Understanding the Milk and Water Mixture Problem

This problem involves a mixture of milk and water where a portion of the mixture is removed and replaced with pure milk. This process is repeated, and we need to find the final ratio of milk and water.

Initial State of the Mixture

We start with a vessel containing 20 litres of a mixture of milk and water. The ratio of milk to water is given as 3 : 2.

• Total volume = 20 litres
• Ratio of Milk : Water = 3 : 2
• Sum of ratio parts = \(3 + 2 = 5\)
• Initial quantity of Milk = \(\frac{3}{5} \times 20 = \frac{60}{5} = 12\) litres
• Initial quantity of Water = \(\frac{2}{5} \times 20 = \frac{40}{5} = 8\) litres

So, initially, the vessel has 12 litres of milk and 8 litres of water.

Step 1: First Removal and Replacement

Ten litres of the mixture are removed. When a portion of the mixture is removed, the milk and water are removed in the same ratio as they exist in the mixture at that time (3:2).

• Amount of mixture removed = 10 litres
• Milk removed = \(10 \times \frac{\text{Current Milk Ratio Part}}{\text{Total Ratio Parts}} = 10 \times \frac{3}{5} = 6\) litres
• Water removed = \(10 \times \frac{\text{Current Water Ratio Part}}{\text{Total Ratio Parts}} = 10 \times \frac{2}{5} = 4\) litres

After removing 10 litres of mixture:

• Remaining Milk = \(12 - 6 = 6\) litres
• Remaining Water = \(8 - 4 = 4\) litres
• Total volume remaining = \(6 + 4 = 10\) litres

Now, 10 litres of pure milk are added to the vessel.

• Milk after adding pure milk = \(6 + 10 = 16\) litres
• Water after adding pure milk = \(4\) litres (since pure milk contains no water)
• Total volume = \(16 + 4 = 20\) litres (The volume is restored to the original 20 litres)

After the first process, the quantity of milk is 16 litres and the quantity of water is 4 litres. The ratio of Milk : Water is \(16 : 4\), which simplifies to \(4 : 1\).

Step 2: Second Removal and Replacement

The process is repeated once again. Now, the mixture has 16 litres of milk and 4 litres of water (total 20 litres). The ratio of Milk : Water is 16 : 4, or 4 : 1.

Ten litres of this current mixture are removed. Milk and water will be removed in the current ratio (4:1).

• Amount of mixture removed = 10 litres
• Current ratio of Milk : Water = 16 : 4 (or 4 : 1)
• Total ratio parts = \(4 + 1 = 5\) (using the simplified ratio) or Total volume = 20 litres (using the actual quantities)
• Proportion of milk in the mixture = \(\frac{16}{20} = \frac{4}{5}\)
• Proportion of water in the mixture = \(\frac{4}{20} = \frac{1}{5}\)
• Milk removed = \(10 \times \frac{16}{20} = 10 \times \frac{4}{5} = 8\) litres
• Water removed = \(10 \times \frac{4}{20} = 10 \times \frac{1}{5} = 2\) litres

After removing 10 litres of mixture:

• Remaining Milk = \(16 - 8 = 8\) litres
• Remaining Water = \(4 - 2 = 2\) litres
• Total volume remaining = \(8 + 2 = 10\) litres

Now, 10 litres of pure milk are added to the vessel.

• Milk after adding pure milk = \(8 + 10 = 18\) litres
• Water after adding pure milk = \(2\) litres
• Total volume = \(18 + 2 = 20\) litres

Final Ratio Calculation

After the second process, the final quantity of milk is 18 litres and the final quantity of water is 2 litres.

The final ratio of Milk : Water is \(18 : 2\).

Simplifying the ratio by dividing both parts by the greatest common divisor (2):

\(\frac{18}{2} : \frac{2}{2} = 9 : 1\)

The final ratio of milk and water is 9 : 1.

Revision Table: Milk and Water Quantities

Additional Information on Mixture Problems

Mixture problems often involve calculating the amounts of components after removing a part of the mixture and adding another substance. A general formula can be used for dilution problems where a pure liquid is added to a mixture repeatedly:

If a container initially contains 'V' units of a mixture, and 'x' units are removed and replaced with pure liquid 'B' (the other component) 'n' times, then the final amount of liquid 'A' (the original component) is given by:

\(\text{Final amount of A} = \text{Initial amount of A} \left(1 - \frac{x}{V}\right)^n\)

In this problem, let liquid 'A' be Water, because pure Milk is added. Initial amount of Water = 8 litres Total Volume V = 20 litres Amount removed and replaced x = 10 litres Number of times repeated n = 2

Final amount of Water = \(8 \left(1 - \frac{10}{20}\right)^2 = 8 \left(1 - \frac{1}{2}\right)^2 = 8 \left(\frac{1}{2}\right)^2 = 8 \times \frac{1}{4} = 2\) litres.

The total volume remains 20 litres after each replacement. Final amount of Milk = Total Volume - Final amount of Water = \(20 - 2 = 18\) litres.

The final ratio of Milk : Water is \(18 : 2 = 9 : 1\). This confirms the step-by-step calculation.

This formula method is very efficient for problems involving repeated dilution by adding one of the pure components.