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

Revision as of 00:24, 15 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 $|x-y|=|42-47|=\boxed{textbf{(E)} 5).$ (Error compiling LaTeX. Unknown error_msg)

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

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.

@@ Line 12: / Line 12: @@
 ~MRENTHUSIASM
+== 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:
+<cmath>x+y+z=96</cmath>
+<cmath>x=6z</cmath>
+<cmath>z=y-40</cmath>
+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>.
+Therefore, we get
+<cmath>6z+(z+40)+z=96</cmath>
+<cmath>8z+40=96</cmath>
+<cmath>8z=56</cmath>
+<cmath>z=7</cmath>
+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>
+So <math>|x-y|=|42-47|=\boxed{textbf{(E)} 5).</math>
 ==Video Solution 1 (Quick and Easy)==

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