Difference between revisions of "2008 AMC 10B Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | {{ | + | 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>? |
+ | |||
+ | <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== | ||
− | {{ | + | Let <math>CD = 1</math>. Then <math>AB = 4(BC + 1)</math> and <math>AB + BC = 9\cdot1</math>. From this system of equations we obtain <math>BC = 1</math>. Adding <math>CD</math> to both sides of the second equation, we obtain <math>AD = AB + BC + CD = 9 + 1 = 10</math>. Thus, <math>\frac{BC}{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}} | ||
+ | {{MAA Notice}} |
Revision as of 12:26, 4 July 2013
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 . Then and . From this system of equations we obtain . Adding to both sides of the second equation, we obtain . Thus,
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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