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

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 $|x-y|=|42-47|=\boxed{(E) 5}$

Video Solution 1 (Quick and Easy)

https://youtu.be/v2eJtm4EUkI

~Education, the Study of Everything

2022 AMC 10A (Problems • Answer Key • Resources)
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2022 AMC 12A (Problems • Answer Key • Resources)
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2022 AMC 10A Problems/Problem 3

Contents

Problem

Solution

Solution 2

Video Solution 1 (Quick and Easy)

See Also