1. Lesson 2-6 Warm-Up

2. “Mixture Problems” (2-6)
How do you solve a percent problem?
Example: 20% of what number is 40 Method 1: Use a proportion table. known relationship: 20% = 40 unknown: 100% = ? 20 • x = 40 • x = 4, x = 200 Method 2: Use an equation. “of” means “x” “what” means “n” “is” means “=“ 20% of what number is 40 means 0.20 x n = x = 40 x = 40  .20 =

3. “Equations and Problem Solving” (2-6)
What is a mixture problem? How do you solve a mixture problem?
Mixture Problem: A problem that involves mixing two or more different things in which you are usually asked to determine the quantities of each thing to end up with the desired mixture. Tip: To solve a mixture problem, you will need to express one quantity in terms of the other (see Lesson 2-5). It would also be very helpful if you 1. organize the quantities into a table similar to the following one in which the amount of item 1 and item 2 is the variable and a quantity in terms of the variable. Then, 2. set up an equation using the “Total Amount” column similar to the following and 3. solve for the variable.
Amount
Cost (or) Percent
Total Amount
(Amount x Cost / %)
Item 1 a $3 3a
Item 2
20 - a $2
2(20 - a)
Mixture 20 $5
20 (5) 3a
2(20 – a)
= 20 (5)

4. Define: Let b = the number of ounces of black tea.
Mixture Problems
LESSON 2-6
Additional Examples
Black tea costs $4.99 per ounce and jasmine tea costs $12.99 per ounce. How many ounces of each should you use to make a 16-oz mixture that costs $6.99 per ounce?
Define: Let b = the number of ounces of black tea.
Then 16 – b = the number of ounces of jasmine tea.
Make a Table:
Black tea b $ b
Amount (oz) Cost Per Ounce Cost (dollars)
Jasmine tea – b $ (16 – b)
Mixture $ (16)

5. 4.99b + 207.84 – 12.99b = 111.84 Use the Distributive Property.
Mixture Problems
LESSON 2-6
Additional Examples
(continued)
Equation:
4.99b (16 – b) = 6.99(16)
4.99b – 12.99b = Use the Distributive Property.
– 8b = Combine like terms.
– 8b – = – Subtract from each side.
–8b = – Simplify
–8b –8
–96 =
Divide each side by –8.
b = Simplify.
You should use 12 ounces of black tea and 4 ounces of jasmine tea.

6. Define: Let a = the number of liters of 30% acid solution.
Mixture Problems
LESSON 2-6
Additional Examples
A chemist has a 30% acid solution and a 70% acid solution. How many liters of each solution does the chemist need to make 500 liters of a 56% acid solution?
Define: Let a = the number of liters of 30% acid solution.
Then 500 – a = the number of liters of 70% acid solution.
Make a Table:
30% Solution a % a
Amount of
Solution (L) Percent Acid Amount of Acid (L)
70% Solution – a % (500 – a)
56% Solution % (500)
Equation:
0.3a + 0.7(500 – a) = 0.56(500)
The amount of acid in the 30% and 70% solutions equals the amount of acid in the mixture.

7. 0.3a + 350 – 0.7a = 280 Use the Distributive Property.
Mixture Problems
LESSON 2-6
Additional Examples
(continued)
0.3a – 0.7a = Use the Distributive Property.
–0.4a = Combine like terms.
–0.4a – 350 = 280 – Subtract 350 from each side.
–0.4a = – Simplify.
–0.4a
–0.4
–70 =
Divide each side by –0.4.
a = Simplify
The chemist needs 175 L of 30% solution and 325 L of 70% solution.
Check: 30% of 175 L is 52.5 L and 70% of 325 L is L.
The total amount of acid in the mixture is , or 280 L.
This is equal to 56% of 500 L.

8. 1. A store sells a mixture of buttons for $6 per pound. Metal
Mixture Problems
LESSON 2-6
Lesson Quiz
1. A store sells a mixture of buttons for $6 per pound. Metal
buttons sell for $12 per pound and plastic buttons sell for $3 per
pound. How many pounds of each should be used to make 25
pounds of the button mixture?
2. A chemist needs a solution that is 65% acid but only has
solutions that are 20% and 80% acid. If the chemist measures
180 L of the 80% acid solution, how many L of the 20% solution
should she add to make a 65% solution?
a. Write an equation to represent the situation.
b. How many liters of the 20% solution should the chemist add?
8 lb of the metal buttons and lb of the plastic buttons 1 3 2
0.8(180) + 0.2x = 0.65(180 + x)
60 L