A vessel contained a solution of acid and water, in which water was 64%. Four litres of the solution was taken out of the vessel and the same quantity of water was added. If the resulting solution contains 30% acid, the quantity (in litres) of the water in the solution, at the beginning in the vessel, was:

\[\begin{array}{*{20}{c}} {}&{{\text{Acid}}}&{{\text{Water}}} \\ {{\text{Old}} \to }&{36\% }&{64\% } \\ {{\text{New}} \to }&{30\% }&{70\% } \end{array}\]

$$\eqalign{ & {\text{Old mixture}} = {\text{Remaining}} + {\text{Taken out}} \cr & = 20{\text{L}} + 4{\text{L}} \cr & = 24{\text{L}} \cr & {\text{Water}} = 24 \times \frac{{64}}{{100}} \Rightarrow 15.36{\text{L}} \cr} $$