1. Algebra 1
Section 3.8

2. Mixture Problems
Mixture problems involve the combination of different substances.
Making a table can be quite helpful.

3. Example 1
The number of bushels, multiplied by the price per bushel, will equal the total cost.
The costs of the corn and barley, added together, will equal the total cost of the mix.

4. Example 1 Let c = number of bushels of corn.
Then 150 – c = number of bushels of barley.
Now it’s time to fill in our chart!

5. Example 1 3.8c + 2.6(150 – c) = 450 Number of Bushels Price per Bushel
Total Cost
Corn
Barley
Mix c
150 – c
150
3.80
2.60
3.00
3.8c
2.6(150 – c)
450
3.8c + 2.6(150 – c) = 450

6. Example 1 3.8c + 2.6(150 – c) = 450 3.8c + 390 – 2.6c = 450 1.2c = 60
50 bushels of corn [c]
100 bushels of barley [150 – c]

7. Example 2 Pure nickel, by definition, is 100% nickel.
Pounds of metal × % nickel = pounds of nickel
The alloy is a combination of nickel silver and pure nickel.
p = number of lb of pure nickel

8. Example 2 0.15(70) + 1p = 0.2(70 + p) Nickel Silver Pure Nickel Alloy
Percent of Nickel
Pounds of Nickel
Pounds of Metal
Nickel Silver
Pure Nickel
Alloy 70 p
70 + p
0.15 1
0.2
15%
100%
20%
0.15(70) 1p
0.2(70 + p)
0.15(70) + 1p = 0.2(70 + p)

9. Example 2 0.15(70) + 1p = 0.2(70 + p) 10.50 + p = 14 + 0.2p
p = lb of pure nickel

10. Solving Mixture Problems
Read the problem carefully.
Plan a strategy.
Solve an equation.
Check.

11. Homework: pp