AMC 8 · 2013 · #18
Grade 5 geometry-3darithmeticProblem
Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?
Pick an answer.
AMC 8 2013 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: Isabella stacks $1$-foot cubical blocks to build a rectangular fort whose outside measures $12$ ft long, $10$ ft wide, and $5$ ft high. The floor and the four side walls are each $1$ ft thick, but the top is open. How many blocks did she use?
Givens: Outer dimensions: $12 \text{ ft} \times 10 \text{ ft} \times 5 \text{ ft}$; Each block is a $1 \times 1 \times 1$ ft cube (so $1$ block $= 1 \text{ ft}^3$); Floor thickness: $1$ ft; each of the $4$ side walls: $1$ ft thick; Top is open (no ceiling); Answer choices: (A) $204$, (B) $280$, (C) $320$, (D) $340$, (E) $600$
Unknowns: The total number of $1$-ft cubical blocks in the fort
Understand
Restated: Isabella stacks $1$-foot cubical blocks to build a rectangular fort whose outside measures $12$ ft long, $10$ ft wide, and $5$ ft high. The floor and the four side walls are each $1$ ft thick, but the top is open. How many blocks did she use?
Givens: Outer dimensions: $12 \text{ ft} \times 10 \text{ ft} \times 5 \text{ ft}$; Each block is a $1 \times 1 \times 1$ ft cube (so $1$ block $= 1 \text{ ft}^3$); Floor thickness: $1$ ft; each of the $4$ side walls: $1$ ft thick; Top is open (no ceiling); Answer choices: (A) $204$, (B) $280$, (C) $320$, (D) $340$, (E) $600$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #17 Visualize Spatial Relationships, #7 Identify Subproblems
Counting wall blocks face-by-face is messy because the corner columns belong to two walls at once and the floor blocks meet the walls along the bottom edge — easy to double-count. Tool #16 (Complement) flips the question: instead of "how many blocks are filled?" ask "how many blocks would fill the whole $12 \times 10 \times 5$ box?" and "how many blocks fit in the empty room inside?" Subtract the second from the first. Tool #17 (Visualize Spatial Relationships) is needed to see that the inner room loses $1$ ft on each of the two long sides, $1$ ft on each of the two short sides, and $1$ ft from the bottom only (no ceiling). Tool #7 (Identify Subproblems) then keeps the three computations — outer volume, inner volume, subtraction — neatly separated.
Execute — Answer: B
5.MD.C.5 Step 1 - Count the blocks the fort would contain if it were solid.
- Each $1$-ft cube fills exactly $1 \text{ ft}^3$, so the solid count equals the outer volume: length $\times$ width $\times$ height.
💡 Filling a rectangular box with unit cubes and multiplying the three side lengths is the Grade 5 volume-as-multiplication idea.
5.MD.C.3 Step 2 - Visualize the inner empty room.
- Looking down from above, the floor takes $1$ ft on the left and $1$ ft on the right, shrinking the $12$-ft length to $12 - 2 = 10$ ft.
- Likewise the two long walls eat $1$ ft each off the $10$-ft width, leaving $10 - 2 = 8$ ft.
💡 Picturing the floor plan and trimming $1$ ft off each wall is spatial reasoning about a unit-cube layout — Grade 5 volume thinking.
5.MD.C.3 Step 3 - The inner height needs more care.
- The floor is $1$ ft thick, but the fort has no ceiling — so the empty room runs all the way to the top.
- Subtract only the floor: $5 - 1 = 4$ ft.
💡 Noticing the asymmetry (floor only, no ceiling) keeps you from over-subtracting — this is the careful spatial step in any "hollow box" problem.
5.MD.C.5 Step 4 - Find how many blocks would fit in the inner empty room — that is, the blocks Isabella did NOT use.
- Multiply the three inner dimensions.
💡 The inner room is itself a rectangular prism, so the same length $\times$ width $\times$ height rule applies.
4.NBT.B.4 Step 5 Apply Tool #16: the blocks Isabella actually used $=$ blocks in the full solid $-$ blocks in the empty inner room.
💡 Counting the complement (empty cubes) and subtracting is faster and safer than tallying corner-sharing walls one by one.
5.MD.C.5 Count the blocks the fort would contain if it were solid. Each $1$-ft cube fills 5.MD.C.3 Visualize the inner empty room. Looking down from above, the floor takes $1$ ft 5.MD.C.3 The inner height needs more care. The floor is $1$ ft thick, but the fort has no 5.MD.C.5 Find how many blocks would fit in the inner empty room — that is, the blocks Isa 4.NBT.B.4 Apply Tool #16: the blocks Isabella actually used $=$ blocks in the full solid $ Review
Reasonableness: The full solid box holds $600$ blocks, and a fort with thin walls should use far less than that. The empty room ($10 \times 8 \times 4 = 320$) is bigger than the walls themselves, so the wall+floor count should be less than half of $600$ — and $280$ is just under half. Also, $280$ blocks for a fort whose outside surface is on the order of $\sim 12 \times 10 = 120$ floor blocks plus four wall slabs of order $\sim 12 \times 5 = 60$ each fits the ballpark. Choices (C) $320$ and (E) $600$ are the two complement values (empty room and full solid) — common trap answers — and (B) $280$ is the only choice equal to $600 - 320$.
Alternative: Tool #7 (Identify Subproblems) without complement: count layer by layer. The bottom layer is a solid floor of $12 \times 10 = 120$ blocks. Each of the $4$ layers above is a hollow frame: outer $12 \times 10 = 120$ minus inner $10 \times 8 = 80$ gives $40$ wall blocks per layer, so $4 \times 40 = 160$ wall blocks above the floor. Total $= 120 + 160 = 280$ blocks — same answer (B). This confirms the complement method without any double-counting.
CCSS standards used (min grade 5)
4.NBT.B.4Fluently add and subtract multi-digit whole numbers using the standard algorithm (Performing the final subtraction $600 - 320 = 280$ to get the block count.)5.MD.C.3Recognize volume as an attribute of solid figures and understand concepts of volume measurement (Treating each $1$-ft cubical block as one unit of volume and visualizing the fort and its inner room as collections of those unit cubes.)5.MD.C.5Relate volume to the operations of multiplication and addition (Computing the outer volume $12 \times 10 \times 5 = 600$ and the inner room volume $10 \times 8 \times 4 = 320$ as length $\times$ width $\times$ height.)
⭐ This AMC 8 problem only needs Grade 5 volume = length $\times$ width $\times$ height — count what the fort would be if it were solid, subtract the empty room inside, and you're done!
⭐ This AMC 8 problem only needs Grade 5 volume = length $\times$ width $\times$ height — count what the fort would be if it were solid, subtract the empty room inside, and you're done!
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