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