Seven years ago Varun's age was five times the square of Swati's age. Three years hence Swati's age will be two-fifths of Varun's age. Find their present ages.
Step-by-Step Solution
Key Concept: 7 years ago: Let Swati $= x$, Varun $= 5x^2$. Present ages: Swati $= x + 7$, Varun $= 5x^2 + 7$. In 3 years: Swati $= x + 10$, Varun $= 5x^2 + 10$. Given $x + 10 = \dfrac{2}{5}(5x^2 + 10) \Rightarrow x + 10 = 2x^2 + 4 \Rightarrow 2x^2 - x - 6 = 0 \Rightarrow (2x + 3)(x - 2) = 0 \Rightarrow x = 2$. Present ages: Swati $= 2 + 7 = 9$ years, Varun $= 5(2)^2 + 7 = 27$ years.
Let Swati's age 7 yrs ago be $x$. Varun's age 7 yrs ago $= 5x^2$. [1.0 Mark]
In 3 years: Swati $= x + 10$, Varun $= 5x^2 + 10$. Given $x + 10 = \dfrac{2}{5}(5x^2 + 10) = 2x^2 + 4$. [1.5 Marks]
$2x^2 - x - 6 = 0 \Rightarrow (2x + 3)(x - 2) = 0 \Rightarrow x = 2$. [1.5 Marks]
Present ages: Swati $= 2 + 7 = 9$ years, Varun $= 5(4) + 7 = 27$ years. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Setting up ages 7 yrs ago as $x$ and $5x^2$: 1.0 Mark
Forming quadratic $2x^2 - x - 6 = 0$: 1.5 Marks
Solving $x = 2$: 1.5 Marks
Calculating present ages $9$ and $27$ years: 1.0 Mark
Correct Answer: