A bucket contains a mixture of two liquids A and B in the proportion 7 : 5. If 9 litres of mixture is replaced by 9 liters of liquid B, then the ratio of the two liquids becomes 7 : 9. How much of the liquid A was there in the bucket ?

Answer: Option A

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively. When 9 litres of mixture are drawn off, quantity of A in mixture left:
$$\begin{array}{rl}& =\left[7x-\frac{7}{12}\times 9\right]\text{litres}\\ & =\left[7x-\frac{21}{4}\right]\text{litres}\\ & \text{Similarly quantity of B in mixture left},\\ & =\left[5x-\frac{5}{12}\times 9\right]\text{litres}\\ & =\left[5x-\frac{15}{4}\right]\text{litres}\\ & \therefore \text{ratio becomes},\\ & \frac{7x-\frac{21}{4}}{\left(5x-\frac{15}{4}\right)+9}=\frac{7}{9}\\ & \Rightarrow \frac{28x-21}{20x+21}=\frac{7}{9}\\ & \Rightarrow 252x-189=140x+147\\ & \Rightarrow 112x=336\\ & \Rightarrow x=3\\ & \text{So the can contained},\\ & =7\times x\\ & =7\times 3\\ & =21\text{litres of A initially}\text{.}\end{array}$$