MATHEMATICS OF INVESTMENT

AN INTRODUCTION

Simple and Discount Interest

In financial transactions, interest is
• the amount earned by an investment after a period of time OR
• the amount paid by a borrower to a lender for the use of money over a period. So it is the COST
of borrowing.
• Interest that is paid as a percent of amount borrowed or invested is called simple interest. The
formula for simple interest is given by the following:

Simple Interest

/ = Prt

Where,

ls = interest earned {owed)

P = principal amount invested (or borrowed) r =

annual rate of interest t = term or time frame in

years

Example

1. Suppose an amount Php 500.00 is invested for 2 years at 6% per year. How much is the earning of
the investment after two years?
Solution:
The following can be obtained from the problem: P = 500, r = 0.06, t = 2.
I = Prt = (500)(0.06)(2) = 60 . From this we conclude that, the investment earned Php60.00.

2. If Php4, 000.00 is borrowed at an annual interest rate of 16% for 9 months. How much is the
interest due in borrowing the amount of money?
Solution:
Given :
9
P = 4,000, r = 0.16, t = = 0.75.

I = Prt = (4,000)(0.16)(0.75) = 480 . From this we conclude that, the interest due is Php480.00.
 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Sometimes if we are the investor, we consider the value of our investment after a given period. In this
case we introduce the concept of future values or accumulated values.

Future Value

F = P(l + rt)

Where,
ls = interest earned {owed)
P = principal amount invested {or borrowed) r
= annual rate of interest t = term or time frame
in years

Present Value

P = F(1+rt)-1 OR P = F + (1+rt)

Example

1. If Php4, 000.00 is borrowed at an annual interest rate of 16% for 0.75 years, what is the value of
the investment after 0.75 years?
Solution:
Since the interest earned by the amount invested for 0.75 years is Php480.00, the value of Php4,000.00
after 0.75 years is Php4,480.00.
2. What is the simple interest rate applied if an investment of Php37,500 accumulates to
Php45,937.00 in the period of 1.5 years?

Solution:
We note that the interest earned by the investment is Php8, 437, that is, I = 8437. From the
I 8 437
formula I = Prt, we have r = — = —1^ = 0.15 = 15 %
Pt (37,500 Xl.5)

3. The repayment on a loan was Php12,100. If the loan was for 15

months at 16.8% interest a year, how much was the principal?
Solution:
Based from the given we have the following: F = 12,100, r = 0.168,
and t = 1.25

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 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Since F = P(1 + rt), we have P F 12,100

10,000 .
(l + 1 + (0.168 Xl-25 )
rt)

Different ways of expressing time/term of a loan or investment.

Sometimes the term of investment is not given in years. The term or time frame given in
certain problems maybe stated in days or months. In cases where the time is expressed in months it is
easy to express it in years. But when the term/time is given in days we use a time factor such as the
following:

t= Actual time
360
Actual time
Ordinary Simple Interest or Bankers Rule
365

t = Exact Simple Interest

Approximate rime
So
t=

Examples:

1. Find the exact interest and the final amount due on P28,000 at 12% for 120 days.

2. Using ordinary interest, determine the final amount due on P10,800 at 15.5% for 100
days.

The Bankers Rule or Ordinary Simple Interest is applied whenever a given

Remark
problem does not specify the time factor to be used.

Sometimes the term or time frame may be drawn from the specified origin and repayment dates. The
following indicated how to compute for the actual time and approximate time.
Actual time - Number of days until the repayment date except the origin date.
Approximate time - Assumes that every month contains 30 days.

Example
Find the actual and approximate time from May 1, 1983 to September 15, 1983.

5L 1_
Actual Time Approximate Time
31-1=30 30-1=29
May 30 May 29
June 30 June 30 3
July 31 July 30
Aug 31 Aug 30
Sept 15 Sept 15
137 134
 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Examples:

1. Find the approximate and actual number of days from March 15, 1993 to December 20 of the
same year.
Year Month Day
1993 12 20
Less: 1993 3 15
0 9 5

Approximate No. of Days: (9 * 30) + 5 = 275 days

Simple Discount Interest.

Similar to simple interest, discount interest is an amount paid for borrowing money. Unlike simple
interest, however, discount interest is charged at the time the loan has been negotiated and executed.
Whereas, simple interest is paid on the maturity date when it is added to the amount of the loan applied
for on the origin date, discount interest is charged in advance and is taken from the amount of the loan
applied for on the origin date.

Discount Interest

Where,

Id = discount interest

F = face value

d = annual rate of

discount interest t =

term or time frame in

Proceeds = F — Id = F (1 — dt)
years
 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Example
Francis borrows Php 10,000 from SSS for a one year term. He was charged a 5% discount interest rate.
Determine his proceeds.
Solution:
Id = Fdt = (10,000)(0.05)(1) = 50 500 P = F - I d = (10,000) - 50 = 9,950 g 5QQ

EXERCISES
1. Determine the Actual and Approximate number of days in the given origin and repayment dates.

Origin Date Repayment Date Actual Time Approximate Time

May 22, 1995 July 09, 1995
January 06, 1997 November 06, 1997
March 03, 2007 October 11, 2007
February 04, 1990 November 05, 1992
March 02, 2005 November 05, 2006

2. Justin borrowed Php 8,600 on November 2, 1992, which is to be repaid on May 21, 1993 at 16.2%
interest per year. Find the amount to be repaid. How much will the interest be at the repayment
date using the following time factors?
a. Bankers Rule
b. Exact Simple Interest
AppraximatB
c. 360
AppraximatB
d 365

3. How much should Mark pay to Michele if he borrowed Php10,000 on June 25, 2008 and if the at
principal and interest are to be paid on November 18, 2008 15% simple interest per year? Use the
following time factors.
a. Bankers Rule
b. Exact Simple Interest
AppraximatB
c. 360
AppraximatB
d. 365

4. At what simple interest rate will a sum of money double itself in 15 years?
5. How long will it take for Php 4,000 to grow Php 14,000 if the simple interest rae is 12.5%?
6. How much will Php 22,500 become if invested at 9% simple interest rate for 45 days?
7. April issues a check for Php 4,950 to settle a 4 month loan of Php 4,500. How much simple interest
rate was she charged?

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 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Promissory Notes

These are written commitments by a person (also called the drawer or borrower) or business to
pay a certain sum to another person (also called the drawee or lender/investor) or business within a
specified time.

A type of promissory note is the SIMPLE INTEREST NOTE which has the following characteristics:
• The note is drawn on the origin date
• The note is redeemed on the maturity date
• The stated value of the note (or the face value ) corresponds to the principal
• The face value plus the interest is called the maturity value.
Example

Php 5, 000.00 Quezon City, Philippines January 11, 2008

Seventy five days after the above date, the undersigned promises to pay to the order of
Justin Cruz Five thousand pesos with simple interest at 16.2% per annum payable at BCC
Bank of QC, Philippines.

Mark O. Tan

From the note we have the following:

Drawer Mark Tan

Drawee Justin Cruz
Face Value (Principal) Php 5,000
Interest Rate 16.2%
Term 75 days
Beginning Date January 11,2008
Maturity Date March 21, 2008 (75 days after January 11, 2008)
Interest (S,000) (0.162) = 160.75
Maturity Value (F) 5,000 +160.75 = 5,163,75

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 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Another type of promissory note is the DISCOUNT NOTE. In this case, the amount stated in the note is
the maturity value on which the interest is computed and charged in advance. The money received by
the drawer on the origin date is the proceeds. In this case, the interest charged in advance is first
computed. Then it is subtracted from the maturity value. The difference is the proceeds.

Example

Php 7,000.00 Manila City, Philippines September 25, 2008

One hundred days after the above date, for value received with interest at 14.4% per
annum discounted to maturity, the undersigned promises to pay to the order of
Justin Cruz Seven thousand pesos payable at BCC Bank of QC, Philippines.

Mark Tan

From the
note we
have the
Drawer Mark Tan following:
Drawee Justin Cruz
Maturity Value, (F) Php 7,000
Discount Interest Rate 14.4%
(d)
Term 100 days
Beginning Date September 25, 2008
Maturity Date January 3, 2009 (100 days after September 25, 2008)
(7,000) (0.144) (^) = 230.00
Interest in advance
(Discount, Id)
Proceeds 7,000 -280 = 6,720.00

Remark:

• If the term in the note is given in months, the day of the origin date and the maturity date
coincide. Thus, a 5 month note with an origin date of May 12, 2008 will mature on October
12, 2008.

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 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Exercises

1. A simple interest note for 120 days at 13.8% per annum has a maturity value of Php 5,753. What
is the face value of this note?
2. Chard signed a Php 2,800 bank discount note on April 16, 2008.If the note was for 10 months a
18% per annum, find the following:
a. Maturity date
b. Interest deducted in advance
c. proceeds
3. Determine the interest and maturity value of a simple interest note signed on October 29, 2008
and due February 11, 2009. The note has a face value of Php 10,000 and an interest rate of
16.5% per annum.

Discounting Notes

Example 1
Mike bought office
furniture from Jay
and gave him a Php
50,000 bank
discount note due
on August 15, 1993,
for the balance. Jay
needs money on
June 16, 1993 and
sells his note to a
bank, which
discounts it a
13.8%. How much
does Jay get?
Solution:

The number of days from June 16, 1993 - August 15, 1993 is 60. Finding the discount interest we
have:
(50,000X0.136) = 1,150

Thus the proceeds of Jay is 50,000 — 1,150 = 48,850.

8
 MATHEMATICS OF INVESTMENT
AN INTRODUCTION

Example 2
Bert holds a Php 7,500, 120-day simple interest note from Nestor Flores. The interest rate is 15% and
the note was made on August 18, 1993. On September 17, 1993, Bert wanted to encash the note. He
found a bank which would buy the note that day at 16% interest, collectible in advance. How much will
Bert receive from the bank on September 17?
Solution:

120
PART 1: The
value of Php
7,500 on the
maturity date
is 7,500 (1 +
(0.15) ( ) ) =
7,875
360
Exercises

Find the proceeds of the following discount notes: PART 2: The

Maturity Value Maturity Date Date of Discount Interest Rate

interest in
a. 30,000 Nov. 25, 1993 July 3, 1993 10.2%
b. 23,200 May 16, 1994 Dec. 15, 1993 12%
C 4,000 Sept. 17, 1993 Feb. 19, 1993 advance
9.75% that
d. 6,600 June 5,1994 Jan. 11, 1994 8%
e. 12,000 April 16, 1993 Oct. 25, 1992 12.6%
the bank will

1. Mia holds a Php 4, 000 note at 12.75%, simple interest, payable in take
a year, from Banco
is de Oro.
In need of cash, Mia sells it back to the bank after 4 months. How much will he receive if the
bank charges 15% interest to be drawn in advance?
2. Caloy receives a 60-day Php 6,200 bank discount note on September 7,875^0,16)
4, 2008. How much will
he received if it is discounted at 9% 15 days later?
= 315
9
Therefore the PROCEEDS that Bert will receive is just 7,875 — 315 = 7, 560