Question

How many ounces of a 1212​% alcohol solution must be mixed with 33 ounces of a 2020​% alcohol solution to make a 1414​% alcohol​ solution? The number of ounces of the 1212​% alcohol solution that needs to be mixed is nothing ounces. ​(Simplify your​ answer.)

Answer 1

99 ounces of the 12​% alcohol solution that needs to be mixed.

Step-by-step 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 =
`33 Oz* (20)/(100)=6.6 Oz`

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 =
`(x + 33) Oz* 0.14`

`(x + 33) Oz* 0.14= (0.12 x +6.6) Oz`

Solving for x :

x = 99 ounces

99 ounces of the 12​% alcohol solution that needs to be mixed.