AMC 10 · 2020 · #2
Easy mode Grade 5Problem
Imagine a pile of small cubes. Carl has cubes, and each one has side length . Kate also has cubes, but each of hers has side length .
What is the total volume of all cubes put together?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: Carl owns $5$ unit cubes (side $1$) and Kate owns $5$ cubes of side $2$. Find the combined volume of all $10$ cubes.
Givens: Carl has $5$ cubes, each with side length $1$; Kate has $5$ cubes, each with side length $2$; Volume of a cube of side $s$ is $s \times s \times s = s^3$; Answer choices: (A) $24$, (B) $25$, (C) $28$, (D) $40$, (E) $45$
Unknowns: Total volume of all $10$ cubes added together
Understand
Restated: Carl owns $5$ unit cubes (side $1$) and Kate owns $5$ cubes of side $2$. Find the combined volume of all $10$ cubes.
Givens: Carl has $5$ cubes, each with side length $1$; Kate has $5$ cubes, each with side length $2$; Volume of a cube of side $s$ is $s \times s \times s = s^3$; Answer choices: (A) $24$, (B) $25$, (C) $28$, (D) $40$, (E) $45$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #10 Create a Physical Representation, #8 Analyze the Units
Tool #7 (Subproblems): split the question into two simple sub-jobs — find Carl's total, find Kate's total, then add. Tool #10 (Physical): if a student can't picture it, stacking unit cubes makes it obvious that a side-$2$ cube holds $8$ unit cubes. Tool #8 (Units): the answer should be in cubic units; tracking units keeps us from confusing side length with volume.
Execute — Answer: E
5.MD.C.5 Step 1 - Subproblem 1 — Carl's total.
- Each of his cubes has volume $1 \times 1 \times 1 = 1$ cubic unit.
- Five such cubes give $5 \times 1 = 5$ cubic units.
💡 Volume of a unit cube is $1$ — Grade 5 “relate volume to multiplication.”
5.MD.C.5 Step 2 - Subproblem 2 — Kate's total.
- Each of her cubes has volume $2 \times 2 \times 2 = 8$ cubic units (imagine stacking $8$ unit cubes into a $2 \times 2 \times 2$ block).
- Five cubes give $5 \times 8 = 40$.
💡 Doubling each side multiplies volume by $2 \times 2 \times 2 = 8$ — building it from unit cubes makes this concrete.
4.OA.A.3 Step 3 Add the two subtotals to get the combined volume: $5 + 40 = 45$ cubic units.
💡 When pieces don't overlap, total volume is just the sum — Grade 4 multi-step problem.
5.MD.C.5 Step 4 - Unit check: side lengths are in units, so each $s^3$ is in cubic units.
- Adding cubic units to cubic units yields cubic units — the answer $45$ is dimensionally consistent.
💡 If our number is in cubic units, the answer is sensible as a “total volume.”
5.MD.C.5 Subproblem 1 — Carl's total. Each of his cubes has volume $1 \times 1 \times 1 = 5.MD.C.5 Subproblem 2 — Kate's total. Each of her cubes has volume $2 \times 2 \times 2 = 4.OA.A.3 Add the two subtotals to get the combined volume: $5 + 40 = 45$ cubic units. 5.MD.C.5 Unit check: side lengths are in units, so each $s^3$ is in cubic units. Adding c Review
Reasonableness: Kate's cubes are $8$ times bigger by volume than Carl's, so they should dominate the total — and indeed $40$ (Kate) vs $5$ (Carl) shows that. Total $45$ is the only answer choice large enough to fit, supporting (E).
Alternative: Tool #2 (Systematic List) — list every cube's volume: $1,1,1,1,1,8,8,8,8,8$ and add them straight: $5 \cdot 1 + 5 \cdot 8 = 45$. Same result, no formula needed.
CCSS standards used (min grade 5)
4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining two volume subtotals with a single addition $5 + 40 = 45$.)5.MD.C.5Relate volume to the operations of multiplication and addition (Using $V = s \times s \times s$ for a cube and adding non-overlapping volumes.)
⭐ This AMC 10 problem only needs Grade 5 “volume of a cube = side × side × side” you already know — $5(1) + 5(8) = 45$.
⭐ This AMC 10 problem only needs Grade 5 “volume of a cube = side × side × side” you already know — $5(1) + 5(8) = 45$.