Mixture problems are word problems in algebra which are solved either using a linear equation in single variable or a system of equations in two variables. Mixture problems are generally formed based on value, money or concentration.
In solvent, solute problems they are related by the formula,
Amount of substance (solute) contained = Concentration x Total volume of the Solution
The related formula in value related problems is,
Value of an item = rate $\times$ weight or quantity of the item.While solving, mixture problems, equations are formed on the amount of substance in the mixture.
The amount of substance in the mixture is equal to the sum of the substances in each of the mixing solutions of the mixture.
C1V1 + C2V2 = CV
where C1, C2 and C are the concentrations corresponding to volumes V1, V2 and V. The letters with subscripts refer to concentration and volume of the mixing solutions, while the letters without subscript refer to the final mixture.
Step 1: Assume variable/s
The first task in solving mixture problems as in any other algebra word problems is to correctly relate the variables assumed to the problem given. Generally the variable will denote that which is required to find. Suppose the given problem reads as follows:
"A grocer mixed grape juice which costs 2.25 dollars per gallon with cranberry juice which costs 1.75 dollars per gallon. How many gallons of each should be used to make 200 gallons of cranberry/grape juice which will cost 2.10 dollars per gallon?"
As we need to find the quantities of each type of juice used in the mixture, we assume as follows:
Let the volume of Grape Juice mixed = x gallons and volume of Cranberry Juice mixed = y gallons.
Step 2: Forming the equation/s
If we have used one variable in our initial assumption, then we will form one equation, often on the basis of amount of substance or value. When there are two variables assumed, then we form a system of two linear equations.
For our problem on juice mixture, we form two equations, one on total volume and the other on value.
Sum of the quantities of two juices used = Quantity of the Mixture got
x + y = 200 (First equation on Quantity)
Sum of the values of two juices used = Value of the Mixture got.
We find a value using the formula, Value of an item = rate x weight or quantity of the item.
2.25x + 1.75y = 200 $\times$ 2.10 ( Second equation on value)
You may also make a table to understand the situation clearer.
| Item || Volume
|| Value = Rate x Volume
|Grape Juice||x gal||2.25|| 2.25 x
|Cranberry Juice|| y gal
||1.75|| 1.75 y
|Mixture||200 gal||2.10|| 200 * 2.10|
Step 3: Solve the equations formed
Using any method you have already learned.
Solving the two equations given in step 2, we get,
Quantity of Grape Juice in the mixture = 140 gallons
quantity of Cranberry Juice in the mixture = 60 gallons.
Problem having one variable:
let us solve one more similar problem related to the values of items in a mix.
Example: A shopkeeper wishes to mix two types of tea together. One type sells at 8 dollars per kg and the second type sells at 12 dollars per kg. He wishes to make 100kg of the mixture to sell at 11 dollars per kg. Find the quantity of the first type of tea mixture contains?
This problem can be solved with an equation in one variable.
Let the quantity of first type of tea in the mixture = x Kg.
Therefore, the quantity of the second type of tea = (100 - x) Kg (The two types together weigh 100 Kg).
The values are calculated multiplying the quantity by selling rate.
Value of first type of tea = 8x
Value of second type of tea = 12(100 - x) = 1200 - 12x
Value of the tea mix = 100 * 11 = 1100
We now form the equation on the value of the tea mix.
Value of first type tea + Value of second type tea = Value of tea Mixture
8x + 1200 - 12x = 1100
1200 - 4x = 1100
-4x = -100
x = 25
Hence the Tea mix contains 25 Kg of first type of tea.
Note: The above problem can also be solved using a linear system, assuming the quantity of the second type of tea in the mix as y Kg. Two equations are then formed one on the total weight and one on total value.
Problem having two variables:
The problems related to two types of investments are also essentially a value mixture problem.
Example: Reggie divided 14,500 dollars in two investments, one paying an interest of 11% and other 8%. If after one year the interest earned in the two investments is equal to 1430 dollars, find the amounts invested in the two investments.
Let the amounts invested in 11% and 8% deposits be correspondingly x dollars and y dollars.
We form two equations, one on the total amount invested and the other on the total interest earned as follows:
x + y = 14500 ---------------(1)
0.11x + 0.08y = 1430. ---------------(2)
Solving the above system, we will get
Amount invested in 11% plan = 9000 dollars
Amount inverted in 8% plan = 5500 dollars.
Let x ml of 48% and y ml 72% Sulfuric acids be mixed to get the required concentration.
We form two equations, one on total volume and one on amount of substance contained.
x + y = 500 ---------------------------------(1)
We use the formula,
Amount of substance contained = Concentration x Volume, where the concentration is expressed as a decimal.
Amount of acid in 48% acid = 0.48x
Amount of acid in 72% acid = 0.72y
Amount of acid in the mix = 500 * 0.6 = 300
Amount of acid in 48% acid + Amount of acid in 72% acid = Amount of acid in the mix
0,48x + 0.72y = 300 -----------------------(2)
The two equations can be solve using substitutions
x + y = 500 ⇒ y = 500 - x
0.48x + 0.72(500 - x) = 300 y = 500 - x is substituted in the equation (2)
0.48x + 360 - 0.72x = 300
- 0.24x = -60
x = 250 ml
y = 500 - 250 = 250 ml
Sally has to mix 250 ml from each type of acid to get a 500 ml acid with 60% concentration.