A dairy container initially contains $80\text{ liters}$ of pure milk. From this container, $8\text{ liters}$ of milk is taken out and replaced with water. This process of drawing $8\text{ liters}$ of liquid and replacing it with water is repeated two more times (for a total of three operations). The final quantity of pure milk remaining in the container in liters is ________ (round off to two decimal places).
Answer Key:
58.32
Step-by-Step Solution & Explanation
**Method 1: Successive Dilution Formula**
When an initial volume $V$ of pure liquid undergoes replacement of volume $x$ with water repeated $n$ times, the remaining quantity of pure liquid is:
$Q_n = V \left(1 - \frac{x}{V}\right)^n$
Here $V = 80\text{ L}$, $x = 8\text{ L}$, and $n = 3$:
$1 - \frac{x}{V} = 1 - \frac{8}{80} = 1 - 0.10 = 0.90$
$Q_3 = 80 \times (0.90)^3 = 80 \times 0.729 = 58.32\text{ liters}$
**Method 2: Step-by-Step Replacement**
- After 1st operation: Milk remaining $= 80 - 8 = 72\text{ L}$.
- In 2nd operation: $8\text{ L}$ removed has milk fraction $\frac{72}{80} = 0.9$. Milk removed $= 8 \times 0.9 = 7.2\text{ L}$. Remaining milk $= 72 - 7.2 = 64.8\text{ L}$.
- In 3rd operation: Milk removed $= 8 \times \frac{64.8}{80} = 8 \times 0.81 = 6.48\text{ L}$. Remaining milk $= 64.8 - 6.48 = 58.32\text{ L}$.