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2003 AMC 10A Problem 3

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2003 AMC 10A Problem 3AMC 10 Step by Step
Name:                             Date: September 13, 2026
Solid Geometry · difficulty 1 of 5 · about 1 min · position P1-10

A solid box is 1515 cm by 1010 cm by 88 cm. A new solid is formed by removing a cube 33 cm on a side from each corner of this box. What percent of the original volume is removed?

  1. A)

    4.5%4.5\%

  2. B)

    9%9\%

  3. C)

    12%12\%

  4. D)

    18%18\%

  5. E)

    24%24\%

Answer

D

Key insight

A box has 8 corners, so 8 cubes of volume 27 are removed from a box of volume 1200.

Solution

Original volume: 15⋅10⋅8=120015 \cdot 10 \cdot 8 = 1200 cubic cm.

Each removed cube has volume 33=273^3 = 27. A rectangular box has 88 corners, and since every edge is at least 8>68 > 6 cm the eight cubes do not overlap, so the removed volume is 8⋅27=2168 \cdot 27 = 216.

The fraction removed is 2161200=18100=18%\dfrac{216}{1200} = \dfrac{18}{100} = 18\%.

The answer is (D) 18%\boxed{\textbf{(D)}\ 18\%}.

Why this works

Percent removed is simply (volume removed)/(original volume). The only geometric facts needed are the volume formulas and the count of corners of a box, which is 88, not 44; picturing the 3-D object rather than a flat rectangle avoids the miscount.

The trap

Removing only 4 cubes (thinking of a rectangle's corners) and answering 9%, choice (B).

Common mistakes

  • Removing only 4 cubes (thinking of a rectangle's corners) and answering 9%, choice (B).
  • Using 32=93^2 = 9 or 3⋅3⋅3⋅83 \cdot 3 \cdot 3 \cdot 8 incorrectly, or dividing by the new volume instead of the original.

Techniques

Set up the equation/formula and compute; no special trick needed