AMC12 2009 A

AMC12 2009 A · Q5

AMC12 2009 A · Q5. It mainly tests Quadratic equations.

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?

一个立方体的一个边长增加 $1$，另一个边长减少 $1$，第三个边长保持不变。新长方体的体积比原立方体小 $5$。原立方体的体积是多少？

(A) 8 8

(B) 27 27

(D) 125 125

(E) 216 216

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let the original cube have edge length $a$. Then its volume is $a^3$. The new box has dimensions $a-1$, $a$, and $a+1$, hence its volume is $(a-1)a(a+1) = a^3-a$. The difference between the two volumes is $a$. As we are given that the difference is $5$, we have $a=5$, and the volume of the original cube was $5^3 = 125\Rightarrow\boxed{\text{(D)}}$.

设原立方体的边长为 $a$，则其体积为 $a^3$。新长方体的三条边分别为 $a-1$、$a$、$a+1$，因此体积为 $(a-1)a(a+1) = a^3-a$。两者体积之差为 $a$。已知体积之差为 $5$，所以 $a=5$，原立方体体积为 $5^3 = 125\Rightarrow\boxed{\text{(D)}}$。

Topics

AL.005 Quadratic equations