General.Mathematics_Simple_Interest.pptx

Simple Interest and Compound Interest

1. Simple Interest General Mathematics O. Oronce

2. Lesson Objectives At the end of the lesson, the students must be able to: define the simple interest; illustrate simple interest; compute interest, maturity value and to be able to identify partial payments in simple interest environment; and solve problems involving simple interest.

3. Converting Percent to Decimal To convert a percent to decimal, drop the percent sign and move the decimal point two places to the left. When the number in the percent is a whole number, the decimal point is understood to be stated at the right of the last digit. For example, to convert 12% to decimal, drop the percent sign and move the decimal point two places to the left to get 0.12.

4. Simple Interest Simple interest is charged only on the loan amount called the principa l. Thus, interest on the interest previously earned is not included. Simple interest is calculated by multiplying the principal by the rate of interest by the number of payment periods in a year.

5. Simple Interest Formula I = Prt (a) (b) (c) where I = interest, P = principal, r = rate of interest, and t = time or term in years or fraction of a year To find the maturity value , simply add interest to the principal.

6. Maturity Value or (Amount or Balance) Formula A = P + I or A = P + Prt or A = P(I + rt ) A = Maturity value P = Principal I = Interest

7. Sample 1 To buy the school supplies for the coming school year, you get a summer job at a resort. Suppose you save ₱4 200.00 of your salary and deposit it into an account that earns simple interest. After 9 months, the balance is ₱4 263.00. What is the annual interest rate?

8. Solution 1 to Sample 1 Use the formula where P= ₱4 200.00, t = 9 months or or year and, I = ₱4 263.00 – ₱4 200.00 = ₱63.00. r= ____ 63 ____=0.02 or 2% The annual interest rate is 2% or

9. Solution 2 to Sample 1 Use the formula A = P(1 + rt ) where A = ₱4 263.00, P = ₱4 200.00, and t = 3 4 year. Solve for r after substituting values for A, P, and t. 4 263 = 4 200 4 263 = 4 200 + 3 150r 63 = 3 150r 0.02 = r The annual interest rate is 2%

10. Ordinary Interest or Banker's Interest – interest based on a 360-day year. Exact Interest – interest based on a 365-day year Sample 2 You get a 180-day ₱200 000.00 loan from a bank at a 10.5% interest. Calculate interest using (a) 360-day and (b) 365-day year.

11. Solution to Sample 2 a. 360-day year: I = Prt where, P = ₱200 000.00, r = 10.5%, and or I = 200 000 × 0.105 × = ₱10 500.00 b. 365-day year: I = Prt = 200 000 × 0.105 x = ₱10 356.16 Note: A 360-day year is favorable for the lender while a 365-day year is favorable for the borrower

12. Partial Payments Calculate interest on principal from date of loan to date of first payment. Remainder of the payment = Amount paid – Interest Portion New Balance = Previous Balance – Principals remainder (portion) of payments

13. Partial Payments 4. In cases of more than one partial payment, calculate interest on new balance from date of previous payment to date of next payment. Perform steps 2 and 3. 5. At maturity, calculate interest on last partial payment. Add this to the new balance to compute the total final payment due.

14. Sample 3 A loan of ₱200,000.00 was made from a bank that charges 9% interest rate and should be repaid after 90 days. If payment of ₱80,000.00 was made after 20 days and the balance on the 90th day, calculate the amount of interest, principal paid for each payment, and the total amount paid .

15. Solution to Sample 3 Payment on Day 20: P = ₱200,000.00, r = 9%, and t = I = Prt = ₱200,000.00 × 0.09 x = ₱1,000.00 Principal's Remainder = ₱80,000.00 – ₱1,000.00 = ₱79,000.00 New Balance = ₱200,000.00 – ₱79,000.00 = ₱121,000.0

16. Solution to Sample 3 Payment on Day 60: P = ₱121 000, r = 9%, and t= (60 days – 20 days = 40 days) I = Prt = ₱121,000.00 × 0.09 x = ₱1,210.00 Total Final Payment Due = ₱1,210.00 + ₱121 000.00 = ₱122 210.00 Principal's Remainder = ₱122,210 – ₱1,210 = – ₱121,000.00 New Balance = ₱121,000.00 – ₱121,000.00 = ₱0.00

17. Note With partial payment I = 986.30 + 1,193.30 I = ₱2,179.60 Without partial payment: I = Prt = 200,000 × 0.09 x I = ₱2,958.90

18. Exercise A Ramil deposited ₱20,000.00 at 4% simple interest for 5 years. At the end of 5 years, his account contains ₱24,000.00. Give the term for each value in relation to the problem. ₱20,000.00 ₱24,000.00 4% 5 year

19. Exercise B Principal (P) Interest Rate (r) Time (t) Maturity Value (A) ₱ 35,600 6% 9 mo. ₱ 140,350 10% 15 mo. ₱ 75,800 8 ½% 2 yr. ₱ 340,200 11% 6 yr. ₱ 1,400,500 9 % 10 yr. Complete the table by finding the maturity value.