2003 AMC 10A Problem 3

Percent removed means (volume cut out) divided by (original volume), so find both volumes.

1. The original box has volume $15 \times 10 \times 8 = 1200$ cubic cm.
2. Each small cube has volume $3 \times 3 \times 3 = 27$ cubic cm.
3. Count the corners of a box: $4$ on the top face and $4$ on the bottom face, so $8$ corners in all.
4. The cubes do not bump into each other: every edge of the box is at least $8$ cm long, and two cubes on the same edge use only $3 + 3 = 6$ cm of it.
5. Volume removed: $8 \times 27 = 216$ cubic cm.
6. Percent removed:
$$\frac{216}{1200} = \frac{18}{100} = 18\%.$$
(Divide top and bottom by $12$: $216 \div 12 = 18$ and $1200 \div 12 = 100$.)

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

A percent of a volume is just a fraction of it, so you only need two volumes. The one fact that matters is that a box has $8$ corners. When a problem is about a solid, picture the solid, not a flat drawing of it.