A milkman takes two parts of a milk water mixture of 60% concentration and one part of a milk water mixture of 90% concentration and takes 100 litres of the new mixture. How much water should he evaporate from 100 litres of the mixture to make it of 80% concentration?

Let's solve the problem step by step:

### Step 1: Calculate the amount of milk and water in the mixture

We are given two mixtures:
- One is 60% milk (40% water).
- The other is 90% milk (10% water).

The milkman takes two parts of the first mixture and one part of the second mixture, so let's determine how much milk and water are in each of these mixtures when combined.

#### First mixture (60% milk):
In 1 part of the first mixture, 60% is milk and 40% is water.
Since he takes two parts of the first mixture, the total milk and water from the first mixture are:
- Milk from first mixture = \( 2 \times 0.60 = 1.2 \) parts.
- Water from first mixture = \( 2 \times 0.40 = 0.8 \) parts.

#### Second mixture (90% milk):
In 1 part of the second mixture, 90% is milk and 10% is water.
So, the total milk and water from the second mixture are:
- Milk from second mixture = \( 1 \times 0.90 = 0.90 \) parts.
- Water from second mixture = \( 1 \times 0.10 = 0.10 \) parts.

### Step 2: Calculate the total amount of milk and water in the new mixture

Now, let's sum up the milk and water from both mixtures:
- Total milk = \( 1.2 + 0.9 = 2.1 \) parts.
- Total water = \( 0.8 + 0.1 = 0.9 \) parts.

Thus, the total mixture is composed of 3 parts (2 from the first mixture + 1 from the second mixture):
- Milk = \( 2.1/3 \) of the mixture.
- Water = \( 0.9/3 \) of the mixture.

So the milk concentration is:
\[
\text{Milk percentage in new mixture} = \frac{2.1}{3} = 70\%.
\]
The new mixture has 70% milk and 30% water.

### Step 3: Find how much water to evaporate

The milkman wants to concentrate the milk to 80%. The initial amount of the mixture is 100 liters. Currently, the mixture contains 70% milk and 30% water. We need to remove enough water to increase the milk concentration to 80%.

Let \( x \) be the amount of water that needs to be evaporated. After evaporating \( x \) liters of water, the new volume of the mixture will be \( 100 - x \) liters, and the amount of milk will remain the same (since evaporating water doesn't affect the amount of milk).

The amount of milk in the original mixture is:
\[
\text{Milk in 100 liters} = 100 \times 0.70 = 70 \text{ liters}.
\]
We want this 70 liters of milk to be 80% of the new mixture:
\[
\frac{70}{100 - x} = 0.80.
\]

### Step 4: Solve for \( x \)
\[
70 = 0.80 \times (100 - x),
\]
\[
70 = 80 - 0.80x,
\]
\[
0.80x = 80 - 70 = 10,
\]
\[
x = \frac{10}{0.80} = 12.5 \text{ liters}.
\]

### Final Answer:
The milkman should evaporate **12.5 liters** of water from the 100 liters of the mixture to make the concentration 80%.

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Would you like to go over any details or ask questions? Here are five related questions you might find interesting:
1. How do you calculate the concentration after mixing multiple solutions?
2. What happens to the concentration if you dilute a solution by adding water?
3. How does the percentage change if you evaporate water from a mixture?
4. How can you calculate the new volume of liquid after adding or removing water?
5. How do you handle similar problems involving different types of mixtures?

**Tip**: When working with mixtures, always track the total volume and the proportion of each component to maintain accurate concentration calculations.

Learn how to solve a milk-water mixture problem where a milkman takes 100 liters of a 70% milk concentration and needs to evaporate water to achieve 80%. This detailed solution involves mixture problems, percentages, and linear equations, suitable for Grades 10-12.

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