2003 AMC 12A Problem 3

Below is the professionally curated solution for Problem 3 of the 2003 AMC 12A, 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 of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:volumepercentage

Difficulty rating: 1130

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

Solution:

The original volume is 15108=1200.15 \cdot 10 \cdot 8 = 1200.

Eight corner cubes are removed, each of volume 33=27,3^3 = 27, totaling 827=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.

Problem 3 in Other Years