a milkman adds 100 ml of water to every 1 liter of milk. he claims to make 25% profit find the actual gain percentage

To find the actual gain percentage, let's break down the problem step by step.

### Given:
- The milkman adds 100 ml of water to every 1 liter (1000 ml) of milk.
- He claims to make a 25% profit.

### Step 1: Determine the total volume of the mixture.
When the milkman adds 100 ml of water to 1 liter (1000 ml) of milk, the total volume becomes:
\[
\text{Total Volume} = 1000 \, \text{ml (milk)} + 100 \, \text{ml (water)} = 1100 \, \text{ml}
\]

### Step 2: Understand the claimed profit.
The milkman claims a 25% profit. This means that he sells the 1100 ml mixture at a price equivalent to 25% more than the price of 1 liter of pure milk.

### Step 3: Determine the cost price (CP) and selling price (SP).
- Let the cost price (CP) of 1 liter of pure milk be \( x \) (in any currency).
- The selling price (SP) according to his claim would be:
\[
\text{SP} = x + 0.25x = 1.25x
\]

### Step 4: Calculate the actual gain percentage.
The milkman is selling 1100 ml of the mixture at the price of 1 liter of pure milk plus 25% profit. Since water is added for free, the cost price of the mixture is still \( x \), but the selling price (SP) is for 1100 ml:
\[
\text{SP for 1100 ml} = 1.25x
\]

The cost price for 1100 ml would just be for 1000 ml (1 liter) of milk because the water is free:
\[
\text{Actual Cost Price for 1100 ml} = x
\]

Now, the gain is the difference between the selling price and the actual cost price:
\[
\text{Gain} = \text{SP} - \text{Actual CP} = 1.25x - x = 0.25x
\]

### Step 5: Calculate the actual gain percentage.
To find the actual gain percentage:
\[
\text{Actual Gain Percentage} = \left(\frac{\text{Gain}}{\text{Actual CP}}\right) \times 100 = \left(\frac{0.25x}{x}\right) \times 100 = 25\%
\]

However, this gain is based on 1000 ml of milk, but since he's selling 1100 ml:

### Step 6: Adjust for the volume of the mixture:
Since the milkman is selling 1100 ml instead of 1000 ml, the actual cost for this volume is \( x \). The selling price remains \( 1.25x \).

The effective profit on the 1 liter (1000 ml) of milk, given that 1100 ml is sold for 1.25x:
\[
\text{Actual Profit Percentage} = \left(\frac{1.25x - x}{x}\right) \times \frac{1000}{1100} \times 100 = \frac{0.25x}{x} \times \frac{1000}{1100} \times 100 = \frac{0.25 \times 100000}{1100} = \frac{25000}{11} \approx 22.73\%
\]

### Final Answer:
The actual gain percentage is approximately **22.73%**.

Would you like more details or have any questions? Here are 5 related questions:

1. How would the gain percentage change if the milkman added 200 ml of water instead?
2. What would be the profit percentage if no water was added?
3. If the selling price was based on 1 liter of the mixture instead of 1 liter of pure milk, how would the profit change?
4. What if the milkman sold the mixture at a 30% profit instead of 25%?
5. How much water should be added to achieve exactly a 25% actual gain?

**Tip:** When dealing with percentages and mixtures, always clarify what the percentage is based on to avoid misunderstandings.

A milkman adds 100 ml of water to every 1 liter of milk. He claims to make 25% profit. Find the actual gain percentage.

Learn how to calculate the actual gain percentage when a milkman adds 100 ml of water to every 1 liter of milk and claims to make a 25% profit. This problem covers key concepts such as mixtures, profit and loss, and percentage calculations. Suitable for students in Grades 9-12.

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