2009 AMC 12A Problems/Problem 5
- The following problem is from both the 2009 AMC 12A #5 and 2009 AMC 10A #11, so both problems redirect to this page.
Problem
One dimension of a cube is increased by , another is decreased by , and the third is left unchanged. The volume of the new rectangular solid is less than that of the cube. What was the volume of the cube?
Solution
Let the original cube have edge length . Then its volume is . The new box has dimensions , , and , hence its volume is . The difference between the two volumes is . As we are given that the difference is , we have , and the volume of the original cube was .
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 4 |
Followed by Problem 6 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.