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# mixture problems - PowerPoint PPT Presentation

MIXTURE PROBLEMS. Prepared for Intermediate Algebra Mth 04 Online by Dick Gill. The following slides give you nine mixture problems to practice. Answers to these problems follow. If some of your answers are

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Prepared for Intermediate Algebra

Mth 04 Online

by Dick Gill

• to the nearest tenth if necessary.

• How many ounces of a solution that is 10% alcohol should be

• mixed with 12 ounces of a solution that is is 24% alcohol to create

• a solution that is 15% alcohol?

• How many liters of a solution that is 20% acid should be added

• to 3 liters of a solution that is 30% acid to create a solution that is

• 24% acid?

• How many liters of a solution that is 50% antifreeze should be

• added to 8 liters of a solution that is 80% antifreeze to create a

• solution that is 60% antifreeze?

to the nearest tenth if necessary.

4. How many ounces of a solution that is 10% alcohol should be added

to a solution that is 28% alcohol to create 30 ounces of a solution that

is 20% alcohol?

5. How many liters of a solution that is 20% acid should be added to

how many liters of a solution that is 38% acid to create 8 liters of a

solution that is 30% acid?

6. How many liters of a solution that is 50% antifreeze should be

added to how many liters of a solution that is 72% antifreeze to

create 2.4 liters of a solution that is 58% antifreeze?

to the nearest hundredth if necessary.

7. Ten liters of a solution that is 30% alcohol is going to be diluted to

24% alcohol by adding water. How much water is needed?

8. A solution that is 30% antifreeze is going to be enriched by adding

pure antifreeze. How much of each is needed to generate 2 gallons of

a solution that is 50% antifreeze?

9. How many gallons should be drained from a 10 gallon tank of

24% alcohol if we are going to replace it with pure alcohol and

create a solution of 35% alcohol?

• 21.6 ounces

• 4.5 liters

• 16 liters

• 13.3 ounces

• 3.6 liters at 20%; 4.4 liters at 38%

• 1.5 liters at 50%; 0.9 liters at 72%

• 2.5 liters

• 1.43 gal at 30%; 0.57 gal at 100%

• 1.45 gal

• Complete solutions follow.

.10x + .24(12) = .15(x + 12)

.10x + 2.88 = .15x + 1.8

1.08 = .05x

21.6 = x

21.6 ounces @ 10%

.20x + .90 = .24x + .72

.18 = .04x

4.5 = x

4.5 liters at 20%

.50x + .80(8) = .60(x + 8)

.50x + 6.40 = .60x + 4.80

1.60 = .10x

16 = x

16 liters at 50%

.10x + .28(30 – x) = .20(30)

.10x + 8.4 - .28x = 6

-.18x = -2.4

x = 13.3

13.3 ounces at 10%

.20x + .38(8 – x) = .30(8)

.20x + 3.04 - .38x = 2.40

-.18x = -.64

x = 3.6; 8 – x = 4.4

3.6 liters at 20%; 4.4 liters at 38%

.50x + .72(2.4 – x) = .58(2.4)

.50x + 1.728 - .72x = 1.392

-.22x = -.336

x = 1.5; 2.4 – x = 0.9

1.5 liters at 50%; 0.9 liters at 72%

.30(10) + 0 = .24(x + 10)

3 = .24x + 2.4

0.6 = .24x

2.5 = x

2.5 liters of water

.30(x) + 1.00(2 – x) = .50(2)

.30x + 2.00 – 1.00x = 1.00

-.70x = -1.00

x = 1.43; 2 – x = 0.57

1.43 gal at 30%; 0.57 gal at 100%

.24(10 – x) + 1.00x = .35(10)

2.4 - .24x + 1.00x = 3.5

.76x = 1.1

x = 1.45

Drain 1.45 gal and replace with 100% alcohol