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 $\boxed{\textbf{(A) }\frac{5}{54}}$ .

Video Solution

https://youtu.be/dqGi7YtApwo ~savannahsolver

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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2014 AMC 8 Problems/Problem 19

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Solution

Video Solution

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