A 20​% dye solution is to be mixed with a 59​% dye solution to get 260 L of a 50​% solution. How many liters of the 20​% and 59​% solutions will be​ needed?

Answer 1

To solve the problem, we can set up a system of equations based on the concentration and the total volume of liquids.

Let \( x \) be the amount (in liters) of the 20% dye solution, and \( y \) be the amount (in liters) of the 59% dye solution. We know the following:

1. The total volume of the mixed solution is 260 liters: \[ x + y = 260 \]

2. We want the final mixture to have a concentration of 50% dye. The amount of dye from the 20% solution is \( 0.20x \) and from the 59% solution is \( 0.59y \). The total amount of dye in the 50% solution is: \[ 0.50 \times 260 = 130 \] Therefore, we have the second equation: \[ 0.20x + 0.59y = 130 \]

Now we have a system of two equations:

1. \( x + y = 260 \) \quad (Equation 1)
2. \( 0.20x + 0.59y = 130 \) \quad (Equation 2)

We can solve Equation 1 for \( y \): \[ y = 260 - x \]

Now substitute this expression for \( y \) into Equation 2: \[ 0.20x + 0.59(260 - x) = 130 \]

Distributing \( 0.59 \) gives: \[ 0.20x + 153.4 - 0.59x = 130 \]

Combine like terms: \[ -0.39x + 153.4 = 130 \]

Now isolate \( x \): \[ -0.39x = 130 - 153.4 \] \[ -0.39x = -23.4 \] \[ x = \frac{-23.4}{-0.39} \approx 60 \]

Now substitute \( x = 60 \) back into Equation 1 to find \( y \): \[ 60 + y = 260 \] \[ y = 260 - 60 = 200 \]

Thus, the amounts needed are:

• \( 60 \) liters of the 20% dye solution
• \( 200 \) liters of the 59% dye solution

Final answer:

• 20% solution: 60 liters
• 59% solution: 200 liters