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?

Correct option is A

Given:

Pipe P can fill the tank in 30 minutes.
Pipe Q can fill the tank in 20 minutes.
Pipe R can fill the tank in 10 minutes.
All three pipes are opened simultaneously and discharge solutions S, T, and U respectively.
5. Time taken to fill the tank = 3 minutes.

Solution:

Pipes P, Q and R can fill a tank in 30 minutes, 20 minutes and 10 minutes respectively.

LCM of 30, 20 and 10 = 60
Efficiency of pipe P = 60/30 = 2 unit
Efficiency of pipe Q = 60/20 = 3 unit
Efficiency of pipe R = 60/10 = 6 unit
Total part of the tank fills in 1 minutes = 11/60 unit
Total part of the tank fills in 3 minutes = $$3×\frac{11}{60 }= \frac{11}{20}$$​​
Total part of tank fills in 3 minutes by pipe C = 3/10
Required proportion = $$\frac{(\frac{3}{10})}{(\frac{11}{20})} =\frac{ 6}{11}$$​​