This is a classic mixture problem involving concentrations of alcohol solutions. We need to determine how many liters of a 5% alcohol solution should be added to 70 liters of a 10% alcohol solution to obtain an 8% alcohol solution.

Let $x$ be the number of liters of the 5% alcohol solution to be added.

Step 1: Write the equation for the total amount of alcohol.

• The amount of alcohol in the 5% solution is $0.05x$ liters.
• The amount of alcohol in the 70 liters of the 10% solution is $0.10 \times 70 = 7$ liters.
• The total volume of the mixture will be $70 + x$ liters, and we want this to be an 8% alcohol solution. So, the amount of alcohol in the final solution will be $0.08(70 + x)$ liters.

Step 2: Set up the equation.

The total amount of alcohol before and after mixing must be the same: $0.05x + 7 = 0.08(70 + x)$

Step 3: Solve the equation.

Now, let's solve for $x$:

$0.05x + 7 = 0.08(70 + x)$

Expanding the right side: $0.05x + 7 = 0.08 \times 70 + 0.08x$ $0.05x + 7 = 5.6 + 0.08x$

Next, subtract $0.05x$ from both sides: $7 = 5.6 + 0.03x$

Subtract 5.6 from both sides: $1.4 = 0.03x$

Finally, divide by 0.03: $x = \frac{1.4}{0.03} = 46.67$

Step 4: Round to the nearest tenth.

$x \approx 46.7$

Thus, the answer is 46.7 liters (Option A).

Would you like any further details or clarification?

Related Questions:

1. How do you calculate the concentration of a solution?
2. What is the general method for solving mixture problems in algebra?
3. How does changing the concentration affect the required amount of the solution?
4. Can you apply this method to mixtures involving substances other than alcohol?
5. How would the problem change if we were mixing more than two solutions?

Tip:

When dealing with mixture problems, always ensure that you set up an equation where the total amount of the substance being mixed (e.g., alcohol) is conserved.