Difference between revisions of "2014 AMC 8 Problems/Problem 19"
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<math> \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} </math> | <math> \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} </math> | ||
==Solution== | ==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 <math>\boxed{\textbf{(A) }\frac{5}{54}}</math>. | + | 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 <math>\boxed{\textbf{(A) }\frac{5}{54}}</math>. |
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/dqGi7YtApwo ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=18|num-a=20}} | {{AMC8 box|year=2014|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 21:34, 24 February 2024
Contents
Problem
A cube with 
-inch edges is to be constructed from 
smaller cubes with 
-inch edges. Twenty-one of the cubes are colored red and 
are colored white. If the -inch cube is constructed to have the smallest possible white surface area showing, what fraction of the surface area is white?
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 
.
Video Solution
https://youtu.be/dqGi7YtApwo ~savannahsolver
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 18 |
Followed by Problem 20 | |
| 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. 
