Today’s Date:

11/16/11

Mixture Word Problems

Notes on Handout

Not given to you, but expected to know:

Water = 0%

“Pure” Acid = 100%

Remember:

a) Define variable

b) Write equation

c) Solve

Mixture Equation: Amount • % = Total

How many kilograms of water should be mixed with 60 kg of a 30% acid solution to make a 10% solution?

Example 1

Amount % Total

Water

Acid

Mix 10(x + 60)

0

x + 60

60

x

30

10

0

(60)(30)

Tells you it’s the mix

Amt of Water + Amt of Acid = Amt of Mix

Asking for amt of water

Multiply Amt ∙ % in each row

Amount % Total

Water

Acid

Mix 10(x + 60)

0

x + 60

60

x

30

10

0

(60)(30)

Example 1 cont…

Get your equation from the last column Total Water + Total Acid = Total Mix 0 + (60)(30) = 10(x + 60)

1800 = 10x + 600 1200 = 10x 120 = x

a) x is amt of waterb) 1800 = 10x + 600c) 120 kg of water

Example 2

How many kilograms of pure salt must be added to 8 kg of a 10% salt solution to obtain a 20% salt solution?

Amount % Total

Pure Salt

Salt

Mix 20(x + 8)

100

x + 8

8

x

10

20

100x

(8)(10)

Tells you it’s the mix

Asking for amt of

“pure” salt

100x + (8)(10) = 20(x + 8)

Example 2 cont…

100x + (8)(10) = 20(x + 8)

100x + 80 = 20x + 160

80x + 80 = 160

80x = 80

x = 1a) x is amt of pure saltb) 100x + 80 = 20x + 160c) 1 kg of pure salt

Example 3

How many kilograms of a 30% acid solution must be added to 80 kg of a 20% solution to produce a 25% solution?

Amount % Total

Solution 1

Solution 2

Mix 25(x + 80)

30

x + 80

80

x

20

25

30x

(80)(20)

Tells you it’s the mix

30x + (80)(20) = 25(x + 80)

Example 3 cont…

30x + (80)(20) = 25(x + 80)

30 x + 1600 = 25x + 2000

5x + 1600 = 2000

5x = 400

x = 80a) x is amt of 30% solutionb) 30x + 1600 = 25x + 2000c) 80 kg of 30% solution

Homework

#9 Two Sided Worksheet:

Mixture ProblemsReview Sum/Difference/Complex

Fractions

Mix Prob #2: Mix Amt = 36

Need to find Milk Amt & Cream Amt

How?