A manufacturer has 600 L of a 12% solution of acid. How many litres of a 30% acid solution must be added to it such that acid content in the resulting mixture will be more than 15% but less than 18%?
more than 120 L but less than 300 L
Solving the Acid Mixture Problem
This problem involves mixing two solutions of different acid concentrations to achieve a resulting mixture with an acid concentration within a specific range. We need to determine the volume of the second solution required.
Understanding the Initial State
We start with 600 L of a solution that contains 12% acid. The amount of acid in this initial solution is calculated as:
\(\text{Amount of acid in initial solution} = 12\% \text{ of } 600 \text{ L}\)
\(\text{Amount of acid in initial solution} = \frac{12}{100} \times 600 \text{ L}\)
\(\text{Amount of acid in initial solution} = 0.12 \times 600 \text{ L} = 72 \text{ L}\)
Adding the Second Solution
Let's say we add \(x\) litres of the 30% acid solution. The amount of acid in this added solution is:
\(\text{Amount of acid in added solution} = 30\% \text{ of } x \text{ L}\)
\(\text{Amount of acid in added solution} = \frac{30}{100} \times x \text{ L}\)
\(\text{Amount of acid in added solution} = 0.30 \times x \text{ L} = 0.3x \text{ L}\)
Calculating the Resulting Mixture
When we mix the two solutions, the total volume of the resulting mixture will be the sum of the initial volume and the added volume:
\(\text{Total volume of resulting mixture} = 600 \text{ L} + x \text{ L} = (600 + x) \text{ L}\)
The total amount of acid in the resulting mixture is the sum of the acid from the initial solution and the acid from the added solution:
\(\text{Total acid in resulting mixture} = 72 \text{ L} + 0.3x \text{ L} = (72 + 0.3x) \text{ L}\)
Setting Up the Inequalities for Acid Concentration
The concentration of acid in the resulting mixture is given by the formula:
\(\text{Concentration} = \frac{\text{Total amount of acid}}{\text{Total volume}} \times 100\%\)
So, the concentration is:
\(\text{Concentration} = \frac{72 + 0.3x}{600 + x} \times 100\%\)
We are told that the acid content in the resulting mixture must be more than 15% but less than 18%. This gives us two inequalities:
\(\frac{72 + 0.3x}{600 + x} \times 100 > 15\)
and
\(\frac{72 + 0.3x}{600 + x} \times 100 < 18\)
Solving the Inequalities
Inequality 1: More than 15%
\(\frac{72 + 0.3x}{600 + x} > \frac{15}{100}\)
\(\frac{72 + 0.3x}{600 + x} > 0.15\)
Since \(x\) represents a volume, \(x > 0\), which means \(600 + x > 0\). We can multiply both sides by \((600 + x)\) without changing the direction of the inequality:
\(72 + 0.3x > 0.15(600 + x)\)
\(72 + 0.3x > 90 + 0.15x\)
Now, let's gather the terms with \(x\) on one side and the constant terms on the other:
\(0.3x - 0.15x > 90 - 72\)
\(0.15x > 18\)
To find \(x\), divide both sides by 0.15:
\(x > \frac{18}{0.15}\)
\(x > \frac{18}{\frac{15}{100}}\)
\(x > \frac{18 \times 100}{15}\)
\(x > \frac{1800}{15}\)
\(x > 120\)
So, the volume of the 30% solution added must be more than 120 L.
Inequality 2: Less than 18%
\(\frac{72 + 0.3x}{600 + x} < \frac{18}{100}\)
\(\frac{72 + 0.3x}{600 + x} < 0.18\)
Again, multiply both sides by \((600 + x)\) since \(600 + x > 0\):
\(72 + 0.3x < 0.18(600 + x)\)
\(72 + 0.3x < 108 + 0.18x\)
Gather the terms with \(x\) on one side and constants on the other:
\(0.3x - 0.18x < 108 - 72\)
\(0.12x < 36\)
To find \(x\), divide both sides by 0.12:
\(x < \frac{36}{0.12}\)
\(x < \frac{36}{\frac{12}{100}}\)
\(x < \frac{36 \times 100}{12}\)
\(x < \frac{3600}{12}\)
\(x < 300\)
So, the volume of the 30% solution added must be less than 300 L.
Conclusion
Combining the results from both inequalities (\(x > 120\) and \(x < 300\)), we find that the amount of the 30% acid solution that must be added is more than 120 L but less than 300 L.
This range is more than 120 L but less than 300 L.
Revision Table: Key Calculations
| Item | Volume (L) | Acid Concentration (%) | Amount of Acid (L) |
|---|---|---|---|
| Initial Solution | 600 | 12% | \(0.12 \times 600 = 72\) |
| Added Solution | \(x\) | 30% | \(0.30 \times x = 0.3x\) |
| Resulting Mixture | \(600 + x\) | >15% and <18% | \(72 + 0.3x\) |
Additional Information: Mixture Problems
Mixture problems like this one often involve setting up equations or inequalities based on the quantities of components being mixed and their properties (like concentration, price, etc.). The key is to track the total amount of the specific component (acid in this case) and the total volume or weight of the mixture. The concentration is the ratio of the component amount to the total mixture amount.
- Concentration: Usually expressed as a percentage or a fraction. It is the amount of solute divided by the total amount of solution (solute + solvent).
- Setting up equations/inequalities: Typically involves equating the total amount of the component in the mixed solution to the sum of the amounts of the component in the individual solutions being mixed. For range problems, inequalities are used.
- Solving: Algebra is used to solve for the unknown variable (the amount of one of the components to be added).
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