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A Mathematical View of Our World. 1 st 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

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a mathematical view of our world

A Mathematical View of Our World

1st ed.

Parks, Musser, Trimpe, Maurer, and Maurer

chapter 13

Chapter 13

Consumer Mathematics:

Buying and Saving

section 13 1 simple and compound interest
Section 13.1Simple 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
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
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:
example 1
Example 1
  • Find the interest on a loan of $100 at 6% simple interest for time periods of:
    • 1 year
    • 2 years
    • 2.5 years
example 1 cont d
Example 1, cont’d
  • Solution: We have P = 100 and r = 0.06.
    • For t = 1 year, the calculation is:
example 1 cont d1
Example 1, cont’d
  • Solution, cont’d: We have P = 100 and r = 0.06.
    • For t = 2 years, the calculation is:
    • For t = 2.5 years, the calculation is:
future value
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:
example 2
Example 2
  • Find the future value of a loan of $400 at 7% simple interest for 3 years.
example 2 cont d
Example 2, cont’d
  • Solution: Use the future value formula with P = 400, r = 0.07, and t = 3.
example 3
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
Example 3, cont’d
  • What was the value of the bond if it was redeemed on November 1, 2005?
  • What was the value of the bond if it was redeemed on July 10, 2004?
example 3 cont d1
Example 3, cont’d
  • 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:
example 3 cont d2
Example 3, cont’d
  • 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:
example 4
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
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:
question
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
  • 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
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
Example 5, cont’d
  • Solution: Use the simple interest formula, measuring time according to ordinary interest conventions.
  • The monthly payments are:
compound interest
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
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
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:
    • The new balance is $1060.00
      • We could also have used the future value formula.
example 6 cont d1
Example 6, cont’d
  • Solution, cont’d:
  • After 2 years the new balance is:
  • After 3 years the new balance is:
example 6 cont d2
Example 6, cont’d
  • Solution, cont’d: The interest earned each year increases because of the increasing principal.
example 6 cont d3
Example 6, cont’d
  • Solution, cont’d: The following table shows the pattern in the calculations for subsequent years.
compound interest cont d
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:
example 7
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:
    • 4.5% simple interest.
    • 4.5% compounded annually.
    • 4.5% compounded every 4 months.
    • 4.5% compounded monthly.
    • 4.5% compounded daily.
example 7 cont d
Example 7, cont’d
  • Solution: We have P = 2457 and t = 3.
    • We have r = 0.045 with simple interest.
    • We have r = 0.045 compounded annually.
example 7 cont d1
Example 7, cont’d
  • Solution, cont’d: We have r = 0.045
    • Compounded every 4 months:
    • Compounded monthly:
    • Compounded daily:
example 7 cont d2
Example 7, cont’d
  • Solution, cont’d: The results are summarized below.
example 8
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:
    • 7.5% simple interest.
    • 7.5% compounded annually.
example 8 cont d
Example 8, cont’d
  • Solution: We have P = 1000, t = 100, and r = 0.075.
    • With simple interest, the future value is:
    • With annually compounded interest, the future value is:
question1
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
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
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
Example 9, cont’d
  • Solution: We know F = 25,000,

r = 0.04, m = 12 and t = 18.

  • Solve for P, the necessary principal.
effective annual rate
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
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
Example 10, cont’d
  • Solution: The balance at the end of 1 year is:
  • Since the account increased by $12.55 in 1 year, the EAR is 12.55%.
effective annual rate cont d
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:
  • Note: The same formula is used for APY.
example 11
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
Example 11, cont’d
  • Solution, cont’d: The effective annual rate for the savings account is:
    • The EAR for the account is about 0.2503%.
example 11 cont d1
Example 11, cont’d
  • Solution: The effective annual rate for the CD is:
    • The EAR for the CD is about 2.1509%.
13 1 initial problem solution
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
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:
section 13 2 loans
Section 13.2Loans
  • Goals
    • Study amortized loans
      • Use an amortization table
      • Use the amortization formula
    • Study rent-to-own
13 2 initial problem
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
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 cont d2
Example 1, cont’d
  • Assuming the billing period is June 10 through July 9, determine each of the following:
  • The average daily balance
  • The daily percentage rate
  • The finance charge
  • The new balance
example 1 cont d3
Example 1, cont’d
  • Solution: The daily balances are shown below.
example 1 cont d4
Example 1, cont’d
  • Solution, cont’d: The average daily balance is:
example 1 cont d5
Example 1, cont’d
  • Solution: The daily percentage rate is:
  • Solution: The finance charge is the simple interest on the average daily balance at the daily rate:
example 1 cont d6
Example 1, cont’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.
amortized loans
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
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.
question2
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 21
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 d1
Example 2, cont’d
  • Solution: Monthly payment #1:
    • The interest owed is
    • The payment toward the principal is

$340 - $10 = $330

    • The new balance is $1000 - $330 = $670.
example 2 cont d2
Example 2, cont’d
  • Solution, cont’d: Monthly payment #2:
    • The interest owed is
    • The payment toward the principal is

$340 - $6.70 = $333.30

    • The new balance is $670 - $333.30 = $336.70
example 2 cont d3
Example 2, cont’d
  • Solution, cont’d: Monthly payment #3:
    • The interest owed is
    • The remaining balance plus the interest is: $336.70 + $3.37 = $340.07.
    • The third and final payment is $340.07.
example 2 cont d4
Example 2, cont’d
  • Solution, cont’d: The amortization schedule for this loan is shown below.
monthly payments
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
Amortization Table
  • An amortization table gives pre-calculated monthly payments for common loan rates and terms.
  • An example of a table is shown on the next slide.
example 31
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 d3
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 d5
Example 3, cont’d
  • Solution, cont’d: The value 22.803486 indicates the couple will pay $22.803486 for each $1000 they borrowed.
  • They will pay $319.14 per month.
example 41
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 d1
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 d3
Example 4, cont’d
  • Solution, cont’d: The interest rate for the amortized loan, according to the table, is approximately 7.75%.
amortization formula
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:
example 51
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 d1
Example 5, cont’d
  • Solution: We have P = 13,995, r = 0.045, and

t = 4.

example 61
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 d4
Example 6, cont’d
  • Solution: Solution: We have P = 7800,

r = 0.0337, and t = 10.

question3
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 71
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 d3
Example 7, cont’d
  • Solution: We know PMT = 250,

r = 0.04, and t = 3.

example 7 cont d4
Example 7, cont’d
  • Solution, cont’d: Solve for P.
  • You can borrow $8468.
rent to own
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 81
Example 8
  • Suppose you can rent-to-own a $500 television for 24 monthly payments of $30.
  • What amount of interest would you pay for the rent-to-own television?
  • What annual rate of simple interest on $500 for 24 months yields the same amount of interest found in part (a)?
example 8 cont d1
Example 8, cont’d
  • 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 d2
Example 8, cont’d
  • Solution:
    • Solve the simple interest formula for r:
    • The equivalent simple interest rate is:
13 2 initial problem solution
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 d1
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 d2
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 d3
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 d4
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
Section 13.3Buying a House
  • Goals
    • Study affordability guidelines
    • Study mortgages
      • Interest rates and closing costs
      • Annual percentage rates
      • Down payments
13 3 initial problem
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
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 13
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 d8
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 more than
affordability guidelines cont d
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 22
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 d5
Example 2, cont’d
  • Solution: The low estimate is 25% of the total monthly income.
  • The high estimate is 38% of the total monthly income.
question4
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
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
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 32
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 d6
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
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 42
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 1-point loan origination fee and a 1-point discount charge.
example 4 cont d4
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 d5
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 d6
Example 4, cont’d
  • Solution, cont’d: Use linear interpolation to determine the APR.
    • The payment is 76.5% of the way from the payment for 6.5% to the payment for 6.75%.
      • The APR is about 6.691%
apr cont d
APR, cont’d
  • Note that the APR is always greater than or equal to the stated annual interest rate.
down payment
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
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 52
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 d2
Example 5, cont’d
  • Solution: The maximum price you can afford to pay is your down payment amount divided by 20%.
    • The most expensive house you can afford is one that is selling for $125,000.
question5
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
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 d5
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 d6
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 d7
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 d8
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.
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