Simple Interest S.Y.Tan.

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Presentation transcript:

Simple Interest S.Y.Tan

Interest (I)– a benefit in the form of a fee that lender received for letting borrower use of his money Origin date (O.D.) – the date on which the borrowed money is received by the borrower Maturity date (M.D.) or repayment date – the date on which the loan (borrowed money) is completely repaid. Term of the loan (t) – the length of time from the origin date to the maturity date S.Y.Tan

Simple Interest is dependent on 3 factors: SIMPLE INTEREST (I) - INTEREST that is computed based ONLY on the original amount of money received by the borrower on the origin date and is added to this amount on maturity date. Simple Interest is dependent on 3 factors: 1) principal (P) – the amount of money borrowed or sum received by borrower on O.D. (in currency) 2) simple interest rate (r) – agreed annual rate of interest (in percentage %) (divide by 100 to convert to decimal) S.Y.Tan

3) term of loan (t)– time in years Simple interest – a type of interest wherein only the original principal earns interest for the duration of the term Formula for simple interest I : S.Y.Tan

Maturity value or Final amount (F)– the amount of money received by the lender at the end of the term; the sum of the principal (P) and the simple interest (I) earned. S.Y.Tan

Principal P Maturity value F Origin date Maturity date Term S.Y.Tan

Ex 1. Find the amount if P800,000 is invested for 2 years at simple interest rate of 14.4% per year. What is the interest earned? P= 800,000 t = 2 years r = 14.4%= 0.144 F = ? I = ? F= 800,000 (1 + 0.144(2)) F = 1,030,400 I = F – P = 230,400 or I = Prt S.Y.Tan

Ex 2. In how much time will P28,000 amount to P29,134 at 16 Ex 2. In how much time will P28,000 amount to P29,134 at 16.2% simple interest rate? P= 28,000 F = 29,134 r = 16.2%= 0.162 t = ? F = P (1 + r t) 29134 = 28000 (1 + (0.162)t) t = 0.25 year = 3 months S.Y.Tan

Ex 3. At what simple interest rate will a sum double itself in 15 years? P F = 2P t = 15 years r = ? F = P (1 + r t) 2P = P (1 + r (15)) r = 0.0667 = 6.67% S.Y.Tan

Ex 4. What principal will amount to P16,856. 10 in 2 years at 10 Ex 4. What principal will amount to P16,856.10 in 2 years at 10.8% simple interest rate? F = 16856.10 t = 2 years r = 0.108 P = ? F = P (1 + r t) 16856.10 = P (1 + 0.108 (2)) P = 13,861.92 S.Y.Tan

If time t is given in months, then it has to be converted to years If time t is given in months, then it has to be converted to years. t = ( months)(1 yr /12months)= years Ex 5. Find the amount if P10,000 is invested for 10 months at 5.04% simple interest rate? P= 10000 t = 10 /12 = 5/6 yr r = 0.0504 F = P (1 + r t) F = 10,420 S.Y.Tan

If time t is given in days D, then it has to be converted to years and this leads to 2 types of interest. t = (D/360) years gives ordinary interest (default) t = (D/365) years gives exact interest Ex 6. Find the amount if P8,000 is invested for 250 days at 14% simple interest rate by using a. ordinary interest b. exact interest P= 8000 t = 250 days r = 0.14 a. b. S.Y.Tan

If term is given in terms of origin date and maturity date, then we get Actual time. That is, we count everyday within the term of the loan except the origin date. Jan, Mar, May, Jul, Aug, Oct, Dec – 31 days Apr, Jun, Sep, Nov – 30 days Feb – 28 days ; 29 days for leap year S.Y.Tan

Note that February has 29 days if it falls on a leap year and a leap year is a year divisible by 4. Dec Oct We can use our knuckles as guide in remembering the number of days for the different months of the year. May Mar Nov Sep Apr Jun Feb Jul Aug Jan Months that fall on knuckles have 31 days while months that fall in between knuckles have 30 days except February which has 28 days. ender S.Y.Tan

2 possible time factors: 1) t = (actual time/360) ordinary interest BANKER’S RULE (default) 2) t = (actual time/365) exact interest If the day of the dates (O.D. & M.D.) coincide with one another or are the same, then we count in MONTHS. S.Y.Tan

Ex 7. Find the amount of P10,000 due on December 15, 2015 if it was invested last March 15, 2015 at 4.03% simple interest rate? P = 10000 O.D. = 3/15/15 M.D.= 12/15/15 r = 0.0403 t = 9 months = 9/12 yr = 3/4 yr F = P (1 + r t) F = 10,302.25 S.Y.Tan

Ex 8. Find the maturity value of P18,000 if it was invested from Feb Ex 8. Find the maturity value of P18,000 if it was invested from Feb. 10, 2012 to Apr 16, 2013 at 15% simple interest rate using i) Banker’s rule ii) exact interest . P = 18000 O.D. = 2/10/12 M.D.= 4/16/13 r = 0.15 2012 is a leap year 2012 Feb (29-10) 19 Oct 31 Mar Nov 30 actual time Apr Dec 431 days May 2013 Jan Jun Feb 28 Jul Aug 16 Sep 198 233 S.Y.Tan

P = 18000 O.D. = 2/10/12 M.D.= 4/16/13 r = 0.15 2012 is a leap year t = 431 days = (431/360) yrs i) Using banker’s rule ii) Using exact interest: t = 431 days = (431/365) yrs S.Y.Tan

Formula for the maturity value F: F is a future value, received at the end of the term. In this context, we say that the principal P is the current or present value of an amount F that is due at some future date . S.Y.Tan

Ex 1. A 5-year investment had a maturity value of P27,500 Ex 1. A 5-year investment had a maturity value of P27,500. If the applied rate was 7.5% simple interest, what was its present value? F = 27500 t = 5 years r = 0.075 P = ? F = P (1 + r t) 27500 = P (1 + 0.075 (5)) S.Y.Tan

Ex 2. At 14% simple interest, find the present value of P9112 Ex 2. At 14% simple interest, find the present value of P9112.50 due in 30 months. F = 9112.50 t = 30/12 yrs r = 0.14 P = ? F = P (1 + r t) 9112.50 = P (1 + 0.14 (30/12)) S.Y.Tan

Ex 3. To pay a debt, Pong offered Bert P1000 now or P1100 three months from now. If saving account interest is 10%, what offer will give greater return for Bert? Values of money can only be compared if they are on the same date. F = 1100 t = 3/12 yr r = 0.10 (value of P1100 now) Option 1 which gives P1000 now is less than P1073.17 which is current value of Option 2. So 2nd offer is better. S.Y.Tan

Ex 4. Find the present value of P100,000, which is an amount due in 200 days, if money's worth is 10.5% simple interest. F = 100,000 t = 200/360 yrs r = 0.105 P = ? F = P (1 + r t) 100,000 = P (1 + 0.105 (200/360)) S.Y.Tan

Ex 5. Susan lends P50,000 to Jane on October 1, 2014 Ex 5. Susan lends P50,000 to Jane on October 1, 2014. She expects Jane to pay the principal and simple interest at 9% to fully settle the debt on March 28, 2015. What amount does Susan receive? 2014 Oct (31-1) 30 Nov Dec 31 2015 Jan Feb 28 Mar 178 S.Y.Tan

Ex 6. Accumulate Php85,000 for 20 months at a simple interest rate of 12%. (Note: To accumulate an amount means to find its maturity value.) S.Y.Tan

Ex 7. At what simple interest rate will P415,000 be the present value of P500,000 for a three years and 4 months transaction? P= 415,000 F = 500,000 r = ? S.Y.Tan

Ex 8. When will Php42,000 earn simple interest of P8000 if it is invested at 11.4%? P= 42,000 I = 8,000 r = 0.114 t = ? S.Y.Tan