Answer
The numbers are $10,40,48$
Work Step by Step
Let the numbers be $x,y,z$
Sum of three numbers is $98$
$x+y+z=98$ Equation $(1)$
Sum of first and second is two more than the third number.
$x+y=2+z$ Equation $(2)$
Second number is four times the first number.
$y=4x$ Equation $(3)$
From Equation $(1)$ and Equation $(2)$
$x+y+z=98$
$2+z+z=98$
$2+2z=98$
$2z=96$
$z=48$
Substituting $z$ value and $y=4x$ in Equation $(2)$, we get,
$x+y=2+z$
$x+4x=2+48$
$5x=50$
$x=10$
Substituting $x$ value in Equation $(3)$, we get,
$y=4x$
$y=4(10)$
$y=40$
The numbers are $10,40,48$
