# 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 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 (Problems • Answer Key • Resources) Preceded byProblem 5 Followed byProblem 7 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 AMC 10 Problems and Solutions
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