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Mathematics

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.

Quadratic Equations

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Answer

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) = 25\dfrac{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\dfrac{11}{2}

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

Substituting the value of x in equation (2),

⇒ 5 ×\times 9 + 9 = 2y

⇒ 45 + 9 = 2y

⇒ 54 = 2y

⇒ y = 542\dfrac{54}{2} = 27

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

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