Q. Reversing the digits of father`s age we get son`s age. One year ago father was twice in age of that of his son? find their current ages?

• Let father's age= 10x+y
  son's age=10y+x (as, it is got by reversing digits of fathers age)
  ATP, (10x+y)-1=2{(10y+x)-1}
  =>x=(19y-1)/8
  let y=3 then x=7 .For any other y value , x value combined with y value doesn't give a realistic age (like father's age 120 etc)
  So, this has to be solution.Hence father's age=73, son's age =37.
• 10 years agoHelpfull: Yes(34) No(2)
• father=73 and son=37
  let
  currently the age of father is f
  currently the age of son is s
  given
  one year ago
  f-1=2(s-1)
  2s-2=f-1
  2s-f=1
• 10 years agoHelpfull: Yes(3) No(2)
• Father=73
  
  Son=37
  
  Eqn is
  
  2s-f=1
• 10 years agoHelpfull: Yes(1) No(4)
• Son s= x
  father f =2(y-1) (1 year ago)
  now,
  1 year ago in father and sin's ages
  2(y-1)=x-1
  2y -x=1
  now assign the x value is =37
  y=73
  here the reverse the digit of father age is = 73 and sun is=37
  Ans : s=37 f=73
  
• 10 years agoHelpfull: Yes(1) No(0)
• Present age of father is 73 and age of his son is 37 ,
  Bcoz (73-1)=2*(37-1)
  i.e 72=2*36
• 10 years agoHelpfull: Yes(1) No(0)
• fathter 73 son 37.
• 10 years agoHelpfull: Yes(0) No(0)
• 73 and 37
  
  fathers is 73
  son is 37
• 10 years agoHelpfull: Yes(0) No(0)
• father's age is 73 and age of his son is 37 ,
  then (73-1(1 year ago))=2*(37-1(1 year ago))
  i.e 72=2*36
• 10 years agoHelpfull: Yes(0) No(0)