chapter 3 copyright © 2015, 2011, 2007 pearson education, inc. chapter 3-1 applications of algebra

Chapter 3

Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-1

Applications of Algebra

2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-2

3.1 – Changing Application Problems into Equations

3.2 – Solving Application Problems

3.3 – Geometric Problems

3.4 – Motion, Money and Mixture Problems

Chapter Sections


Motion, Money and Mixture Problems


Motion Problems

A motion problem is one in which an object is moving at a specified rate for a specified period of time.

distance = rate · time


Motion ProblemsExample:Maryanne and Paul Justinger and their son Danny are on a canoe trip on the Erie Canal. Danny is in one canoe and Paul and Maryanne are in a second canoe. Both

canoes start at the same time from the same point and travel in the same direction. The parents paddle their canoe at 2 miles per hour and their son paddles his canoe at 4 miles per hour. In how many hours will the two canoes be 5 miles apart?

Continued.


Motion Problems

distance = rate · time

Let t = time when canoes are 5 miles apart

- = 5 miles 4t – 2t = 5

t = 2.5

After 2.5 hours the two canoes will be 5 miles apart

Example continued:

distance traveled by faster canoe

distance traveled by slower canoe


Motion Problems

Example:Two highway paving crews are 20

miles apart working toward each other. One crew paves 0.4 mile of road per day more than the other crew, and the two crews meet after 10 days. Find the rate at which each crew paves the road.

Continued.


Motion Problems

Let r = the time it takes the glasses to fall

Then r + 0.4 = rate of the faster crew

The total distance covered by both crews is 20 miles. Since the crews are moving in opposite directions, the total miles of road paved is found by adding the two distances.

Continued.


Distance ProblemsExample continued:

= 20 miles

10r + 10r + 4 = 20 20r + 4 = 20 20r = 16

r = 0.8

The slower crew paves 0.8 miles of road per day and the faster crew paves r + 0.4 or 0.8 + 0.4 = 1.2 miles of road per day.

distance covered by slower crew

distance covered by faster crew

-


Money Problems

interest = principal · rate · time

Example:Olga Harrison has $13,000 to invest and

wishes to invest in two investments. The first investment is a loan to a small business that pays her 5% simple interest for one year. The second investment is a 1-year certificate of deposit (CD) that pays 2% simple interest. Olga wishes to earn a total of $500 in interest in one year from the two investments. How much money should she put into each investment?

Continued.


Money Problems

principal · rate · time = interestAccoun

tPrincip

alRate Tim

eInterest

Loan x 0.05 1 0.05x

CD 13,000 – x

0.02 10.02(13,000

– x)

Let x = the amount of money invested at 5%Then 13,000 – x = amount invested at 2%

Continued.

Example continued:

Interest from 5%

loan

Interest from 2% CD+ = total

interest


Money Problems

0.05x + 0.02(13000 – x) = 500 0.05x + 260 – 0.2x = 500 0.03x = 240 x = 8,000

Thus $8,000 should be invested in the loan at 5%. The amount to be invested in the CD at 2% is 13,000 – x = 13,000-8,000 = 5,000

Therefore, $5,000 should be invested in the CD. The total amount invested is $8,000 + $5,000 = $13,000, which checks with the information given.

Example continued:


Mixture Problems

Any problem in which two or more quantities are combined to produce a single quantity or a single quantity is separated into two or more quantities may be considered a mixture problem.


Mixture Problems

Quantity · Price (per unit) = Value of item

Type 1– Mixing SolidsWhen working with mixture problems involving solids, we

generally use the fact that the value (or cost) of one part of the mixture plus the value (or cost) of the second part of the mixture is equal to the total value (or total cost) of the mixture.


Mixture Problems

Example:Troy’s Family grass seed sells for $2.75 per pound, and Troy’s Spot Filler grass seed sells for $2.35 per pound. How many pounds of each should be mixed to get a 10-pound mixture that sells for $2.50 per pound?

Continued.


Mixture Problems

Type of Seed

Number of Pounds

Cost per Pound

Cost of Seed

Family x 2.75 2.75xSpot Filler 10 – x 2.35 2.35(10 – x)

Mixture 10 2.50 2.50(10)

Let x = number of pounds of Family grass seed

Example continued:

Continued.

Then 10 - x = number of pounds of Spot Filler grass seed


Mixture Problems

Thus, 3.75 pounds of the Family grass seed must be mixed with 10 – x or 10 – 3.75 = 6.25 pounds of Spot Filler grass seed to make a mixture that sells for $2.50 per pound.

Example continued:Cost of Family Seed

Cost of Spot Filler Seed+ = Cost of

mixture2.75x + 2.35(10 – x) = 2.50(10)

2.75x + 23.5 – 2.35x = 25.0

0.40x + 23.5 = 25.0

x = 3.75


Mixture Problems

Quantity · Strength = Amount of Substance

Type 2– Mixing SolutionsWhen working with solutions, we use the following

formula: amount of substance in the solution = quantity of solution x strength of solution (in percent written as a decimal.


Mixture Problems

Example:Mr. Dave Lumsford needs a 10% acetic acid solution for a chemistry experiment. After checking

the store room, he finds that there are only 5% and 20% acetic acid solutions available. Mr. Lumsford decides to make the 10% solution by combining the 5% and 20% solutions. How many liters of the 5% solution must he add to 8 liters of the 20% solution to get a solution that is 10% acetic acid?

Continued.


Mixture Problems

Continued.

Let x = number of liters of 5% acetic acid solution


Mixture Problems

Solution

Liters

Strength Amount of Acetic Acid

5% x 0.05 0.05x20% 8 0.20 0.20(8)Mixture x + 8 0.10 0.10(x + 8)

Example continued:

Continued.

Amount of acid in 5% solution

Amount of acid in 20%

solution

+ =Amount of acid in 10% mixture


Mixture Problems

0.5x + 0.20(8) = 10(x + 8) 0.5x + 1.6 = 0.10x + 0.8 0.5x + 0.8 = 0.10x 0.8 = 0.05x 16 = x

Sixteen liters of 5% acetic acid solution must be added to the 8 liters of 20% acetic acid solution to get 10% acetic acid solution. The total number of liters that will be obtained is 16 + 8 or 24.

Example continued:


Mixture Problems

Example:Nicole Pappas, a medical researcher, has 40% and 5% solutions of

phenobarbital. How much of each solution must she mix to get 0.6 liter of 20% phenobarbital solution?

Continued.


Mixture Problems

Continued.

Let x = number of liters of the 40% solution.Then 0.6 – x = number of liters of the 5% solution.


Mixture Problems

Solution

Liters

Strength Amount of Phenobarbita

l40% x 0.40 0.40x5% 0.6 - x 0.05 0.05(0.6 - x)Mixture 0.6 0.20 0.6(0.20)

Example continued:

Continued.

Amount of phenobarbital

in 40% solution

Amount of phenobarbital in 5% solution

+ =Amount of

phenobarbital in mixture


Mixture Problems

0.40x + 0.05(0.6 - x) = (0.6)(0.20) 0.40x + 0.03 - 0.05x = 0.12 0.35x + 0.03 = 0.12 0.35x = 0.09 x = 0.26

Since the answer was less than 0.6 liter, the answer is reasonable. About 0.26 liter of the 40% solution must be mixed with about 0.6 – x = 0.60 – 0.26 = 0.34 liter of the 5% solution to get 0.6 liter of the 20% mixture.

Example continued:

chapter 3 copyright © 2015, 2011, 2007 pearson education, inc. chapter 3-1 applications of algebra

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