2. 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

3. 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

4. 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

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

6. 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

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

8. Example:
Julie bought a leather sofa that was on sale for 35% off the original price of $ 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. 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%.

10. 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%.

12. I Introduction:
$2.35
II Body:

13. II Body:
III Conclusion:

14. 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

15. 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

16. Eliminate the parentheses.
II Body
6n + 8(144  n) = 144(7.5)
6n  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

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

18. Example

19. Example

20. Example

21. Example