Difference between revisions of "2009 AMC 12A Problems/Problem 5"

Latest revision as of 17:53, 16 February 2014

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 $1$ , 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?

$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 27 \qquad \textbf{(C)}\ 64 \qquad \textbf{(D)}\ 125 \qquad \textbf{(E)}\ 216$

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 $5^3 = 125\Rightarrow\boxed{\text{(D)}}$ .

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.

@@ Line 14: / Line 14: @@
 == See Also ==
+{{AMC10 box|year=2009|ab=A|num-b=10|num-a=12}}
 {{AMC12 box|year=2009|ab=A|num-b=4|num-a=6}}
-{{AMC10 box|year=2009|ab=A|num-b=10|num-a=12}}
 [[Category:Introductory Geometry Problems]]
 [[Category:3D Geometry Problems]]
 {{MAA Notice}}

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Difference between revisions of "2009 AMC 12A Problems/Problem 5"

Latest revision as of 17:53, 16 February 2014

Problem

Solution

See Also