2006 AMC 8 Problems/Problem 18

Problem

A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?

$\textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27}$

Solution

The surface area of the cube is $6(3)(3)=54$ . Each of the eight black cubes has 3 faces on the outside, making $3(8)=24$ black faces. Therefore there are $54-24=30$ white faces. To find the ratio, we evaluate $\frac{30}{54}= \boxed{\textbf{(D)}\ \frac{5}{9}}$ .

Solution 2

We can notice that each face is the same, so each face is $\boxed{\textbf{(D)}\ \frac{5}{9}}$ white.

Video Solution by WhyMath

https://youtu.be/nDkpGeJcTd8

2006 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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.

Page

Toolbox

Search

2006 AMC 8 Problems/Problem 18

Contents

Problem

Solution

Solution 2

Video Solution by WhyMath

See Also