AMC 8 · 2002 · #13
Grade 6 geometry-3dProblem
For his birthday, Bert gets a box that holds 125 jellybeans when filled to capacity. A few weeks later, Carrie gets a larger box full of jellybeans. Her box is twice as high, twice as wide and twice as long as Bert's. Approximately, how many jellybeans did Carrie get?
Pick an answer.
AMC 8 2002 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: Bert's box holds exactly $125$ jellybeans. Carrie's box has the same shape but twice the length, twice the width, and twice the height. About how many jellybeans does Carrie's box hold?
Givens: Bert's box holds $125$ jellybeans when filled; Carrie's box is twice as long, twice as wide, and twice as high as Bert's; Jellybean count is proportional to box volume; Answer choices: (A) $250$, (B) $500$, (C) $625$, (D) $750$, (E) $1000$
Unknowns: The approximate number of jellybeans in Carrie's box
Understand
Restated: Bert's box holds exactly $125$ jellybeans. Carrie's box has the same shape but twice the length, twice the width, and twice the height. About how many jellybeans does Carrie's box hold?
Givens: Bert's box holds $125$ jellybeans when filled; Carrie's box is twice as long, twice as wide, and twice as high as Bert's; Jellybean count is proportional to box volume; Answer choices: (A) $250$, (B) $500$, (C) $625$, (D) $750$, (E) $1000$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems
We are not told Bert's actual dimensions, only that the count is $125$. Tool #9 (Easier Related Problem) lets us pick convenient dimensions that match: a $5 \times 5 \times 5$ cube of jellybeans gives exactly $125$. Once the easy case is concrete, Carrie's box becomes $10 \times 10 \times 10$ and we can multiply directly. Tool #7 (Identify Subproblems) handles the three independent doublings — length doubles, width doubles, height doubles — so the volume factor is $2 \times 2 \times 2 = 8$, which explains why $125 \times 8 = 1000$ no matter what shape Bert's box really is.
Execute — Answer: E
5.MD.C.5 Step 1 - Choose convenient dimensions for Bert's box.
- Since $125 = 5 \times 5 \times 5$, model Bert's box as a $5 \times 5 \times 5$ cube of jellybeans.
- The volume in jellybeans is $125$, matching the problem.
💡 Grade 5 volume work: a rectangular prism's volume equals length $\times$ width $\times$ height, so any triple whose product is $125$ will do — a cube is the simplest.
5.MD.C.5 Step 2 - Double every dimension to build Carrie's box.
- Each side of Bert's $5 \times 5 \times 5$ box doubles to $10$, giving a $10 \times 10 \times 10$ box.
💡 Each dimension is its own subproblem: length, width, and height all double independently.
5.MD.C.5 Step 3 Compute Carrie's volume directly and read off the jellybean count.
💡 Same volume formula, bigger numbers — the answer matches choice (E).
6.RP.A.3 Step 4 - Confirm the scaling factor so the answer does not depend on Bert's true dimensions.
- Doubling three independent factors multiplies the product by $2 \times 2 \times 2 = 8$, so Carrie's box always holds $8$ times as many jellybeans as Bert's.
💡 Grade 6 ratio reasoning: scaling each side by $2$ scales the volume by the cube of $2$. Any box, not just a cube, gets multiplied by $8$.
5.MD.C.5 Choose convenient dimensions for Bert's box. Since $125 = 5 \times 5 \times 5$, 5.MD.C.5 Double every dimension to build Carrie's box. Each side of Bert's $5 \times 5 \t 5.MD.C.5 Compute Carrie's volume directly and read off the jellybean count. 6.RP.A.3 Confirm the scaling factor so the answer does not depend on Bert's true dimensio Review
Reasonableness: Two checks agree. First, the concrete $5 \times 5 \times 5 \to 10 \times 10 \times 10$ model gives $1000$. Second, the scaling argument gives $125 \times 8 = 1000$ without needing to know Bert's shape. The other choices are easy to rule out: $250$ corresponds to doubling only one dimension, $500$ to doubling two, and $625 = 125 \times 5$ has no geometric meaning here. Choice (E) $1000$ is the only count consistent with doubling all three dimensions.
Alternative: Tool #13 (Convert to Algebra): let Bert's box have dimensions $L \times W \times H$ with $L \cdot W \cdot H = 125$. Carrie's box is $2L \times 2W \times 2H$, so its volume is $(2L)(2W)(2H) = 8 \cdot L W H = 8 \cdot 125 = 1000$. Same answer (E), reached by letting variables stand in for the unspecified dimensions.
CCSS standards used (min grade 6)
5.MD.C.5Relate volume to the operations of multiplication and addition and solve real world and mathematical problems involving volume (Modeling Bert's $125$-jellybean box as a $5 \times 5 \times 5$ cube and Carrie's box as $10 \times 10 \times 10$, then multiplying length $\times$ width $\times$ height to count jellybeans.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Recognizing that doubling each of three dimensions multiplies the volume by $2 \times 2 \times 2 = 8$, so Carrie's count is $8 \times 125 = 1000$ for any box shape.)
⭐ Doubling one side doubles the volume, doubling two sides quadruples it, doubling all three sides multiplies it by $8$. That is why Carrie's box holds $125 \times 8 = 1000$ jellybeans.
⭐ Doubling one side doubles the volume, doubling two sides quadruples it, doubling all three sides multiplies it by $8$. That is why Carrie's box holds $125 \times 8 = 1000$ jellybeans.
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