Question

Six years hence a man’s age will be three times the age of his son and three years ago he was nine times as old as his son. Find their present ages.

Answer 1

Answer:

Father's age= 30

Son's age= 6

Step-by-step explanation:

Let their current age be x and y

∴x+6 = 3(y+6)

⇒x+6 = 3y+18

⇒x = 3y+12     ..............(1)

and, x-3 = 9(y-3)

   ⇒x-3 = 9y-27

   ⇒x = 9y-24     .............(2)

Equating (1) and (2), we get,

3y+12 = 9y-24

6y = 36

y = 6

Substituting the value of y in (1), we get,

x = 3×6+12

  = 18+12

  = 30

Answer 2

Answer:

Father's age= 30

Son's age= 6

Step-by-step explanation:

Let their current age be x and y

∴x+6 = 3(y+6)

⇒x+6 = 3y+18

⇒x = 3y+12     ..............(1)

and, x-3 = 9(y-3)

  ⇒x-3 = 9y-27

  ⇒x = 9y-24     .............(2)

Equating (1) and (2), we get,

3y+12 = 9y-24

6y = 36

y = 6

Substituting the value of y in (1), we get,

x = 3×6+12

 = 18+12

 = 30

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