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

1⁵/₁₁
2⁶/₁₁
3⁷/₁₁
4⁸/₁₁

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Part filled by (A + B + C) in 3 minutes \\ = 3 (\frac{1}{30} + \frac{1}{20} + \frac{1}{10}) \\ = (3 \times \frac{11}{60}) = \frac{11}{20} \\ Part filled by C in 3 minutes = \frac{3}{10} \\ ∴ Required ratio \\ = (\frac{3}{10} \times \frac{20}{11}) = \frac{6}{11} \end{aligned}

$\eqalign{ & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,\left( {A + B + C} \right)\,{\text{in}}\,{\text{3}}\,{\text{minutes}} \cr & = 3\left( {\frac{1}{{30}} + \frac{1}{{20}} + \frac{1}{{10}}} \right) \cr & = \left( {3 \times \frac{{11}}{{60}}} \right) = \frac{{11}}{{20}} \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,{\text{C}}\,{\text{in}}\,{\text{3}}\,{\text{minutes}} = \frac{3}{{10}} \cr & \therefore {\text{Required}}\,{\text{ratio}} \cr & = \left( {\frac{3}{{10}} \times \frac{{20}}{{11}}} \right) = \frac{6}{{11}} \cr}$