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Copyright © 2011 Pearson Education, Inc. Investment and Mixture 3.5 1.Use a table to solve problems involving two investments. 2.Use a table to solve problems.

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Presentation on theme: "Copyright © 2011 Pearson Education, Inc. Investment and Mixture 3.5 1.Use a table to solve problems involving two investments. 2.Use a table to solve problems."— Presentation transcript:

1 Copyright © 2011 Pearson Education, Inc. Investment and Mixture 3.5 1.Use a table to solve problems involving two investments. 2.Use a table to solve problems involving mixtures.

2 Understanding Quantities += ABC 23_____ x3 112 5_____9 x 20 5 4 11 x + 3 20 – x 9 – 5 12 – 1

3 Understanding Quantities += x6_____ x14 x_____13 ABC x + 6 14 – x 13 – x

4 Objective 1 Use a table to solve problems involving two investments. Interest = Principal ∙ Rate ∙ Time Change percents to decimals for calculations Interest = Principal ∙ Rate Time = 1 year I = Pr

5 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 Interest from Plan 2 + = Principal ∙ Rate Investment Problems AccountsPrincipalRateInterest Plan 1 Plan 2 Total I = Pr 6% =.06 9% =.09 x 12,000 – x Total Interest 12,000.06x.09(12,000 – x).06x.09(12,000 – x) += $828

6 AccountsPrincipalRateInterest Plan 1 Plan 2 Total 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 What did we find?Did we answer the question? Plan 2: 12,000 – x 12,000 – 8400 3600 Plan 1: $8400 Plan 2: $3600 6% =.06 9% =.09 x 12,000 – x 12,000.06x.09(12,000 – x) $828 = $8400

7 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? AccountsPrincipalRateInterest Plan 1 Plan 2 Total I = Pr.08.12 x x + 650 328.08x.12(x + 650) Interest from Plan 1 Interest from Plan 2 + = Total Interest What did we find? Did we answer the question? = 1250 Plan 2: x + 650 1250 + 650 1900 $1250 at 8% $1900 at 12%

8 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? AccountsPrincipalRateInterest Plan 1 Plan 2 Total I = Pr.06.07 x 4000 – x 264.06x.07(4000 – x) Interest from Plan 1 Interest from Plan 2 + = Total Interest What did we find? Did we answer the question? = 1600 Plan 2: 4000 – x 4000 – 1600 2400 $1600 at 6% $2400 at 7% 4000

9 Slide 1- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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

10 Slide 1- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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

11 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? + = Mixture Problems 26% 35% 30% x 300 - x 300

12 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%30% x300 - x300 Types% ConcentrationQuantityTotal 26% 35% 30%.26.35.30 x 300 – x 300.26x.35(300 – x).30(300) What did we find? Did we answer the question? 35%:

13 + = 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? 15%20%18% 80x 80 + x Types% ConcentrationQuantityTotal 15% 20% 18%.15.20.18 80 x 80 + x.15(80).20x.18(80 + x) What did we find? Did we answer the question? = 120 120 ml of the 20% solution

14 + = 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?.601.20.80 x115 – x 115 Types% ConcentrationQuantityTotal $.60 $1.20 $.80.60 1.20.80 x 115 – x 115.60x 1.20(115 – x ).80(115) What did we find? Did we answer the question?

15 Slide 1- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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

16 Slide 1- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 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


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