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Difference between revisions of "2008 AMC 8 Problems/Problem 6"

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<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math>
 
<math> \textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1 </math>
  
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==Solution==
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Dividing the gray square into four smaller squares, there are <math>6</math> gray tiles and <math>10</math> white tiles, giving a ratio of <math>\boxed{\textbf{(D)}\ 3:5}</math>.
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=5|num-a=7}}
 
{{AMC8 box|year=2008|num-b=5|num-a=7}}

Revision as of 22:02, 24 December 2012

Problem

In the figure, what is the ratio of the area of the gray squares to the area of the white squares? [asy] size((70)); draw((10,0)--(0,10)--(-10,0)--(0,-10)--(10,0)); draw((-2.5,-7.5)--(7.5,2.5)); draw((-5,-5)--(5,5)); draw((-7.5,-2.5)--(2.5,7.5)); draw((-7.5,2.5)--(2.5,-7.5)); draw((-5,5)--(5,-5)); draw((-2.5,7.5)--(7.5,-2.5)); fill((-10,0)--(-7.5,2.5)--(-5,0)--(-7.5,-2.5)--cycle, gray); fill((-5,0)--(0,5)--(5,0)--(0,-5)--cycle, gray); fill((5,0)--(7.5,2.5)--(10,0)--(7.5,-2.5)--cycle, gray); [/asy] $\textbf{(A)}\ 3:10 \qquad\textbf{(B)}\ 3:8 \qquad\textbf{(C)}\ 3:7 \qquad\textbf{(D)}\ 3:5 \qquad\textbf{(E)}\ 1:1$

Solution

Dividing the gray square into four smaller squares, there are $6$ gray tiles and $10$ white tiles, giving a ratio of $\boxed{\textbf{(D)}\ 3:5}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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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