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6-6. Simple Interest. Course 2. Warm Up. Problem of the Day. Lesson Presentation. 6-6. Simple Interest. Course 2. Warm Up Multiply. 1. 800  0.04  3 2. 1,700  2  0.25 3. 900  0.05  4 4. 1,200  6  0.35. 96. 850. 180. 2,520. 6-6. Simple Interest. 75. 3

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6-6

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  1. 6-6 Simple Interest Course 2 Warm Up Problem of the Day Lesson Presentation
  2. 6-6 Simple Interest Course 2 Warm Up Multiply. 1.800  0.04  3 2. 1,700  2  0.25 3. 900  0.05  4 4. 1,200  6  0.35 96 850 180 2,520
  3. 6-6 Simple Interest 75 3 4 3 40 = ; the others equal . 1,000 Course 2 Problem of the Day Which number does not belong in the group? 0.75 75% 3 4 6 8 75 1,000
  4. 6-6 Simple Interest Course 2 Learn to solve problems involving simple interest.
  5. 6-6 Simple Interest Course 2 Insert Lesson Title Here Vocabulary interest simple interest principal
  6. 6-6 Simple Interest Course 2 When you keep money in a savings account, your money earns interest. Interest is an amount that is collected or paid for the use of money. For example, the bank pays you interest to use your money to conduct its business. Likewise, when you borrow money from the bank, the bank collects interest on its loan to you.
  7. 6-6 Simple Interest Course 2 One type of interest, called simple interest, is money paid only on the principal. The principal is the amount of money deposited or borrowed. To solve problems involving simple interest, you can use the following formula. Rate of interest per year (as a decimal) Interest I= P · r · t Time in years that the money earns interest Principal
  8. 6-6 Simple Interest Course 2 Additional Example 1A: Using the Simple Interest Formula Find the missing value. A. I = , P =$575, r =8%, t = 3 years I =P · r · t I =575 · 0.08 · 3 Substitute. Use 0.08 for 8%. I = $138 Multiply. The simple interest is $138.00.
  9. 6-6 Simple Interest 10,200r 10,200 204 10,200 = Course 2 Additional Example 1B: Using the Simple Interest Formula Find the missing value. B. I = $204, P = $1,700, r = , t = 6 years I =P · r · t 204= 1,700 · r · 6 Substitute. 204 = 10,200r Multiply. Divide by 10,200 to isolate the variable. 0.02 = r The interest rate is 2%
  10. 6-6 Simple Interest Course 2 Try This: Example 1A Find the missing value. A. I = , P =$525, r =7%, t = 2 years I =P · r · t I =525 · 0.07 · 2 Substitute. Use 0.07 for 7%. I = $73.50 Multiply. The simple interest is $73.50.
  11. 6-6 Simple Interest 6,000r 6,000 600 6000 = Course 2 Try This: Example 1B Find the missing value. B. I = $600, P = $2,000, r = , t = 3 years I =P · r · t 600= 2,000 · r · 3 Substitute. 600 = 6,000r Multiply. Divide by 10,200 to isolate the variable. 0.1 = r The interest rate is 10%
  12. 6-6 Simple Interest 1 Understand the Problem Course 2 Insert Lesson Title Here Additional Example 2: Problem Solving Application Avery deposits $6,000 in an account that earns 4% simple interest. How long will it take for his account balance to reach $6,800? Rewrite the question as a statement: • Find the number of years it will take for Avery’s account to reach $6,800. List the important information: •The principal is $6,000. • The interest rate is 4%. • His account balance will be $6,800.
  13. 6-6 Simple Interest Make a Plan 2 Course 2 Insert Lesson Title Here Additional Example 2: Problem Solving Application Avery deposits $6,000 in an account that earns 4% simple interest. How long will it take for his account balance to reach $6,800? Avery’s account balance A includes the principal plus the interest: A = P + I. Once you solve for I, you can use I = P · r · t to find the time.
  14. 6-6 Simple Interest 3 Solve –6,000 – 6,000 800 240 240t 240 = Course 2 Additional Example 2:Problem Solving Application A = P + I Substitute. 6,800 = 6,000 + I Subtract to isolate the variable. 800 = I I = P · r · t 800 = 6,000 · 0.04 · t Substitute. Use 0.04 for 4%. Multiply. 800 = 240t Divide to isolate the variable. 3.33 t Round to the nearest hundredth. 1 3 It will take 3 years.
  15. 6-6 Simple Interest 4 Course 2 Additional Example 2:Problem Solving Application Look Back 1 3 After exactly 3 years, Avery’s money will have earned $800 in simple interest and his account balance will be $6,800. 1 3 = 800 I = 6,000 · 0.04 · 3 1 3 So it will take 3 years to reach $6,800.
  16. 6-6 Simple Interest 1 Understand the Problem Course 2 Insert Lesson Title Here Try This: Example 2 Linda deposits $10,000 in an account that earns 8% simple interest. How long will it take for the total amount in her account to reach $12,000? Rewrite the question as a statement: • Find the number of years it will take for Linda’s account to reach $12,000. List the important information: •The principal is $10,000. • The interest rate is 8%. • Her account balance will be $12,000.
  17. 6-6 Simple Interest Make a Plan 2 Course 2 Insert Lesson Title Here Try This: Example 2 Linda deposits $10,000 in an account that earns 8% simple interest. How long will it take for the total amount in her account to reach $12,000? Linda’s account balance A includes the principal plus the interest: A = P + I. Once you solve for I, you can use I = P · r · t to find the time.
  18. 6-6 Simple Interest 3 Solve –10,000 –10,000 2,000 800 800t 800 = Course 2 Try This: Example 2 A = P + I Substitute. 12,000 = 10,000 + I Subtract to isolate the variable. 2,000 = I I = P · r · t 2,000 = 10,000 · 0.08 · t Substitute. Use 0.08 for 8%. Multiply. 2,000 = 800t Divide to isolate the variable. 2.5 = t 1 2 It will take 2 years.
  19. 6-6 Simple Interest 4 Course 2 Try This: Example 2 Look Back 1 2 After exactly 2 years, Linda’s money will have earned $2,000 in simple interest and her account balance will be $12,000. 1 2 = 12,000 I = 10,000 · 0.08 · 2 1 2 So it will take 2 years to reach $12,000.
  20. 6-6 Simple Interest Course 2 Insert Lesson Title Here Lesson Quiz Find each missing value. 1.I = 2.I = $18, P = $150, r = 3.I = $640, P = 4.I = $120, P = $600, r = 5% t = , P = $800, r = 10%, t = 3 years $240 6% , t = 2 years $2,000 , r = 4%, t = 8 years 4 years 5. Dennis deposits $6,000 in an account that earns 5.5% simple interest. How long will it take before the total amount is $8,000? a little over 6 years
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