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Simple Interest and the Time Value of Money

Learn about simple interest and how it relates to the time value of money, including determining the principal, interest rates, and term of a loan. Practice exercises included.

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Simple Interest and the Time Value of Money

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  1. Chapter 1 Simple Interest START EXIT 1-1

  2. 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 Chapter Summary Chapter Exercises

  3. 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.

  4. 1.1 Simple Interest and Time Value of Money 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.

  5. 1.1 Simple Interest and Time Value of Money Example 1.1.2 • Problem • 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.

  6. 1.1 Simple Interest and Time Value of Money 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.

  7. 1.1 Simple Interest and Time Value of Money 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.

  8. 1.1 Simple Interest and Time Value of Money 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.

  9. 1.1 Simple Interest and Time Value of Money 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.

  10. 1.1 Simple Interest and Time Value of Money Example 1.1.3 • Problem • 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?

  11. 1.1 Simple Interest and Time Value of Money 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)

  12. 1.1 Simple Interest and Time Value of Money Example 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 Example 1.1.5 • Problem • Convert 18.25% to a decimal. • Solution • 18.25% = 0.1825 Example 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

  13. 1.1 Simple Interest and Time Value of Money Example 1.1.7 • Problem • 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

  14. 1.1 Simple Interest and Time Value of Money Example 1.1.8 • Problem • 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)

  15. 1.1 Simple Interest and Time Value of Money 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

  16. 1.1 Simple Interest and Time Value of Money Example 1.1.9 • Problem • Rewrite 7 13/16% as a decimal. • Solution • 13/16 = 0.8125 • 7 13/16% = 7.8125% = 0.078125

  17. 1.1 Simple Interest and Time Value of Money 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?

  18. FORMULA 1.1 The Simple Interest Formula I = PRT where I represents the amount of simple INTEREST for a loan P represents the amount of money borrowed or PRINCIPAL R rate T represents the TERM of the loan

  19. 1.1 Simple Interest and Time Value of Money Example 1.1.10 • Problem • Suppose that Bill 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

  20. 1.1 Simple Interest and Time Value of Money Example 1.1.11 • Problem • 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 0.05125 x 2 • Interest = $1,896.25

  21. 1.1 Simple Interest and Time Value of Money 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.

  22. 1.1 Simple Interest and Time Value of Money Example 1.1.12 • Problem • Jake deposited $2,318.29 into a 2-year CD paying 5.17% simple interest. 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

  23. Section 1.1 Exercises

  24. Problem 1 • Linda borrowed $5,000 and paid back a total of $5,845. How much interest did she pay? CHECK YOUR ANSWER

  25. 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

  26. Problem 2 • Sam loaned Andrew $8,900 for 6 months. Andrew agreed to pay $600 interest. How much will Andrew pay back? CHECK YOUR ANSWER

  27. Solution 2 • Sam 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

  28. Problem 3 • Laura loaned Bill $390 for 1 year at a simple interest rate of 8 3/8%. How much interest will Bill have to pay? CHECK YOUR ANSWER

  29. Solution 3 • Laura loaned Bill $390 for 1 year at a simple interest rate of 8 3/8%. How much interest will Bill 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

  30. Problem 4 • Timothy has agreed to loan Kate $2,350 for 2 years at 9% simple interest. How much will Kate have to repay in 2 years? CHECK YOUR ANSWER

  31. Solution 4 • Timothy has agreed to loan Kate $2,350 for 2 years at 9% simple interest. How much will Kate 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 • Total = $2,350 + $423 = $2,773 BACK TO GAME BOARD

  32. 1.2 The Term of a Loan • Example 1.2.1 • Problem • If Sarah 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

  33. 1.2 The Term of a Loan • Example 1.2.2 • Problem • 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 = Principal x Interest Rate x Time • 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

  34. 1.2 The Term of a Loan • Example 1.2.3 • Problem • Anthony deposited $2,719.00 in an account paying 4.6% simple interest for 20 months. Find the total interest he earned. • Solution • Interest = Principal x Interest Rate x Time • Interest = $2,719.00 x 4.6% x 20/12 • Interest = $208.46

  35. 1.2 The Term of a Loan • Example 1.2.4 • Problem • 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 • Interest = $18.62 • Total = $1,600 + $18.62 = $1,618.62

  36. 1.2 The Term of a Loan • 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

  37. 1.2 The Term of a Loan • 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.

  38. 1.2 The Term of a Loan • Example 1.2.6 • Problem • 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

  39. 1.2 The Term of a Loan • Example 1.2.7 • Problem • 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 = $286.67

  40. 1.2 The Term of a Loan • 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

  41. Section 1.2 Exercises

  42. Problem 1 • If Grace loans Anthony $800 for 9 months at 6 ½% simple interest, how much interest will Anthony pay? CHECK YOUR ANSWER

  43. 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

  44. 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

  45. 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 • Total = $5,239 + $473.66 = $5,712.66 BACK TO GAME BOARD

  46. Problem 3 • Use the bankers’ rule to solve this problem. • 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? CHECK YOUR ANSWER

  47. 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

  48. 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

  49. 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

  50. 1.3 Determining Principal Interest Rates and Time • 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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