A solution of milk and water has 75% milk. 20% of solution is removed and replaced with milk. Again 20% of solution is removed and replaced with milk. Find the percentage of milk in the solution now

Let's solve this problem step by step.

### Initial Setup:
You have a solution with **75% milk**. Let the initial volume of the solution be **100 units** for simplicity.

- Initially, there is \(75\%\) of milk, so the volume of milk is:
  \[
  75\ \text{units of milk} \text{ and } 25\ \text{units of water}.
  \]

### Step 1: First removal and replacement
- **20% of the solution** is removed. The total volume removed is \(20\%\) of 100 units, which equals:
  \[
  20\ \text{units}.
  \]
  
  Out of this, \(75\%\) is milk and \(25\%\) is water. So, the removed quantities are:
  \[
  \text{Milk removed} = 75\%\ \text{of}\ 20\ = 15\ \text{units}.
  \]
  \[
  \text{Water removed} = 25\%\ \text{of}\ 20\ = 5\ \text{units}.
  \]

- After removal, the remaining amounts of milk and water are:
  \[
  \text{Milk remaining} = 75 - 15 = 60\ \text{units},
  \]
  \[
  \text{Water remaining} = 25 - 5 = 20\ \text{units}.
  \]

- **Replacement**: The 20 units removed are replaced entirely by milk. So, the new volume of milk becomes:
  \[
  \text{Milk after replacement} = 60 + 20 = 80\ \text{units}.
  \]
  The total solution is still 100 units, so the water remains at 20 units.

### Step 2: Second removal and replacement
- Again, **20% of the solution** is removed. This is 20 units, which contains the new percentage of milk and water:
  - The percentage of milk after the first replacement is:
    \[
    \text{Percentage of milk} = \frac{80}{100} = 80\%.
    \]
  - So, in the 20 units removed:
    \[
    \text{Milk removed} = 80\%\ \text{of}\ 20 = 16\ \text{units},
    \]
    \[
    \text{Water removed} = 20\%\ \text{of}\ 20 = 4\ \text{units}.
    \]

- After the second removal, the remaining amounts of milk and water are:
  \[
  \text{Milk remaining} = 80 - 16 = 64\ \text{units},
  \]
  \[
  \text{Water remaining} = 20 - 4 = 16\ \text{units}.
  \]

- **Replacement**: The 20 units removed are again replaced by milk. So, the new volume of milk becomes:
  \[
  \text{Milk after replacement} = 64 + 20 = 84\ \text{units}.
  \]
  The total solution is still 100 units, with the water remaining at 16 units.

### Final Answer:
The total amount of milk is 84 units out of 100, so the percentage of milk in the solution is:
\[
\frac{84}{100} \times 100 = 84\%.
\]

Thus, the percentage of milk in the solution after the two replacements is **84%**.

---

Would you like further details or have any questions?

### Related Questions:
1. How would the result change if only 10% of the solution were removed and replaced?
2. What would happen if the initial percentage of milk was 50%?
3. How does the process differ if water is added instead of milk?
4. Can you generalize the formula for \( n \) repetitions of this process?
5. What is the effect of removing and replacing different percentages each time?

### Tip:
Always start with simple values (like 100 units) when dealing with percentages to make the calculations easier.

A solution of milk and water has 75% milk. 20% of the solution is removed and replaced with milk. Again 20% of the solution is removed and replaced with milk. Find the percentage of milk in the solution now.

Learn how to find the percentage of milk in a solution after sequentially removing and replacing 20% of the solution with milk. This step-by-step solution covers mixture problems and percentage calculations, suitable for students in Grades 9-10.

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