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Difference between revisions of "2008 AMC 10B Problems/Problem 6"

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==Solution==
 
==Solution==
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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)
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Let <math>\overline{CD} = 1</math>. Then <math>\overline{AB} = 4(\overline{BC}+1)</math>, and <math>\overline{AB}+\overline{BC} = 9\cdot1</math>. From this system of equations we obtain <math>\overline{BC} = 1</math>. Adding <math>\overline{CD}</math> to both sides of the second equation, we obtain <math>\overline{AB}+\overline{BC}+\overline{CD} = 9+1 = 10 = \overline{AD}</math>. Thus, <math>\frac{\overline{BC}}{\overline{AD}} = \frac{1}{10} \implies\text{(C)}</math>
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=5|num-a=7}}
 
{{AMC10 box|year=2008|ab=B|num-b=5|num-a=7}}

Revision as of 11:57, 29 October 2009

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, $\frac{\overline{BC}}{\overline{AD}} = \frac{1}{10} \implies\text{(C)}$

See also

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