A man buys milk at $\text{Rs. }60$ per litre, adds water to it, and then sells the mixture at $\text{Rs. }50$ per litre. What is the ratio of milk and water if his profit in the deal is $20\%$?

Given:
* Cost Price ($\text{CP}$) of pure milk = $\text{Rs. }60/\text{litre}$
* Cost Price ($\text{CP}$) of water = $\text{Rs. }0/\text{litre}$
* Selling Price ($\text{SP}$) of the mixture = $\text{Rs. }50/\text{litre}$
* Profit percentage = $20\%$

First, find the Cost Price ($\text{CP}$) of the mixture:

$${\text{CP}}_{\text{mixture}}=\frac{\text{SP}}{1+\frac{\text{Profit}\%}{100}}=\frac{50}{1+0.20}=\frac{50}{1.2}=\frac{500}{12}=\text{Rs. }\frac{125}{3}/\text{litre}$$

Using the rule of allegation:

* Component 1 (Milk CP) = $60$
* Component 2 (Water CP) = $0$
* Mean Price (${\text{CP}}_{\text{mixture}}$) = $\frac{125}{3}$

$$\text{Ratio of Milk : Water}=\frac{\text{Mean}-\text{Water CP}}{\text{Milk CP}-\text{Mean}}$$

$$\text{Ratio}=\frac{\frac{125}{3}-0}{60-\frac{125}{3}}=\frac{\frac{125}{3}}{\frac{180-125}{3}}=\frac{125}{55}$$

Simplifying by dividing both numbers by $5$:

$$\text{Ratio}=\frac{25}{11}=25:11$$