10 8 Mixture Problems Goal To solve problems

10. 8 Mixture Problems Goal: To solve problems involving the mixture of substances

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

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

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

One solution is 80% acid another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? Let x = y= Amount of • Percent solution Acid = Amount of pure Acid 1 st Solution x 0. 80(x) 2 nd Solution y 0. 30(y) 3 rd Solution 200 0. 62(200) 124

One solution is 80% acid another is 30% acid. How much of each is required to make 200 L of solution that is 62% acid? Y= 200 -x Y = 200 -128 1 st 8 x + 3 y =1240 8 x + 3 (200 -x) =1240 Amount of • Percent 8 x +600 -3 x =1240 solution Acid x 0. 80 Y = 72 L Solution 2 nd Solution 3 rd Solution y 5 x +600 =1240 5 x = 640 200 0. 30 X= 128 L 0. 62 = Amount of Acid. 80(x) . 30(y). 62(200) 124

A chemist has one solution that is 60% acid another that is 30% acid. How much of each solution is needed to make a 750 ml solution that is 50% acid? Amount of • Percent solution Acid = Amount of Acid 1 st Solution x 0. 60(x) 2 nd Solution y 0. 30(y) 3 rd Solution 750 0. 50(750) 375

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 • Percent solution Acid = Amount of Acid 1 st Solution x 0. 28(x) 2 nd Solution y 0. 40 . 4(y) 3 rd Solution 300 0. 36(300) 108

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?

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? Amount of Solution (gallons) Percent of Salt Amount of Salt Solution 1 x . 50 . 5 x Solution 2 60 . 15 9 Final Solution y . 40 . 4 y

System x + 60 =y 150 + 60 = y 210 gallons =y 0. 5 x + 9 = 0. 4 y 5 x +90 = 4 y 5 x + 90 = 4 (x +60) 5 x + 90 = 4 x + 240 x + 90 =240 x =150 gallons

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?

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? Pounds of Coffee $ per pound total cost Coffee mix 1 X $2. 20 x Coffee mix 2 2 $1. 40 2. 80 Final Coffee mix Y $2. 04 y

System X + 2 =y 2. 20 x +2. 80 = 2. 04 y

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

Assignment: Page 462 (1 -9) odd

Amount of • Percent lake salt = Amount of salt 1963 1 . 20 2 nd Solution x . 00(x) x+1 . 06 . 20 1984

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