The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water in the mixture?

Given:

• Initial wine = 8 liters
• Initial water = 32 liters
• Total mixture = 8 + 32 = 40 liters
• New wine concentration = 30%

Let x = liters of mixture to be replaced with pure wine.

After removing x liters of mixture:

• Remaining mixture = 40 - x liters
• Water left = (32 / 40) × (40 - x) = 32 - (32x / 40)

We replace x liters with pure wine, so total wine becomes:

• 8 - (8x / 40) + x = 8 - (x / 5) + x = 8 + (4x / 5)

Now, new mixture = 40 liters and new wine quantity = 8 + (4x / 5)

Set up the equation:

(8 + 4x / 5) / 40 = 30 / 100 = 3 / 10

Step-by-step calculation:

(8 + 4x/5) / 40 = 3 / 10
=> (8 + 4x/5) = 12
=> 4x/5 = 4
=> x = 5

Answer: 5 liters of mixture should be replaced with pure wine.