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 SolutionMixture 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 
result 
concentration 

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 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
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 
result 
concentration 

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 
result 
concentration 
0.8 
1 
0.7 
amount 
20 
x 
20 – x 
Step 3: Multiply down each column.

original 
removed 
result 
concentration 
0.8 
1 
0.7 
amount 
20 
x 
20 – x 
multiply 
0.8 × 20 
1 × x 
0.70(20 – x) 
Step 4: Since the water is removed, we need to subtract
original – removed = result
0.8 × 20 – 1 × x = 0.70(20 – x)
16 – x = 14 – 0.7x
Isolate variable x
x – 0.7x = 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 
result 
concentration 

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 
result 
concentration 
0.15 
0.15 
0.8 
0.7 
amount 
10 
x 
x 
10 
Step 3: Multiply down each column.

original 
removed 
added 
result 
concentration 
0.15 
0.15 
0.8 
0.7 
amount 
10 
x 
x 
10 
multiply 
0.15 × 10 
0.15 × x 
0.8 × x 
0.7 × 10 
Step 4: Since the alcohol solution is replaced, we need to subtract and add.
original – removed + added = result
0.15 × 10 – 0.15 × x + 0.8 × x = 0.7 × 10
1.5 – 0.15x + 0.8x = 7
Isolate variable x
0.8x – 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 
result 
cost 

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 
result 
cost 
0.9 
1.2 
1 
amount 
10 
x 
x + 10 
Step 3: Multiply down each column.

original 
added 
result 
cost 
0.9 
1.2 
1 
amount 
10 
x 
x + 10 
multiply 
0.9 × 10 
1.2 × x 
1 × (x + 10) 
Step 4: original + added = result
0.9 × 10 + 1.2 × x = 1 × (x + 10)
9 + 1.2x = x + 10
Isolate variable x
1.2x – x = 10  9
0.2x = 1
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