To solve the problem, let's define:
- \( x \): the volume of the 50% acid solution to be mixed (in liters).
- \( y \): the volume of the 15% acid solution to be mixed (in liters).
We know that the total volume of the mixture must equal 280 liters: \[ x + y = 280 \quad \text{(1)} \]
Now, we also need to consider the amount of acid in the solutions. The amount of acid in each solution can be expressed as follows:
- The amount of acid in the 50% solution: \( 0.50x \).
- The amount of acid in the 15% solution: \( 0.15y \).
The total amount of acid in the final 280 L mixture must be 40% of 280 L: \[ 0.40 \times 280 = 112 \text{ liters of acid} \quad \text{(2)} \]
Now, we can set up the equation for total acid: \[ 0.50x + 0.15y = 112 \quad \text{(3)} \]
Now, we have a system of two equations:
- \( x + y = 280 \)
- \( 0.50x + 0.15y = 112 \)
We can solve this system. Let's solve equation (1) for \( y \): \[ y = 280 - x \quad \text{(4)} \]
Now substitute equation (4) into equation (3): \[ 0.50x + 0.15(280 - x) = 112 \]
Expand this equation: \[ 0.50x + 42 - 0.15x = 112 \]
Combine like terms: \[ (0.50 - 0.15)x + 42 = 112 \] \[ 0.35x + 42 = 112 \]
Now, isolate \( x \): \[ 0.35x = 112 - 42 \] \[ 0.35x = 70 \] \[ x = \frac{70}{0.35} = 200 \]
Now that we have \( x \), we can find \( y \) using equation (4): \[ y = 280 - 200 = 80 \]
Finally, we conclude:
- The volume of the 50% acid solution needed is 200 liters.
- The volume of the 15% acid solution needed is 80 liters.
Therefore, you need 200 liters of the 50% acid solution.