Question

10 years ago father was 12 times as old as his son and after 10 years father will be twice as old as his son find the present age of father is

Answer 1

Answer:

father's present age = x = 34 years

son's present age = y = 12 years

Step-by-step explanation:

let the father's present age be x

10 years ago father's age was x - 10

let the son's present age be y

10 years ago son's age was y - 10

10 years ago father's age was 12 times age of son

x - 10 = 12(y - 10)

x - 10 = 12y - 120

x - 12y = - 120 + 10

x - 12y = - 110 equ (1)

10 years later father will be twice the age of son

x + 10 = 2(y + 10)

x + 10 = 2y + 20

x - 2y = 20 - 10

x - 2y = 10 equ (2)

subtract equ (1) - (2)

x - 12y - ( x - 2y) = - 110 - 10

x - 12y - x + 2y = - 120

- 10y = - 120

10y = 120

y = 120/10

y = 12

substitute y= 12 in equ (2)

x - 2y = 10

x - 2(12) = 10

x - 24 = 10

x = 10 + 24

x = 34

therefore father's present age = x = 34 years

son's present age = y = 12 years

hope you get your answer