 A mixture of 50 litres of milk and water contains 12% water. If an equal quantity of water is added twice to make it 20% in the new mixture, then how many litres of water was added each time?

We are given the following information:

- A 50-litre mixture of milk and water contains 12% water.
- Water is added twice in equal quantities.
- After the second addition, the water content becomes 20%.

We are to determine how many litres of water were added each time.

### Step-by-step solution:

1. **Initial water content:**
   - The initial mixture contains 12% water.
   - Therefore, the amount of water in the 50-litre mixture is:
     \[
     \text{Water} = 12\% \text{ of } 50 = \frac{12}{100} \times 50 = 6 \text{ litres}.
     \]
     Thus, the initial mixture contains 6 litres of water and \( 50 - 6 = 44 \) litres of milk.

2. **Water added the first time:**
   - Let \( x \) be the amount of water added each time.
   - After adding \( x \) litres of water the first time, the new total quantity of the mixture is \( 50 + x \) litres.
   - The amount of water in this mixture becomes \( 6 + x \) litres.
   - The percentage of water in the new mixture is still unknown after the first addition, so we proceed to the second addition.

3. **Water added the second time:**
   - After adding \( x \) litres of water the second time, the new total quantity of the mixture becomes \( 50 + 2x \) litres.
   - The amount of water now becomes \( 6 + 2x \) litres.
   - We are told that after the second addition, the water content becomes 20%.
   - Therefore, we set up the equation for the percentage of water:
     \[
     \frac{6 + 2x}{50 + 2x} = \frac{20}{100}.
     \]

4. **Solve the equation:**
   We solve the equation:
   \[
   \frac{6 + 2x}{50 + 2x} = 0.2.
   \]
   Cross-multiply to eliminate the fraction:
   \[
   6 + 2x = 0.2 \times (50 + 2x).
   \]
   Expanding both sides:
   \[
   6 + 2x = 10 + 0.4x.
   \]
   Move all the terms involving \( x \) to one side:
   \[
   2x - 0.4x = 10 - 6.
   \]
   Simplify:
   \[
   1.6x = 4.
   \]
   Solving for \( x \):
   \[
   x = \frac{4}{1.6} = 2.5 \text{ litres}.
   \]

### Final Answer:
Each time, 2.5 litres of water was added.

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Would you like any further details or explanations? Here are some related questions to deepen your understanding:

1. What would happen if instead of adding water twice, it was added three times?
2. How would the solution change if the initial water content was 15% instead of 12%?
3. What would the final water percentage be if 3 litres of water were added each time instead of 2.5 litres?
4. Can we calculate the final amount of milk in the mixture after both additions of water?
5. How does the concept of ratios help in solving problems involving mixtures like this one?

**Tip:** Always double-check your calculations for consistency, especially when dealing with percentages and mixtures!

A mixture of 50 litres of milk and water contains 12% water. If an equal quantity of water is added twice to make it 20% in the new mixture, then how many litres of water was added each time?

Learn how to solve a problem involving the addition of water to a mixture to achieve a specific water percentage. This problem covers key algebraic concepts, including percentages and proportions, and is suitable for students in Grades 8-10.

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