Simple Interest
FDNBUSF
Outline
1.1 Simple Interest
1.2 Exact and Ordinary Interest
1.3 Actual and Approximate Time
1.4 Simple Discount
1.5 Promissory Notes
Learning Outcomes
•
•
•
•
Explain the concept of simple interest
Derive the formula of simple interest
Solve problems about simple interest
Why do we study simple interest
Definition of Terms
Interest – a certain sum of money that the lender
charges the borrower for the use of the funds.
Lender / Creditor – the person or institution that makes
the funds available to those who need it.
Borrower / Debtor – the person or institution that
avails of the funds from the lender.
The interest earned when a loan or
investment is repaid in a lump sum.
Simple Interest
•
Understanding simple interest is one of the most
important and fundamental concepts for mastering
your finances.
•
With an understanding of how interest works, you
become empowered to make better financial
decisions that save you money.
•
The simple interest calculation provides a very basic
way of looking at interest. It’s an introduction to the
concept of interest in general. In the real world,
your interest—whether you’re paying it or earning
it.
L1: Simple
Interest
Three Factors:
• Principal
• Rate of Interest
• Time or Term of the loan
/ investment
Formula:
I=Prt
I = Interest
P = Principal
r = rate
t = term of the loan in
years
• Principal – is the sum of
money borrowed or
invested.
• Interest Rate – is the rate
charged by the lender or
rate of increase of the
investment.
– Expressed in decimals or
percentages
• Time or Term of the loan –
the number of years the
sum of money was
borrowed or invested.
Simple Interest
Three Factors:
• Principal
• Rate of Interest
• Time or Term of the loan
/ investment
Formula of Interest:
I=Prt
I = Interest
P = Principal
r = rate
t = term of the loan in
years
How much interest is charged when
P10,000 is borrowed for 2 years at an
interest rate
of 3%?
Given:
P = P10,000
r = 0.03
t=2
Solution: I=Prt
I = (10,000)(0.03)(2)
I = P600
Answer:
The interest charged for borrowing
P10,000 for 2 years is P600.
SANITY CHECK:
• Akiko is saving a little extra money to pay for
her car insurance next year. If she invests
$1,000 for 18 months at 4%, how much
interest can she earn?
– $60
• Habib is going to borrow $2,000 for 42
months at 7% . What will be the amount of
interest owed?
– $490
Exact and Ordinary Interest
Calculation of Simple Interest and Maturity
Value
1. Calculate simple interest and maturity value
for months and years.
2. Calculate simple interest and maturity value
by (a) exact interest and (b) ordinary interest.
Exact and Ordinary Interest
• Exact interest is a type of simple interest that is
computed assuming there are 365 days in a year.
(366 in a leap year).
Time =
Exact number of days
365
• Ordinary interest is a type of simple interest that is
computed on a 360-day period. (Banker’s Rule)
Exact number of days
Time =
360
Simple Interest
Two categories:
•Exact Interest
•Ordinary Interest
Determine the simple interest earned if
P3,500 is invested at 15% interest rate
in 245 days, (a) using exact interest;
(b) using ordinary interest.
Given:
number of days
te =
365
number of days
to =
360
P = P3,500
r = 0.15
t = 245
I=?
Solution (a):
Ie = P r te
æ 245 ö
Ie = 3,500 (0.15) ç
÷
è 365 ø
Ie = P352.40
Answer:
The simple interest earned is P352.40
using exact interest.
Simple Interest
Two categories:
•Exact Interest
•Ordinary Interest
Determine the simple interest earned if
P3,500 is invested at 15% interest rate
in 245 days, (a) using exact interest;
(b) using ordinary interest.
Given:
number of days
te =
365
number of days
to =
360
P = P3,500
r = 0.15
t = 245
I=?
Solution (b):
Io = P r t0
æ 245 ö
Io = 3,500 (0.15) ç 360 ÷
è
ø
Io = P357.29
Answer:
The simple interest earned is P357.29
using ordinary interest.
Maturity Value or
Future Amount
Two categories:
•Exact Interest
•Ordinary Interest
number of days
te =
365
number of days
to =
360
How much will the maturity value of
P5,000 be in 48 days if interest rate is at
20%, (a) using exact interest and
(b) using ordinary interest.
Given:
P = P5,000
r = 0.20
F=?
t = 48
Solution (a):
F = P ( 1 + r te )
é
æ 48 öù
F = 5,000 ê1 + (0.2)ç
÷ú
è 365 øû
ë
F = P5,131.51
Answer:
P5,000 will accumulate to P5,131.51
using exact interest.
Maturity Value or
Future Amount
Two categories:
•Exact Interest
•Ordinary Interest
number of days
te =
365
number of days
to =
360
How much will the maturity value of
P5,000 be in 48 days if interest rate is at
20%, (a) using exact interest and
(b) using ordinary interest.
Given:
P = P5,000
r = 0.20
F=?
t = 48
Solution (b):
F = P ( 1 + r to )
é
æ 48 öù
F = 5,000 ê1 + (0.2)ç
÷ú
è 360 øû
ë
F = P5,133.33
Answer:
P5,000 will accumulate to P5,133.33
using ordinary interest.
LESSON 1.3
ACTUAL AND
APPROXIMATE TIME
Lesson 1.3
• Origin date is the date when the loan or
investment is made
• Maturity date is the date when the loan is
paid or investment is terminated
L3: Actual Time
and Approximate
Time
Find the actual time between April 15
and December 21 of the same year.
Given: Origin date
: April 15
• Origin date
Maturity date : Dec. 21
•Maturity date
Actual time = ?
Actual time – is obtained Solution (a):
by counting the actual
number of days between
the two given dates.
Approximate time – is
obtained by counting the
actual number of days
between the two given
dates
but
on
the
assumption that each
month has 30 days.
Month
Apr
May
Jun
Jul
Aug
No. of
days
30 – 15
= 15
31
30
31
31
Month
Sep
Oct
Nov
Dec
Total
No. of
days
Answer:
30
31
30
21
250
There are 250 actual days from April 15
to December 21.
Actual Time and
Approximate Time
• Origin date
• Maturity date
Find the approximate time between
April 15 and December 21 of the same
year.
Given: Origin date : April 15
Maturity date : Dec. 21
Approximate time = ?
Solution (b):
Month
Apr
May
Jun
Jul
Aug
No. of
days
30 – 15
= 15
30
30
30
30
Month
Sep
Oct
Nov
Dec
Total
No. of
days
30
30
30
21
246
Answer:
There are approximately 246 days from
April 15 to December 21.
Actual Time and
Approximate Time
• Origin date
•Maturity date
Actual time – is obtained
by counting the actual
number of days between
the two given dates.
Using Actual or Approximate time as
numerator to obtain the TERM, you can
compute interest based on four
possible scenarios.
a.Exact interest for the approximate
time;
b.Exact interest for the actual time;
c.Ordinary interest for the approximate
Approximate time – is time;
obtained by counting the
actual number of days d.Ordinary interest for the actual time
between the two given
dates
but
on
the
assumption that each
month has 30 days.
PROBLEM: PARTICIPATION
I = Prt
A= P + I
Mr. Aguirre borrowed P10,000 on
June 25, 2019. If the maturity
value is to be paid on November
18 of the same year at 15%
interest, how much should he pay
given each set of conditions
below?
a.Exact interest for the approximate
time;
b.Ordinary interest for the approximate
time;
c.Exact interest for the actual time;
d.Ordinary interest for the actual time?
ANSWER
We can do this!!!
Mr. Aguirre borrowed P10,000 on June 25,
2012. If the maturity value is to be paid on
November 18 of the same year at 15%
interest, how much should he pay given
each set of conditions below?
Given:
Origin date: June 25, 2012
Maturity date: Nov. 18, 2012
P = P10,000
r = 0.15 F = ?
Approximate time
Month
No. of
days
Jun
Jul Aug Sep Oct Nov
30 – 25 30
=5
30
30
30
18
Total
143
Actual time
Month
No. of
days
Jun
Jul Aug Sep Oct Nov
30 – 25 31
=5
31
30
31
18
Total
146
Actual Time and
Approximate Time
• Origin date
•Maturity date
Approx. time: 143
days
Actual time :146 days
Mr. Aguirre borrowed P10,000 on June 25,
2012. If the maturity value is to be paid on
November 18 of the same year at 15% interest,
how much should he pay given each set of
conditions below?
Given:
Origin date: June 25, 2012
Maturity date: Nov. 18, 2012
P = P10,000
r = 0.15 F = ?
Solution (a):Exact interest for the approx. time?
A = P ( 1 + r te )
é
æ 143 öù
1
+
(
0
.
15
)
ç
÷ú
= 10,000ê
è 365 øû
ë
= P10,587.67
Actual Time and
Approximate Time
• Origin date
•Maturity date
Approx. time: 143 days
Actual time :146 days
A=P+I
Mr. Aguirre borrowed P10,000 on June 25,
2012. If the maturity value is to be paid on
November 18 of the same year at 15% interest,
how much should he pay given each set of
conditions below?
Given:
Origin date: June 25, 2012
Maturity date: Nov. 18, 2012
P = P10,000
r = 0.15
Required: Maturity
Solution (b): Ordinary interest for the approx. time?
A= P + Io
é
æ 143 öù
1
+
(
0
.
15
)
ç
÷ú
ê
360
è
øû
A = 10,000 ë
A= P10,595.83
FINDING P, r, and t
Formula: I= PRT
In some cases, we will be
asked to find the;
• Principal?
P= I/rt
• Rate?
R= I/pt
• Time?
T= I/pr
Find the Principal, Rate or Time
Using the Simple Interest Formula
The Simple Interest Formula
Find the principal using
the simple interest formula
HOW TO:
The Simple Interest Formula
Judy paid $108 in interest on a loan that she had for 6
months. The interest rate was 12%.
How much was the principal?
Substitute the known values and solve.
I
P=
RT
$108
P=
(0.12)(0.5)
P = $1,800
Find the rate using
the simple interest formula
HOW TO:
The Simple Interest Formula
Sam wants to borrow $1,500 for 15 months
and will have to pay $225 in interest.
What is the rate he is being charged?
Substitute the known values and solve.
I
R=
PT
$225
R=
($1,500)(1.25)
R = .12 or 12%
The rate Sam will pay is 12%.
Find the time using
the simple interest formula
HOW TO:
The Simple Interest Formula
Shelby borrowed $10,000 at 8%
and paid $1,600 in interest.
What was the length of the loan?
Substitute the known values and solve.
I
T=
PR
$1,600
T=
($10,000)(0.08)
Length of the
loan was two years.
Maturity Value or
Accrued Amount
- The sum of the principal
and the interest
Formula:
F=P+I
or
F = P (1 + rt)
Granger borrowed P40,000 from a
lending firm that charges 6% per year.
How much will she pay the lending firm
after 5 years?
Given:
P = P40,000
r = 0.06
F=?
t=5
Solution: F = P +I
F = 40,000+( 40000)(0.06)(5)
= 40,000+12000
= P52,000
or F = 40,000(1.30)
= P52,000
Answer:
Granger will have to pay P52,000 after
5 years.
Another Example:
SOLUTION:
I = $40,000 ´.04 ´
Layla the marksman
borrowed $40,000 for
office furniture. The
loan was for 6 months
at an annual interest
rate of 4%. What are
Layla interest and
maturity value?
6
= $800 interest
12
F = $40,000 + $800
= $40,800 maturity value
or
F = $40,000(1.02)
= $40,800 maturity value
Actual Time and
Approximate Time
• Origin date
•Maturity date
Approx. time: 143 days
Actual time :146 days
Mr. Aguirre borrowed P10,000 on June 25,
2013. If the maturity value is to be paid on
November 18 of the same year at 15% interest,
how much should he pay given each set of
conditions below?
Given:
Origin date: June 25, 2013
Maturity date: Nov. 18, 2013
P = P10,000
r = 0.15 F = ?
Solution (c): Exact interest for the actual time?
A= P + Ie
é
æ 146 öù
1
+
(
0
.
15
)
ç
÷ú
= 10,000 ê
è 365 øû
ë
= P10,600
Actual Time and
Approximate Time
• Origin date
•Maturity date
Approx. time: 143 days
Actual time :146 days
Mr. Aguirre borrowed P10,000 on June 25,
2013. If the maturity value is to be paid on
November 18 of the same year at 15% interest,
how much should he pay given each set of
conditions below?
Given:
Origin date: June 25, 2013
Maturity date: Nov. 18, 2013
P = P10,000
r = 0.15 F = ?
Solution (d): Ordinary interest for the actual time?
A = P + Ie
é
146 öù
÷ú
è 360 øû
= 10,000 ê1 + (0.15)æç
ë
= P10,608.33
End of Presentation
for Simple Interest
KNOWLEDGE TEST