# 2003 AMC 12A Problems/Problem 3

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The following problem is from both the 2003 AMC 12A #3 and 2003 AMC 10A #3, so both problems redirect to this page.

## 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? $\mathrm{(A) \ } 4.5\%\qquad \mathrm{(B) \ } 9\%\qquad \mathrm{(C) \ } 12\%\qquad \mathrm{(D) \ } 18\%\qquad \mathrm{(E) \ } 24\%$

## Solution

The volume of the original box is $15\cdot10\cdot8=1200$

The volume of each cube that is removed is $3\cdot3\cdot3=27$

Since there are $8$ corners on the box, $8$ cubes are removed.

So the total volume removed is $8\cdot27=216$.

Therefore, the desired percentage is $\frac{216}{1200}\cdot100 = \boxed{\mathrm{(D)}\ 18\%}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 