# Difference between revisions of "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, $\frac{\overline{BC}}{\overline{AD}} = \frac{1}{10} \implies\text{(C)}$