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2008 AMC 10B Problem 6

Problem 6 of 25EasierArithmetic

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=4 BDAB=4\,BD and AB+BD=AD,AB+BD=AD, we get 5 BD=AD,5\,BD=AD, so BD=15AD.BD=\tfrac15 AD. Since AC=9 CDAC=9\,CD and AC+CD=AD,AC+CD=AD, we get 10 CD=AD,10\,CD=AD, so CD=110AD.CD=\tfrac{1}{10}AD. Then BC=BD−CDBC=BD-CD =15AD−110AD=\tfrac15 AD-\tfrac{1}{10}AD =110AD.=\tfrac{1}{10}AD. Thus, the correct answer is C.
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