Objective: To Solve Mixture Problems Mixture problems involve combining two mixtures to make one final mixture. We are going to use buckets to help solve these problems. 1st Mixture 2nd Mixture Final Mixture + =

Buckets

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? STEP #1: Draw buckets + =

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Step #1: Draw the buckets STEP #2: Label the buckets Pure OJ Orange Drink Final Mixture + =

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Step #1: Draw the buckets Step #2: Label the buckets STEP #3: Price or Percent goes on top + = Pure OJ Orange Drink Final Mixture 10% 40% 100%

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Step #1: Draw the buckets Step #2: Label the buckets Step #3: Price or Percent goes on top STEP #4: Change Percents to decimals + = Pure OJ Orange Drink Final Mixture 10% 40% = 0.4 100% = 1 = 0.1

1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Step #1: Draw the buckets Step #2: Label the buckets Step #3: Price or Percent goes on top Step #4: Change Percents to decimals STEP #5: Weight goes on the bottom + = Pure OJ Orange Drink Final Mixture 10% 40% = 0.4 100% = 1 = 0.1 x 5 x + 5

+ = Pure OJ STEP #6: Set up and Solve Orange Drink Final Mixture x 5 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? Step #1: Draw the buckets STEP #6: Set up and Solve Step #2: Label the buckets Step #3: Price or Percent goes on top Step #4: Change Percents to decimals Step #5: Weight goes on the bottom + = Pure OJ Orange Drink Final Mixture 10% 40% = 0.4 100% = 1 = 0.1 x 5 x + 5

+ = 1x + 0.1(5) = 0.4(x+5) 1.0x + 0.5 = 0.4x + 2 -0.4x -0.4x 1. An advertisement for an orange drink claims that the drink contains 10% orange juice. How much pure orange juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% orange juice? STEP #6: Set up and Solve + = Pure OJ Orange Drink Final Mixture 100% 10% 40% = 1 = 0.1 = 0.4 x 5 x + 5 1x + 0.1(5) = 0.4(x+5) 1.0x + 0.5 = 0.4x + 2 -0.4x -0.4x 0.6x + 0.5 = 2.0 -0.5 -0.5 2.5 quarts of pure orange juice would have to be added to the 5 quarts of the drink. 0.6x = 1.5 0.6 0.6 x = 2.5

+ = Final Mixture 2. How much water would you have to add to 3 liters of lake water that is 30 percent salt to get a solution that is only 12 percent salt? Step #1: Draw buckets Step #2: Label the buckets Step #3: Percent or price goes on top Step #4: Change percents to decimals Step #5: Amount goes on bottom Step #6: Set up and solve Lake Water Water Final Mixture + = 0% = 0 30% = 0.3 12% = 0.12 x 3 x +3

+ = Final Mixture Lake Water Water 4.5 liters of water should 0% 30% 12% = 0 = 0.3 = 0.12 x 3 x +3 #2 0(x) + 0.3(3) = 0.12(x+3) 0 + 0.9 = 0.12x + 0.36 0.9 = 0.12x + 0.36 4.5 liters of water should be added to the lake water -0.36 -0.36 0.54 = 0.12x + 0 0.54 = 0.12x 0.12 0.12 x = 4.5

+ = Final Mixture 3. Sterling Silver is 92.5% pure silver. How many grams of pure silver and sterling silver must be mixed to obtain 100g of a 94% Silver alloy? Step #1: Draw buckets Step #2: Label the buckets Step #3: Percent or price goes on top Step #4: Change percents to decimals Step #5: Amount goes on bottom Step #6: Set up and solve Pure Silver Sterling Silver Final Mixture + = 92.5% = 0.925 100% = 1 94% = 0.94 x 100 - x 100

+ = Final Mixture Pure Silver Sterling Silver #3 92.5% 100% 94% = 1 = 0.925 = 1 = 0.94 x 100 - x 100 0.925x + 1 (100-x) = 0.94(100) 0.925x + 100 – 1x = 94 80 grams of the sterling silver and 20 grams of the pure silver are needed. -0.075x + 100 = 94 -100 -100 -0.075x = -6 -0.075 -0.075 x = 80 100 – 80 = 20 100 - x

+ = Final Mixture 4. A mixture containing 6% boric acid is to be mixed with 2 quarts of a mixture which is 15% boric acid in order to obtain a solution which is 12% boric acid. How much of the 6% solution must be used? Step #1: Draw buckets Step #2: Label the buckets Step #3: Percent or price goes on top Step #4: Change percents to decimals Step #5: Amount goes on bottom Step #6: Set up and solve Boric Acid Mixture Boric Acid Mixture Final Mixture + = 6% = 0.06 15% = 0.15 12% = 0.12 x 2 x + 2

+ = Final Mixture Boric Acid Mixture 6% 15% 12% = 0.15 = 0.12 x 2 = 0.06 = 0.15 = 0.12 x 2 x + 2 #4 0.06x + 0.15(2) = 0.12(x+2) 0.06x + 0.3 = 0.12x + 0.24 -0.06x -0.06x 1 quart of the 6% boric acid mixture should be added. 0.3 = 0.06x + 0.24 -0.24 -0.24 0.06 = 0.06x 0.06 0.06 1 = x

+ = Final Mixture 5. ClearShine window cleaner is 12% alcohol and Sunstream cleaner is 30% alcohol. How much of each should be used to make 90 oz. (ounces) of a cleaner that is 20% alcohol? Step #1: Draw buckets Step #2: Label the buckets Step #3: Percent or price goes on top Step #4: Change percents to decimals Step #5: Amount goes on bottom Step #6: Set up and solve Sunstream Clearshine Final Mixture + = 12% = 0.12 30% = 0.3 20% = 0.20 x 90-x 90

+ = Final Mixture Sunstream Clearshine 12% 30% 20% = 0.3 = 0.20 x 90-x = 0.12 = 0.3 = 0.20 x 90-x 90 #5 0.12x + 0.3(90-x) = 0.20(90) 0.12x + 27 - 0.3x = 18 -0.18x + 27 = 18 50 ounces of Clearshine window cleaner and 40 ounces of Sunstream window cleaner. -27 -27 -0.18x = -9 -0.18 -0.18 x = 50