Simple and compound interesmmmmmmmt.pptx

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1. Learning Outcome(s): - At the end of the lesson, the learners are able to compute interest, maturity value and present value in simple interest environment and solve problems involving simple interest. Lesson Outline: Compute simple interest Compute maturity value Compute unknown principal, rate and time SIMPLE INTEREST

2. ANNUAL SIMPLE INTEREST I s = Prt where: I s = simple interest P = principal, or the amount invested or borrowed r = simple interest rate t = term or time in years

3. Example 1 : A bank offers 0.25% annual simple interest rate for a particular deposit. How much interest will be earned if 1 million pesos is deposited in this savings account for 1 year? Given: P = 1,000,000 r = 0.25% = 0.0025 t = 1 year Find: I s Solution: I s = Prt I s = (1,000,000)(0.0025)(1) I s = 2,500 Answer: The interest earned is P 2,500

4. Example 2 : How much interest is charged when ₱ 50,000 is borrowed for 9 months at an annual interest rate of 10%? Given: P = 50,000 r = 10% = 0.10 t = 9/12 = 0.75 year Find: I s Solution: I s = Prt I s = (50,000)(0.10)(9/12) I s = (50,000)(0.10)(0.75) I s = 3,750 Answer: The simple interest charged is ₱ 3,750

5. NOTE: when the term is expressed in months (M), it should be converted in years by: t =

6. Required Formulas Interest ( I ) I = Prt Final Amount (F) F = P + I F = P( 1 + rt) Principal (P) or P = F – I Time ( t ) Rate of interest ( r ) Required Formulas Interest ( I ) I = Prt Final Amount (F) F = P + I F = P( 1 + rt) Principal (P) Time ( t ) Rate of interest ( r ) Formulas in solving problems involving simple interest

7. Sophia borrowed Php 25,000 from a lending corporation that charges 12% interest with an agreement to pay the principal and the interest at the end of the term. If she paid Php 35, 500 at the end of the term, for how long did she use the money ? Example 3:

8. 1. For academic problems (like your example) Usually, we round off to 2 decimal places or express the time in years and months. Example: t=3.5 years → we can write as 3 years and 6 months (no rounding needed). If it was t=3.33 years → that’s 3 years and 4 months (since 0.33 × 12 ≈ 4 months). Take Note:

9. 2. For real-life financial situations If the lender charges per full year/month → you usually round up (because even a partial period may be charged as a whole). If exact days/months are considered → you keep the exact decimal or convert it properly into months/days. Take Note:

10. 3. For compound interest Time is often given in whole years, half-years, quarters, or months depending on the compounding period. If your answer is not exact, you usually round to the nearest compounding period. Take Note:

11. Julianne needs Php 57,300 to buy cooking equipment for her new house. She is willing to pay the interest of Php 7,735.50 if she borrowed the said amount from the bank. If she intends to pay her obligation within 30 months, what must be the interest rate of her loan? Example 4:

12. Maria borrowed ₱10,000 from a cooperative at a rate of 6% simple interest per year. She plans to pay it back after 3 years. How much interest will she pay? What is the total amount she has to pay after 3 years? Example 5:

13. Jun borrowed ₱15,000 from a lending company at 5% simple interest per year. If he pays it back after 4 years, how much interest does he pay and what is the total amount due? Example 6:

14. A fisherman from Zambales borrowed ₱8,000 from a cooperative to buy a new fishing net. The cooperative charges 10% simple interest per year. If he pays the loan after 18 months (1.5 years), how much will he pay in total? Example 7:

15. A farmer from Zambales borrowed ₱10,000 at 6% simple interest per year. He already paid ₱3,600 interest. How long did he keep the money before paying it back? Example 9:

16. Julianne borrowed ₱57,300 to buy cooking equipment. She paid an interest of ₱7,735.50 after 30 months (2.5 years). What was the interest rate of her loan? Example 10:

17. A fisherman borrowed ₱20,000 from a cooperative and paid ₱6,000 interest after 5 years. What was the simple interest rate charged? Example 10:

18. Complete the table below by finding the unknown. Prinicipal (P) Rate (r) Time (t) Interest (a) 2.5% 4 1,500 36,000 (b) 1.5 4,860 250,000 0.5% (c) 275 500,000 12.5% 10 (d)

19. Solution: (a) The unknown principal can be obtained by: P = P = P = 15,000 (b) The unknown rate can be computed by: r = r = r = 0.09 = 9%

20. (c) The unknown time can be calculated by: t = t = t = 0.22 years (d) The unknown simple interest is given by: I s = Prt I s = (500,000)(0.125)(10) I s = 625,000

21. Example 4 : When invested at an annual interest rate of 7%, the amount earned ₱ 11,200 of simple interest in two years. How much money was originally invested? Given: r = 7% = 0.07 t = 2 years I s = 11,200 Find: Amount invested or principal P Solution: P = P = P = 80,000 Answer: The amount invested is P 80,000

22. Example 5 : If an entrepreneur applies for a loan amounting to ₱ 500,000 in a bank, the simple interest of which is ₱ 157,500 for 3 years, what interest rate is being charged? Given: P = 500,000 I s = 157,500 t = 3 years Find: r Solution: r = r = r = 0.105 = 10.5%

23. Example 6 : How long will a principal earn an interest equal to half of it at 5% simple interest? Given: P r = 5% = 0.05 I s = 1/2P = 0.5P Find: t Solution: t = t = t = 10 years Answer: It will take 10 years for a principal to earn half of its value at 5% simple annual interest rate.

24. Maturity (Future) Value: F = P + I s where: F = maturity (future) value P = principal I s = simple interest Substituting I s by Prt gives F = P (1 + rt )

25. Maturity (Future) Value: F = P (1 + rt ) where: F = maturity (future) value P = principal r = interest rate t = term/time in years

26. Example 7 : Find the maturity value if 1 million pesos is deposited in a bank at an annual simple interest rate of 0.25% after (a) 1 year (b) 5 years ? Given: P = 1,000,000 r = 0.25% = 0.0025 Solution: (a) when t = 1, the simple interest is given by: Method 1: I s = Prt I s = (1,000,000)(0.0025)(1) I s = 2,500

27. The maturity or future value is given by: F = P+ F = (1,000,000)(2,500) F = 1,002,500 Method 2:To directly solve the future value F, F = P (1 + rt ) F = 1,000,000 (1 + (0.0025)(1)) F = 1,002,500 Answer: The future or maturity value after 1 year is P1,002,500

28. (b) when t = 5, Method 1: I s = Prt I s = (1,000,000)(0.0025)(5) I s = 12,500 F = P + I s F = 1,000,000 + 12,500 F = 1,012,500