There are three vessels A, B and C, Vessel A and B filled with mixture of milk and Water in the ratio of 5 : 4.and 5:3 respectively. 25% of mixture from vessel A taken out and mixed in vessel C, which contains 45 l pure milk. If in resulting mixture milk is 250% more than water in vessel C and initial quantity of mixture in vessel B is 20 l less than that of mixture in vessel A, then find the quantity of milk in vessel B?

Options

A. A)180 litres
B. B)120 litres
C. C)80 litres
D. D)100 litres

Answer

To solve the problem step-by-step, we will break down the information provided and use it to find the quantity of milk in vessel B.

### Step 1: Define the quantities in vessels A and B

Let the quantity of mixture in vessel A be \( A \) liters. Since the quantity of mixture in vessel B is 20 liters less than that in vessel A, we can express the quantity in vessel B as:
\[ B = A - 20 \]

### Step 2: Determine the composition of mixtures in vessels A and B

For vessel A, the ratio of milk to water is 5:4. Therefore, the quantity of milk and water in vessel A can be expressed as:
- Milk in A = \( \frac{5}{9} A \)
- Water in A = \( \frac{4}{9} A \)

For vessel B, the ratio of milk to water is 5:3. Thus, the quantities can be expressed as:
- Milk in B = \( \frac{5}{8} B \)
- Water in B = \( \frac{3}{8} B \)

### Step 3: Calculate the mixture taken from vessel A

25% of the mixture from vessel A is taken out and mixed into vessel C. The quantity taken out from vessel A is:
\[ \text{Quantity taken out} = 0.25A \]

The composition of this quantity (milk and water) is:
- Milk taken from A = \( 0.25 \times \frac{5}{9} A = \frac{5}{36} A \)
- Water taken from A = \( 0.25 \times \frac{4}{9} A = \frac{4}{36} A = \frac{1}{9} A \)

### Step 4: Determine the composition of vessel C after mixing

Vessel C initially contains 45 liters of pure milk. After adding the mixture from vessel A:
- Total milk in C = \( 45 + \frac{5}{36} A \)
- Total water in C = \( \frac{1}{9} A \)

### Step 5: Set up the relationship between milk and water in vessel C

According to the problem, the amount of milk in vessel C is 250% more than the amount of water. This can be expressed as:
\[ \text{Milk in C} = \text{Water in C} + 2.5 \times \text{Water in C} \]
\[ 45 + \frac{5}{36} A = 3.5 \times \frac{1}{9} A \]

### Step 6: Simplify and solve for A

Substituting the water quantity:
\[ 45 + \frac{5}{36} A = \frac{3.5}{9} A \]

To eliminate fractions, multiply through by 36:
\[ 36 \times 45 + 5A = 14A \]
\[ 1620 + 5A = 14A \]
\[ 1620 = 14A - 5A \]
\[ 1620 = 9A \]
\[ A = \frac{1620}{9} = 180 \text{ liters} \]

### Step 7: Calculate the quantity in vessel B

Now, substituting \( A \) back to find \( B \):
\[ B = A - 20 = 180 - 20 = 160 \text{ liters} \]

### Step 8: Calculate the quantity of milk in vessel B

Using the ratio of milk in vessel B:
\[ \text{Milk in B} = \frac{5}{8} B = \frac{5}{8} \times 160 = 100 \text{ liters} \]

### Final Answer

The quantity of milk in vessel B is **100 liters**.

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