which is the $4$ minutes of phase 2.
💡 Christen only worked during phase 2, so her share is her rate times the phase-2 duration — nothing more, nothing less.
5.NBT.B.5 Subproblem 1: count what Homer peeled in the first $4$ minutes alone. Multiply h 4.OA.A.3 Find how many potatoes are still in the pile when Christen joins. Subtract Homer 5.NBT.B.5 Subproblem 2: combine the two rates. Once Christen joins, they peel from the sam 6.RP.A.3 Find how long phase 2 lasts. Divide the remaining $32$ potatoes by the combined 5.NBT.B.5 Count Christen's potatoes. Multiply her rate by the time she actually peeled, wh Review
Reasonableness: Cross-check the totals. Homer peeled $12$ alone plus $3 \times 4 = 12$ more during phase 2, so Homer peeled $24$. Christen peeled $20$. Together that is $24 + 20 = 44$ potatoes — the whole pile, exactly as required. The trap answers fit common slips. (C) $32$ is the leftover pile, not Christen's share. (B) $24$ is Homer's total. (D) $33$ matches splitting the original $44$ by the rate ratio $5$:$3$ as if Christen had peeled the whole time. (E) $40$ comes from giving Christen all $8$ minutes at $5$/minute, ignoring her $4$-minute late start. Choosing (A) means the late start was handled correctly.
Alternative: Tool #2 (Make a Systematic List) by minute: minutes $1$–$4$ Homer peels $3, 6, 9, 12$ (Christen $0$ each). At minute $5$, Homer at $15$, Christen at $5$, pile $20$. Minute $6$: Homer $18$, Christen $10$, pile $16$. Minute $7$: Homer $21$, Christen $15$, pile $9$. Minute $8$: Homer $24$, Christen $20$, pile $0$ — done. Christen's column ends at $20$, the same answer (A).
CCSS standards used (min grade 6)
5.NBT.B.5Fluently multiply multi-digit whole numbers using the standard algorithm (Computing the rate-times-time products $3 \times 4 = 12$ and $5 \times 4 = 20$ that count Homer's solo work and Christen's share.)4.OA.A.3Solve multi-step word problems with whole numbers using the four operations (Subtracting $44 - 12 = 32$ to find the potatoes still in the pile when Christen joins, linking phase 1 to phase 2.)6.RP.A.3Use ratio and rate reasoning to solve real-world problems (Adding the rates $3 + 5 = 8$ potatoes/minute and dividing $32 \div 8 = 4$ minutes to find how long phase 2 lasts.)
⭐ Split the story into two phases — Homer alone, then both together. Subtract what Homer peeled alone, add the rates for the joint phase, and Christen's $4$ minutes at $5$/minute give $20$ potatoes, answer (A).
⭐ Split the story into two phases — Homer alone, then both together. Subtract what Homer peeled alone, add the rates for the joint phase, and Christen's $4$ minutes at $5$/minute give $20$ potatoes, answer (A).
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