Lesson 3.3: Solve Mixture Applications

New Concept

This lesson introduces mixture problems, where items of different values are combined. You will learn to organize information and write equations to solve real-world problems involving coins, tickets, investments, and other mixtures with a specific total value.

What’s next

Next, you’ll work through interactive examples that break down the setup process. Then, you'll tackle a series of practice cards to master each type of mixture problem.

Property

For the same type of coin, the total value of a number of coins is found by using the model:

To find the total value of a pile of different coins, you separate the coins by type, find the total value of each type, and then add all the values together.

Examples

• Adela has 3.45 dollars in dimes and nickels. She has six more nickels than dimes. Let $d$ be the number of dimes. The equation is $0.10d + 0.05(d+6) = 3.45$, which gives $d=21$ dimes and $27$ nickels.

• Ben has 5.40 dollars in quarters and dimes. He has twice as many dimes as quarters. Let $q$ be the number of quarters. The equation is $0.25q + 0.10(2q) = 5.40$, which gives $q=12$ quarters and $24$ dimes.

Property

The mixture model, $\operatorname{number} \cdot \operatorname{value} = \operatorname{total value}$, also applies to items like tickets and stamps. When the total quantity of items sold is known, you can define the number of each type algebraically, such as $x$ for one type and $(\text{total quantity} - x)$ for the other.

Examples

• A school play sold 620 dollars in tickets. Adult tickets cost 10 dollars and child tickets cost 6 dollars. The number of adult tickets was 10 more than twice the child tickets. Let $c$ be the number of child tickets. $10(2c+10) + 6c = 620$, so $c=20$ child tickets and $50$ adult tickets were sold.

• A carnival sold 550 tickets for a total of 2300 dollars. Adult tickets were 5 dollars and child tickets were 3 dollars. Let $a$ be the number of adult tickets. The equation $5a + 3(550-a) = 2300$ gives $a=325$ adult tickets and $225$ child tickets.

Property

To solve mixture problems, we use the principle that the sum of the values of the individual components equals the value of the final mixture. The equation is structured as:
(Amount of Item 1 $\cdot$ Price of Item 1) + (Amount of Item 2 $\cdot$ Price of Item 2) = (Total Amount of Mixture $\cdot$ Final Price of Mixture).

Examples

• A coffee shop wants to make a 20-pound blend of coffee costing 15 dollars per pound by mixing beans costing 12 dollars per pound with beans costing 16 dollars per pound. Let $x$ be the pounds of 12-dollar beans. $12x + 16(20-x) = 20(15)$, so they need 5 pounds of 12-dollar beans and 15 pounds of 16-dollar beans.

• A candy store mixes gummy worms at 3 dollars a pound with chocolate drops at 5 dollars a pound to create 10 pounds of a mix that sells for 4.50 dollars a pound. Let $w$ be pounds of worms. $3w + 5(10-w) = 10(4.50)$, so they use 2.5 pounds of worms and 7.5 pounds of drops.

Property

The mixture model applies to simple interest investment problems. The interest from each investment adds up to the total interest. The formula is:

Here, the amount invested is the principal, and the interest rate is the 'value'.

Examples

• Javier invested 30,000 dollars in two accounts, one paying 2% interest and the other 5%. He wants to earn 4% overall. Let $x$ be the amount at 2%. $0.02x + 0.05(30000-x) = 0.04(30000)$, so he invested 10,000 dollars at 2% and 20,000 dollars at 5%.

• Sarah has 12,000 dollars to invest between a bond paying 1.5% and a stock fund paying 6% to get an average return of 4%. Let $x$ be the amount in the bond. $0.015x + 0.06(12000-x) = 0.04(12000)$, so she should invest about 5333.33 dollars in bonds and 6666.67 dollars in stocks.