A vessel contains mixture of milk and water. The quantity of milk is 6 liters more than water in the vessel and the ratio of milk to water in the vessel is 3 : 2. If 50% mixture is taken out from the vessel, then find the quantity of milk remaining in the final mixture.

Correct option is A

Given :

Quantity of milk is 6 liters more than water in the vessel

The ratio of milk to water in the vessel is 3 : 2

Formula Used:

$$\frac{\text{initial quantity of A} ± \text{Removed/added quantity of the A} }{\text{initial quantity of B}± \text{Removed/added quantity of the B}} = \frac{\text{Final quantity of A}}{\text{Final quantity of B}}$$

Explanation:
Let quantity of water in the vessel initially = x liters
So, quantity of milk in the vessel initially = x + 6 liters
ATQ –

$$\frac{x+6}{x}=\frac{3}{2} \\12 = x$$​

So, quantity of milk in the vessel initially = 12 + 6 = 18 liters
Required milk remaining in the final mixture = 18 - (12+18)$$\times \frac{50}{100} \times \frac{3}{5} = 18 - 9 = 9 \ liters$$​​