Solve the problem.
-How many liters of a $10 \%$ alcohol solution must be mixed with 80 liters of a $50 \%$ solution to get a $20 \%$ solution?
A) 32 liters
B) 320 liters
C) 240 liters
D) 24 liters

Question

Quizplus · Accepted Answer

To find the amount of 10% solution needed, set up the equation: $0.10x + 0.50(80) = 0.20(x + 80)$. Solving for $x$, we get $x = 240$ liters.

Solve the Problem 10%10 \%10% Alcohol Solution Must Be Mixed with 80 Liters of A Alcohol

Solve the Problem $10 \%$ Alcohol Solution Must Be Mixed with 80 Liters of A Alcohol