2008 AMC 10B Problem 6

Below is the professionally curated solution for Problem 6 of the 2008 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10B solutions, or check the answer key.

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Concepts:ratio and proportion fraction

Difficulty rating: 960

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 $\overline{AD}?$

$\dfrac{1}{36}$

$\dfrac{1}{13}$

$\dfrac{1}{10}$

$\dfrac{5}{36}$

$\dfrac{1}{5}$

Solution:

Since $AB=4\,BD$ and $AB+BD=AD,$ we get $5\,BD=AD,$ so $BD=\tfrac15 AD.$

Since $AC=9\,CD$ and $AC+CD=AD,$ we get $10\,CD=AD,$ so $CD=\tfrac{1}{10}AD.$

Then $BC=BD-CD=\tfrac15 AD-\tfrac{1}{10}AD=\tfrac{1}{10}AD.$

Thus, the correct answer is C.

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