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:

The time taken by pipes P, Q, and R to fill the tank are 30 minutes, 20 minutes, and 10 minutes respectively.

The solutions they discharge are S, T, and U respectively.

Concept Used:
The part of the tank filled by each pipe in 1 minute is:

​$$\text{Part filled by P in 1 min} = \frac{1}{30}\\\ \\\text{Part filled by Q in 1 min} = \frac{1}{20}\\\ \\\text{Part filled by R in 1 min} = \frac{1}{10}\\\ \\$$​

Solution:

In 3 minutes, the parts of the tank filled by P, Q, and R are:

​$$\text{Part filled by P in 3 min} = 3 \times \frac{1}{30} = \frac{1}{10}\\\ \\\text{Part filled by Q in 3 min} = 3 \times \frac{1}{20} = \frac{3}{20}\\\ \\\text{Part filled by R in 3 min} = 3 \times \frac{1}{10} = \frac{3}{10}\\\ \\$$​​

The total part of the tank filled in 3 minutes is:

​$$\text{Total part filled} = \frac{1}{10} + \frac{3}{20} + \frac{3}{10}\\\ \\= \frac{2}{20} + \frac{3}{20} + \frac{6}{20} = \frac{11}{20}$$​​

The proportion of solution U (discharged by pipe R) is the part filled by R divided by the total part filled:

​$$\text{Proportion of U} = \frac{\frac{3}{10}}{\frac{11}{20}}\\\ \\= \frac{3}{10} \times \frac{20}{11} = \frac{6}{11}$$​​

Therefore, the proportion of solution U in the contents of the tank after 3 minutes is 6/11