2008 AMC 10B Problems/Problem 6

Problem

Points $B$ and $C$ lie on $\overline{AD}$ . The length of $\overline{AB}$ is $4$ times the length of $\overline{BD}$ , and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}$ . The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$ ?

$\textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qquad\textbf{(E)}\ \frac{1}{5}$

Solution

Let $\overline{CD} = 1$ . Then $\overline{AB} = 4(\overline{BC}+1)$ , and $\overline{AB}+\overline{BC} = 9\cdot1$ . From this system of equations we obtain $\overline{BC} = 1$ . Adding $\overline{CD}$ to both sides of the second equation, we obtain $\overline{AB}+\overline{BC}+\overline{CD} = 9+1 = 10 = \overline{AD}$ . Thus, $\frac{\overline{BC}}{\overline{AD}} = \frac{1}{10} \implies\text{(C)}$

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2008 AMC 10B Problems/Problem 6

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