Mixture problems involve mixing two different concentrations to obtain some concentration in between. Often these problems are stated as alcohol or acid solutions, but there are many more types. For example, you might want to know how many pure peanuts should be mixed with a 40% peanut mixture to obtain a 50% peanut mixture. You might have a two-cycle engine requiring a particular oil and gas mixture. Or, you might have a recipe calling for 1% fat milk and all you have on hand is 2% fat milk and 1 2% fat milk. These problems can be solved using the method illustrated below.
There will be three quantities—the two
concentrations being mixed together and the final concentration. One of the
three quantities will be a fixed number. Let the variable represent one of the
two concentrations being mixed The other unknown quantity will be written as
some combination of the variable and the fixed quantity. If one of the
quantities being mixed is known, then let x represent the other quantity being
mixed and the final solution will be ‘‘x þ known quantity.’’ If the final
solution is known, again let x represent one of the quantities being mixed, the
other quantity being mixed will be of the form ‘‘final solution quantity – x.’’
For example, in the following problem, the
amount of one of the two concentrations being mixed will be known.
‘‘How many liters of 10% acid solution
should be mixed with 75 liters of 30% acid solution to yield a 25% acid
solution?’’ Let x represent the number of liters of 10% acid solution. Then x +
75 will represent the number of liters of the final solution. If the problem
were stated, ‘‘How many liters of 10% acid solution and 30% solution should be
mixed together with to produce 100 liters of 25% solution?’’ We can let x represent
either the number of liters of 10% solution or 30% solution. We will let x represent
the number of liters of 10% solution. How do we represent the number of liters
of 30% solution? For the moment, let ‘‘?’’ represent the number of liters of
30% solution. We know that the final solution must be 100 liters, so the two
amounts have to sum to 100: x + ? = 100.
Now we see
that 100 – x represents the number of liters of 30% solution.
Draw three boxes. Write the percentages
given above the boxes and the volume inside the boxes. Multiply the percentages
(converted to decimal numbers) and the volumes. Write these quantities below
the boxes, this will give you the equation to solve. Incidentally, when you
multiply the percent by the volume, you are getting the volume of pure acid/ alcohol/milk-fat/etc.
Examples
How much 10%
acid solution should be added to 30 liters of 25% acid solution to achieve a
15% solution?
Let x represent the amount of 10% solution.
Then the total amount of solution will be 30 + x.
(There are
0.10x liters of pure acid in the 10% mixture, 0.25(30) liters of pure acid in
the 25% mixture, and 0.15(x + 30) liters of pure acid in the 15% mixture.)
Add 60
liters of 10% acid solution to 30 liters of 25% acid solution to achieve a 15%
acid solution.
How much 10%
acid solution and 30% acid solution should be mixed together to yield 100
liters of a 25% acid solution?
Let x represent the amount of 10% acid
solution. Then 100 – x represents the amount of 30% acid solution.
Add 25 liters
of 10% solution to 100 – x = 100 – 25 = 75 liters of 30% solution to obtain 100
liters of 25% solution.
How much
pure alcohol should be added to six liters of 30% alcohol solution to obtain a
40% alcohol solution?
Think of water as a ‘‘0% solution.’’
Add 4.5
liters of water to weaken 9 liters of 45% solution to a 30% solution.
How much
pure acid and 30% acid solution should be mixed together to obtain 28 quarts of
40% acid solution?
Add 4 quarts
of pure acid to 28 – x = 28 – 4 = 24 quarts of 30% acid solution to yield 28
quarts of a 40% solution.
Practice
1. How
much 60% acid solution should be added to 8 liters of 25% acid solution to
produce a 40% acid solution?
2. How
many quarts of 1 2% fat milk should be added to 4 quarts of 2% fat milk to
produce 1% fat milk?
3. How
much 30% alcohol solution should be mixed with 70% alcohol solution to produce
12 liters of 60% alcohol solution?
4. How
much 65% acid solution and 25% acid solution should be mixed together to
produce 180 ml of 40% acid solution?
5. How
much water should be added to 10 liters of 45% alcohol solution to produce a
30% solution?
6. How
much decaffeinated coffee (assume this means 0% caffeine) and 50% caffeine
coffee should be mixed to produce 25 cups of 40% caffeine coffee?
7. How
much pure acid should be added to 18 ounces of 35% acid solution to produce 50%
acid solution?
8. How
many peanuts should be mixed with a nut mixture that is 40% peanuts to produce
36 ounces of a 60% peanut mixture?
Solutions
1.
Add
6 liters of 60% solution to 8 liters of 25% solution to produce a 40% solution.
2.
Add 8 quarts of ½ % fat milk to 4 quarts of 2% milk to produce 1% milk.
3.
Add 3 liters of 30% alcohol solution to 12 _ x ¼ 12 _ 3 ¼ 9 liters of 70%
alcohol solution to produce 12 liters of 60% alcohol solution.
4.
Add 67.5 ml of 65% acid solution to 180 – x = 180 – 67.5 = 112.5 ml of
25% solution to produce 180 ml of 40% acid solution.
5.
Add 5 liters of water to 10 liters of 45% alcohol solution to produce a
30% alcohol solution.
6.
Mix 5 cups of decaffeinated coffee with 25 – x = 25 – 5 = 20 cups of 50%
caffeine coffee to produce 25 cups of 40% caffeine coffee.
7.
Add 5.4 ounces of pure acid to 18 ounces of 35% acid solution to produce
a 50% acid solution.
8.
Add 12 ounces of peanuts to 36 – x = 36 – 12 = 24 ounces of a 40% peanut
mixture to produce 36 ounces of a 60% peanut mixture.
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