Mixture Replacement: Calculating Remaining Milk

This problem involves calculating the final amount of a substance (original milk) remaining in a container after a series of removal and replacement operations.

Formula for Repeated Replacements

When a fixed amount of mixture is removed from a container and replaced with another liquid, the amount of the original substance remaining after n such operations can be calculated using the formula:

$ \text{Remaining Amount} = V_0 \times \left(1 - \frac{v}{V_0}\right)^n $

Where:

• $V_0$ is the initial volume of the substance (pure milk).
• $v$ is the volume of mixture removed and replaced in each step.
• $n$ is the number of times the operation is repeated.

Step-by-Step Calculation

Given values:

• Initial Volume ($V_0$) = 30 litres
• Volume Removed/Replaced ($v$) = 6 litres
• Number of Repetitions ($n$) = 4

Substitute these values into the formula:

$ \text{Remaining Milk} = 30 \times \left(1 - \frac{6}{30}\right)^4 $

Simplify the fraction inside the parenthesis:

$ 1 - \frac{6}{30} = 1 - \frac{1}{5} = \frac{4}{5} $

Now, calculate the remaining milk:

$ \text{Remaining Milk} = 30 \times \left(\frac{4}{5}\right)^4 $

$ \text{Remaining Milk} = 30 \times \frac{4^4}{5^4} $

$ \text{Remaining Milk} = 30 \times \frac{256}{625} $

$ \text{Remaining Milk} = \frac{30 \times 256}{625} = \frac{7680}{625} $

Simplify the fraction by dividing the numerator and denominator by 5:

$ \text{Remaining Milk} = \frac{1536}{125} \text{ litres} $

Convert the improper fraction to a mixed number:

$ \frac{1536}{125} = 12 \text{ with a remainder of } 36 $

So, the exact remaining amount is $12 \frac{36}{125}$ litres.

Approximation and Final Answer

The calculated amount is $12 \frac{36}{125}$ litres. To compare with the options, we can convert this to a decimal:

$ \frac{36}{125} = 0.288 $

Therefore, the remaining milk is approximately $12.288$ litres.

Let's examine the options provided:

• Option 1: $15\frac{78}{255}$ ≈ 15.31 L
• Option 2: $17\frac{28}{255}$ ≈ 17.11 L
• Option 3: $12\frac{74}{255}$ ≈ 12.29 L
• Option 4: $15\frac{58}{255}$ ≈ 15.23 L

The calculated value of approximately $12.288$ litres is closest to Option 3, $12\frac{74}{255}$ litres.