There are two containers A and B. Container A is empty while container B is full with a mixture of milk and water in the ratio 5:3. If 50% of mixture of B is poured in-container-A then A is `28 4/7 %` filled. Now, If A contains 10 L milk then find the capacity of A.

Options

A. A)48 lit.
B. B)54 lit
C. C)56 lit
D. D)36 lit

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the Mixture Ratio
Container B has a mixture of milk and water in the ratio of 5:3. This means for every 5 parts of milk, there are 3 parts of water.

**Hint:** Remember that the total parts in the mixture is the sum of the parts of milk and water.

### Step 2: Calculate the Total Parts in the Mixture
The total parts of the mixture in container B is:
\[ 5 \text{ (milk)} + 3 \text{ (water)} = 8 \text{ parts} \]

**Hint:** Always add the parts of each component to find the total parts in a mixture.

### Step 3: Determine the Amount of Mixture Poured into Container A
50% of the mixture from container B is poured into container A. Therefore, if we denote the total volume of container B as \( V_B \), the volume poured into container A is:
\[ \text{Volume poured} = 0.5 \times V_B \]

**Hint:** When calculating percentages, remember to convert them into decimal form.

### Step 4: Find the Amount of Milk and Water in the Poured Mixture
Since the mixture is in the ratio of 5:3, the amount of milk and water in the poured mixture can be calculated as follows:
- Milk in the poured mixture:
\[ \text{Milk} = \frac{5}{8} \times (0.5 \times V_B) = \frac{5V_B}{16} \]
- Water in the poured mixture:
\[ \text{Water} = \frac{3}{8} \times (0.5 \times V_B) = \frac{3V_B}{16} \]

**Hint:** Use the ratio to find the individual components in the mixture.

### Step 5: Total Volume in Container A
According to the problem, after pouring the mixture, container A is filled to \( 28 \frac{4}{7} \% \). This can be converted to a fraction:
\[ 28 \frac{4}{7} \% = \frac{200}{7} \% = \frac{200}{700} = \frac{2}{7} \]

This means that:
\[ \frac{2}{7} \text{ of the capacity of A} = 0.5 \times V_B \]

**Hint:** Converting percentages to fractions can simplify calculations.

### Step 6: Relate the Volume of Milk to the Capacity of Container A
We know that container A contains 10 L of milk. From the previous steps, we have:
\[ \text{Milk in A} = \frac{5V_B}{16} \]
Setting this equal to 10 L gives us:
\[ \frac{5V_B}{16} = 10 \]

**Hint:** Equating the amount of milk helps us find the total volume of the mixture.

### Step 7: Solve for \( V_B \)
To solve for \( V_B \):
\[ 5V_B = 10 \times 16 \]
\[ 5V_B = 160 \]
\[ V_B = \frac{160}{5} = 32 \text{ L} \]

**Hint:** Isolate the variable to find its value.

### Step 8: Find the Capacity of Container A
Now, substituting \( V_B \) back into the equation for the capacity of container A:
\[ \frac{2}{7} \text{ of capacity of A} = 0.5 \times 32 \]
\[ \frac{2}{7} \text{ of capacity of A} = 16 \]
Let the capacity of A be \( C_A \):
\[ \frac{2}{7} C_A = 16 \]
\[ C_A = 16 \times \frac{7}{2} = 56 \text{ L} \]

**Hint:** To find the total capacity, multiply by the reciprocal of the fraction.

### Final Answer
The capacity of container A is \( 56 \text{ L} \).