MIXTURE PROBLEMS

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MIXTURE PROBLEMS. Prepared for Intermediate Algebra Mth 04 Online by Dick Gill. The following slides are designed to help you organize mixture problems, form the necessary equation and solve that equation. The problems in this module will involve

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MIXTURE PROBLEMS

Prepared for Intermediate Algebra

Mth 04 Online

by Dick Gill

The following slides are designed to help you organize mixture problems, form the necessary equation and solve that equation. The problems in this module will involve

chemical solutions.

Example 1. How many liters of a solution that is 20% alcohol

should be combined with 10 liters of a solution that is 50%

alcohol to create a solution that is 30% alcohol? (Print this frame

now so that you can keep track of the problem.)

The primary technique that we will develop here is to trace

the target ingredient in the problem. In this problem the

target ingredient is alcohol. In future problems it might be

antifreeze or acid. The equation will state that the amount

of alcohol in the first container and the amount of alcohol

in the second will equal the amount of alcohol in the final

mixture.

+

=

have filled in the grid you are looking at the equation in the last

column.

Fill the third column with the total amounts in each

container. We will use x for the amount in the second

container.

10

.50

.20

x

.30

x + 10

columns. Focus on the row for the first container. If you

multiply the percent strength by the total amount, you will have

the amount of alcohol in the first container.

.50

10

5

.20

x

.20x

.30

x + 10

.30(x+10)

alcohol in the second container equals the amount of alcohol in the

final mixture. The equation is in the final column.

.50

10

5

.20

x

.20x

.30

x + 10

.30(x+10)

Now we solve the equation.

5 + .20x = .30(x + 10).

5 + .20x = .30x + 3

5 + .20x - .20x = .30x + 3 - .20x

5 = .10x + 3

5 – 3 = .10x + 3 – 3

2/.10 = .10x/.10

20 = x

It takes 20 liters of a 20% solution to dilute 10 liters of a 50%

solution to a final solution of 30%.

should be added to how many liters of a solution that is pure

antifreeze to create 5 liters of a solution that is 50% antifreeze?

Round your answer to the nearest tenth of a liter. (Print this frame.)

Notice a couple of differences between this Example 1 and

Example 2. In the first place, what do you think we are going

to use as the percentage strength for pure antifreeze? Think

before you click.

If you guessed 100%, pat yourself on the back. Secondly, if we

let x be the number of liters of the 30% solution, what do you

think we should use as the number of liters at 100%? Think

before you click. Hint: the total number of liters has to be 5.

If you guessed 5 – x, congratulations!

have filled in the grid you are looking at the equation in the last

column. Print this frame and fill in the grid as best you can before

you click to the next frame.

.30

1.00

.50

of the 30% solution. We also know that the mixture will total 5 liters.

.30

x

1.00

.50

5

the number of liters of pure antifreeze? Since the total number of

liters has to be 5, subtract the number of liters at 30% and get:

.30

x

1.00

5 - x

.50

5

amount of the target ingredient (antifreeze).

The third column is your equation. Solve it before you click.

.30

x

.30x

1.00

5 - x

1(5 – x)

.50

5

.50(5)

Just in case you are peeking

the best way to learn

is by working

problems—

not by

watching

someone

else

work

problems.

Now back to our regular program.

From the third column, the equation is:

.30x + 1.00(5 – x) = .5(5)

.30x + 5 – 1.00x = 2.5

5 - .70x = 2.5

-.70x = -2.5

x = 3.6 liters (to the nearest tenth)

5 – x = 1.4 liters

How much water would you have to add to dilute 4 liters of

a 70% acid solution down to a 35% acid solution? Think before

you click.

Congratulate yourself if you guessed 4 liters. Another 4 liters

would dilute the original 4 liters to half of its original strength.

See if you can work this problem on your own before you click

to the solution. Guess first, then use the grid to solve algebraically.

Example 3. How much water would you have to add to dilute

4 liters of a 75% acid solution down to a 25% acid solution?

water. Did you use 0% for the strength of the second container?

water. Did you use 0% for the strength of the second container?

two terms. This means that the amount of acid in the first container

Is also the amount of acid in the final mixture.

.75(4) = .25(4 + x)

3 = 1 + .25x

2 = .25x

8 = x

It takes 8 liters of water to dilute 4 liters of 75% acid down to 25%

strength.

The following slides give you nine mixture problems to practice.

• 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?

Answers to mixture problems 1 – 9.

• 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.

1.

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

.10x + 2.88 = .15x + 1.8

1.08 = .05x

21.6 = x

21.6 ounces @ 10%

2.

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

.18 = .04x

4.5 = x

4.5 liters at 20%

3.

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

.50x + 6.40 = .60x + 4.80

1.60 = .10x

16 = x

16 liters at 50%

4.

.10x + .28(30 – x) = .20(30)

.10x + 8.4 - .28x = 6

-.18x = -2.4

x = 13.3

13.3 ounces at 10%

5.

.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%

6.

.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%

7.

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

3 = .24x + 2.4

0.6 = .24x

2.5 = x

2.5 liters of water

8.

.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%

9.

.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