2009 AMC 12A Problem 5

Below is the professionally curated solution for Problem 5 of the 2009 AMC 12A, 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.

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Concepts:volumedifference of squares

Difficulty rating: 1100

5.

One dimension of a cube is increased by 1,1, another is decreased by 1,1, and the third is left unchanged. The volume of the new rectangular solid is 55 less than that of the cube. What was the volume of the cube?

88

2727

6464

125125

216216

Solution:

Let the cube have side length x.x. The new solid has dimensions x+1,x + 1, x1,x - 1, and x,x, so its volume is x(x+1)(x1)=x3x.x(x+1)(x-1) = x^3 - x.

Setting this equal to x35x^3 - 5 gives x3x=x35,x^3 - x = x^3 - 5, so x=5.x = 5.

The cube's volume is 53=125.5^3 = 125.

Thus, the correct answer is D.

Problem 5 in Other Years