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

Latest revision as of 21:07, 27 May 2023

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 $35-25=\boxed{\textbf{(B) } 10}$ .

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/2BrH2Dugkjg

~Education, the Study of Everything

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-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?
+== Problem ==
+A line with slope <math>2</math> intersects a line with slope <math>6</math> 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>\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50</math>
+== Solution 1 ==
+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>.
-===Solution 1===
+== 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=\boxed{\textbf{(B) } 10}</math>.
-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 gives the <math>x</math>-intercepts
+==Video Solution (HOW TO THINK CRITICALLY!!!)==
+https://youtu.be/2BrH2Dugkjg
+~Education, the Study of Everything
+==See Also==
+{{AMC12 box|year=2018|ab=B|num-b=2|num-a=4}}
+{{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

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