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

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== 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</math>, <math>y</math>, and <math>z</math> be the three numbers, respectively. We get three equations:
−
<cmath>x+y+z=96</cmath>
+
<math>x+y+z=96</math>
−
<cmath>x=6z</cmath>
+
<math>x=6z</math>
−
<cmath>z=y-40</cmath>
+
<math>z=y-40</math>
−
Rewriting the third equation gives us <cmath>y=z+40</cmath>, so we can substitute <math>x</math> as <math>6z</math> and <math>y</math> as <cmath>z+40</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
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<cmath>6z+(z+40)+z=96</cmath>
+
<math>6z+(z+40)+z=96</math>
−
<cmath>8z+40=96</cmath>
+
<math>8z+40=96</math>
−
<cmath>8z=56</cmath>
+
<math>8z=56</math>
−
<cmath>z=7</cmath>
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<math>z=7</math>
  
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Substituting 7 in for <cmath>z</cmath> gives us <cmath>x=6z=6(7)=42</cmath> and <cmath>y=z+40=7+40=47</cmath>
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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>
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So <math>|x-y|=|42-47|=\boxed{textbf{(E)} 5).</math>
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So <math>|x-y|=|42-47|=\boxed{textbf{(E)} 5}</math>
  
 
==Video Solution 1 (Quick and Easy)==
 
==Video Solution 1 (Quick and Easy)==

Revision as of 23:26, 14 November 2022

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

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: $x+y+z=96$ $x=6z$ $z=y-40$ Rewriting the third equation gives us $y=z+40$, so we can substitute $x$ as $6z$ and $y$ as $z+40$.

Therefore, we get $6z+(z+40)+z=96$ $8z+40=96$ $8z=56$ $z=7$

Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47$ So $|x-y|=|42-47|=\boxed{textbf{(E)} 5}$

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

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

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