Solve the problem.
-How many liters of a $10 \%$ alcohol solution must be mixed with 60 liters of a $70 \%$ solution to get a $60 \%$ solution?
A) $12 \mathrm{~L}$
B) $72 \mathrm{~L}$
C) $7.2 \mathrm{~L}$
D) $1.2 \mathrm{~L}$

Question

Quizplus · Accepted Answer

To solve this, we use the equation representing the total amount of alcohol before and after the mixture. Let $x$ be the amount of $10\%$ solution to be added. The amount of alcohol in the $10\%$ solution is $0.10x$ liters, and in the $70\%$ solution, it's $0.70 \times 60$ liters. The total volume of the solution after mixing is $60 + x$ liters, and we want this to be a $60\%$ alcohol solution, so the amount of alcohol in the final solution should be $0.60(60 + x)$ liters. Setting up the equation:$$0.10x + 0.70 \times 60 = 0.60(60 + x)$$Solving this equation for $x$ gives $x = 12$ liters.

Solve the Problem 10%10 \%10% Alcohol Solution Must Be Mixed with 60 Liters of A Alcohol

Solve the Problem $10 \%$ Alcohol Solution Must Be Mixed with 60 Liters of A Alcohol