Vessels A and B contain 60 litres of alcohol and 60 litres of water, respectively. A certain volume is taken out from A and poured into B. After stirring, the same volume is taken out from B and poured into A. If the resultant ratio of alcohol and water in A is 15 : 4, then the volume, in litres, initially taken out from A is

Vessel A contains $$60$$ litres of pure alcohol, and Vessel B contains $$60$$ litres of pure water. Let the amount taken out from vessel A be $$x$$ litres. The contents in Vessel A now are $$60-x$$ litres of alcohol, and in Vessel B now are $$60$$ litres of water and $$x$$ litres of alcohol.

After mixing, when the same quantity $$x$$ is taken out from Vessel B, the alcohol that would be taken out would be $${\left(\dfrac{x}{60+x}\right)}^{\text{th}}$$ of $$x$$, which is $$\dfrac{x^2}{60+x}$$ litres.

The initial total quantity of $$60$$ litres has been restored in A after replacement. The total amount of alcohol in Vessel A after the process is complete is:

$$60 - x + \dfrac{x^2}{60+x} = \dfrac{(60+x)(60-x) + x^2}{60+x} = \dfrac{3600}{60+x}$$

The total quantity in vessel A is $$60$$ litres where the ratio of alcohol to total quantity is $$\dfrac{15}{15+4} = \dfrac{15}{19}$$

Therefore, 

$$\dfrac{3600\div (60+x)}{60} = \dfrac{15}{19}$$

$$\Rightarrow \dfrac{60}{60+x} = \dfrac{15}{19}$$

$$\Rightarrow x = 16$$

Thus, the quantity taken out and replaced in the two iterations is $$16$$ litres.