QUE.4:From a 500ml milk container 50ml is removed and 100ml water is added. Again 50ml is removed from the resultant solution after the previous operation and 100ml of water is added. Third time again 50ml is removed from the present solution and 100ml of water is added. What is the quantity of the milk in the solution now.

Let's solve this step-by-step:

### Initial Condition:
- **Milk in the container**: 500 ml
- **Water in the container**: 0 ml

### Step 1: Remove 50 ml of the original solution
- Since the solution only contains milk initially, removing 50 ml of solution means removing 50 ml of milk.

- **Milk left**: \( 500 - 50 = 450 \) ml
- **Water left**: 0 ml

### Step 2: Add 100 ml of water
- **Milk**: 450 ml
- **Water**: 0 + 100 = 100 ml

### Step 3: Remove 50 ml of the new solution
- The concentration of milk in the solution is \(\frac{450}{450 + 100} = \frac{450}{550} \approx 0.8182 \).

- Amount of milk removed: \( 50 \times 0.8182 = 40.91 \) ml (approximately).
- Amount of water removed: \( 50 - 40.91 = 9.09 \) ml (approximately).

- **Milk left**: \( 450 - 40.91 = 409.09 \) ml
- **Water left**: \( 100 - 9.09 = 90.91 \) ml

### Step 4: Add 100 ml of water
- **Milk**: 409.09 ml
- **Water**: 90.91 + 100 = 190.91 ml

### Step 5: Remove 50 ml of the new solution
- The concentration of milk in the solution is \(\frac{409.09}{409.09 + 190.91} = \frac{409.09}{600} \approx 0.6818 \).

- Amount of milk removed: \( 50 \times 0.6818 = 34.09 \) ml (approximately).
- Amount of water removed: \( 50 - 34.09 = 15.91 \) ml (approximately).

- **Milk left**: \( 409.09 - 34.09 = 375 \) ml
- **Water left**: \( 190.91 - 15.91 = 175 \) ml

### Step 6: Add 100 ml of water
- **Milk**: 375 ml
- **Water**: 175 + 100 = 275 ml

### Final Quantity of Milk in the Solution:
After all operations, the quantity of milk left in the solution is **375 ml**.

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### Do you want more details or have any questions?

### Related Questions:
1. What is the final concentration of milk in the solution?
2. If 50 ml is removed one more time, what will be the new quantity of milk?
3. How much water would be needed to make the final concentration of milk exactly 50%?
4. If the same procedure is repeated 2 more times, what would be the quantity of milk left?
5. How does this process affect the concentration of milk in each step?

### Tip:
Always calculate the new concentration after each addition and removal step to keep track of the changes in solution composition.

From a 500ml milk container 50ml is removed and 100ml water is added. Again 50ml is removed from the resultant solution after the previous operation and 100ml of water is added. Third time again 50ml is removed from the present solution and 100ml of water is added. What is the quantity of the milk in the solution now?

Solve a step-by-step problem involving sequential removal and replacement of milk with water. This problem illustrates key concepts such as ratios, concentrations, and how repeated operations affect the solution composition. Suitable for students in Grades 9-10.

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