Three pipes A, B and C fill a tank in 30 minutes, 20 minutes and 10 minutes respectively. When the tank is empty, all the three pipes are opened. If A, B and C discharge chemical solutions P, Q and R respectively, the part of solution R in the liquid in the tank after 3 minutes is

Work done by pipe A to fill the tank in 1 minute = $$\frac{1}{30}$$

Similarly, work done by pipe B and C to fill the tank in 1 minute = $$\frac{1}{20}$$ and $$\frac{1}{10}$$ respectively

Work done by all three pipes together in 1 minute = $$\frac{1}{30}+\frac{1}{20}+\frac{1}{10}=\frac{11}{60}$$

Work done by all three pipes in 3 minutes = $$\frac{11}{60} \times 3$$ (or) $$\frac{11}{20}$$

Part of chemical solution R discharged by C in 3 minutes = $$\frac{1}{10} \times 3$$ (or) $$\frac{3}{10}$$ 

Part of chemical solution R discharged in the liquid in 3 minutes = $$(\frac{3}{10}) / (\frac{11}{20}) = \frac{6}{11}$$

Hence, option C is the correct answer.