2014 AMC 8 Problems/Problem 19

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?

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

Solution

For the least possible surface area that is white, we should have 1 cube in the center, and the other 5 with only 1 face exposed. This gives 5 square inches of white surface area. Since the cube has a surface area of 54 square inches, our answer is $\boxed{\textbf{(A) }\frac{5}{54}}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS