1. **Problem statement:** A drum contains chemicals P, Q, and R in the ratio 3:5:2. Let the initial total quantity be $x$ litres. 2. The quantities of P, Q, and R initially are: $$P = \frac{3}{10}x, \quad Q = \frac{5}{10}x, \quad R = \frac{2}{10}x$$ 3. When 30 litres of the blend are removed, the quantities of each chemical removed are proportional to their ratios: $$P_{removed} = \frac{3}{10} \times 30 = 9$$ $$Q_{removed} = \frac{5}{10} \times 30 = 15$$ $$R_{removed} = \frac{2}{10} \times 30 = 6$$ 4. After removal, the quantities left are: $$P_{left} = \frac{3}{10}x - 9$$ $$Q_{left} = \frac{5}{10}x - 15$$ $$R_{left} = \frac{2}{10}x - 6$$ 5. Then, 12 litres of P and 8 litres of Q are added: $$P_{final} = \left(\frac{3}{10}x - 9\right) + 12 = \frac{3}{10}x + 3$$ $$Q_{final} = \left(\frac{5}{10}x - 15\right) + 8 = \frac{5}{10}x - 7$$ 6. Given that the final quantity of Q is 20 litres more than P: $$Q_{final} = P_{final} + 20$$ Substitute the expressions: $$\frac{5}{10}x - 7 = \frac{3}{10}x + 3 + 20$$ 7. Simplify the equation: $$\frac{5}{10}x - 7 = \frac{3}{10}x + 23$$ $$\frac{5}{10}x - \frac{3}{10}x = 23 + 7$$ $$\frac{2}{10}x = 30$$ 8. Solve for $x$: $$x = \frac{30 \times 10}{2} = 150$$ **Final answer:** The initial total quantity of the mixture in the drum was **150 litres**.