1.1 Simple Interest

Simple Interest

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1.1 Simple Interest

• 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

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1.1 Simple Interest

• 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)

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1.1 Simple Interest

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 :PrtI

years)(in term- t

form) decimal(in year per rate - r

currency)any (in principal - P

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1.1 Simple Interest

• 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.

IPF

tPrPF

)rt1(PF

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1.1 Simple Interest

Origin date Maturity date

Term

Principal P Maturity value F

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1.1 Simple Interest

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

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1.1 Simple Interest

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

tr

1P

F

t162.0

128000

29134

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1.1 Simple Interest

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%

rt

1P

F

t15

1P

P2

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1.1 Simple Interest

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

P)rt1(

F

P))2(108.01(

10.16856

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1.1 Simple Interest

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

)))(0504.0(1(10000F1210

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1.1 Simple Interest

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.

78.8777)))(14.0(1(8000F360250

12.8767)))(14.0(1(8000F365250

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1.1 Simple Interest

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

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1.1 Simple Interest

Note that February 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 February

which has 28 days.

Jan

MarMay

Jul

Aug

OctDec

Feb

Apr Jun

Nov

Sep

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1.1 Simple Interest

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.

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1.1 Simple Interest

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

)))(0403.0(1(10000F129

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1.1 Simple Interest

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 year2012 Feb (29-10) 19 Oct 31

Mar 31 Nov 30 actual time

Apr 30 Dec 31 431 days

May 31 2013 Jan 31

Jun 30 Feb 28

Jul 31 Mar 31

Aug 31 Apr 16

Sep 30 198

233

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1.1 Simple Interest

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

431360

F 18000 (1 (0.15) ) 21,232.50

431365

F 18000 (1 (0.15) ) 21188.21918 21,188.22

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1.1 Simple Interest

• 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 .

IPF )rt1(PF or

)rt1(

FP

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1.1 Simple Interest

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))

rt1

FP

)5)(075(.1

500,27

000,20P

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1.1 Simple Interest

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))

rt1

FP

))(14(.1

50.9112

1230

6750P

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1.1 Simple Interest

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.

17.1073))(10(.1

1100P

123

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1.1 Simple Interest

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))

P F

1 rt

100,000

1 (.105) 200360

4,488.199P

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1.1 Simple Interest

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?

)rt1(PF

360178)09(.1000,50F

225,52F

2014 Oct (31-1) 30

Nov 30

Dec 31

2015 Jan 31

Feb 28

Mar 28

178

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1.1 Simple Interest

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.)

)rt1(PF

1220)12(.1000,85F

000,102F

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1.1 Simple Interest

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 = ?

years33t3

1031

124

))(000,415(

000,415000,500

310

.14%60614.0r

Pt

I

Pt

PFr

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1.1 Simple Interest

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 = ?

Pr

It

)114)(.000,42(

8000

months 8 andyear 1years 67.1t

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