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Difference between revisions of "2008 AMC 10B Problems/Problem 6"
 
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<math> \textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qquad\textbf{(E)}\ \frac{1}{5} </math>
 
<math> \textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qquad\textbf{(E)}\ \frac{1}{5} </math>
  
−
==Solution==
+
==Solution 1==
−
Let <math>CD = 1</math>. Then <math>AB = 4(BC + 1)</math> and <math>AB + BC = 9\cdot1</math>. From this system of equations we obtain <math>BC = 1</math>. Adding <math>CD</math> to both sides of the second equation, we obtain <math>AD = AB + BC + CD = 9 + 1 = 10</math>.  Thus, <math>\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}</math>
+
Let <math>CD = 1</math>. Then <math>AB = 4(BC + 1)</math> and <math>AB + BC = 9\cdot1</math>. From this system of equations, we obtain <math>BC = 1</math>. Adding <math>CD</math> to both sides of the second equation, we obtain <math>AD = AB + BC + CD = 9 + 1 = 10</math>.  Thus, <math>\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}</math>
 +
 
 +
==Solution 2==
 +
Let <math>x = BD</math> and <math>y = CD</math>. Therefore, <math>AB = 4x</math> and <math>AC = 9y</math>, as shown in the diagram(the labels on the bottom are for that line segment while the labels on the top are from one point to the left to one point to the right). 
 +
<center><asy>
 +
dot((0,0));
 +
label("A", (0,0), S);
 +
dot((5,0));
 +
label("B", (5,0), S);
 +
dot((10,0));
 +
label("C", (10,0), S);
 +
dot((15,0));
 +
label("D", (15,0), S);
 +
draw((0,0)--(5,0));
 +
draw((5,0)--(10,0));
 +
draw((10,0)--(15,0));
 +
draw((0,0)--(10,0));
 +
draw((10,0)--(15,0));
 +
label("$4x$", (0,0)--(5,0), S);
 +
label("$9y$", (0,0)--(10,0), N);
 +
label("$y$", (10,0)--(15,0), S);
 +
label("$x$", (5,0)--(15,0), N); 
 +
</asy></center>
 +
 
 +
From this, we can see that <math>AD = 10y = 5x</math>, and since <math>BC = BD - CD = x-y</math>. Now, our ratio is <math>\frac{x-y}{AD}</math>. We can split this into 2 fractions: <math>\frac{x}{AD} - \frac{y}{AD} = \frac{x}{5x} - \frac{y}{10y} = \frac{1}{5} - \frac{1}{10} =  \boxed{\textbf{(C)}\ \frac{1}{10}}  </math>
 +
 
 +
~idk12345678
  
 
==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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:09, 12 April 2024

Problem

Points $B$ and $C$ lie on $\overline{AD}$. The length of $\overline{AB}$ is $4$ times the length of $\overline{BD}$, and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}$. The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$?

$\textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{13}\qquad\textbf{(C)}\ \frac{1}{10}\qquad\textbf{(D)}\ \frac{5}{36}\qquad\textbf{(E)}\ \frac{1}{5}$

Solution 1

Let $CD = 1$. Then $AB = 4(BC + 1)$ and $AB + BC = 9\cdot1$. From this system of equations, we obtain $BC = 1$. Adding $CD$ to both sides of the second equation, we obtain $AD = AB + BC + CD = 9 + 1 = 10$. Thus, $\frac{BC}{AD} = \frac{1}{10} \implies\text{(C)}$

Solution 2

Let $x = BD$ and $y = CD$. Therefore, $AB = 4x$ and $AC = 9y$, as shown in the diagram(the labels on the bottom are for that line segment while the labels on the top are from one point to the left to one point to the right).

[asy] dot((0,0)); label("A", (0,0), S); dot((5,0)); label("B", (5,0), S); dot((10,0)); label("C", (10,0), S); dot((15,0)); label("D", (15,0), S); draw((0,0)--(5,0)); draw((5,0)--(10,0)); draw((10,0)--(15,0)); draw((0,0)--(10,0)); draw((10,0)--(15,0)); label("$4x$", (0,0)--(5,0), S); label("$9y$", (0,0)--(10,0), N); label("$y$", (10,0)--(15,0), S); label("$x$", (5,0)--(15,0), N); [/asy]

From this, we can see that $AD = 10y = 5x$, and since $BC = BD - CD = x-y$. Now, our ratio is $\frac{x-y}{AD}$. We can split this into 2 fractions: $\frac{x}{AD} - \frac{y}{AD} = \frac{x}{5x} - \frac{y}{10y} = \frac{1}{5} - \frac{1}{10} = \boxed{\textbf{(C)}\ \frac{1}{10}}$

~idk12345678

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

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