Question

ten years ago , a father was 12 times as old as his son and ten years hence , hence he will be twice as old as his son will be then . find their present ages

Answer 1

Ten years ago,
Let the age of the son be x years.
So,
The age of father was 12*x = 12x years

Now,
A/q,
x+20/12x+20 = 1/2
After cross multiplication,
2(x+20) = 1(12x+20)
=> 2x+40 = 12x+20
=> 40-20 = 12x-2x
=> 20 = 10x
=> x = 20/10
=> x = 2

So,
10 years ago,
Age of son = x = 2 years
Age of father = 12x = 12*2 = 24 years

Hence,
The present age of son = 2 + 10 = 12 years
The present age of father = 24+10 = 34 years


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Answer 2

Hello !

Present ages :-
Father's age = x years
Son's age = y years

Ages ten years before ,
Father's age = x  -10
Son's age = y -10

x -10 = 12(y-10)
x - 10 = 12y - 120
x - 12y = -120 + 10
x - 12y = -110                     -----> (1)

--------------------------------------------------------

Age after 10 yrs ,

Father's age = x + 10
Son's age = y + 10

x + 10 = 2(y+10)
x + 10 = 2y + 20
x - 2y = 10                   -----> (2)

Subtracting equation 2 from 1


            x - 12y = -110    
  -         x - 2y = 10  
==================
             -10y = -120 

y = -120/-10
   = 12

x - 2y = 10
x - 2 × 12 = 10
x = 10 + 24
x  = 34

====================
Age of father = x = 34 years
Age of son = y = 12 years