Three pipes P, Q and R can fill a tank in 30 minutes, 20 minutes and 10 minutes. When the tank is empty, all three pipes are opened while they discharge three chemical solutions S, T and U respectively. What is the proportion of solution U in the contents in the tank after 3 minutes?

P filled the tank = 30 min

Q filled the tank = 20 min

R filled the tank = 10 min

work done = lcm (30,20,10)= 60 min

P's efficiency = $$\frac{60}{30}$$ = 2

Q's efficiency = $$\frac{60}{20}$$ = 3

R's efficiency = $$\frac{60}{10}$$ = 6

total time taken by all the pipes to fill the tank = $$\frac{60}{11}$$ min

in 1 min this much part of tank is filled = $$\frac{11}{60}$$ 

in 3 min  this much part of tank is filled= $$\frac{11}{60}3$$ 

                                                              $$\frac{11}{20}$$ 

part of tank filled by pipe C =  $$\frac{1}{10}3$$

so required proportion =  $$\frac{\frac{3}{10}}{\frac{11}{20}}$$ min

                                    $$\frac{6}{11}$$