Question

A man is twice as old as his son. 20 years ago his age was 12 times the age of his son. Find their present ages.

Answer 1

Answer:

Son's age = 22

Father's age = 44

Step-by-step explanation:

let son be x years

father = 2x

..

20 years ago

son = x-20

father = 2x-20

..

2x-20=12*(x-20)

2x-20 = 12x -240

add 20 to both sides

2x = 12x-240+20

2x=12x-220

add -12x to both sides

2x-12x= 12x-12x-220

-12x+2x= -240+20

-10x= -220

divide by -10

x= -220/-10

x=22 son's age

Father's age = 2*22 = 44 years.