1. Percent Mixture Multiple Choice

2. 1.Jose has 20 ounces of a 20% salt solution. How much salt should he add to make it a 25% salt solution? a.He should add ¾ ounces of salt. b.He should add 21 ounces of salt. c.He should add 4/3 ounces of salt d.He should add 1 ounce of 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 20%100%25% = 0.20 = 1= 0.25 20 x x + 20

3. + = 20%100%25% = 0.20 = 1= 0.25 20 x x + 20 0.2(20) + 1x = 0.25(x+20) 4 + 1x = 0.25x + 5 -0.25x –0.25x 4 + 0.75x = 5 -4 -4 0.75x = 1 0.75 x = 1.333…or 4/3 You need 4/3 ounces of salt. C #1

4. 2.If you mix 3 liters of can A which contains 10% oil with 6 liters of can B which contains 40% oil, how much oil is in the mixture? a.20% oilb. 30% oil c. 5% oild. 12% oil 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 + = 10%40%x % = 0.10 = 0.4 3 6 9

5. + = 10%40%X % = 0.10 = 0.4 3 6 9 0.10(3) + 0.4(6) = 9x 0.3 + 2.4 = 9x 2.7 = 9x 9 0.3 = x 30% oil is in the final mixture. B #2

6. 3. How much water should Jazmin add to 3 gallons of a 40% salt solution to get a solution that is 25% salt? a.1.8 gallons of water b.7.8 gallons of water c.0.8 gallons of water d. 16.8 gallons of water 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 0%40%25% = 0 = 0.4= 0.25 x 3 x + 3

7. + = 0%40%25% = 0 = 0.4= 0.25 x 3 x + 3 0x + 0.4(3) = 0.25(x+3) 0 + 1.2 = 0.25x + 0.75 1.2 = 0.25x + 0.75 -0.75 0.45 = 0.25x 0.25 1.8 = x a. 1.8 gallons of water

8. 4. A 20% alcohol solution is added to a 45% alcohol solution. How much of each solution is needed to make a 40 milliliter solution that 32.5% solution? a.20 of each solution b.15 milliliters of the 20% alcohol solution and 25 milliliters of the 45% alcohol solution c.25 milliliters of the 20% alcohol solution and 15 milliliters of the 45% alcohol solution d.30 milliliters of the 20% alcohol solution and 10 milliliters of the 45% alcohol solution + = 20%45%32.5% = 0.2 = 0.45 = 0.325 x 40-x 40

9. + = 20%45%32.5% = 0.2 = 0.45 = 0.325 x 40-x 40 0.2x + 0.45(40-x) = 0.325(40) 0.2x + 18 – 0.45x = 13 -0.25x +18 = 13 -18 -0.25x = -5 -0.25 x = 20 40 –x 40 – 20 = 20 a.20 of each solution 4.

10. 5.How many liters of a 14 percent alcohol solution must be mixed with 20 liters of a 50 percent alcohol solution to get a 20 percent alcohol solution? a.100 liters must be added to the 14% alcohol solution b.150 liters must be added to the 14% alcohol solution c.120 liters must be added to the 14% alcohol solution d. 75 liters must be added to the 14% alcohol solution + = 14%50%20% = 0.14 = 0.5= 0.2 x 20x + 20

11. + = 14%50%20% = 0.14 = 0.5= 0.2 x 20x + 20 5. 0.14x + 0.5(20) = 0.2(x + 20) 0.14x + 10 = 0.2x + 4 -0.2x 0.20x - 0.14x -0.06x -0.06x + 10 = 4 -10 -10 -0.06x = -6 -0.06 x = 100 a.100 liters must be added to the 14% alcohol solution

12. 6.Milk that has 5% butterfat is mixed with milk that has 2% butterfat. How much of each is needed to obtain 60 gallons of milk that has 3% butterfat? a. 50 liters of the milk with 5% butterfat and 10 liters of the milk with 2% butterfat b. b. 40 liters of the milk with 2% butterfat and 1.20 liters of the milk with 5% butterfat 5.c. 10 liters of the milk with 5% butterfat and 1.50 liters of the milk with 2% butterfat 6.d. 20 liters of the milk with 5% butterfat and 1.40 liters of the milk with 2% butterfat+ = 5%2% 3% = 0.05 = 0.02= 0.03 x 60-x60

13. + = 5%2% 3% = 0.05 = 0.02= 0.03 x 60-x60 0.05x + 0.02(60-x) = 0.03 (60) 0.05x + 1.2 – 0.02x = 1.8 0.03x + 1.2 = 1.8 -1.2 -1.2 0.03x = 0.6 0.03 x = 20 60 –20 = 40 d.20 liters of the milk with 5% butterfat and 40 liters of the milk with 3% butterfat

14. 7.Jesse has 3 gallons of a solution that is 30 percent antifreeze, which he wants to use to winterize his car. How much pure antifreeze should he add to this solution so that the new solution will be 65 percent antifreeze? a.2 gallons of pure antifreeze should be added b.3 gallons of pure antifreeze should be added c.4 gallons of pure antifreeze should be added 5.5 gallons of pure antifreeze should be added + = 30%100%65% = 0.3 = 1= 0.65 3 x3 + x

15. + = 30%100%65% = 0.3 = 1= 0.65 3 x3 + x 0.3(3) + 1x = 0.65(3 + x) 0.9 + 1x = 1.95 + 0.65x -0.65x 0.9 + 0.35x = 1.95 -0.9 -0.90 0.35x = 1.05 0.35 x = 3 b.3 gallons of the pure antifreeze should be added