This is a classic mixture problem, where we are asked to mix two solutions of different acid concentrations to obtain a desired concentration. Let’s define the variables:

• Let $x$ be the amount (in liters) of the 50% acidic solution.
• Let $10 - x$ be the amount (in liters) of the 25% acidic solution, since the total volume is 10 liters.

The amount of acid in each solution is:

• From the 50% solution: $0.50x$ liters of acid.
• From the 25% solution: $0.25(10 - x)$ liters of acid.

The total amount of acid in the mixture should equal 40% of the total 10 liters, i.e., $0.40 \times 10 = 4$ liters.

We now form the equation: $0.50x + 0.25(10 - x) = 4$

Solving the equation:

1. Distribute the $0.25$ on the left-hand side: $0.50x + 0.25(10) - 0.25x = 4$ $0.50x + 2.5 - 0.25x = 4$

2. Combine like terms: $(0.50x - 0.25x) + 2.5 = 4$ $0.25x + 2.5 = 4$

3. Subtract 2.5 from both sides: $0.25x = 1.5$

4. Solve for $x$: $x = \frac{1.5}{0.25} = 6$

So, you need 6 liters of the 50% acidic solution.

Since the total is 10 liters, the amount of the 25% acidic solution is: $10 - 6 = 4 \text{ liters}.$

Conclusion:

• 6 liters of the 50% acidic solution.
• 4 liters of the 25% acidic solution.

Would you like more details or have any questions?

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Here are 5 related questions to expand your understanding:

1. How would the amounts change if you needed 20 liters of 40% acidic solution instead of 10 liters?
2. What would happen if the required concentration was 35% instead of 40%?
3. Can you solve the problem using a system of equations?
4. How much of each solution would you need to mix if you had only 5 liters of the 50% solution available?
5. What is the general formula for solving mixture problems like this one?

Tip: When solving mixture problems, always define variables carefully and write an equation based on the total amount of the substance being mixed (in this case, acid).