Transcript Document

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.
+
=
The purpose of the grid is to lead you to the equation. Once you
have filled in the grid you are looking at the equation in the last
column.
Percent
strength
First
container
Second
container
mixture
Amount of Amount
solution
of target
ingredient
Fill the second column with the percent strengths of each
container.
Percent
strength
First
container
.50
Second
container
.20
mixture
.30
Amount of Amount
solution
of target
ingredient
Fill the third column with the total amounts in each
container. We will use x for the amount in the second
container.
Percent
strength
Amount of Amount
solution
of target
ingredient
First
container
.50
10
Second
container
.20
x
.30
x + 10
mixture
Fill the last column with the products of the second and third
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.
Percent
strength
Amount of Amount
solution
of target
ingredient
First
container
.50
10
5
Second
container
.20
x
.20x
mixture
.30
x + 10 .30(x+10)
The amount of alcohol in the first container plus the amount of
alcohol in the second container equals the amount of alcohol in the
final mixture. The equation is in the final column.
Percent
strength
Amount of Amount
solution
of target
ingredient
First
container
.50
10
5
Second
container
.20
x
.20x
mixture
.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%.
Example 2. How many liters of a solution that is 30% antifreeze
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!
The purpose of the grid is to lead you to the equation. Once you
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.
Percent
strength
First
container
.30
Second
container
1.00
mixture
.50
Amount of Amount
solution
of target
ingredient
As we discussed before, we are going to let x be the number of liters
of the 30% solution. We also know that the mixture will total 5 liters.
Percent
strength
First
container
.30
Second
container
1.00
mixture
.50
Amount of Amount
solution
of target
ingredient
x
5
The big question: what are we going to use as an algebraic name for
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:
Percent
strength
Amount of Amount
solution
of target
ingredient
First
container
.30
x
Second
container
1.00
5-x
mixture
.50
5
Now multiply percent strength by the amount of solution and get the
amount of the target ingredient (antifreeze).
The third column is your equation. Solve it before you click.
Percent
strength
Amount of Amount
solution
of target
ingredient
First
container
.30
x
Second
container
1.00
5-x
1(5 – x)
5
.50(5)
mixture
.50
.30x
Just in case you are peeking
instead of solving, remember:
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
We will introduce our final example with a little common sense.
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?
The target ingredient will be acid. The second container will be the
water. Did you use 0% for the strength of the second container?
Percent
strength
First
container
Second
container
mixture
Amount of Amount
solution
of target
ingredient
The target ingredient will be acid. The second container will be the
water. Did you use 0% for the strength of the second container?
Percent
strength
First
.75
container
Amount of Amount
solution
of target
ingredient
4
.75(4)
Second 0
container
x
0
Mixture
4+x
.25(4 + x)
.25
Since there is no acid in the second container, the equation has only
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.
Answers to these problems follow. If some of your answers are
wrong, the complete solutions will follow the answer slide.
Three mixture problems at the beginners level. Round your answers
to the nearest tenth if necessary.
1. 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?
2. 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?
3. 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?
Three mixture problems at the intermediate level. Round your answers
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?
Three mixture problems at the advanced level. Round your answers
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.
1. 21.6 ounces
2. 4.5 liters
3. 16 liters
4. 13.3 ounces
5. 3.6 liters at 20%; 4.4 liters at 38%
6. 1.5 liters at 50%; 0.9 liters at 72%
7. 2.5 liters
8. 1.43 gal at 30%; 0.57 gal at 100%
9. 1.45 gal
Complete solutions follow.
1.
Percent Amount of Amount of
solution
target
1st
container
.10
x
.10x
2nd
container
.24
12
.24(12)
Mixture
.15
x + 12
.15(x + 12)
.10x + .24(12) = .15(x + 12)
.10x + 2.88 = .15x + 1.8
1.08 = .05x
21.6 = x
21.6 ounces @ 10%
2.
Percent Amount of Amount of
solution
target
1st
container
.20
x
.20x
2nd
container
.30
3
.30(3)
Mixture
.24
x+3
.24(x + 3)
.20x + .90 = .24x + .72
.18 = .04x
4.5 = x
4.5 liters at 20%
3.
Percent Amount of Amount of
solution
target
1st
container
.50
x
.50x
2nd
container
.80
8
.80(8)
Mixture
.60
x+8
.60(x + 8)
.50x + .80(8) = .60(x + 8)
.50x + 6.40 = .60x + 4.80
1.60 = .10x
16 = x
16 liters at 50%
4.
Percent Amount of Amount of
solution
target
1st
container
.10
x
.10x
2nd
container
.28
30 - x
.28(30 – x)
Mixture
.20
30
.20(30)
.10x + .28(30 – x) = .20(30)
.10x + 8.4 - .28x = 6
-.18x = -2.4
x = 13.3
13.3 ounces at 10%
5.
Percent Amount of Amount of
solution
target
1st
container
.20
X
.20x
2nd
container
.38
8-x
.38(8 – x)
Mixture
.30
8
.30(8)
.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.
Percent Amount of Amount of
solution
target
1st
container
.50
x
.50x
2nd
container
.72
2.4 - x
.72(2.4 – x)
Mixture
.58
2.4
.58(2.4)
.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.
Percent Amount of Amount of
solution
target
1st
container
.30
10
.30(10)
2nd
container
0
x
0(x)
Mixture
.24
x + 10
.24(x + 10)
.30(10) + 0 = .24(x + 10)
3 = .24x + 2.4
0.6 = .24x
2.5 = x
2.5 liters of water
8.
Percent Amount of Amount of
solution
target
1st
container
.30
x
.30(x)
2nd
container
1.00
2-x
1.00(2 – x)
Mixture
.50
2
.50(2)
.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.
Percent Amount of Amount of
solution
target
1st
container
.24
10 - x
.24(10 – x)
2nd
container
1.00
x
1.00x
Mixture
.35
10
.35(10)
.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