# 2022 AMC 10A Problems/Problem 3

## 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: $$x+y+z=96$$ $$x=6z$$ $$z=y-40$$ 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 $|x-y|=|42-47|=\boxed{\textbf{(E) } 5}$

~alexdapog A-A

## Video Solution 1 (Quick and Easy)

~Education, the Study of Everything

## Video Solution 2

~Charles3829

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