Question

How many liters of a 50​% acid solution must be mixed with a 15​% acid solution to get 280 L of a 40​% acid​ solution?

Answer 1

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:
1. \( x + y = 280 \)
2. \( 0.50x + 0.15y = 112 \)

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.