We are given that a milkman mixes two parts of a milk-water mixture with 60% concentration and one part with 90% concentration to form 100 liters of a new mixture. We need to determine how much water should be evaporated from this mixture to obtain a final concentration of 80%.

Step 1: Initial Mixture Calculation

Let’s first calculate the concentration of milk in the initial mixture.

• Mixture A (60% milk): Two parts of this mixture.
• Mixture B (90% milk): One part of this mixture.

The average concentration of the combined mixture can be calculated using the weighted average formula:

$\text{Concentration of new mixture} = \frac{2 \times 60 + 1 \times 90}{2 + 1}$ $\text{Concentration of new mixture} = \frac{120 + 90}{3} = \frac{210}{3} = 70\%$

So, the new mixture has 70% milk and 30% water.

Step 2: Quantity of Milk and Water in the 100 Liters of Mixture

Now, calculate the amount of milk and water in the 100 liters of mixture.

• Milk in the mixture: Since the mixture is 70% milk: $\text{Milk} = 0.7 \times 100 = 70 \text{ liters}$
• Water in the mixture: Since the mixture is 30% water: $\text{Water} = 0.3 \times 100 = 30 \text{ liters}$

Step 3: Final Concentration of 80% Milk

We are asked to evaporate water until the concentration of milk becomes 80%. Let’s assume $x$ liters of water need to be evaporated.

After evaporating $x$ liters of water, the amount of milk will remain the same (70 liters), but the total volume of the mixture will reduce. The new total volume will be $100 - x$ liters, and the concentration of milk should be 80%. Hence, the equation for the final concentration is:

$\frac{70}{100 - x} = 0.8$

Step 4: Solve for $x$

Solving the above equation for $x$:

$70 = 0.8 \times (100 - x)$ $70 = 80 - 0.8x$ $0.8x = 80 - 70$ $0.8x = 10$ $x = \frac{10}{0.8} = 12.5$

Step 5: Conclusion

The milkman needs to evaporate 12.5 liters of water from the 100-liter mixture to increase the concentration to 80%.

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Do you need more details on any step? Here are 5 related questions:

1. How would the process change if the final concentration target was 85% instead of 80%?
2. What if the milkman started with equal parts of both mixtures? How would the calculations change?
3. How can we use this method for other similar mixture problems?
4. What if the initial concentrations were in volume ratios other than 2:1?
5. Can you explain how we derive the weighted average for mixtures in general?

Tip: Always ensure the concentration units (percentage) and volumes are consistent when solving mixture problems.