Copyright © 2011 Pearson Education, Inc.

Investment and Mixture3.53.5

1. Use a table to solve problems involving two investments.2. Use a table to solve problems involving mixtures.

Understanding Quantities

+ =A B C5

4

11

x + 3

20 – x

9 – 5

12 – 1

Understanding Quantities

+ =

x 6 _____

_____ x 14

x _____ 13

A B C

x + 6

14 – x

13 – x

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

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 fromPlan 1

Interest fromPlan 2+ =

Principal ∙ Rate Principal ∙ Rate

Investment ProblemsInvestment Problems

Accounts Principal Rate Interest

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

$828

Accounts Principal Rate Interest

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

828000120906 x,.x. 800820001296 ,x,x

8008290001086 ,x,x 800820001083 ,,x

200253 ,x 8400x

What did we find? Did we answer the question?

Plan 2: 12,000 – x

12,000 – 8400

3600

Plan 1: $8400Plan 2: $3600

6% = .06

9% = .09

x

12,000 – x

12,000

.06x

.09(12,000 – x)

$828

= $8400

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?

Accounts Principal Rate Interest

Plan 1

Plan 2

Total

I = Pr

.08

.12

x

x + 650

328

.08x

.12(x + 650)

Interest fromPlan 1

Interest fromPlan 2+ = Total

Interest

3286501208 x.x. 32800650128 xx

328007800128 xx32800780020 x

2500020 x1250x

What did we find?

Did we answer the question?

= 1250

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?

Accounts Principal Rate Interest

Plan 1

Plan 2

Total

I = Pr

.06

.07

x

4000 – x

264

.06x

.07(4000 – x)

Interest fromPlan 1

Interest fromPlan 2+ = Total

Interest

26440000706 x.x.

What did we find?

Did we answer the question?

= 1600

Plan 2: 4000 – x

4000 – 1600

2400

$1600 at 6%$2400 at 7%

4000

26400400076 xx264007280006 xx

2640028000 x1600 x

1600x

Slide 1- 9Copyright © 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

Slide 1- 10Copyright © 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

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 ProblemsMixture Problems

26% 35% 30%

x 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% 30%

x 300 - x 300

Types % Concentration Quantity Total

26%

35%

30%

.26

.35

.30

x

300 – x

300

.26x

.35(300 – x)

.30(300)

300303003526 .x.x. 300303003526 xx

9000351050026 xxx9000105009 x

15009 x

91500

x32

166

What did we find?

Did we answer the question?

35%:

32

166300

x300

31

133

milk 35%of pounds 31

133

milk 26%of pounds 32

166

+ =

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%

80 x 80 + x

Types % Concentration Quantity Total

15%

20%

18%

.15

.20

.18

80

x

80 + x

.15(80)

.20x

.18(80 + x)

x.x.. 8018208015 xx 8018208015

xx 181440201200 2402 x

120x

What did we find?

Did we answer the question?

= 120

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 1.20 .80

x 115 – x 115

Types % Concentration Quantity Total

$.60

$1.20

$.80

.60

1.20

.80

x

115 – x

115

.60x

1.20(115 – x ).80(115)

1158011520160 .x.x. 1158011512060 xx

92001201380060 xx92001380060 x

460060 x

What did we find?

Did we answer the question?

32

7660

4600x

candy $.60of pounds 32

76

Slide 1- 15Copyright © 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

Slide 1- 16Copyright © 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