Difference between revisions of "2008 AMC 10B Problems/Problem 6"

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==Solution==
 
==Solution==
{{solution}}
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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==
 
==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}}

Revision as of 17:07, 11 January 2009

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?

A) 1/36 B) 1/13 C) 1/10 D) 5/36 E) 1/5

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 (ProblemsAnswer KeyResources)
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