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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.


Effective annual rate cont d1

Effective Annual Rate, cont’d


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 12

Example 1


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.


Example 1 cont d7

Example 1, cont’d


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.


Amortization table cont d

Amortization Table, cont’d


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 d4

Example 3, cont’d


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 d2

Example 4, cont’d


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. Yesb. 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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