1 MateriTerbuka Simple Interest

(1)

Simple InterestInterest

6

6 - 1

Chap

ter

6

II

(2)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 3

Borrower

Example:

Loan

Lender

Parties

Lends the

Principal

Borrower

OWES (Debt)

to Lender

Borrower

OWES (Debt)

to

Lender

LO-1

(4)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 7

Up to this point we have taken months to

represent 1/12

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)

Simple InterestInterest

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 InterestInterest

6

6 - 9

t

r

P

I

“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)

Principal

Amount

(loan or

investment)

Annual Rate

of SI

Annual Rate

of SI

Amount

(paid or

received)

Amount

(paid or

received)

Time period

…expressed

as a fraction

or a multiple

of a year

Time period

…expressed

as a fraction

or a multiple

of a year

(10)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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

!

There are

three methods

that can be used to

calculate the number of days:

(13)

Simple InterestInterest

6

6 - 13

Method 1.

What is

the

interest earned

on

$5000

invested

from

O

ct. 11

to

Dec.

29 at

4.5%

?

Calculating the

Number of Days

Oct

₁₁

to end of month =

20 Days

Nov

Total month =

30 Dec

From 1

_{of month =}

to 29

₂₉

79

(14)

Simple InterestInterest

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)

Simple InterestInterest

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 InterestInterest

6

6 - 16 Day of

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 2 33 61 92 122 153 183 214 245 275 306 336 3 34 62 93 123 154 184 215 246 276 307 337 4 35 63 94 124 155 185 216 247 277 308 338 5 36 64 95 125 156 186 217 248 278 309 339 6 37 65 96 126 157 187 218 249 279 310 340 7 38 66 97 127 158 188 219 250 280 311 341 8 39 67 98 128 159 189 220 251 281 312 342 9 40 68 99 129 160 190 221 252 282 313 343 10 41 69 100 130 161 191 222 253 283 314 344 11 42 70 101 131 162 192 223 254 284 315 345 12 43 71 102 132 163 193 224 255 285 316 346 13 44 72 103 133 164 194 225 256 286 317 347 14 45 73 104 134 165 195 226 257 287 318 348 15 46 74 105 135 166 196 227 258 288 319 349 16 47 75 106 136 167 197 228 259 289 320 350 17 48 76 107 137 168 198 229 260 290 321 351 18 49 77 108 138 169 199 230 261 291 322 352 19 50 78 109 139 170 200 231 262 292 323 353 20 51 79 110 140 171 201 232 263 293 324 354 21 52 80 111 141 172 202 233 264 294 325 355 22 53 81 112 142 173 203 234 265 295 326 356 23 82 113 143 174 204 235 266 296 327 357 24 55 83 114 144 175 205 236 267 297 328 358 25 56 84 115 145 176 206 237 268 298 329 359 26 57 85 116 146 177 207 238 269 299 330 360 27 58 86 117 147 178 208 239 270 300 331 361 28 59 87 118 148 179 209 240 271 301 332 362 29 88 119 149 180 210 241 272 302 333 363 30 89 120 150 181 211 242 273 303 334 364 31 90 151 212 243 304 365 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

363

284 Oct 11

Dec 29

= 79 Days

Method 2.

(17)

Simple InterestInterest

6

6 - 17

12.2902

Us

ing

…

i

Texas Instruments

_{BAII PLUS}

Days Between Dates

D

ays

B

etween

D

ates

2nd

CPT

Enter

Method 3.

Example 1.

Calculate…

the

interest

earned

on

$5000

invested

from

Oct. 11

to

Dec. 29

,

02 at

4.5%

.

Start

DBD =

79 _Date

(18)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 19

Method 1.

What is

the

interest earned

on

$5000

invested

from

Nov 30

, 02

to

Jan

6 , 03

at

4.5%

?

Nov

₃₀

to end of month =

0 Days

Dec

Total month =

31 Jan

From 1

_{of month =}

to 6

₆

37 Includes the last date

(20)

Simple InterestInterest

6

6 - 20 Day of

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

6

365

334 Jan 6

=

37 Days

Method 2.

Nov 30

Dec 31

(21)

McGraw-Hill Ryerson©

Simple InterestInterest

6

6 - 21

01.0603

Us

ing

…

i

Texas Instruments

_{BAII PLUS}

Days Between Dates

D

ays

B

etween

D

ates

2nd

CPT

Enter

Method 3.

Example 1.

Calculate…

the

interest

earned

on

$5000

invested

from

Nov.

30

th,

, 02

_to

Jan

6 . 03

at

4.5%

.

Start

DBD =

37 _Date

(22)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 23

Method 1.

What is

the

interest earned

on

$5000

invested

from

Oct 11

, 02

to

Mar

11 , 03

at

4.0%

?

Oct

₁₁

to end of month = 20

Days

Total = 61

Total = 59

151 Jan & Feb

Mar

To 11

of month =

11 Nov & Dec

30

31

28

31 Includes the last date

(24)

Simple InterestInterest

6

6 - 24 Day of

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 Oct 11

Dec 31

₃₆₅

284 ₌

=

₈₁

81 _Days

Days

Mar 11

₇₀

₌

₇₀

_Days

=

151 Days

(25)

McGraw-Hill Ryerson©

Simple InterestInterest

6

6 - 25

03.1103

Us

ing

…

i

Texas Instruments

_{BAII PLUS}

Days Between Dates

D

ays

B

etween

D

ates

2nd

CPT

Enter

Method 3.

Example 1.

Calculate…

the

interest

earned

on

$5000

invested

from

Oct.

11

th,

, 02

_to

Mar

11 . 03

at

4.0%.

Start

DBD =

151 _Date

(26)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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

…Indicates the steps needed

to implement the solution

Using

a

Time Line

(31)

McGraw-Hill Ryerson©

Simple InterestInterest

6

6 - 31

Calculate the interest earned at 4% on $5000

invested from Oct. 11

_{to March 11}

Calculate the

interest

earned at

4%

on

$5000

invested from

Oct. 11

_to

_{March 11}

Draw a line for

the entire period

Step 1

Oct 11

March 11

$5000

Start

Finish

Enter the key

dates on the line

Step 2

Dec 31

Enter the

number of days

between

each date and

add

Step 4

81 days

70 days

151 days

Enter the

Investment

Step 3

The calculation can now be

restated as follows…

284

365

70 Look up

Using

a

Time Line

(32)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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%

.

Find the

initial

investment

.

(35)

Simple InterestInterest

6

6 - 35

.0525

*

150 /

365 =

/

195 =

$9038.10

1/x

9038.10

(36)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 37

5000

*

180 /

365 =

/

200 =

8.11%

1/x

0.08111

/

($5000

*

180/365

)

(38)

Simple InterestInterest

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

-

$2000

I

_I

(39)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 40

100 2000

/

0.056 =

365 =

326 days

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

/ (

$2000

*

.056

)

$100

t

=

/

0.8928 Years

(41)

Simple InterestInterest

6

(42)

Simple InterestInterest

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

&

S

um =

P

+

I

We can combine them as follows:

(43)

Simple InterestInterest

6

6 - 43

S

um =

P

(1+

r

t

)

I place $17000

in a 150 day

term deposit on

Jan. 6 paying

6.5%.

I place

$17000

in a

150 day

term

deposit

on

Jan. 6 paying

6.5%.

How much will

the bank pay

me on the

maturity date

?

How much will

the bank pay

me on the

maturity date

?

S

=

$

17

454.11 S

=

$

17000

(

1.0267123

)

S

=

$

17000

_{1+ .065}

₍

150 ₎

(44)

McGraw-Hill Ryerson©

Simple InterestInterest

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 InterestInterest

6

6 - 45

Day of

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 Determine the

Maturity

Date

Determine the

Maturity

Date

I place $17000 in a 150 day term

deposit on Jan. 6 paying 6.5%

_deposit

I place

_{on Jan. 6 paying}

$17000

in a

150 day term

_6.5%

pa

_pa

.

_.

= 6 days

+

150 = 156 days

Find

156 Find

156 The Term Deposit will

mature on June 5

th

_.

(46)

Simple InterestInterest

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

$5,000

Mar 17

$3,000

July 1

$2,000

What Total amount will I have on December 29

th

?

(The second and third amounts were

added to the Feb 14

amount.)

(47)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 48

Step

1 Draw a

time

line

Feb 14

March 17

July 1

Dec 29

45

76

182

363 $500

0 $300

0 $200

0 363 – 45 = 318 Days

363 – 76 = 287 Days

Step

2 Determi

ne the

Time

363 – 182 = 181 Days

(49)

McGraw-Hill Ryerson©

Simple InterestInterest

6

6 - 49

S = $10308.28

Total Amount

S = $

10

308.28 Total Amount

Step

3 Calculat

e the

Interest

amount

s,

and

add

togethe

r

.

Calculat

e the

Interest

amount

s,

and

add

togethe

r

.

Formula

I =

P

r

t

I

₁

=

5000

*

.04

*

318/365

$174.25

I

₂

=

3000

*

.04

*

287/365

I

₃

=

2000

*

.04

*

182/365

94.36

39.67 $

308.28 $10000 +

$308.2

8 $

308.2

8

(50)

McGraw-Hill Ryerson©

Simple InterestInterest

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

make a calculation up to

that point.

(51)

Simple InterestInterest

6

6 - 51

I

invest $1,000

on Feb. 14th at

6%

.

The changes in interest rates to July 4

are as follows:

Investment

Date

rate

$1,000

Feb 14

6%

April 20

6.8%

May 18

7.1%

(52)

Simple InterestInterest

6

6 - 52

Look up

for Days

… by finding the number of days between

each rate change!

Investment

Date

rate

$1,000

Feb 14

6%

April 20

6.8%

May 18

7.1%

(53)

Simple InterestInterest

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

April 20

May 18

July 4th

_{= 47 Days}

Interest Earned

= 138

(54)

Simple InterestInterest

6

6 - 54

Feb 14

April 20

65 Days

Formula

I =

P

r

t

I

₁

=

1000

*

.06

*

65/365

=

$10.68

May 18

28 Days

I

=

1000

*

.068

*

28/365

I

₃

=

1000

*

.071

*

47/365

= 5.22

July 4th

47 Days

= 9.14

$25.04

Interest

Earned

Interest

Earned

(55)

Simple InterestInterest

6

(56)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

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)

Simple InterestInterest

6

6 - 61

LO-2

Simple InterestInterest

6

(62)

Simple InterestInterest

6

6 - 62

Futur

e

Value

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?

Step

1 Step

2 Draw a

Timeline

July 4

September 15

$2000

2000

[1+

.06

(

69/365

)]

=

$2022.68

$

2022.68

S

um =

69 Days

(63)

Simple InterestInterest

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)

Simple InterestInterest

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?

B)

90 days before

the scheduled

payment?

(65)

Simple InterestInterest

6

6 - 65

Step

1 Draw a

Timeline

April 27

P

resent

V

alue

30 days

90 days

$3000

Step

(66)

Simple InterestInterest

6

6 - 66

A

P

=

3000

/[1+

.04

(

30/365

)]

P

=

$2990.17

B

P

=

3000

/[1+

.04

(

90/365

)]

P

=

$2970.17

(67)

Simple InterestInterest

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 InterestInterest

6

6 - 68

Data

$1234 =

P

resent

V

alue =

P

₁

$

1432

=

F

uture

V

alue =

S

₂

P

₂

=

1432

/[1+.

06 (

.25

)]

Tip:

Find the

PV of the

future

payment!

Tip:

Find the

PV of the

future

payment!

P

₂

=

$1410.84

Money saved ($176.84) by paying now!

Which

formula

should

you use?

Which

formula

should

you use?

$1234.00

P

=

Sum/(1+rt)

S

um/(1+

r

t

)

3 months

= 1/4

3 months

= 1/4

(69)

Simple InterestInterest

6

6 - 69

This completes Chapter 6

(1)

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?

(2)

Simple Interest

6

6 - 65

Step 1

Draw a

Timeline

April 27

Present Value

30 days

90 days

$3000

Step

(3)

A

P

=

3000

/[1+

.04

(

30/365

)]

P

=

$2990.17

B

P

=

3000

/[1+

.04

(

90/365

)]

(4)

Simple Interest

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, which option should you choose?

(5)

Data

$

1432

=

F

uture

V

alue =

S

₂

P

₂

=

1432

/[1+.

06 (

.25

)]

Tip:

Find the PV of the

future payment!

Tip:

Find the PV of the

future

payment!

P

₂

=

$1410.84

Which

formula

should

you use?

Which

formula

should

you use?

$1234.00

P

=

Sum/(1+rt)

Sum/(1+r

t

)

3 months = 1/4 3 months = 1/4

(6)

Simple Interest

6

6 - 69

This completes Chapter 6