1. Mixture Problems

3. Ex# 1: Emily mixed together 12 gallons of Brand A fruit drink and with 8 gallons of water. Find the percent of fruit juice in Brand A if the mixture contained 30% fruit juice. Mixture A=amountP=percent p12 12p 08 0 0.320 6 Juice Water

4. 12p Mixture A=amountP=percent p12 12p 08 0 0.320 6 Juice Water + 0 = 6

5. 12p + 0 = 6 p = 0.5 12p = 6 p = 50%

6. Ex #2: How many gallons of a 3% acid solution must be mixed with 60 gallons of 10% acid solution to obtain an acid solution that is 8%? 3% 10% 8% Mix.03x.03x.1060 6.08x + 60.08x+ 4.8

7. .03x 3% 10% 8% Mix.03x.03x.1060 6.08x + 60.08x+ 4.8 + 6 =.08x + 4.8

8. .03x + 6 =.08x + 4.8 + 600 =3x 100 8x+ 480 600 = 5x + 480 120 = 5x 24 = x 24 gal

9. Ex #3: How many mg of a metal containing 40% nickel must be combined with 6 mg of pure nickel to form an alloy containing 80% nickel? 40% 100% 80% Mix.4x.4x 16 6.8x + 6.8x+ 4.8

10. .4x + 40% 100% 80% Mix.4x.4x 16 6.8x + 6.8x+ 4.8 6 =.8x + 4.8

11. .4x + 6 =.8x + 4.8 + 60 =4x 10 8x+ 48 60 = 4x + 48 12 = 4x 3 = x 3 mg

12. Example 4: How many gallons each of a 5% and 10% salt solutions should be mixed to obtain 50 gallons of a 8% solution? 5% 10% 8% mix.05x.05x.1050 – x 5.0850 –.10x 4

13. .05x + 5 –.10x = 4 5% 10% 8% mix.05x.05x.1050 – x 5.0850 4 –.10x

14. .05x + 5 –.10x = 4 100 + 5005x– 10x= 400 -5x + 500 = 400 -5x = -100 x = 20 20 gal30 gal

15. Example 5: A 35% acid solution was mixed with a 15% acid solution to make 10L of a 29% acid solution. Find the percent concentration of each solution. 0.35x 0.1510 – x 0.2910 29% mix 35% 15%.35x 1.5–.15x 2.9

16. 0.35x 0.1510 – x 0.2910 29% mix 35% 15%.35x 1.5 2.9 –.15x.35x + 1.5 –.15x = 2.9

17. 100 + 15035x– 15x= 290 20x + 150 = 290 20x = 140 x = 7 7 L 3 L