2006 AMC 8 Problems/Problem 18

Revision as of 16:11, 8 March 2019 by G1zq (talk | contribs) (Solution)

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.

See Also

2006 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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