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

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==Problem==
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== Problem ==
 
 
 
A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines?  
 
A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines?  
  
 
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math>
 
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math>
  
 
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== Solutions ==
==Solution 1==
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=== Solution 1 ===
 
 
 
 
 
Using the slope-intercept 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}
 
Using the slope-intercept 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}
 
\indent</math>
 
\indent</math>
  
==Solution 2==
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=== Solution 2 ===
 
 
 
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}
 
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}
 
\indent</math>
 
\indent</math>

Revision as of 13:20, 19 January 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?

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

Solutions

Solution 1

Using the slope-intercept 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 $35-25 = 10 \Rightarrow \boxed{(\text{B}) 10} \indent$

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 $35-25=10 \Rightarrow \boxed{(\text{B}) 10} \indent$

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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