Question

how many liters of a 30% alcohol solution must be mixed with 8 liters of pure water (0% alcohol) to make a 6% solution?
1 liter
2 liters
3 liters
5 liters

Explanation:

Step1: Set up the equation

Let $x$ be the number of liters of the 30% alcohol solution. The amount of alcohol in the 30% solution is $0.3x$, and the amount of alcohol in the 8 - liter pure - water (0% alcohol) is $0$. The total volume of the final mixture is $x + 8$ liters, and the concentration of the final mixture is 6%, so the amount of alcohol in the final mixture is $0.06(x + 8)$. Then we have the equation $0.3x+0 = 0.06(x + 8)$.

Step2: Expand the right - hand side

Expand $0.06(x + 8)$ using the distributive property: $0.3x=0.06x+0.48$.

Step3: Subtract $0.06x$ from both sides

$0.3x−0.06x=0.06x + 0.48−0.06x$, which simplifies to $0.24x=0.48$.

Step4: Solve for $x$

Divide both sides of the equation $0.24x = 0.48$ by $0.24$: $x=\frac{0.48}{0.24}=2$.

Answer:

B. 2 liters