Chapter 3 Equivalence A Factor Approach

1 49 Chapter 3 Equivalence A Factor Approach 3-1 If you had $1,000 now and invested it at 6%, how much would it be worth 12 years from now? Solution F = 1,000(F/P, 6%, 12) = $2, 3-2 Mr. Ray deposited $200,000 in the Old and Third National Bank. If the bank pays 8% interest, how much will he have in the account at the end of 10 years? Solution F = 200,000(F/P, 8%, 10) = $431,800 3-3 If you can earn 6% interest on your money, how much is $1,000 paid to you 12 years in the future worth to you now? Solution P = 1,000(P/F, 6%, 12) = $ 3-4 Determine the value of P using the appropriate Factor . Solution P = F(P/F, 6%, 5) = $500( ) = $ 3-5 Downtown is experiencing an explosive population growth of 10% per year.

2 At the end of 2005 F = $500 i = 6% P 0 1 2 3 4 5 50 Chapter 3 Equivalence A Factor Approach the population was 16,000. If the growth rate continues unabated, at the end of how many years will it take for the population to triple? Solution Use i = 10% to represent the growth rate. 48,000 = 16,000(F/P, 10%, n) (F/P, 10%, n) = 48,000/16,000 = From the 10% table n is 12 Note that population would not have tripled after 11 years. 3-6 If the interest rate is 6% compounded quarterly, how long (number of quarters) does it take to earn $100 interest on an initial deposit of $300? Solution i = 6%/4 = 1 % 400 = 300(F/P, 1 %, n) (F/P, 1 %, n) = 400/300 = From the 1 % table n = 20 quarters 3-7 The amount of money accumulated in five years with an initial deposit of $10,000, if the account earned 12% compounded monthly the first three years and 15% compounded semi-annually the last two years is closest to a.

3 $18,580 b. $19,110 c. $19,230 d. $1,034,285 Solution F = [10,000(F/P, 1%, 36)](F/P, , 4) = 10,000( )( )4 = $19, 3-8 One thousand dollars is deposited into an account that pays interest monthly and allowed to remain in the account for three years. If the annual interest rate is 6%, the balance at the end of the three years is closest to a. $1,180 Chapter 3 Equivalence A Factor Approach 51 b. $1,191 c. $1,197 d. $2,898 Solution i = 6/12 = % n = (12)(3) = 36 F = P(1 + i)n = 1,000( )36 = $1, or using interest tables F = 1,000(F/P, %, 36) = 1,000( ) = $1,197 The answer is c. 3-9 On July 1 and September 1, Abby placed $2,000 into an account paying 3% compounded monthly.

4 How much was in the account on October 1? Solution i = 3/12 = % F = 2,000(1 + .0025)3 + 2,000(1 + .0025)1 = $4, or F = 2,000(F/P, %, 3) + 2,000(F/P, %, 1) = $4, 3-10 The Block Concrete Company borrowed $20,000 at 8% interest, compounded semi-annually, to be paid off in one payment at the end of four years. At the end of the four years, Block made a payment of $8,000 and refinanced the remaining balance at 6% interest, compounded monthly, to be paid at the end of two years. The amount Block owes at the end of the two years is nearest to a. $21,580 b. $21,841 c. $22,020 d. $34,184 Solution i1 = 8/2 = 4% n1 = (4)(2) = 8 i2 = 6/2 = % n2 = (12)(2) F = [20,000(F/P, 4%, 8) -8,000](F/P, %, 24) = $21, The answer is b.

5 52 Chapter 3 Equivalence A Factor Approach 3-11 How much should Abigail invest in a fund that will pay 9%, compounded continuously, if she wishes to have $60,000 in the fund at the end of 10 years? Solution r = n = 10 P = Fe-rn = 60, (10) = $24, 3-12 Given : Find: a) F b) ieff Solution a) F = 350(F/P, .75%, 6) = $ b) ieff = (1 + i)m - 1 = ( )12 - 1 = 3-13 Five hundred dollars is deposited into an account that pays 5% interest compounded continuously. If the money remains in the account for three years the account balance is nearest to a. $525 b. $578 c. $580 d. $598 Solution F = ern = (3) = $ The answer is c.

6 3-14 The multistate Powerball Lottery, worth $182 million, was won by a single individual who had purchased five tickets at $1 each. The winner was given two choices: Receive 26 payments of $7 million each, with the first payment to be made now and the rest to be made at the end of each of the next twenty five years; or receive a single lump-sum payment now that would be equivalent to Jan Feb Mar Apr May June F = ? P = $350 i = 9% Chapter 3 Equivalence A Factor Approach 53 the 26 payments of $7 million each. If the state uses an interest rate of 4% per year, the amount of the lump sum payment is closest to a.

7 $109,355,000 b. $111,881,000 c. $116,354,000 d. $182,000,000 Solution P = 7,000,000 + 7,000,000(P/F, 4%, 25) = $116,354,000 The answer is c. 3-15 The future worth (in year 8) of $10,000 deposited at the end of year 3, $10,000 deposited at the end of year 5, and $10,000 deposited at the end of year 8 at an interest rate of 12% per year is closest to a. $32,100 b. $39,300 c. $41,670 d. $46,200 Solution F = 10,000(F/P, 12%, 5) + 10,000(F/P, 12%, 3) + 10,000 = $41,670 The answer is c. 3-16 A woman deposited $10,000 into an account at her credit union. The money was left on deposit for 10 years. During the first five years the woman earned 9% interest, compounded monthly.

8 The credit union then changed its interest policy so that the second five years the woman earned 6% interest, compounded quarterly. a. How much money was in the account at the end of the 10 years? b. Calculate the rate of return that the woman received. Solution a) at the end of 5 years: F = 10,000 (F/P, %, 60)* = $15, * i = 9/12 = % n = (12)(5) = 60 at the end of 10 years: F = 15,660(F/P, 1 %, 20)** = $21, ** i = 6/4 = 1 % n = (4)(5) = 20 b) 10,000(F/P, i, 10) = 21, (F/P, i, 10) = 54 Chapter 3 Equivalence A Factor Approach try i = 7% (F/P, 7%, 10) = try i = 8% (F/P, 8%, 10) = 7% < i < 8% interpolate i = 3-17 A young engineer wishes to buy a house but only can afford monthly payments of $500.

9 Thirty-year loans are available at 6% interest compounded monthly. If she can make a $5,000 down payment, what is the price of the most expensive house that she can afford to purchase? Solution i = 6/12 = % n = (30)(12) = 360 P* = 500(P/A, %, 360) = 83, P = 83, + 5,000 P = $88,396 3-18 A person borrows $15,000 at an interest rate of 6%, compounded monthly to be paid off with payments of $ a. What is the length of the loan in years? b. What is the total amount that would be required at the end of the twelfth month to payoff the entire loan balance? Solution a) P = A(P/A, i%, n) 15,000 = (P/A, %, n) (P/A, %, n) = 15,000 = From the % interest table n = 36 months = 6 years.

10 B) + (P/A, %, 24) = $10, 3-19 A $50,000 30-year loan with a nominal interest rate of 6% is to be repaid with payments of $ The borrower wants to know how many payments, N*, he will have to make until he owes only half of the amount she borrowed initially. Solution The outstanding principal is equal to the present worth of the remaining payments when the payments are discounted at the loan's effective interest rate. Chapter 3 Equivalence A Factor Approach 55 Therefore, let N' be the remaining payments. (50,000) = (P/A, %, N') (P/A, %, N') = N' = 108 From i = % table So, N* = 360 - N' = 252 payments 3-20 Homeowner has just bought a house with a 20-year, 9%, $70,000 mortgage on which he is paying $ per month.