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

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

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

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

Let n = the cost of the meal. 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? I Introduction Let n = the cost of the meal. .2n = tip, 20% of the cost 66 = total amount spent

II Body Cost of meal n + Tip, 20% of the cost = $66 n + 20% of n = $66 n + 0.2n = 66 III Conclusion Mark and Peggy can spend up to $55 on the meal itself.

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.

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 = new amount – original amount = 17,280 – 16,000 = 1280 The car’s cost increased by 8%.

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%.

I Introduction: $2.35 II Body:

II Body: III Conclusion:

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? I Introduction 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

Use a table to summarize the information. II Body Use a table to summarize the information. Number of Pounds Price per Pound Value of Candy $6 candy n 6 6n $8 candy 144  n 8 8(144  n) $7.50 candy 144 7.50 144(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

Eliminate the parentheses. II Body 6n + 8(144  n) = 144(7.5) 6n + 1152  8n = 1080 Eliminate the parentheses. 1152  2n = 1080 Combine like terms. 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. (144  n) = 144  36 = 108

III Conclusion She should use 36 pounds of the $6 per pound candy and 108 pounds of the $8 per pound candy.

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