Chapter 1 Simple Interest START EXIT 1 1

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Chapter 1 Simple Interest START EXIT 1 -1 Mc. Graw-Hill/Irwin Copyright © 2008 by

Chapter 1 Simple Interest START EXIT 1 -1 Mc. Graw-Hill/Irwin Copyright © 2008 by The Mc. Graw-Hill Companies, Inc. All rights reserved.

Chapter Outline 1. 1 Simple Interest and the Time Value of Money 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 1 -2

1. 1 Simple Interest and Time Value of Money Definition 1. 1. 1 Interest

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. 1 -3

1. 1 Simple Interest and Time Value of Money Example 1. 1. 1 •

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. 1 -4

1. 1 Simple Interest and Time Value of Money Example 1. 1. 2 •

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. 1 -5

1. 1 Simple Interest and Time Value of Money Definition 1. 1. 3 The

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. 1 -6

1. 1 Simple Interest and Time Value of Money Example • Problem – Sam

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. 1 -7

1. 1 Simple Interest and Time Value of Money Example • Problem – Sam

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. 1 -8

1. 1 Simple Interest and Time Value of Money Example • Problem – Sam

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. 1 -9

1. 1 Simple Interest and Time Value of Money Example 1. 1. 3 •

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? 1 -10

1. 1 Simple Interest and Time Value of Money Working with Percents • When

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) 1 -11

1. 1 Simple Interest and Time Value of Money Example 1. 1. 4 •

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 1 -12

1. 1 Simple Interest and Time Value of Money Example 1. 1. 7 •

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 1 -13

1. 1 Simple Interest and Time Value of Money Example 1. 1. 8 •

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) 1 -14

1. 1 Simple Interest and Time Value of Money Mixed Number and Fractional Percents

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 1 -15

1. 1 Simple Interest and Time Value of Money Example 1. 1. 9 •

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 1 -16

1. 1 Simple Interest and Time Value of Money The Impact of Time •

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? 1 -17

FORMULA 1. 1 The Simple Interest Formula I = PRT where I represents the

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 1 -18

1. 1 Simple Interest and Time Value of Money Example 1. 1. 10 •

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 1 -19

1. 1 Simple Interest and Time Value of Money Example 1. 1. 11 •

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 1 -20

1. 1 Simple Interest and Time Value of Money Loans in Disguise • Sometimes

1. 1 Simple Interest and Time Value of Money Loans in Disguise • Sometimes interest is paid in situations we might 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. 1 -21

1. 1 Simple Interest and Time Value of Money Example 1. 1. 12 •

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 1 -22

Section 1. 1 Exercises Problem 1: Interest as Difference Problem 2: Determining Repayment Amount

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) 1 -23

Problem 1 • Linda borrowed $5, 000 and paid back a total of $5,

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

Solution 1 • Linda borrowed $5, 000 and paid back a total of $5,

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 1 -25

Problem 2 • Sam loaned Andrew $8, 900 for 6 months. Andrew agreed to

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 1 -26

Solution 2 • Sam loaned Andrew $8, 900 for 6 months. Andrew agreed to

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 1 -27

Problem 3 • Laura loaned Bill $390 for 1 year at a simple interest

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 1 -28

Solution 3 • Laura loaned Bill $390 for 1 year at a simple interest

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 1 -29

Problem 4 • Timothy has agreed to loan Kate $2, 350 for 2 years

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 1 -30

Solution 4 • Timothy has agreed to loan Kate $2, 350 for 2 years

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 1 -31

1. 2 The Term of a Loan Example 1. 2. 1 • • Problem

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 1 -32

1. 2 The Term of a Loan Example 1. 2. 2 • • Problem

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: Operation Result 3412. 59*. 0525*7/12 104. 51056875 – Interest = $104. 51 1 -33

1. 2 The Term of a Loan Example 1. 2. 3 • • Problem

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 Operation Result 2719*0. 046*20/12 208. 45677777 – Interest = $208. 46 1 -34

1. 2 The Term of a Loan Example 1. 2. 4 • • Problem

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 1 -35

1. 2 The Term of a Loan Example 1. 2. 5 • However, not

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 1 -36

1. 2 The Term of a Loan • Interest that is calculated on the

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. 1 -37

1. 2 The Term of a Loan Example 1. 2. 6 • • Problem

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 1 -38

1. 2 The Term of a Loan Example 1. 2. 7 • • Problem

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 1 -39

1. 2 The Term of a Loan Example 1. 2. 8 • • While

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 1 -40

Section 1. 2 Exercises Problem 1: Loans with Terms in Months Problem 2: Terms

Section 1. 2 Exercises 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 1 -41

Problem 1 • If Grace loans Anthony $800 for 9 months at 6 ½%

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

Solution 1 • If Grace loans Anthony $800 for 9 months at 6 ½%

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 1 -43

Problem 2 • Annabelle borrowed $5, 239 at 8 ¼% for 400 days. How

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 1 -44

Solution 2 • Annabelle borrowed $5, 239 at 8 ¼% for 400 days. How

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 1 -45

Problem 3 • Use the bankers’ rule to solve this problem. • Mia agreed

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 1 -46

Solution 3 • Mia agreed to loan Abigail $2, 300 for 60 days. Assuming

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 1 -47

Problem 4 • How much interest would you earn if you deposited $500 in

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 1 -48

Solution 4 • How much interest would you earn if you deposited $500 in

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 1 -49

1. 3 Determining Principal Interest Rates and Time • You can now calculate the

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? 1 -50

1. 3 Determining Principal Interest Rates and Time • Finding Principal – Problem •

1. 3 Determining Principal Interest Rates and Time • 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? – Solution I = 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 1 -51

1. 3 Determining Principal Interest Rates and Time • The Balance Principle – When

1. 3 Determining Principal Interest Rates and Time • 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 1 -52

1. 3 Determining Principal Interest Rates and Time • We need to get rid

1. 3 Determining Principal Interest Rates and Time • 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. 1 -53

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 1 • •

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 1 • • Problem – 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%. 1 -54

1. 3 Determining Principal Interest Rates and Time • Finding The Simple Interest Rate

1. 3 Determining Principal Interest Rates and Time • 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? – Solution I = PRT I = Total Amount – Principal = $525 -- $500 = $25 = $500 x Rate x 90/360 $25 = $125(R) 0. 2 = R or R = 0. 2 = 20% Therefore, the interest rate is 20%. 1 -55

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 2 • •

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 2 • • Problem – 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 = $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%. 1 -56

1. 3 Determining Principal Interest Rates and Time • Finding Time – Problem •

1. 3 Determining Principal Interest Rates and Time • 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 = $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 1 -57

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 3 • •

1. 3 Determining Principal Interest Rates and Time Example 1. 3. 3 • • Problem – 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 1 -58

Section 1. 3 Exercises Problem 1: Finding Principal Problem 2: Finding the Simple Interest

Section 1. 3 Exercises Problem 1: Finding Principal Problem 2: Finding the Simple Interest Rate Problem 3: Finding Time Problem 4: Grab Bag 1 -59

Problem 1 • How much does Nina need in her investment account for 18

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 1 -60

Solution 1 • How much does Nina need in her investment account for 18

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 1. 5 $200 = P x 0. 045 P = $4, 444. 44 BACK TO GAME BOARD 1 -61

Problem 2 • Tony lent James $150. Five months later, James gave Tony a

Problem 2 • Tony lent James $150. Five months later, James gave Tony a total of $160 to pay off the loan. What rate of simple interest did James pay? CHECK YOUR ANSWER 1 -62

Solution 2 • Tony lent James $150. Five months later, James gave Tony a

Solution 2 • Tony lent James $150. Five months later, James gave Tony 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 R R = 0. 16 R = 16% BACK TO GAME BOARD 1 -63

Problem 3 • Sam and Tanya deposited $10, 000 into a bank account paying

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 $10, 100? CHECK YOUR ANSWER 1 -64

Solution 3 • Sam and Tanya deposited $10, 000 into a bank account paying

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 T T = 0. 44247787610619469026548672566372 x 365 T = 161. 50442477876106194690265486726 T = 162 days BACK TO GAME BOARD 1 -65

Problem 4 • XYZ Community College wants to establish a new scholarship that would

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 1 -66

Solution 4 • XYZ Community College wants to establish a new scholarship that would

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 1 -67

Chapter 1 Summary • The Concept of Interest • Simple Interest as a Percent

Chapter 1 Summary • The Concept of Interest • Simple Interest as a Percent • Calculating Simple Interest • Loans with Terms in Months • The Exact Method • Bankers’ Rule • Loans with Terms in Weeks • Finding Principal • Finding the Interest Rate • Finding Time 1 -68

Chapter 1 Exercises Section 1. 1 Section 1. 2 Section 1. 3 Section 1.

Chapter 1 Exercises Section 1. 1 Section 1. 2 Section 1. 3 Section 1. 4 Section 1. 5 $100 $100 $200 $200 EXIT 1 -69

Section 1. 1 -- $100 • Michelle borrowed $80 and paid $90 when the

Section 1. 1 -- $100 • Michelle borrowed $80 and paid $90 when the loan came due. How much interest did she pay? CHECK YOUR ANSWER 1 -70

Section 1. 1 -- $100 • Michelle borrowed $80 and paid $90 when the

Section 1. 1 -- $100 • Michelle borrowed $80 and paid $90 when the loan came due. How much interest did she pay? • $90 -- $80 = $10 BACK TO GAME BOARD 1 -71

Section 1. 1 -- $200 • Aldon deposited $2, 000 in a 5 -year

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 -72

Section 1. 1 -- $200 • Aldon deposited $2, 000 in a 5 -year

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. BACK TO GAME BOARD 1 -73

Section 1. 2 -- $100 • If Kelsey borrows $700 for 5 months at

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 -74

Section 1. 2 -- $100 • If Kelsey borrows $700 for 5 months at

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 BACK TO GAME BOARD 1 -75

Section 1. 2 -- $200 • The Fresh Flowers Fast, Inc. borrowed $300, 000

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 1 -76

Section 1. 2 -- $200 • The Fresh Flowers Fast, Inc. borrowed $300, 000

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 BACK TO GAME BOARD 1 -77

Section 1. 3 -- $100 • Aaron lent Cindy $350 for 100 days. If

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 -78

Section 1. 3 -- $100 • Aaron lent Cindy $350 for 100 days. If

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% BACK TO GAME BOARD 1 -79

Section 1. 3 -- $200 • Leslie borrowed $3, 000 at a simple interest

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 1 -80

Section 1. 3 -- $200 • Leslie borrowed $3, 000 at a simple interest

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 BACK TO GAME BOARD 1 -81