Answer & Solution
Answer: Option C
Solution:
$$\eqalign{
& {\text{Suppose}}\,{\text{the}}\,{\text{can}}\,{\text{initially}}\,{\text{contains}}\, \cr
& 7x\,{\text{and}}\,5x\,{\text{of}}\,{\text{mixtures}}\,{\text{A}}\,{\text{and}}\,{\text{B}}\,{\text{respectively}}. \cr
& {\text{Quantity}}\,{\text{of}}\,{\text{A}}\,{\text{in}}\,{\text{mixture}}\,{\text{left}} \cr
& = \left( {7x - \frac{7}{{12}} \times 9} \right)\,{\text{litres}} \cr
& = \left( {7x - \frac{{21}}{4}} \right)\,{\text{litres}} \cr
& {\text{Quantity}}\,{\text{of}}\,{\text{B}}\,{\text{in}}\,{\text{mixture}}\,{\text{left}} \cr
& = \left( {5x - \frac{5}{{12}} \times 9} \right)\,{\text{litres}} \cr
& = \left( {5x - \frac{{15}}{4}} \right)\,{\text{litres}} \cr
& \therefore \frac{{\left( {7x - \frac{{21}}{4}} \right)}}{{\left( {5x - \frac{{15}}{4}} \right) + 9}} = \frac{7}{9} \cr
& \Rightarrow \frac{{28x - 21}}{{20x + 21}} = \frac{7}{9} \cr
& \Rightarrow 252x - 189 = 140x + 147 \cr
& \Rightarrow 112x = 336 \cr
& \Rightarrow x = 3 \cr} $$
So, the can contained 21 litres of A