AMC 10 · 2003 · #3

Grade 6 geometry-3d

Archetype Solid Volume by Decomposition └ Sub-archetype Fraction or Difference of a Bounding Solid

volume-rectangular-prismpercentagespatial-visualization identify-subproblems ↑ Prerequisites: volume-rectangular-prism

📏 Medium solution 💡 2 insights

Problem

A solid box is $15$ cm by $10$ cm by $8$ cm. A new solid is formed by removing a cube $3$ cm on a side from each corner of this box. What percent of the original volume is removed?

Pick an answer.

(A)

$4.5\%$

(B)

$9\%$

(C)

$12\%$

(D)

$18\%$

(E)

$24\%$

AMC 10 2003 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.

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Toolkit + CCSS Solution

Understand

Restated: A solid box measures $15$ cm by $10$ cm by $8$ cm. A $3$ cm cube is cut away from each corner of the box. Find what percent of the box's original volume is cut away.

Givens: The box is a rectangular solid: $15$ cm long, $10$ cm wide, $8$ cm tall; A cube $3$ cm on each side is removed from every corner; Answer choices: (A) $4.5\%$, (B) $9\%$, (C) $12\%$, (D) $18\%$, (E) $24\%$

Unknowns: The percent of the original volume that is removed

Understand

Restated: A solid box measures $15$ cm by $10$ cm by $8$ cm. A $3$ cm cube is cut away from each corner of the box. Find what percent of the box's original volume is cut away.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #17 Visualize Spatial Relationships, #8 Analyze the Units

The word "percent" hides three smaller jobs, so Tool #7 (Identify Subproblems) breaks it into: find the whole box's volume, find the volume taken away, then turn the ratio into a percent. The one place a solver can slip is counting the corners, so Tool #17 (Visualize Spatial Relationships) pins down that a box has exactly $8$ corners — not $4$ or $6$. Tool #8 (Analyze the Units) keeps every volume in cm$^3$ so the final ratio is a clean number-over-number percent.

Execute — Answer: D

#7 Identify Subproblems 5.MD.C.5 Step 1

Volume of the whole box

The box is a rectangular solid, so its volume is length times width times height.
Multiply $15 \times 10 \times 8$.
Do it in two steps: $15 \times 10 = 150$, then $150 \times 8 = 1200$.
The box holds $1200$ cubic centimeters.

$$V_{\text{box}} = 15 \times 10 \times 8 = 1200 \text{ cm}^3$$

💡 The amount of space in a box is just its three side lengths multiplied together.

#17 Visualize Spatial Relationships 5.MD.C.5 Step 2

Count the corners and one cube

Picture the box like a room.
A rectangular box has $8$ corners — four on the top face and four on the bottom face.
From each corner you remove a cube $3$ cm on a side, and the volume of one such cube is $3 \times 3 \times 3 = 27$ cubic centimeters.

$$8 \text{ corners}, \qquad V_{\text{cube}} = 3^3 = 27 \text{ cm}^3$$

💡 A box has one corner for each of the four bottom edges and four top edges — eight in all.

#7 Identify Subproblems 5.NBT.B.5 Step 3

Total volume removed

All $8$ cut-out cubes are the same size, so the total removed is $8$ copies of $27$.
Multiply $8 \times 27 = 216$.
So $216$ cubic centimeters are taken away in all.

$$V_{\text{removed}} = 8 \times 27 = 216 \text{ cm}^3$$

💡 Eight identical pieces removed means eight times the size of one piece.

#8 Analyze the Units 6.RP.A.3 Step 4

Turn the ratio into a percent

The percent removed is the removed volume over the original volume, written out of $100$.
That is $\dfrac{216}{1200}$.
Divide: $216 \div 1200 = 0.18$, which is $18\%$.
So the amount cut away is $18\%$ of the original box, choice (D).

$$\dfrac{V_{\text{removed}}}{V_{\text{box}}} = \dfrac{216}{1200} = 0.18 = 18\% \;\Rightarrow\; \textbf{(D)}$$

💡 "What percent" means the part divided by the whole, then read as a number out of $100$.

[1] #7 5.MD.C.5 The box is a rectangular solid, so its volume is length times width times height

[2] #17 5.MD.C.5 Picture the box like a room. A rectangular box has $8$ corners — four on the top

[3] #7 5.NBT.B.5 All $8$ cut-out cubes are the same size, so the total removed is $8$ copies of $

[4] #8 6.RP.A.3 The percent removed is the removed volume over the original volume, written out

Review

Reasonableness: The removed volume $216$ cm$^3$ is a small slice of the $1200$ cm$^3$ box, so the answer should be well under a quarter — and $18\%$ is. A quick sanity check: $\tfrac{216}{1200}$ is a bit under $\tfrac{240}{1200} = \tfrac{1}{5} = 20\%$, so $18\%$ sits right where it should. If a solver had wrongly used $6$ corners they would get $\tfrac{162}{1200} = 13.5\%$ (no such choice), and using $4$ corners gives $9\%$ (choice B) — both are traps for miscounting corners, which confirms getting the count right is the whole game.

Alternative: Work with percents per cube instead of adding first. One cube is $\dfrac{27}{1200} = 2.25\%$ of the box. Eight of them give $8 \times 2.25\% = 18\%$, the same answer (D).

CCSS standards used (min grade 6)

5.MD.C.5 Relate volume to the operations of multiplication and addition (Finding the box volume as $15 \times 10 \times 8$ and each corner cube as $3^3$.)
5.NBT.B.5 Fluently multiply multi-digit whole numbers (Multiplying $8 \times 27$ to get the total volume removed.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Writing the removed-to-original ratio $\tfrac{216}{1200}$ as a percent, $18\%$.)

⭐ "What percent is removed" is just the volume taken away divided by the whole volume — and remember a box has eight corners, not four.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Volume of the whole box

Count the corners and one cube

Total volume removed

Turn the ratio into a percent

Review

CCSS standards used (min grade 6)

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