Related Topics: More Algebra Word Problems

In these lessons, we will learn how to solve mixture word problems using algebra.

In these lessons, we will learn how to solve mixture word problems using algebra.

Mixture problems are word problems where items or quantities of different values are mixed together.

Sometimes different liquids are mixed together changing the concentration of the mixture as shown in example 1, example 2 and example 3. At other times, quantities of different costs are mixed together as shown in example 4.

We recommend using a table to organize your information for mixture problems. Using a table allows you to think of one number at a time instead of trying to handle the whole mixture problem at once.

We will show you how it is done by the following examples of mixture problems:

Adding to the SolutionRemoving from the Solution

Replacing the Solution

Mixing Quantities of Different Costs

Mixture Problems: Example 1:

John has 20 ounces of a 20% of salt solution. How much salt should he add to make it a 25% solution?

Solution:

Set up a table for salt using the information from the question.

Mixture Problems: Example 2:

John has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution?

Solution:

Set up a table for water. The water is removed from the original solution.

Mixture Problems: Example 3:

A tank has a capacity of 10 gallons. When it is full, it contains 15% alcohol. How many gallons must be replaced by an 80% alcohol solution to give 10 gallons of 70% solution?

Solution:

Set up a table for alcohol. The alcohol is replaced: an amount of 15% alcohol is removed and the same amount of 80% alcohol is added.

Mixture Problems: Example 4:

How many pounds of chocolate worth $1.20 a pound must be mixed with 10 pounds of chocolate worth 90 cents a pound to produce a mixture worth $1.00 a pound?

Solution:

Set up a table for the chocolates with different costs.

Example: The mad scientist has one solution that is 30% acid and another solution that is 18% acid. How much of each should he use to get 300 L of a solution that is 21% acid? Example: How much pure acid must be mixed with 200 mL of 5% acid to get a 25% acid?

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