Replacement of solution

Let initial volume be V, final be F for milk.

The formula is given by : $$F\ =\ V\cdot\left(1-\frac{K}{V}\right)^n$$ n is the number of times the milk is drawn and replaced.

so we get $$F=\ V\left(1-\frac{K}{V}\right)^{^2}$$
here K =9
we get
$$\frac{16}{25}V\ =\ V\ \left(1-\frac{9}{V}\right)^{^2}$$
we get $$1-\frac{9}{V}=\ \frac{4}{5}or\ -\frac{4}{5}$$

If considering $$1-\frac{9}{V}=-\frac{4}{5}$$

V =5, but this is not possible because 9 liters is drawn every time.

Hence : $$1-\frac{9}{V}=\frac{4}{5},\ V\ =\ 45\ liters$$

Given that in the final mixture, the ratio of coffee and cocoa is 16:9

Let us assume coffee is 16 units and cocoa is 9 units.

=> Initially, there are 25 units of coffee and 0 units of cocoa

Let's say x units of the mixture is removed and replaced with cocoa

=> Now, we have (25-x) coffee and x units of cocoa. => Mixture P

Now, if x units of the mixture is removed:

Amount of coffee present = (25-x) - $$\dfrac{\left(25-x\right)}{25}\times\ x$$

=> $$\left(25-x\right)\left(1-\dfrac{x}{25}\right)=16$$

=> $$\left(25-x\right)^2=16\times\ 25$$

=> 25 - x = 20 => x = 5.

In mixture P, cocoa = x = 5

In mixture Q, cocoa = 9 units.

=> Required ratio = 5:9

Let's assume that after n iteration, the volume of the milk will be less than 50%, which is less than 20 liters.

Initially, the amount of milk is 40 liters, after the first iteration, the volume of milk is$$40\cdot\frac{9}{10}$$

After the second iteration, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^2}$$

Similarly, after the n iterations, the volume of milk is $$40\times\left(\frac{9}{10}\right)^{^n}$$

Now,

$$40\times\left(\frac{9}{10}\right)^{^n}\ \le\ 20$$

=> $$\left(\frac{9}{10}\right)^{^n}\ \le\ \frac{1}{2}$$

=> $$n\ge\ 7$$

Hence, the correct answer is 7