A vessel contains a solution of two liquids A and B in the ratio 5 : 3. When 10 litres of the solution is taken out and replaced by the same quantity of B, the ratio of A and B in the vessel becomes 10 : 11. The quantity (in litres) of the solution, in the vessel was ........

The ratio of A and B in the vessel = 5 : 3

Let the liquid A be 5x and liquid B be 3x.

10 litres of the solution is taken out so,

Remaining quantity of A = 5x - 10 $$\times \frac{5}{8} = 5x - \frac{25}{4}$$

Remaining quantity of B = 3x - 10 $$\times \frac{3}{8} = 3x - \frac{15}{4}$$

10 litres quantity of B are mixed so,

Quantity of B = 3x - $$\frac{15}{4}$$ + 10

Now, ratio of A to B = 10 : 11

5x - $$\frac{25}{4} : 3x - \frac{15}{4}$$ + 10 = 10 : 11

20x - 25 : 12x - 15 + 40 = 10 : 11

$$\frac{20x - 25}{12x + 25} = \frac{10}{11}$$

220x - 275 = 120x + 250

100x = 525

x = 45.25

Total solution = 5x + 3x = 8x = 8 $$\times$$ 5.25 = 42 liters