Difference between revisions of "2006 AMC 8 Problems/Problem 18"
(→Solution) |
|||
| Line 5: | Line 5: | ||
==Solution== | ==Solution== | ||
| − | The surface area of the cube is <math>6(3)(3)=54</math>. Each of the eight black cubes has 3 faces on the outside, making <math>3(8)=24</math> black faces. Therefore there are <math>54-24=30</math> white faces. To find the | + | The surface area of the cube is <math>6(3)(3)=54</math>. Each of the eight black cubes has 3 faces on the outside, making <math>3(8)=24</math> black faces. Therefore there are <math>54-24=30</math> white faces. To find the ratio, we evaluate <math>\frac{30}{54}= \boxed{\textbf{(D)}\ \frac{5}{9}}</math>. |
| + | |||
| + | |||
| + | Solution 2 | ||
| + | We can notice that each face is the same, and that each face is 5/9 white. | ||
==See Also== | ==See Also== | ||
Revision as of 19:22, 10 November 2018
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?
Solution
The surface area of the cube is 
. Each of the eight black cubes has 3 faces on the outside, making 
black faces. Therefore there are 
white faces. To find the ratio, we evaluate 
.
Solution 2
We can notice that each face is the same, and that each face is 5/9 white.
See Also
| 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. 
