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# MIXTURE PROBLEMS

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

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1. 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.

2. 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. + =

3. 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.

4. Fill the second column with the percent strengths of each container. .50 .20 .30

5. 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

6. 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. .50 10 5 .20 x .20x .30 x + 10 .30(x+10)

7. 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. .50 10 5 .20 x .20x .30 x + 10 .30(x+10)

8. 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%.

9. 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!

10. 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. .30 1.00 .50

11. 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. .30 x 1.00 .50 5

12. 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: .30 x 1.00 5 - x .50 5

13. 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. .30 x .30x 1.00 5 - x 1(5 – x) .50 5 .50(5)

14. 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.

15. 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

16. 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?

17. The target ingredient will be acid. The second container will be the water. Did you use 0% for the strength of the second container?

18. The target ingredient will be acid. The second container will be the water. Did you use 0% for the strength of the second container?

19. 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.

20. Three mixture problems at the beginners level. Round your answers • 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?

21. 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?

22. 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?

23. 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.

24. 1. .10x + .24(12) = .15(x + 12) .10x + 2.88 = .15x + 1.8 1.08 = .05x 21.6 = x 21.6 ounces @ 10%

25. 2. .20x + .90 = .24x + .72 .18 = .04x 4.5 = x 4.5 liters at 20%

26. 3. .50x + .80(8) = .60(x + 8) .50x + 6.40 = .60x + 4.80 1.60 = .10x 16 = x 16 liters at 50%

27. 4. .10x + .28(30 – x) = .20(30) .10x + 8.4 - .28x = 6 -.18x = -2.4 x = 13.3 13.3 ounces at 10%

28. 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%

29. 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%

30. 7. .30(10) + 0 = .24(x + 10) 3 = .24x + 2.4 0.6 = .24x 2.5 = x 2.5 liters of water

31. 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%

32. 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

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