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

2 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter – Changing Application Problems into Equations 3.2 – Solving Application Problems 3.3 – Geometric Problems 3.4 – Motion, Money and Mixture Problems Chapter Sections

3 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-3 Motion, Money and Mixture Problems

4 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-4 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

5 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-5 Motion Problems Example: 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.

6 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-6 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

7 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-7 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.

8 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-8 Motion Problems Let r = the time it takes the glasses to fall Then r = 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.

9 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-9 Distance Problems Example 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 or = 1.2 miles of road per day. distance covered by slower crew distance covered by faster crew -

10 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-10 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.

11 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-11 Money Problems principal · rate · time = interest AccountPrincipalRateTimeInterest Loan x x CD 13,000 – x (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

12 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-12 Money Problems 0.05x (13000 – x) = x – 0.2x = x = 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:

13 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-13 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.

14 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-14 Mixture Problems Quantity · Price (per unit) = Value of item Type 1– Mixing Solids When 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.

15 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-15 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.

16 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-16 Mixture Problems Type of Seed Number of Pounds Cost per Pound Cost of Seed Family x x Spot Filler 10 – x (10 – x) Mixture (10) Let x = number of pounds of Family grass seed Example continued: Continued. Then 10 - x = number of pounds of Spot Filler grass seed

17 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-17 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 mixture 2.75x (10 – x) = 2.50(10) 2.75x – 2.35x = x = 25.0 x = 3.75

18 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-18 Mixture Problems Quantity · Strength = Amount of Substance Type 2– Mixing Solutions When 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.

19 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-19 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.

20 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-20 Mixture Problems Continued. Let x = number of liters of 5% acetic acid solution

21 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-21 Mixture Problems SolutionLitersStrengthAmount of Acetic Acid 5% x x 20% (8) Mixture x (x + 8) Example continued: Continued. Amount of acid in 5% solution Amount of acid in 20% solution += Amount of acid in 10% mixture

22 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-22 Mixture Problems 0.5x (8) = 10(x + 8) 0.5x = 0.10x x = 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 or 24. Example continued:

23 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-23 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.

24 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-24 Mixture Problems Continued. Let x = number of liters of the 40% solution. Then 0.6 – x = number of liters of the 5% solution.

25 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-25 Mixture Problems SolutionLitersStrengthAmount of Phenobarbital 40% x x 5% x (0.6 - x) Mixture (0.20) Example continued: Continued. Amount of phenobarbital in 40% solution Amount of phenobarbital in 5% solution += Amount of phenobarbital in mixture

26 Copyright © 2015, 2011, 2007 Pearson Education, Inc. Chapter 3-26 Mixture Problems 0.40x (0.6 - x) = (0.6)(0.20) 0.40x x = x = x = 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: