Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10A Problems/Problem 3"

(→‎Solution 2)
 
(13 intermediate revisions by 7 users not shown)
Line 4: Line 4:
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math>
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5</math>
  
−
== Solution ==  
+
== Solution 1 ==  
 
Let <math>x</math> be the third number. It follows that the first number is <math>6x,</math> and the second number is <math>x+40.</math>
 
Let <math>x</math> be the third number. It follows that the first number is <math>6x,</math> and the second number is <math>x+40.</math>
  
Line 14: Line 14:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Solve this using a system of equations. Let <math>x</math>, <math>y</math>, and <math>z</math> be the three numbers, respectively. We get three equations:
+
Solve this using a system of equations. Let <math>x,y,</math> and <math>z</math> be the three numbers, respectively. We get three equations:
−
<math>x+y+z=96</math>
+
<cmath>\begin{align*}
−
<math>x=6z</math>
+
x+y+z&=96, \\
−
<math>z=y-40</math>
+
x&=6z, \\
−
Rewriting the third equation gives us <math>y=z+40</math>, so we can substitute <math>x</math> as <math>6z</math> and <math>y</math> as <math>z+40</math>.
+
z&=y-40.
 +
\end{align*}</cmath>
 +
Rewriting the third equation gives us <math>y=z+40,</math> so we can substitute <math>x</math> as <math>6z</math> and <math>y</math> as <math>z+40.</math>
  
 
Therefore, we get
 
Therefore, we get
−
<cmath>6z+(z+40)+z=96</cmath>
+
<cmath>\begin{align*}
−
<cmath>8z+40=96</cmath>
+
6z+(z+40)+z&=96 \\
−
<cmath>8z=56</cmath>
+
8z+40&=96 \\
−
<cmath>z=7</cmath>
+
8z&=56 \\
 +
z&=7.
 +
\end{align*}</cmath>
 +
Substituting 7 in for <math>z</math> gives us <math>x=6z=6(7)=42</math> and <math>y=z+40=7+40=47.</math>
  
−
Substituting 7 in for <math>z</math> gives us <math>x=6z=6(7)=42</math> and <math>y=z+40=7+40=47</math>
+
So, the answer is <math>|x-y|=|42-47|=\boxed{\textbf{(E) } 5}.</math>
−
So <math>|x-y|=|42-47|=\boxed{(E) 5}</math>
+
 
 +
~alexdapog A-A
 +
 
 +
== Solution 3 ==
 +
In accordance with Solution 2, <cmath>y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}.</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
 
==Video Solution 1 (Quick and Easy)==
 
==Video Solution 1 (Quick and Easy)==
Line 33: Line 43:
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
 +
==Video Solution 2==
 +
https://youtu.be/B-5zSnDFVXs
 +
 +
~Charles3829
 +
 +
==Video Solution 3 (2 minutes) ==
 +
https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322
 +
 +
~Math-x
 +
 +
==Video Solution 4==
 +
https://youtu.be/BmWgwtKJExw
  
 
== See Also ==
 
== See Also ==

Latest revision as of 10:35, 24 March 2024

Problem

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?

$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution 1

Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$

We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$

Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{\textbf{(E) } 5}.$

~MRENTHUSIASM

Solution 2

Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$

Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$

So, the answer is $|x-y|=|42-47|=\boxed{\textbf{(E) } 5}.$

~alexdapog A-A

Solution 3

In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{\textbf{(E) } 5}.\] vladimir.shelomovskii@gmail.com, vvsss

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

Video Solution 2

https://youtu.be/B-5zSnDFVXs

~Charles3829

Video Solution 3 (2 minutes)

https://youtu.be/7yAh4MtJ8a8?si=Xpc2h85yyqEMOnVb&t=322

~Math-x

Video Solution 4

https://youtu.be/BmWgwtKJExw

See Also

2022 AMC 10A (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 10 Problems and Solutions
2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10A_Problems/Problem_3&oldid=217364"