1. Problem 1: Three friends have their marks in the ratio $3 : 4 : 5$ in the mid-term examination. They increased their marks by $10\%$, $5\%$, and $20\%$ respectively in the final term. We need to find their new ratio. 2. Let their original marks be $3x$, $4x$, and $5x$. 3. After increase, their marks become: - First friend: $3x + 10\%$ of $3x = 3x + 0.1 \times 3x = 3x \times 1.1 = 3.3x$ - Second friend: $4x + 5\%$ of $4x = 4x + 0.05 \times 4x = 4x \times 1.05 = 4.2x$ - Third friend: $5x + 20\%$ of $5x = 5x + 0.2 \times 5x = 5x \times 1.2 = 6x$ 4. New ratio of marks is $3.3x : 4.2x : 6x$. 5. Dividing all terms by $0.3x$ to simplify: $$\frac{3.3x}{0.3x} : \frac{4.2x}{0.3x} : \frac{6x}{0.3x} = 11 : 14 : 20$$ 6. Therefore, the new ratio of marks is $\boxed{11 : 14 : 20}$. --- 7. Problem 2: Given $$p = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \quad\text{and}\quad q = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$ We need to find the value of $$\frac{1}{p^2} + \frac{1}{q^2}$$ 8. Note that $q = \frac{1}{p}$ since $$q = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{1}{p}$$ 9. Therefore, $$\frac{1}{p^2} + \frac{1}{q^2} = \frac{1}{p^2} + \frac{1}{(1/p)^2} = \frac{1}{p^2} + p^2$$ 10. Calculate $p^2$: $$p = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}$$ $$p^2 = \left( \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \right)^2 = \frac{(\sqrt{3} - \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})^2}$$ 11. Evaluate numerator and denominator: $$ (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6} $$ $$ (\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} $$ 12. So, $$ p^2 = \frac{5 - 2\sqrt{6}}{5 + 2\sqrt{6}} $$ 13. Rationalize the denominator: $$ p^2 = \frac{5 - 2\sqrt{6}}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{(5 - 2\sqrt{6})^2}{25 - (2\sqrt{6})^2} $$ 14. Calculate numerator and denominator: $$ (5 - 2\sqrt{6})^2 = 25 - 20\sqrt{6} + 24 = 49 - 20\sqrt{6} $$ $$ 25 - (2\sqrt{6})^2 = 25 - 4 \times 6 = 25 - 24 = 1 $$ 15. Thus, $$ p^2 = 49 - 20\sqrt{6} $$ 16. Therefore, $$ \frac{1}{p^2} + p^2 = \frac{1}{49 - 20\sqrt{6}} + 49 - 20\sqrt{6} $$ 17. Rationalize $\frac{1}{49 - 20\sqrt{6}}$: Multiply numerator and denominator by $49 + 20\sqrt{6}$: $$ \frac{1}{49 - 20\sqrt{6}} = \frac{49 + 20\sqrt{6}}{(49)^2 - (20\sqrt{6})^2} = \frac{49 + 20\sqrt{6}}{2401 - 400 \times 6} = \frac{49 + 20\sqrt{6}}{2401 - 2400} = 49 + 20\sqrt{6} $$ 18. Now sum: $$ (49 + 20\sqrt{6}) + (49 - 20\sqrt{6}) = 49 + 49 + 20\sqrt{6} - 20\sqrt{6} = 98 $$ 19. Hence, $$ \frac{1}{p^2} + \frac{1}{q^2} = 98 $$ 20. The answer is $\boxed{98}$.