Difference between revisions of "2008 AMC 10B Problems/Problem 6"
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==Problem== | ==Problem== | ||
| − | Points B and C lie on AD. The length of AB is 4 times the length of BD, and the length of AC is 9 times the length of CD. The length of BC is what fraction of the length of AD? | + | Points <math>B</math> and <math>C</math> lie on <math>\overline{AD}</math>. The length of <math>\overline{AB}</math> is <math>4</math> times the length of <math>\overline{BD}</math>, and the length of <math>\overline{AC}</math> is <math>9</math> times the length of <math>\overline{CD}</math>. The length of <math>\overline{BC}</math> is what fraction of the length of <math>\overline{AD}</math>? |
| − | A) 1 | + | <math> \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} </math> |
==Solution== | ==Solution== | ||
Revision as of 12:51, 29 October 2009
Problem
Points 
and 
lie on 
. The length of 
is 
times the length of 
, and the length of 
is 
times the length of 
. The length of 
is what fraction of the length of ?
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 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 | ||
