Answer:
99 ounces of the 12% alcohol solution that needs to be mixed.
Explanation:
Let the volume of 12% alcohol solution be x.
Volume of alcohol added = 0.12 x
Volume of 20% alcohol solution = 33 ounces
Volume of alcohol in 20% of alcohol solution = [tex]33 Oz\times \frac{20}{100}=6.6 Oz[/tex]
Total volume of alcohol after addition = (0.12 x +6.6) Oz
Total volume of alcohol solution after addition = x + 33 Oz
Percentage of alcohol solution after addition = 14%
Also ,Total volume of alcohol after addition = [tex](x + 33) Oz\times 0.14[/tex]
[tex](x + 33) Oz\times 0.14= (0.12 x +6.6) Oz[/tex]
Solving for x :
x = 99 ounces
99 ounces of the 12% alcohol solution that needs to be mixed.
