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Difference between revisions of "2009 AMC 12A Problems/Problem 5"
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Let the original cube have edge length <math>a</math>. Then its volume is <math>a^3</math>.
 
Let the original cube have edge length <math>a</math>. Then its volume is <math>a^3</math>.
 
The new box has dimensions <math>a-1</math>, <math>a</math>, and <math>a+1</math>, hence its volume is <math>(a-1)a(a+1) = a^3-a</math>.  
 
The new box has dimensions <math>a-1</math>, <math>a</math>, and <math>a+1</math>, hence its volume is <math>(a-1)a(a+1) = a^3-a</math>.  
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The difference between the two volumes is <math>a</math>. As we are given that the difference is <math>5</math>, we have <math>a=5</math>, and the volume of the original cube was <math>5^3 = \boxed{125}</math>.
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The difference between the two volumes is <math>a</math>. As we are given that the difference is <math>5</math>, we have <math>a=5</math>, and the volume of the original cube was <math>5^3 = 125\Rightarrow\boxed{(D)}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 19:24, 2 November 2012

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

See Also

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
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
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