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Difference between revisions of "2018 AMC 12B Problems/Problem 3"

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<math>\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50</math>
 
<math>\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50</math>
  
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== Solutions ==
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== Solution 1 ==
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=== Solution 1 ===
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Using point slope form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25=\boxed{\textbf{(B) } 10}</math>.
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Using point slope form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10}
 
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\indent</math>
 
  
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=== Solution 2 ===
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== Solution 2 ==
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In order for the line with slope <math>2</math> to travel "up" <math>30</math> units (from <math>y=0</math>), it must have traveled <math>30/2=15</math> units to the right. Thus, the <math>x</math>-intercept is at <math>x=40-15=25</math>. As for the line with slope <math>6</math>, in order for it to travel "up" <math>30</math> units it must have traveled <math>30/6=5</math> units to the right. Thus its <math>x</math>-intercept is at <math>x=40-5=35</math>. Then the distance between them is <math>35-25=10 \Rightarrow \boxed{(\text{B}) 10}
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In order for the line with slope <math>2</math> to travel "up" <math>30</math> units (from <math>y=0</math>), it must have traveled <math>30/2=15</math> units to the right. Thus, the <math>x</math>-intercept is at <math>x=40-15=25</math>. As for the line with slope <math>6</math>, in order for it to travel "up" <math>30</math> units it must have traveled <math>30/6=5</math> units to the right. Thus its <math>x</math>-intercept is at <math>x=40-5=35</math>. Then the distance between them is <math>35-25=\boxed{\textbf{(B) } 10}</math>.
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\indent</math>
 
  
 
==See Also==
 
==See Also==

Revision as of 17:39, 18 September 2021

Problem

A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$. What is the distance between the $x$-intercepts of these two lines?

$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$

Solution 1

Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$. Simplifying, we get $6x-y=210$ and $2x-y=50$. Letting $y=0$ in both equations and solving for $x$ gives the $x$-intercepts: $x=35$ and $x=25$, respectively. Thus the distance between them is $35-25=\boxed{\textbf{(B) } 10}$.

Solution 2

In order for the line with slope $2$ to travel "up" $30$ units (from $y=0$), it must have traveled $30/2=15$ units to the right. Thus, the $x$-intercept is at $x=40-15=25$. As for the line with slope $6$, in order for it to travel "up" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$-intercept is at $x=40-5=35$. Then the distance between them is $35-25=\boxed{\textbf{(B) } 10}$.

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 12 Problems and Solutions

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