2008 AMC 10B Problems/Problem 6

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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 CD = 1. Then AB = 4(BC+1), and AB+BC = 9*1. From this system of equations we obtain BC = 1. Adding CD to both sides of the second equation, we obtain AB+BC+CD = 9+1 = 10 = AD. BC/AD = 1/10 (C)

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AMC 10 Problems and Solutions
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