A Mathematical View
of Our World
1st ed.
Parks, Musser, Trimpe,
Maurer, and Maurer
Chapter 13
Consumer Mathematics:
Buying and Saving
Section 13.1
Simple and Compound Interest
• Goals
• Study simple interest
• Calculate interest
• Calculate future value
• Study compound interest
• Calculate future value
• Compare interest rates
• Calculate effective annual rate
13.1 Initial Problem
• Suppose you discover you are the only
direct descendant of a man who loaned the
Continental Congress $1000 in 1777 and
was never repaid.
• Using an interest rate of 6% and a
compounding period of 3 months, how much
should you demand from the government?
• The solution will be given at the end of the
section.
Simple Interest
• If P represents the principal, r the
annual interest rate expressed as a
decimal, and t the time in years, then
the amount of simple interest is:
I  Prt
Example 1
• Find the interest on a loan of $100 at
6% simple interest for time periods of:
a) 1 year
b) 2 years
c) 2.5 years
Example 1, cont’d
• Solution: We have P = 100 and
r = 0.06.
a) For t = 1 year, the calculation is:
I  Prt   $100 0.061  $6
Example 1, cont’d
• Solution, cont’d: We have P = 100 and
r = 0.06.
b) For t = 2 years, the calculation is:
I  Prt   $100 0.06 2  $12
c) For t = 2.5 years, the calculation is:
I  Prt   $100 0.06 2.5  $15
Future Value
• For a simple interest loan, the future value of
the loan is the principal plus the interest.
• If P represents the principal, I the interest,
r the annual interest rate, and t the time in
years, then the future value is:
F  P  I  P  Prt  P 1  rt 
Example 2
• Find the future value of a loan of $400
at 7% simple interest for 3 years.
Example 2, cont’d
• Solution: Use the future value formula
with P = 400, r = 0.07, and t = 3.
•
F  P 1  rt 
 400 1  0.07  3  $484
Example 3
•
In 2004, Regular Canada Savings Bonds
paid 1.25% simple interest on the face
value of bonds held for 1 year.
•
If the bond is cashed early, the investor
receives the face value plus interest for
every full month.
•
Suppose a bond was purchased for $8000
on November 1, 2004.
Example 3, cont’d
a) What was the value of the bond if it
was redeemed on November 1,
2005?
b) What was the value of the bond if it
was redeemed on July 10, 2004?
Example 3, cont’d
a) Solution: If the bond was redeemed
on November 1, 2005, it had been
held for 1 year.
• The future value of the bond after 1 year
is:
F  P 1  rt 
 8000 1  0.0125 1  $8100
Example 3, cont’d
b) Solution: If the bond was redeemed
on July 10, 2004, it had been held for
7 full months.
• The future value of the bond after 7/12 of
a year is:
F  P 1  rt 

 8000 1  0.0125  7
 $8058.33

12
Example 4
• What is the simple interest on a $500
loan at 12% from June 6 through
October 12 in a non-leap year?
Example 4, cont’d
• Solution: The time must be converted
to years.
• (30 - 6) + 31 + 31 + 30 + 12 = 128 days
• A non-leap year has 365 days.
• The interest will be:

I  Prt   500  0.12  128
 $21.04
365

Question:
What is the simple interest on a
$2000 loan at 8% from March 19th
through August 15th in a leap year?
a. $65.32
b. $653.15
c. $651.37
d. $65.14
Ordinary Interest
• Ordinary interest simplifies calculations
by using 2 conventions:
• Each month is assumed to have 30 days.
• Each year is assumed to have 360 days.
Example 5
• A homeowner owes $190,000 on a
4.8% home loan with an interest-only
option.
• An interest-only option allows the
borrower to pay only the ordinary interest,
not the principal, for the first year.
• What is the monthly payment for the
first year?
Example 5, cont’d
• Solution: Use the simple interest
formula, measuring time according to
ordinary interest conventions.
• The monthly payments are:
1
I  Prt  190000  0.048   
 12 
 $760
Compound Interest
• Reinvesting the interest, called
compounding, makes the balance grow
faster.
• To calculate compound interest, you
need the same information as for
simple interest plus you need to know
how often the interest is compounded.
Example 6
• Suppose a principal of $1000 is
invested at 6% interest per year and
the interest is compounded annually.
• Find the balance in the account after 3
years.
Example 6, cont’d
• Solution: We must calculate the
interest at the end of each year and
then add that interest to the principal.
• After 1 year:
• The interest is: I  1000 0.061  $60
• The new balance is $1060.00
• We could also have used the future value
formula.
Example 6, cont’d
• Solution, cont’d:
• After 2 years the new balance is:
F  P 1  rt 
 1060 1  0.06 1  $1123.60
• After 3 years the new balance is:
F  P 1  rt 
 1123.6 1  0.06 1  $1191.02
Example 6, cont’d
• Solution, cont’d: The interest earned each
year increases because of the increasing
principal.
Example 6, cont’d
• Solution, cont’d: The following table shows
the pattern in the calculations for
subsequent years.
Compound Interest, cont’d
• If P represents the principal, r the
annual interest rate expressed as a
decimal, m the number of equal
compounding periods per year, and
t the time in years, then the future value
of the account is:
mt
r

F  P 1  
 m
Example 7
•
Find the future value of each account at the
end of 3 years if the initial balance is $2457
and the account earns:
a) 4.5% simple interest.
b) 4.5% compounded annually.
c) 4.5% compounded every 4 months.
d) 4.5% compounded monthly.
e) 4.5% compounded daily.
Example 7, cont’d
• Solution: We have P = 2457 and t = 3.
a) We have r = 0.045 with simple interest.
•
F  2457 1  0.045  3  $2788.70
b) We have r = 0.045 compounded
annually.
•
13
F  P 1  r m   2457 1  0.045 1
mt
 $2803.85
Example 7, cont’d
• Solution, cont’d: We have r = 0.045
c) Compounded every 4 months:
•
 33
F  2457 1  0.045 3
 $2809.31
d) Compounded monthly:
•
123
 $2811.42
 3653
 $2812.10
F  2457 1  0.045 12
e) Compounded daily:
•
F  2457 1  0.045 365
Example 7, cont’d
• Solution, cont’d: The results are
summarized below.
Example 8
• Find the future value of each account
at the end of 100 years if the initial
balance is $1000 and the account
earns:
a) 7.5% simple interest.
b) 7.5% compounded annually.
Example 8, cont’d
• Solution: We have P = 1000, t = 100,
and r = 0.075.
a) With simple interest, the future value is:
•
F  1000 1  0.075 100  $8500
b) With annually compounded interest, the
future value is:
•

F  1000 1  0.075
1

 
1100
 $1,383,000
Question:
If you loan your friend $100 at 3%
interest compounded daily, how
much will she owe you at the end of
1 year?
a. $103.05
b. $103.00
c. $103.33
d. $103.02
Interest, cont’d
• Simple interest exhibits arithmetic
growth.
• The same amount is added each year.
• Compound interest exhibits
exponential growth.
• The same amount is multiplied each
year.
Example 9
• How much money should be
invested at 4% interest compounded
monthly in order to have $25,000
eighteen years later?
Example 9, cont’d
• Solution: We know F = 25,000,
r = 0.04, m = 12 and t = 18.
• Solve for P, the necessary principal.
•
25000  P 1  0.04 12 
P
25000
1  0.04 12 
 216 
1218
 $12,183.39
Effective Annual Rate
• The effective annual rate (EAR) or annual
percentage yield (APY) is the simple interest
that would give the same result in 1 year.
• The stated rate is called the nominal rate.
• This provides a basis for comparing different
savings plans.
• APY is used only for savings accounts.
• EAR is used in any context.
Example 10
• Find the effective annual rate by
computing what happens to $100
over 1 year at 12% annual interest
compounded every 3 months.
Example 10, cont’d
• Solution: The balance at the end of
 41
1 year is: F  100 1  0.12 4   $112.55
• Since the account increased by
$12.55 in 1 year, the EAR is
12.55%.
• F  100 1  0.1255 1  $112.55
Effective Annual Rate, cont’d
• If r represents the annual interest rate
expressed as a decimal and m is the number
of equal compounding periods per year, then
the effective annual rate is:
m
r

EAR  1    1
 m
• Note: The same formula is used for APY.
Effective Annual Rate, cont’d
Example 11
• A bank offers a savings account with an
interest rate of 0.25% compounded daily,
with a minimum deposit of $100.
• The same bank offers an 18-month CD with
an interest rate of 2.13% compounded
monthly, with deposits less than $10,000.
• Find the effective annual rate for each
option.
Example 11, cont’d
• Solution, cont’d: The effective annual rate for
the savings account is:
EAR  1  r m   1
m
 1  0.0025 365
365
1
 0.00250312
• The EAR for the account is about 0.2503%.
Example 11, cont’d
• Solution: The effective annual rate for the
CD is:
EAR  1  r m   1
m
 1  0.0213 12   1
12
 0.02150918
• The EAR for the CD is about 2.1509%.
13.1 Initial Problem Solution
• Suppose you discover you are the only direct
descendant of a man who loaned the
Continental Congress $1000 in 1777 and was
never repaid.
• Using an interest rate of 6% and a
compounding period of 3 months, how much
should you demand from the government?
Initial Problem Solution, cont’d
• We have P = $1000, r = 0.06, m = 4
and t = 223.
• The value of your ancestor’s loan is:
F  1000 1  0.06 4 
 $585, 746, 479
 4223
Section 13.2
Loans
• Goals
• Study amortized loans
• Use an amortization table
• Use the amortization formula
• Study rent-to-own
13.2 Initial Problem
• Home mortgage rates have decreased and
Howard plans to refinance his home.
• He will refinance $85,000 at either 5.25%
for 15 years or 5.875% for 30 years.
• In each case, what is his monthly payment
and how much interest will he pay?
• The solution will be given at the end of the section.
Simple Interest Loans
• The interest on a simple interest loan
is simple interest on the amount
currently owed.
• The simple interest each month is
called the finance charge.
• Finance charges are calculated using an
average daily balance and a daily interest
rate.
Example 1
Example 1, cont’d
• Assuming the billing period is June 10
through July 9, determine each of the
following:
a) The average daily balance
b) The daily percentage rate
c) The finance charge
d) The new balance
Example 1, cont’d
a) Solution: The daily balances are shown
below.
Example 1, cont’d
a) Solution, cont’d: The average daily
balance is:
2  287.84   6  333.44   4 183.44   11 203.44   7  281.94 
2  6  4  11  7
7521.50

 $250.72
30
Example 1, cont’d
b) Solution: The daily percentage rate is:
21%
 0.057534%
365
c) Solution: The finance charge is the
simple interest on the average daily
balance at the daily rate:
I   250.72 0.21 36530   $4.33
Example 1, cont’d
d) Solution: The new balance is the sum
of the previous balance, any new
charges, and the finance charge,
minus any payments:
287.84 + 144.10 + 4.33 – 150.00 =
286.27
• The new balance is $286.27.
Example 1, cont’d
Amortized Loans
•
Amortized loans are simple interest loans
with equal periodic payments over the
length of the loan.
•
The important variables for an amortized
loan are:
•
Principal
•
Interest rate
•
Term (length) of the loan
•
Monthly payment
Amortized Loans, cont’d
• Each payment includes the interest
due since the last payment and an
amount paid toward the balance.
• The amount paid each month is
constant, but the split between principal
and interest varies.
• The amount of the last payment may be
slightly more or less than usual.
Question:
When paying off an amortized loan,
the percent of the monthly payment
going toward the interest will:
a. increase as time goes by.
b. decrease as time goes by.
c. remain the same every month.
Example 2
• Chart the history of an amortized loan
of $1000 for 3 months at 12% interest
with monthly payments of $340.
Example 2, cont’d
• Solution: Monthly payment #1:
• The interest owed is
 
I  1000  0.12  1
 $10
12
• The payment toward the principal is
$340 - $10 = $330
• The new balance is $1000 - $330 = $670.
Example 2, cont’d
• Solution, cont’d: Monthly payment #2:
• The interest owed is
 
I   670 0.12  1
 $6.70
12
• The payment toward the principal is
$340 - $6.70 = $333.30
• The new balance is $670 - $333.30 =
$336.70
Example 2, cont’d
• Solution, cont’d: Monthly payment #3:
• The interest owed is
 
I   336.70  0.12  1
 $3.37
12
• The remaining balance plus the interest
is: $336.70 + $3.37 = $340.07.
• The third and final payment is $340.07.
Example 2, cont’d
• Solution, cont’d: The amortization
schedule for this loan is shown below.
Monthly Payments
• The monthly payments for an
amortized loan can be determined in
one of three ways:
• Using an amortization table.
• Using a formula.
• Using financial software or an online
calculator.
Amortization Table
• An amortization table gives precalculated monthly payments for
common loan rates and terms.
• An example of a table is shown on the
next slide.
Amortization Table, cont’d
Example 3
• A couple is buying a vehicle for
$20,995.
• They pay $7000 down and finance the
remainder at an annual interest rate of
4.5% for 48 months.
• Use the amortization table to
determine their monthly payment.
Example 3, cont’d
• Solution: The amount being financed is
$20,995 – $7000 = $13,995.
• In the table, find the row corresponding
to 4.5% and the column corresponding
to 4 years.
• This entry is highlighted on the next slide.
Example 3, cont’d
Example 3, cont’d
• Solution, cont’d: The value 22.803486
indicates the couple will pay
$22.803486 for each $1000 they
borrowed.
•
 22.80348613.995  319.13479
• They will pay $319.14 per month.
Example 4
• The couple in the previous example
borrowed $13,995 to buy a car and will
pay the loan over 4 years.
• If their payments are $340.02, what
interest rate are they being charged?
Example 4, cont’d
• Solution: They are paying $340.02 a
month for $13,995, or approximately
$24.295820 per $1000.
• Look at the amortization table to see to
which interest rate this payment amount
corresponds.
Example 4, cont’d
Example 4, cont’d
• Solution, cont’d: The interest rate for
the amortized loan, according to the
table, is approximately 7.75%.
Amortization Formula
• If P is the amount of the loan, r is the annual
interest rate expressed as a decimal, and t
is the length of the loan in years, then the
monthly payment for an amortized loan is:
12 t
r 
r 
P   1  
12  12 
PMT 
12 t
r 

1    1
 12 
Example 5
• Use the monthly payment formula to
determine the monthly car payment for
a loan of $13,995 at 4.5% annual
interest for 48 months.
Example 5, cont’d
• Solution: We have P = 13,995, r = 0.045, and
t = 4.
124
 0.045   0.045 
13995  
  1 

12  
12 

PMT 
124
 0.045 
1 
 1
12 

 $319.14
Example 6
• Suppose a student accumulated $7800
in student loans which she must now
pay over 10 years.
• Determine her monthly payment
amount using an interest rate of 3.37%
Example 6, cont’d
• Solution: Solution: We have P = 7800,
r = 0.0337, and t = 10.
1210
 0.0337   0.0337 
 7800   
  1 

12  
12 

PMT 
1210
 0.0337 
1 
 1
12 

 $76.66
Question:
Use the amortization formula to
determine the amount of the
monthly payment for a loan of
$30,000 at 5% for 3 years.
a. $527.38
b. $124.00
c. $722.46
d. $899.13
Example 7
• Suppose you can afford car payments
of $250 per month.
• If a 3-year loan at 4% interest is
available, how much can you finance?
Example 7, cont’d
• Solution: We know PMT = 250,
r = 0.04, and t = 3.
123
 0.04   0.04 
P 
  1 

12  
12 

250 
123
 0.04 
1 
 1
12 

Example 7, cont’d
• Solution, cont’d: Solve for P.
  0.04 

250  1 
 1



12



P
 $8468
36
 0.04   0.04 

  1 

12 
 12  
36
• You can borrow $8468.
Rent-to-Own
• In a rent-to-own transaction, you rent
the item at a monthly rate, but after a
contracted number of payments, the
item becomes yours.
• The difference between the retail price
of the item and the total of your
monthly payments is the interest.
Example 8
•
Suppose you can rent-to-own a $500
television for 24 monthly payments of $30.
a) What amount of interest would you pay for
the rent-to-own television?
b) What annual rate of simple interest on
$500 for 24 months yields the same
amount of interest found in part (a)?
Example 8, cont’d
a) Solution:
• The total of your monthly payments will
be 24($30) = $720.
• You will pay $720 - $500 = $220 in
interest over the 2 years.
Example 8, cont’d
b) Solution:
• Solve the simple interest formula for r:
I
r
Pt
• The equivalent simple interest rate is:
220
r
 0.22  22%
 500 2
13.2 Initial Problem Solution
• Home mortgage rates have decreased
and Howard plans to refinance his
home. He will refinance $85,000 at
either 5.25% for 15 years or 5.875%
for 30 years.
• In each case, what is his monthly
payment and how much interest will
he pay?
Initial Problem Solution, cont’d
• The 15-year loan has an interest rate of
5.25%.
• According to the amortization table, the
monthly payment per $1000 would be
$8.038777.
• Under this loan, Howard’s monthly
payment would be $8.038777(85)
which is approximately $683.30.
Initial Problem Solution, cont’d
• For the 15-year loan, Howard will pay a
total of ($683.30)(12)(15) = $122,994.
• The amount spent on interest is
$122,994 - $85,000 = $37,994.
Initial Problem Solution, cont’d
• The 30-year loan has an interest rate of
5.875%, which is not found in the table.
• Using the amortization formula, we find
a monthly payment amount of $502.81.
Initial Problem Solution, cont’d
• For the 30-year loan, Howard will pay a
total of ($502.81)(12)(30) =
$181,011.60.
• The amount spent on interest is
$181,011.60 - $85,000 = $96,011.60
Section 13.3
Buying a House
• Goals
• Study affordability guidelines
• Study mortgages
• Interest rates and closing costs
• Annual percentage rates
• Down payments
13.3 Initial Problem
• Suppose you have saved $15,000 toward a down
payment on a house and your total yearly income is
$45,000. What is the most you could afford to pay
for a house?
• Assume you pay 0.5% of the value for insurance,
you pay 1.5% of the value for taxes, your closing
costs will be $2000, and you can obtain a fixed-rate
mortgage for 30 years at 6% interest.
• The solution will be given at the end of the section.
Affordability Guidelines
• The 2 most common guidelines for
buying a house are:
• The maximum house price is 3 times your
annual gross income.
• Your maximum monthly housing expenses
should be 25% of your gross monthly
income.
Example 1
• If your annual gross income is $60,000,
what do the guidelines tell you about
purchase price and monthly expenses
for your potential home purchase?
Example 1, cont’d
• Solution:
• The purchase price should be no more
than 3($60,000) = $180,000.
• The monthly expenses for mortgage
payments, property taxes, and
homeowner’s insurance should be no
more0.25
than1 12 60000   $1250.
Affordability Guidelines, cont’d
• Some lenders allow monthly expenses
up to 38% of the buyer’s monthly
income.
• We call the 25% level the low maximum
monthly housing expense estimate.
• We call the 38% level the high maximum
monthly housing expense estimate.
Example 2
• Suppose Andrew and Barbara both
have jobs, each earning $24,000 a year,
and they have no debts.
• What are the low and high estimates of
how much they can afford to pay for
monthly housing expenses?
Example 2, cont’d
• Solution: The low estimate is 25% of the
total monthly income.
•
 48, 000 
0.25 
  $1000
 12 
• The high estimate is 38% of the total
monthly income.
•
 48, 000 
0.38 
  $1520
 12 
Question:
If you make $28,000 per year, can
you afford to buy a house for
$83,000 with monthly housing
expenses of $650?
a. Yes, according to the low
maximum guideline.
b. Yes, according to the high
maximum guideline.
c. No
Mortgages
• A mortgage is a loan that is guaranteed
by real estate.
• The interest rate of a fixed-rate
mortgage is set for the entire term.
• The interest rate of an adjustable-rate
mortgage (ARM) can change.
Mortgages, cont’d
• The finalizing of a house purchase is called
the closing.
• Points are fees paid to the lender at the time
of the closing.
• Loan origination fees
• Discount charges
• Points and any other expenses paid at the
time of the closing are called closing costs.
Example 3
• Suppose you will borrow $80,000 for a
home at 6.5% interest on a 30-year
fixed-rate mortgage.
• The loan involves a one-point loan
origination fee and a one-point discount
charge. What are your added costs?
• Note: One point is equal to 1 percent of the
loan amount.
Example 3, cont’d
• Solution: Each fee will cost you 1% of
$80,000, or $800.
• Your total added fees are $1600.
Annual Percentage Rate
• The annual percentage rate (APR)
helps borrowers compare the true cost
of a loan.
• The APR includes the annual interest
rate, any points, and any other loan
processing or private mortgage
insurance fees.
Example 4
• Suppose you plan to borrow $80,000 for
a home at 6.5% interest on a 30-year
fixed-rate mortgage.
• Determine the loan’s APR if there is a 1point loan origination fee and a 1-point
discount charge.
Example 4, cont’d
• Solution: The loan is $80,000 plus
points totaling $1600, for a total of
$81,600.
• According to the amortization table, the
total monthly payment will be
81.6($6.320680) or about $515.77, if
the fees were paid monthly instead of at
the time of closing.
Example 4, cont’d
• Solution, cont’d: The interest rate that
corresponds to a loan of $80,000 at
6.5% for 30 years with a monthly
payment of $515.77 is found:
• Divide $515.77 by 80 to find the amount
per $1000, which is $6.44713.
• In the table, this value corresponds to a
rate between 6.5% and 6.75%.
Example 4, cont’d
• Solution, cont’d: Use linear interpolation to
determine the APR.
• 6.44713  6.320680
100%  76.5%
6.485981  6.320680
• The payment is 76.5% of the way from the
payment for 6.5% to the payment for 6.75%.
•
6.5  07.65  6.75  6.5  6.691%
• The APR is about 6.691%
APR, cont’d
• Note that the APR is always greater than or
equal to the stated annual interest rate.
Down Payment
• A down payment on a house is the amount of
cash the buyer pays at closing, minus any
points and fees.
• Traditionally a down payment is 20% of the
value, but can be lower.
• The loan to value ratio of a mortgage is the
percent of the home’s value that is not paid for
by the down payment.
• For example, a down payment of 20% results in
an 80% loan to value ratio.
Down Payment, cont’d
• Another guideline for the maximum
price you can afford when buying a
home is to find your maximum price
by dividing the amount you have for
a down payment by the percent of
the value of the house that amount
represents.
Example 5
• If you have $25,000 for a down
payment, what is the highest-priced
home you can afford if a 20% down
payment is required?
Example 5, cont’d
• Solution: The maximum price you can
afford to pay is your down payment
amount divided by 20%.
25,
000
•
 $125, 000
0.20
• The most expensive house you can afford
is one that is selling for $125,000.
Question:
Suppose you have $5000 for a down
payment on a house that is selling
for $82,000. If the lender requires a
5% down payment for first-time
homebuyers, is this house within
your price range?
a. Yes
b. No
13.3 Initial Problem Solution
• Suppose you have saved $15,000 toward a
down payment on a house and your total
yearly income is $45,000. What is the most
you could afford to pay for a house?
• Assume you pay 0.5% of the value for
insurance, you pay 1.5% of the value for
taxes, your closing costs will be $2000, and
you can obtain a fixed-rate mortgage for 30
years at 6% interest.
Initial Problem Solution, cont’d
• Your total income is $45,000
• You have $15,000 saved for the
purchase
• $2000 will be used for closing costs.
• This leaves $13,000 for a down
payment.
Initial Problem Solution, cont’d
• The first affordability guideline says you
can spend at most 3($45,000) = $135,000
on a house.
• Next, consider your monthly expenses:
• You would be financing $122,000 at 6% for 30
years.
• The monthly mortgage payments, from the
table, would be 122($5.995505) = $732.
Initial Problem Solution, cont’d
• The insurance and taxes are 2% of the
home’s value annually.
• This adds $225 to the monthly expenses, for a
total monthly expense of $957.
• According to the second affordability
guideline you can only afford monthly
expenses of at most $938.
• The monthly expenses for this house are
above your maximum. You cannot afford it.
Initial Problem Solution, cont’d
• A house priced $135,000 is slightly
out of your reach, so your options are:
• Wait for interest rates to fall.
• Increase your income.
• Come up with a larger down payment.
• Choose a less expensive house.