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Mixture problems often involve a mixture composed of a ratio of different parts, solutions, or liquids. Generally speaking, the problem will describe some way in which the mixture is changed. Perhaps you add a completely different liquid to the mixture, or you add more of a specific liquid already in the mixture. Regardless, your task will likely be to identify what you are trying to solve, translate the problem into an algebraic expression, and solve for some property of the final mixture. Take extra care to be consistent when you define what the mixtures, solutions, and/or liquids are when you create your equations.

These problems generally differ from question to question, so the best way to prepare will be to practice. Try your hand at these example questions.

If Rhea likes her coffee-to-milk ratio at $9$ to $1$, and Asia likes her coffee-to-milk ratio at $3$ to $1$, what ratio of Rhea’s to Asia’s drinks is required to make a mixture with a $4$ to $1$ ratio of coffee to milk?

A. $1$ to $3$

B. $1$ to $2$

C. $1$ to $1$

D. $2$ to $1$

E. $3$ to $1$

(spoiler)

Answer: D. $2$ to $1$

Clearly, one needs to mix some of Asia’s drink with some of Rhea’s drink to make the ratio $4$ parts coffee for 1 part milk because $4$**:**$1$ is between their preferences of $9$ to $1$ and $3$ to $1$. The question asks for the specific ratio of their drinks needed to make a $4$ to $1$ coffee-to-milk ratio.

Begin by writing out an equation. One interesting way to write the equation is to set the amount of milk (you could also choose coffee instead) equal. In other words, the amount of milk in their individual drinks must be the same as the amount of milk in their combined drink. For example, the total milk in Asia’s drink is $25%$ of the whole, that is because there are $3$ parts coffee and $1$ part milk, making $4$ total parts with one of them being milk. So the total amount of milk is just $.25$ multiplied by Asia’s total, which we will just call $a$.

$a(.25)$

The same can be done for Rhea. For Rhea, milk makes up only $10%$ of the whole drink. Let $r$ represent the total amount of liquid in the drink.

$r(.1)$

Adding the total milk between them would look like this.

$a(.25)+r(.1)$

Now, set that equal to the amount of milk in their mixed drink. This should be $20$ of the combined total $(a+r)$ because the ratio of coffee to milk in their combined drink would be $4$ to $1$, making $5$ total parts, with $1$ of them being milk.

$a(.10)+r(.25).1a+.25r.05rr/ar/a =.2(a+r)=.2a+.2r=.1a=.1/.05=2/1 $

The ratio of Rhea to Asia would then be $2$ to $1$.

**Using Weighted Averages**

A common method to help solve mixture problems involves using the weighted average equation. Feel free to revisit the “mean, median, mode, and range” chapter to review weighted averages! Try using your knowledge of weighted averages to solve the problem below.

A very hot mixture consists of three solutions, each containing a specific concentration of a melted rare metal.

Solution Rare Metal Concentration Level A 5% B 15% C 50%

If $15%$ of the mixture is made up of Solution A, $25%$ is made up of Solution B, and $60%$ is made up of Solution C, approximately what is the concentration of the rare metal in the mixture? Round to the nearest percent.

(spoiler)

Answer: $35%$

Make a weighted average equation. Remember, all the mixture percentages for each solution should add up to $100%$, NOT the concentrations.

Let’s pull out the key information from the question:

- Percent of mixture that’s $A$ = $.15$
- Concentration of $A$ = $.05$
- Percent of mixture that’s $B$ = $.25$
- Concentration of $B$ = $.15$
- Percent of mixture that’s $C$ = $.6$
- Concentration of $C$ = $.5$

$mixture’s cocentration =(.15)(.05)+(.25)(.15)+(.6)(.5)=.0075+.0375+.3=.345 $

The concentration of the rare metal in the mixture is approximately $35%$.

Notice how $35$ is between the minimum and maximum concentrations of the individual solutions? You should implement this important fail-safe when you find your final solution. Always check if your weighted average is between your inputs. If it is outside of your inputs’ minimum or maximum, there must have been a mistake. There is no way that an average could be greater or less than all of the elements that it is derived from.

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