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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

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## A Mathematical View of Our World

1st ed.

Parks, Musser, Trimpe, Maurer, and Maurer

## Chapter 13

Consumer Mathematics:

### 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

• 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:

### 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

• Solution: We have P = 100 and r = 0.06.

• For t = 1 year, the calculation is:

### 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

• 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

• 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.

### 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

• 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’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’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

• 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:

### 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:

### 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:

• 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:

• After 3 years the new balance is:

### 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:

### 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

• 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’d

• Solution, cont’d: We have r = 0.045

• Compounded every 4 months:

• Compounded monthly:

• Compounded daily:

### 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:

• 7.5% simple interest.

• 7.5% compounded annually.

### 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:

### 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.

### 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 1 year is:

• Since the account increased by \$12.55 in 1 year, the EAR is 12.55%.

### 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:

• Note: The same formula is used for APY.

### 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:

• The EAR for the account is about 0.2503%.

### 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

• 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:

### Section 13.2Loans

• 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, 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’d

• Solution: The daily balances are shown below.

### Example 1, cont’d

• Solution, cont’d: The average daily balance is:

### 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’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 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

• 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

• 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

• 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 pre-calculated monthly payments for common loan rates and terms.

• An example of a table is shown on the next slide.

### 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

• 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 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

• 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:

### 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.

### 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.

### 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.

### Example 7, cont’d

• Solution, cont’d: Solve for P.

• 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.

• 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’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’d

• Solution:

• Solve the simple interest formula for r:

• The equivalent simple interest rate is:

### 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

• 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 more than

### 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.

• The high estimate is 38% of the total monthly income.

### 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.

### 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 1-point 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.

• 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

• 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%.

• 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. Yesb. 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.