How many liters of a $30 \%$ - alcohol solution must be mixed with 40 liters of a solution that is $60 \%$ alcohol to get a solution that is $50 \%$ alcohol?
A) $6 \mathrm{~L}$
B) $2 \mathrm{~L}$
C) $60 \mathrm{~L}$
D) $20 \mathrm{~L}$

Question

Quizplus · Accepted Answer

To find the amount of the $30\%$ alcohol solution needed, we can set up an equation representing the total amount of alcohol before and after the mixture. Let $x$ be the amount of $30\%$ alcohol solution to be added. The total alcohol from the $30\%$ solution is $0.30x$ liters, and from the $60\%$ solution is $0.60 \times 40$ liters. The final mixture has a $50\%$ alcohol concentration, so the equation is $0.30x + 0.60 \times 40 = 0.50(x + 40)$. Solving for $x$ gives $x = 20$ liters.

How Many Liters of A 30%30 \%30% - Alcohol Solution Must Be Mixed with 40 Liters of Must

How Many Liters of A $30 \%$ - Alcohol Solution Must Be Mixed with 40 Liters of Must