Percent-Mixture Solutions Problems
(Std. 15.0 Students apply algebraic techniques to solve rate problems,
work problems and percent-mixture problems.)
1. Grain Mixture for Horses. Brenda is a barn manager at a horse
stable. She needs to calculate the correct mix of grain and hay to feed
her horse. On the basis of her horse’s age, weight, and workload, she
determines that he needs to eat 15 lb. of feed per day, with an average
protein content of 8%. Hay contains 6% protein, whereas grain has a 12%
protein content. How many pounds of hay and grain should she feed her
horse each day?
Hay Solution
Grain Solution
Mixture
Equation
Amount of
Solution
Percent of
Protein
Amount of
Protein in
Solution
2. Paint Mixtures. At a local “paint swap”, Jaime found large supplies
of Skylite Pink (12.5% red pigment) and MacIntosh Red (20% red pigment).
How many gallons of each color should Jaime pick up in order to mix a
gallon of Summer Rose (17% red pigment)?
Skylite Pink
Solution
MacIntosh Red
Solution
Summer Rose
Mixture
Equation
Amount of
Solution
Percent of
red pigment
Amount of red
pigment in
Solution
3. Mixture of Solutions (Acid). Solution A is 50% acid and solution B
is 80% acid. How many liters of each should be used in order to make
100 L (liters) of a solution that is 68% acid?
Solution A
Amount of
Solution
Percent of
Acid
Amount of Acid
in Solution
Solution B
Mixture
Equation
4. Mixture of Solutions (Alcohol). Solution A is 30% alcohol and
solution B is 75% alcohol. How much of each should be used in order to
make 100 L (liters) of a solution that is 50% alcohol?
Hay Solution
Grain Solution
Mixture
Equation
Amount of
Solution
Percent of
Protein
Amount of
Protein in
Solution
PERCENT-MIXTURE PROBLEMS (CONT’D)
PAGE
2
5. Production. ClearShine window cleaner is 12% alcohol and Sunstream
window cleaner is 30% alcohol. How much of each should be used to make
90 oz. (ounces) of a cleaner that is 20% alcohol?
Clearshine
Sunstream
Mixture
Equation
cleaner
cleaner
Amount of
Solution
Percent of
Alcohol
Amount of
Alcohol in
Solution
6. Mixed Nuts. A customer has asked
(pounds) of nuts, 60% of which are to
available mixtures of 70% cashews and
each mixture should be used?
Cashew
Cashew
mixture 1
mixture
Amount of
Solution
Percent of
Cashews
Amount of
Cashews in
Solution
a caterer to provide 60 lb.
be cashews. The caterer has
45% cashews. How many pounds of
Mixture
Equation
2
7. Cough Syrup. Dr. Duarte’s cough syrup is 2% alcohol. Vitabrite
cough syrup is 5% alcohol. How much of each type of cough syrup should
be used in order to prepare an 80-oz. (ounces) batch of cough syrup that
is 3% alcohol?
Dr. Duarte
cough syrup
Vitabrite
cough syrup
Mixture
Equation
Amount of
Solution
Percent of
Alcohol
Amount of
Alcohol in
Solution
8. Milk Mixture. A farmer has 100 L (liters) of milk that is 4.6%
butterfat. How much skim milk (no butterfat) should be mixed with it in
order to make milk that is 3.2% butterfat?
Butterfat
Skim milk
Mixture
Equation
milk
Amount of
Solution
Percent of
Butterfat
Amount of
Butterfat in
Solution
Percent-Mixture Solution Problems
(Std. 15.0 Students apply algebraic techniques to solve rate problems,
work problems and percent-mixture problems.)
1. Grain Mixture for Horses. Brenda is a barn manager at a horse
stable. She needs to calculate the correct mix of grain and hay to feed
her horse. On the basis of her horse’s age, weight, and workload, she
determines that he needs to eat 15 lb. of feed per day, with an average
protein content of 8%. Hay contains 6% protein, whereas grain has a 12%
protein content. How many pounds of hay and grain should she feed her
horse each day?
xxxxxxxxxxxxxxxxxxxx Hay
Solution
Amount of Solution
H = 10
lbs.
Percent of Protein
.06 H
Amount of Protein in .06 (10)
Solution
.6 lbs.
=
Grain
Solution
G = 5 lbs.
Mixture
Equation
15 lbs.
H + G = 15
.12 G
.08 (15)
6H + 12G = 120
.12 (5) =
.6 lbs.
1.2 lbs of
mixture is
protein
xxxxxxxxxxxxxxxxxx
System of linear equations:
H + G = 15 
h = 15 – g
(15 – G) + 12G = 120
Substitute into 2nd equation:
Simplify:
6
90 – 6G + 12G =
120
Combine “like” terms:
90 + 6G
= 120
Solve for variable G:
6G = 30

G = 5 lbs.
Go back and substitute into above table, solve for H and G and find out
how much protein is in the solution.
2. Paint Mixtures. At a local “paint swap”, Jaime found large supplies
of Skylite Pink (12.5% red pigment) and MacIntosh Red (20% red pigment).
How many gallons of each color should Jaime pick up in order to mix a
gallon of Summer Rose (17% red pigment)?
xxxxxxxxxxxxxxxxxxx Skylite
Pink
Solution
Amount of Solution
P = .4
gallon
Percent of red
.125 P
pigment
Amount of red
.125 (.4) =
pigment in Solution .05 gallon
System of linear equations:
P + R = 1 
R = 1 – P
200 (1 – P) = 170
MacIntosh Red
Solution
Summer Rose
Mixture
Equation
R = .6 gallon
1 gallon
P + R = 1
.20 R
.17 (1)
125P + 200R = 170
.20 (.6) =
.120 gallon
.17 (17%)
of a gallon
is red
pigment
xxxxxxxxxxxxxxxxx
Substitute into 2nd equation:
Simplify:
125P +
125P + 200 – 200P
= 170
Combine “like” terms:
-75P +
200 = 170
Solve for variable P:
-75P =
-30  P = .4 gallon
Go back and substitute into above table, solve for P and R and find out
how much red pigment is in the solution.
3. Mixture of Solutions (Acid). Solution A is 50% acid and solution B
is 80% acid. How many liters of each should be used in order to make
100 L (liters) of a solution that is 68% acid?
xxxxxxxxxxxxxxxxxxxx Solution A
Solution B
Mixture
Equation
Amount of Solution
B = 60
liters
.80 B
100 Liters
A + B = 100
Percent of Acid
A = 40
liters
.50 A
.68 (100)
50A + 80B = 6800
Amount of Acid in
Solution
.50 (40) =
20.00
.80 (60) =
48.00
68%
xxxxxxxxxxxxxxxxxxx
System of linear equations:
A + B = 100 
B = 100 - A
50A + 80 (100 – A) = 6800
Substitute into 2nd equation:
Simplify:
50A + 8000 – 80A =
6800
Combine “like” terms:
-30A +
8000 = 6800
Solve for variable A:
-30A =
-1200  A = 40 liters
Go back and substitute into above table, solve for A and B and find out
how much acid is in the solution.
4. Mixture of Solutions (Alcohol). Solution A is 30% alcohol and
solution B is 75% alcohol. How much of each should be used in order to
make 100 L (liters) of a solution that is 50% alcohol?
xxxxxxxxxxxxxxxxxx Solution A
Solution B
Mixture
Equation
Amount of Solution A = 55.5
liters
Percent of Alcohol .30 A
B = 44.5
liters
.75 B
100 liters
A + B = 100
.50 (100)
30A + 75B = 5000
Amount of Alcohol
in Solution
.75 (44.5) =
33.35
50%
xxxxxxxxxxxxxxxxx
.30 (55.5) =
16.65
System of linear equations:
A + B = 100 
B = 100 - A
30A + 75 (100 – A) = 5000
Substitute into 2nd equation:
Simplify:
30A + 7500 – 75A =
5000
Combine “like” terms:
-45A +
7500 = 5000
Solve for variable A:
-45A =
-2500 
A = 55.5 liters
Go back and substitute into above table, solve for A and B and find out
how much alcohol is in the solution.
5. Production. ClearShine window cleaner is 12% alcohol and Sunstream
window cleaner is 30% alcohol. How much of each should be used to make
90 oz. (ounces) of a cleaner that is 20% alcohol?
xxxxxxxxxxxxxxxxxxx Clearshine
cleaner
Amount of Solution
C = 50
ounces
Percent of Alcohol
.12 C
Sunstream
cleaner
S = 40
ounces
.30 S
Mixture
Equation
90 ounces
C + S = 90
.20 (90)
12C + 30S = 1800
Amount of Alcohol
in Solution
.30 (40) =
12
18%
xxxxxxxxxxxxxxxx
.12 (50) =
6
System of linear equations:
C + S = 90 
C = 90 - S
12 (90 – S) + 30S = 1800
Substitute into 2nd equation:
Simplify:
1080 – 12S + 30S =
1800
Combine “like” terms:
1080 +
18S = 1800
Solve for variable S:
18S =
720  S = 40 ounces
Go back and substitute into above table, solve for C and S and find out
how much alcohol is in the solution.
6. Mixed Nuts. A customer has asked a caterer to provide 60 lb.
(pounds) of nuts, 60% of which are to be cashews. The caterer has
available mixtures of 70% cashews and 45% cashews. How many pounds of
each mixture should be used?
xxxxxxxxxxxxxxxxxxx Cashew
Cashew
Mixture
Equation
mixture A
mixture B
Amount of Solution
A = 36
B = 24
60 pounds
A + B = 60
pounds
pounds
Percent of Cashews
.70 A
.45 B
.60 (60)
70A + 45B = 3600
Amount of Cashews
in Solution
.70 (36)
=
25.2%
.45 (24)
=
10.8%
System of linear equations:
A + B = 60 
A = 60 - B
70 (60 – B) + 45B = 3600
36
xxxxxxxxxxxxxxxxxx
Substitute into 2nd equation:
Simplify:
4200 – 70B + 45B =
3600
Combine “like” terms:
4200 –
25B = 3600
Solve for variable A:
-25B =
-600  B = 24 pounds
Go back and substitute into above table, solve for A and B and find out
how much cashews are in the solution.
7. Cough Syrup. Dr. Duarte’s cough syrup is 2% alcohol. Vitabrite
cough syrup is 5% alcohol. How much of each type of cough syrup should
be used in order to prepare an 80-oz. (ounces) batch of cough syrup that
is 3% alcohol?
xxxxxxxxxxxxxxxxxxx Dr.
Duarte
cough
syrup
Amount of Solution
D = 53.3
ounces
Percent of Alcohol .02 D
Vitabrite
Mixture
cough syrup
Equation
V = 26.7
ounces
.05 V
80 ounces
D + V = 80
.03 (80)
2D + 5V = 240
Amount of Alcohol
in Solution
.05 (26.7)
= 1.334
2.4
xxxxxxxxxxxxxxxxx
.02
(53.3) =
1.066
System of linear equations:
D + V = 80 
D = 80 - V
(80 – V) + 5V = 240
Substitute into 2nd equation:
Simplify:
2
160 – 2V + 5V =
240
Combine “like” terms:
160 + 3V
Solve for variable V:
3V = 80
= 240
 V = 26.7 ounces
Go back and substitute into above table, solve for D and V and find out
how much alcohol is in the solution.
8. Milk Mixture. A farmer has 100 L (liters) of milk that is 4.6%
butterfat. How much skim milk (no butterfat) should be mixed with it in
order to make milk that is 3.2% butterfat?
xxxxxxxxxxxxxxxxxxx Butterfat
Skim milk Mixture
Equation
milk
Amount of Solution
100
S = 43.75 M = 143.75
100 + S = M
Percent of
.046 (100)
Butterfat
Amount of Butterfat 4.6
in Solution
System of linear equations:
4600 = 32M  M = 143.75
= 143.75
.00 S
.032 M
4600 = 32M
0.0
.032
(143.75) =
4.6
xxxxxxxxxxxxxxxx
Substitute into 2nd equation:
Combine “like” terms:
100 + S
100 – 100
+ S = 143.75 - 100
Solve for variable S:
S =
43.75
Go back and substitute into above table, solve for B and S and find out
how much protein is in the solution.