Solve the problem.
-How many liters of a $30 \%$ alcohol solution must be mixed with 60 liters of a $70 \%$ solution to get a $40 \%$ solution?
A) $18 \mathrm{~L}$
B) $180 \mathrm{~L}$
C) $240 \mathrm{~L}$
D) $24 \mathrm{~L}$

Question

Quizplus · Accepted Answer

To solve this, we can set up an equation based on the amount of pure alcohol in each solution. Let $x$ be the amount of $30\%$ solution needed. The amount of alcohol in the $30\%$ solution is $0.30x$ liters, and in the $70\%$ solution, it's $0.70 \times 60$ liters. The final mixture has a $40\%$ alcohol concentration, so the equation is $0.30x + 0.70 \times 60 = 0.40(x + 60)$. Solving for $x$ gives $x = 180$ liters.

Solve the Problem 30%30 \%30% Alcohol Solution Must Be Mixed with 60 Liters of A Alcohol

Solve the Problem $30 \%$ Alcohol Solution Must Be Mixed with 60 Liters of A Alcohol