10L water is added to vessel containing 40L of pure milk. 25L of the resultant mixture was then sold and some more quantity of pure milk and water was added to the vessel in the ratio 2:1. If the resultant respective ratio of milk and water in the vessel was 3:1, then what was the quantity of pure milk added in the vessel?

Let's break down the problem step by step:

### Step 1: Initial setup
- The vessel contains **40 liters of pure milk**.
- **10 liters of water** is added to this vessel.
- Therefore, the total volume of the mixture becomes:
  \[
  \text{Total mixture} = 40 \, \text{liters of milk} + 10 \, \text{liters of water} = 50 \, \text{liters of mixture}
  \]
- The ratio of milk to water is:
  \[
  \text{Milk : Water} = \frac{40}{10} = 4:1
  \]

### Step 2: Selling 25 liters of the mixture
- **25 liters** of this mixture is sold.
- Since the ratio of milk to water is **4:1**, out of the 25 liters, the quantities of milk and water sold are:
  - Milk sold = \(\frac{4}{5} \times 25 = 20 \, \text{liters}\)
  - Water sold = \(\frac{1}{5} \times 25 = 5 \, \text{liters}\)

### Step 3: Remaining mixture in the vessel
- After selling 25 liters, the remaining quantities of milk and water are:
  - Remaining milk = \(40 - 20 = 20 \, \text{liters}\)
  - Remaining water = \(10 - 5 = 5 \, \text{liters}\)

Thus, after selling the mixture, the remaining liquid in the vessel consists of 20 liters of milk and 5 liters of water.

### Step 4: Adding more milk and water in the ratio 2:1
- Let the amount of pure milk added be \(2x\) liters and the amount of water added be \(x\) liters. 
- So, the total quantity of milk becomes \(20 + 2x\) liters, and the total quantity of water becomes \(5 + x\) liters.

### Step 5: Final ratio of milk to water is 3:1
- According to the problem, after adding the milk and water, the ratio of milk to water in the vessel becomes 3:1. Therefore, we can set up the equation:
  \[
  \frac{20 + 2x}{5 + x} = 3
  \]
  
### Step 6: Solve for \(x\)
- Cross-multiply to solve for \(x\):
  \[
  20 + 2x = 3(5 + x)
  \]
  \[
  20 + 2x = 15 + 3x
  \]
  \[
  20 - 15 = 3x - 2x
  \]
  \[
  5 = x
  \]

### Step 7: Calculate the quantity of pure milk added
- Since \(x = 5\), the amount of pure milk added is:
  \[
  2x = 2 \times 5 = 10 \, \text{liters}
  \]

Thus, the quantity of pure milk added to the vessel is **10 liters**.

---

Would you like any further details or clarification?

Here are 5 related questions you might find interesting:
1. How do you calculate ratios when quantities change?
2. How does the ratio of components in a mixture change when some part is removed?
3. What are other methods to solve mixture and ratio problems?
4. How do changes in one component of a mixture affect the whole system?
5. What are other real-life applications of mixture problems in different fields?

**Tip**: In mixture problems, it's often helpful to break down each step clearly and use equations to maintain the relationships between the different components (like milk and water here).

Solve the mixture problem where water and milk are added and sold from a vessel. Learn how to calculate the amount of pure milk added using algebra and ratios, with a final milk-to-water ratio of 3:1. Suitable for students in Grades 9-11.

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