To solve this problem, we need to determine how much water should be added to 60 liters of milk to change the price ratio from 3/2 liters per rupee to 15/8 liters per rupee.

Step 1: Establish the initial conditions.

• Initial quantity of milk = 60 liters
• Initial price of milk = 3/2 liters per rupee

This means for every 1 rupee, you get $\frac{3}{2}$ liters of milk. So, the cost per liter of milk initially is:

$\text{Cost per liter of milk} = \frac{1 \text{ rupee}}{\frac{3}{2} \text{ liters}} = \frac{2}{3} \text{ rupees per liter}.$

Step 2: Establish the desired condition.

The final mixture should be sold at the rate of $\frac{15}{8}$ liters per rupee, which means for every 1 rupee, you get $\frac{15}{8}$ liters. So, the cost per liter of the mixture is:

$\text{Cost per liter of mixture} = \frac{1 \text{ rupee}}{\frac{15}{8} \text{ liters}} = \frac{8}{15} \text{ rupees per liter}.$

Step 3: Set up the equation.

Let $x$ be the amount of water to be added. Since water is free, it has no cost, and the price per liter of the mixture will be the weighted average of the price per liter of milk and water.

Using the concept of alligation:

$\frac{\text{Cost per liter of mixture} - \text{Cost per liter of water}}{\text{Cost per liter of milk} - \text{Cost per liter of mixture}} = \frac{\text{Quantity of milk}}{\text{Quantity of water}}.$

Substituting the values:

$\frac{\frac{8}{15} - 0}{\frac{2}{3} - \frac{8}{15}} = \frac{60}{x}.$

Step 4: Solve the equation.

First, simplify the numerator and the denominator:

$\frac{8}{15} - 0 = \frac{8}{15},$

$\frac{2}{3} - \frac{8}{15} = \frac{10}{15} - \frac{8}{15} = \frac{2}{15}.$

Now, the equation becomes:

$\frac{\frac{8}{15}}{\frac{2}{15}} = \frac{60}{x},$

$\frac{8}{2} = \frac{60}{x},$

$4 = \frac{60}{x},$

$x = \frac{60}{4} = 15 \text{ liters}.$

Final Answer:

You need to add 15 liters of water to the 60 liters of milk.

---

Would you like more details, or do you have any questions?

Here are 5 related questions to expand your understanding:

1. How is the alligation method used to mix two quantities of different prices?
2. What is the significance of the price per liter in such problems?
3. How can the ratio of two solutions' prices be calculated using proportions?
4. Can you explain how the weighted average is applied in mixing problems?
5. How does adding water affect the overall cost of a solution?

Tip: In mixing problems, water is often treated as a zero-cost ingredient, which makes its effect on price calculations unique.