Vessel A contains a 360 liters mixture of milk and water in the ratio of 5:4 and vessel B contains the mixture of milk and water in the ratio of 7:5. Vessel C contains 60 liters of milk and 40 liters of water. If vessel A and B mixtures are mixed with vessel C, then the ratio of the milk to water in the resultant mixture is 4:3. Find the initial total quantity of vessel B.

We are tasked with finding the initial total quantity of the mixture in vessel B. Let's break down the given information:

- **Vessel A** contains a mixture of 360 liters of milk and water in the ratio \(5:4\). 
- **Vessel B** contains a mixture of milk and water in the ratio \(7:5\). We need to find the total amount of the mixture in vessel B.
- **Vessel C** contains 60 liters of milk and 40 liters of water.
- The resultant mixture of milk and water from all three vessels has a ratio of \(4:3\).

### Step 1: Calculate the amount of milk and water in vessel A
The total quantity of the mixture in vessel A is 360 liters, and the ratio of milk to water is \(5:4\). This means that out of every 9 parts (5 parts of milk + 4 parts of water), 5 parts are milk and 4 parts are water.

The total parts are 9. Therefore, the amount of milk and water in vessel A is:

- **Milk in vessel A** = \(\frac{5}{9} \times 360 = 200 \, \text{liters}\),
- **Water in vessel A** = \(\frac{4}{9} \times 360 = 160 \, \text{liters}\).

### Step 2: Calculate the amount of milk and water in vessel C
From the problem, vessel C contains:
- 60 liters of milk,
- 40 liters of water.

### Step 3: Let the total amount of mixture in vessel B be \(x\)
The ratio of milk to water in vessel B is \(7:5\). Therefore, the amounts of milk and water in vessel B are:

- **Milk in vessel B** = \(\frac{7}{12} \times x\),
- **Water in vessel B** = \(\frac{5}{12} \times x\).

### Step 4: Set up the equation for the resultant mixture
When vessels A, B, and C are mixed together, the ratio of milk to water in the resultant mixture is \(4:3\).

The total amount of milk in the resultant mixture is the sum of the milk in all three vessels:
\[
\text{Milk total} = 200 + \frac{7}{12}x + 60
\]

The total amount of water in the resultant mixture is the sum of the water in all three vessels:
\[
\text{Water total} = 160 + \frac{5}{12}x + 40
\]

The ratio of milk to water is given as \(4:3\), so:
\[
\frac{\text{Milk total}}{\text{Water total}} = \frac{4}{3}
\]

Substitute the expressions for milk and water:
\[
\frac{200 + \frac{7}{12}x + 60}{160 + \frac{5}{12}x + 40} = \frac{4}{3}
\]

### Step 5: Solve the equation
Simplify the equation:
\[
\frac{260 + \frac{7}{12}x}{200 + \frac{5}{12}x} = \frac{4}{3}
\]

Cross-multiply:
\[
3 \left( 260 + \frac{7}{12}x \right) = 4 \left( 200 + \frac{5}{12}x \right)
\]

Expand both sides:
\[
780 + \frac{21}{12}x = 800 + \frac{20}{12}x
\]

Multiply through by 12 to eliminate fractions:
\[
9360 + 21x = 9600 + 20x
\]

Simplify and solve for \(x\):
\[
21x - 20x = 9600 - 9360
\]
\[
x = 240
\]

### Step 6: Conclusion
The total quantity of the mixture in vessel B is \(240\) liters.

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Do you need any additional details or further clarification?

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Here are 5 related questions to expand on this:
1. How would the result change if the ratio of milk to water in vessel B were different?
2. What if vessel C had a different amount of milk and water? How would that affect the final ratio?
3. How can we generalize this approach for any number of vessels mixed together?
4. If the total final volume of the mixture was given instead of the ratio, how would you approach this problem?
5. What other methods could be used to solve this type of ratio and mixture problem?

**Tip**: Always break down mixture problems into parts (milk and water, in this case) and solve using ratio principles step by step.

Solve a problem involving three vessels containing milk and water mixtures, using algebra and ratios. Vessel A contains 360 liters with a 5:4 ratio, vessel B has a 7:5 ratio, and vessel C contains 60 liters of milk and 40 liters of water. Find the total quantity of vessel B such that the final mixture's ratio is 4:3.

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