1. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 9 Equations, Inequalities and Problem Solving

2. Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. 9.6 Percent and Mixture Problem Solving

3. Martin-Gay, Developmental Mathematics, 2e 33 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Strategy for Problem Solving General Strategy for Problem Solving 1.UNDERSTAND the problem. Read and reread the problem. Choose a variable to represent the unknown. Construct a drawing. Propose a solution and check. 2.TRANSLATE the problem into an equation. 3.SOLVE the equation. 4.INTERPRET the result: Check proposed solution in problem. State your conclusion.

4. Martin-Gay, Developmental Mathematics, 2e 44 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Percent Problem A percent problem has three different parts: 1. When we do not know the amount: n = 10% · 500 Any one of the three quantities may be unknown. amount = percent · base 2. When we do not know the base: 50 = 10% · n 3. When we do not know the percent: 50 = n · 500

5. Martin-Gay, Developmental Mathematics, 2e 55 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Percent Problem: Amount Unknown What is 9% of 65? 5.85 is 9% of 65 Example:

6. Martin-Gay, Developmental Mathematics, 2e 66 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Percent Problem: Base Unknown 36 is 6% of what? 36 is 6% of 600 Example:

7. Martin-Gay, Developmental Mathematics, 2e 77 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Percent Problem: Percent Unknown 24 is what percent of 144? Example:

8. Martin-Gay, Developmental Mathematics, 2e 88 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Markup Problems Mark is taking Peggy out to dinner. He has $66 to spend. If he wants to tip the server 20%, how much can he afford to spend on the meal? Let n = the cost of the meal. Cost of meal n+tip of 20% of the cost=$66 100% of n+20% of n=$66 120% of n=$66 Mark and Peggy can spend up to $55 on the meal itself. Example:

9. Martin-Gay, Developmental Mathematics, 2e 99 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Discount Problems Julie bought a leather sofa that was on sale for 35% off the original price of $1200. What was the discount? How much did Julie pay for the sofa? Julie paid $780 for the sofa. Discount = discount rate  list price = 35%  1200 = 420 The discount was $420. Amount paid = list price – discount = 1200 – 420 = 780 Example:

10. Martin-Gay, Developmental Mathematics, 2e 10 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Increase Problems The cost of a certain car increased from $16,000 last year to $17,280 this year. What was the percent of increase? Amount of increase = original amount – new amount The car’s cost increased by 8%. = 17,280 – 16,000 = 1280 Example:

11. Martin-Gay, Developmental Mathematics, 2e 11 Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall. Solving Decrease Problems Patrick weighed 285 pounds two years ago. After dieting, he reduced his weight to 171 pounds. What was the percent of decrease in his weight? Amount of decrease = original amount – new amount Patrick’s weight decreased by 40%. = 285 – 171 = 114 Example:

12. Martin-Gay, Developmental Mathematics, 2e 12 Solving Mixture Problems The owner of a candy store is mixing candy worth $6 per pound with candy worth $8 per pound. She wants to obtain 144 pounds of candy worth $7.50 per pound. How much of each type of candy should she use in the mixture? Let n = the number of pounds of candy costing $6 per pound. Let 144 – n = candy costing $8 per pound. Example: continued Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

13. Martin-Gay, Developmental Mathematics, 2e 13 continued Use a table to summarize the information. Number of Pounds Price per PoundValue of Candy $6 candy n66n6n $8 candy 144  n 8 8(144  n) $7.50 candy 1447.50144(7.50) 6n + 8(144  n) = 144(7.5) # of pounds of $6 candy # of pounds of $8 candy # of pounds of $7.50 candy continued Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

14. Martin-Gay, Developmental Mathematics, 2e 14 continued 6n + 8(144  n) = 144(7.5) 6n + 1152  8n = 1080 1152  2n = 1080  2n =  72 n = 36 She should use 36 pounds of the $6 per pound candy. She should use 108 pounds of the $8 per pound candy. (144  n) = 144  36 = 108 continued Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.

15. Martin-Gay, Developmental Mathematics, 2e 15 continued Check: Will using 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy yield 144 pounds of candy costing $7.50 per pound? State: She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy. 6(36) + 8(108) = 144(7.5) ? 216 + 864 = 1080 ? 1080 = 1080 ? Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall.