Question

Three years ago my age was 5 times the age of my son then six years ago my age was twice the square the age of my son then find the present age of my son?​

Answer 1

Solution is done below :

Let the current age of son = x years

and the age of the father = y years

Three years ago, son's age was

  = (x - 3) years

and that the father was

  = (y - 3) years

Three years ago father's age

= 5 * (son's age before three years)

  ⇒ y - 3 = 5 (x - 3)

  ⇒ y - 3 = 5x - 15

  ⇒ y = 5x - 12 .....(i)

Six years ago, son's age was

  = x - 6 years

and that of father was

  = y - 6 years

Six years ago father's age

  = 2 * (age of my son before six years)²

  ⇒ y - 6 = 2 (x - 6)²

  ⇒ 5x - 12 - 6 = 2 (x² - 12x + 36)

  ⇒ 5x - 18 = 2x² - 24x + 72

  ⇒ 2x² - 29x + 90 = 0

  ⇒ x² - $\frac{29}{2}$ x + $\frac{90}{2}$ = 0

(dividing both sides by 2)

  ⇒ x² - ($\frac{9}{2}$ + 10) x + $\frac{90}{2}$ = 0

  ⇒ x² - $\frac{9}{2}$ x - 10x + $\frac{90}{2}$ = 0

  ⇒ x (x - $\frac{9}{2}$) - 10 (x - $\frac{9}{2}$) = 0

  ⇒ (x - $\frac{9}{2}$) (x - 10) = 0

Either, x - $\frac{9}{2}$ or, x - 10 = 0

  ⇒ x = $\frac{9}{2}$ , x = 10

Therefore, son's present age can be $\frac{9}{2}$ years or 10 years.