Mixture Problems

Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 55 cents, how many of each coin does she have?

Since we have 2 types of coins, let's call the number of nickels $x$ and the number of dimes $y$. We are given two key pieces of information to make our equations: the number of coins and their value.

# of coins equation: $x+y = 7$ (number of nickels) + (number of dimes)

value equation: $5x + 10y = 55$ (since nickels are worth $5c$ and dimes $10$.

We can quickly rearrange the first equation to isolate $x$.

Janine has 3 nickels and 4 dimes.

Sometimes a question asks you to determine (from concentrations) how much of a particular substance to use. The substance in question could be something like coins as above, or it could be a chemical in solution, or even heat. In such a case, you need to know the amount of whatever substance is in each part. There are several common situations where to get one equation you simply add two given quantities, but to get the second equation you need to use a product. Three examples are below.

For example, when considering mixing chemical solutions, we will most likely need to consider the total amount of solute in the individual parts and in the final mixture. (A solute is the chemical that is dissolved in a solution. An example of a solute is salt when added to water to make a brine.) To find the total amount, simply multiply the amount of the mixture by the fractional concentration. To illustrate, let's look at an example where you are given amounts relative to the whole.

A chemist needs to prepare 500 ml of copper-sulfate solution with a 15% concentration. She wishes to use a high concentration solution (60%) and dilute it with a low concentration solution (5%) in order to do this. How much of each solution should she use?

To set this problem up, we first need to define our variables. Our unknowns are the amount of concentrated solution ($x$) and the amount of dilute solution ($y$). We will also convert the percentages (60%, 15% and 5%) into decimals (0.6, 0.15 and 0.05). The two pieces of critical information are the final volume (500 ml) and the final amount of solute ((15% of 500 ml=75 ml). Our equations will look like this:

Volume equation: $x + y = 500$

Solute equation: $0.6x+0.05y=75$

To isolate a variable for substitution, we can see it's easier to start with equation 1:

So the chemist should mix 91 ml of the 60% solution with 409 ml of the 5% solution.

A coffee company makes a product which is a mixture of two coffees, using a coffee that costs $10.20 per pound and another coffee that costs $6.80 per pound. In order to make 20 pounds of a mixture that costs $8.50 per pound, how much of each type of coffee should it use?

Let $m$ be the amount of the $10.20 coffee, and let $n$ be the amount needed of the $6.80 coffee. Since we want 20 pounds of coffee that costs $8.50 per pound, the total cost for all 20 pounds is $20⋅$8.50=$170$. The cost for the 20 pounds of mixture is equal to the cost of each type of coffee added together: $10.20⋅m+6.8⋅n=170.$

Also, the amount of each type of coffee added together equals 20 pounds: $m+n=20$.

The system is: $\left\{\begin{array}{c} m+n=20 \\ 10.20 \cdot m+6.8 \cdot n=170 \end{array}\right.$

We can isolate one variable and use substitution to solve the system:

$m=20−n$

Now solve for $n$.

$\begin{aligned} 10.20(20-n)+6.8 n &=170 & \\ 204-10.20 n+6.8 n &=0170 && \text {   Distributive Property } \\ 204-3.4 n &=170 && \text {   Add like terms. } \\ -3.4 n &=-34 && \text {   Subtract 204. } \\ n &=10 && \text {   Divide by }-3.4 \end{aligned}$

Since $n = 10$, we can plug that into $m + n = 20$.

$m+10=20 \Rightarrow m=10$

The coffee company needs to use 10 pounds of each type of coffee in order to have a 20 pound mixture that costs $8.50 per pound.

A light green latex paint that is 20% yellow paint is combined with a darker green latex paint that is 45% yellow paint. How many gallons of each paint must be used to create 15 gallons of a green paint that is 25% yellow paint?

Let $x$ be the number of gallons of the 20% yellow paint and let $y$ be the number of gallons of the 40% yellow paint. This means that we want those two numbers to add up to 15: $x+y=15$.

Now if we want 15 gallons of 25% yellow paint, that means we want $0.25⋅15=3.75$ gallons of pure yellow pigment. The expression $0.20⋅x$ represents the amount of pure yellow pigment in the $x$ gallons of 20% yellow paint. The expression $0.45⋅y$ represents the amount of pure yellow pigment in the $y$ gallons of 45% yellow paint. Combing the last two adds up to the 3.75 gallons of pure pigment in the final mixture:

$0.20x+0.40y=3.75$

The system is: $\left\{\begin{aligned} x+y &=15 \\ 0.20 x+0.45 y &=3.75 \end{aligned}\right.$

We can isolate one variable and use substitution to solve the system:

$x=15−y$

Now solve for $y$.

$\begin{aligned} 0.20(15-y)+0.45 y &=3.75 & & \\ 3-0.20 y+0.45 y &=3.75 & & \text { Distributive Property } \\ 3+0.2 y &=3.75 & & \text { Add like terms. } \\ 0.25 y &=0.75 & & \text { Subtract } 3 . \\ y &=3 & & \text { Divide by } 0.25 \end{aligned}$

Now we can plug in $y=3$ into $x+y=15$:

$x+y=15 \Rightarrow x+3=15 \Rightarrow x=12$

This means 12 gallons of 20% yellow paint should be mixed with 3 gallons of 45% yellow paint in order to get 15 gallons of 25% yellow paint.

1.

a. 1.6lbs peanuts and 3.4lbs cashews

b. No, minimum price for 5lbs is $11

c. 6.81lb (all peanuts)

2.

a. 0.8L of 10% solution and 0.2L of 35% solution

b. 1.6L of 10% solution and 0.4L of 35% solution

3. The bracelet density is 16.37 g/cc

4. 82 purchased in advance and 38 at the door

5. 10 gal of 40% and 5 gal of 100%

6. 25 multiple choice and 10 short answer

7. 23 multiple choice and 8 short answer

8. 4 multiple choice and 1 short answer

9. 1st way: $𝑥 + 𝑦 = 5, 2𝑥 + 5𝑦 = 13$ 2nd way: $𝑥 + 𝑦 = 30, 2𝑥 + 5𝑦 = 87$

10. Calculate what Kwan missed rather than what she got right: $𝑥 + 𝑦 = 4, 2𝑥 + 5𝑦 = 14$