Examples of Setting Up "Mixture" Word Problems
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The solutions to mixture problems tend to be generated by linear equations; that is, solving the set-up for mixture exercises tends to be pretty easy. It's the set-up that can be a bear.
What follows are some examples showing that set-up.
These examples demonstrate how to take the information you've been given, and how to use that to pick variables, create a table to organize everything, and then create the equations. In the exercises below are a few more problems, together with their grids (but not their solutions; I've left that part for you to do).
- How many liters of a 70% alcohol solution must be added to 50 liters of a 40% alcohol solution to produce a 50% alcohol solution?
You will be mixing a stronger solution with a weaker solution. The number of liters of the stronger solution is currently unknown; I'll use s to stand for the number of liters of the stronger solution. Then the number of liters of the mixture you're making will be s + 50.
Make a table, with the columns showing the numbers of liters of solutions, the percentage of alcohol in each solution (this is your "rate"), and the number of liters of actual alcohol in each solution.
| liters sol'n |
percent alcohol |
total liters alcohol |
|
|---|---|---|---|
| 70% sol'n | s | 0.70 | 0.70s |
| 40% sol'n | 50 | 0.40 | 0.40(50) = 20 |
| 50% sol'n | s + 50 | 0.50 | 0.50(s + 50) |
From the last column, you get the equation:
0.7s + 20 = 0.50(s + 50)
Solve for the value of the variable. Remember to put appropriate units (in this case, "liters") on your hand-in answer.
Forty percent is closer to 50% than is 70%. So you would logically expect that you'll end up using fewer liters of 70% solution than the 50 liters of 40% solution.
- How many ounces of pure water must be added to 50 ounces of a 15% saline solution to make a saline solution that is 10% salt?
I'll use w to stand for the number of ounces of water that are needed. And, since pure water has no salt in it, then the percentage of water that is salt is zero, and the number of ounces of salt is also zero.
| ounces liquid |
percent salt |
ounces salt |
|
|---|---|---|---|
| water | w | 0 | 0 |
| 15% sol'n | 50 | 0.15 | 0.15(50) = 7.5 |
| 10% sol'n | 50 + w | 0.10 | 0.10(50 + w) |
From the last column, you get the equation:
7.5 = 0.10(50 + w)
Solve for the value of w. Remember to put the appropriate units (in this case, "ounces") on your hand-in answer.
Note: This exercise is typical of algebra courses, and is used to teach problem-solving skills in algebra classes. But, in real life, the actual process is different. (Example) If you're in a chemistry lab, expect the measuring and mixing to work in a different way.
- Find the selling price per pound of a coffee mixture made from 8 pounds of coffee that sells for $9.20 per pound and 12 pounds of coffee that costs $5.50 per pound.
This exercise asks us to make the implicit assumption that the selling price of the mix is based only on the selling price (and amounts) of the inputs. Of course, in real life, the selling price of the mix would be a markup on the cost of the mix, and the cost of the mix would be related to the costs of the inputs, plus the extra costs involved in mixing and re-bagging. But this is algebra, not real life.
The price per pound is the "rate" for this exercise. The sum of the prices of the inputs is assumed to be equal to the total price for the mixture.
| pounds of coffee |
cost per pound |
total cost of coffee |
|
|---|---|---|---|
| pricy | 8 | 9.20 | 8(9.20) = 73.60 |
| cheapo | 12 | 5.50 | 12(5.50) = 66 |
| mixture | 8 + 12 = 20 | ? | 73.60 + 66 = 139.60 |
From the last row, you see that you have 20 pounds of coffee mixture. This mixture will sell for $139.60. To find the selling price per pound of the mixture, divide ($139.60) by (20 pounds). Simplify the division to find the unit rate.
Remember to put appropriate units (in this case, "dollars per pound") on your hand-in answer.
Note that, in this case, no variable was actually necessary. But if you'd picked a variable to replace the query-mark (that is, the "?" in the bottom row), this would also have been fine.
- How many pounds of lima beans that cost $0.90 per pound must be mixed with 16 pounds of corn that costs $0.50 per pound to make a mixture of vegetables that costs $0.65 per pound?
The cost per pound is the "rate" for this exercise.
You are given the number of pounds of corn, but not the number of pounds of beans. I'll use b to stand for this amount.
| pounds of veggies |
cost per pound |
total cost of veggies |
|
|---|---|---|---|
| beans | b | 0.90 | 0.90b |
| corn | 16 | 0.50 | 16(0.50) = 8 |
| mix | b + 16 | 0.65 | (b + 16)(0.65) |
The cost of the inputs add to the cost of the mix, which (from the far right column) gives the equation:
0.90b + 8 = 0.65(b + 16)
Solve for the value of the variable. Remember to put the appropriate units (in this case, "pounds") on your hand-in answer.
- Two hundred liters of a punch that contains 35% fruit juice is mixed with 300 liters (L) of another punch. The resulting fruit punch is 20% fruit juice. Find the percent of fruit juice in the 300 liters of punch.
The percentage of the punch that comes from actual fruit is the "rate" for this exercise. Since the exercise is asking for a percentage, I will use the variable f.
| punch, L | % juice | juice, L | |
|---|---|---|---|
| 35% juice | 200 | 0.35 | 0.35(200) = 70 |
| other juice | 300 | p | 300p |
| mixture | 200 + 300 = 500 | 0.20 | 0.20(500) = 100 |
The sum of the input amounts of juice will equal the total amount of juice in the mixture. You can use the last column to create the equation:
70 + 300p = 100
Or you can just eyeball the amounts and notice that 300p must equal 30. Either way, do the division to find the value of the variable. Remember that you're looking for a percentage, so you'll need to convert the decimal solution into percentage form.
- Ten grams of sugar are added to a 40-g serving of a breakfast cereal that is 30% sugar. What is the percent concentration of sugar in the resulting mixture?
Note that, because sugar is 100% sugar, the percentage of sugar in what is added to the bowl, in decimal form, is 1.00.
I will use the variable s to stand for the percentage of sugar in the mixture.
| grams in bowl |
percent sugar |
grams sugar |
|
|---|---|---|---|
| sugar | 10 | 1.00 | 10(1.00) = 10 |
| cereal | 40 | 0.30 | 40(0.30) = 12 |
| mixture | 10 + 40 = 50 | s | 10 + 12 = 22 |
From the bottom row, you see that there are 22 grams of sugar in the 50 grams in the bowl, or 22/50. Convert the decimal value to percentage form.
As it turns out in this case, the variable wasn't actually necessary. But there's no harm in developing the habit of defining one.
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