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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?

1 Answer

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Answer:

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.

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