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Lesson 2-6 Warm-Up
ALGEBRA 1
“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 • 100
20x = 4,000
20
20
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 = 40
0.20 x = 40
x = 40  .20 = 200
0.20
0.20
ALGEBRA 1
“Equations and Problem
Solving” (2-6)
What is 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.
How do you solve a
mixture problem?
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.
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)
Then, 2. set up an equation using the “Total Amount” column similar
to the following and 3. solve for the variable.
3a
+
2(20 – a)
= 20 (5)
ALGEBRA 1
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
Amount (oz)
b
Cost Per Ounce
$4.99
Cost (dollars)
4.99b
Jasmine tea
16 – b
$12.99
12.99(16 – b)
Mixture
16
$6.99
6.99(16)
ALGEBRA 1
Mixture Problems
LESSON 2-6
Additional Examples
(continued)
Equation:
4.99b + 12.99(16 – b) = 6.99(16)
4.99b + 207.84 – 12.99b = 111.84
207.84 – 8b = 111.84
Use the Distributive Property.
Combine like terms.
207.84 – 8b – 207.84 = 111.84 – 207.84 Subtract 207.84 from each side.
–8b = –96
Simplify
–8b
Divide each side by –8.
–8
=
–96
–8
b = 12
Simplify.
You should use 12 ounces of black tea and 4 ounces of jasmine tea.
ALGEBRA 1
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:
Amount of
Solution (L)
Percent Acid Amount of Acid (L)
30% Solution
a
30%
0.3a
70% Solution
500 – a
70%
0.7(500 – a)
56% Solution
500
56%
0.56(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.
ALGEBRA 1
Mixture Problems
LESSON 2-6
Additional Examples
(continued)
0.3a + 350 – 0.7a = 280
–0.4a + 350 = 280
–0.4a + 350 – 350 = 280 – 350
–0.4a = –70
–70
–0.4a
=
–0.4
–0.4
a = 175
Use the Distributive Property.
Combine like terms.
Subtract 350 from each side.
Simplify.
Divide each side by –0.4.
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 227.5 L.
The total amount of acid in the mixture is 52.5 + 227.5, or 280 L.
This is equal to 56% of 500 L.
ALGEBRA 1
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?
8 1 lb of the metal buttons and 16 2 lb of the plastic buttons
3
3
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.
0.8(180) + 0.2x = 0.65(180 + x)
b. How many liters of the 20% solution should the chemist add?
60 L
ALGEBRA 1