Sample FM Questions

8/22/2019 Sample FM Questions

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1 Exam FM sample questions, from the Society of Actu-

aries and the Casualty Actuarial Society

1. Bruce deposits 100 into a bank account. His account is credited interest at a nominal rate

of interest of 4% convertible semiannually. At the same time, Peter deposits 100 into a

separate account. Peters account is credited interest at a force of interest of. After 7.25

years, the value of each account is the same. Calculate .

(A) 0.0388 (B) 0.0392 (C) 0.0396 (D) 0.0404 (E) 0.0414

2. Kathryn deposits 100 into an account at the beginning of each 4year period for 40 years.

The account credits interest at an annual effective interest rate of i. The accumulated

amount in the account at the end of 40 years is X, which is 5 times the accumulated

amount in the account at the end of 20 years. Calculate X.

(A) 4695 (B) 5070 (C) 5445 (D) 5820 (E) 6195

3. Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of

i, compounded semiannually. Mike deposits 2X into a different savings account at time 0,

which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of

interest during the last 6 months of the 8-th year. Calculate i.

(A) 9.06% (B) 9.26% (C) 9.46% (D) 9.66% (E) 9.86%

4. John borrows 10,000 for 10 years at an annual effective interest rate of 10%. He can repay

this loan using the amortization method with payments of 1,627.45 at the end of each year.Instead, John repays the 10,000 using a sinking fund that pays an annual effective interest

rate of 14%. The deposits to the sinking fund are equal to 1,627.45 minus the interest on

the loan and are made at the end of each year for 10 years. Determine the balance in the

sinking fund immediately after repayment of the loan.

(A) 2130 (B) 2180 (C) 2230 (D) 2300 (E) 2370

5. An association had a fund balance of 75 on January 1 and 60 on December 31. At the

end of every month during the year, the association deposited 10 from membership fees.

There were withdrawals of 5 on February 28, 25 on June 30, 80 on October 15, and 35 on

October 31. Calculate the dollarweighted rate of return for the year.

(A) 9.0% (B) 9.5% (C) 10.0% (D) 10.5% (E) 11.0%

6. A perpetuity costs 77.1 and makes annual payments at the end of the year. The perpetuity

pays 1 at the end of year 2, 2 at the end of year 3, . . . n at the end of year (n + 1). After

year (n + 1), the payments remain constant at n. The annual effective interest rate is

10.5%. Calculate n.

(A) 17 (B) 18 (C) 19 (D) 20 (E) 21

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7. 1000 is deposited into Fund X, which earns an annual effective rate of 6%. At the end

of each year, the interest earned plus an additional 100 is withdrawn from the fund. At

the end of the tenth year, the fund is depleted. The annual withdrawals of interest and

principal are deposited into Fund Y, which earns an annual effective rate of 9%. Determine

the accumulated value of Fund Y at the end of year 10.(A) 1519 (B) 1819 (C) 2085 (D) 2273 (E) 2431

8. You are given the following table to interest rates.

Calendar year Investment year rates Portfolio

of original (in %) rates

investment (in %)

y iy1 iy2 i

y3 i

y4 i

y5 i

y+5

1992 8.25 8.25 8.4 8.5 8. 8.351993 8.5 8.7 8.75 8.9 9.0 8.6

1994 9.0 9.0 9.1 9.1 9.2 8.85

1995 9.0 9.1 9.2 9.3 9.4 9.1

1996 9.25 9.35 9.5 9.55 9.6 9.35

1997 9.5 9.5 9.6 9.7 9.7

1998 10.0 10.0 9.9 9.8

1999 10.0 9.8 9.7

2000 9.5 9.5

2001 9.0

A person deposits 1000 on January 1, 1997. Let the following be the accumulated value of

the 1000 on January 1, 2000:

P: under the investment year method

Q: under the portfolio yield method

R: where the balance is withdrawn at the end of every year and is reinvested at the new

money rate Determine the ranking of P, Q, and R.

(A) P > Q > R (B) P > R > Q (C) Q > P > R (D) R > P > Q (E)R > Q > P

9. A 20year loan of 1000 is repaid with payments at the end of each year. Each of the first

ten payments equals 150% of the amount of interest due. Each of the last ten payments is

X. The lender charges interest at an annual effective rate of 10%. Calculate X.

(A) 32 (B) 57 (C) 70 (D) 97 (E) 117

10. A 10,000 par value 10-year bond with 8% annual coupons is bought at a premium to yield

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an annual effective rate of 6%. Calculate the interest portion of the 7th coupon.

(A) 632 (B) 642 (C) 651 (D) 660 (E) 667

11. A perpetuityimmediate pays 100 per year. Immediately after the fifth payment, the

perpetuity is exchanged for a 25year annuityimmediate that will pay X at the end of

the first year. Each subsequent annual payment will be 8% greater than the preceding

payment. The annual effective rate of interest is 8Calculate X .

(A) 110 (B) 120 (C) 130 (D) 140 (E) 150

12. Jeff deposits 10 into a fund today and 20 fifteen years later. Interest is credited at a nominal

discount rate of d compounded quarterly for the first 10 years, and at a nominal interest

rate of 6% compounded semiannually thereafter. The accumulated balance in the fund at

the end of 30 years is 100. Calculate d.

(A) 4.33% (B) 4.43% (C) 4.53% (D) 4.63% (E) 4.73%

13. Ernie makes deposits of 100 at time 0, and X at time 3. The fund grows at a force of

interest t =t2

100 , t > 0. The amount of interest earned from time 3 to time 6 is X.

Calculate X.

(A) 385 (B) 485 (C) 585 (D) 685 (E) 785

14. Mike buys a perpetuityimmediate with varying annual payments. During the first 5

years, the payment is constant and equal to 10. Beginning in year 6, the payments start

to increase. For year 6 and all future years, the current years payment is K% larger than

the previous years payment. At an annual effective interest rate of 9.2%, the perpetuity

has a present value of 167.50. Calculate K, given K < 9.2.

(A) 4.0 (B) 4.2 (C) 4.4 (D) 4.6 (E) 4.8

15. A 10year loan of 2000 is to be repaid with payments at the end of each year. It can be

repaid under the following two options:

(i) Equal annual payments at an annual effective rate of 8.07%.

(ii) Installments of 200 each year plus interest on the unpaid balance at an annual effective

rate of i.

The sum of the payments under option (i) equals the sum of the payments under option

(ii). Determine i.

(A) 8.75% (B) 9.00% (C) 9.25% (D) 9.50% (E) 9.75%

16. A loan is amortized over five years with monthly payments at a nominal interest rate of

9% compounded monthly. The first payment is 1000 and is to be paid one month from

the date of the loan. Each succeeding monthly payment will be 2% lower than the prior

payment. Calculate the outstanding loan balance immediately after the 40th payment is

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made.

(A) 6751 (B) 6889 (C) 6941 (D) 7030 (E) 7344

17. To accumulate 8000 at the end of 3n years, deposits of 98 are made at the end of each of

the first n years and 196 at the end of each of the next 2 n years. The annual effective rate

of interest is i. You are given (l + i)n = 2.0. Determine i.

(A) 11.25% (B) 11.75% (C) 12.25% (D) 12.75% (E) 13.25%

18. Olga buys a 5year increasing annuity for X. Olga will receive 2 at the end of the first

month, 4 at the end of the second month, and for each month thereafter the payment

increases by 2. The nominal interest rate is 9% convertible quarterly. Calculate X.

(A) 2680 (B) 2730 (C) 2780 (D) 2830 (E) 2880

19. You are give the following information about two funds:

Account K

Fund value

Date before activity Deposits Withdrawals

January 1, 1999 100

July 1, 1999 125 x

October 1, 1999 110.0 2x

December 31, 1999 125.0

Account L

Fund value

Date before activity Deposits Withdrawals

January 1, 1999 100

July 1, 1999 125 x

December 31, 1999 105.8

During 1999, the dollarweighted (moneyweighted) return for investment account K

equals the time-weighted return for investment account L, which equals i. Calculate i.

(A) 10% (B) 12% (C) 15% (D) 18% (E) 20%

20. David can receive one of the following two payment streams:

(i) 100 at time 0, 200 at time n, and 300 at time 2n

(ii) 600 at time 10

At an annual effective interest rate of i, the present values of the two streams are equal.

Given n = 0.75941, determine i.

(A) 3.5% (B) 4.0% (C) 4.5% (D) 5.0% (E) 5.5%

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21. Payments are made to an account at a continuous rate of (8k + tk), where 0 t 10.Interest is credited at a force of interest t =

18+t

. After 10 years, the account is worth

20, 000. Calculate k.

(A) 111 (B) 116 (C) 121 (D) 126 (E) 131

22. You have decided to invest in two bonds. Bond X is an n-year bond with semi-annual

coupons, while bond Y is an accumulation bond redeemable in n2

years. The desired yield

rate is the same for both bonds. You also have the following information:

Bond X

Par value is 1000. The ratio of the semi-annual bond rate to the desired semi-annual yield rate, ri is

1.03125.

The present value of the redemption value is 381.50.

Bond Y

Redemption value is the same as the redemption value of bond X. Price to yield is 647.80.

What is the price of bond X?

(A) 1019 (B) 1029 (C) 1050 (D) 1055 (E) 1072

23. Project P requires an investment of 4000 at time 0. The investment pays 2000 at time 1

and 4000 at time 2. Project Q requires an investment of X at time 2. The investment pays

2000 at time 0 and 4000 at time 1. The net present values of the two projects are equal at

an interest rate of 10%. Calculate X.

(A) 5400 (B) 5420 (C) 5440 (D) 5460 (E) 5480

24. A 20year loan of 20, 000 may be repaid under the following two methods:

(i) amortization method with equal annual payments at an annual effective rate of 6 .5%

(ii) sinking fund method in which the lender receives an annual effective rate of 8% and

the sinking fund earns an annual effective rate of j

Both methods require a payment of X to be made at the end of each year for 20 years.

Calculate j .

(A) j 6.5% (B) 6.5% < j 8.0% (C) 8.0% < j 10.0% (D) 10.0% < j 12.0%(E) j > 12.0%

25. A perpetuityimmediate pays X per year. Brian receives the first n payments, Colleen

receives the next n payments, and Jeff receives the remaining payments. Brians share of

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the present value of the original perpetuity is 40%, and Jeffs share is K. Calculate K.

(A) 24% (B) 28% (C) 32% (D) 36% (E) 40%

26. Seth, Janice, and Lori each borrow 5000 for five years at a nominal interest rate of 12%,

compounded semiannually. Seth has interest accumulated over the five years and pays all

the interest and principal in a lump sum at the end of five years. Janice pays interest at the

end of every six-month period as it accrues and the principal at the end of five years. Lori

repays her loan with 10 level payments at the end of every six-month period. Calculate

the total amount of interest paid on all three loans.

(A) 8718 (B) 8728 (C) 8738 (D) 8748 (E) 8758

27. Bruce and Robbie each open up new bank accounts at time 0. Bruce deposits 100 into

his bank account, and Robbie deposits 50 into his. Each account earns an annual effective

discount rate of d. The amount of interest earned in Bruces account during the 11th yearis equal to X. The amount of interest earned in Robbies account during the 17th year is

also equal to X. Calculate X.

(A) 28.0 (B) 31.3 (C) 34.6 (D) 36.7 (E) 38.9

28. Ron has a loan with a present value of an|. The sum of the interest paid in period t plus

the principal repaid in period t + 1 is X. Calculate X.

(A) 1 + nt

i (B) 1 +nt

d (C) 1 + nti (D) 1 + ntd (E) 1 + nt

29. At an annual effective interest rate of i, i% > 0, the present value of a perpetuity paying

10 at the end of each 3year period, with the first payment at the end of year 3, is 32. At

the same annual effective rate of i, the present value of a perpetuity paying 1 at the end

of each 4month period, with first payment at the end of 4 months, is X. Calculate X.

(A) 31.6 (B) 32.6 (C) 33.6 (D) 34.6 (E) 35.6

30. As of 12/31/03, an insurance company has a known obligation to pay $1,000,000 on

12/31/2007. To fund this liability, the company immediately purchases 4-year 5% an-

nual coupon bonds totaling $822,703 of par value. The company anticipates reinvestment

interest rates to remain constant at 5Under the following reinvestment interest rate move-

ment scenarios effective 1/1/2004, what best describes the insurance companys profit or

(loss) as of 12/31/2007 after the liability is paid?

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interest rate drop by 1/2% interest rate increase by 1/2%

(A) +6606 +11147

(B) (14757) +14418

(C) (18911) +19185

(D) (1313) +1323(E) Break even Break even

31. An insurance company has an obligation to pay the medical costs for a claimant. Average

annual claims costs today are $5,000, and medical inflation is expected to be 7% per year.

The claimant is expected to live an additional 20 years. Claim payments are made at yearly

intervals, with the first claim payment to be made one year from today. Find the present

value of the obligation if the annual interest rate is 5%.

(A) 87,932 (B) 102,514 (C) 114,611 (D) 122,634 (E) Cannot be determined

32. An investor pays $100,000 today for a 4-year investment that returns cash flows of $60,000

at the end of each of years 3 and 4. The cash flows can be reinvested at 4.0% per annum

effective. If the rate of interest at which the investment is to be valued is 5.0%, what is the

net present value of this investment today?

(A) 1398 (B) 699 (C) 699 (D) 1398 (E) 2, 629

33. You are given the following information with respect to a bond:

par amount: 1000

term to maturity: 3 yearsannual coupon rate 6% payable annually

term Annual Spot Interest Rates

1 7%

2 8%

3 9%

Calculate the value of the bond.

(A) 906 (B) 926 (C) 930 (D) 950 (E) 1000

34. You are given the following information with respect to a bond:

par amount: 1000

term to maturity: 3 years

annual coupon rate 6% payable annually

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term Annual Spot Interest Rates

1 7%

2 8%

3 9%

Calculate the annual effective yield rate for the bond if the bond is sold at a price equal to

its value.

(A) 8.1% (B) 8.3% (C) 8.5% (D) 8.7% (E) 8.9%

35. The current price of an annual coupon bond is 100. The derivative of the price of the bond

with respect to the yield to maturity is 700. The yield to maturity is an annual effectiverate of 8%. Calculate the duration of the bond.

(A) 7.00 (B) 7.49 (C) 7.56 (D) 7.69 (E) 8.00

36. Calculate the duration of a common stock that pays dividends at the end of each year intoperpetuity. Assume that the dividend is constant, and that the effective rate of interest is

10%.

(A) 7 (B) 9 (C) 11 (D) 19 (E) 27

37. Calculate the duration of a common stock that pays dividends at the end of each year into

perpetuity. Assume that the dividend increases by 2% each year and that the effective rate

of interest is 5%. (A) 27 (B) 35 (C) 44 (D) 52 (E) 58

38. Eric and Jason each sell a different stock short at the beginning of the year for a price

of 800. The margin requirement for each investor is 50% and each will earn an annual

effective interest rate of 8% on his margin account. Each stock pays a dividend of 16 at the

end of the year. Immediately thereafter, Eric buys back his stock at a price of (800 2X)and Jason buys back his stock at a price of (800 + X). Erics annual effective yield, i, on

the short sale is twice Jasons annual effective yield. Calculate i.

(A) 4% (B) 6% (C) 8% (D) 10% (E) 12%

39. Jose and Chris each sell a different stock short for the same price. For each investor, the

margin requirement is 50% and interest on the margin debt is paid at an annual effective

rate of 6%. Each investor buys back his stock one year later at a price of 760. Joses stock

paid a dividend of 32 at the end of the year while Chriss stock paid no dividends. During

the 1-year period, Chriss return on the short sale is i, which is twice the return earned by

Jose. Calculate i.

(A) 12% (B) 16% (C) 18% (D) 20% (E) 24%

40. Bill and Jane each sell a different stock short for a price of 1000. For both investors, the

margin requirement is 50%, and interest on the margin is credited at an annual effective

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rate of 6%. Bill buys back his stock one year later at a price of P. At the end of the year,

the stock paid a dividend of X. Jane also buys back her stock after one year, at a price of

(P 25). At the end of the year, her stock paid a dividend of 2X. Both investors earnedan annual effective yield of 21% on their short sales. Calculate P.

(A) 800 (B) 825 (C) 850 (D) 875 (E) 900

41. On January 1, 2005, Marc has the following options for repaying a loan:

Sixty monthly payments of 100 beginning February 1, 2005.

A single payment of 6000 at the end of K months.

Interest is at a nominal annual rate of 12% compounded monthly.

The two options have the same present value.

Determine K.

(A) 29.0 (B) 29.5 (C) 30.0 (D) 30.5 (E) 31.0

42. You are given an annuity-immediate with 11 annual payments of 100 and a final payment

at the end of 12 years. At an annual effective interest rate of 3.5%, the present value at

time 0 of all payments is 1000. Calculate the final payment.

(A) 146 (B) 151 (C) 156 (D) 161 (E) 166

43. A 10,000 par value bond with coupons at 8%, convertible semiannually, is being sold

3 years and 4 months before the bond matures. The purchase will yield 6% convertible

semiannually to the buyer. The price at the most recent coupon date, immediately after the

coupon payment, was 5640. Calculate the market price of the bond, assuming compoundinterest throughout.

(A) 5500 (B) 5520 (C) 5540 (D) 5560 (E) 5580

44. A 1000 par value 10year bond with coupons at 5%, convertible semiannually, is selling for

1081.78. Calculate the yield rate convertible semiannually.

(A) 1.00% (B) 2.00% (C) 3.00% (D) 4.00% (E) 5.00%

45. You are given the following information about an investment account:

Date Value Immediately Before Deposit Deposit

January 1 10

July 1 12 X

December 31 X

Over the year, the timeweighted return is 0%, and the dollar-weighted return is Y. Cal-

culate Y.

(A) 25% (B) 10% (C) 0% (D) 10% (E) 25%

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46. Seth borrows X for four years at an annual effective interest rate of 8%, to be repaid with

equal payments at the end of each year. The outstanding loan balance at the end of the

second year is 1076.82 and at the end of the third year is 559.12. Calculate the principal

repaid in the first payment.

(A) 444 (B) 454 (C) 464 (D) 474 (E) 484

47. Bill buys a 10-year 1000 par value 6% bond with semi-annual coupons. The price assumes

a nominal yield of 6%, compounded semi-annually. As Bill receives each coupon payment,

he immediately puts the money into an account earning interest at an annual effective rate

of i . At the end of 10 years, immediately after Bill receives the final coupon payment and

the redemption value of the bond, Bill has earned an annual effective yield of 7% on his

investment in the bond. Calculate i.

(A) 9.50% (B) 9.75% (C) 10.00% (D) 10.25% (E) 10.50%

48. A man turns 40 today and wishes to provide supplemental retirement income of 3000 at the

beginning of each month starting on his 65th birthday. Starting today, he makes monthly

contributions ofX to a fund for 25 years. The fund earns a nominal rate of 8% compounded

monthly. Each 1000 will provide for 9.65 of income at the beginning of each month starting

on his 65th birthday until the end of his life. Calculate X.

(A) 324.73 (B) 326.89 (C) 328.12 (D) 355.45 (E) 450.65

49. Happy and financially astute parents decide at the birth of their daughter that they will

need to provide 50,000 at each of their daughters 19th, 20th and 21st birthdays to fundher college education. They plan to contribute X at each of their daughters 1st through

17th birthdays to fund the four 50,000 withdrawals. If they anticipate earning a constant

5% annual effective rate on their contributions, which the following equations of value can

be used to determine X, assuming compound interest?

(A) X[1.05 + 2.05 + + 17.05] = 50, 000[1.05 + + 4.05].

(B) X[(1.05)16 + (1.05)15 + + (1.05)1] = 50, 000[1 + + 3.05].(C) X[(1.05)17 + (1.05)16 + + 1] = 50, 000[1 + + 3.05].(D) X[(1.05)17 + (1.05)16 +

+ (1.05)1] = 50, 000[1 +

+ 3.05].

(E) X[1 + 1.05 + 2.05 + + 17.05] = 50, 000[1 + + 22.05].

50. A 1000 bond with semi-annual coupons at i(2) = 6% matures at par on October 15, 2020.

The bond is purchased on June 28, 2005 to yield the investor i(2) = 7%. What is the

purchase price? Assume simple interest between bond coupon dates and note that:

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Date Day of the Year

April 15 105

June 28 179

October 15 288

(A) 906 (B) 907 (C) 908 (D) 919 (E) 925

51. Joe must pay liabilities of 1,000 due 6 months from now and another 1,000 due one year

from now. There are two available investments: a 6-month bond with face amount of 1,000,

a 8% nominal annual coupon rate convertible semiannually, and a 6% nominal annual

yield rate convertible semiannually; and a one-year bond with face amount of 1,000, a 5%

nominal annual coupon rate convertible semiannually, and a 7% nominal annual yield rate

convertible semiannually.

How much of each bond should Joe purchase in order to exactly (absolutely) match theliabilities?

Bond I Bond II

(A) 1 .97561

(B) .93809 1

(C) .97561 .94293

(D) .93809 .97561

(E) .98345 .97561

52. (use the information on the previous question) What is Joes total cost of purchasing the

bonds required to exactly (absolutely) match the liabilities?

(A) 1894 (B) 1904 (C) 1914 (D) 1924 (E) 1934

53. (use the information on the previous question) What is the annual effective yield rate for

investment in the bonds required to exactly (absolutely) match the liabilities?

(A) 6.5% (B) 6.6% (C) 6.7% (D) 6.8% (E) 6.9%

54. Matt purchased a 20-year par value bond with semiannual coupons at a nominal annual

rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at

par value X on any coupon date starting at the end of year 15 after the coupon is paid.

The price guarantees that Matt will receive a nominal annual rate of interest convertible

semiannually of at least 6%. Calculate X. (A) 1400 (B) 1420 (C) 1440 (D)

1460 (E) 1480

55. Toby purchased a 20year par value bond with semiannual coupons at a nominal annual

rate of 8% convertible semiannually at a price of 1722.25. The bond can be called at par

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value 1100 on any coupon date starting at the end of year 15. What is the minimum

yield that Toby could receive, expressed as a nominal annual rate of interest convertible

semiannually?

(A) 3.2% (B) 3.3% (C) 3.4% (D) 3.5% (E) 3.6%

56. Sue purchased a 10year par value bond with semiannual coupons at a nominal annual rate

of 4% convertible semiannually at a price of 1021.50. The bond can be called at par value

X on any coupon date starting at the end of year 5. The price guarantees that Sue will

receive a nominal annual rate of interest convertible semiannually of at least 6%. Calculate

X. (A) 1120 (B) 1140 (C) 1160 (D) 1180 (E) 1200

57. Mary purchased a 10year par value bond with semiannual coupons at a nominal annual

rate of 4% convertible semiannually at a price of 1021.50. The bond can be called at

par value 1100 on any coupon date starting at the end of year 5. What is the minimumyield that Mary could receive, expressed as a nominal annual rate of interest convertible

semiannually? (A) 4.8% (B) 4.9% (C) 5.0% (D) 5.1% (E) 5.2%

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2 Answers

Answers:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

C E C A E C C D D B A C E A B

16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

B C B C A A D D E D D E D B D

31 32 33 34 35 36 37 38 39 40 41 42 43 44 45

D C B E C C B B B E A B D D A

46 47 48 49 50 51 52 53 54 55 56 57

A B A D D D B D C A E B

3 Solutions

The solutions use the notation and formulas in the Binghamton Manual.

1. Since both balances are equal, 100(1 + 0.042

)(2)(7.25) = 100e(7.25), (1 + 0.042 )2 = e and

= 0.0396.

2. The cashflow is

Contributions 100 100 100 100 0

Time 0 4 8 36 40The 4year effective rate of interest is (1 + i)4 1. So, the accumulated amount in theaccount at the end of 40 years is

X = 100s10|(1+i)41 = 100(1 + i)4((1 + i)40 1)

(1 + i)4 1 ,

and the accumulated amount in the account at the end of 20 years is

100s5|(1+i)41 = 100(1 + i)4((1 + i)20 1)

(1 + i)4

1

.

So, 5 = (1 + i)20 + 1 and (1 + i)4 = 1.3195. So,

X = 100(1 + i)4((1 + i)40 1)

(1 + i)4 1 = 100(1.3195)(42 1)

1.3195 1 = 6194.68.

3. Erics interest is

x

1 +

i

2

16 x

1 +

i

2

15= x

1 +

i

2

15i

2.

Mikes interest is 2xi12 = xi. So, (1 +i2)

15 = 2 and i = 9.45988%.

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4. We find P solving 1000 = P a10|10%, we get that P = 1627.5. The deposits to the sinking

fund are of 627.5. The value of the sinking fund after the last deposit is 627.5s10|14% =

2133.33. So, the balance in the sinking fund immediately after repayment of the loan is

2133.3.

5. We have that V0 = 75 and F V = 60. We have deposits of 10 at times112

, 212

, . . . , 1212

, and

withdrawals of 5, 25, 80, 35, at respective times 212

, 12

, 9.512

, 1012

. We have that

I = F V V0 n

j=1

Cj = 60 75 (10)(12) + 5 + 25 + 80 + 35 = 10

and

V0t +n

j=1(t tj)Cj

= (75)(1) + (10)12

j=1

12j

12 5 1 1

12 25 1

1

2 80 1

9.5

12 35 1

10

12

= 75 + 1012(11)(12)

2 51012 2512 802.512 35 212 = 90.8347.

So,

i =I

V0t +n

j=1(t tj)Cj=

10

90.8347= 11%.

6. The cashflow is

Payments 0 1 2 n 1 n n Time 1 2 3

n n + 1 n + 2

The present value at time 0 of the perpetuity is

77.1 = (Ia)n|i + n+1n

1

i=

an|i(1 + i) nni

+n+1n

i=

an|ii

.

So, 8.0955 = an|10.5% and n = 19.

7. The balances along time in the Fund X are:

Balance 1000 900 800 0Time 0 1 2 10

Hence, the deposits at Fund Y are:

Balance 100 + (6) (10) 100 + (6) (9) 100 + (6) (8) 100 + (6) (1)Time 1 2 3 10

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The accumulated value of Fund Y at the end of year 10 is

100s10|9% + 6 (Ds)10|9% = 100s10|9% +6(10(1+0.09)10s

10

|9%)

0.09

= 1519.29 + 6(8.4807)0.09 = 2084.67.

8. Under the investment year method, the accumulated value on January 1, 2000 is

P = (1000)(1.095)(1.095)(1.096) = 1314.131.

Under the portfolio yield method, the accumulated value on January 1, 2000 is

Q = (1000)(1.0835)(1.086)(1.0885) = 1280.817.

When the balance is withdrawn and reinvested, the accumulated value on January 1, 2000

is

R = (1000)(1.095)(1.10)(1.10) = 1324.95.

So, R > P > Q.

9. Suppose that the principal payments are 150% of the amount of interest due. If the

balance at the end of one year is B, then the interest accrued in one year is B(0.10),

the amount paid to principal is B(0.15). So, the outstanding balance becomes B(0.05).

Hence, the outstanding balance reduces by 5% each year. Hence, the balance at time 10 is

(1000)(0.95)10 = 598.73693. Since this balance is paid by 10 payments of x, 598.7369 =

xa10|10% and x = 97.44168.

10. We have that F = C = 10000, F r = 800, i = 6% and n = 10. So,

I7 = Ci + (F r Ci)(1 n+1k) = 600 + (800 600)(1 (1.06)4) = 641.5813.

11. The value of the perpetuity immediately after the fifth payment is 100i = 1250. The

payments of the 25year annuityimmediate are

Inflows x x(1.08) x(1.08)2

x(1.08)24

Time 1 2 3 25

Since i = r, the present value at time 0 of this annuity is 1250 = 25x(1.08)1. Hence,

25x = 50(1.08) and x = 54. TO CHECK.

12. The following series of payments has present value zero:

Contributions 10 20 100Time 0 15 30

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Since the rate changes at 10, we find the present value of the payments at this time. We

have at time 10, the following series of payments have the same value

Contributions 10

Time 0

Contributions 20 100Time 15 30

The present value at time 10 of the payments at time 15 and at time 30 is

20(1.03)10 + 100(1.03)40 = 14.8818 + 30.6557 = 15.7738.

This equals the present value at time 10 of the initial deposit, i.e. 10

1 d440

= 15.7738.

So, d = 4.5318%.

13. We have that

a(t) = eRt0 s ds = e

Rt0

s2

100ds = e

t3

300 .

The accumulation of the investments at time t = 3 is 100a(3) + x. The accumulation ofthe investments at time t = 6 is a(6)a(3) (100a(3) + x). The amount of interest earned from

time 3 to time 6 is

x =a(6)

a(3)(100a(3) + x) (100a(3) + x) = a(6) a(3)

a(3) (100a(3) + x) .

Solving for x, we get that

x = 100a(3)(a(6)a(3))2a(3)a(6)

= 100e33

300 (e63

300e33

300 )

2e33300e

63300

= (100)(1.094174)(2.0544331.094174)(2)(1.094174)2.054433 = 784.0595.

14. Le r = k100 . The cashflow of the perpetuity is

Payments 10 10 10 10 10 10(1 + r) 10(1 + r)2 10(1 + r)3 Time 1 2 3 4 5 6 7 8

The present value at time 0 of this perpetuity is

167.5 = 10a4|i + 104 1

i r = 32.2555 + 7.03251

0.092 r .

So, r = 0.092 7.0325167.532.2555 = 4%.

15. Let P be the amount of equal annual payments made under the first option. Then,

P a10|0.0807 = 2000. So, P = 299. The total payments are 2990. Under the second op-

tion, at time j, the loan is 2000 200j = (200)(10 j). Hence, the payment at time j is(200)(10 (j 1))i = (200)(11 j)i. So, the total payments under the second option are

2990 =10

j=1(200 + (200)(11 j)i) = 2000 + 200i10

j=1(11 j) = 2000 + 200i10

j=1j =

= 2000 + 200i10112 = 2000 + 11000i.

So, i = 9%.

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16. The interest per period is i = (9/12)% = 0.75%. Let r = 0.02 and let P = 1000. Theoutstanding balance immediately after the 40th payment is

B40 = P(1 + r)k1ank| ir

1+r= 1000(0.98)39a20| 0.0075+0.02

0.98= 454.7963a20|0.028061 = 6889.06.

17. We have that

8000 = 98(1 + i)2nsn|i + 196s2n|i = 98(1 + i)2n (1+i)

n1i

+ 196 (1+i)2n1i

= (98)22 21i

+ 196221i

= 980i

So, i = 9808000 = 0.1225 = 12.25%.

18. The cashflow is:

Payments 2 4 6

120

Time (in months) 1 2 3 60

We have i(4) = 9%. So, i = 9.3083% and i(12) = 8.933%. We use the formula for the

increasing annuities with n = 60 and i = 8.933%/12 = 0.74444%:

2 (Ia)n|i = 2 an|i nn

i=

2(10.158740)

0.00744= 2729.21.

19. The dollarweighted return for investment account K is

125 100 + x 2x100 x/2 + 2x/4 = 25 x100 .

The time-weighted return for investment account L is

125

100

105.8

125 x 1 =132.25

125 x 1 =7.25 + x

125 x .

So, 25x100

= 7.25+x125x

, x2 150x + (25)(125) = 725 + 100x, x2 250x 4100 = 0, andx =

250

2502(4)(4100)

(2)= 10. Hence, i = 2510

125= 15%.

20. Since the equations of value of the two streams of payments at time zero are equal,

100 + 200n + 3002n = 60010.

Using that n = 0.75941, we have that

600(1 + i)10 = 60010 = 100 + 200(0.75941) + 300(0.75941)2 = 424.89.

So, i = 3.51%.

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21. We have that a(t) = eRt0

18+s

ds = eln(8+t)ln 8 = 8+t8

. The future value of the continuous

cashflow is

20000 =100 C(s)

a(10)a(s) ds =

100 k(8 + s) 8+108+s ds =

100 18k ds = 180k.

So, k =20000180 = 111.11.

22. Let C be the redemption value of the bonds are. Let i be the semiannual effective interest

rate. Let = (1 + i)1. Then, 381.50 = C2n and 647.80 = Cn. So, 381.50647.80 = n and

(1 + i)n = 1.6980. The present value of the coupons in bond X is

F ra2n|i = F r1 2n

i= (1000)(1.03125)(1 (1.6980)2) = 673.59.

So, the total present value of the bond X is 381.50 + 673.59 = 1055.09.

23. For Project P, the net present value of the investment is

2000(1.1)1 + 4000(1.1)2 4000 = 1123.967.

For Project Q, the net present value of the investment is

2000 + 4000(1.1)1 x(1.1)2 = 5636.364 x(1.1)2

So, 1123.967 = 5636.364 x(1.1)2 and x = (1.1)2(5636.364 1123.967) = 5460.

24. Let P be the payment under the first option Then, 20000 = P a20|0.065 and P = 1815.12.

The annual interest payment under the second option is 20000(0.08) = 1600. So, the

deposit into the sinking fund is 1815.12 1600 = 215.12. So, 20000 = 215.12s20|j andj = 14.18%.

25. The total value of the perpetuityimmediate is xa| =xi

. The present value of Brian

payments is xan|i =x(1n)

i. So, Brians proportion of the perpetuity is 1 n = 0.40. So,

n = 0.6. The present value of Jeffs payments is 2nx1i. So, Jeffs share is 2n = 0.36 =

36%.

26. Seths interest is 5000((1.06)10 1) = 3954.23. Janices interest is(10)(5000)(0.06) = 3000. Loris semiannual payment is 5000

a10|0.06

= 679.34. So, Loris interest

is (10)(679.34) 5000 = 1793.4. The total amount of interest paid is 3954.23 + 3000 +1793.4 = 8747.63.

27. Since each account earns an annual effective discount rate ofd, each account earns the same

annual effective interest rate i. Bruces interest in the 11th year is 100((1+i)11(1+i)10) =100i(1 + i)10. Robbies interest in the 17th year is 50((1 + i)17 (1 + i)16) = 50i(1 + i)16.So, 100i(1 + i)10 = 50i(1 + i)16, (1 + i)6 = 2, i = 12.24% and x = 100i(1 + i)10 =

100(0.1224)(1 + 0.1224)10 = 38.85.

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28. The interest paid in period t is 1 n+1t. The principal repaid in period t + 1 is nt. So,

x = 1 n+1t + nt = 1 + nt(1 ) = 1 + ntd.

29. The present value of the first perpetuity is 32 = 10(1+i)31

. So,i

= 9.4979%. The present

value of the second perpetuity is 1(1+i)1/31

= 1(1.094979)1/31

= 32.59.

30. Note that if the rates remain at 5%,

(0.05)(8227093)s4|5% + (8227093) = 1000000

the company pays the liability. If the interest rates drop to 4.5%, the accumulation value

of the investment is

(0.05)(8227093)s4

|4.5% + (8227093) = 998687

and there is a loss of 1000000998687 = 1313. If the interest rates drop to 4.5%, the priceof the bond is

(0.05)(8227093)a4|5.5% + (8227093)(1.05)4 = 1001323

and there exists a gain of 1001323 1000000 = 1323.

31. The cashflow is

Payments 5000(1.07) 5000(1.07)2 5000(1.07)3 5000(1.07)20Time 1 2 3 20

The present value of this obligation is 5000a20|(1.05/1.07)1 = 122634.

32. The investor return from his investment is (60000)(1.04) + (60000) = 122400. At a rate

of interest at which the investment is to be valued is 5.0%, the net present value of this

investment today is (122400)(1.05)4 100000 = 698.7829.

33. The price of the bond is (60)(1.07)

1

+ (60)(1.08)

2

+ 1060(1.09)

3

= 926.0296.

34. Since the price of the bond is 926.0296. To find the yield rate, we solve for i in 926.0296 =

60a3|i + 1000(1 + i)3 and get i = 8.9181%.

35. The duration is d = P(i)(1+i)P(i) =

700(1.08)100 = 7.56.

36. Here, P(i) = Di and P(i) = Di2 . So, the duration is d =

1+ii

= 1.10.1 = 11.

37. The cashflow is

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Contributions D D(1.02) D(1.02)2 Time 1 2 3

The present of the cashflow under an effective rate of interest i, i > 0.02, is

P(i) =k=1

D(1.02)k1(1 + i)j =D(1 + i)1

1 1.021+i=

Di 0.02 .

The duration is

d =(1 + i)P(0.05)

P(0.05)=

(1.05) D(0.050.02)2D

0.050.02

=1.05

0.05 0.02 = 35.

38. For Eric, P = 800 (800 2X) = 2X, M = (800)(0.50) = 400, I = (400)(0.08) = 32and D = 16. Erics net profit is P + I

D = 2X + 32

16 = 2X + 16. Erics interest is

i = P+IDM =2X+16400 =

X+8200 . For Jason, P = 800(800+X) = X, M = (800)(0.50) = 400,

I = (400)(0.08) = 32 and D = 16. Jasons net profit is P+ID = X+3216 = X+16.Jasons interest is i = P+IDM =

X+16400 . Solving for X in

X+8200 = 2

X+16400 , we get that X = 4

and i = X+8200

= 4+8200

= 6%.

39. Let x be the price at which Jose and Chris each sell a different stock short. For Jose,

P = 760 x, M = x(0.50), I = x(0.50)(0.06) = (0.03)x and D = 32. Joses yield isx760+(0.03)x32

(0.5)x . For Chris, P = 760 x, M = x(0.50), I = x(0.50)(0.06) = (0.03)x andD = 0. Chris yield is i = x760+(0.03)x(0.5)x . We have

i =x 760 + (0.03)x

(0.5)x= 2 x 760 + (0.03)x 32

(0.5)x

or

2 1520x

+ 0.06 = 4 3040x

+ 0.12 128x

which implies (1520 + 3040 + 128) 1x = 4 + 0.12 2 0.06 and x = 1520+3040+1284+0.1220.06 = 800.The interest is i = 800760+(0.03)(800)

(0.5)(800)= 16%.

40. For Bill, profit= 1000

P, M = (0.5)(1000) = 500, I = 500(0.06) = 30 and D = x.

Bills yield is 0.21 = 1000P+30x500 . So, P + x = 925. For Jane, profit= 1000 (P 25) =1025 P, M = (0.5)(1000) = 500, I = 500(0.06) = 30 and D = 2x. Janes yield is0.21 = 1025P+302x500 . So, P+2x = 950. From the equations, P+x = 925 and P+2x = 950,

we get x = 25 and P = 900.

41. We have that (100)a60|1% = 4495.504 = 6000(1.1)k and k = 29.01226977.

42. The cashflow is

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Contributions 100 100 100 2xTime 1 2 11 12

We solve the equation

1000 = 100a11|3.5% + x(1.035)12 = 100a12|3.5% + (x 100)(1.035)12.

Pressing in the calculator, 12 N -1000 PV 3.5 I/Y 100 PMT CPT FV

We get that x 100 = 50.87 and x = 150.87.

43. Using the theoretical method, the market price of the bond is

Bmk+t = Bk(1 + i)t F r

(1 + i)t 1

i

.

Since 6 months is a period, 2 months is t =13 periods. So,

Bmk+t = (5640)(1.03)1/3 (400)

(1.03)1/3 1

0.03

= 5563.82.

44. We have F = C = 1000, r = 2.5%, F r = 25 and n = 20. We solve the equation

1081.78 = 25a20|i% + 1000(1 + i)20

and get that i = 2%. The annual yield rate convertible semiannually is i(2) = 4%.

45. Since the timeweighted return is 0%, 1210

x12+x

= 1. So, x = 60. The dollarweighted return

y satisfies

y =10 + 60 60

(10)(1) + (60)(1/2)= 0.25 = 25%.

46. Let P be the amount of each payment. The outstanding loan balance at the end of the

second year is 1076.82 = P a2|0.08. So, P = 603.85. The principal repaid in the first

payment is P n = 603.85(1.08)4 = 443.85.

47. Since Bill got an effective annual rate of interest of 7% in his investment, Bill got at the

end 1000(1.07)10 = 1967.15. The coupon payments are 1000(.03) = 30. So, 1967.15 =

30s20|i(2)/2 + 1000 and 967.15 = 30s20|i(2)/2. So, i(2)/2 = 4.7597% and i = 9.7459%.

48. To get an income of $1, the man needs 10009.65 . So, to get an income of $3000, the man needs(3000)(1000)

9.65 = 310880.83. We solve for x in the equation xs(25)(12)|8%/12 = xs300|0.33333% =

310880.83 and we get x = 324.72.

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49. In (A), X[1.05+2.05+ +17.05] is the PV at time 0 of the investment payments. 50, 000[1.05+

+ 4.05] is the PV at time 17 of the liability payments. (A) is not correct.In (B), X[1.05 +

2.05 + + 17.05] has only 16 payments. The parents made 17 payments.

(B) is not correct.

In (C), X[(1.05)17 + (1.05)16 + + 1] has 18 payments. The parents made 17 payments.(C) is not correct.

In (D), X[(1.05)17+(1.05)16+ +(1.05)1] is the PV at time 18 of the investment payments.50, 000[1 + + 3.05] is the PV at time 18 of the liability payments. (D) is correct.In (E), X[1 + 1.05 +

2.05 + + 17.05] has 18 payments. The parents made 17 payments. (E)

is not correct.

50. We have F = C = 1000, r = 3%, F r = 30 and i = 3.5%. On April 15, 2005, there are 31

periods left. So, the price of the bond on April 15, 2005 is

30a31|3.5% + 1000(1.035)31 = 906.3186.

Assuming simple interest the purchase price of the bond is

906.3186

1 +

179 105365

0.07

= 918.1808.

51. (i) The cash flow of liabilities is

Liabilities 1000 1000Time (in months) 6 12

Let x be the amount of the six-month bond, which Joe buys. Let y be the amount of the

one-year bond, which Joe buys. The cash flow of assets is

Assets 1040x+25y 1025y

Time (in months) 6 12

To match the liabilities, x and y must satisfy

1040x + 25y = 1000, 1025y = 1000,

i.e. y = 10001025 = 0.9756098 and x =100025y

1040= 0.9380863.

52. The price of the six-month bond is (1040)(1.03)1 = 1009.709. The price of the one-year

bond is (25)(1.035)1 + (1025)(1.035)2 = 981.0031. Joes total cost of purchasing the

bonds is

(0.9380863)(1009.709) + (0.9756098)(981.0031) = 1904.27.

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53. We solve for i(2) in the equation

1904.27 = (1000)

1 +

i(2)

2

1+ (1000)

1 +

i(2)

2

2

to get i(2)

= 6.6664% and i = 6.7775%.

54. We have F = C = x, P = 1722.25, r = 4%, i = 3.5% and n = 40. Since r > i, the bond

was sold at premium. We assume that the redemption value is as soon as possible. The

price of the bond is

1722.25 = (0.04)xa30|3% + x(1.03)30 = 1.196004x

and x = 1440.004.

55. We have F = C = 1100, P = 1722.25, r = 4% and F r = 44. Since P > C, the bond wasbought at premium. We assume that the redemption value is as soon as possible. From

the equation

1722.25 = (44)a30|i + 1100(1 + i)30

we get that i = 1.60824% and i(2) = 3.20165%.

56. We have F = C = x, P = 1021.50, r = 2%, i = 3%, F r = (0.02)x and n = 20. Since

i > r, the bond was bought at discount. We assume that the redemption value is as late

as possible. From the equation

1021.50 = (0.02)xa20|3% + x(1.03)20 = (0.8512253)x

we get that x = 1200.035.

57. We have F = C = 1100, P = 1021.50, r = 2%, F r = 22 and n = 20. Since C > P, the

bond was bought at discount. We assume that the redemption value is as late as possible.

From the equation

1021.50 = 22a20|i + 1100(1 + i)20

we get that i = 2.4558 and i(2) = 4.9117%.

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