© the mcgraw-hill companies, inc., 2008 mcgraw-hill/irwin 1-1 chapter 1 simple interest startexit

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© The McGraw-Hill Companies, Inc., 2008McGraw-Hill/Irwin

Chapter 1

Simple Interest

START EXIT

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Chapter Outline

1.1 Simple Interest and the Time Value of Money

1.2 The Term of a Loan

1.3 Determining Principal, Interest Rates, and Time

1.4 Promissory Notes

1.5 Non-Annual Interest Rates

Chapter Summary

Chapter Exercises

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1.1 Simple Interest and Time Value of Money

Definition 1.1.1

Interest is what a borrower pays a lender for the temporary use of the lender’s money.

Definition 1.1.2

Interest is the “rent” that a borrower pays a lender to use the lender’s money.

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Example 1.1.1• Problem

– Sam loans Danielle $500 for 100 days. Danielle agrees to pay her $80 interest for the loan. How much will Danielle pay Sam in total?

• Solution– Interest is added onto the amount borrowed. – $500 + $80 = $580 – Therefore, Danielle will pay Sam a total of $580 at the

end of the 100 days.

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– Tom borrows $200 from Larry, agreeing to repay the loan by giving Larry $250 in 1 year. How much interest will Tom pay?

• Solution– The interest is the difference between what Tom

borrows and what he repays. – $250 -- $200 = $50– So Tom will pay a total of $50 in interest.

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Definition 1.1.3

The principal of a loan is the amount borrowed.

Definition 1.1.4

A debtor is someone who owes someone else money.

A creditor is someone to whom money is owed.

Definition 1.1.5

The amount of time for which a loan is made is called its term.

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Example• Problem

– Sam loans Danielle $500 for 100 days. Danielle agrees to pay her $80 interest for the loan. What is the principal of this loan?

• Solution– The principal of a loan is the amount

borrowed. – Therefore, the principal of this loan is $500.

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Example• Problem

– Sam loans Danielle $500 for 100 days. Danielle agrees to pay her $80 interest for the loan. Who is a debtor and who is a creditor?

• Solution– Danielle is Sam’s debtor because she owes money.– Sam is Danielle’s creditor because Danielle owes her

money.

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Example• Problem

– Sam loans Danielle $500 for 100 days. Danielle agrees to pay her $80 interest for the loan. What is the term of this loan?

• Solution– The term of a loan is the amount of time for

which a loan is made.– Therefore, the term of this loan is 100 days.

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– Let’s consider Example 1.1.2 again. Tom borrows $200 from Larry, agreeing to repay the loan by giving Larry $250 in 1 year. Therefore, Tom will pay $50 in interest.

– Suppose that now Tom wants to borrow $1,000 from Larry for 1 year. How much interest would Larry charge him?

• Solution– The interest Larry was charging Tom was ¼ of the amount

borrowed or 25%.– Therefore, Larry would charge $250 or 25% of $1,000.– Of course, this calculation was easy. What if we had to work

with less friendly numbers?

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Working with Percents• When we talk about percents, we usually are

taking a percent (or portion) of something.• So, to find 25% of $1,000, we found a portion of

$1,000.• We do this by multiplying percent rate by the

amount: 25% x $1,000.• However, first we need to convert a percent to a

decimal: 25% = 0.25 (25/100 or move decimal point two places to the left)

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1.1 Simple Interest and Time Value of MoneyExample 1.1.4• Problem

– Convert 30% to a decimal.• Solution

– By dividing: 30% = 30/100 = 0.30– By moving the decimal point: 30% = 0.30


– Convert 18.25% to a decimal.• Solution

– 18.25% = 0.1825Example 1.1.6• Problem

– Convert 5.79% to a decimal• Solution

– Here, simply moving the decimal point two places to the left won’t work because we don’t have enough numbers. We will deal with this problem by placing a 0 in front of 5.

– 5.79% = 0.0579

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– Let’s rework Example 1.1.3, this time by converting the interest rate percent to a decimal and using it.

– Suppose that now Tom wants to borrow $1,000 from Larry for 1 year. How much interest would Larry charge him?

• Solution– 25% = 0.25– Interest = Principal x Interest Rate as a decimal– Interest = $1,000 x 0.25– Interest = $250

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– Suppose Bruce loans Jamal $5,314.57 for 1 year. Jamal agrees to pay 8.72% interest for the year. How much will he pay Bruce when the year is up?

• Solution– Interest = Principal x Interest Rate as a decimal– Interest = $5,314.57 x 8.72%– Interest = $5,314.57 x 0.0872– Interest = $463.43 (actually, the answer is 463.4305;

however, money is measured in dollars and cents so we should round the final answer to two decimal places)

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Mixed Number and Fractional Percents• It’s not unusual for interest rates to be

expressed as mixed numbers or fractions, such as 5 ¾%.

• Some of these, such as 4 ½%, are easily converted to decimal format – 4.5%.

• However, to convert a more difficult fraction, such as 9 5/8%,– divide 5/8 to obtain 0.625– replace the fraction in the mixed number with its

decimal equivalent to obtain 9.625%– Move the decimal point two places to obtain 0.09625

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Example 1.1.9

• Problem– Rewrite 7 13/16% as a decimal.

• Solution– 13/16 = 0.8125– 7 13/16% = 7.8125% = 0.078125

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The Impact of Time• Let’s return again to Tom and Larry. Suppose

Tom is paying the loan back in 2 years instead of 1 year. Could he reasonably expect to still pay the same amount of interest, even though the loan is now taken out for twice as long?

• Of course, not!• If the loan is for twice as long, it seems

reasonable that Tom would pay twice as much interest, right?

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FORMULA 1.1

The Simple Interest Formula

I = PRTwhere

I represents the amount of simple INTEREST for a loan

P represents the amount of money borrowed or PRINCIPAL

T represents the TERM of the loan

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– Suppose that Raeshawn loans Dianne $4,200 at a simple interest rate of 8 ½% for 3 years. How much interest will Dianne pay?

• Solution– Interest = Principal x Interest Rate x Time– Interest = $4,200 x 8 ½% x 3– Interest = $4,200 x 8.5% x 3– Interest = $4,200 x 0.085 x 3– Interest = $1,071

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– Heather borrows $18,500 at 5 1/8% simple interest for 2 years. How much interest will she pay?

• Solution– Interest = P x R x T– Interest = $18,500 x 5 1/8% x 2– Interest = $18,500 x 0.05875 x 2– Interest = $2,173.75

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Loans in Disguise• Sometimes interest is paid in situations we might not

normally think of as loans.• For example, if you deposit money in a savings account,

you expect to be paid interest.• Checking and savings accounts are known as demand

accounts because you can withdraw money at any time you want.

• Another common type of account is a certificate of deposit or CD.

• CDs require customers to keep money on deposit for a fixed period of time.

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– Jake deposited $2,318.29 into a 2-year CD paying 5.17% simple interest per annum. How much will his account be worth at the end of the term?

• Solution– Interest = P x R x T– Interest = $2,318.29 x 5.17% x 2– Interest = $2,318.29 x 0.0517 x 2– Interest = $239.71– Total Value = Principal + Interest– Total Value = $2,318.29 + $239.71– Total Value = $2,558.00

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Section 1.1 Exercises

Problem 1:

Interest as Difference

Problem 2:

Determining Repayment Amount

Problem 3:

Interest as a Percent (One Year Loans)

Problem 4:

Interest as a Percent (Multiple-Year Loans)

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Problem 1

• Linda borrowed $5,000 and paid back a total of $5,845. How much interest did she pay?

CHECK YOUR ANSWER

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Solution 1

• Linda borrowed $5,000 and paid back a total of $5,845. How much interest did she pay?

• Interest = Total Amount – Principal• Interest = $5,845 -- $5,000• Interest = $845

BACK TO GAME BOARD

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Problem 2

• Salam loaned Andrew $8,900 for 6 months. Andrew agreed to pay $600 interest. How much will Andrew pay back?

CHECK YOUR ANSWER

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Solution 2

• Salam loaned Andrew $8,900 for 6 months. Andrew agreed to pay $600 interest. How much will Andrew pay back?

• Total Amount = Principal + Interest• Interest = $8,900 + $600• Interest = $9,500

BACK TO GAME BOARD

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Problem 3

• Latasha loaned Jose $390 for 1 year at a simple interest rate of 8 3/8%. How much interest will Jose have to pay?

CHECK YOUR ANSWER

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Solution 3

• Latasha loaned Jose $390 for 1 year at a simple interest rate of 8 3/8%. How much interest will Jose have to pay?

• Interest = Principal x Interest Rate• Interest = $390 x 8 3/8%• Interest = $390 x 8.375%• Interest = $390 x 0.08375• Interest = $32.66

BACK TO GAME BOARD

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Problem 4

• Timothy has agreed to loan Katrina $2,350 for 2 years at 9% simple interest. How much will Katrina have to repay in 2 years?

CHECK YOUR ANSWER

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Solution 4

• Timothy has agreed to loan Katrina $2,350 for 2 years at 9% simple interest. How much will Katrina have to repay in 2 years?

• Interest = Principal x Interest Rate x Time• Interest = $2,350 x 9% x 2• Interest = $2,350 x 0.09 x 2• Interest = $423

BACK TO GAME BOARD

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– If Sarai borrows $5,000 for 6 months at 9% simple interest, how much will she need to pay back?

• Solution

– Interest = Principal x Interest Rate x Time

– However, we can’t just plug in T = 6 because time is expressed in months, not years.

– Since a year contains 12 months, 6 months is equal to 6/12 of a year, or ½ of a year.

– Interest = $5,000 x 9% x ½

– Interest = $5,000 x 0.09 x ½

– Interest = $225

– Total Amount = Principal + Interest

– Total Amount = $5,000 + $225

– Total Amount = $5,225

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– Zachary deposited $3,412.59 in a bank account paying 5 ¼% simple interest for 7 months. How much interest did he earn?

• Solution


– Interest = $3,412.59 x 5 ¼% x 7/12• Oops, a problem! 7/12 does not come out evenly, so can it be rounded?

In general, it’s better to use all decimal places given by your calculator.• On most calculators, you can simply enter the whole expression:

– Interest = $104.51

Operation Result

3412.59*.0525*7/12 104.51056875

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– Yvonne deposited $2,719.00 in an account paying 4.6% simple interest for 20 months. Find the total interest she earned.

• Solution


– Interest = $2,719.00 x 4.6% x 20/12


Operation Result

2719*0.046*20/12 208.45677777

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– Nick deposited $1,600 in a credit union CD with a term of 90 days and a simple interest rate of 4.72%. Find the value of his account at the end of its term.

• Solution

– Interest = Principal x Interest Rate x Time• Oops, time is expressed in days, not in months or years!• Since there are 365 days in a year, we divide the term of the loan by 365.

– Interest = $1,600 x 4.72% x 90/365

– Interest = $1,600 x 0.0472 x 90/365


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Example 1.2.5• However, not every year has exactly 365 days since leap years have an

extra day and, therefore, 366 days. Although there are several methods to calculate term of the loan expressed in days, the difference is very small.

• Problem

– Calculate the simple interest due on a 120-day loan of $1,000 at 8.6% simple interest in three different ways: assuming there are 365, 366, or 365.25 days in the year.

• Solution

I = PRT = $1,000 x 8.6% x 120/365 = $1,000 x 0.086 x 120/365 = $28.27

I = PRT = $1,000 x 8.6% x 120/366 = $1,000 x 0.086 x 120/366 = $28.20

I = PRT = $1,000 x 8.6% x 120/365.25 = $1,000 x 0.086 x 120/365.25 = $28.26

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• Interest that is calculated on the basis of the actual number of days in the year is called exact interest; calculating interest in this way is known as the exact method.

• Always using a 365-day year may be referred to as the simplified exact method.– Unless otherwise specified, calculate interest using the simplified exact

method while solving problems in this tutorial.

• Under bankers’ rule, we assume that the year consists of 12 months having 30 days each, for a total of 360 days in the year.

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– Using the simplified exact method, calculate the simple interest due on a 150-day loan of $120,000 at 9.45% simple interest.

• Solution

I = PRT

I = $120,000 x 9.45% x 150/365

I = $120,000 x 0.0945 x 150/365

I = $4,660.27

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– Using the bankers’ method, calculate the simple interest due on a 120-day loan of $10,000 at 8.6% simple interest.

• Solution

I = PRT

I = $10,000 x 8.6% x 120/360

I = $10,000 x 0.086 x 120/360

I = $28.67

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Example 1.2.8• While such situations are far less common, it is possible to measure the

term of the loan with units other than years, months, or days.

• Problem– Bridget borrows $2,000 for 13 weeks at 6% simple interest. Find the total

interest she will pay.

• Solution– Since the term is in weeks, we divide by 52 since there are 52 weeks per year.

I = PRT

I = $2,000 x 6% x 13/52

I = $2,000 x 0.06 x 13/52

I = $30

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Problem 1:

Loans with Terms in Months

Problem 2:

Terms in Days:

Exact Method

Problem 3:

Terms in Days: Bankers’ Rule

Problem 4:

Grab Bag

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Problem 1

• If Grace loans Anthony $800 for 9 months at 6 ½% simple interest, how much interest will Anthony pay?

CHECK YOUR ANSWER

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Solution 1

• If Grace loans Anthony $800 for 9 months at 6 ½% simple interest, how much interest will Anthony pay?

• I = PRT• I = $800 x 6 ½% x 9/12• I = $800 x 6.5% x 9/12• I = $800 x 0.065 x 9/12• I = $39

BACK TO GAME BOARD

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Problem 2

• Annabelle borrowed $5,239 at 8 ¼% for 400 days. How much will she need in total to pay the loan back?

CHECK YOUR ANSWER

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Solution 2

• Annabelle borrowed $5,239 at 8 ¼% for 400 days. How much will she need in total to pay the loan back?

• I = PRT• I = $5,239 x 8 ¼% x 400/365• I = $5,239 x 8.25% x 400/365• I = $5,239 x 0.0825 x 400/365• I = $473.66

BACK TO GAME BOARD

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Problem 3

• Use the bankers’ rule to solve this problem.• Mia agreed to loan Abigail $300 for 60 days.

Assuming the simple interest rate is 17%, how much will Mia earn from this loan?

CHECK YOUR ANSWER

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Solution 3

• Mia agreed to loan Abigail $2,300 for 60 days. Assuming the simple interest rate is 17%, how much will Mia earn from this loan?

• I = PRT• I = $2,300 x 17% x 60/360• I = $2,300 x 0.17 x 60/360• I = $65.17

BACK TO GAME BOARD

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Problem 4

• How much interest would you earn if you deposited $500 in a certificate of deposit (CD) paying 2 ½% simple interest for 8 months?

CHECK YOUR ANSWER

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Solution 4

• How much interest would you earn if you deposited $500 in a certificate of deposit (CD) paying 2 ½% simple interest for 8 months?

• I = PRT• I = 500 x 2 ½% x 8/12• I = 500 x 2.5% x 8/12• I = 500 x 0.025 x 8/12• I = $8.33

BACK TO GAME BOARD

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• You can now calculate the amount of interest due when given principal, rate, and time.

• However, what if you already know the amount of interest and need to calculate other quantities?

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• Finding Principal– Problem

• A retiree hopes to be able to generate $1,000 income per month from an investment account that earns 4.8% simple interest. How much money would he need in the account to achieve this goal?

– SolutionI = PRT$1,000 = P x 4.8% x 1/12$1,000 = P x 0.048 x 1/12$1,000 = P x 0.004


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• The Balance Principle– When we write an equation, we are making the claim

that the things on the left side of the “=“ sign have the exact same value as the things on the other side.

– You can make any change you like to one side of an equation as long as you make an equivalent change to the other side.

$1,000 P x 0.004

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• We need to get rid of the (0.004) multiplied by P on the right side.

• Since any number divided by itself is 1, we can eliminate 0.004 if we divide both sides by 0.004.

$250,000 = P which is the same as P = $250,000• Therefore, our retired friend needs to have $250,000 in his

account in order to generate $1,000 per month at 4.8% simple interest.

004.0

)004.0)((

004.0

000,1$ P

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– How much would you need to have in an investment account to earn $2,000 simple interest in 4 months, assuming that the simple interest rate is 5.9%?

• Solution

I = PRT

$2,000 = P x 5.9% x 4/12

$2,000 = P x 0.059 x 4/12

$2,000 = P x 0.01966667

P = $101,694.92

• Therefore, you would need to invest $101,694.92 in order to earn $2,000 simple interest in 4 months at 5.9%.

01966667.0

)01966667.0(

01966667.0

000,2$ P


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• Finding The Simple Interest Rate– Problem

• Jim borrowed $500 from his brother-in-law and agreed to pay back $525 ninety (90) days later. What rate of simple interest is Jim paying for this loan, assuming that they agreed to calculate the interest with bankers’ rule?

– SolutionI = PRTI = Total Amount – Principal = $525 -- $500 = $25$25 = $500 x Rate x 90/360$25 = $125(R)

0.2 = R or R = 0.2 = 20%Therefore, the interest rate is 20%.

125$

)(125$

125$

25$ R


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– Calculate the simple interest rate for a loan of $9,764.55 if the term is 125 days and the total required to repay the loan is $10,000.

• Solution

I = PRT

I = Total Amount – Principal = $10,000 -- $9,764.55 = $235.45

$235.45 = $9,764.55 x Rate x 125/365

$235.45 = $3,344.02397270(R)

R = 0.0704091842

R = 7.04091842% or approximately 7.04%

Therefore, the interest rate is 7.04%.


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• Finding Time– Problem

• Maria deposited $9,750 in a savings account that pays 5 ¼% simple interest. How long will it take for her account to grow to $10,000?

– Solution

I = PRT

I = Total Amount – Principal = $10,000 -- $9,750 = $250

$250 = $9,750 x 5 ¼% x T

$250 = $9,750 x 0.0525 x T

T = 0.4884004884

– But what does this number mean??? It means that time of the loan is 0.4884004884 years.

– To convert this number into days, we need to multiply it by 365:

T = 0.4884004884 x 365 = 178.2661783 days or approximately 178 days


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– Suppose that you deposit $3,850 in an account paying 4.65% simple interest. How long will it take to earn $150 in interest?

• Solution

I = PRT

$150 = ($3,850)(0.0465)T

$150 = ($179.025)T

T = 0.837871806

T = (0.837871806)365 = 306 days


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Problem 1:

Finding Principal

Problem 2:

Finding the Simple Interest Rate

Problem 3:

Finding Time

Problem 4:

Grab Bag

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Problem 1

• How much does Nina need in her investment account for 18 months if she wants to earn $200 in interest to cover the travel expenses? The current simple interest rate is 3%.

CHECK YOUR ANSWER

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Solution 1• How much does Nina need in her investment

account for 18 months if she wants to earn $200 in interest to cover the travel expenses? The current simple interest rate is 3%.

I = PRT$200 = P x 3% x 18/12$200 = P x 0.03 x 18/12$200 = P x 0.03 x 1.5$200 = P x 0.045

P = $4,444.44

BACK TO GAME BOARD

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Problem 2

• Rinat lent James $150. Five months later, James gave Rinat a total of $170 to pay off the loan. What rate of simple interest did James pay?

CHECK YOUR ANSWER

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Solution 2• Rinat lent James $150. Five months later,

James gave Rinat a total of $160 to pay off the loan. What rate of simple interest did James pay?

I = PRT$10 = $150 x R x 5/12$10 = $62.5 x RR = 0.16

R = 16%

BACK TO GAME BOARD

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Problem 3

• Sam and Tanya deposited $10,000 into a bank account paying 2.26% simple interest. How long will it take for the account value to grow to $12,000?

CHECK YOUR ANSWER

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Solution 3

• Sam and Tanya deposited $10,000 into a bank account paying 2.26% simple interest. How long will it take for the account value to grow to $10,100?I = PRT$100 = $10,000 x 2.26% x T$100 = $10,000 x 0.0226 x T$100 = $226 x TT = 0.44247787610619469026548672566372 x 365T = 161.50442477876106194690265486726T = 162 days

BACK TO GAME BOARD

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Problem 4

• XYZ Community College wants to establish a new scholarship that would honor an outstanding Nursing Program student with the highest GPA. In order to do this, the college needs to raise enough funds to earn $250 a year in interest, assuming a simple interest rate of 5%. How much money need to be raised?

CHECK YOUR ANSWER

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Solution 4• XYZ Community College wants to establish a new

scholarship that would honor an outstanding Nursing Program student with the highest GPA. In order to do this, the college needs to raise enough funds to earn $250 a year in interest, assuming a simple interest rate of 5%. How much money need to be raised?

I = PRT$250 = P x 5% x 1$250 = P x 0.05

P = $5,000

BACK TO GAME BOARD

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Definition 1.4.1 A promissory note (or just note for short) is a written agreement

between a borrower and a lender that sets out the terms of a loan.Definition 1.4.2 The date (or loan date) of a note is the day it is signed and the loan is

made.

The face value of a note is the amount of the loan for which it is written.

On the day that the loan is to be repaid, we say the loan comes due and its note matures.

The date on which this happens is the maturity date and the total amount that must be repaid is called the maturity value.

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Promissory Note

I, Jane Doe, acknowledge a loan of $1,000 made to me today, April 22, 2004, by the Seventh National Bank of Idaho Falls, and agree to repay this loan in 6 months together with simple interest calculated at a rate of 12%.

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Finding the Term of the Loan


– Find the number of days between April 7, 2003, and September 23, 2003. Use Table 1.3.1 on page 34 of your textbook.

• Solution– Looking in the table, we see that April 7 is Day 97,

and September 23 is Day 266. So there are 266 – 97 = 169 days between those dates.

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– Use the abbreviated day of the year table (Table 1.3.2, page 34) to find the number of days between August 19, 2006, and October 1, 2006.

• Solution– The ordinal date for August 19 is 212 + 19 = 231– The ordinal date for October 1 is 273 + 1 = $274– Subtracting to find the difference, we find there are

274 – 231 = 43 days.

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– Find the maturity value for the following promissory note:

Promissory NoteI, Cameron Mills, acknowledge a loan of $2,500 made to me today, January 19, 2005, by William Burgh, and agree to repay this loan on June 30, 2005, together with simple interest calculated at a rate of 10%.

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Example 1.4.3• Solution

– June 30 is day 181, and January 19 is day 19, so the term of this loan is 181 – 19 = 162 days.

I = PRT

I = $2,500 x 10% x 162/365

I = $110.96

Maturity Value = Principal + Interest

Maturity Value = $2,500 + $110.96 = $2,610.96

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Example 1.4.4• Leap years throw a wrench into the problems because the 60th day of the year is

February 29, not March 1; obviously, that throws off all dates thereafter.• A simple way of dealing with this problem is just to add one to the value shown

in the table for all dates after February 28 if it’s the leap year.• Any year between 1900 and 2100 that is divisible by 4 is a leap year.• Problem

– Find the number of days (a) between January 6, 2004, and May 1, 2004, and (b) between September 3, 2004, and November 16, 2004.

• Solution– Since 2004 is a leap year, we need to make adjustments.– January 6 fall before the leap day so no adjustment is necessary; it’s Day 6.– For May 1, we add one to the table value, so it becomes Day 122.– 122 – 6 = 116 days– In a non-leap year, September 3 is Day 246; since it’s a leap year, we add one to the

value, so it becomes Day 247.– Same for November 16 – in a leap year, it’s Day 321.– 321 – 247 = 74 days

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Finding Loan Dates and Maturity DatesExample 1.4.5• Problem

– Find the maturity date of a 135-day note signed on March 7, 2005.

• Solution– March 7 is Day 66. If we start at Day 66 and go

forward 135 days, we arrive at Day 201, because 66 + 135 = 201

– According to the table, Day 201 is July 20, so that is the maturity date of the note.

– Therefore, the maturity date is July 20, 2005.

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– Find the date of a 200-day note that matures on November 27, 2006.

• Solution– November 27 is Day 331.– If we go backward 200 days, we arrive at Day 131

since 331 – 200 = 131– According to the table, Day 131 is May 11.– Therefore, the note day is May 11, 2006.

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– A 100-day note was signed on January 3, 2008. When does the note mature?

• Solution– January 3 is Day 3, and so 100 day after that is Day

103.– However, 2008 is a leap year so we actually need to

find Day 102 in the table.– Therefore, the note’s maturity date is April 12, 2008.

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– Find the term of a note dated June 7, 2004, that matures on March 15, 2006.

• Solution– June 7 is Day 158 and March 15 is Day 74. – Since they fall in different years, we split the term of the

note by year into the parts that fall in 2004, 2005, and 2006.

– Thus, the term of the note is 207 + 365 + 74 = 644 days2004 2005 2006 207 days 365 days 74 days

6/7/04 3/16/06

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Example 1.4.9

• Problem

– Find the loan date for a 500-day note that matured on February 26, 2003.

• Solution

Running Tally

500 days

2003 -- 57 days

2002 -- 365 days

2001 -- 78 days

• The year ends on Day 365, so moving 78 days backward takes us to Day 287.

• According to the table, it’s October 14, 2001.

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Problem 1:

Promissory Notes

Problem 2:

Finding Terms

Problem 3:

Finding Loan and Maturity Dates

Problem 4:

Dealing with Leap Years

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Problem 1

• Identify the loan’s (a) date, (b) face value, (c) maturity date, (d) maturity value, and (e) term.

I, Robert Watkins, acknowledge a loan of $10,000 made to me by Ford Motor Company on April 11, 2007. I agree to repay this loan together with $2,400 in interest in 48 months, on April 11, 2011. CHECK YOUR ANSWER

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Solution 1• Date

– April 11, 2007• Face Value

– $10,000• Maturity Date

– April 11, 2011• Maturity Value

– $12,400• Term

– 48 months

BACK TO GAME BOARD

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Problem 2

• Savannah loaned her friend, Jerry, $250 on April 7, 2007. Jerry signed a note, agreeing to pay the loan back, with interest, on December 1, 2007. Find the term of the loan.

CHECK YOUR ANSWER

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Solution 2• Savannah loaned her friend, Jerry, $250 on April

7, 2007. Jerry signed a note, agreeing to pay the loan back, with interest, on December 1, 2007. Find the term of the loan.

• April 7 is Day 97 and December 1 is Day 335• 335 – 97 = 238 days

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Problem 3

• Nancy signed a 150-day note dated February 1, 2007. When does the note mature?

CHECK YOUR ANSWER

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Solution 3• Nancy signed a 150-day note dated February

1, 2007. When does the note mature?• February 1 is Day 32• Adding 150 days brings us to Day 182• Day 182 is July 1, 2007

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1-87

Problem 4

• If you take out a 250-day loan on January 1, 2008, when will the note come due?

CHECK YOUR ANSWER

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Solution 4• If you take out a 250-day loan on January 1,

2008, when will the note come due?• January 1 is Day 1• Adding 250 days would take us to Day 251• However, 2008 is a leap year, so we actually

need Day 250, which is September 7, 2008.

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1-89


Example 1.5.1

• Problem– Suppose that my credit card charges a daily

simple interest rate of 0.05%. How much interest would I owe for 1 day, on which my balance was $1,800?

• SolutionI = PRT = $1,800 x 0.05% x 1 = $0.90 or 90 cents

1-90


Example 1.5.2

• Problem– For the same credit card used in Example 1.5.1,

how much interest would I owe for two weeks (14 days) during which my balance was $2,000?

• SolutionI = PRT = $2,000 x 0.05% x 14 = $14.00

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Example 1.5.3• However, the interest rate used in two previous examples is much

lower that the rates you are probably accustomed to seeing. It’s small because it covers only a short period of time.

• What would the interest rate look like if it were expressed as a rate per year?

• Problem– Convert the rate 0.05% per day into equivalent simple interest rate per

year.• Solution

– For the simplified exact method, there are 365 days in a year.– Thus, the interest for a full year would be 365 times the interest for a single

day.– 365 x 0.05% = 18.25%– Therefore, 0.05% per day is equivalent to 18.25% per year.

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Example 1.5.4

• Problem– Calculate the interest due for 10 days on a

principal of $1,200 using (a) simple interest rate of 0.05% per day and (b) a simple interest rate of 18.25% annually.

• SolutionI = PRT = $1,200 x 0.05% x 10 = $6.00

I = PRT = $1,200 x 18.25% x 10/365 = $6.00

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Example 1.5.5

• Problem– Convert an 18% annual simple interest rate to a

rate per month.

• Solution– There are 12 months in a year.– Therefore, 18% ÷ 12 = 1.5% or 1 ½%

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Example 1.5.6

• Problem– Convert an annual simple interest rate of 9.75%

into a daily rate.

• Solution

9.75% ÷ 365 = 0.02671%

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Example 1.5.7

• Problem– Convert a simple interest rate of 0.045% per

day into a rate per month. Assume that bankers’ rule is being used.

• Solution– Under bankers’ rule, we assume that each

month has 30 days.

0.045% x 30 = 1.35% per month

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Problem 1:

Using Non-Annual Rates

Problem 2:

Converting Non-Annual to Annual Rates

Problem 3:

Converting Annual to Non-Annual Rates

Problem 4:

Grab Bag

1-97

Problem 1

• How much interest would Tammy owe on a loan of $3,500 for four weeks if the simple interest rate is 0.0328% per day?

CHECK YOUR ANSWER

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Solution 1• How much interest would Tammy owe on a loan

of $3,500 for four weeks if the simple interest rate is 0.0328% per day?

• I = PRT• I = $3,500 x 0.0328% x 28 = $32.14

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1-99

Problem 2

• Convert 0.0734% daily rate to the annual simple interest rate.

CHECK YOUR ANSWER

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Solution 2• Convert 0.0734% daily rate to the annual simple

interest rate.• 0.0734% x 365 = 26.79%

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1-101

Problem 3

• Convert 15% simple interest rate per year to (a) monthly rate and (b) daily rate.

CHECK YOUR ANSWER

1-102

Solution 3• Convert 16% simple interest rate per year to

(a) monthly rate and (b) daily rate.• 15% ÷ 12 = 1.33%• 15% ÷ 365 = 0.04384%

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1-103

Problem 4

• Suppose you are offered a credit card with a daily interest rate of 0.0825%. What would be the corresponding annual rate?

CHECK YOUR ANSWER

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Solution 4

• Suppose you are offered a credit card with a daily interest rate of 0.0825%. What would be the corresponding annual rate?

• 0.0825% x 365 = 30.11%

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Chapter 1 Summary

• The Concept of Interest• Simple Interest as a Perc

ent• Calculating Simple Intere

st• Loans with Terms in Mont

hs• The Exact Method• Bankers’ Rule• Loans with Terms in Wee

ks• Finding Principal

• Finding the Interest Rate• Finding Time• Finding the Term of a Not

e• Finding Maturity Dates• Finding Loan Dates• Finding Terms and Dates

Across Multiple Years• Using Non-Annual Interes

t Rates• Converting Between Non-

Annual and Annual Rates

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Chapter 1 Exercises

Section 1.1 Section 1.2 Section 1.3 Section 1.4 Section 1.5

$100 $100 $100 $100 $100

$200 $200 $200 $200 $200

EXIT

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Section 1.1 -- $100

• Michelle borrowed $80 and paid $90 when the loan came due. How much interest did she pay?

CHECK YOUR ANSWER

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Section 1.1 -- $100

• Michelle borrowed $80 and paid $90 when the loan came due. How much interest did she pay?

• $90 -- $80 = $10

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1-109

Section 1.1 -- $200

• Aldon deposited $2,000 in a 5-year certificate of deposit paying 2.75% simple interest. What will be the value of his account in 5 years?

CHECK YOUR ANSWER

1-110

Section 1.1 -- $200

• Aldon deposited $2,000 in a 5-year certificate of deposit paying 2.75% simple interest. What will be the value of his account in 5 years?

• I = PRT

I = $2,000 x 2.75% x 5 = $275• Therefore, the value of Aldon’s account will be

$2,000 + $275 = $2,275.

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1-111

Section 1.2 -- $100

• If Kelsey borrows $700 for 5 months at 3.99% simple interest, how much interest will she pay?

CHECK YOUR ANSWER

1-112

Section 1.2 -- $100

• If Kelsey borrows $700 for 5 months at 3.99% simple interest, how much interest will she pay?

• I = PRT

I = $700 x 3.99% x 5/12 = $11.64

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1-113

Section 1.2 -- $200

• The Fresh Flowers Fast, Inc. borrowed $300,000 for 60 days at 9% simple interest. Find the total amount of interest the company will have to pay.

• Use the bankers’ rule for this problem.

CHECK YOUR ANSWER

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Section 1.2 -- $200

• The Fresh Flowers Fast, Inc. borrowed $300,000 for 60 days at 9% simple interest. Find the total amount of interest the company will have to pay. Use the bankers’ rule for this problem.

• I = PRT

I = $300,000 x 9% x 60/360 = $4,500

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1-115

Section 1.3 -- $100

• Aaron lent Cindy $350 for 100 days. If Cindy paid back $400, what rate of simple interest did Aaron charge?

CHECK YOUR ANSWER

1-116

Section 1.3 -- $100

• Aaron lent Cindy $350 for 100 days. If Cindy paid back $370, what rate of simple interest did Aaron charge?

• I = PRT

$20 = $350 x R X 100/365

R = 0.10428 = 10.4%

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1-117

Section 1.3 -- $200

• Leslie borrowed $3,000 at a simple interest rate of 8.99%. She paid $67.43 simple interest for this loan. What was the term of the loan? Use the bankers’ rule for this problem.

CHECK YOUR ANSWER

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Section 1.3 -- $200• Leslie borrowed $3,000 at a simple interest rate of

8.99%. She paid $67.43 simple interest for this loan. What was the term of the loan? Use the bankers’ rule for this problem.

• I = PRT

$67.43 = $3,000 x 8.99% x T

T = 0.25001853911753800519095291064145 x 360

T = 90 days

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1-119

Section 1.4 -- $100

• Leslie borrowed $50 from her friend on April 3, 2007. She agreed to pay the loan back, with interest, on May 1, 2007. Find the term of the note.

CHECK YOUR ANSWER

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Section 1.4 -- $100

• Leslie borrowed $50 from her friend on April 3, 2007. She agreed to pay the loan back, with interest, on May 1, 2007. Find the term of the note.

• April 3 is Day 93• May 1 is Day 121• 121 – 93 = 28 days

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1-121

Section 1.4 -- $200

• Simpson Bookkeeping Services, Inc. signed a note on March 11, 2007. The loan’s principal was $4,500, the simple interest rate was 8.4%, and the note’s maturity value was $4,750. What will be the note’s maturity date?

CHECK YOUR ANSWER

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Section 1.4 -- $200• Simpson Bookkeeping Services, Inc. signed a note

on March 11, 2007. The loan’s principal was $4,500, the simple interest rate was 8.4%, and the note’s maturity value was $4,750. What will be the note’s maturity date?

• I = PRT$250 = $4,500 x 8.4% x TT = 241 days

• March 11 is Day 70• 70 + 241 = 311• Day 311 is November 7, 2007

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1-123

Section 1.5 -- $100

• Kyle’s credit card company is charging 0.0431% daily rate. What is the corresponding annual rate?

CHECK YOUR ANSWER

1-124

Section 1.5 -- $100

• Kyle’s credit card company is charging 0.0431% daily rate. What is the corresponding annual rate?

• 0.0431% x 365 = 15.73%

BACK TO GAME BOARD

1-125

Section 1.5 -- $200

• Shalonda received a credit card offer with a 7.9% simple interest rate per year. What is a corresponding daily rate?

CHECK YOUR ANSWER

1-126

Section 1.5 -- $200

• Shalonda received a credit card offer with a 7.9% simple interest rate per year. What is a corresponding daily rate?

• 7.9% ÷ 365 = 0.02164%

BACK TO GAME BOARD

© the mcgraw-hill companies, inc., 2008 mcgraw-hill/irwin 1-1 chapter 1 simple interest startexit

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