Difference between revisions of "2008 AMC 10B Problems/Problem 6"
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==Solution== | ==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== | ==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 16:07, 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 (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 | ||
