10.8 Mixture Problems
Goal: To solve problems involving the mixture of substances
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?
Steps to Solve Mixture Problems
• Set up a chart (4x4)
Amount of Solution
Percent of_______
Amount of _____
Solution 1
Solution 2
Final Solution
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
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?
Amount of solution
• Percent Acid
= Amount of pure Acid
1st Solution
2nd Solution
3rd Solution
Let x = y =
x .80(x)0.80
y 0.30
.30(y)
200 0.62 .62(200)124
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?
Amount of solution
• Percent Acid
= Amount of Acid
1st Solution
2nd Solution
3rd Solution
x .80(x)0.80
y 0.30
.30(y)
200 0.62 .62(200)
200 yx 1243.8. yx
124
Y= 200-x8x + 3y =1240
8x + 3 (200-x) =1240
8x +600 -3x =1240
5x +600 =12405x = 640
X= 128 L
Y = 200 -128
Y = 72 L
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 .60(x)0.60
y 0.30
.30(y)
750 0.50 .50(750)375
750 yx 3753.6. yxyx 750 3753.7506. yy
3753.6.450 yy753. y
250y
500x
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 .28(x)0.28
y 0.40
.4(y)
300 0.36 .36(300)108
300x y .28 .40 108x y 300x y .28 300 .40 108y y
84 .28 .40 108y y .12 24y
200y
100x
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?
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
Systemx + 60 =y 0 .5x + 9 = 0.4y
5x +90 = 4y
5x + 90 = 4 (x +60)
5x + 90 = 4x + 240
x + 90 =240
x =150 gallons
150 + 60 = y
210 gallons =y
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?
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
System
2.20x +2.80 = 2.04 yX + 2 =y
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
Assignment:Page 462 (1 -9) odd
Amount of lake
• Percent salt
= Amount of salt
1963
2nd Solution
1984
1 .20.20
x .00 .00(x)
x+1 .06 .20
20.0106.0 x20.006.06. x
14.006. xliters 33.2x
Vocabulary
• Mixture- two substances combined
• Concentrate or Solution- how much non-water is mixed (juice)
• 10% solution -10% concentration and 90% water