A dishonest milkman dilutes milk by mixing water in it and then sells the diluted milk at a price of 20% higher than the price at which he purchased the milk. lf he makes an overall profit of 50% in this manner, then how many ml of water does he add to every liter of milk?

Let the cost price of 1 liter of pure milk be \( C \).

### Step 1: Define variables
- Let the amount of water added to 1 liter of milk be \( x \) ml. 
- Therefore, the total volume of the mixture is \( (1000 + x) \) ml.
- The selling price of the mixture is 20% more than the cost price, so:
  \[
  \text{Selling price of 1 liter of pure milk} = 1.2C
  \]

### Step 2: Calculate the overall profit
- The milkman achieves a 50% profit overall. The total cost price of the milk is \( C \), but he earns 50% profit, meaning the selling price of the mixture is:
  \[
  \text{Selling price of the mixture} = 1.5C
  \]

### Step 3: Relate the selling price to the diluted mixture
The milkman sells \( (1000 + x) \) ml of diluted milk at a price that gives him a profit of 50%. However, he is selling the milk at the rate of \( 1.2C \) per 1000 ml of pure milk. Therefore, the selling price for the diluted mixture is:
  \[
  \text{Selling price of the mixture} = \frac{1000}{1000 + x} \times 1.2C
  \]

### Step 4: Equate the selling price expressions
We now equate the two expressions for the selling price:
  \[
  1.5C = \frac{1000}{1000 + x} \times 1.2C
  \]

Canceling \( C \) from both sides:
  \[
  1.5 = \frac{1000}{1000 + x} \times 1.2
  \]

### Step 5: Solve for \( x \)
Multiply both sides by \( (1000 + x) \) to clear the denominator:
  \[
  1.5(1000 + x) = 1200
  \]

Expand the left-hand side:
  \[
  1500 + 1.5x = 1200
  \]

Subtract 1500 from both sides:
  \[
  1.5x = 1200 - 1500
  \]
  \[
  1.5x = -300
  \]

Solve for \( x \):
  \[
  x = \frac{-300}{1.5} = -200
  \]

Thus, \( x = 200 \) ml.

### Conclusion:
The milkman adds **200 ml of water** to every liter (1000 ml) of milk.

Would you like more details or have any questions?

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Here are 5 questions to expand on the concept:

1. How would the profit change if the milkman added 300 ml of water instead?
2. What if the milkman sells the diluted milk at 30% higher than the cost price?
3. How can the problem change if the milkman used different quantities of water and milk mixtures?
4. What is the maximum amount of water that can be added to achieve a 100% profit?
5. How does changing the percentage of profit affect the total selling price?

**Tip:** In problems involving profit, it's essential to link the selling price and cost price with the profit percentage.

A dishonest milkman dilutes milk by mixing water in it and then sells the diluted milk at a price of 20% higher than the price at which he purchased the milk. If he makes an overall profit of 50% in this manner, then how many ml of water does he add to every liter of milk?

Solve a problem involving a dishonest milkman who dilutes milk with water and sells it at a 20% higher price. Learn how he makes a 50% profit and how to calculate the amount of water added. This step-by-step solution covers profit and loss, percentage calculations, and algebraic reasoning. Suitable for students in Grades 9-10.

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