1. Objective: To Solve Mixture Problems
Mixture problems involve combining two mixtures to make one final mixture.
We are going to use buckets to help solve these problems.
1st Mixture
2nd Mixture
Final Mixture + =

2. Buckets

3. 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
STEP #1: Draw buckets + =

4. 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
Step #1: Draw the buckets
STEP #2: Label the buckets
Pure OJ
Orange Drink
Final Mixture + =

5. 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
Step #1: Draw the buckets
Step #2: Label the buckets
STEP #3: Price or Percent goes on top + =
Pure OJ
Orange Drink
Final Mixture
10%
40%
100%

6. 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
Step #1: Draw the buckets
Step #2: Label the buckets
Step #3: Price or Percent goes on top
STEP #4: Change Percents to decimals + =
Pure OJ
Orange Drink
Final Mixture
10%
40%
= 0.4
100%
= 1
= 0.1

7. 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
Step #1: Draw the buckets
Step #2: Label the buckets
Step #3: Price or Percent goes on top
Step #4: Change Percents to decimals
STEP #5: Weight goes on the bottom + =
Pure OJ
Orange Drink
Final Mixture
10%
40%
= 0.4
100%
= 1
= 0.1 x 5
x + 5

8. + = Pure OJ STEP #6: Set up and Solve Orange Drink Final Mixture x 5
1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
Step #1: Draw the buckets
STEP #6: Set up and Solve
Step #2: Label the buckets
Step #3: Price or Percent goes on top
Step #4: Change Percents to decimals
Step #5: Weight goes on the bottom + =
Pure OJ
Orange Drink
Final Mixture
10%
40%
= 0.4
100%
= 1
= 0.1 x 5
x + 5

9. + = 1x + 0.1(5) = 0.4(x+5) 1.0x + 0.5 = 0.4x + 2 -0.4x -0.4x
1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice?
STEP #6: Set up and Solve + =
Pure OJ
Orange Drink
Final Mixture
100%
10%
40%
= 1
= 0.1
= 0.4 x 5
x + 5
1x (5) = (x+5)
1.0x = x + 2
-0.4x x
0.6x = 2.0
2.5 quarts of pure
orange juice would
have to be added to
the 5 quarts of the drink.
0.6x = 1.5
x = 2.5

10. + = Final Mixture 2. How much water would you have to add to 3 liters
of lake water that is 30 percent salt to get a solution
that is only 12 percent salt?
Step #1: Draw buckets
Step #2: Label the buckets
Step #3: Percent or price goes on top
Step #4: Change percents to decimals
Step #5: Amount goes on bottom
Step #6: Set up and solve
Lake Water
Water
Final Mixture + = 0%
= 0
30%
= 0.3
12%
= 0.12 x 3
x +3

11. + = Final Mixture Lake Water Water 4.5 liters of water should0%
30%
12%
= 0
= 0.3
= 0.12 x 3
x +3 #2
0(x) (3) = (x+3)
= 0.12x
0.9 = 0.12x
4.5 liters of water should
be added to the lake water
0.54 = x
0.54 = 0.12x
x = 4.5

12. + = Final Mixture 3. Sterling Silver is 92.5% pure silver. How many
grams of pure silver and sterling silver must be mixed
to obtain 100g of a 94% Silver alloy?
Step #1: Draw buckets
Step #2: Label the buckets
Step #3: Percent or price goes on top
Step #4: Change percents to decimals
Step #5: Amount goes on bottom
Step #6: Set up and solve
Pure Silver
Sterling Silver
Final Mixture + =
92.5%
= 0.925
100%
= 1
94%
= 0.94 x
100 - x
100

13. + = Final Mixture Pure Silver Sterling Silver #3 92.5% 100% 94% = 1
= 0.925
= 1
= 0.94 x
100 - x
100
0.925x (100-x) = (100)
0.925x – 1x = 94
80 grams of the
sterling silver
and 20 grams
of the pure silver
are needed.
-0.075x = 94
-0.075x = -6
x = 80
100 – 80 = 20
100 - x

14. + = Final Mixture 4. A mixture containing 6% boric acid is to be mixed
with 2 quarts of a mixture which is 15% boric acid in
order to obtain a solution which is 12% boric acid.
How much of the 6% solution must be used?
Step #1: Draw buckets
Step #2: Label the buckets
Step #3: Percent or price goes on top
Step #4: Change percents to decimals
Step #5: Amount goes on bottom
Step #6: Set up and solve
Boric Acid Mixture
Boric Acid Mixture
Final Mixture + = 6%
= 0.06
15%
= 0.15
12%
= 0.12 x 2
x + 2

15. + = Final Mixture Boric Acid Mixture 6% 15% 12% = 0.15 = 0.12 x 2
= 0.06
= 0.15
= 0.12 x 2
x + 2 #4
0.06x (2) = (x+2)
0.06x = x
-0.06x x
1 quart of the 6%
boric acid mixture
should be added.
= x
= 0.06x
1 = x

16. + = Final Mixture 5. ClearShine window cleaner is 12% alcohol and
Sunstream cleaner is 30% alcohol. How much of each
should be used to make 90 oz. (ounces) of a cleaner
that is 20% alcohol?
Step #1: Draw buckets
Step #2: Label the buckets
Step #3: Percent or price goes on top
Step #4: Change percents to decimals
Step #5: Amount goes on bottom
Step #6: Set up and solve
Sunstream
Clearshine
Final Mixture + =
12%
= 0.12
30%
= 0.3
20%
= 0.20 x
90-x 90

17. + = Final Mixture Sunstream Clearshine 12% 30% 20% = 0.3 = 0.20 x 90-x
= 0.12
= 0.3
= 0.20 x
90-x 90 #5
0.12x (90-x) = (90)
0.12x x = 18
-0.18x = 18
50 ounces of Clearshine
window cleaner and 40
ounces of Sunstream
window cleaner.
-0.18x = -9
x = 50