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Word Problems About Mixtures Help

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By — McGraw-Hill Professional
Updated on Sep 27, 2011

Introduction to Mixture Word Problems

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 Increasing/Decreasing by a Percent Solutions fat milk. These problems can be solved using the method illustrated below.

Setting up Mixture Word Problems

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 + 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 .”

Solving Mixture Word Problems

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.

x + ? = 100

- x    – x

? = 100 – x

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

Example 1:

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 .

Increasing/Decreasing by a Percent Examples

(There are 0.10 x 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.)

Increasing/Decreasing by a Percent Examples

Add 60 liters of 10% acid solution to 30 liters of 25% acid solution to achieve a 15% acid solution.

Example 2:

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.

Increasing/Decreasing by a Percent Examples

Increasing/Decreasing by a Percent Examples

Add 25 liters of 10% solution to 100 – x = 100 – 25 = 75 liters of 30% solution to obtain 100 liters of 25% solution.

Example 3:

How much pure alcohol should be added to six liters of 30% alcohol solution to obtain a 40% alcohol solution?

Increasing/Decreasing by a Percent Examples

Increasing/Decreasing by a Percent Examples

Add one liter of pure alcohol to six liters of 30% alcohol solution to obtain a 40% alcohol solution.

Example 4:

How much water should be added to 9 liters of 45% solution to weaken it to a 30% solution?

Think of water as a “0% solution.”

Increasing/Decreasing by a Percent Examples

Increasing/Decreasing by a Percent Examples

Add 4.5 liters of water to weaken 9 liters of 45% solution to a 30% solution.

Example 5:

How much pure acid and 30% acid solution should be mixed together to obtain 28 quarts of 40% acid solution?

Increasing/Decreasing by a Percent Examples

Increasing/Decreasing by a Percent Examples

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.

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