Let's analyze the given problem step by step.

Problem:

• A jar contains x ml of water and 500 ml of alcohol.
• 20% of the mixture is removed and replaced with water.
• This process is repeated once more.
• After these two steps, the volume of water in the mixture becomes 305 ml.

Key Points:

• Initially, the total volume of the mixture is $x + 500$ ml.
• In each step, 20% of the mixture is removed, and the same volume of water is added back.
• After two iterations, the water content is given as 305 ml.

Step 1: First iteration

• Total volume of the mixture = $x + 500$ ml.
• 20% of the mixture is removed, which is $0.2(x + 500)$ ml.
• The amount of water removed in this process is $0.2x$ ml (because 20% of the water will be removed).
• The remaining water after this step is $x - 0.2x = 0.8x$ ml.
• The volume removed is replaced with water, so we add back $0.2(x + 500)$ ml of water. Now, the new water content is: $0.8x + 0.2(x + 500)$ Simplifying: $0.8x + 0.2x + 100 = x + 100 \, \text{ml of water after the first step}.$

Step 2: Second iteration

• Again, 20% of the mixture is removed. This time, the water content is $x + 100$ ml, and the total volume of the mixture is still $x + 500$ ml.
• The amount of water removed in this process is $0.2(x + 100)$ ml.
• So, the remaining water after the second removal is: $(x + 100) - 0.2(x + 100) = 0.8(x + 100) = 0.8x + 80 \, \text{ml}.$
• We again add the same volume of water back, which is $0.2(x + 500) = 100$ ml of water. Now, the total amount of water after the second step is: $0.8x + 80 + 100 = 0.8x + 180 \, \text{ml}.$

Step 3: Using the final water content

• It is given that after the second iteration, the volume of water is 305 ml. Therefore: $0.8x + 180 = 305.$
• Solving for $x$: $0.8x = 305 - 180 = 125,$ $x = \frac{125}{0.8} = 156.25 \, \text{ml}.$

Thus, the initial amount of water in the jar was 156.25 ml.

The total volume of liquid left in the jar now is $156.25 \, \text{ml (water)} + 500 \, \text{ml (alcohol)} = 656.25 \, \text{ml}$.

Final Answer:

• The volume of liquid left in the jar now is 656.25 ml.

Would you like further details on any part of the process or have any other questions?

Here are 5 related questions:

1. What if the replacement was done 3 times instead of 2?
2. How would the process change if 30% of the mixture was removed each time?
3. What would happen if instead of water, alcohol was added in the second step?
4. How would the solution change if the amount of alcohol initially was different?
5. What is the mathematical formula to generalize such processes?

Tip: In mixture problems, tracking the proportions of each substance after each step simplifies solving for unknowns.