a vessel has a milk solution in which milk and water are in the ratio 4:1. by addition of water to it, milk solution with milk and water in the ratio 4:3 was formed. on replacing 14l of this solution with pure milk the ratio of milk and water changed to 5:3. what is the volume of water added


We are given a vessel containing a milk solution with an initial ratio of milk to water as \( 4:1 \). After adding water, the ratio of milk to water changes to \( 4:3 \). Further, after replacing 14 liters of the new solution with pure milk, the ratio becomes \( 5:3 \). We need to determine the volume of water added during the process.

### Step-by-Step Solution:

#### Step 1: Initial composition
Let the initial volume of the solution be \( V \). The initial ratio of milk to water is \( 4:1 \), meaning:

- Milk volume = \( \frac{4}{5}V \)
- Water volume = \( \frac{1}{5}V \)

#### Step 2: Composition after adding water
After water is added, the ratio of milk to water becomes \( 4:3 \). Let the volume of water added be \( x \). So, the total water volume becomes \( \frac{1}{5}V + x \), and the total volume of the solution becomes \( V + x \).

Since the new ratio of milk to water is \( 4:3 \), we can set up the equation:

\[
\frac{\frac{4}{5}V}{\frac{1}{5}V + x} = \frac{4}{3}
\]

Cross-multiplying, we get:

\[
3 \times \frac{4}{5}V = 4 \times \left( \frac{1}{5}V + x \right)
\]

Simplifying:

\[
\frac{12}{5}V = \frac{4}{5}V + 4x
\]

Subtract \( \frac{4}{5}V \) from both sides:

\[
\frac{8}{5}V = 4x
\]

Solving for \( x \):

\[
x = \frac{8}{20}V = \frac{2}{5}V
\]

So, the volume of water added is \( \frac{2}{5}V \).

#### Step 3: After replacing 14 liters of the solution with milk
After 14 liters of the new solution is replaced with pure milk, the ratio of milk to water becomes \( 5:3 \).

At this stage:
- Total volume of the solution is still \( V + x = V + \frac{2}{5}V = \frac{7}{5}V \).
- Before the replacement, the amount of milk was \( \frac{4}{7} \times \frac{7}{5}V = \frac{4}{5}V \), and the water was \( \frac{3}{7} \times \frac{7}{5}V = \frac{3}{5}V \).

Now, we replace 14 liters of this solution with pure milk. Since the ratio is maintained in the replacement, \( \frac{4}{7} \times 14 = 8 \) liters of milk and \( \frac{3}{7} \times 14 = 6 \) liters of water are removed.

Thus, after replacing 14 liters:
- Milk volume becomes \( \frac{4}{5}V - 8 + 14 = \frac{4}{5}V + 6 \),
- Water volume becomes \( \frac{3}{5}V - 6 = \frac{3}{5}V - 6 \).

The new ratio of milk to water is given as \( 5:3 \). Hence:

\[
\frac{\frac{4}{5}V + 6}{\frac{3}{5}V - 6} = \frac{5}{3}
\]

Cross-multiplying:

\[
3 \times \left( \frac{4}{5}V + 6 \right) = 5 \times \left( \frac{3}{5}V - 6 \right)
\]

Simplifying:

\[
\frac{12}{5}V + 18 = \frac{15}{5}V - 30
\]

\[
\frac{12}{5}V - \frac{15}{5}V = -30 - 18
\]

\[
-\frac{3}{5}V = -48
\]

Solving for \( V \):

\[
V = \frac{48 \times 5}{3} = 80
\]

Thus, the initial volume of the solution is \( V = 80 \) liters.

#### Step 4: Volume of water added
We found that the volume of water added was \( \frac{2}{5}V \). Substituting \( V = 80 \):

\[
\text{Volume of water added} = \frac{2}{5} \times 80 = 32 \text{ liters}
\]

### Final Answer:
The volume of water added is **32 liters**.

---

Do you want further clarification or details on any part of the solution? 

Here are five related questions to explore:
1. How would the solution change if the initial ratio were different, say \(3:2\)?
2. What if instead of replacing the solution with pure milk, pure water was added?
3. How is the overall volume of the solution affected by continuous addition and replacement processes?
4. Can this problem be generalized to any initial and final ratios of milk and water?
5. How can this method be applied to real-life problems of mixing solutions with specific concentrations?

**Tip:** For problems involving ratios, it's helpful to express all components in terms of a single variable to simplify the equations.

A vessel has a milk solution in which milk and water are in the ratio 4:1. By addition of water to it, a milk solution with milk and water in the ratio 4:3 was formed. On replacing 14 liters of this solution with pure milk, the ratio of milk and water changed to 5:3. What is the volume of water added?

Solve a milk and water ratio problem where a vessel contains a solution initially in the ratio 4:1. Learn how adding water changes the ratio to 4:3 and further how replacing 14 liters of the solution affects the milk-to-water ratio. Understand the detailed step-by-step process and key algebraic concepts involved.

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