A container is full of milk, `1/4`th of the milk is taken out and is replaced with water, and this process is repeated 3 times and 297 l of milk is finally left in the container, find the capacity of container? (in l)

Options

A. A)`528`
B. B)`644`
C. C)`740`
D. D)`704`

Answer

To solve the problem, we need to determine the capacity of the container after a series of operations where milk is removed and replaced with water. Here’s a step-by-step solution:

### Step 1: Understand the Process
Initially, the container is full of milk. When `1/4` of the milk is taken out, `3/4` of the milk remains in the container. This process is repeated 3 times.

### Step 2: Calculate the Remaining Milk After Each Operation
Let the initial volume of milk in the container be \( V \) liters.

1. **After the first operation:**
- Milk remaining = \( \frac{3}{4}V \)

2. **After the second operation:**
- Milk taken out = \( \frac{1}{4} \times \frac{3}{4}V = \frac{3}{16}V \)
- Milk remaining = \( \frac{3}{4} \times \frac{3}{4}V = \frac{9}{16}V \)

3. **After the third operation:**
- Milk taken out = \( \frac{1}{4} \times \frac{9}{16}V = \frac{9}{64}V \)
- Milk remaining = \( \frac{3}{4} \times \frac{9}{16}V = \frac{27}{64}V \)

### Step 3: Set Up the Equation
After 3 operations, we know that the remaining milk is 297 liters:
\[ \frac{27}{64}V = 297 \]

### Step 4: Solve for \( V \)
To find the total capacity \( V \):
\[ V = 297 \times \frac{64}{27} \]

### Step 5: Calculate the Value
Now, we simplify the calculation:
\[ V = 297 \times \frac{64}{27} = 297 \div 27 \times 64 \]
Calculating \( 297 \div 27 \):
\[ 297 \div 27 = 11 \]
Now, substituting back:
\[ V = 11 \times 64 = 704 \]

### Final Answer
The capacity of the container is \( 704 \) liters.

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