0.2 × 20 + 1 × x = 0.25(20 + x)
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
Answer: He should evaporate 6.67 ounces of water.
Replacing The 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:
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
Answer: 8.46 gallons of alcohol solution needs to be replaced.
Mixing Quantities Of Different Costs
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
Answer: 5 pounds of the $1.20 chocolate needs to be added.
Share with your friends: