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

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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2022 AMC 10A Problems/Problem 3

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