Question

A chef wants to make 1 gallon (128 ounces) of a 15% vinegar-to-oil salad dressing. He only has pure vinegar and a mild 4% vinegar-to-oil salad dressing on hand. How many ounces of each should he mix to make the desired dressing

Answer 1

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:

$v$ the amount of vinegar in the salad

$o$ the amount of oil in the salad

This means that

$\frac{v}{o}=\frac{15}{100}$

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 $v=0.04o$ (4% of the amount of oil)

This means that the total amount of vinegar in the final salad will be:

$v'=v+0.04o$ (1a)

Where v is the amount of pure vinegar added and $o$ is the amount of oil in the 4% salad

While the amount of oil needed is $o'=o$

So we have, since the final salad has 15% concentration of vinegar to oil:

$\frac{v+0.04o}{o}=\frac{15}{100}$ (1)

Moreover, we know that the final volume must be 128 ounces, so

$v'+o=128$ (2)

From eq.(2) and (1a) we get

$v+0.04o+o=128\\v=128-1.04o$

Substituting into (1) and solving for $o$,

$\frac{128-1.04o+0.04o}{o}=\frac{15}{100}\\100(128-o)=15o\\20(128-o)=3o\\2560-20o=3o\\o=\frac{2560}{23}=111.3$

Therefore the amount of vinegar must be

$v=128-1.04o=128-(1.04)(111.3)=12.2$

So, the amount of each that should be added is:

- Pure vinegar: 12.2 ounces

- 4% vinegar-to-oil mix: $1.04o=(1.04)(111.3)=115.8$ ounces