Question

A container contains only milk. 2/3 of the mixture is removed and replaced with water. This process is done once and then repeated another 3 times. What is the final ratio of milk and water?

Hint:

The formula \(V_{final} = V_{initial} \times (1 - x)^n\) is a powerful shortcut for all repeated replacement problems. Identify the initial amount, the fraction replaced (x), and the number of repetitions (n) to solve quickly.

Answer 1

Step 1: Start with milk at \(100\%\) of the container. Each replacement keeps \(\tfrac13\) (about \(33.33\%\)) of whatever milk is currently present, since \(\tfrac23\) is removed each time.
Step 2: Track the milk percentage round by round: after round 1, \(33.33\%\); after round 2, \(11.11\%\); after round 3, \(3.70\%\); after round 4, \(1.2346\%\) (each is the previous value divided by \(3\)).
Step 3: As a clean fraction, \(1.2346\%=\tfrac{1}{81}\) of the original volume is milk, so water makes up the remaining \(\tfrac{80}{81}\).
Final Answer: milk : water \[ =\frac1{81}:\frac{80}{81}=\boxed{1:80} \]