Question

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?

$\dfrac{5}{54}$

$\dfrac{1}{9}$

$\dfrac{5}{27}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

Solution:

To minimize visible white area, place one white cube in the center of the large cube, where no faces are visible. Place each of the other five white cubes at the center of a face of the large cube, where each contributes only one visible white square.

The visible white surface area is therefore $5$ square inches. The total surface area of the $3$ -inch cube is $6\cdot3^2=54$ square inches, so the fraction that is white is $\dfrac{5}{54}$ .

Thus, A is the correct answer.

2014 AMC 8 Problem 19

Solution:

Problem 19 in Other Years