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, 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>\textbf{(A) }\frac{5}{54}</math>.
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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>\textbf{(A) }\frac{5}{54}</math>.
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==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}}

Revision as of 15:30, 1 October 2017

Problem

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?

$\textbf{(A) }\frac{5}{54}\qquad\textbf{(B) }\frac{1}{9}\qquad\textbf{(C) }\frac{5}{27}\qquad\textbf{(D) }\frac{2}{9}\qquad\textbf{(E) }\frac{1}{3}$

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 $\textbf{(A) }\frac{5}{54}$.

See Also

2014 AMC 8 (ProblemsAnswer KeyResources)
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

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