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

Related Topics: More Algebra Word Problems

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:

Step 1: Set up a table for salt.

original

added

resultconcentration

amount

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 decimalsLet

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

original

added

resultconcentration

0.2

1

0.25

amount

20

x20 +

x

Step 3: Multiply down each column.

original

added

resultconcentration

0.2

1

0.25

amount

20

x20 +

xmultiply

0.2 × 20

1 ×

x0.25(20 +

x)

Step 4: original + added = result

Answer: He should add ounces of salt.0.2 × 20 + 1 ×

x= 0.25(20 +x)

4 +x= 5 + 0.25xIsolate variable

x– 0.25

xx= 5 – 4

0.75x= 1

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:

Step 1: Set up a table for water. The water is removed from the original.

original

removed

resultconcentration

amount

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

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

The original concentration of water is 100% – 20% = 80%

The resulted concentration of water is 100% – 30% = 70%

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

Let

x= amount of water evaporated. The result would be 20 –x.

original

removed

resultconcentration

0.8

1

0.7

amount

20

x20 –

x

Step 3: Multiply down each column.

original

removed

resultconcentration

0.8

1

0.7

amount

20

x20 –

xmultiply

0.8 × 20

1 ×

x0.70(20 –

x)

Step 4: Since the water is removed, we need to subtract

Answer: He should evaporate 6.67 ounces of water.original – removed = result

0.8 × 20 – 1 ×x= 0.70(20 –x)

16 –x= 14 – 0.7xIsolate variable

x– 0.7

xx= 16 – 14

0.3x= 2

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:

Step 1: Set up a table for alcohol. The alcohol is replaced i.e. removed and added.

original

removed

added

resultconcentration

amount

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

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?

Change all the percent to decimals.

Let

x= amount of alcohol solution replaced.

original

removed

added

resultconcentration

0.15

0.15

0.8

0.7

amount

10

x

x10

Step 3: Multiply down each column.

original

removed

added

resultconcentration

0.15

0.15

0.8

0.7

amount

10

x

x10

multiply

0.15 × 10

0.15 ×

x0.8 ×

x0.7 × 10

Step 4: Since the alcohol solution is replaced, we need to subtract and add.

Answer: 8.46 gallons of alcohol solution needs to be replaced.original – removed + added = result

0.15 × 10 – 0.15 ×x+ 0.8 ×x= 0.7 × 10

1.5 – 0.15x+ 0.8x= 7Isolate variable

x0.8

x– 0.15x= 7 – 1.5

0.65x= 5.5

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:

Step 1: Set up a table for different types of chocolate.

original

added

resultcost

amount

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

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?

Let

x= amount of chocolate added.

original

added

resultcost

0.9

1.2

1

amount

10

x

x+ 10

Step 3: Multiply down each column.

original

added

resultcost

0.9

1.2

1

amount

10

x

x+ 10multiply

0.9 × 10

1.2 ×

x1 × (

x+ 10)

Step 4: original + added = result

Answer: 5 pounds of the $1.20 chocolate needs to be added.0.9 × 10 + 1.2 ×

x= 1 × (x+ 10)

9 + 1.2x=x+ 10Isolate variable

x1.2

x–x= 10 - 9

0.2x= 1

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 6% acid to get a 25% acid?

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