Question 958413
Let x,y,z be three numbers.
x+y+z=50------------(1)
y=3x==>x=y/3
z=2y
substituting x=y/3,z=2y in (1)
(y/3)+y+ 2y = 50
=>(y+3y+6y)/3 = 50 =>(y+3y+6y)=150
=>10y=150 =>y=15
x=y/3=15/3=5=>x=5
z=2y=>z=2(15)=30=>z=30
x=5, y=15, z=30