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2008 AMC 10B Problem 6

2008 AMC 10B Problem 6AMC 10 Step by Step

Name: Date: September 13, 2026

Ratios, Percents & Averages · difficulty 2 of 5 · about 2 min · position P1-10

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)
$1/5$

Answer

Key insight

Both conditions are measured from D: BD is 1/5 of AD and CD is 1/10 of AD, so BC = BD - CD = 1/10 of AD.

Solution

Since $AB = 4\,BD$ , the whole segment is $AD = AB + BD = 5\,BD$ , so $BD = \tfrac{1}{5} AD$ . Likewise $AC = 9\,CD$ gives $AD = 10\,CD$ , so $CD = \tfrac{1}{10} AD$ .

Both $B$ and $C$ are measured from $D$ , and $C$ is closer to $D$ than $B$ is. Therefore
$BC = BD - CD = \tfrac{1}{5} AD - \tfrac{1}{10} AD = \tfrac{1}{10} AD.$

Concretely, take $AD = 10$ : then $BD = 2$ , $CD = 1$ , and $BC = 1$ .

The answer is $\boxed{\textbf{(C)}\ 1/10}$ .

Why this works

" $AB$ is $4$ times $BD$ " splits $AD$ into $4 + 1 = 5$ equal parts, not $4$ . Converting each ratio into a fraction of the same whole ( $AD$ ) puts both points on one ruler, after which $BC$ is a subtraction. Choosing a convenient length ( $AD = 10$ , a common multiple of $5$ and $10$ ) makes the arithmetic instant.

The trap

Turning 'AB is 4 times BD' into BD = AD/4 instead of AD/5 (forgetting that AD = AB + BD).

Common mistakes

Turning 'AB is 4 times BD' into BD = AD/4 instead of AD/5 (forgetting that AD = AB + BD).
Adding $BD + CD$ instead of subtracting, or placing $B$ between $C$ and $D$ .

Techniques

Test small/specific values or special cases to find or verify the answer