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Difference between revisions of "2005 AMC 10A Problems/Problem 11"
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A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?
 
A wooden cube <math>n</math> units on a side is painted red on all six faces and then cut into <math>n^3</math> unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is <math>n</math>?
  
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<math> \mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 4\qquad \mathrm{(C) \ } 5\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 7 </math>
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<math> \textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7 </math>
  
 
==Solution==
 
==Solution==
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Since there are <math>n^2</math> little [[face]]s on each face of the big wooden [[cube]], there are <math>6n^2</math> little faces painted red.  
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Since there are <math>n^2</math> little [[face]]s on each face of the big wooden [[cube (geometry) | cube]], there are <math>6n^2</math> little faces painted red.  
  
 
Since each unit cube has <math>6</math> faces, there are <math>6n^3</math> little faces total.  
 
Since each unit cube has <math>6</math> faces, there are <math>6n^3</math> little faces total.  
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<math>\frac{1}{n}=\frac{1}{4}</math>
 
<math>\frac{1}{n}=\frac{1}{4}</math>
  
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<math>n=4\Longrightarrow \mathrm{(B)}</math>
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<math>n=\boxed{\textbf{(B) }4}</math>
  
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==See Also==
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==See also==
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*[[2005 AMC 10A Problems]]
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{{AMC10 box|year=2005|ab=A|num-b=10|num-a=12}}
  
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*[[2005 AMC 10A Problems/Problem 10|Previous Problem]]
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{{MAA Notice}}
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*[[2005 AMC 10A Problems/Problem 12|Next Problem]]
 
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[[Category:Introductory Geometry Problems]]
 

Latest revision as of 11:40, 13 December 2021

Problem

A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^3$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?

$\textbf{(A) } 3\qquad \textbf{(B) } 4\qquad \textbf{(C) } 5\qquad \textbf{(D) } 6\qquad \textbf{(E) } 7$

Solution

Since there are $n^2$ little faces on each face of the big wooden cube, there are $6n^2$ little faces painted red.

Since each unit cube has $6$ faces, there are $6n^3$ little faces total.

Since one-fourth of the little faces are painted red,

$\frac{6n^2}{6n^3}=\frac{1}{4}$

$\frac{1}{n}=\frac{1}{4}$

$n=\boxed{\textbf{(B) }4}$

See also

2005 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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