# Mixture Word Problems and Solutions

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

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.

Related Topics:
More Mixture Word Problems

More Algebra Topics

Mixture Problems: Example:

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

Solution:

Step 1: Set up a table for salt.

Step 2: Fill in the table with information given in the question.

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

The salt added is 100% salt, which is 1 in decimal.
Change all the percent to decimals

Let x = amount of salt added. The result would be 20 + x.

 original added result concentration 0.2 1 0.25 amount 20 x 20 + x

Step 3: Multiply down each column.

 original added result concentration 0.2 1 0.25 amount 20 x 20 + x multiply 0.2 × 20 1 × x 0.25(20 + x)

Step 4: original + added = result

0.2 × 20 + 1 × x = 0.25(20 + x)
4 + x = 5 + 0.25x

Isolate variable x
x
– 0.25x = 5 – 4
0.75x = 1

## Videos

Mixture Problems
Some word problems using systems of equations involve mixing two quantities with different prices. To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables. Other types of word problems using systems of equations include rate word problems and work word problems.

Algebra - Word Problems - Mixture Problems
In this video, we're going to cover the method used to obtain our desired concentration of an acid solution from two solutions of different concentrations.
Example: How many gallons of a 30% acid-solution must be added to how many gallons of a 70% acid-solution to make 50 gallons of a 40% acid-solution?
Example: A 10% alcohol solution and a 40% alcohol solution are mixed to form 80ml of a 35% solution. How much of each mixture was needed?

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