Simple Interest. S.Y.Tan. 1.1 Simple Interest

Transcription

1 Simple Interest

2 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

3 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)

4 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 ormula for simple interest I : I Prt P - principal (in any currency) r t - - rate per year (in decimal form) term(in years)

5 Maturity value or inal amount () 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. P I P Prt P(1 rt)

6 Principal P Maturity value Origin date Maturity date Term

7 Ex 1. ind 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%= =? I =? = 800,000 ( (2)) = 1,030,400 I = P = 230,400 or I = Prt

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

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

10 Ex 4. What principal will amount to P16, in 2 years at 10.8% simple interest rate? = t = 2 years r = P =? = P (1 + r t) = P ( (2)) (1 rt) P ( (2)) P P = 13,861.92

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

12 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. ind 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 (1 (0.14)( 360)) (1 (0.14)( 365))

13 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 eb 28 days ; 29 days for leap year

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

15 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.

16 Ex 7. ind 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 = O.D. = 3/15/15 M.D.= 12/15/15 r = t = 9 months = 9/12 yr = 3/4 yr = P (1 + r t) (1 (0.0403)( 12)) = 10,302.25

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

18 P = O.D. = 2/10/12 M.D.= 4/16/13 r = is a leap year t = 431 days = (431/360) yrs i) Using banker s rule (1 (0.15) ) 21, ii) Using exact interest: t = 431 days = (431/365) yrs (1 (0.15) ) ,

19 ormula for the maturity value : P I or P(1 rt) 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 that is due at some future date. P (1 rt)

20 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? = t = 5 years r = P =? = P (1 + r t) = P ( (5)) P 1 rt P 20,000 27,500 1 (.075)(5)

21 Ex 2. At 14% simple interest, find the present value of P due in 30 months. = t = 30/12 yrs r = 0.14 P =? = P (1 + r t) = P ( (30/12)) P 1 rt P (.14)( )

22 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. = 1100 t = 3/12 yr r = 0.10 P (.10)( 3 12 ) (value of P1100 now) Option 1 which gives P1000 now is less than P which is current value of Option 2. So 2 nd offer is better.

23 Ex 4. ind the present value of P100,000, which is an amount due in 200 days, if money's worth is 10.5% simple interest. = 100,000 t = 200/360 yrs r = P =? = P (1 + r t) 100,000 = P ( (200/360)) P 100,000 1 rt 1 (.105) P 94,488.19

24 Ex 5. Susan lends P50,000 to Jane on October 1, She expects Jane to pay the principal and simple interest at 9% to fully settle the debt on March 28, What amount does Susan receive? P(1 rt) 50,0001 (.09) 52, Oct (31-1) 30 Nov 30 Dec Jan 31 eb 28 Mar

25 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.) P(1 rt) 85,0001 (.12) ,000

26 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? 4 t years P= 415,000 = 500,000 r =? r r P Pt I Pt % 500, , (415,000)( ) 3

27 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 = t =? t I Pr 8000 (42,000)(.114) t 1.67 years 1 year and 8 months