1. Percent and Mixture Problem Solving § 2.6

2. Strategy for Problem Solving General Strategy for Problem Solving • UNDERSTAND the problem. • Read and reread the problem. • Choose a variable to represent the unknown. • Construct a drawing, whenever possible. • Propose a solution and check. • TRANSLATE the problem into an equation. • SOLVE the equation. • INTERPRET the result. • Check the proposed solution in problem. • State your conclusion.

3. Solving a Percent Equation A percent problem has three different parts: amount = percent · base Any one of the three quantities may be unknown. 1. When we do not know the amount: n = 10% · 500 2. When we do not know the base: 50 = 10% ·n 3. When we do not know the percent: 50 = n· 500

4. Solving a Percent Equation: Amount Unknown n = 9% 65 · n = (0.09) (65) n = 5.85 amount = percent · base What is 9% of 65? 5.85 is 9% of 65

5. Solving a Percent Equation: Base Unknown 36 = 6% n · amount = percent ·base 36 is 6% of what? 36 = 0.06n 36 is 6% of 600

6. Solving a Percent Equation: Percent Unknown · 24 = n 144 amount = percent· base 24 is what percent of 144?

7. Solving Markup Problems Example: 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.

8. Solving Discount Problems Example: 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? Discount = discount rate  list price = 35%  1200 = 420 The discount was $420. Amount paid = list price – discount = 1200 – 420 = 780 Julie paid $780 for the sofa.

9. Solving Percent Increase Problems Example: 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 = 17,280 – 16,000 = 1280 The car’s cost increased by 8%.

10. Solving Percent Decrease Problems Example: 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 = 285 – 171 = 114 Patrick’s weight decreased by 40%.

11. Solving Mixture Problems Example: 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? 1.) UNDERSTAND Let n = the number of pounds of candy costing $6 per pound. Since the total needs to be 144 pounds, we can use 144  n for the candy costing $8 per pound. Continued

12. # of pounds of $6 candy # of pounds of $8 candy # of pounds of $7.50 candy Solving Mixture Problems Example continued 2.) TRANSLATE Use a table to summarize the information. 6n + 8(144  n) = 144(7.5) Continued

13. (144  n) = 144  36 = 108 Solving Mixture Problems Example continued 3.) SOLVE 6n + 8(144  n) = 144(7.5) Eliminate the parentheses. 6n + 1152  8n = 1080 Combine like terms. 1152  2n = 1080 2n = 72 Subtract 1152 from both sides. n = 36 Divide both sides by 2. She should use 36 pounds of the $6 per pound candy. She should use 108 pounds of the $8 per pound candy. Continued

14. ? 6(36) + 8(108) = 144(7.5) ? 216 + 864 = 1080 ? 1080 = 1080 Solving Mixture Problems Example continued 4.) INTERPRET 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. 