Intermediate Algebra – 4
HFCC Math Lab
SOLVING MIXTURE PROBLEMS
PERCENET MIXTURE PROBLEMS
Percent mixture problems generally involve two or more solutions made up of mixture of
two or more ingredients. As in most application problems, the solution involves setting up
and then solving an equation. For all mixture problems the following rule should be
followed:
Of the given ingredients, choose one of them, and setup the equation in terms of the
amount of this ingredient.
Example 1:
Two antifreeze solutions are available, the first with 75% strength, and the second with a
20% strength. How much of each solution should be used in order to produce 20 quarts of a
50% solution of antifreeze?
Solution: We choose the amount of antifreeze as the basis of the equation (using our rule).
We begin with a table listing the important data for this problem. At this stage, we introduce
the variable (the unknown). The table should have three columns and as many rows as
necessary depending on the number of solutions being used.
The first column will always have the representation of the amount of each solution. The
unknown will appear in one or more places in this column. The data for the second column is
usually given in the statement of the problem. All percents must be changed to decimals or
fractions. The entries in the third column are found by multiplying the corresponding
elements in column 1 and column 2. The required equation is derived from the entries in the
third column. See the table generated for example 1 below.
Amount of Solution
Solution 1
Solution 2
Mixture
x
20 - x
20
Percent of
Antifreeze
0.75
0.20
0.50
Amount of Pure
Antifreeze
0.75x
0.20(20-x)
20*0.50 = 10
Now, the amount of pure antifreeze from solution 1 added to the amount of antifreeze from
solution 2 should equal the amount of antifreeze in the mixture.
Therefore, the required equation is:
0.75x 0.20(20 - x) 0.50(20)
Revised 11/09
1
Solve the equation for the variable, x
0.75 x 0.20(20 - x) 0.50(20)
0.75 x 4 0.2 x 10
0.55 x 4 10
0.55 x 4 4 10 4 [subtract 4 from both sides to solve for x]
6
0.55 x 6 x
10.9
0.55
Therefore, we should have 10.9 quarts of solution 1 and 20 – 10.9 = 9.1 quarts of solution 2
in order to make a mixture of 20 quarts that is 50% antifreeze.
Example 2: How much water should be evaporated from 600 gallons of a 10% salt solution
to make the solution 16% salt?
Solution: Choose an ingredient from among the two that we have (salt and water). In this
case, we choose the salt ingredient.
Set up a table and use it to generate the equation
Percent of Salt
Starting Solution
Amount of Solution
(gal)
600
0.10
Amount of Salt
(gal)
0.10*600 = 60
Evaporated Amount
Final Solution
x
600 - x
0
0.16
0
0.16(600 – x)
Now, the amount of salt before evaporating must equal the amount of salt after evaporating
water. No change in the amount of salt. The only change is in the amount of water
remaining. Therefore, the required equation is:
0.1(600) 0 x 0.16(600 x)
Solve the equation for x.
60 96 0.16 x
60 96 96 0.16 x 96
36
0.16 x [divide both sides by -0.16]
36
x
225
0.16
Therefore 225 gallons of water should be evaporated to produce the required salt
concentration.
Revised 11/09
2
Example 3: Fred wants to raise the strength of antifreeze in his truck radiator from 38% to
50% by draining off a part of the present 38% strength solution and replacing with pure
antifreeze. The capacity of the radiator is 30 quarts. How much should he drain and replace.
Solution: We choose to base our equation on the amount of antifreeze. Then we set up the
table that is used to generate the equation.
Starting Solution
Amount of Solution
(qt.)
30
Percent of
Antifreeze
0.38
Amount of
Antifreeze (qt.)
0.38(30)
Remove
Add
Final Solution
x
x
30
0.38
1.00
0.50
0.38(30)
1.0(x)
0.50(30) = 15
Therefore, the required equation is:
0.38(30)
0 38 x x 0.50(30)
Solve the equation for x.
11.40 0.62 x 15
0.62 x 15 11.40 [subtract 11.40 from both sides of the equation]
0.62 x 3.60 [divide both sides by 0.62]
x
5.8
Therefore, he should drain 5.8 quarts of existing solution and replace with 5.8 quarts of pure
antifreeze.
Revised 11/09
3
EXERCISES
For exercises 1 – 5, use the following problem. How much pure alcohol should be added to
30 quarts of a 15% solution of alcohol to give a solution that is 60% alcohol?
1. In setting this problem up, how many solutions are there? How many ingredients?
2. List all the solutions
3. Assuming that we wish to base our equation on the amount of alcohol, set up threecolumn table and supply all table entries
4. Write the equation in terms of alcohol.
5. Solve the equation
6. How much water should be added to 50 gallons of a 15% salt solution to give a 7%
solution?
7. One acid solution is 25% strength and the second is 10% strength. How much of each
should used to give a 30 oz. of a 15% strength solution?
8. How much iodine should be added to 20 oz. of an 8% iodine solution to give a solution
that 15% iodine?
9. A radiator contains 40 quarts of a solution that is 25% antifreeze. How much should be
drained from the radiator and replaced by pure antifreeze to give a solution that is 50%
antifreeze?
10. A radiator contains 40 quarts of a solution that is 25% antifreeze. How much should be
drained from the radiator and replaced with pure water to give a solution that is 10%
antifreeze?
11. A druggist must make 20 oz. of 12% argyrols from his supply of 5% and 15% solutions.
How much of the 5% solution should be used?
12. A student has 50 cc of 30% sulfuric acid. How much of this solution must be removed
and replaced with pure sulfuric acid to bring the acid strength to 58% strength?
13. A nurse must make 20% alcohol solution by diluting 10 oz. of a 25% solution. How
much water must be added?
14. Two alloys, one 20% silver and 80% copper and the other 75% silver and 25% copper
are mixed to give 50 pounds of an alloy that is 30% silver. How much of each should be
used?
15.Two alloys, one 1 part silver and 5 parts copper and the other 3 parts silver and 1 part
copper are mixed to form 350 pounds of an alloy that is equal in silver and copper
content. How many pounds of each should be used?
Revised 11/09
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ANSWERS AND SOLUTIONS
1. Three solutions, two ingredients.
2. Pure alcohol, 15% solution, 60% solution
3. Let x be the amount of pure alcohol in quarts.
Percent of Alcohol
Alcohol
Amount of Solution
(qt.)
x
1.00
Amount of Alcohol
(qt.)
x
15% Solution
Mixture
30
x + 30
0.15
0.60
0.15*30 = 4.5
0.60( x + 30)
Percent of Salt
Water
Amount of Solution
(gal)
x
0
Amount of Salt
(gal)
0
15% Solution
Mixture
50
x + 50
0.15
0.07
0.15*50 = 7.5
0.07( x + 50)
4. The equation: x + 4.5 = 0.60(x + 30)
5. Solving the equation:
x 4.5 0.60 x 0.60 30
x 4.5 0.6 x 18
x 0.6 x 18 4.5
13.5
0.4 x 13.5 x
33.75 qt.
0.4
6. Let x be the number of gallons of water.
0 7.5 0.07( x 50)
7.5 0.07 x 3.5
7.5 3.5 0.07 x [subtract 3.5 from both sides]
4
57.14 gal.
0.07
Therefore, 57.14 gallons of water should be added to make the required solution.
0.07 x
4
x
7. 10 oz. of the 25% solution and 20 oz. of the 10% solution.
Revised 11/09
5
8. Let x = the amount of pure iodine in ounces.
Percent of Iodine
Iodine
Amount of Solution
(oz)
x
1.0
Amount of Iodine
(oz)
x
8% Solution
Mixture
20
x + 20
0.08
0.15
0.08*20 = 1.6
0.15( x + 20)
x 1.6 0.15( x 20)
x 1.6 0.15 x 3.0
x 0.15 x 3 1.6
1.4
1.65 oz. approximately.
0.85
Therefore, 1.65 oz. of iodine should be added to make the required solution.
0.85 x 1.4
x
9. 13.33 quarts approximately.
10. Let x be the number of quarts drained and replaced.
Starting Solution
Amount of
Solution
40
Percent of
Antifreeze
0.25
Amount of
Antifreeze
0.25(40)
Drain
Add
Final Solution
x
x
40
0.25
0.0
0.10
0.25x
0
0.10(40)
0.25(40) 0.25 x 0 x
0.10(40)
10 0.25 x 4
10 4 0.25 x
6
24 qt.
0.25
Therefore, 20 quarts of the existing solution should be drained and replaced with 20
quarts of water.
0.25 x
6
x
11. 6 ounces.
Revised 11/09
6
12. Let x be the number of cc removed (and replaced)
Starting Solution
Amount of
Solution
50
Percent of
Antifreeze
0.30
Amount of
Antifreeze
0.30(50)
x
0.30
0.3x
Remove
x
1.0
1x
Add
50
0.58
0.58(50) = 29
Final Solution
0.30(50) 0.3 x x 29
15 0.3 x x 29
15 0.7 x 29
14
0.7 x 14 x
20 cc.
0.7
Therefore, 20 cc of the existing solution should be removed and replaced with 20 cc of
pure sulfuric acid.
13. 2.5 ounces.
14. Choose silver as the ingredient upon which to base the equation.
Let x be the number of pounds.
1st Alloy
nd
2 Alloy
Mixture
Amount of Alloy
Percent of Silver
Amount of Silver
x
0.20
0.20x
50 - x
50
0.75
0.30
0.75(50- x)
0.30(50) = 15
0.20 x 0.75(50 x) 15
0.2 x 37.50 0.75 x 15
0.55 x 15 37.5
22.5
22.5
40.9 lb. approximately
0.55
Therefore, 40.9 lbs of silver alloy and 9.1 lbs of copper alloy should be used.
0.55 x
22.5
Revised 11/09
x
7
15. Set up the equation based on silver ingredient.
Let x be the number of pounds of 1st alloy.
Amount of Alloy
Percent of Silver
Amount of Silver
1st Alloy
x
1/6
(1/6)x
2nd Alloy
Mixture
350 - x
350
3/4
1/2
(3/4)(350- x)
(1/2)(350) = 175
1
6
x
3
4
(350 x)
1
2
350
1
6
x
3
4
(350 x)
1
2
350 [Multiply both sideds by 12 to eliminate the denominators]
2x 9(350 x)
2 x 3150 - 9 x
7x
2100
2100
2100 3150
1050
1050
150
7
150 lbs of the first alloy and 200 lbs of the second alloy should be used.
x
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NOTE: You can get additional instruction and practice by going to the
following websites:
http://www.purplemath.com/modules/mixture.htm
Demonstrates, step-by-step and with worked examples, how to set up and
solve mixture word problems.
http://www.analyzemath.com/math_problems/mixture_problems.html
More mixture problems with worked out solutions.
Revised 11/09
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