AMC 8 · 2013 · #18

Grade 5 geometry-3darithmetic

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Rectilinear Frame / Box Border

volume-rectangular-prismspatial-visualizationmulti-digit-arithmetic area-differenceidentify-subproblems ↑ Prerequisites: volume-rectangular-prismmulti-digit-arithmetic

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

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?

Pick an answer.

(A)

204

(B)

280

(C)

320

(D)

340

(E)

600

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

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 × 10 × 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.

1STEP 1

If the fort were solid, its block count is just the outer volume: 12 × 10 × 5 = 600.

V_outer = 12 × 10 × 5 = 600 ft³ = 600 blocks

2STEP 2

From above, the walls trim 1 ft off each end: inner length 12 − 2 = 10 ft, inner width 10 − 2 = 8 ft.

Inner length = 12 - 2 = 10 ft, inner width = 10 - 2 = 8 ft

3STEP 3

No ceiling, so subtract only the floor from the height: inner height 5 − 1 = 4 ft.

Inner height = 5 - 1 = 4 ft

4STEP 4

The empty inner room is a box too, so the unused blocks = 10 × 8 × 4 = 320.

V_inner = 10 × 8 × 4 = 320 ft³ = 320 blocks of empty space

5STEP 5

Blocks used = solid − empty room = 600 − 320 = 280, which is (B).

600 - 320 = 280 blocks → (B)

Answer

280

The full solid box holds 600 blocks, and a fort with thin walls should use far less than that. The empty room (10 × 8 × 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 ∼ 12 × 10 = 120 floor blocks plus four wall slabs of order ∼ 12 × 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.

💡Key takeaway

This AMC 8 problem only needs Grade 5 volume = length × width × height — count what the fort would be if it were solid, subtract the empty room inside, and you're done!

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