Let \( x \) be the number of liters of the 10% acid solution, and let \( y \) be the number of liters of the 30% acid solution.
We have two main conditions to satisfy:
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The total volume of the solution: \[ x + y = 22 \quad (1) \]
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The total amount of acid in the solution: The 10% solution contributes \( 0.10x \) liters of acid, and the 30% solution contributes \( 0.30y \) liters of acid. We need the total acid to equal 25% of 22 liters, which is: \[ 0.25 \times 22 = 5.5 \text{ liters of acid} \] This gives us the second equation: \[ 0.10x + 0.30y = 5.5 \quad (2) \]
Now we can solve these equations simultaneously. From equation (1), we can express \( y \) in terms of \( x \): \[ y = 22 - x \]
Now, substitute \( y \) into equation (2): \[ 0.10x + 0.30(22 - x) = 5.5 \] Expanding this: \[ 0.10x + 6.6 - 0.30x = 5.5 \] Combining like terms: \[ -0.20x + 6.6 = 5.5 \] Subtracting 6.6 from both sides: \[ -0.20x = 5.5 - 6.6 \] \[ -0.20x = -1.1 \] Dividing both sides by -0.20: \[ x = \frac{-1.1}{-0.20} = 5.5 \]
Now substituting \( x \) back into the equation for \( y \): \[ y = 22 - 5.5 = 16.5 \]
Thus, the chemist should mix:
- 5.5 liters of the 10% acid solution
- 16.5 liters of the 30% acid solution