1. 10.8 Mixture Problems Goal: To solve problems involving the mixture of substances

2. Mixture Problems One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid?

3. Steps to Solve Mixture Problems Set up a chart (4x4) Amount of Solution Percent of _______ Amount of _____ Solution 1 Solution 2 Final Solution

4. Steps to Solve Mixture Problems Convert the percentages to decimals and fill out the chart Multiply going across the chart Add going down the chart Set up 2 equations with 2 variables (system) Solve the system by substitution or addition

5. One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? Amount of solution Percent Acid = Amount of pure Acid 1 st Solution 2 nd Solution 3 rd Solution Let x = y = x.80(x)0.80 y0.30.30(y) 2000.62.62(200) 124

6. One solution is 80% acid and another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? Amount of solution Percent Acid = Amount of Acid 1 st Solution 2 nd Solution 3 rd Solution x.80(x)0.80 y0.30.30(y) 2000.62.62(200) 124 Y= 200-x 8x + 3y =1240 8x + 3 (200-x) =1240 8x +600 -3x =1240 5x +600 =1240 5x = 640 X= 128 L Y = 200 -128 Y = 72 L

7. A chemist has one solution that is 60% acid and another that is 30% acid. How much of each solution is needed to make a 750ml solution that is 50% acid? Amount of solution Percent Acid = Amount of Acid 1 st Solution 2 nd Solution 3 rd Solution x.60(x)0.60 y0.30.30(y) 7500.50.50(750) 375

8. A chemist has one solution that is 28% oil and another that is 40% oil. How much of each solution is needed to make a 300 L solution that is 36% oil? Amount of solution Percent Acid = Amount of Acid 1 st Solution 2 nd Solution 3 rd Solution x.28(x)0.28 y0.40.4(y) 3000.36.36(300) 108

9. Try to make your own chart How many gallons of a 50% salt solution must be mixed with 60 gallons of a 15% solution to obtain a solution that is 40% salt?

10. Amount of Solution (gallons) Percent of Salt Amount of Salt Solution 1x.50.5x Solution 260.159 Final Solutiony.40.4y

11. System x + 60 =y 0.5x + 9 = 0.4y 5x +90 = 4y 5x + 90 = 4 (x +60) 5x + 90 = 4x + 240 x + 90 =240 x =150 gallons 150 + 60 = y 210 gallons =y

12. Coffee Beans How many pounds of coffee beans selling for $2.20 per pound should be mixed with 2 pounds of coffee beans selling for $1.40 per pound to obtain a mixture selling for $2.04 per pound?

13. Pounds of Coffee $ per pound total cost Coffee mix 1X$2.202.20 x Coffee mix 22$1.402.80 Final Coffee mix Y$2.042.04 y

14. System 2.20x +2.80 = 2.04 yX + 2 =y

16. Your Turn Come up with your own mixture word problem. Make it interesting! Remember to include: Amount of SolutionWieght of Object % of (acid /water / oil/salt/etc)Cost per weight Amount of (acid / water /oil/salt/etc) Total cost Solution 1 and 22 objects Final SolutionMixture of 2 objects

17. Assignment: Page 462 (1 -9) odd

19. Amount of lake Percent salt = Amount of salt 1963 2 nd Solution 1984 1.20 x.00.00(x) x+1.06.20

20. Vocabulary Mixture- two substances combined Concentrate or Solution- how much non- water is mixed (juice) 10% solution -10% concentration and 90% water