1 MateriTerbuka Simple Interest

(1)

Simple InterestInterest

6

6 - 1

Chap

ter

6

II

(2)

6

6 - 2

Calculate

Learning

Objectives

Learning

Objectives

After completing this chapter, you will be able to:

…

i

nterest

,

m

aturity

v

alue

,

f

uture

v

alue

, and

p

resent

v

alue

in a

simple interest

environment

…

details of the amount

and

timing of payments in a time

diagram

…

the

equivalent value

on any date

of a

single

payment

or a

stream of payments,

and

Present

LO-1

LO-2

(3)

6

6 - 3

Borrower

Example:

Loan

Lender

Parties

Lends the

Principal

Borrower

OWES (Debt)

to Lender

Borrower

OWES (Debt)

to

Lender

LO-1

(4)

6

6 - 4

Borrower

Lender

Earns

(Income)

from Borrower

i.e.

Interest

on

the Principal

Earns

(Income)

from

Borrower

i.e.

Interest

on

the Principal

Borrower

pays

Interest

to

Lender

Borrower

pays

Interest

to

Lender

Rate

of

Interest:

_Simple

_Interest

…

Calculated

on an ANNUAL or

per annum

(pa) basis

Simple Interest

…

Calculated

on an ANNUAL or

per annum

(pa) basis

(5)

6

6 - 5

Examples

Invest $1000

at

10%

simple interest

for

one year.

Interest earned is

?

Principal

X

Interest Rate

_$1000

*

10%

=

$100

Invest $1000

at

10%

simple interest

for

six months.

Interest earned is

?

Principal

X

Interest Rate

$1000

*

10%

/

2 =

$50

=

(6)

6

6 - 6

Invest $1000

at

10%

simple interest

for

three months

Interest earned is

?

Principal

X

Interest Rate

=

$1000

_X

10%

$25

Invest $1000

at

10%

simple interest

for

one month.

Interest earned is

?

Principal

X

Interest Rate

=

$1000

X

10%

/

12 =

$8.33

/

4 =

(7)

6

6 - 7

Up to this point we have taken months to

represent 1/12

th

of a year,

i.e. each month is treated

as having the same number of days!

Would it not be more accurate to

calculate the interest due or payable

based on the actual

number of days in

(8)

6

6 - 8

Yes, it would!

In fact, interest continues

to

accumulate

as each day passes!

Invest $1000

at

10%

simple interest

for

30 days!

Interest earned is

?

Example

Principal

x

Interest Rate

=

$1000

*

10%

*

30 Year = 365 days or 366 in a Leap Year

Year

=

365 days

or

366 in a Leap Year

$8.22

=

(9)

Simple

“What is the formula that can

be used to calculate SI?”

Principal

Interest Rate

Time

Four Elements

are involved …

Interest

Formula

Principal

Amount

(loan or

investment)

(10)

6

6 - 10

Calculate the

Interest

earned

on

$5000

invested

at 4%

for

7 months.

Formula

I =

P

r

t

I =

P *

r *

t

$5000

*

.04

*

7 /12

(11)

6

6 - 11

“I may need to invest or

need a loan for a number

of days rather than a

complete

month.”

“How do I calculate the

time

between the

starting date

and the

(12)

6

6 - 12

Add up

the days according to the

days covered by the specific months.

Include

either

the

FIRST

date

or the

LAST

date when doing

this calculation!

Inc

lude

either

the

FIRST

date

or the

LAST

date when doing

this calculation!

Use

a

Number of Days

TABLE

Use

the

TI BAII Plus

calculator

!

(13)

6

6 - 13

Method 1.

What is

the

interest earned

on

$5000

invested

from

O

ct. 11

th

to

Dec.

29

th

at

4.5%

?

Calculating the

Number of Days

Oct

₁₁

th

to end of month =

20 Days

Nov

Total month =

30 Dec

From 1

_{of month =}

st

to 29

th

₂₉

79

(14)

6

6 - 14

Jan

Mar

May

July

Feb

Apr

Jun

Spaces

are

ALL 30 days

except for

February

which can be either

28 or

29 days in a Leap

year!

Aug

Oct

Dec

Nov

Sept

The “

Hand

Calculator!”

The “

Hand

Calculator!”

(15)

6

6 - 15

Thirty days

has

September

,

April

,

June

and

November

.

All the rest

have

31 ,

but

February

with just

28 days clear

(16)

Simple

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov DecThe Serial Number of Each Day of the Year

T

ABLE

6.2

1 32 60 91 121 152 182 213 244 274 305 335

Method 2.

(17)

Simple

Method 3.

Example 1.

(18)

6

6 - 18

Formula

I =

P

r

t

Calculate the

Interest

earned

on

$5000 invested

at

4.5%

for

?

.

79 Days

I =

$5000

*

.045

*

79/365

(19)

6

6 - 19

Method 1.

What is

the

interest earned

on

$5000

invested

from

Nov 30

th

, 02

to

Jan

6

th

, 03

at

4.5%

?

Nov

₃₀

th

to end of month =

0 Days

Dec

Total month =

31 Jan

From 1

_{of month =}

st

to 6

th

₆

37

37 Includes the last date

(20)

Simple

T

ABLE

6.2

1 32 60 91 121 152 182 213 244 274 305 335

Method 2.

Nov 30

Dec 31

(21)

McGraw-Hill Ryerson©

Method 3.

Example 1.

(22)

6

6 - 22

Formula

I =

P

r

t

I =

$5000

*

.045

*

37/365

I = $22.81

Calculate the

Interest

earned

on

$5000 invested

at

4.5%

for

?

.

(23)

6

6 - 23

Method 1.

What is

the

interest earned

on

$5000

invested

from

Oct 11

th

, 02

to

Mar

11

th

, 03

at

4.0%

?

Oct

₁₁

th

to end of month = 20

Days

Total = 61

Total = 59

151

151 Jan & Feb

Mar

To 11

th

of month =

11 Nov & Dec

30

31

28

31 Includes the last date

(24)

Simple

T

ABLE

6.2

1 32 60 91 121 152 182 213 244 274 305 335

(25)

Method 3.

Example 1.

(26)

6

6 - 26

Formula

I =

P

r

t

I =

$5000

*

.04

*

151/365

I = $82.74

Calculate the

Interest

earned

on

$5000 invested

at

4.0%

for

?

.

(27)

6

6 - 27

Formula

I =

P

r

t

We can

‘reorganize’

the formula to also get

each of the following separately:

Principal

Rate

Time

(28)

6

6 - 28

I

P

r

t

Formula

I =

P

r

t

To

help

remember this…

we can place the formula into a triangle as follows…

The

Tria

ngle

…another useful

non

-calculator!

Where

variables

are

BESIDE

EACH OTHER

this

means to

MULTIPLY!

Using this

tool!

**Prt**

P*

r*

t

I

P

r

t

Where a

variable is

ABOVE

ANOTHER

this means to

(29)

6

6 - 29

If you want to find

P

_then

I

/

r

t

If you want to find

r

then

I

/

P

r

then

I

/

P

t

If you want to find

t

Using this

tool!

I

P

r

t

(30)

6

6 - 30

What is

a Time

Line?

A

Time Line

is, as the name suggests, a line

that shows the various points of time along

which a loan or investment travels to maturity.

It is used for diagramming problems involving

multiple payments or investments.

Two Benefits

…Helps organize data

…Indicates the steps needed

to implement the solution

(31)

Calculate the interest earned at 4% on $5000

invested from Oct. 11

th

_{to March 11}

th

Calculate the

interest

earned at

4%

on

$5000

invested from

Oct. 11

th

_to

_{March 11}

th

Draw a line for

the entire period

Step 1

Oct 11

March 11

$5000

The calculation can now be

restated as follows…

284

365

70

(32)

6

6 - 32

Calculate the interest earned at 4% on $5000

invested for 151 days.

Calculate the interest earned at

4%

on

$5000

invested for

151 days

.

I

=

5000

*

Using this

tool!

I

P

r

t

₌

_$

_82.74

I =

P

r

t

.04

*

151/365

(33)

6

6 - 33

You can now use this

tool!

I

P

r

t

In the next few examples you have to…

(a)

First identify

which variable you are

being asked to solve for, and

In the next few examples you have to…

(a)

First

identify

which variable you are

being asked to solve

for

,

and

(b)

Reorganize the formula

in order to

meet the requirement in (a).

(b)

Reorganize

the formula

in order to

(34)

6

6 - 34

Calculating

the

Principal

Formula

P=

I

r

t

P =

$

195 .

0525

*(

150 /

365 )

P =

$

195 P =

$9,038.10

.0121575342

$195

interest

is

earned

on a

150 day

GIC

at 5.25%

.

(35)

6

6 - 35

.0525

*

150 /

365 =

/

195 =

$9038.10

1/x

9038.10

(36)

6

6 - 36

Calculating the

Rate

What Rate of Interest is

needed to earn $200 on a

$5000 investment invested for

180 days?

What Rate

of Interest

is

needed to

earn

$200

on a

$5000

investment

invested for

180 days?

r

=

Formula

r

=

I

P

t

/

($5000

*

180/365

)

$200

0.081111

or

8.11%

/2465.75

r

=

(37)

6

6 - 37

5000

*

180 /

365 =

/

200 =

8.11%

1/x

0.08111

/

($5000

*

180/365

)

(38)

6

6 - 38

Calculating the

Time

What is the length of Time

required for $2000 to grow to

$2100 when invested at

5.6%?

What is the length of

Time

required for

$2000

to grow to

$

2

100 when invested at

5.6%?

Formula

t

=

I

P

r

Step 1

Find the amount of Interest

Find the amount of

Interest

$

2

100 P

_P

=

$100

More

-

$2000

I

_I

(39)

6

6 - 39

What is the length of Time

required for $2000 to grow to

$2100 when invested at 5.6%?

What is the length of

Time

required for

$2000

to grow to

$2100 when invested at 5.6%?

t =

**/ ($2000*.056)**

/ (

$2000

*

.056

)

t =

$100

/ 112

t =

0.8928 Years

t =

326 days

Step 2

Calculate

Step 2

Calculate

Calculating the

Time

Formula

t

=

I

P

r

(40)

6

6 - 40

100 2000

/

0.056 =

365 =

326 days

325.89 **/ ($2000*.056)**

/ (

$2000

*

.056

)

$100

t

=

/

0.8928 Years

(41)

6

(42)

6

6 - 42

Formula

I =

P

r

t

P

+

I

S

um =

Up to this point we have used two

formulae:

Now, we substitute for

I

with

P

rt

Step 1

P

+

S

um =

Collecting like terms

Formula

Future Value

F

uture

V

alue

S

um

=

P

(

1 +

r

t

)

Step 2

(43)

Simple

term deposit on

(44)

6

6 - 44

.065

Date of Maturity

*

150 /

365 =

+

1 =

$17454.11

$

17

454.11 *

17000

S

=

$17000

_{1+ .065}

₍

150 ₎

365 17454.11

(45)

Simple

T

ABLE

6.2

1 32 60 91 121 152 182 213 244 274 305 335

Determine the

Maturity

Date

Determine the

Maturity

Date

156 The Term Deposit will

mature on June 5

th

_.

(46)

6

6 - 46

5,000

8,000

10,000

During the year, I invested the

following funds at a constant 4% p.a.

Feb 14

th

$5,000

Mar 17

th

$3,000

July 1

st

$2,000

What Total amount will I have on December 29

th

?

(47)

6

6 - 47

Step

1 Three Steps

are required to solve this problem:

Draw a time line

,

including the dates

and dollar amounts.

Step

2 Determine the time

between each of the

dates

Step

3 Calculate the

interest

amounts

,

and

add

(48)

(49)

(50)

6

6 - 50

“During the year I made an

investment that had changes in

the

rate of interest

.

How do I determine the total

interest earned

during the

period under review?”

Every time the interest rate

changes

,

you

must stop

and

(51)

6

6 - 51

I

invest $1,000

on Feb. 14th at

6%

.

The changes in interest rates to July 4

th

are as follows:

Investment

Date

rate

$1,000

Feb 14

th

6%

April 20

th

6.8%

May 18

th

7.1%

(52)

6

6 - 52

Look up

for Days

… by finding the number of days between

each rate change!

Investment

Date

rate

$1,000

Feb 14

th

6%

April 20

th

6.8%

May 18

th

7.1%

(53)

6

6 - 53

= 45

= 110

= 65 Days

= 28 Days

Number

of Days

Number

of Days

Table

reading

Table

reading

Look up

for Days

Feb 14

th

April 20

th

May 18

th

July 4th

_{= 47 Days}

Interest Earned

= 138

(54)

6

6 - 54

Feb 14

th

April 20

65 Days

th

Formula

I =

P

r

t

I

₁

=

1000

*

.06

*

65/365

=

$10.68

May 18

th

28 Days

I

2

=

1000

*

.068

*

28/365

I

₃

=

1000

*

.071

*

47/365

= 5.22

July 4th

47 Days

= 9.14

$25.04

Interest

Earned

(55)

6

(56)

6

6 - 56

What are we being asked

to provide

?

“How much

must I

invest

…

to grow…”

“How much

must I

invest

…

to grow…”

…this suggests finding the

P

rincipal

How

much

must

I invest

in order for it

to

grow to $5000

within

6 months

(57)

6

6 - 57

What data do

we need

?

We need the

appropriate data

to be

able to use the

appropriate formula

…

We need the

appropriate data

to be

able to use the

appropriate formula

…

Formulae

I

=

P

r

t

&

S

um =

P

(1+

r

t

)

How

much

must

I invest

in order for it

to

grow to $5000

within 6 months, @ 4.4% simple interest?

How

much

must

I invest

in order for it

to

grow to $5000

(58)

6

6 - 58

r

=

4.4%

t

=

6 months

=

.5

S

um =

$5000

How

much

must

I invest

in order for it

to

grow to $5000

within 6 months, @ 4.4% simple interest?

How

much

must

I invest

in order for it

to

grow to $5000

within

6 months

,

@ 4.4% simple interest

?

What data do

we have

?

(59)

6

6 - 59

As we know the

S

um, the formula now becomes…

How

much

must

I invest

in order for it

to

grow to $5000

within 6 months, @ 4.4% simple interest?

How

much

must

I invest

in order for it

to

grow to $5000

within

6 months

,

@ 4.4% simple interest

?

Formulae

I

=

P

r

t

&

S

um =

P

(1+

r

t

)

Using

S

um =

P

(1+

r

t

)

(60)

6

6 - 60

P

=

5000

/ 1 +

0.044 (

.5

)

P

= $4892.37

=

$4892.37

How

much

must

I invest

in order for it

to

grow to $5000

within 6 months, @ 4.4% simple interest?

How

much

must

I invest

in order for it

to

grow to $5000

within

6 months

,

@ 4.4% simple interest

?

(61)

6

6 - 61

LO-2

(62)

Simple

What is the equivalent value on

September 15

of a $2000 payment on July

4, if money is worth 6% pa?

What is the

equivalent value

on

September 15

of a $2000 payment on July

4, if money is worth 6% pa?

July 4

September 15

(63)

6

6 - 63

Presen

t

Value

What is the equivalent value

on May 18th

of a $2000 payment due on the following

December 15th if money can earn 5.2%?

What is the

equivalent value

on May 18th

of a $2000 payment due on the following

December 15th if money can earn 5.2%?

Step

1 Step

2 Draw a

Timeline

May 18

December 15

$2000

=

2000

/[1+

.052

(

211/365

)]

211 Days

=

$1941.63

$

1941.63

(64)

6

6 - 64

Heather owes Mark $3000 payable

on April 27.

If money can

earn 4%,

what

amount should Mark accept in

settlement of the debt:

A)

30 days before

the scheduled

payment?

(65)

6

6 - 65

Step

1 Draw a

Timeline

April 27

P

resent

V

alue

30 days

90 days

$3000

Step

(66)

6

6 - 66

A

P

=

3000

/[1+

.04

(

30/365

)]

P

=

$2990.17

B

P

=

3000

/[1+

.04

(

90/365

)]

P

=

$2970.17

(67)

6

6 - 67

You can prepay $1234 tuition for a course

or delay payment for 3 months and pay

$1432.

If you can

earn 6%

on your money,

(68)

Simple

payment!

P

₂

=

$1410.84

Money saved ($176.84) by paying now!

(69)

6

6 - 69

This completes Chapter 6