2003 AMC 12A Problem 3

Attempt Problem 3 of the 2003 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

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?
4.54.5
99
1212
1818
2424
Answer: D
Concepts:volumepercentage
Small hint:
The box has 88 corners
Big hint:
Each removed cube has volume 333^3
Solution:
The original volume is 15⋅10⋅8=1200.15 \cdot 10 \cdot 8 = 1200.
Eight corner cubes are removed, each of volume 33=27,3^3 = 27, totaling 8⋅27=216.8 \cdot 27 = 216.
The fraction removed is 2161200=0.18,\dfrac{216}{1200} = 0.18, which is 18%.18\%.
Thus, the correct answer is D.
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