A bucket contains a mixture of two liquids A & B in the proportion 5 : 3. If 16 litres of the mixture is replaced by 16 litres of liquid B, then the ratio of the two liquids becomes 3 : 5. How much of the liquid B was there in the bucket?

Explanation:-

Let bucket contains 5x and 3x of liquids A and B respectively.
When 16 litres of mixture are drawn off, quantity of A in mixture left:
$$\eqalign{ & \left[ {5x - \left( {\frac{5}{8}} \right) \times 16} \right] = \left( {5x - 10} \right) \cr & {\text{Similarly quantity of B in mixture left}}, \cr & \left[ {3x - \left( {\frac{3}{8}} \right) \times 16} \right] = \left( {3x - 6} \right) \cr & {\text{Now the ratio becomes,}} \cr & \frac{{\left( {5x - 10} \right)}}{{\left( {3x - 6} \right)}} = \frac{3}{5} \cr & \Rightarrow 25x - 50 = 9x - 18 \cr & \Rightarrow 16x = 32 \cr & \Rightarrow x = 2 \cr & {\text{So, quantity of liquid B initially}}, \cr & = 3 \times 2 = 6\,{\text{litres}} \cr}$$