Difference between revisions of "2008 AMC 12A Problems/Problem 8"

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What is the [[volume]] of a [[cube]] whose [[surface area]] is twice that of a cube with volume 1?  
 
What is the [[volume]] of a [[cube]] whose [[surface area]] is twice that of a cube with volume 1?  
  
<math>\textbf{(A)} \sqrt{2} \qquad \textbf{(B)} 2 \qquad \textbf{(C)} 2\sqrt{2} \qquad \textbf{(D)} 4 \qquad \textbf{(E)} 8 </math>
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<math>\mathrm{(A)}\ \sqrt{2}\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 2\sqrt{2}\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 8</math>
  
 
==Solution==
 
==Solution==
 
A cube with volume <math>1</math> has a side of length <math>\sqrt[3]{1}=1</math> and thus a surface area of <math>6 \cdot 1^2=6</math>.  
 
A cube with volume <math>1</math> has a side of length <math>\sqrt[3]{1}=1</math> and thus a surface area of <math>6 \cdot 1^2=6</math>.  
  
A cube whose surface area is <math>6\cdot2=12</math> has a side of length <math>\sqrt{\frac{12}{6}}=\sqrt{2}</math> and a volume of <math>(\sqrt{2})^3 = 2\sqrt{2} \Rightarrow C</math>.  
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A cube whose surface area is <math>6\cdot2=12</math> has a side of length <math>\sqrt{\frac{12}{6}}=\sqrt{2}</math> and a volume of <math>(\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}</math>.  
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2008|ab=A|num-b=7|num-a=9}}

Revision as of 00:34, 26 April 2008

Problem

What is the volume of a cube whose surface area is twice that of a cube with volume 1?

$\mathrm{(A)}\ \sqrt{2}\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 2\sqrt{2}\qquad\mathrm{(D)}\ 4\qquad\mathrm{(E)}\ 8$

Solution

A cube with volume $1$ has a side of length $\sqrt[3]{1}=1$ and thus a surface area of $6 \cdot 1^2=6$.

A cube whose surface area is $6\cdot2=12$ has a side of length $\sqrt{\frac{12}{6}}=\sqrt{2}$ and a volume of $(\sqrt{2})^3=2\sqrt{2}\Rightarrow\mathrm{(C)}$.

See Also

2008 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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