1. § 2.6 Percent and Mixture Problem Solving

2. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 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, whenever possible Propose a solution and check 2)TRANSLATE the problem into an equation 3)SOLVE the equation 4)INTERPRET the result Check the proposed solution in problem State your conclusion

3. Martin-Gay, Beginning and Intermediate Algebra, 4ed 33 Solving a Percent Equation 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

4. Martin-Gay, Beginning and Intermediate Algebra, 4ed 44 Solving a Percent Equation: Amount Unknown amount = percent · base What is 9% of 65? n=9% · 65 n=(0.09)(65) n=5.85 5.85 is 9% of 65

5. Martin-Gay, Beginning and Intermediate Algebra, 4ed 55 Solving a Percent Equation : Base Unknown amount = percent · base 36 is 6% of what? n =6% · 36 36 = 0.06n 36 is 6% of 600

6. Martin-Gay, Beginning and Intermediate Algebra, 4ed 66 Solving a Percent Equation : Percent Unknown amount = percent · base n =144 · 24 24 is what percent of 144?

7. Martin-Gay, Beginning and Intermediate Algebra, 4ed 77 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:

8. Martin-Gay, Beginning and Intermediate Algebra, 4ed 88 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:

9. Martin-Gay, Beginning and Intermediate Algebra, 4ed 99 Solving Percent 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:

10. Martin-Gay, Beginning and Intermediate Algebra, 4ed 10 Solving Percent 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:

11. Martin-Gay, Beginning and Intermediate Algebra, 4ed 11 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? 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. Example: Continued

12. Martin-Gay, Beginning and Intermediate Algebra, 4ed 12 Solving Mixture Problems Example continued 2.) TRANSLATE Continued Use a table to summarize the information. Number of PoundsPrice 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

13. Martin-Gay, Beginning and Intermediate Algebra, 4ed 13 Solving Mixture Problems Example continued Continued 3.) SOLVE 6n + 8(144  n) = 144(7.5) 6n + 1152  8n = 1080 1152  2n = 1080  2n =  72 Eliminate the parentheses. Combine like terms. 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. (144  n) = 144  36 = 108

14. Martin-Gay, Beginning and Intermediate Algebra, 4ed 14 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. 6(36) + 8(108) = 144(7.5) ? 216 + 864 = 1080 ? 1080 = 1080 ?