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Formulas and Mixtures Word Problems Study Guide (page 3)

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Updated on Aug 24, 2011

Mixture Problems

A mixture problem is just like what it sounds—when two quantities are mixed together. For example, a salt solution is a mixture of salt and water. We can change the solution by adding or removing salt or water. Or we could mix a solution that has a lot of salt in it with a solution that has only a little salt in it to make a new solution.

We solve mixture problems by making tables to organize the data. Let's look at an example.

Example

Miguel has 10 ounces of a solution of salt and water that is 30% salt. He has 10 ounces of this solution, but he wants the solution to be 35% salt. How much salt must he add to the solution?

Our table will have three rows and three columns. We will have a row for our original solution, a row for what we are adding to the solution, and a row for our new, final solution. We will have a column to hold the percent concentration of salt in each solution, a column to hold the total quantity of each solution, and a column to hold the salt quantity of each solution:

The original solution is 10 ounces and 30% salt. Enter these figures in the table after converting 30% to a decimal. The number of ounces of salt in the solution is equal to the total number of ounces multiplied by the percent of ounces that are salt (10 × 0.30):

To this solution, we must add salt to make a solution that is 35% salt. We don't know how many ounces of salt to add, so we represent that quantity with x. The percent concentration of salt that we are adding is 100%, since we are adding pure salt to the solution. The salt quantity of what we are adding is equal to 100% of x, or 1 × x:

The total quantity in ounces of the final solution is equal to the total quantity in ounces of the original solution plus the quantity of salt in ounces that was added: 10 + x. The salt quantity of the final solution is equal to 0.35 multiplied by the total quantity, 10 + x:

The salt quantity of the final solution is also equal to the salt quantity of the original solution plus the salt quantity that was added: (0.3 × 10) + (1 × x) = 3 + x. Since the salt quantity of the final solution is equal to 3 + x and 0.35(10 + x), we can set these two values equal to each other and solve for x, the number of ounces of salt added:

3 + x = 0.35(10 + x)

3 + x = 3.5 + 0.35x

3 – 3 + x = 3.5 – 3 + 0.35x

x = 0.5 + 0.35x

0.65x = 0.5

x =

Miguel should add ounces of salt to the solution.

We can also remove water from a solution to change the concentration of the solution.

Example

Eli has 50 ounces of an alcohol solution that is 20% alcohol. How many ounces of water must be evaporated to make the concentration 40% alcohol?

Our table again will have three rows and three columns. Since we are looking for how much water will be evaporated, our table will be about the percent and quantity of water. Since the solution is 20% alcohol, it is 100% – 20% = 80% water, and we are looking for a final solution that is only 100% – 40% = 60% water:

Remember, we are removing water, so the total quantity will be 50 – x, not 50 + x:

The final water quantity is equal to 0.60(50 – x) and (0.8 × 50) – (1 × x), which is the water quantity of the original solution minus the quantity that was removed (evaporated). We can set these equations equal to each other and solve for x:

0.60(50 – x) = (0.8 × 50) – (1 × x)

30 – 0.6x = 40 – x

0.4x = 10

x = 25

To make the concentration of the solution 40% alcohol, 25 ounces of water must be evaporated.

Let's look at one last type of mixture problem: mixing two different solutions together.

Example

Dawn has 15 liters of a 20% alcohol solution. How many liters of 50% alcohol solution must be added to create a 30% alcohol solution?

We have 15 liters of a 0.20 solution, which means that the alcohol quantity of that solution is 0.2 × 15. We are adding an unknown quantity of a 0.50 solution, which has an alcohol quantity of 0.5 × x, and our final solution will be a 0.30 alcohol solution, with an alcohol quantity of 0.30 times the total quantity, 15 + x:

The variable x represents how many liters of the 50% solution we must add. Just as in the last examples, look at the final column. The sum of the first two rows, 0.2 × 15 and 0.5 × x, is equal to the product of the two columns of the last row, 0.30 and 15 + x, since both expressions are equal to the alcohol quantity of the final solution:

(0.2 × 15) + (0.5 × x) = 0.3(15 + x)

3 + 0.5x = 4.5 + 0.3x

0.2x = 1.5

x = 7.5

Dawn must add 7.5 liters of 50% alcohol solution to the 20% alcohol solution to create a 30% alcohol solution.

Pace Yourself

Try making a mixture yourself. Fill a beaker with 8 ounces of water, and add 2 ounces of salt. The beaker is now a 20% salt solution. How much salt must you add to make a 50% salt solution?

Summary

We can solve some word problems using formulas. After reviewing some common formulas, we made a small change to our eightstep process to look for words that signal which of these formulas to use. To save time, we learned to substitute values into the formula rather than using the entire eight-step process. We also learned how to rearrange formulas to solve for each part of a formula. Mixture problems are easier to solve with a table than with the eight-step process, so we learned how to set up a specific kind of table to solve this type of problem.

Find practice problems and solutions for these concepts at Formulas and Mixtures Word Problems Practice Problems.

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