2008 AMC 12B Problems/Problem 8

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

Since $\overline{AB}=4\overline{BD}$ and $\overline{AB}+\overline{BD}=\overline{AD}$ , $\overline{AB}=\frac{4}{5}\overline{AD}$ .

Since $\overline{AC}=9\overline{CD}$ and $\overline{AC}+\overline{CD}=\overline{AD}$ , $\overline{AC}=\frac{9}{10}\overline{AD}$ .

Thus, $\overline{BC}=\overline{AC}-\overline{AB}=\left(\frac{9}{10}-\frac{4}{5}\right)\overline{AD} = \frac {1}{10}\overline{AD} \Rightarrow C$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 8

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