Difference between revisions of "2008 AMC 8 Problems/Problem 25"

m (Solution)
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==Solution==
 
==Solution==
<cmath>\begin{tabular}{c|cc}
+
<cmath>\begin{array}{c|cc}
 
\text{circle #} & radius & area \\ \hline
 
\text{circle #} & radius & area \\ \hline
 
1 & 2 & 4\pi \\
 
1 & 2 & 4\pi \\
Line 30: Line 30:
 
5 & 10 & 100\pi \\
 
5 & 10 & 100\pi \\
 
6 & 12 & 144\pi
 
6 & 12 & 144\pi
\end{tabular}</cmath>
+
\end{array}</cmath>
  
 
The entire circle's area is <math>144\pi</math>. The area of the black regions is <math>(100-64)\pi + (36-16)\pi + 4\pi = 60\pi</math>. The percentage of the design that is black is <math>\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}</math>.
 
The entire circle's area is <math>144\pi</math>. The area of the black regions is <math>(100-64)\pi + (36-16)\pi + 4\pi = 60\pi</math>. The percentage of the design that is black is <math>\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}</math>.

Revision as of 20:06, 10 March 2015

Problem

Margie's winning art design is shown. The smallest circle has radius 2 inches, with each successive circle's radius increasing by 2 inches. Approximately what percent of the design is black?

[asy] real d=320; pair O=origin; pair P=O+8*dir(d); pair A0 = origin; pair A1 = O+1*dir(d); pair A2 = O+2*dir(d); pair A3 = O+3*dir(d); pair A4 = O+4*dir(d); pair A5 = O+5*dir(d); filldraw(Circle(A0, 6), white, black); filldraw(circle(A1, 5), black, black); filldraw(circle(A2, 4), white, black); filldraw(circle(A3, 3), black, black); filldraw(circle(A4, 2), white, black); filldraw(circle(A5, 1), black, black); [/asy] $\textbf{(A)}\ 42\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 48\qquad$

Solution

\[\begin{array}{c|cc}
\text{circle #} & radius & area \\ \hline
1 & 2 & 4\pi \\
2 & 4 & 16\pi \\
3 & 6 & 36\pi \\
4 & 8 & 64\pi \\
5 & 10 & 100\pi \\
6 & 12 & 144\pi
\end{array}\] (Error compiling LaTeX. ! You can't use `macro parameter character #' in restricted horizontal mode.)

The entire circle's area is $144\pi$. The area of the black regions is $(100-64)\pi + (36-16)\pi + 4\pi = 60\pi$. The percentage of the design that is black is $\frac{60\pi}{144\pi} = \frac{5}{12} \approx \boxed{\textbf{(A)}\ 42}$.

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Problem
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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