Question

Seven years ago Rohit's age was five times the square of Geeta's age. 3 years hence, Geeta's age will be two fifths of Rohit's age. Find their ages.

Answer 1

Let the age of Geeta be x years and that of Rohit be y years.

It is given that 7 years ago, Rohit's age was five times the square of Geeta's age.

⇒ y − 7 = 5(x − 7)²

⇒ y − 7 = 5(x² + 49 − 14x)

⇒ y − 7 = 5x² + 245 − 70x

⇒ y = 5x² + 245 − 70x + 7

⇒ y = 5x² + 252 − 70x   ...... (1)

And, 3 years hence, Geeta's age will be two fifths of Rohit's age.

⇒ (x + 3) = `2/5`​ (y + 3)

⇒ 5(x + 3) = 2(y + 3)

⇒ 5x + 15 = 2y + 6

⇒ 5x + 15 − 6 = 2y

⇒ 5x + 9 = 2y   ...... (2)

Substituting the value of y in the above equation,

⇒ 5x + 9 = 2(5x² + 252 − 70x)

⇒ 5x + 9 = 10x² + 504 − 140x

⇒ 10x² + 504 − 140x − 5x − 9 = 0

⇒ 10x² − 145x + 495 = 0

⇒ 2x² − 29x + 99 = 0

⇒ 2x² − 18x − 11x + 99 = 0

⇒ 2x(x − 9) − 11(x − 9) = 0

⇒ (x − 9)(2x − 11) = 0

⇒ (x − 9) = 0 or (2x − 11) = 0

⇒ x = 9 or 2x = 11

⇒ x = 9 or x = 112

Since, age cannot be in fraction. So, the age of Geeta = 9 years.

Substituting the value of x in equation (2),

⇒ 5 × 9 + 9 = 2y

⇒ 45 + 9 = 2y

⇒ 54 = 2y

⇒ y = `54/2​` = 27

Thus, the age of Rohit = 27 years and that of Geeta = 9 years.