Transcript Document

MIXTURE PROBLEMS
Prepared for Intermediate Algebra
Mth 04 Online
by Dick Gill
The following slides give you nine mixture problems to practice.
Answers to these problems follow. If some of your answers are
wrong, the complete solutions will follow the answer slide.
Three mixture problems at the beginners level. Round your answers
to the nearest tenth if necessary.
1. How many ounces of a solution that is 10% alcohol should be
mixed with 12 ounces of a solution that is is 24% alcohol to create
a solution that is 15% alcohol?
2. How many liters of a solution that is 20% acid should be added
to 3 liters of a solution that is 30% acid to create a solution that is
24% acid?
3. How many liters of a solution that is 50% antifreeze should be
added to 8 liters of a solution that is 80% antifreeze to create a
solution that is 60% antifreeze?
Three mixture problems at the intermediate level. Round your answers
to the nearest tenth if necessary.
4. How many ounces of a solution that is 10% alcohol should be added
to a solution that is 28% alcohol to create 30 ounces of a solution that
is 20% alcohol?
5. How many liters of a solution that is 20% acid should be added to
how many liters of a solution that is 38% acid to create 8 liters of a
solution that is 30% acid?
6. How many liters of a solution that is 50% antifreeze should be
added to how many liters of a solution that is 72% antifreeze to
create 2.4 liters of a solution that is 58% antifreeze?
Three mixture problems at the advanced level. Round your answers
to the nearest hundredth if necessary.
7. Ten liters of a solution that is 30% alcohol is going to be diluted to
24% alcohol by adding water. How much water is needed?
8. A solution that is 30% antifreeze is going to be enriched by adding
pure antifreeze. How much of each is needed to generate 2 gallons of
a solution that is 50% antifreeze?
9. How many gallons should be drained from a 10 gallon tank of
24% alcohol if we are going to replace it with pure alcohol and
create a solution of 35% alcohol?
Answers to mixture problems 1 – 9.
1. 21.6 ounces
2. 4.5 liters
3. 16 liters
4. 13.3 ounces
5. 3.6 liters at 20%; 4.4 liters at 38%
6. 1.5 liters at 50%; 0.9 liters at 72%
7. 2.5 liters
8. 1.43 gal at 30%; 0.57 gal at 100%
9. 1.45 gal
Complete solutions follow.
1.
Percent Amount of Amount of
solution
target
1st
container
.10
x
.10x
2nd
container
.24
12
.24(12)
Mixture
.15
x + 12
.15(x + 12)
.10x + .24(12) = .15(x + 12)
.10x + 2.88 = .15x + 1.8
1.08 = .05x
21.6 = x
21.6 ounces @ 10%
2.
Percent Amount of Amount of
solution
target
1st
container
.20
x
.20x
2nd
container
.30
3
.30(3)
Mixture
.24
x+3
.24(x + 3)
.20x + .90 = .24x + .72
.18 = .04x
4.5 = x
4.5 liters at 20%
3.
Percent Amount of Amount of
solution
target
1st
container
.50
x
.50x
2nd
container
.80
8
.80(8)
Mixture
.60
x+8
.60(x + 8)
.50x + .80(8) = .60(x + 8)
.50x + 6.40 = .60x + 4.80
1.60 = .10x
16 = x
16 liters at 50%
4.
Percent Amount of Amount of
solution
target
1st
container
.10
x
.10x
2nd
container
.28
30 - x
.28(30 – x)
Mixture
.20
30
.20(30)
.10x + .28(30 – x) = .20(30)
.10x + 8.4 - .28x = 6
-.18x = -2.4
x = 13.3
13.3 ounces at 10%
5.
Percent Amount of Amount of
solution
target
1st
container
.20
X
.20x
2nd
container
.38
8-x
.38(8 – x)
Mixture
.30
8
.30(8)
.20x + .38(8 – x) = .30(8)
.20x + 3.04 - .38x = 2.40
-.18x = -.64
x = 3.6; 8 – x = 4.4
3.6 liters at 20%; 4.4 liters at 38%
6.
Percent Amount of Amount of
solution
target
1st
container
.50
x
.50x
2nd
container
.72
2.4 - x
.72(2.4 – x)
Mixture
.58
2.4
.58(2.4)
.50x + .72(2.4 – x) = .58(2.4)
.50x + 1.728 - .72x = 1.392
-.22x = -.336
x = 1.5; 2.4 – x = 0.9
1.5 liters at 50%; 0.9 liters at 72%
7.
Percent Amount of Amount of
solution
target
1st
container
.30
10
.30(10)
2nd
container
0
x
0(x)
Mixture
.24
x + 10
.24(x + 10)
.30(10) + 0 = .24(x + 10)
3 = .24x + 2.4
0.6 = .24x
2.5 = x
2.5 liters of water
8.
Percent Amount of Amount of
solution
target
1st
container
.30
x
.30(x)
2nd
container
1.00
2-x
1.00(2 – x)
Mixture
.50
2
.50(2)
.30(x) + 1.00(2 – x) = .50(2)
.30x + 2.00 – 1.00x = 1.00
-.70x = -1.00
x = 1.43; 2 – x = 0.57
1.43 gal at 30%; 0.57 gal at 100%
9.
Percent Amount of Amount of
solution
target
1st
container
.24
10 - x
.24(10 – x)
2nd
container
1.00
x
1.00x
Mixture
.35
10
.35(10)
.24(10 – x) + 1.00x = .35(10)
2.4 - .24x + 1.00x = 3.5
.76x = 1.1
x = 1.45
Drain 1.45 gal and replace with 100% alcohol