Difference between revisions of "2014 AMC 8 Problems/Problem 19"
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| − | 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? | + | 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 19:22, 26 November 2014
A cube with 
-inch edges is to be constructed from 
smaller cubes with 
-inch edges. Twenty-one of the cubes are colored red and 
are colored white. If the -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
