Mixture problems are word problems where quantities of different values are mixed. Many times different liquids are mixed to change the concentration of the mixture. Other times, amounts of different costs are mixed.

Here, we will look at a brief summary of mixture problems. Also, we will look at several mixture examples with answers to learn how to solve these types of problems.

## Summary of mixture problems

Mixture problems consist of combining two or more things and determining some characteristic of either the ingredients or the resulting mixture. For example, we may want to find the amount of water we have to add to dilute a saline amount, or we may want to find the percentage of lemon concentration in a lemonade.

To solve these types of problems, it is important to think of mixtures as a type of ratio or proportion. Any situation in which two or more different variables are combined to determine a third is a type of ratio. Speed and time combine to give us distance. Wages and hours worked produce earnings.

Thus, the way to solve mixture problems is to treat them like other ratio and proportion problems. That is, we identify the variables, create equations, and form tables if necessary to organize the information and solve the problem.

## Examples with answers of mixture problems

Practice solving mixture problems with the following examples. The examples have their respective solution to improve the understanding of the process used. It is recommended that you try to solve the problems yourself before looking at the answer.

**EXAMPLE 1**

How many liters of a 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?

##### Solution

We can use *x* to represent the amount of the 20% alcohol solution that has to be added to the 40 liters of 50% alcohol solution and we can use *y* to represent the final amount of the 30% solution. Therefore, we can form the equation:

Now, we have to form an equation to represent the amount of alcohol in *x* liters plus the amount of alcohol in the 40 liters which is equal to the amount of alcohol in *y* liters. We remember that the amount of alcohol is represented in percentages:

Now, we can substitute *y* for to get:

Now, we change all the percentages to fractions:

We can multiply all the terms by 100 to eliminate the fractions and solve for *x*:

Then, 80 liters of 20% alcohol are added to 40 liters of 50% alcohol to form a 30% solution.

**EXAMPLE 2**

If we want to form a 100 ml solution of 5% alcohol by mixing a quantity of a 2% alcohol solution with a 7% alcohol solution, what quantities of each solution do we have to use?

##### Solution

Let us represent with *x* the quantity of the 2% alcohol solution and let us represent with *y* the quantity of the 7% alcohol solution. Then, we can form the equation:

Now, we have to form an equation to indicate that the amount of alcohol in *x* ml plus the amount of alcohol in *y* ml equals the amount of alcohol in 100 ml:

If we rearrange the first equation, we get . Now, we can plug this into the second equation:

We multiply by 100 and simplify the percentages:

Now, we solve for *x*:

ml

Now, we substitute in the first equation to find *y*:

ml

**EXAMPLE 3**

Sterling silver is made up of 92.5% pure silver. How many grams of sterling silver must be mixed with a 90% silver alloy to obtain 500g of a 91% silver alloy?

##### Solution

Let us represent with *x* the quantity of sterling silver and let us represent with *y* the quantity of the 90% silver alloy. Then, we can form the equation that represents these amounts to form 500g of a 91% alloy:

The number of grams of pure silver in *x* plus the number of grams of pure silver in *y* equals the number of grams of pure silver in the 500 grams. We represent these using percentages:

Rearranging the first equation, we get . Now, we can plug this into the second equation:

We multiply by 100 and simplify the percentages:

Now, we solve for *x*:

ml

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**EXAMPLE 4**

How many kilograms of pure water do we have to add to 100 kilograms of a 30% saline solution to form a 10% saline solution?

##### Solution

Let us represent with *x* the quantity in kilograms of pure water and let us represent with *y* the quantity of the 10% saline solution. Then, we have:

Now, we form an equation to represent that the amount of salt in pure water, which is 0, plus the amount of salt in the 30% saline solution is equal to the amount of salt in the final 10% saline solution.:

We substitute the expression in the second equation:

We multiply by 100 and simplify the percentages:

Now, we solve for *x*:

kilograms

**EXAMPLE 5**

A 50 ml substance contains 30% alcohol and is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?

##### Solution

The final amount in the mixture is given by the following equation:

The amount of alcohol is equal to the amount of alcohol in pure water, which is 0, plus the amount of alcohol in the 30% solution. Using *x* to represent the percentage of alcohol in the final solution, we have:

Multiplying by 100 to eliminate the percentage and solving for *x*, we have:

## Examples of mixture problems – Practice

Practice solving mixture problems with the following exercises and test your knowledge on this topic. Choose an answer and check it to see if you chose the correct one. You can look at the solved exercises above if you have problems with these exercises.

## See also

Interested in learning more about algebraic topics? Take a look at these pages:

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