1. 10.8 Mixture Problems
Goal: To solve problems involving the mixture of substances

2. Mixture Problems
One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid?

3. Steps to Solve Mixture Problems
Set up a chart (4x4)
Amount of Solution
Percent of
_______
Amount of _____
Solution 1
Solution 2
Final Solution

4. Steps to Solve Mixture Problems
Convert the percentages to decimals and fill out the chart
Multiply going across the chart
Add going down the chart
Set up 2 equations with 2 variables (system)
Solve the system by substitution or addition

5. One solution is 80% acid and another is 30% acid
One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid?
Let x = y =
Amount of solution
• Percent Acid
= Amount of pure Acid
1st Solution
2nd Solution
3rd Solution x
0.80
.80(x) y
0.30
.30(y)
.62(200)
200
0.62
124

6. One solution is 80% acid and another is 30% acid
One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid?
8x + 3y =1240
Y= 200-x
8x + 3 (200-x) =1240
Amount of solution
• Percent Acid
= Amount of Acid
1st Solution
2nd Solution
3rd Solution
Y =
8x x =1240 x
0.80
.80(x)
5x +600 =1240
Y = 72 L
5x = 640 y
0.30
.30(y)
X= 128 L
.62(200)
200
0.62
124

7. A chemist has one solution that is 60% acid and another that is 30% acid. How much of each solution is needed to make a 750ml solution that is 50% acid?
Amount of solution
• Percent Acid
= Amount of Acid
1st Solution
2nd Solution
3rd Solution x
0.60
.60(x) y
0.30
.30(y)
.50(750)
750
0.50
375

8. A chemist has one solution that is 28% oil and another that is 40% oil
A chemist has one solution that is 28% oil and another that is 40% oil. How much of each solution is needed to make a 300 L solution that is 36% oil?
Amount of solution
• Percent Acid
= Amount of Acid
1st Solution
2nd Solution
3rd Solution x
0.28
.28(x) y
0.40
.4(y)
.36(300)
300
0.36
108

9. Try to make your own chart
How many gallons of a 50% salt solution must be mixed with 60 gallons of a 15% solution to obtain a solution that is 40% salt?

10. Amount of Solution (gallons)
How many gallons of a 50% salt solution must be mixed with 60 gallons of a 15% solution to obtain a solution that is 40% salt?
Amount of Solution (gallons)
Percent of Salt
Amount of Salt
Solution 1 x
.50
.5x
Solution 2 60
.15 9
Final Solution y
.40
.4y

11. System
x + 60 =y 0 .5x + 9 = 0.4y 5x +90 = 4y 5x + 90 = 4 (x +60) 5x + 90 = 4x x + 90 =240 x =150 gallons = y 210 gallons =y

12. Coffee Beans
How many pounds of coffee beans selling for $2.20 per pound should be mixed with 2 pounds of coffee beans selling for $1.40 per pound to obtain a mixture selling for $2.04 per pound?

13. How many pounds of coffee beans selling for $2
How many pounds of coffee beans selling for $2.20 per pound should be mixed with 2 pounds of coffee beans selling for $1.40 per pound to obtain a mixture selling for $2.04 per pound?
Pounds of Coffee
$ per pound
total cost
Coffee mix 1 X
$2.20
2.20 x
Coffee mix 2 2
$1.40
2.80
Final Coffee mix Y
$2.04
2.04 y

14. System
X + 2 =y
2.20x = 2.04 y

16. Your Turn
Come up with your own mixture word problem. Make it interesting!
Remember to include:
Amount of Solution
Wieght of Object
% of (acid /water / oil/salt/etc)
Cost per weight
Amount of (acid / water /oil/salt/etc)
Total cost
Solution 1 and 2
2 objects
Final Solution
Mixture of 2 objects

17. Assignment: Page 462 (1 -9) odd

19. 1 .20 .20 x .00 .00(x) .20 x+1 .06 Amount of lake • Percent salt
= Amount of salt
1963
2nd Solution
1984 1
.20
.20 x
.00
.00(x)
.20
x+1
.06

20. Vocabulary Mixture- two substances combined
Concentrate or Solution- how much non-water is mixed (juice)
10% solution -10% concentration and 90% water