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:

Assume a convenient starting volume for the milk, a number divisible by 3 several times over, such as 81 litres. This keeps every round of removal and replacement in whole numbers, making it easy to track the milk volume step by step instead of working with fractions throughout.

Answer 1

Approach: Each replacement scales the remaining milk by the same factor, so after \(n\) rounds the milk fraction is just that factor to the \(n\)th power. "Done once and repeated another 3 times" means \(n = 4\).

Step 1: Take the container volume as 1. Removing \(\tfrac23\) of the mixture and topping up with water keeps a fraction \(1 - \tfrac23 = \tfrac13\) of whatever milk was present.

Step 2: After 4 such operations, milk remaining \[ = \left(\frac13\right)^4 = \frac{1}{81}. \]

Step 3: Total volume stays 1, so water \(= 1 - \dfrac{1}{81} = \dfrac{80}{81}\).

Step 4: Ratio milk : water \(= \dfrac{1}{81} : \dfrac{80}{81} = 1 : 80\).

Final Answer: Final milk : water \(= \boxed{1 : 80}\).

Answer 2

Approach: Instead of deriving the exponent from scratch, apply the standard replacement-dilution result directly: if a fraction \(f\) of a solution is repeatedly removed and replaced with the other liquid, the fraction of the original substance remaining after \(n\) rounds is \((1-f)^n\).

Here, \(f=\tfrac23\) is removed each time (and replaced with water), so the fraction of milk surviving after \(n=4\) rounds (one round, then three more) is \[ \left(1-\frac23\right)^4=\left(\frac13\right)^4=\frac1{81}. \]
So milk \(:\) total \(=1:81\), meaning water \(=81-1=80\) parts. The final ratio of milk to water is therefore \[ \boxed{1:80} \]

Answer 3

Concept:

• Instead of deriving a general formula, assume a convenient concrete starting volume for the milk, one that stays a whole number every time it is divided by 3, such as $81$ litres (since $3^4=81$).
• Track the actual volume of milk, not a fraction, through each of the four rounds directly, removing $\dfrac23$ of the current mixture each time and topping up with water back to the original volume.

Step 1: Choose a convenient starting volume and do round 1.
Start with $81$ litres, all milk. Removing $\dfrac23$ of the mixture removes $54$ litres of milk, leaving $27$ litres of milk; topping up with $54$ litres of water restores the volume to $81$ litres (milk $=27$, water $=54$).

Step 2: Round 2.
Milk is now $27$ out of $81$ litres. Removing $\dfrac23$ of the $81$-litre mixture removes $\dfrac23\times27=18$ litres of milk, leaving $27-18=9$ litres of milk. Topping up with water restores the total to $81$ litres (milk $=9$, water $=72$).

Step 3: Round 3 and round 4.
Round 3: milk is $9$ out of $81$; removing $\dfrac23\times9=6$ litres of milk leaves $9-6=3$ litres of milk (milk $=3$, water $=78$).
Round 4: milk is $3$ out of $81$; removing $\dfrac23\times3=2$ litres of milk leaves $3-2=1$ litre of milk (milk $=1$, water $=80$).

Step 4: Read off the final ratio.
After all four rounds, milk $=1$ litre and water $=80$ litres out of the original $81$ litres.

Final Answer: Milk : Water $=1:80$