To complete the table and find the volumes of the 20% and 59% solutions needed to create 260 liters of a 50% solution, we can use the following equations based on the information provided.
Part 1: Completing the table
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For the 20% solution:
- Liters of solution: x
- Percent (as a decimal): 0.20
- Liters of pure dye: 0.20x
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For the 59% solution:
- Liters of solution: y
- Percent (as a decimal): 0.59
- Liters of pure dye: 0.59y
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For the final solution:
- Total liters of solution: 260
- Percent (as a decimal): 0.50
- Liters of pure dye: 0.50 × 260 = 130
Part 2: Setting up the equations
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The total amount of solution equation: \[ x + y = 260 \]
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The total amount of pure dye equation: \[ 0.20x + 0.59y = 130 \]
Solving the equations
From the first equation, isolate \( y \): \[ y = 260 - x \]
Substituting \( y \) in the second equation: \[ 0.20x + 0.59(260 - x) = 130 \]
Expanding the equation: \[ 0.20x + 153.4 - 0.59x = 130 \] \[ -0.39x + 153.4 = 130 \] \[ -0.39x = 130 - 153.4 \] \[ -0.39x = -23.4 \] \[ x = \frac{-23.4}{-0.39} \approx 60 \]
Now substituting back to find \( y \): \[ y = 260 - 60 = 200 \]
Final Results
- Liters of the 20% solution (x): 60 liters
- Liters of the 59% solution (y): 200 liters
So in response to your inquiry:
- Enter 60 liters of the 20% solution.
- Enter 200 liters of the 59% solution.
Completed Table
| Liters of solution | Percent (as a decimal) | Liters of pure dye | |--------------------|-----------------------|---------------------| | x | 0.20 | 0.20x (12) | | y | 0.59 | 0.59y (118) | | 260 | 0.50 | 130 |
(Simplified to integers or decimals as required.)
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