Question

10 years ago father was 12 times as old as his son and 10 years hence he will be twice as old as his son.find their oresent ages

Answer 1

let father present age be x

let son present age be y

given

x-10=12(y-10)

x+10=2(y+10)

solving this we get

x=34(father's present age is 34)

y=12(son's present age is 12)

Answer 2

$\textbf{Answer:}$


$\bf{let \: the \: present \: age \: of \: father \: be \: \: x, \: and \: the \: present \: age \: of \: son \: be \: y} \\ \\x - 10 = 12(y - 10) \\ \\ = > x - 10 = 12y - 120 \\ \\ = > x - 12y = - 110 \: \: \: \: (i) \\ \\ \bf{now.} \\ \\ x + 10 = 2(y + 10) \\ \\ = > x + 10 = 2y + 20 \\ \\ = > x - 2y = 20 \\ \\ = > \times = 10 + 2y \: \: \: (ii) \\ \\ \bf{by \: (ii) \: and \: (i)} \\ \\ x - 12y = - 110 \\ \\ = > 10 + 2y - 12y = - 110 \\ \\ = > - 10y = - 120 \\ \\ = > y = \frac{ - 120}{ - 10} = 12 \: years \\ \\ \bf{finally.} \\ \\ x = 10 + 2y \\ \\ = 10 + 2y = 34 \: years$

$\textbf{Hope it helps you!}$