Question 253475
Let the three numbers be x,y and z.
Given,The first of three numbers exceeds twice the second number by 4
           x = 2y+4  ...(1)
      The third number is twice the first.
           z = 2x
         2x-z= 0  ...(2)
The sum of the numbers is 54
      x+y+z = 54  ...(3)
(2)+(3)=> 3x+y = 54   ...(4)
           Put x = 2y+4 in (4)
          3(2y+4)+y = 54
          6y+12+y   = 54
                 7y = 54-12
                 7y = 42
                  y = 42/7
                  y = 6
Substituting in (1), we get
                 x = 2*6+4
                 x = 12+4
                 x = 16
Substituting in (2), we get
        2*16-z = 0
            32 = z  
Thereforethe solution is x=16, y=6 and z = 32