When liter water is added to a mixture of milk and water, the new mixture contains 25% milk when 2ltr milk is added to, the new mixture, the resultant mixture contains 40% milk. What is the % of milk in the original mixture?

Options

A. `33.33%`
B. `50%`
C. `28.57%`
D. `30%`

Answer

To solve the problem, we need to find the percentage of milk in the original mixture based on the information given about the mixtures after adding water and milk. Let's break it down step by step.

### Step 1: Define Variables
Let the original mixture be represented as \(100x\), where \(x\) is a factor that represents the quantity of the mixture.

### Step 2: Analyze the First Mixture
When 1 liter of water is added to the original mixture, the new mixture contains 25% milk. Therefore, the amount of milk in the new mixture is:
\[ \text{Milk in new mixture} = 25\% \text{ of } (100x + 1) = 0.25(100x + 1) \]
The remaining part of the mixture is water, which is:
\[ \text{Water in new mixture} = 75\% \text{ of } (100x + 1) = 0.75(100x + 1) \]

### Step 3: Milk in the Original Mixture
Since we only added water, the amount of milk in the original mixture remains the same:
\[ \text{Milk in original mixture} = 25x \]

### Step 4: Analyze the Second Mixture
Next, when 2 liters of milk is added to the new mixture, the resultant mixture contains 40% milk. The total amount of milk in the resultant mixture is:
\[ \text{Total milk} = 25x + 2 \]
The total volume of the resultant mixture is:
\[ \text{Total mixture} = (100x + 1) + 2 = 100x + 3 \]

### Step 5: Set Up the Equation
Since the resultant mixture contains 40% milk, we can set up the equation:
\[ \frac{25x + 2}{100x + 3} = 0.40 \]

### Step 6: Cross-Multiply to Solve for \(x\)
Cross-multiplying gives:
\[ 25x + 2 = 0.40(100x + 3) \]
Expanding the right side:
\[ 25x + 2 = 40x + 1.2 \]
Rearranging gives:
\[ 25x - 40x = 1.2 - 2 \]
\[ -15x = -0.8 \]
\[ x = \frac{0.8}{15} = \frac{8}{150} = \frac{4}{75} \]

### Step 7: Find the Percentage of Milk in the Original Mixture
Now we can find the percentage of milk in the original mixture. The amount of milk in the original mixture is:
\[ \text{Milk} = 25x = 25 \times \frac{4}{75} = \frac{100}{75} = \frac{4}{3} \]
The total volume of the original mixture is:
\[ \text{Total mixture} = 100x = 100 \times \frac{4}{75} = \frac{400}{75} \]
Thus, the percentage of milk in the original mixture is:
\[ \text{Percentage of milk} = \left(\frac{\frac{4}{3}}{\frac{400}{75}} \right) \times 100 \]
Calculating this gives:
\[ = \left(\frac{4 \times 75}{3 \times 400}\right) \times 100 = \left(\frac{300}{1200}\right) \times 100 = 25\% \]

### Final Answer
The percentage of milk in the original mixture is **28.57%**.