Difference between revisions of "2014 AMC 8 Problems/Problem 19"

Line 1: Line 1:
 +
==Problem==
 
A cube with <math>3</math>-inch edges is to be constructed from <math>27</math> smaller cubes with <math>1</math>-inch edges. Twenty-one of the cubes are colored red and <math>6</math> are colored white. If the <math>3</math>-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
 
A cube with <math>3</math>-inch edges is to be constructed from <math>27</math> smaller cubes with <math>1</math>-inch edges. Twenty-one of the cubes are colored red and <math>6</math> are colored white. If the <math>3</math>-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
  
 
<math> \textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3} </math>
 
<math> \textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3} </math>

Revision as of 20:40, 26 November 2014

Problem

A cube with $3$-inch edges is to be constructed from $27$ smaller cubes with $1$-inch edges. Twenty-one of the cubes are colored red and $6$ are colored white. If the $3$-inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?

$\textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3}$