Final Acid Percentage Calculation

The question asks for the final percentage of acid in a 60-litre mixture after a specific operation is performed twice. The operation involves removing 20% of the mixture and replacing it with water.

Initial Acid Concentration

First, determine the initial amount and concentration of acid.

• Total initial mixture: 60 L
• Initial ratio of Acid to Water: 7:3
• Total ratio parts: $7 + 3 = 10$

Calculate the initial quantity of acid:

Initial Acid Quantity = $\frac{\text{Acid parts}}{\text{Total parts}} \times \text{Total Volume}$

Initial Acid Quantity = $\frac{7}{10} \times 60 \text{ L} = 42 \text{ L}$

Calculate the initial concentration of acid:

Initial Acid Concentration = $\frac{\text{Initial Acid Quantity}}{\text{Total Volume}} \times 100\%$

Initial Acid Concentration = $\frac{42 \text{ L}}{60 \text{ L}} \times 100\% = 0.70 \times 100\% = 70\%$

Acid Concentration After Replacements

Each operation involves removing 20% of the current mixture and adding water. This means 80% of the existing mixture remains.

The fraction of the acid remaining after one such operation is $(1 - 0.20) = 0.80$.

This operation is repeated once more, meaning it occurs a total of two times (N=2).

The formula to find the final concentration of the component (acid) is:

$ \text{Final Concentration} = \text{Initial Concentration} \times (1 - \text{Fraction Removed})^N $

Calculation

Using the formula with Initial Concentration = 0.70, Fraction Removed = 0.20, and N = 2:

$ \text{Final Acid Concentration} = 0.70 \times (1 - 0.20)^2 $

$ \text{Final Acid Concentration} = 0.70 \times (0.80)^2 $

$ \text{Final Acid Concentration} = 0.70 \times 0.64 $

$ \text{Final Acid Concentration} = 0.448 $

Convert the final concentration to a percentage:

$ \text{Final Acid Percentage} = 0.448 \times 100\% = 44.8\% $

Therefore, the final percentage of acid in the mixture is 44.8%.