Difference between revisions of "2012 AMC 8 Problems/Problem 21"

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==Solution==
 
==Solution==
The surface area of this cube is <math> 600 </math> square feet. Maria's paint can cover a total of <math> 300 </math> square feet. Since all the faces of the cube have an identical design, Maria's paint covers half of every face of the cube. The surface area of one of the faces of the cube is <math> 100 </math> square feet. Then, her paint will cover <math> 50 </math> square feet of every face. Thus, the area of one of the white squares will be <math> 100 - 50 = \boxed{\textbf{(D)}\ 50} </math> square feet.
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If Marla evenly distributes her <math>300</math> square feet of paint between the 6 faces, each face will get <math>300\div6 = 50</math> square feet of paint. The surface area of one of the faces of the cube is <math>10^2 = 100 </math> square feet. Therefore, there will be <math>100-50 = \boxed{\textbf{(D)}\ 50} </math> square feet of white on each side.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2012|num-b=20|num-a=22}}
 
{{AMC8 box|year=2012|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:54, 25 November 2013

Problem

Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?

$\textbf{(A)}\hspace{.05in}5\sqrt2\qquad\textbf{(B)}\hspace{.05in}10\qquad\textbf{(C)}\hspace{.05in}10\sqrt2\qquad\textbf{(D)}\hspace{.05in}50\qquad\textbf{(E)}\hspace{.05in}50\sqrt2$

Solution

If Marla evenly distributes her $300$ square feet of paint between the 6 faces, each face will get $300\div6 = 50$ square feet of paint. The surface area of one of the faces of the cube is $10^2 = 100$ square feet. Therefore, there will be $100-50 = \boxed{\textbf{(D)}\ 50}$ square feet of white on each side.

See Also

2012 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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