Mixture Applications
15
Mixture Applications Example 1: Susan would like to mix a 10% acid solution with a 40% acid solution to make 20 ounces of a 30% acid solution. How much of each should she mix.
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Mixture Applications. Example 1:. Susan would like to mix a 10% acid solution with a 40% acid solution to make 20 ounces of a 30% acid solution. How much of each should she mix. 1) Variable declaration:. Let x represent the amount of the 10% solution. - PowerPoint PPT Presentation
Transcript of Mixture Applications
Mixture Applications
Example 1:
Susan would like to mix a 10% acid solution with a 40% acid solution to make 20 ounces of a 30% acid solution. How much of each should she mix.
10% 40% 30%
1) Variable declaration:
Let x represent the amount of the 10% solution
The desired amount of 30% solution is 20 ounces.
The amount of 40% solution is (20 – x) ounces.
x 2020 x
10% 40% 30%
2) Write the equation
x 2020 x
amount of
acid in 10%
solution
amount of
acid in 40%
solution
amount of
acid in 30%
solution
10%
x
amount of
acid in 10%
solution
amount of
acid in 40%
solution
amount of
acid in 30%
solution
0.10 x0.10 x
40%
20 x
amount of
acid in 30%
solution
0.40 20 x 0.40 20 x
0.10 xamount of
acid in 40%
solution
30%
20
amount of
acid in 30%
solution
0.30 200.30 20
0.10 x 0.40 20 x
3) Solve the equation:
0.10 0.40(20 ) 0.30 20x x
10 40(20 ) 30 20x x
10 800 40 600x x
30 800 600x
30 200x
6.666...x
6.7x
x = amount of the 10% solution
4) Write an answer in words, explaining the meaning in light of the application
Susan needs approximately 6.7 oz. of the 10% solution.
10%
x6.7x
20 - x = amount of the 40% solution
Susan needs approximately 13.3 oz. of the 40% solution.
20 6.7 13.3
40%
20 x
Mixture Applications
Example: 2
Al would like an 8% iodine solution, but only has a 5% and a 20% solution on hand. How much of the 5% solution should he mix with 6 ml of the 20% solution to get the 8% solution.
Make a drawing of the situation:
5% 20% 8%
1) Variable declaration:
Let x represent the amount of the 5% solution
The given amount of 20% solution is 6 ml.
The amount of 8% solution is (x+6) ml.
x 6 6x
5% 20% 8%
x 6 6x
2) Write the equation
amount of
iodine in 5%
solution
amount of
iodine in 20%
solution
amount of
iodine in 8%
solution
0.05 x 0.20 6 0.08 6x
0.05 0.20 6 0.08 6x x
3) Solve the equation:
5 120 8 48x x
5 20 6 8 6x x
3 120 48x
3 72x
24x
x = amount of the 5% solution
4) Write an answer in words, explaining the meaning in light of the application
Al needs 24 ml of the 5% iodine solution to mix with the 6 ml of 20% iodine solution to get an 8% solution.
5%
x24x
