A lab technician needs to mix a 20% fungicide solution with a 25% fungicide solu

Question

A lab technician needs to mix a 20% fungicide solution with a 25% fungicide solution to obtain a 50 liter mixture consisting of 23% fungicide. How many liters of each of the fungicide solutions must be used? Begin by filling in the missing entries in the table shown below. Then use the fact that the amount of fungicide in the 20% solution plus the amount of fungicide in the 25% solution must equal the amount of fungicide in the 23% mixture Number of Liters Percent of FungicideAmount of Fungicide 0.20 0.25 0.23 0.20x Fungicide Solution 5% Fungicide Solution 3% Fungicide Mixture 50 First, complete the table shown below Number of Liters |×| Percent of Fungicide-Amount of Fungicide % Fungicide Solution 5% Fungicide Solution 3% Fungicide Mixture 0.20 0.25 0.23 0.20x 0.25y 0.23 50 (Do not simplify.) 50 The lab technician needs to mix (Simplity your answers.) liters o 20% fungicide solution with liters of 25% ungicide solution to make 50 liters o a 23% ung cide solution.

Explanation / Answer

for solute ( fungicide here )

x&(20%) + y*(25%) = 50 *23%

20x + 25y = 1150 ------------------(1)

for solvent

x*(100-20)% +y*(100-25)% = 50*(100-23)%

80x + 75y = 3850 ------------------(2)

3*equation(1) - equation(2) gives

(60-80)x = 3*1150 - 3850

20x = 400

x = 20

using equation 1

20*20 + 25*y =1150

25y = 1150-400

25y = 750

y = 30

x = 20 liters , y = 30 liters (answer)

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