2009 AMC 12A Problem 5

Attempt Problem 5 of the 2009 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 2009 AMC 12A solutions, or check the answer key.

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

One dimension of a cube is increased by

1,

another is decreased by

1,

and the third is left unchanged. The volume of the new rectangular solid is

5

less than that of the cube. What was the volume of the cube?

8

27

64

125

216

Let the cube have side length

x

and write the new volume as a product

The new solid has volume

x(x+1)(x-1) = x^3 - x

Let the cube have side length

x.

The new solid has dimensions

x + 1,

x - 1,

and

x,

so its volume is

x(x+1)(x-1) = x^3 - x.

Setting this equal to

x^3 - 5

gives

x^3 - x = x^3 - 5,

x = 5.

The cube’s volume is

5^3 = 125.

Thus, the correct answer is D.

Problem 5 in other years