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 $\textbar42-47\textbar=(\mathbf{E}) 5$. 
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) 

(a) $$1$$ 
(b) $$2$$ 
(c) $$3$$ 
(d) $$4$$ 
(e) $$5$$ 