Calculating Milk Percentage in Mixture After Replacements

This problem involves a mixture of milk and water, where a portion of the mixture is repeatedly withdrawn and replaced with milk. We need to determine the final percentage of milk after this process is carried out multiple times.

Initial State of the Mixture

We start with a 100 litres mixture of milk and water in the ratio 2 ∶ 3.

• Total volume of mixture = 100 litres
• Ratio of Milk ∶ Water = 2 ∶ 3
• Total parts in the ratio = 2 + 3 = 5 parts
• Initial quantity of Milk = \(\frac{2}{5} \times 100 = 40\) litres
• Initial quantity of Water = \(\frac{3}{5} \times 100 = 60\) litres

Understanding the Replacement Process

In each step, 10 litres of the mixture are withdrawn, and then 10 litres of milk are added back. This process changes the composition of the mixture. The total volume of the mixture remains constant at 100 litres throughout the process.

The process is done a total of 3 times (initial withdrawal and replacement + 2 more times).

Applying the Formula for Repeated Replacement

When a quantity 'x' is withdrawn from a total volume 'V' of a mixture, and the same quantity 'x' of a pure substance (in this case, milk) is added back, and this process is repeated 'n' times, the amount of the other substance (water in this case) remaining after 'n' operations can be calculated using the formula:

\( \text{Amount of substance remaining} = \text{Initial amount} \times \left(1 - \frac{\text{Quantity withdrawn}}{\text{Total volume}}\right)^\text{Number of operations} \)

In this problem, we can calculate the amount of water remaining, as water is the substance that is only removed (as part of the mixture withdrawal) and not added back.

• Initial amount of Water (\(W_{initial}\)) = 60 litres
• Quantity withdrawn in each step (\(x\)) = 10 litres
• Total volume of the mixture (\(V\)) = 100 litres
• Number of operations (\(n\)) = 3

Using the formula, the amount of water remaining after 3 operations (\(W_{final}\)) is:

\( W_{final} = W_{initial} \times \left(1 - \frac{x}{V}\right)^n \)

\( W_{final} = 60 \times \left(1 - \frac{10}{100}\right)^3 \)

\( W_{final} = 60 \times \left(1 - \frac{1}{10}\right)^3 \)

\( W_{final} = 60 \times \left(\frac{9}{10}\right)^3 \)

\( W_{final} = 60 \times \frac{9^3}{10^3} \)

\( W_{final} = 60 \times \frac{729}{1000} \)

\( W_{final} = \frac{60 \times 729}{1000} \)

\( W_{final} = \frac{43740}{1000} \)

\( W_{final} = 43.74 \) litres

Calculating Final Amount of Milk

After the 3 operations, the total volume of the mixture is still 100 litres. The final mixture consists of the remaining water and the final amount of milk.

• Total final mixture volume = 100 litres
• Final amount of Water = 43.74 litres

Final amount of Milk = Total final mixture volume - Final amount of Water

Final amount of Milk = \(100 - 43.74\)

Final amount of Milk = \(56.26\) litres

Calculating Percentage of Milk in Final Mixture

The percentage of milk in the final mixture is calculated as:

\( \text{Percentage of Milk} = \left(\frac{\text{Final amount of Milk}}{\text{Total final mixture volume}}\right) \times 100 \)

\( \text{Percentage of Milk} = \left(\frac{56.26}{100}\right) \times 100 \)

\( \text{Percentage of Milk} = 56.26\% \)

Thus, the percentage of milk in the final mixture is 56.26 percent.

Revision Table: Mixture Replacement Concepts

Understanding the key concepts in mixture problems with replacement is crucial.

• Mixture: A substance formed by mixing two or more components.
• Ratio: Expresses the relative amount of different components in a mixture.
• Withdrawal: Removing a portion of the mixture; the removed quantity has the same composition ratio as the current mixture.
• Replacement: Adding a pure substance or another mixture back to restore the original volume (or change the volume as specified).
• Repeated Replacement: Performing the withdrawal and replacement process multiple times. This changes the concentration of the original components.

Additional Information: Mixture Problems and Formulas

Mixture problems often involve calculating concentrations or quantities after mixing or replacing components. The formula used in this solution is particularly useful for repeated replacement where a pure substance is added back.

• If 'x' quantity is withdrawn and replaced with another substance 'n' times from a total volume 'V', the final quantity of the original component is given by: \( Q_{final} = Q_{initial} \times \left(1 - \frac{x}{V}\right)^n \).
• This formula applies to the component that is *not* used for replacement. If the replacement is done with one of the original components, this formula is used for the *other* component.
• The total volume often remains constant in replacement problems, simplifying the calculation of the final percentage once the quantity of one component is known.