Question

The age of a father 8 yrs back was 5 times that of his son. After 8 yrs, his age will be 8 yrs more than double the age of his son. Find their present ages

Answer 1

Hiiii friend,


Let the present age of father and his son be X and Y year's.


Before 8 years age of father = (X-8) years


Before 8 years age of son = (Y-8) years



According to question,

(X-8) = 5(Y-8)


X - 8 = 5Y - 40


X - 5Y = -40 + 8


X - 5Y = -32------------(1)



After 8 years age of father = (X+8) years


After 8 years age of son = (Y+8) years



According to question,


(X+8) = 2(Y+8) + 8


X + 8 = 2Y + 16 +8


X + 8 = 2Y + 24


X - 2Y = 24 -8

X - 2Y = 16--------(2)


From equation (1) we get,

X - 5Y = -32


X = -32 + 5Y----------(3)


Putting the value of X in equation (2)


X - 2Y = 16


-32 + 5Y - 2Y = 16


3Y = 16 + 32


3Y = 48


Y = 48/3

Y = 16 years


Putting the value of Y in equation (3)


X = -32+5Y => -32 + 5 × 16 => -32 + 80


X = 48 years


Age of father = X = 48 years


And,


Age of son = Y = 16 years




HOPE IT WILL HELP YOU....... :-)

Answer 2

let , father's age as X 
 and son's age as Y

A/Q 
     X - 8 = 5Y - 8 }----- (1)
     
     X + 8 = 8 + 2Y}------(2)

     thn , 
   eq.n (1) 
       X = 5Y - 8 +8 

 hence , +8 will get cancelled by -8

    X = 5Y}------(3) 
.....................
putting equation (3) in equation (2)

       X + 8 = 8 + 2Y    {X=5Y as per equation (3)}
     =>  5Y + 8 = 8 + 2Y
    => 5Y = 8-8+2Y
    => 5Y = 0 + 2Y
    => 5Y - 2Y = 0 
    => 3Y = 0 
    => Y = 0 /3 
    
hence , 
       Y = 0/3 
and 
       X = 5Y = 5 x 0/3 = 5/3