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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? (2022 AMC 10A Problems, Question \#3) 
Overseas Competition->Knowledge Point->Calculation Modules
Let $x$ be the third number. It follows that the first number is $6 x$, and the second number is $x+40$.  We have $$ 6 x+(x+40)+x=8 x+40=96, $$ from which $x=7$.  Therefore, the first number is $42$ , and the second number is $47$ . Their absolute value of the difference is $\textbar42-47\textbar=(\mathbf{E}) 5$. 