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2008 AMC 12B Problem 8

Problem 8 of 25EasierAlgebra

Points BB and CC lie on AD.\overline{AD}. The length of AB\overline{AB} is 44 times the length of BD,\overline{BD}, and the length of AC\overline{AC} is 99 times the length of CD.\overline{CD}. The length of BC\overline{BC} is what fraction of the length of AD?\overline{AD}?

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Solution

Since AB=4BDAB = 4\,BD and AB+BD=AD,AB + BD = AD, we have 5BD=AD,5\,BD = AD, so BD=15AD.BD = \tfrac{1}{5}AD. Likewise AC=9CDAC = 9\,CD with AC+CD=ADAC + CD = AD gives CD=110AD.CD = \tfrac{1}{10}AD. Because BB and CC both measure from A,A,   BC=BDCD\;BC = BD - CD =15AD110AD= \tfrac{1}{5}AD - \tfrac{1}{10}AD =110AD.= \tfrac{1}{10}AD. Thus, the correct answer is C.

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Concepts: ratio and proportion

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.