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Difference between revisions of "2008 AMC 10B Problems/Problem 6"
 
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==Problem==
 
==Problem==
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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?
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Points <math>B</math> and <math>C</math> lie on <math>\overline{AD}</math>. The length of <math>\overline{AB}</math> is <math>4</math> times the length of <math>\overline{BD}</math>, and the length of <math>\overline{AC}</math> is <math>9</math> times the length of <math>\overline{CD}</math>. The length of <math>\overline{BC}</math> is what fraction of the length of <math>\overline{AD}</math>?
  
−
A) 1/36 B) 1/13 C) 1/10 D) 5/36 E) 1/5
+
<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>
  
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==Solution==
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==Solution 1==
−
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)
+
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);
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dot((5,0));
 +
label("B", (5,0), S);
 +
dot((10,0));
 +
label("C", (10,0), S);
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dot((15,0));
 +
label("D", (15,0), S);
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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}}

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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