Two numbers differ by 40.

False

Given, one number $=x$

and other number $=40-x$

Let $(40-x)>x$

Then, according to the question,

$40-x+8=3(x+8)$

$\Rightarrow \quad 48-x=3 x+24$

$\Rightarrow \quad-x-3 x=24-48 \quad$ [transposing $3 x$ to LHS and 48 to RHS]

$\Rightarrow$ $-4 x=-24$

$\Rightarrow$ $x=-24 \times\left(-\frac{1}{4}\right)$

$\therefore$ $x=6$

Hence, one number $=x=6$

and other number $=40-x=40-6=34$

Now, difference between numbers $=34-6=28 \neq 40$ which is not satisfy the condition given in question.