Mixing task
Mixing problems or mixing equations are a subgroup of mathematical problems in which several substances with different properties are mixed in order to obtain a substance with predetermined, new properties.
The properties of the various substances can e.g. B. alcohol content, price, fat or water content, etc.
Mixing tasks often arise in the chemical industry when a mixed product with a defined concentration is to be produced from two preliminary products or basic materials with specified concentrations .
Examples
Mixture of solids
A mixture of a total of 60 kg of walnuts and peanuts should cost 3.05 euros per kilogram. The walnuts cost 3.20 euros per kilogram, the peanuts 2.60 euros per kilogram. How much of the two should you take to get the desired price for the mixture?
W = weight of walnuts
E = weight of the peanuts
1) 60 kg = W + E → W = 60 kg - E
2) Endpreis · "Endmenge" = (Kilopreis-W)· (Menge-W) + (Kilopreis-E) · (Menge-E)
Rechnung 3,05 € · 60 kg = 3,20 € · W + 2,60 € · E // Einsetzen 1) in 2) 3,05 € · 60 kg = 3,20 € · (60 kg - E) + 2,60 € · E // Klammer auflösen 3,05 € · 60 kg = 3,20 € · 60 kg - 3,20 € · E + 2,60 € · E // Zusammenfassen 3,05 € · 60 kg = 3,20 € · 60 kg - 0,60 € · E // nach E umstellen E = (3,20 € · 60 kg - 3,05 € · 60 kg) / 0,6€ E = 15 kg // Einsetzen in 1) W = 60 kg - 15 kg W = 45 kg
Result:
You have to mix 15 kg of peanuts and 45 kg of walnuts to get 60 kg of nut mix at a price of € 3.05 / kg.
Fluid mixture
A mixed beer drink in a 0.33 l bottle has an alcohol content of 2.9%. The original beer has 4% alcohol content. How much cola and how much beer are mixed in?
B = amount of beer in liters
C = amount of cola in liters
1) 0,33 l = B + C → C = 0,33 l - B
2) 0,33 l · 2,9 % = B · 4 % + C · 0 % // C · 0 % = 0 0,33 l · 2,9 % = B · 4 % // Nach B auflösen
B = (0,33 l · 2,9 %) / 4 % B = 0,24 l // Einsetzen in 1)
C = 0,33 l - 0,24 l C = 0,09 l
Result:
To obtain 0.33 l mixed beer drink with 2.9% alcohol, mix 0.24 l beer with 4% alcohol and 0.09 l cola.