Chapter 1 - site.iugaza.edu.pssite.iugaza.edu.ps/nbarakat/files/2010/02/Chapter-11.pdf · The...

Mathematics of FinanceIslamic University of Gaza Dr. Nafez M. Barakat

1

Chapter 1

Simple Interest


2

Simple interest is defined as the product of principal, rate, and time.

This definition leads to the simple interest formula.

I = P. r. t

I : simple interest in (dollars) or other monetary unit)

P: principal in dollars

r: interest rate

t: time in units that correspond to the rate

table 1

No. Month days

1 January 31

2 February 28/29

3 March 31

4 April 30

5 May 31

6 June 30

7 July 31

8 August 31

9 September 30

10 October 31

11 November 30

12 December 31

Example (1)

A bank pays 8% per annum on saving accounts. A

person opens an account with a deposit of $300 on

January 1. how much interest will the person receive

on April 1.


3

Solution : January 1

February

March

April 1

Time 3 months

I = P. r. t

0.612

308.0300 I $

Example (2) a couple buys a home and gets a loan for

$ 50000. the annual interest rate is 12%. The term of

the loan is 30 years, and the monthly payments is

$514.31. find the interest for the first month and the

amount of the house purchased with the first

payment.

Solution :

Substituting p= 50000, r= 0.12, and t= 1/12 in the

next formula

I = P. r. t

0.50012

112.0500000 I $

So the 514.31 payment buys only 514.31-500.0 = $14.31

worth of house

Example (3) the interest paid on a loan of $500.0 for 2

months was $ 12.5. what was the interest rate.


4

Solution :

Substituting p= 500, I= 12.50, and t= 2/12 in the

next formula

I = P. r. t

12

25005.12 r $

r = 15.0%

Example (4)

A person gets $ 63.75 every 6 months from an

investment that pays 6% interest. How much money is

invested?

Solution:

Substituting r= 0.06, I= 63.75, and t= 6/12 in the

next formula

I = P. r. t

12

606.075.63 P

P= $2125.0

Example (5)

How long will it take $ 5000 to earn $ 50 interest at 6%.

Solution:

Substituting r= 0.06, I= $50, P= $5000 , and t= ? in

the next formula

I = P. r. t


5

t 06.050000.50

t= 1/6 years or 2 months

example (6)

a woman borrows $ 2000 from a credit union. Each month she is

to pay $ 100 on the principal. She also pays interest at rate of

1% a month on the unpaid balance at the beginning of the

month. Find the total interest.

Solution : note that the rate is monthly rate.

The first month's interest is

0.20101.02000 I

The total payment for the first month is $120. and the new

unpaid balance is $ 1900.

For the second month the interest is

0.19101.01900 I

After 19 payments the debt is down to $100, and the interest

payment is

0.1101.0100 I

The interest payments are $20, $19, $18, …, $1.0

Total interest = 20 + 19 + 18 + … + 1

According to arithmetic progression

Sum = 0.210)10.20(2

20)(

21 naa

n$


6

Amount :

The sum of the principal and the interest is called the amount,

designated by the symbols S.

S = P + I

= P + P. r . t = P ( 1 + r . t )

Example

A man borrows $ 350 for 6 months at 15%. What a

mount must he repay?

Solution :

Substituting r= 0.15, I= ?, P= $350 , and t= 6/12 in

the next formula

I = P. r. t

25.2612/615.0350 I

S == P + I = 350.0 + 26.25 = $376.25

Another solution:

S = P ( 1 + r . t ) = 350 ( 1 + 0.15 * 6/12) = $376.25


7

Example 1 page 21: The current annual dividend rate )معدل الرباد ( of a savings and

loan association is 5.5%. dividend are credited to a persons

account on June 30 and December 31. Money put in the 10th

of

the month earns dividends for the entire month. If money is put

in after the 10th

. it starts earning dividends the following

month. A person opens an account on January 7 with 450$. On

February 25, $300 is added, and on June 10, $ 240 is placed in

the association. What is the amount in the account on June 30?

Solution:

S = 400 (1+0.055*6/12) + 300 (1+0.055*4/12)

+ 240 ( 1+.005*1/*12) = $957

Exercise 1 a: page(25)

1, 5,7, 8, 9, 10, 11,12, 13, 14, 15,16, 17, 18, 19 , 20,21, 22


8

Exact and ordinary interest

When the time is in days and the rate is an annual rate , it is

necessary to convert the dayes to a fractional part of a year when

substituting in the simple interest formulas

Interest computed using a divisor 360 is called ordinary

interest.

Interest computed using a divisor 365 or 366 is called

exact interest.

Example: figure the ordinary and exact interest on a 60 days

loan of $ 300 if the rate is 15%.

Solution:

Substituting r= 0.15, I= ?, P= $300 , and t= 60.0 days

in the next formula

I = P. r. t

Ordinary inertest 50.7360/6015.0300 I

Exact inertest 40.7365/6015.0300 I

Note that the ordinary interest is greater than exact intertest.

Exact and approximate time:

Exact time includes all days except the first


9

Approximate time based on the assumption that all the full

months contain 30 days

Example:

Find the exact and approximate time between 5 March and 28

September .

Exact time :

3 5- March 31- 5 = 26

4 April 30

5 May 31

6 June 30

7 July 31

8 August 31

9 28 -September 28

Total 207 days

Approximate time : we count the number of months from 5

March to 5 September which is equal 6 months, and equal 6 *

30 = 180 days, and we add the 23 days from September 5 to

September 28, so the total approximate time equal 203 days.

Commercial Practice)الممارسات التجارية(

There are four ways to compute simple interest:

1- Ordinary interest and exact time ( BANKERS Rule's)


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2- Exact interest and exact time

3- Ordinary interest and approximate

4- Exact interest and approximate time

Example:

On November 15 , 1993, a woman borrowed $500 at 15 %. The

debt is repaid on February20, 1994. find the simple interest

using the four methods.

Solution :

First we get the exact and the approximate time:

Exact time

Month days 15 -November 30-15=15

December 31

January 1994 31

20- February 20

TOTAL 97 DAYS

Approximate time :

From 15 November to 15 February

there is three months which is equal 90 days.

And from 15 to 20 February there is 5 days

Total time ………………………….. 95 days

1. Ordinary interest and exact time ( BANKERS Rule's)

I = P. r. t

21.20360

9715.0500 I

2. Exact interest and exact time


11

93.19365

9715.0500 I

3.Ordinary interest and approximate

79.19360

9515.0500 I

4.Exact interest and approximate time

79.19365

9515.0500 I

Example :

The builder of an apartment building obtained an $800000

construction loan at an annual rate of 15%. The money was

advanced as follows:

March 1, 1994 $ 300000

June 1, 1994 $ 200000

October 1, 1994 $ 200000

December 1, 1994 $ 100000

The building was completed in February of 1995 and the loan

repaid on march 1, 1995. find the amount using ordinary interest

and approximate time.

Solution :

Time Month days

1

yea r 1- March 1944 31

April1994 30


12

May1994 31

9 m

on

ths

1-June 1944 30

July1994 31

August1994 31

September1994 30

5 m

on

ths

1- October 1994 31

Novembe1994r 30

3 m

on

ths 1- December 1994 31

January 1955 30

February 1955 28

1- March 1995 30

Interest of each part of the loan is

45000115.0300000 I

2250012

915.0200000 I

12500012

515.0200000 I

375012

315.0100000 I

Total interest = $ 83750

Amount of loan = 800000 + 83750 = $883750

Exercise 1b (q12. page 34)

On may 4,1991, a person borrows $1850 and promises to repay

the debits in 120 day’s with interest at 12%. If the loan is not

paid on time the contract requires the borrower to pay 10% on

the unpaid amount for the time after the due date. Determine

how much this person must pay to settle the debt on December

15, 1991.

Solution :

No. Month days


13

5 4-May 31-4=27

6 June 30

7 July 31

8 August 31

9 September 30

10 October 31

11 November 30

12 15-December 15

Total 225

Time at second period 225-120=105

days

S1 = P(1 + r . t ) = 1850 ( 1 + 0.12 * 120/360) = $1924

S2 = P(1 + r . t ) = 1924 ( 1 + 0.10 * 105/360) = $1980

H.W: 1-14 page 34

Present value at simple interest

If we know the amount and we want to obtain the principal,

we solve the formula for P.

trSP

1

Example :

If money is worth 5 % , what is the present value of %105

due in 1 year?

Solution:


14

Substituting r = 0.05, S= 105 , P= ? , and t= 1 year in

the next formula

0.100105.01

1051

trSP

Example:

A person can buy a piece of property for $5000

cash or $54000 in a year. the prospective buyer has

cash and invest it in at 7%. Which method of

payment is better and by how much now?

Solution :

73.5046107.01

54001

trSP

This mean that the buyer would have to invest $5046.73 now

at 7% to have $5400 in a year.

So by paying cash the buyer save $46.73 now.

If another rate of return on the money was available, the

decision might be different. For example:

Rate of

return

Present value

of $5400 due

in 1 year

Better plan

7% $5046.73 Save $ 46.73 now by paying cash

8% $5000.00 Planes are equivalent

9% $4954.13 Save $45.87 now by paying $5400

Exercise 1c: 1-10 page 38


15

Present value of interest –bearing debt

Example : if we want to find the current value of an interest-

bearing debt in the future, we must find the maturity

value of the debt, using the stated interest rate for the

term of the ;loan. Then we compute the present value

of this maturity value for the time between the day it

is discounted and the due date.

Example :

A debtor signs a note for $2000 due in 6 months with

interest at 9%. One month after the debt is

contracted, the holder of the note sells it to a thirty

party, who determine the present value at 12%. How

much is received for the note?

Solution :


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Step 1: the maturity value is obtained at 9%

S = P ( 1 + r . t ) = 2000 ( 1 + 0.09 * 6/12) = $2090

Step 2: the maturity value is discounted for 5 months at 12%

48.1990)12/5(12.01

20901

trSP

Equations of value

There are two ways to move the money backward and

forward, lock at any time diagram. If a sum is to be moved

forward use an amount formula, and if backward use a

present value formula.

Note : always bring obligations to the same point using the

specified rate before combining them. This common point is

called a focal date.

Example :


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A person owes $200 due in 6 months and $300 due in 1 year.

The creditor will accept a cash settlement of both debts

using a simple interest rate of 18% and putting the focal

date now. Determine the size of the cash settlement.

Solution :

We set the equation of value :

73.437$24.25449.183

118.01300

12

618.01

200

x

Example 2:

Solve the preceding problem using 12 months hence as the

focal date.



18

98.438$18.1

518

51830021818.1

300)12

618.01(200)118.01(

x

x

x

Example 3:

A person owes $ 1000 due in 1 year with interest at 14% .

two equal payments in 3 and 9 months, respectively, will be

used to discharge this obligation. What will be the size of

these payments if the person and the creditor agree to use an

interest rate of 14% and a focal date 1 year hence.

Solution :



19

71.532$14.2

1140

14.21140$

035.1105.11140$

)12

314.01()

12

914.01()114.01(1000

x

x

xx

xx

Example 4:

A person borrows $ 2000 at 15% interest on June 1,

1996. the debt will be repaid with two equal payments,

one on December 1 , 1996 and the other on June 1, 1997.

put the focal date on June 1, 1996 and find the size of the

payments



21

24.1111$

200079979.1

2000869565.0930233.0

200015.1075.1

2000115.01

12

615.01

x

x

x

xx

xx

Example 5:

Work example 4 with the focal date on June 1, 1997.

Solution :

The equation of value is


21

43.1108$075.2

2300

2300075.2

2300075.1

)115.01(2000)2

115.01(

x

x

xx

xx

Example 6: A person borrowed $ 6000 on September 15, 1992

agreeing to pay $2000 on January 15, 1993, and $ 2000

on may 15, 1993. if the interest rate was 18% , how

much paid on September 15, 1993 to settle the debt ( put

the focal date September 15, 1993.

Solution :

The equation of value is

2720$212022407080

212022407080

)3

118.01(2000)

3

218.01(2000)118.01(6000

x

x

x

Exercise 1 d: 1-10


22

Partial payments

There are two common ways to allow interest credit

on short-term transactions: Merchant Rule and

United state rule.

Example 1:

A debt of $1000 is due in 1 year with interest at 15%

the debtor pays $300 in 4 months and $200 in 10

months. Find the balance due in1 year using

Merchant Rule and United state rule.

Solution:

a) By Merchant Rule

Put the focal date at the final settlement

1000(1+0.15*1)= 300(1+0.15*8/12)


23

+ 200(1+0.15*2/12) + x

X= 615$

b) by United state rule

I1 = 1000*0.05*4/12 = 50 < 300

S1 = 1000+50-300 = 750$

I2 = 750*0.05*6/12 = 56.25 < 200

S2 = 750+56.25-200 = 606.25

I3 = 606.25*0.15* 2/12 = 15.18

S3 = 606.25+15.18 = 621.41$

Example2 :

On June 15, 1995 a Pearson borrows $500 at 165.

Payments are made as follows : $2000 on July 10,

1995, $50 on November 201995; $1000 on January


24

12, 1996. What is the balance due on march 10 ,1996,

by united state rule?

Solution:

I1 = 5000*0.16*25/360=55.56<2000

S1 = 5000+55.56-2000= 3055.56

I2 = 3055.56*0.16*133/360 = 180.62 >50

I2* = 3055.56*0.16*186/360= 252.56<1000

S2 = 3055.56+252.56-(1000+50)=2258.15$

I3 = 2258.15*0.16*58/360= 58.21

S3 = 2258.15+58.21=2316.36$


25

Example3:

A couple gets an $80000, 30-year, 12$ loan. The

monthly payments is $822.90. how much of the first

two payments goes to interest and how much to

principle?

Solution:

I= 80000*0.12*1/12 = $800.0

Payments to principal = $822.90 -$800.0 = $22.90

Balance at the end of the first month = 80000-22.9 =

$79977.10

I2 = 79977.1*0.12*1/12 = $799.7

Payments to principal = 822.9-799.77 = $23.13

Total interest at the first two months = 800 + 799.7 =

$1599.7 and (22.9+23.13 = 46.03) goes to principal.

WH : Page 60-61 : Problems (1-10)

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