A milkman has 500 litres of milk with him. He also has 2 helpers H1 and H2. All three of them decide to cheat customer by mixing water. Each one of them first removes 100 L of the solution and add 200 L of water in it one after the other. Find the purity % of the milk in the final mixture.
I went on solving this question by focusing on what scenario is being created at each step for both milk and water in the container.
However in the textbook, there is a far more efficient solution which I want to learn, what they did was
Purity of Milk = $100* (400/600) * (500/700) * (600/800)$%
Please explain on how did these ratios came up $400/600 , 500/700 , 600/800$
We start with $100\%$ concentration of milk. Each person removes one hundred liters of milk and adds two hundred liters of water. The concentration of the milk doesn't change when the initial one hundred liters is removed. It changes when the two hundred liters of water is added.
If $V_0$ is the volume of the mixture and $C_0$ is the concentration of milk then the new concentration $$C_1=\frac{C_0(V_0-100)}{V_0+100}$$
With $V_0=100$ and $C_0=100\%$ we get $$C_1=100\%\frac{400}{600}$$
Notice that $V_1=V_0+100$. On each iteration, the volume increases by one hundred.
Continuing in this manner we get $$C_2=(100\%\frac{400}{600})\frac{500}{700}$$ because we get $C_1$ applied to $V_1-100$ liters and the total volume increased by a hundred.
Finally, $$C_3=C_2\frac{600}{800}=(100\%)\frac{400}{600}\frac{500}{700}\frac{600}{800}$$ and the pattern repeats if we continue to remove a hundred liters of solution, followed by an addition of two hundred liters of water.