Transcript Section 3.5
3.5 Investment and Mixture 1. Use a table to solve problems involving two investments. 2. Use a table to solve problems involving mixtures. Copyright © 2011 Pearson Education, Inc. Understanding Quantities A + B = C 2 3 5 _____ x 3 x+3 _____ 1 12 11 12 – 1 _____ 5 4 9–5 _____ 9 x 20 – x _____ 20 Understanding Quantities A + B = C x 6 x_____ +6 14 – x _____ x 14 x 13 – x _____ 13 Objective 1 Use a table to solve problems involving two investments. Interest = Principal ∙ Rate ∙ Time Interest = Principal ∙ Rate Time = 1 year I = Pr Change percents to decimals for calculations Investment Problems Marvin invests a total of $12,000 in two plans. Plan 1 is at an APR (annual percentage rate) of 6% and Plan 2 is at an APR of 9%. If the total interest earned after one year is $828, what principal was invested in each plan? Interest from Plan 1 + Principal ∙ Rate Interest from Plan 2 = Total Interest Principal ∙ Rate .06x + .09(12,000 – x) = $828 I = Pr Accounts Plan 1 Principal Rate Interest x 6% = .06 .06x Plan 2 12,000 – x 9% = .09 .09(12,000 – x) Total 12,000 $828 What did we find? Did we answer the question? Marvin invests a total of $12,000 in two plans. Plan 1 is at an APR (annual percentage rate) of 6% and Plan 2 is at an APR of 9%. If the total interest earned after one year is $828, what principal was invested in each plan? I = Pr Accounts Principal Rate Interest Plan 1 x = $8400 6% = .06 .06x Plan 2 12,000 – x 9% = .09 .09(12,000 – x) Total 12,000 .06 x .0912 ,000 x 828 6 x 912 ,000 x 82 ,800 6 x 108 ,000 9 x 82 ,800 3x 108 ,000 82 ,800 3 x 25 ,200 x 8400 $828 Plan 2: 12,000 – x 12,000 – 8400 3600 Plan 1: $8400 Plan 2: $3600 Jon invests in a plan that has an APR of 8%. He invests $650 more than what he invested in the 8% account in a 12% APR account. If the total interest after one year from the investments is $328, how much was invested in each plan? I = Pr Accounts Principal Plan 1 Plan 2 x = 1250 x + 650 Rate Interest What did we find? .08 .08x .12 .12(x + 650) 328 Did we answer the question? Total Interest from Plan 1 + Interest from Plan 2 .08 x .12 x 650 328 8 x 12 x 650 32800 8 x 12 x 7800 32800 20 x 7800 32800 20 x 25000 x 1250 = Total Interest Plan 2: x + 650 1250 + 650 1900 $1250 at 8% $1900 at 12% Sam has $4000. She put some of the money into savings that pays 6% and the rest in an account that pays 7%. If her total interest for the year is $264, how much did she invest at each rate? I = Pr Accounts Principal Rate Interest What did we find? Plan 1 .06 .06x Plan 2 x = 1600 4000 – x .07 Total 4000 .07(4000 – x) 264 Did we answer the question? Interest from Plan 1 + Interest from Plan 2 = .06 x .074000 x 264 6 x 74000 x 26400 6 x 28000 7 x 26400 x 28000 26400 x 1600 x 1600 Total Interest Plan 2: 4000 – x 4000 – 1600 2400 $1600 at 6% $2400 at 7% Lisa invests a total of $6000 in two different accounts. The first account earns 8% while the second account earns 3%. If the total interest earned is $390 after one year, what amount is invested at 8%? a) $1800 b) $2100 c) $4200 d) $4800 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 9 Lisa invests a total of $6000 in two different accounts. The first account earns 8% while the second account earns 3%. If the total interest earned is $390 after one year, what amount is invested at 8%? a) $1800 b) $2100 c) $4200 d) $4800 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 10 Mixture Problems The dairy is making a 30% buttermilk cream. If it mixes a 26% buttermilk cream with a 35% buttermilk cream, how much of each does it need to use to produce 300 pounds of 30% buttermilk cream? 26% = + x 30% 35% 300 - x 300 The dairy is making a 30% buttermilk cream. If it mixes a 26% buttermilk cream with a 35% buttermilk cream, how much of each does it need to use to produce 300 pounds of 30% buttermilk cream? 26% 35% = + x Types 30% 300 - x What did we find? Did we answer the question? 300 % Concentration Quantity 26% .26 x 35% .35 .30 300 – x 300 30% Total .26x .35(300 – x) .30(300) .26 x .35300 x .30300 35%: 300 x 2 1 26 x 35300 x 30300 300 166 133 26 x 10500 x 35 x 9000 3 3 9 x 10500 9000 2 166 pounds of 26% milk 9 x 1500 3 1 2 1500 133 pounds of 35% milk 166 x 3 3 9 Ken has 80 milliliters of 15% acid solution. How much of a 20% acid solution must be added to create a solution that is 18% acid? What did we find? 15% 80 Types + 20% x = 18% Did we answer the question? 80 + x % Concentration Quantity 15% .15 80 20% .20 .18 x = 120 80 + x 18% .1580 .20 x .1880 x 1580 20 x 1880 x 1200 20 x 1440 18 x 2 x 240 x 120 Total .15(80) .20x .18(80 + x) 120 ml of the 20% solution The Candy Shoppe wants to mix 115 pounds of candy to sell for $.80 per pound. How many pounds of $.60 candy must be mixed with a candy costing $1.20 per pound to make the desired mix? .60 x Types + What did we find? 1.20 115 – x = .80 Did we answer the question? 115 % Concentration Quantity Total $.60 .60 x .60x $1.20 1.20 .80 115 – x 115 1.20(115 – x ) .80(115) $.80 .60 x 1.20115 x .80115 60 x 120115 x 80115 60 x 13800 120 x 9200 60 x 13800 9200 60 x 4600 4600 2 x 76 60 3 2 76 pounds of $.60 candy 3 Martin has a bottle containing 120 milliliters of 30% HCl solution and a bottle of 15% HCl solution. He wants a 25% HCl solution. How much of the 15% solution must be added to the 30% solution so that a 25% concentration is created? a) 30 milliliters b) 45 milliliters c) 60 milliliters d) 75 milliliters Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 15 Martin has a bottle containing 120 milliliters of 30% HCl solution and a bottle of 15% HCl solution. He wants a 25% HCl solution. How much of the 15% solution must be added to the 30% solution so that a 25% concentration is created? a) 30 milliliters b) 45 milliliters c) 60 milliliters d) 75 milliliters Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 1- 16