a father is twice as old his son. 8 years ago their ages were in the ratio 8:3. find their present ages?

• F=2 S
  (F-8)/(S-8) = 8/3
  (2S-8)/(S-8)=8/3
  6S-24 =8S-64
  2S= 40
  S=20
  F=40
• 12 years agoHelpfull: Yes(7) No(1)
• Present age of Father - 40
  Present age of Son - 20
  
  Explanation:
  Let x be father age and y be son's age.
  x = 2y ->1
  (x-8)/(y-8) = 8/3 ->2
  Solving 2, 3x-24 = 8y-64
  => x = (8y-40)/3
  Substituting in 1, (8y-40)/3 = 2y
  => 8y-40 = 6y
  => 2y = 40
  => y = 20
  Substituting in 1, x = 2*20 = 40
• 12 years agoHelpfull: Yes(2) No(3)