Question

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?

Answer 1

Another eight cups of pure water will be required.

Step-by-step explanation:

Start by calculating the amount (in cups) of vinegar in these six cups:

• One cup of
  `5\%` vinegar-water solution would contain
  `0.05` cups of vinegar.
• Two cups of
  `10\%` vinegar-water solution would contain
  `2 * 0.10 = 0.20` cups of vinegar.
• Three cups of
  `15\%` vinegar-water solution would contain
  `3 * 0.15 = 0.45` cups of vinegar.

Therefore, if these six cups of solutions were mixed, the mixture would contain
`0.05 + 0.20 + 0.45 = 0.70` 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:

`\displaystyle (0.70\; \rm cups)/(6\; \rm cups) \approx 12\%`.

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
`5\%`.

The goal is to dilute this solution with
`0.70` cups of vinegar to a concentration of
`5\%`. 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:
  `0.70` cups.
• Concentration of vinegar in this solution:
  `5\% = 0.05`.

After dilution, the volume of this solution would be:

`\displaystyle (0.70\; \rm cups)/(0.05) = 14\; \rm cups`.

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