Lecture Two
Section 2.1 – Simple and Compound Interest
Simple Interest simply means money is not compounded only once between the beginning and the end
of the note (or investment). With simple interest, the interest is calculated only at the end of the time
period {rather than periodically as is done with compounding.}
I  Pr t
A  P(1  rt )  P 
A
1  rt
A: Amount is the balance of the account (including the interest).
P: Principal is the initial amount of principal that is borrowed or invested.
I: Interest is the fee that is applied to the note or investment.
r: Rate is the interest rate
t: Time is time in years.
Example
Find the total amount due on a loan of $500 at 12% simple interest at the end of 30 months.
Solution
A  P(1  rt )

 500 1  .12 30
12

 $650
500 * (1 + .12(30/12))
Example
To buy furniture for a new apartment, you borrowed $5,000 at rate of 8% simple interest for 11 months.
How much should you pay?
Solution
I  Pr t
 
 5000(.08) 11
12
5000 *.08 *11 / 12
 $366.67
1
Example
T-Bills are one of the instrument of the U.S Treasury Department uses to finance the public debit. If you
buy a 180-day T-bill with a maturity value of $10,000 for $9,828.74, what annual simple interest rate will
you earn?
Solution
t  180  .5
365
A  P(1  rt )
10000  9828.74(1  .5r )
10000  1  .5r
9828.74
10000  1  .5r
9828.74
10000 1
9828.74
r
.5
(10000 / 9828.74 – 1) / .5
r  .03485 or 3.485%
Example
Find the maturity value for a loan of $2500 to be repaid in 8 months with interest of 9.2%.
Solution
Given:
P  2, 500 r  0.092 t  8  2
12
A  P(1  rt )

3
 
 2500 1  .092 2
3
 $2,653.33
2
Compounded Interest

A  P 1 

r
m 
mt
A: Amount in the account (also called future value)
P: Amount invested ($)
r: Annual simple interest rate
t: Time in years
m: Number of times a year the interest is compounded





Daily: m = 365
Monthly: m = 12
Quarterly: m = 4
Semi-annually: m = 2
Annually: m = 1
Compounded Continuously
A  Pert
Example
If $1,000 is invested at 6% compounded over an 8-year period.
Solution
a) Annually


A  1000 1 
1(8)
.06 

1 
 $1,593.85
1000(1+.06/1)^(1*8)
b) Semiannually


.06 

2 
2(8)


.06 

4 
4(8)


.06 

12 
A  1000 1 
 $1,604.71
c) Quarterly
A  1000 1 
 $1,610.32
d) Monthly
A  1000 1 
12(8)
 $1,614.14
3
Example
What amount will an account have after 1.5 years, if $8,000 is invested at annual rate of 9%
a) Compounded Weekly


A  8000 1 
52(1.5)
.09 

52 
 $9,155.23
8000(1+.09/52) ^ (52*1.5)
b) Compounded Continuously
A  8000e.09(1.5)  $9,156.29
8000 e ^ (.09*1.5)
(e: 2nd ln)
Example
How much should new parents invest now at 8% to have $80,000 toward their child’s college education
in 17 years if compounded?
a) Semiannually: m = 2
Given: r = 8% = 0.08
r
A  P  1  
 m
A = 80,000
t = 17
mt
80,000  P1  0.08 
2 

2(17)
80,000
P
2
(
17
)
1 0.08 
2 

P

80,000

2(17)
1 0.08
2
 $21, 084.17
80000 / (1+ .08/2) ^ (2*17)
b) Continuously : A  Pert
80,000  Pe (.08)(17)
P
80,000
e(.08)(17)
 $20, 532.86
80000 / e ^ (.08*17)
4
Annual Percentage Yield (APY)
r

Annual  t  1  A  P 1  
 m
mt


 P 1 
r

m
m
Amount @ simple interest = Amount @ Compound interest


m
P(1  APY )  P 1  r
Divide P both side
m
m
1  APY  1  r
m
m
APY  1  r  1
m




APY  re  er  1
Continuously:
Effective Rate
The effective rate corresponding to a started rate of interest r compounded m times per year is


m
re  1  r
1
m
APY is also referred to as effective rate or true interest rate.
Example
How much should you invest now at 10% to have $8,000 toward the purchase of a car in 5 years if
compounded?
Solution
Given: r = 10% = 0.1
A = 8000
t=5
a) Quarterly: m = 4
A  P  1 

r

m
8000  P 1 
P

8000
 0.1 
1 

4 

mt
0.1 

4 
20
4(5)
 $4882.17
8000 / (1  0.1 / 4)  20
b) Continuously : A  Pert
8000  Pe(0.1)(5)
P  8000  $4852.25
e0.5
8000 / (e  (.1* 5))
5
Example
A $10,000 investment in a particular growth mutual fund over a recent 10-year period would have grown
to $126,000. What annual nominal rate would produce the same growth if
Solution
a) Annually: m = 1
1(10)
 r
126000  10000  1  
 1
126000  1  r 10
10000
12.6  1  r 
10
1  r 10  12.6
1  r  10 12.6
r  (12.6)1/10  1  .28836 or 28.84%
(12.6)  (1 / 10)  1
b) Continuously : A  Pert
126000  10000e10r
12.6  e10r
ln 12.6  ln e10r
ln12.6  10r ln e
ln12.6  10r
r  ln12.6  .25337 or
10
25.337%
ln(12.6) / 10
Example
A bank pays interest of 4.9% compounded monthly. Find the effective rate.
Solution
Given:
r  0.049, m  12
 m
12
 1  .049   1
12
m
re  1  r
1
1  .049 / 12 ^ 12  1
 0.0501156
Or re  5.01%
6
Example
You need to borrow money. Bank A charges 8% compounded semiannually. Another bank B charges
7.9% compounded monthly. At which bank will you pay the lesser amount of interest?
Solution


m
 1  1  .08
2


m
 1  1  .079
12
Bank A: re  1  r
m
Bank B: re  1  r
m



2

 1  .0816  re  8.16%
12
 1  .0819  re  8.19%
Bank A has the lower effective rate, although it has a higher stated rate.
7
Exercises
Section 2.1 – Simple and Compound Interest
1.
If you want to earn an annual rate of 10% on your investments, how much should you pay for a note
that will be worth $5,000 in 6 month?
2.
How much should you deposit initially in an account paying 10% compounded semiannually in
order to have $1,000,000 in 30 years? b) Compounded monthly? c) Compounded daily?
3.
You have $7,000 toward the purchase of a $10,000 automobile. How long will it take the $7000 to
grow to the $10,000 if it is invested at 9% compounded quarterly? (Round up to the next highest
quarter if not exact.)
4.
How long, to the nearest tenth of a year, will it take $1000 to grow to $3600 at 8% annual interest
compounded quarterly?
5.
Jennifer invested $4,000 in her savings account for 4 years. When she withdrew it, she had
$4,350.52. Interest was compounded continuously. What was the interest rate on the account? Round
to the nearest tenth of a percent.
6.
An actuary for a pension fund need to have $14.6 million grows to $22 million in 6 years. What
interest rate compounded annually does he need for this investment to growth as specified? Round
your answer to the nearest hundredth of a percent.
7.
Which is the better investment: 9% compounded monthly or 9.1% compounded quarterly?
8.
Sun Kang borrowed $5200 from his friend to pay for remodeling work on his house. He repaid the
loan 10 months later with simple interest at 7%. His friend then invested the proceeds in a 5-year
CD paying 6.3% compounded quarterly. How much will his friend have at the end of the 5 years?
9.
The consumption of electricity has increased historically at 6% per year. If it continues to increase at
this rate indefinitely, find the number of years before the electric utilities will need to double their
generating capacity. Round up to the next highest year.
10.
In the New Testament, Jesus commends a widow who contributed 2 mites (roughly ¼ cent) to the
temple treasury. Suppose the temple invested those mites at 4% compounded quarterly. How much
would the money be worth 2000 years later?
11.
If $1,000 is invested in an account that earns 9.75% compounded annually for 6 years, find the
interest earned during each year and the amount in the account at the end of each year. Organize
your results in a table.
12.
If $2,000 is invested in an account that earns 8.25% compounded annually for 5 years, find the
interest earned during each year and the amount in the account at the end of each year. Organize
your results in a table.
8
13.
If an investment company pays 6% compounded semiannually, how much you should deposit now
to have $10,000
a) 5 years from now?
b) 10 years from now?
14.
If an investment company pays 8% compounded quarterly, how much you should deposit now to
have $6,000
a) 3 years from now?
b) 6 years from now?
15.
What is the annual percentage yield (APY) for money invested at
a) 4.5% compounded monthly?
b) 5.8% compounded quarterly?
16.
What is the annual percentage yield (APY) for money invested at
a) 6.2% compounded semiannually?
b) 7.1% compounded monthly?
17.
A newborn child receives a $20,000 gift toward a college education from her grandparents. How
much will the $20,000 be worth in 17 years if it is invested at 7% compounded quarterly?
18.
A person with $14,000 is trying to decide whether to purchase a car now, or to invest the money at
6.5% compounded semiannually and then buy more expensive car. How much will be available for
the purchase of a car at the end of 3 years?
19.
You borrowed $7200 from a bank to buy a car. You repaid the bank after 9 months at an annual
interest rate of 6.2%. Find the total amount you repaid. How much of this amount is interest?
20.
An account for a corporation forgot to pay the firm’s income tax of $321,812.85 on time. The
government changed a penalty based on an annual interest rate of 13.4% for the 29 days the money
was late. Find the total amount (tax and penalty) that was paid. (Use 365 days a year.)
21.
A bond with a face value of $10,000 in 10 years can be purchased now for $5,988.02. What is the
simple interest rate?
22.
A stock that sold for $22 at the beginning of the year was selling for $24 at the end of the year. If the
stock paid a dividend of $0.50 per share, what is the simple interest rate on an investment in this
stock?
23.
The Frank Russell Company is an investment fund that tracks the average performance of various
groups of stocks. On average, a $10,000 investment in midcap growth funds over a recent 10-year
period would have grown to $63,000. What annual nominal rate would produce the same growth if
9
Solution
Section 2.1 – Simple and Compound Interest
Exercise
If you want to earn an annual rate of 10% on your investments, how much should you pay for a note that
will be worth $5,000 in 6 month?
Solution
A  P(1  rt )
 12 
5000  P 1  .1 
2
5000  P 1  .1( 6 )
P  5000  $4761.90
1  .1
2


5000 / (1 + .1 / 2)
Exercise
a) How much should you deposit initially in an account paying 10% compounded semiannually in order
to have $1,000,000 in 30 years?
b) Compounded monthly?
c) Compounded daily?
Solution
r
A  P  1  
 m
a) P 
mt
P
1000000
( 1  0.10 / 2 )60
A
 1 r 


 m
mt
r

 or  A  1  
 m
 $53, 535.52
or 1000000(1  0.10 / 2)60  $53, 535.52
360  $50, 409.83
b) P  1000000(1  0.10 / 12)
10950  $49,807.53
c) P  1000000(1  0.10 / 365)
1
 mt
Exercise
You have $7,000 toward the purchase of a $10,000 automobile. How long will it take the $7000 to grow
to the $10,000 if it is invested at 9% compounded quarterly? (Round up to the next highest quarter if not
exact.)
Solution
10, 000  7, 000(1  0.09 / 4)n
10 / 7  1.0225n
x
Log Power Property: lna = xlna
 ln(10 / 7)  ln1.0225n
 ln(10 / 7)  n ln1.0225
n
ln(10 / 7)
 16.03 or 17 quarters
ln1.0225
Exercise
How long, to the nearest tenth of a year, will it take $1000 to grow to $3600 at 8% annual interest
compounded quarterly?
Solution
 
4t
3600  1000 1  0.08 
4
A  P 1 r
n
3.6  1.02 
nt
4t
ln 3.6  ln 1.02 
4t
ln 3.6  4t ln 1.02 
ln 3.6  t
4 ln1.02
 t  16.2 yr
2
Exercise
Jennifer invested $4,000 in her savings account for 4 years. When she withdrew it, she had $4,350.52.
Interest was compounded continuously. What was the interest rate on the account? Round to the nearest
tenth of a percent.
Solution
A  Pert
4350.52  4000er 4
4350.52  e4r
4000


ln  4350.52   4r
4000
 r  1 ln  4350.52   .021
4
4000
ln 4350.52  ln e4r
4000
Inverse Property: ln e x  x
 r  2.1%
Exercise
An actuary for a pension fund need to have $14.6 million grow to $22 million in 6 years. What interest
rate compounded annually does he need for this investment to growth as specified. Round your answer to
the nearest hundredth of a percent.
Solution
 1
22  14.6 1  r
(1)(6)
22  14.6(1  r )6
22  (1  r )6
14.6
22
 1 r
14.6
1/6
r   22   1
14.6
1/6
 .0707
 r  7.07%
3
Exercise
Which is the better investment: 9% compounded monthly or 9.1% compounded quarterly?
Solution
12
 0.09 
For 9%: APY  re  1 

12 

 1  9.38%
4
 0.091 
For 9.1%: re  1 
  1  9.42%
4 

9.1% is better
Exercise
Sun Kang borrowed $5200 from his friend to pay for remodeling work on his house. He repaid the loan
10 months later with simple interest at 7%. His friend then invested the proceeds in a 5-year CD paying
6.3% compounded quarterly. How much will his friend have at the end of the 5 years?
Solution
For 7%: A  P 1  rt 
1


A  5200 1  0.07 10  $5503.33
1
12

For 6.3%: A  5503.33 1  0.063
2
4

20
 $7,522.50
Exercise
The consumption of electricity has increased historically at 6% per year. If it continues to increase at this
rate indefinitely, find the number of years before the electric utilities will need to double their generating
capacity. {Round up to the next highest year}
Solution

2 P  P 1  0.06
1

n
 2  1.06n
ln both sides
ln 2  ln 1.06n
ln 2  n ln1.06
n  ln 2
ln 1.06
 12 years
4
Exercise
In the New Testament, Jesus commends a widow who contributed 2 mites (roughly ¼ cent) to the temple
treasury. Suppose the temple invested those mites at 4% compounded quarterly. How much would the
money be worth 2000 years later?
Solution
Given:
P  1 c  .25c $1  $0.0025
100c
4
r  0.04 m  4 t  2000
r mt

A  P 1  
 m

A  0.0025 1  0.04
4

4(2000)
0.00251 0.04/4 ^(4*2000)
 $9.3 x 1031
Exercise
If $1,000 is invested in an account that earns 9.75% compounded annually for 6 years, find the interest
earned during each year and the amount in the account at the end of each year. Organize your results in a
table.
Solution
Given:
P  1, 000 r  .0975 m  1 t  6
r mt

A  P 1  
 m
st

2
nd

1(1)
A  1000 1  .0975
 $1, 097.50
1
1
Interest  $1, 097.50  $1, 000  $97.50
1 year: t  1 
year: t  2 
A  1000 1  .0975  $1, 204.51
2
2
Interest  $1, 204.51  $1, 097.50  $107.01
rd
3 year: t  3 
th
4 year: t  4 
th
5 year: t  5 
th
6 year: t  6 
A  1000 1  .0975  $1, 321.95
3
3
A  1000 1  .0975  $1, 450.84
Period
Amount
Interest
0
$1,000.00
1
$1,097.50
$97.50
2
$1,204.51
$107.01
3
$1,321.95
$117.44
4
$1,450.84
$128.89
5
$1,582.29
$141.46
6
$1,747.54
$155.25
4
4
A  1000 1  .0975  $1, 592.29
5
5
A  1000 1  .0975  $1, 747.54
6
6
5
Exercise
If $2,000 is invested in an account that earns 8.25% compounded annually for 5 years, find the interest
earned during each year and the amount in the account at the end of each year. Organize your results in a
table.
Solution
P  2, 000 r  8.25%  .08.25 m  1 t  5
Given:
r mt

A  P 1  
 m

2
nd

1(1)
A  1000 1  .0825
 $2,165.00
1
1
Interest  $2,165.00  $2, 000  $165.00
st
1 year: t  1 
A  2000 1  .0825  $2, 343.61
2
year: t  2 
2
Interest  $2, 343.61  $2,165.00  $178.61
rd
3 year: t  3 
A  2000 1  .0825  $2, 536.96
3
3
A  2000 1  .0825  $2, 746.26
Period
Amount
Interest
0
$2,000.00
1
$2,165.00
$165.00
2
$2,343.61
$178.61
3
$2,536.96
$193.35
4
$2,746.26
$209.30
5
$2,972.83
$226.57
4
th
4 year: t  4 
4
A  2000 1  .0825  $2, 972.83
5
th
5 year: t  5 
5
Exercise
If an investment company pays 6% compounded semiannually, how much you should deposit now to
have $10,000
a) 5 years from now?
b) 10 years from now?
Solution
A  10, 000 r  .06 m  2
Given:
 
2(5)
10, 000  P 1  .06 
2
a) t  5 A  P 1  r
m
mt
10, 000  P 1.0310
P
10,000
1.0310
 $7, 440.94
b) t  10
6
10, 000  P 1.032(10)
P
10,000
 $5, 536.76
1.0320
Exercise
If an investment company pays 8% compounded quarterly, how much you should deposit now to have
$6,000
a) 3 years from now?
b) 6 years from now?
Solution
Given:
A  6, 000 r  8%  .05 m  4

a) t  3 A  P 1  r
m

6, 000  P 1  .08
4


mt
4(3)
6, 000  P 1.02 12
P
b) t  6
6,000
1.0212

 $4, 730.96
6, 000  P 1  .08
4

6000 / 1.02 ^ 12
4(6)
6, 000  P 1.02 24
P
6,000
1.0224
 $3, 730.33
7
Exercise
What is the annual percentage yield (APY) for money invested at:
a) 4.5% compounded monthly?
b) 5.8% compounded quarterly?
Solution
r  0.045 m  12
a) Given:

APY  1  .045
12

12
1  .045 / 12  ^ 12  1
 1  0.04594
APY : 4.594%
r  0.058 m  4
b) Given:

APY  1  .058
4

4
1  .058 / 4  ^ 4  1
 1  0.05927
APY : 5.927%
Exercise
48.
What is the annual percentage yield (APY) for money invested at
a) 6.2% compounded semiannually?
b) 7.1% compounded monthly?
Solution
r  6.2%  0.062 m  2
a) Given:

APY  1  .062
2

2
1  .062 / 2  ^ 2  1
 1  0.06296
APY : 6.296%
r  7.1%  0.071 m  12
b) Given:

APY  1  .071
12

12
1  .071 / 12  ^ 12  1
 1  0.07336
APY : 7.336%
8
Exercise
A newborn child receives a $20,000 gift toward a college education from her grandparents. How much
will the $20,000 be worth in 17 years if it is invested at 7% compounded quarterly?
Solution
Given:
P  20, 000 r  7%  .07 m  4 t  17

4(17)
 20, 000 1  .07 
4

mt
A  P 1 r
m
200001.07/4 ^  4*17 
 $65, 068.44
Exercise
A person with $14,000 is trying to decide whether to purchase a car now, or to invest the money at 6.5%
compounded semiannually and then buy more expensive car. How much will be available for the
purchase of a car at the end of 3 years?
Solution
Given: P  14,000 r  6.5%  .065 m  2 t  3


mt
A  P 1 r
m

 14, 000 1  .065
2
 $16, 961.66

2(3)
14000 1  .065 / 2  ^  2 * 3
Exercise
You borrowed $7200 from a bank to buy a car. You repaid the bank after 9 months at an annual interest
rate of 6.2%. Find the total amount you repaid. How much of this amount is interest?
Solution
Given:
P  7, 200 r  6.2%  .062 t  9
12
A  P 1  rt 
 
 7200 1  .062 9 

12 
 $7, 534.80
7200 1  .07  9 / 12  
9
Exercise
An account for a corporation forgot to pay the firm’s income tax of $321,812.85 on time. The government
changed a penalty based on an annual interest rate of 13.4% for the 29 days the money was late. Find the
total amount (tax and penalty) that was paid. (Use 365 days a year.)
Solution
Given:
P  321,812.85 r  13.4%  .13.4 t  29
365
A  P 1  rt 
 
 321,812.85 1  .134 29 

365 
 $325, 239.05
7200 1  .07  9 / 12  
Exercise
A bond with a face value of $10,000 in 10 years can be purchased now for $5,988.02. What is the simple
interest rate?
Solution
The interest earned is: $10,000  $5,988.02 = $4011.98
Given:
P  5988.02 I  4011.98 t  10
I  Pr t
4011.98  5988.02 r 10 
r
4011.98 /  5988.02 *10 
4011.98  0.067
5988.02 10 
The interest rate was about 6.7%
Exercise
A stock that sold for $22 at the beginning of the year was selling for $24 at the end of the year. If the
stock paid a dividend of $0.50 per share, what is the simple interest rate on an investment in this stock?
Solution
The interest earned is:  $24  $22  $0.50  $2.50
Given:
P  22 I  2.50 t  1
I  Pr t
2.50  22 r 1
r  2.5  0.11364
22
The interest rate was about 11.36%
10
Exercise
The Frank Russell Company is an investment fund that tracks the average performance of various groups
of stocks. On average, a $10,000 investment in midcap growth funds over a recent 10-year period would
have grown to $63,000. What annual nominal rate would produce the same growth if
a) Annually
b) Continuously
Solution
a) Annually: m = 1
1(10)
63,000  10,0001  r 
 1
63000  1  r 10
10000
6.3  1  r 10
10 6.3  1  r
r  10 6.3  1  0.20208 or 20.208%
rt
b) Continuously : A  Pe
63,000  10,000e r (10)
6.3  e10r
ln6.3  lne10r
ln e x  x
ln6.3  10r
r  ln6.3  0.18405
10lne
or 18.405%
11