Answer:
Pure vinegar: 12.2 ounces, 4% mix salad: 115.8 ounces
Explanation:
The chef wants to male a 15% vinegar-to-oil salad.
If we call:
[tex]v[/tex] the amount of vinegar in the salad
[tex]o[/tex] the amount of oil in the salad
This means that
[tex]\frac{v}{o}=\frac{15}{100}[/tex]
In order to get this salad, the chef has to mix:
- Pure vinegar, which has 100% concentration of vinegar
- A 4% vinegar-to-oil salad: this means that the amount of vinegar in this salad is [tex]v=0.04o[/tex] (4% of the amount of oil)
This means that the total amount of vinegar in the final salad will be:
[tex]v'=v+0.04o[/tex] (1a)
Where v is the amount of pure vinegar added and [tex]o[/tex] is the amount of oil in the 4% salad
While the amount of oil needed is [tex]o'=o[/tex]
So we have, since the final salad has 15% concentration of vinegar to oil:
[tex]\frac{v+0.04o}{o}=\frac{15}{100}[/tex] (1)
Moreover, we know that the final volume must be 128 ounces, so
[tex]v'+o=128[/tex] (2)
From eq.(2) and (1a) we get
[tex]v+0.04o+o=128\\v=128-1.04o[/tex]
Substituting into (1) and solving for [tex]o[/tex],
[tex]\frac{128-1.04o+0.04o}{o}=\frac{15}{100}\\100(128-o)=15o\\20(128-o)=3o\\2560-20o=3o\\o=\frac{2560}{23}=111.3[/tex]
Therefore the amount of vinegar must be
[tex]v=128-1.04o=128-(1.04)(111.3)=12.2[/tex]
So, the amount of each that should be added is:
- Pure vinegar: 12.2 ounces
- 4% vinegar-to-oil mix: [tex]1.04o=(1.04)(111.3)=115.8[/tex] ounces
