2014 AMC 8 Problems/Problem 19: Difference between revisions
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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 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? | ||
<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 \ | <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> | ||
==See Also== | |||
{{AMC8 box|year=2014|num-b=18|num-a=20}} | |||
{{MAA Notice}} | |||
Revision as of 19:22, 26 November 2014
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?
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 | ||
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