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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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.

KnowledgeBoat · Accepted 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) = (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 = 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 = = 27 Thus, the age of Rohit = 27 years and that of Geeta = 9 years.

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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