$\text{Let the first number is x, the second number is y & the third number is z}$
$\text{According to the question}$
$x+y=45 ...(1)$
$y+z=55 ...(2)$
$z+3x=90...(3)$
from (1) & (2)
$x-z=-10...(4)$
from (3) & (4):
$x=20$
from equation (1):
$y=25$
from equation (3):
$z=30$
So option $(C)$ is correct.