Acid with 25% and 40% concentration are mixed to get 60 litres of 30% concentration. How many litres of each kind is needed?

Answer

To solve the problem of mixing acids with different concentrations, we can set up a system of equations based on the information provided.

### Step-by-Step Solution:

1. **Define Variables**:
Let \( x \) be the amount of 25% acid in liters, and \( y \) be the amount of 40% acid in liters.

2. **Set Up the Equations**:
We know that the total volume of the mixture is 60 liters:
\[ x + y = 60 \]

3. **Set Up the Concentration Equation**:
The total amount of acid in the mixture must equal the amount of acid from each component. The total acid from the 25% solution is \( 0.25x \) and from the 40% solution is \( 0.4y \). The total acid in the 30% solution is \( 0.3 \times 60 = 18 \) liters. Therefore, we can write:
\[ 0.25x + 0.4y = 18 \]

4. **Substitute for \( y \)**:
From the first equation \( x + y = 60 \), we can express \( y \) in terms of \( x \):
\[ y = 60 - x \]

5. **Substitute \( y \) in the Concentration Equation**:
Substitute \( y \) in the concentration equation:
\[ 0.25x + 0.4(60 - x) = 18 \]

6. **Simplify the Equation**:
Distributing \( 0.4 \):
\[ 0.25x + 24 - 0.4x = 18 \]
Combine like terms:
\[ -0.15x + 24 = 18 \]

7. **Solve for \( x \)**:
Subtract 24 from both sides:
\[ -0.15x = 18 - 24 \]
\[ -0.15x = -6 \]
Divide by -0.15:
\[ x = \frac{-6}{-0.15} = 40 \]

8. **Find \( y \)**:
Substitute \( x \) back into the equation for \( y \):
\[ y = 60 - x = 60 - 40 = 20 \]

9. **Conclusion**:
Therefore, the amounts of each acid needed are:
- \( x = 40 \) liters of 25% acid
- \( y = 20 \) liters of 40% acid

### Final Answer:
- **25% Acid**: 40 liters
- **40% Acid**: 20 liters