Let's denote the amount of 5% horseradish dip that Tom will use as \( x \) ounces. Consequently, the amount of the 20% horseradish dip that he will use will be \( 24 - x \) ounces since he wants a total of 24 ounces of dip.
Now we can set up an equation based on the desired concentration of horseradish. The total amount of horseradish in the mixture should equal the amount from each component.
The amount of horseradish from the 5% dip is: \[ 0.05x \]
The amount of horseradish from the 20% dip is: \[ 0.20(24 - x) \]
The total amount of horseradish in the final 24-ounce dip should be: \[ 0.10 \times 24 = 2.4 \text{ ounces} \]
Now, we can set up the equation: \[ 0.05x + 0.20(24 - x) = 2.4 \]
Expanding the equation: \[ 0.05x + 4.8 - 0.20x = 2.4 \]
Combining like terms: \[ -0.15x + 4.8 = 2.4 \]
Now, subtract \( 4.8 \) from both sides: \[ -0.15x = 2.4 - 4.8 \] \[ -0.15x = -2.4 \]
Divide both sides by \(-0.15\): \[ x = \frac{-2.4}{-0.15} = 16 \]
Thus, Tom should use \( \boxed{16} \) ounces of the 5% horseradish dip.
To verify:
- The amount of 20% dip used is \( 24 - 16 = 8 \) ounces.
- The total horseradish from 5% dip: \( 0.05 \times 16 = 0.8 \) ounces.
- The total horseradish from 20% dip: \( 0.20 \times 8 = 1.6 \) ounces.
- Total horseradish: \( 0.8 + 1.6 = 2.4 \) ounces, which is indeed 10% of 24 ounces.
Thus, the calculation is correct, and Tom should use \( \boxed{16} \) ounces of the 5% dip.
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