Tom has 1 cup of a 5% vinegar-water solution, 2 cups of a 10% vinegar-water solution, and 3 cups of a 15% vinegar-water solution. He needs at least 8 cups of a 5% vinegar-water solution for his chemistry project. What is the number of cups of pure water he should add to the total of all his vinegar-water solutions to complete the project?

One cup of [tex]5\%[/tex] vinegar-water solution would contain [tex]0.05[/tex] cups of vinegar.
Two cups of [tex]10\%[/tex] vinegar-water solution would contain [tex]2 \times 0.10 = 0.20[/tex] cups of vinegar.
Three cups of [tex]15\%[/tex] vinegar-water solution would contain [tex]3 \times 0.15 = 0.45[/tex] cups of vinegar.

Therefore, if these six cups of solutions were mixed, the mixture would contain [tex]0.05 + 0.20 + 0.45 = 0.70[/tex] cups of vinegar. Also, the mixture is supposed to have a volume of six cups. Therefore, the concentration of this six-cup mixture would be:

[tex]\displaystyle \frac{0.70\; \rm cups}{6\; \rm cups} \approx 12\%[/tex].

In other words, if these six cups of vinegar-water solutions are mixed, it would be necessary to dilute (by adding more water) before reaching the concentration of [tex]5\%[/tex].

The goal is to dilute this solution with [tex]0.70[/tex] cups of vinegar to a concentration of [tex]5\%[/tex]. Consider: what would be the volume the solution after dilution?

The volume of a solution can be found from the quantity and concentration of its solute:

Quantity of vinegar in this solution: [tex]0.70[/tex] cups.
Concentration of vinegar in this solution: [tex]5\% = 0.05[/tex].

After dilution, the volume of this solution would be:

[tex]\displaystyle \frac{0.70\; \rm cups}{0.05} = 14\; \rm cups[/tex].

At this moment, this solution only has a volume of six cups. Therefore, [tex]14 - 6 = 8[/tex] cups of pure water will be required.