Asked by Gianna
Complete the table below.
Liters of solution
Percent (as a decimal)
Liters of pure dye
x
0.20
0.20 x
y
0.59
0.59 y
260
0.50
130
(Simplify your answers. Use integers or decimals for any numbers in the expression.)
Part 2
To get 260 L of a 50% solution,
enter your response here liters of the 20% solution and
enter your response here liters of the 59% solution are needed.
Liters of solution
Percent (as a decimal)
Liters of pure dye
x
0.20
0.20 x
y
0.59
0.59 y
260
0.50
130
(Simplify your answers. Use integers or decimals for any numbers in the expression.)
Part 2
To get 260 L of a 50% solution,
enter your response here liters of the 20% solution and
enter your response here liters of the 59% solution are needed.
Answers
There are no human answers yet.
Answered by
GPT-4o mini
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
1. For the **20% solution**:
- Liters of solution: **x**
- Percent (as a decimal): **0.20**
- Liters of pure dye: **0.20x**
2. For the **59% solution**:
- Liters of solution: **y**
- Percent (as a decimal): **0.59**
- Liters of pure dye: **0.59y**
3. 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
1. The total amount of solution equation:
\[
x + y = 260
\]
2. 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.)
### Part 1: Completing the table
1. For the **20% solution**:
- Liters of solution: **x**
- Percent (as a decimal): **0.20**
- Liters of pure dye: **0.20x**
2. For the **59% solution**:
- Liters of solution: **y**
- Percent (as a decimal): **0.59**
- Liters of pure dye: **0.59y**
3. 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
1. The total amount of solution equation:
\[
x + y = 260
\]
2. 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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