Contemporary Business Mathematics with Canadian Applications 12E Sieg A. Hummelbrunner Test Bank.pd

Mont2 3000 -1000 = 2000 Loan 10.00 h 3 A20m0o0u -n10t 0o0w =i Interest for Month 5.00 1 n1g0 00 3000.00 Total Inter1e5st.0 =0 $30.00 44) M onthly payment = 3900.00 ÷ 4 = $975.00 Monthly rate of interest = .0956 ÷ 12 = .00796666667 Month 1 2 3 4

Loan Amount Owing $3,900.00 3900.00 - 975.00 = 2925.00 2925.00 - 975.00 = 1950.00 1950.00 - 975.00 = 975.00 Total Interest

Interest for Month $31.07 23.3 15.51 7.77 $77.68

45) R epayment schedule Balance Payment number 0 1 2 3 4 5 Totals

Balance

before payment

Amount paid

Interest paid

Principal repaid

2900.00 2376.88 1855.00 1329.91

535.00 535.00 535.00 535.00

11.88 13.12

802.25

806.68 2946.68

523.12 521.88 525.09 527.66 802.25 2900.00

9.91 7.34 4.43 46.68

after payment $2900.00 2376.88 1855.00 1329.91 802.25

46) Repayment Schedule Payment number 0 1 2 3 4 5 6 Totals

Balance before payment 1200.00 972.20 742.08 509.62 274.80 37.59

Amount paid

Interest paid

Principal repaid

240.00 240.00 240.00 240.00 240.00

12.20 9.88 7.54 5.18 2.79 .38

227.80 230.1 232.46 234.82 237.21

37.97 1237.97

Balance after payment 1200.00 972.20 742.08 509.62 274.80 37.59

37.59 1200.00

37.97

a) $509.62 47)

b) $1237.97 Repayment Schedule Payment number 0 1 2 3 4

Balance before payment

Amount paid

Interest paid

Principal repaid

5200 4287.7 3368.79 2443.21

950 950 950 950

37.70 31.09 24.42 17.71

912.30 918.91 925.58 932.29

Balance after payment 5200 4287.70 3368.79 2443.21 1510.92

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5

Totals

1510.92

950 5326 .0 21 0.9 51 26. 0 29 39.0552571.87 00 .0 0

571.87

576.02

48) D 49) C 50) D 51) D 52) B 53) C 54) B 55) A 56) D 57) A 58) E 59) B 60) A 61) A 62) D 63) D 64) D

4.15

571.87

0.00

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65) P resent value of $50,000 discounted at 1.04% for 180 days =

= $49 744.87

Days remaining to maturity = 180 - 39 = 141 days Selling price =

= $49 782.69

Effective interest on initial investment = $49 782.69 - $49 744.87 = $37.82 Rate of return =

=

= 0.00712 = 0.712%

66) C onstruct a repayment schedule, listing chronological order of all dates on which a transaction or an event affecting the loan occurs. Use I = Prt to calculate interest at an interest rate of 4% Number of Date July 15 August 15 August 20 September 10 September 15 October 1 October 15 November 1 November 15

Days 0 31 5 21 5 15 14 17 14

Total interest accumulated = $658.19 67) B 68) D 69) B 70) B

Payment / Interest 0 $152.88 $24.66 $121.97 $34.52 $110.47 $73.64 $89.42 $50.63

(Advance) ($45000) 0 ($8000) ($10000) 0 $15,000 0 $15,000 0

Balance $45000 $45000 $53000 $63000 $63000 $48000 $48000 $33000 $33000

71) C 72) I nterest paid to the credit card = $14 990 × 0.28 ×

= $160.99

Total payment to the credit card using secured line of credit = $14 990 + $160.99 = $15 150.99 Interest paid to the secured line of credit = $15 150.99 × 0.04 ×

= $278.94

Total interest paid on the purchase of the painting = $160.99 + $278.94 = $439.93 73) I nterest paid to the credit card = $14 990 × 0.28 × = $160.99 Total payment to the credit card using secured line of credit = $14 990 + $160.99 = $15 150.99 Interest rate for first 90 days = 3.5% Interest rate for (168 - 90) = 78 days = 4% + $15 150.99 × 0.04 ×

Interest paid to the secured line of credit = $15 150.99 × 0.035 ×

= $260.26

Total interest paid on the purchase of the painting = $160.99 + $260.26 = $421.25 74) I nterest paid after first 4 months = $5000 × 0.06 × = $100 Let p be the first installment, then balance after first four months Interest paid after next 4 months = ($5100 - p) × 0.06 ×

= $5000 - p + 100 = $5100 -p = $102 - 0.02p

Total interest paid = $100 + $102 - 0.02p = 202 - 0.02p Balance after 8 months = $5000 - 2p + 202 -0.02p = 0 ⇒ 5202 - 2.02p = 0 ⇒p=

= $2575.25

75) B 76) A 77) T here are 153 days from the issue date until the legal due date. By August 21, 67 of the days have passed and 86 days remain. Since it is a non-interest bearing note, the future value remains the same as the present value. Hence 86 days earlier, at a discount rate of 12.5%: = $24 576.18

Proceeds =

78) S = $4520; r = 2%; t = P=

=1

= $4431.37

Amount paid by Sally = $4431.37 1.13 = $5007.45 79) A 80) B 81) C 82) B 83) A 84) B 85) D 86) Payment Balance Before No. Payment

Amount Paid

Interest Paid

Principal Repaid

Balance After Payment

0 1 2 3 4

5 6 7 8

$11,2 48.07 $3,000.00 $37.49 $2,9 62.51 $ 23,000 $.080,2 85.56 $23,000 $.080,2 85.5 6$ 3,000 $.030,0 00.00 $76.67 $ 27.62 $2,923 $.323,9 72.38 $ 20,076 $.657,3 13.18 $20,076 $.657,3 13.1 8$ 3,000 $.030,0 00.00 $66.92 $ 17.71 $2,933 $.028,9 82.29 $ 17,143 $.529,3 30.89 $17,143 $.529,3 30.8 9$ 3,000 $.020,3 38.66 $57.15 $7.77 $2,942 $.825,3 30.89 $$ 14,200.73 $14,200.73 $3,000.00 $47.34 $2,952.66 $11,248.07

87) Payment No. 0 1 2 3 4 5 6 7 8

Balance Before Payment

Amount Paid Interest Paid

$1,450.55 $1,258.41 $1,065.22 $870.99 $675.71 $479.37 $281.97 $83.50

$200.00 $200.00 $200.00 $200.00 $200.00 $200.00 $200.00 $83.95

Principal Repaid

$7.86 $6.82 $5.77 $4.72 $3.66 $2.60 $1.53 $0.45

$192.14 $193.18 $194.23 $195.28 $196.34 $197.40 $198.47 $83.50

Balance After Payment $1,450.55 $1,258.41 $1,065.22 $870.99 $675.71 $479.37 $281.97 $83.50 $0.00

88) L egal due date is October 21, 2013.

Interest period April 20, 2019 to October 21, 2019 is 184 days.

Interest = 5500(0.085)

= 235.67

89) a ) Since the legal due date is June 21, the interest period is 131 days. b) S = 890

= $912.20

90) S = 100 000; r = 0.0305; t = P=

= $99 245.33

91) S = 75 000.00; r = 0.0306; t = P=

= $74 283.82

92) Interest Period July 23 - July 31 August 1 - August 11 August 12 - August 31 September 1 - September 4 September 5 - September 15

Principal 3200.00 3200.00 2600.00 2600.00 1550.00

Rate 0.0696 0.0696 0.0696 0.0696 0.0696

93) Payment Balance before number payment 0 1 9100 2 7902.33

Time 0.02465753 0.030136986 0.054794521 0.010958904 0.071232877

Interest 5.4 9 6.7 1 9.9 2 Total interest = $31.79 1.9 8 Principal Balance after repaid payment 7.6 9100 8

Amount paid

Interest paid

1250

52.33

1197.67

1250

45.44

1204.56

7902.33 6697.76

3 4 5 6 7 8 Totals

6697.76 5486.28 4267.82 3042.36 1809.86 570.26

1250 1250 1250 1250 1250

38.51 31.55 24.54 17.49 10.41

1211.49 1218.45 1225.46 1232.51 1239.59

5486.28 4267.82 3042.36 1809.86 570.26

573.54 9323.54

3.28 223.54

570.26 9100.00

0.00

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) F ind the compound amount of $5750.00 at 13.2% p.a. for seven years compounded 1) _____________ monthly.

2) D etermine the accumulated value of $4100.00 compounded semi-annually at 8% p.a. for seven years.

2) _____________

3) C alculate the accumulated value of $3000.00 at 8% compounded quarterly for fifteen years. How much of the amount is interest?

3) _____________

4) H ow much will a registered retirement savings deposit of $11 750.00 be worth in 10 years at 8.44% compounded semi-annually? How much of the amount is interest?

4) _____________

5) H ow much will a registered retirement savings deposit of $10 000.00 be worth in 15 years at 6.00% compounded quarterly? How much of the amount is interest?

5) _____________

6) D arcy's parents made a trust deposit of $3500.00 on October 31, 2002, to be withdrawn on Darcy's eighteenth birthday on July 31, 2016. To what will the deposit amount on that date at 13.48% compounded quarterly?

6) _____________

7) T o what future value will a principal of $7100.00 amount in three years at 7.6% p.a. compounded: a) annually b) semi-annually c) quarterly d) monthly

7) _____________

8) F ind the future value of and the compound interest on $4575.00 invested at 12.24% compounded monthly for: a) 3.5 years b) 7 years c) 12.5 years

8) _____________

9) L eridian Credit Union expects an average annual growth rate of 16% for the next four

9) _____________

years. If the assets of the credit union currently amount to $16.3 million, what will the forecasted assets be in four years?

10) G BC Credit Union expects an average annual growth rate of 10% for the next 10 years. If the assets of the credit union currently amount to $50 million, what will the forecasted assets be in ten years?

10) _____________

11) U se the exact method to determine the accumulated value of $3875.00 due in 61 months compounded annually at 9.75% p.a.

11) _____________

12) S uppose $4320.00 is invested for five years, eight months at 8.25% compounded annually. What is the compounded amount?

12) _____________

13) F ind the maturity value of a promissory note for $1400.00 dated February 29, 2016, and

due on August 29,

2021, if 13) interest is 7.64% compou nded quarterly .

___ ___ ___ ___ _

14) A $5725.00 investment matures in three years, seven months. Find the maturity value if interest is 9.16% p.a. compounded quarterly.

14) _____________

15) A $8000.00 investment matures in five years, three months. Find the maturity value if interest is 12% p.a. compounded quarterly.

15) _____________

16) A n investment of $21 700 is accumulated at 5.24% compounded quarterly for three and

16) _____________

one-half years. At that time the interest rate is changed to 6.12% compounded monthly.

How much is the investment worth two years after the change in interest rate?

17) D etermine the maturity value of $5400 due in 91 months compounding annually at 8.75%.

17) _____________

18) D etermine the maturity value of $2000 due in 63 months compounding annually at 10%.

18) _____________

19) A t Alpha Financial Services, Michael is offered three-year term deposits at 9.125% compounded annually. Beta Bank offers such deposits at 9.0% compounded semi-annually. If he has $3000 to invest, what is the maturity value of Michael's deposit? a) at Alpha Financial Services? b) at Beta Bank?

19) _____________

20) A lternative Savings offers five-year term deposits at 10% compounded annually while your credit union offers such deposits at 9.6% compounded quarterly. If you have $1000 to invest, what is the maturity value of your deposit a) at Alternative Savings? b) at your credit union?

20) _____________

21) T he Rob U Blind Bank advertises capital savings at 7.128% compounded semi-annually while Take Your Money Trust offers premium savings at 7.1% compounded monthly.

Suppose you have $4400.00 to invest for two years. a) Which deposit will earn more interest? b) What is the difference in the amount of interest?

21) _____________

22) B oston Holdings offers a savings account at 1.2% compounded monthly while Albany Secure Savings offers premium savings at 1.236% compounded yearly. Suppose you have $8100.00 to invest for two years. a) Which deposit will earn more interest? b) What is the difference in the amount of interest?

22) _____________

23) T wo years after Sean deposited $5000 in a savings account that earned interest at 6% compounded monthly, the rate of interest was changed to 6.4% compounded semi-annually. How much was in the account fifteen years after the deposit was made?

23) _____________

24) B etty deposited $17 150.00 in an RRSP on March 1, 2010, at 6.4% compounded quarterly.

Subsequently the

interest 24) rate was changed to 6.6% compou nded monthly on Septemb er 1, 2012, and to 6.8% compou nded semi-ann ually on June 1, 2014.

What was the value of the RRSP deposit on Decembe r 1, 2016, if no further changes in interest were made?

___ ___ ___ ___ _

25) F ahira invested $5000.00 at 4.8% compounded quarterly on January 1, 2012. On July 1, 2013 the rate changed to 5.6% p.a. compounded semi-annually. Calculate the amount Fahira will have on January 1, 2005.

25) _____________

26) A n investment of $2500.00 accumulates interest at 9.25% compounded quarterly. After 18 months the rate changed to 9.75% compounded semi-annually. Calculate the accumulated value three years after the initial investment.

26) _____________

27) A n investment of $4300.00 earns interest at 9.96% p.a. compounded monthly for three and a half years. At that time the interest rate is changed to 9% compounded semi-annually. How much will the accumulated value be six years after the change?

27) _____________

28) A n investment of $5000.00 earns interest at 6% p.a. compounded monthly for two years.

At that time the interest rate is changed to 8% compounded semi-annually. How much will the accumulated value be three and a half years after the change?

28) _____________

29) S ix years after Mr. Robertson deposited $7110.00 in a savings account that earned interest at 6.8% p.a., compounded semi-annually, the interest was changed to 7.2% p.a.,

co ounded mpquarterly. How

much 29) was in the account ten years after the deposit was made?

___ ___ ___ ___ _

30) A ccumulated $1720.00 at 8.4% p.a. compounded monthly from March 1, 2011, to July 1, 2013, and thereafter at 8.88% p.a. compounded quarterly. What is the amount on April 1, 2015?

30) _____________

31) Y ou borrowed $1700.00 at 12.36% p.a. compounded monthly, and repaid $800.00 after three years and $950.00 after five years. How much do you owe at the end of the nine years?

31) _____________

32) A llison started an RRSP on March 1, 2011, with a deposit of $2470.00. She added $1900.00 on December 1, 2012, and $1850.00 on September 1, 2013. What is the accumulated value of her account on December 1, 2015, if interest is 7.24% compounded quarterly?

32) _____________

33) F ind the compound interest earned by $1910 invested at 7.5% compounded semi-annually for seven years.

33) _____________

34) F ind the compound interest earned by $5000 invested at 3.2% compounded semi-annually for 18 years.

34) _____________

35) S ix years after Ellen deposited $4500 in a savings account that earned interest at 4.68% compounded monthly, the rate of interest was changed to 6.4% compounded semi-annually. How much was in the account thirteen years after the deposit was made?

35) _____________

36) S ean started an RRSP on March 1, 2007, with a deposit of $2000.00. He added $1500.00 on March 1, 2008. What is the accumulated value of his account on December 1, 2008, if interest is 6% compounded quarterly?

36) _____________

37) Y ou have a line of credit loan with the Bank of Hong Kong. The initial loan balance was $7000.00. Payments of $3000.00 and $2500.00 were made after four months and ten months respectively. At the end of one year, you borrowed an additional $4250.00.

Seven months later, the line of credit loan was converted into a collateral mortgage loan.

What was the amount of the mortgage if the line of credit interest was 8.52% compounded monthly?

37) _____________

38) A variable rate demand loan showed an initial balance of $20 000.00, payments of $5000.00 after six months, $10000.00 after one year, and a final payment after two years.

Interest was 6% compounded semi-annually for the first year and 12% compounded monthly for the remaining time. How much was the size of the final payment?

38) _____________

39) W hat is the present value of $7800.00 payable in six years if the current interest rate is 7.6% p.a., compounded quarterly?

39) _____________

40) D etermine the sum of money that will grow to $18 530 in four years, eleven months at 5.2% compounded quarterly.

40) _____________

41) D etermine the proceeds of $14 000 three years and three months before the due date if interest is 7.6% compounded quarterly.

41) _____________

42) F ind the present value and the compound discount of $6100.00 due in 7.5 years if interest is 8.26% compounded semi-annually.

42) _____________

43) D etermine the proceeds of $5000 four years and nine months before the due date if interest is 8% compounded semi-annually.

43) _____________

44) C alculate the present value of $12 500.00 due in two years and nine months if interest is 7.8% p.a. compounded semi-annually.

44) _____________

45) F ind the present value and the compound discount of $6 600.00 due in seven years, three months, if interest is 7.2% compounded quarterly.

45) _____________

46) W hat sum of money invested at 5.95% p.a., compounded semi-annually, will grow to $28 000.00 in 19.5 years?

46) _____________

47) W hat sum of money invested at 5.2% p.a., compounded biweekly, will grow to $10 000.00 in 12.5 years?

47) _____________

48) W hat sum of money will grow to $23 000.00 in seven years at 9.612% compounded monthly?

48) _____________

49) H ow much would you have to deposit in an account today to have $37 000.00 in a three-year term deposit at maturity if interest is 7.775% compounded annually?

49) _____________

50) C ompute the discounted value of $5125.00 due in three years, eight months if money is worth 8.24% compounded quarterly.

50) _____________

51) F ind the principal which will grow to $7258.00 at 6.58% compounded semi-annually in four years and five months.

51) _____________

52) R ichard wants to have $114 150.00 in six years, seven months from now. How much would he have to invest right now to achieve his goal, if he is getting 7.6% compounded quarterly rate of return on his investment?

52) _____________

53) C alculate the proceeds of $8956.00 due in seven years, eleven months discounted at 7.5% compounded semi-annually.

53) _____________

54) A $41 200.00, non-interest-bearing note due August 1, 2003, discounted on March 1, 2001, at 7.32% compounded monthly. Find the proceeds.

54) _____________

55) A n eleven-year promissory note discounted after six years at 9.2% compounded quarterly has a maturity value of $71 500. Find the proceeds.

55) _____________

56) A twenty-year promissory note discounted after seven years at 10% compounded quarterly has a maturity value of $40 000. Find the proceeds.

56) _____________

57) A $5000.00, six-year note bearing interest at 8.24% compounded quarterly, discounted three and a half years after the date of issue at 6.6% compounded monthly. Find the proceeds.

57) _____________

58) A ten-year promissory note dated April 1, 2001, with a face value of $4700.00 bearing interest at 7.2% compounded semi-annually, discounted seven years later when money was worth 9.92% compounded monthly. Find the proceeds.

58) _____________

59) A seven-year, non-interest-bearing note for $6532.00 is discounted three years and two months before its due date at 9.12% compounded quarterly. Find the proceeds of the note.

59) _____________

60) U se the exact method to compute the proceeds of a non-interest-bearing note for $5640.00 six years and seven months before the due date, if money is worth 7.25% p.a. compounded annually.

60) _____________

61) A $4300.00 promissory note issued without interest for nine years on September 30, 2001, is discounted on July 31, 2006, at 8.32% compounded quarterly. Find the compound discount.

61) _____________

62) F our years and five months after its date of issue, a 7-year promissory note for $8900.00 bearing interest at 5.04% compounded monthly is discounted at 6.5% compounded semi-annually. Find the proceeds of the note using the exact method.

62) _____________

63) O n July 28, 2018, a ten-year note dated January 28, 2015, is discounted at 8.2% compounded quarterly. If the face value of the note is $4000.00 and interest is 7.2% compounded quarterly, find the compound discount.

63) _____________

64) A ten-year note for $2220.00 bearing interest at 6.6% compounded monthly is discounted at 8.92% compounded quarterly four years and four months before maturity. Find the proceeds of the note.

64) _____________

65) A twenty-year note for $10 000.00 bearing interest at 12% compounded monthly is discounted at 8% compounded quarterly four years and six months before maturity.

Find the proceeds of the note.

65) _____________

66) A seven-year promissory note for $3100 dated May 2, 2016, bearing interest at 8% compounded quarterly, is discounted on September 2, 2019, at 8.5% compounded semi-annually. Determine the proceeds of the note.

66) _____________

67) A note dated May 1, 2011 promises the payment of $5660.00 with interest at 6.5% p.a. compounded semi-annually on November 1, 2015. Find the proceeds of the sale of the note on May 1, 2013 if money was then worth 7.2% p.a. compounded monthly.

67) _____________

68) C alculate the proceeds of $5000.00 due in five years, nine months discounted at 8.0% compounded semi-annually.

68) _____________

69) A debt can be repaid by payments of $4125.00 today, $3770.00 in two years and $5600.00 in five years. What single payment would settle the debt three years from now if money is worth 9.88% p.a. compounded quarterly?

69) _____________

70) A debt can be repaid by payments of $1000.00 today, and $3000.00 in two years. What

single payment

would 70) settle the debt four years from now if money is worth 1.6% p.a. compou nded semi-ann ually?

71) L oans of $1600.00, $8300.00, and $12 100.00 are due now, in four years, and in seven

___ ___ ___ ___ _

71) _____________

years respectively. What is the equivalent single sum of money due three and a half years from now if interest is 10.8% compounded monthly?

72) D etermine the discounted value now of $7700.00 due in forty-four months at 7.5% compounded quarterly.

72) _____________

73) D ebts of $400.00, $450.00 and $500.00 are due in one year, eighteen months and thirty months from now respectively. Determine the single payment now that would settle the debts if interest is 8% p.a. compounded quarterly.

73) _____________

74) S cheduled debt payments of $1500.00 due seven months ago, $1200.00 due two months

74) _____________

ago, and $1800.00 due in five months are to be settled by two equal payments now and three months from now respectively. Determine the size of the equal replacement payments at 9% p.a. compounded monthly.

75) A debt of $5000.00 is to be repaid by payments of $2000.00 after two years, $2500.00 after three years and a final payment after five years. Determine the size of the final payment if interest is 10% p.a. compounded semi-annually.

75) _____________

76) D ebts of $2800.00 due three months from now and $4600.00 due twenty-one months from now are to be settled by two equal payments due in three months and nine months from now respectively. Determine the size of the equal replacement payments if interest is 5.5% p.a. compounded quarterly.

76) _____________

77) D orian owes his friend Katrina after she helped him pay for a used car. She originally asked Dorian to pay $1600.00 due today, and a second payment of $7200.00 due in 15 months with interest at 9.6% p.a. compounded quarterly. Dorian's financial position is currently disastrous, however, so he renegotiated with Katrina and now owes $3400.00 nine months from now and one more final payment in 21 months. What's the size of Dorian's final payment if the money is worth 12.12% p.a. compounded monthly?

77) _____________

78) D ebts of $850 due in six months, $700 due in sixteen months, and $1100 due in three years are to be settled by a single payment one year from now. What is the size of that single payment if interest is 7.5% compounded monthly?

78) _____________

79) A debt payment in the amount of $2000.00 due today, is to be settled by a payment of $1500.00 nine months from now and a final payment in 18 months. Determine the size of

the l payment if the finamoney is worth

12% p.a. 79) compou nded quarterly .

Use the Banker's Method.

___ ___ ___ ___ _

80) A loan of $2500.00 made today is to be repaid in three equal installments due in one year, two years, and four years respectively. What is the size of the equal installments if money is worth 8.4% compounded monthly?

80) _ ____________

81) A n obligation of $5150.00 due in 2.5 years is to be settled by four equal payments due today, nine months from now, 21 months from now, and 27 months from now, respectively. What is the size of the equal payment at 8.88% p.a. compounded monthly?

81) _ ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

82) ______ 82) Y ou invest $10 000 in a savings account that pays interest of 8% compounded monthly. What is the value of your account after 14 months?

A) $ 10 004.88 B) $ 10 970.88 C) $ 10 974.88 D) $ 10 904.88 E) $ 10 074.88 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

83) _ ____________ 83) K aren started a registered retirement savings plan on February 1, 2000, with a deposit of $2210. She added $2000 on February 1, 2001, and $1475 on February 1, 2004. What is the accumulated value of her RRSP account on August 1, 2005, if interest is 10.44% compounded quarterly?

84) J enni started a registered retirement savings plan on January 1, 2008, with a deposit of $2000. She added $3000 on January 1, 2009, and $2000 on January 1, 2010. What is the accumulated value of her RRSP account on July 1, 2010, if interest is 12% compounded quarterly?

84) _ ____________

85) F ind the present value and the compound discount of $7500 due in five years and six months if interest is 4.8% compounded quarterly.

85) _ ____________

86) F ind the proceeds of a non-interest-bearing promissory note for $175 500 discounted 54 months before maturity at 9.75% compounded semi-annually.

86) _ ____________

87) A three-year, $12 000 promissory note bearing interest at 11.6% compounded quarterly

87) _ ____________

is discounted two years after the date of issue at 9.64% compounded semi-annually.

What are the proceeds of the note?

88) A n obligation of $17 320 is due two years from now with interest at 9.2% compounded semi-annually. The obligation is to be settled by a payment of $7250 in seven months and a final payment in sixteen months. What is the size of the second payment if interest is now 10.8% compounded monthly?

88) _ ____________

89) D ebt payments of $1400.00 due today, $5100.00 due in twenty-one months, and $1900.00

due in 4.5 years are

to be 89) combine d into a single payment due three years from now.

What is the size of the single payment if interest is 8.04% p.a. compou nded quarterly ?

___ ___ ___ ___ _

90) A loan of $8000.00 taken out two years ago is to be repaid by three equal installments due now, three years from now, and six years from now respectively. What is the size of the equal installments if interest on the debt is 7.32% p.a. compounded monthly?

90) _____________

91) A $17 200 debt will accumulate for six years at 11.6% compounded semi-annually. For

91) _____________

how much will the debt sell four years after it was incurred if the buyer of the debt charges 9% compounded quarterly?

92) D etermine the compound discount on $18 875 due in 7.75 years if interest is 2.84% compounded monthly.

92) _____________

93) A note dated July 1, 2000, promises to pay $9000 with interest at 12.4% compounded quarterly on January 1, 2007. Find the proceeds from the sale of the note on July 1, 2002, if money is then worth 8.64% compounded semi-annually.

93) _____________

94) A debt of $8125 due today is to be settled by three equal payments due three months from now, 18 months from now, and 39 months from now respectively. What is the size of the equal payments at 6.8% compounded quarterly?

94) _____________

95) A debt of $4000 due today is to be settled by two equal payments due three months from now, and 9 months from now respectively. What is the size of the equal payments at 8% compounded quarterly?

95) _____________

96) P ayments of $10 200.00 due one year ago and $13 000.00 due nine months ago are to be replaced by a payment of $5800.00 now, a second payment of $10 000.00 fifteen months from now, and a final payment twenty-four months from now. What is the size of the final payment if interest is 9.2% compounded quarterly?

96) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

97) F ind m for the investment of $1000.00 for 12 years at 4% compounded quarterly.

A) 1 % B) 1 6% C) 4 8

97) ______

D) 3

98) F ind i if an amount is invested at 6.00% compounded every 3 months.

A) 3 .00% B) 0 .02 C) 6 .00% D) 2 .00% 99) D etermine n if an amount is invested for 3.5 years at 2.25% compounded quarterly.

A) 5 6.25 B) 1 4 C) 0 .5625% D) 5

98) ______ E) 1 .50% 99) ______ E) 1 0.5

100) Y ou invest $8500 in a savings account that pays interest of 4.8% compounded monthly. What is the value of your account after 19 months?

A) $ 9999.79 B) $ 9119.79 C) $ 9169.79 D) $ 9219.79 E) $ 9669.79

100) _____

101) I f $3000.00 is invested for seven years and seven months at 6% p.a. compounded quarterly, calculate the maturity value.

A) $ 3776.32 B) $ 4712.57 C) $ 3763.18 D) $ 4745.43 E) $ 4882.57

101) _____

102) C alculate the accumulated value of $1000.00 at 8% compounded monthly for 12 years.

A) $ 2003.39 B) $ 603.39 C) $ 2603.39 D) $ 2600.39

102) _____

103) C alculate the future value of $4200 if it is invested at an interest rate of 8.6% compounded quarterly for 3 years and 10 months.

A) $ 5875.91 B) $ 5891.75 C) $ 5888.75 D) $ 5819.75

E) $ 1603.39 103) _____ E) $ 5919.75

104) C alculate the future value of $5000 if it is invested at an interest rate of 16% compounded quarterly for 5 years and 6 months.

A) $ 11 049.59 B) $ 5849.59 C) $ 11 809.59 D) $ 10 849.59 E) $ 11 849.59

104) _____

105) Y ou invest $6780 in a floating rate guaranteed investment certificate. For the first 30 months you

105) _____

earn 4.9% compounded semi-annually. For the next 8 months you earn 4.32% compounded monthly. What is the maturity value of the certificate?

A) $ 7888.44 B) $ 7875.44 C) $ 7785.44 D) $ 7857.44

E) $ 7775.44

106) Y ou invest $8000 in a floating rate guaranteed investment certificate. For the first 24 months you earn 6% compounded semi-annually. For the next 6 months you earn 12% compounded monthly. What is the maturity value of the certificate?

A) $ 8558.00 B) $ 9558.02 C) $ 9585.00 D) $ 9888.00 E) $ 9558.00

106) _____

107) Y ou have an investment that will mature in 20 months with the value of $2500. You need some quick cash and decide to sell it today at a discount rate of 10% compounded quarterly. What is the cash value?

A) $ 2299.63 B) $ 659.23 C) $ 2219.63 D) $ 2009.63 E) $ 2120.54

107) _____

108) Y ou want to retire with $370 000 in the bank and you are able to earn 4.84% compounded quarterly for the next 25 years. You currently have $175 000 saved up. How much more do you have to invest in 15 years in order to achieve your goal?

A) $ 110 649.58 B) $ 130 649.58

108) _____

C) $ 100 649.58 D) $ 0.00 E) $ 120 649.58 109) F ind the principal that will grow to $11 019.45 at 6% compounded semi-annually in 5 years and five months.

A) $ 7980.32 B) $ 1630.33 C) $ 8000.00 D) $ 8661.65 E) $ 788.57

109) _____

110) Y ou want to retire with $400 000 in the bank and you are able to earn 6% compounded quarterly for the next 25 years. How much money do you have to invest today in order to achieve your goal?

A) $ 90 251.77 B) $ 90 201.77 C) $ 90 001.78 D) $ 90 001.77 E) $ 90 251.78

110) _____

111) W hat is the cash value of $5000 if it is discounted for 5 years at 10% compounded yearly?

A) $ 3104.61 B) $ 3143.28 C) $ 3100.61 D) $ 3228.43

111) _____ E) $ 3000.61

112) Y ou have an investment that will mature for $6825 in 57 months. You sell the investment 21 months before maturity. The discount rates used are 5.6% compounded quarterly for the first nine months of the discount period (from the date of maturity) and then 4.92% compounded monthly for the remaining discount period. How much did you sell the investment for?

A) $ 6262.54 B) $ 6223.54 C) $ 6232.54 D) $ 6666.54 E) $ 6632.54

112) _____

113) C alculate the cash value of a bond that will mature with a value of $16 500 in 7 years and 5 months. The bond is discounted at 5.8% compounded semi-annually.

A) $ 10 979.48 B) $ 11 117.48 C) $ 10 997.48 D) $ 10 779.48 E) $ 10 797.48

113) _____

114) A 9-month non-interest bearing promissory note is sold 2 months after it was issued. The face value of the note is $8500 and it is discounted at a rate of 5.2% compounded annually. What are the proceeds?

A) $ 8255.33 B) $ 8525.33 C) $ 8252.33 D) $ 8552.33 E) $ 8522.33

114) _____

115) Y ou invest $12 300 of your money in a locked-in savings account that earns 4.92% compounded monthly for 7 months. You have a family emergency and need to withdraw your funds 2 months early. You sell the savings account at a 5.6% compounded quarterly interest rate. How much did you sell the account for?

A) $ 11 569.61 B) $ 12 956.61 C) $ 12 666.61 D) $ 12 659.61 E) $ 12 540.61

115) _____

116) Y ou owe $4510 due in 7 months. In addition you owe $3780 due in 13 months and $5125 due in 21 months. You are paying 8.64% compounded monthly on your loan. What single amount three months from now will pay off the entire loan of the three future payments?

A) $ 12 040.86 B) $ 12 440.86 C) $ 12 884.86 D) $ 12 404.86 E) $ 12 444.86

116) _____

117) Y ou lend a friend $800 and they agree to make quarterly payments for 1 year. You charge your friend 8.52% compounded quarterly. What is the size of the payments?

117) _____

A) $ 220.76

B) $ 212.76

C) $ 120.76

D) $ 200.76

E) $ 210.76

118) Y ou borrow $4750 at a rate of 6.72% compounded monthly. You make a partial payment of $1400 in 3 months. You also agree to make two equal payments 3 and 5 months after the partial payment. What is the size of the equal payments? Use a focal date of 8 months.

A) $ 1754.02 B) $ 1735.84 C) $ 1784.35 D) $ 1573.84 E) $ 1788.02

118) _____

119) Y ou will need three amounts of $14 200 in each year for four years in order to go to school. You are planning on going to school starting in 5 years and ending in 8 years (years 5, 6, 7, 8). You are able to earn 9.64% compounded quarterly. How much money do you have to have today in order to be able to go to school?

A) $ 30 475.34 B) $ 30 574.35 C) $ 30 754.35 D) $ 30 457.35 E) $ 33 354.35

119) _____

120) Y ou owe $4510 due in 7 months. In addition you owe $3780 due in 13 months and $5125 due in 21 months. You are paying 8.64% compounded monthly on your loan. What single amount three months from now will pay off the entire loan of the three future payments?

A) $ 12 884.86 B) $ 12 444.86 C) $ 12 440.86 D) $ 12 040.86 E) $ 12 404.86

120) _____

121) Y ou have an investment that will mature in 33 months and have a value of $4300. You need some quick cash and decide to sell today at a discount rate of 8.2% compounded quarterly. What is the cash value?

A) $ 3439.74 B) $ 4339.74 C) $ 3539.74 D) $ 3339.74 E) $ 3669.74

121) _____

122) D etermine the nominal rate of interest if the periodic rate is 1.75% per quarter.

A) 3 .5% B) 1 .75% C) 2 1% D) 5 .25%

122) _____ E) 7 %

123) A lbert made investments of $1000 each for two years. The first one at an interest of 4.8% compounded monthly and the second at 4.9% compounded annually. Calculate the total value of his investment in two years.

A) $ 2100.55 B) $ 2000 C) $ 1100.55 D) $ 2200.95 E) $ 1100.40

123) _____

124) V aleri invested $5000 at 9.25% compounded quarterly. After 18 months, the rate changed to 9.75% compounded semi-annually. What amount will Valeri have 3 years after the initial investment?

A) $ 6754.33 B) $ 6578.31 C) $ 6615.44 D) $ 6652.76 E) $ 5735.12

124) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

125) K eith took a loan of $20 000 to buy a car at an interest of 4% compounded quarterly, 125) ____________ with no monthly payments. On first, second and third anniversary of the loan, he made payments of $5000. What payment on the fourth anniversary will eliminate the debt?

126) N ader makes semi-annual payments of $3000 to pay off his loan. Because of unemployment, he defaulted on his last two payments. What total payment is he expected to make during his third payment, if the defaulted payments are compounded

monrate of 6%? thly at a

126)

____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

127) S haun wants to invest his money such that he accumulates $10 000 after 3 years at a rate of 4% 127) _____ compounded monthly? How much money should he invest today?

A) $ 9615.38 B) $ 9884.20 C) $ 8695.61

D) $ 8717.33

E) $ 11 500.06

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

128) A lex's property taxes are $4768.00 for his house valued at $349 900 in Whitby due on 128) ____________ July 1 of 2014. How much discount should city give, if Alex pays his property tax 8 months in advance, when the city can earn 3% compounded monthly on the surplus fund?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

129) T wo payments of $49 000 each must be made 3 year and 5 year from now. If money can earn 129) _____ 4.9% compounded monthly, what single payment 3 years from now would be equivalent to the two scheduled payments?

A) $ 93 435 B) $ 98 000 C) $ 88 869 D) $ 101 177 E) $ 105 742 130) T wo payments of $49 000 each must be made 3 year and 5 year from now. If money can earn 4.9% compounded monthly, what single payment 5 years from now would be equivalent to the two scheduled payments?

A) $ 101 177 B) $ 103 034 C) $ 98 117 D) $ 98 000 E) $ 105 742

130) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

131) T wo payments of $49 000 each must be made 3 year and 5 year from now. If money can 131) ____________ earn 4.9% compounded monthly, what single payment 4 years from now would be equivalent to the two scheduled payments?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

132) D oris borrowed $5000 from a finance company at an interest of 8% quarterly. The loan is to be 132) _____ repaid in 3 equal payments, annually. What is the amount of the equal payment?

A) $ 1419.51 B) $ 1856.77 C) $ 1666.67 D) $ 1787.42 E) $ 1948.69 133) W hat periodic payment does Imran receive from a $100 000, 3-year, monthly payment GIC earning a nominal rate of 2.25% payable monthly?

A) $ 187.50 B) $ 208.33 C) $ 191.75 D) $ 185.50 E) $ 180.50

133) _____

134) I CICI quotes nominal annual interest rate of 6.6% compounded annually, 6.5% compounded semi-annually and 6.4% compounded monthly on 5 year compound interest GICs. What rate should an investor choose?

A) 6 .6% compounded annually B) 6 .4% compounded monthly C) 6 .5% compounded semi-annually D) A ny, as all alternatives are same E) I nsufficient data to make a decision

134) _____

135) A $100 000 face value strip bond has 6 years remaining until maturity. If the current market return rate is 4% compounded semi-annually, what is the fair market value of the strip bond?

A) $ 126 824 B) $ 78 849 C) $ 88 797 D) $ 79 031 E) $ 100 000

135) _____

136) A five year promissory note with a face value of $5000, bearing interest at 6% compounded semi-annually, was sold 18 months after its issue date to yield the buyer 4% compounded quarterly. What amount was paid for the note?

A) $ 6719.58 B) $ 4753.16 C) $ 5463.64 D) $ 5845.79 E) $ 4349.82

136) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

137) O ntario's recommended discount rate is 2.5% per year, which is the same as rate of 137) ____________ inflation in Ontario. What annual income will be needed 17 years from now, to have the same purchasing power as a $134 000 annual income today?

138) ____________

138) O ntario passed a law in 1996 that the names of all public servants earning $100 000 or more will be published starting 1997. What is the equivalent of that amount in 2013 assuming Ontario's recommended discount rate is 2.5% per year?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

139) O ntario passed a law in 1996 that the names of all public servants earning $100 000 or more will 139) _____ be published starting 1997. This year (2013), Ontario is planning to revise the amount to an equivalent purchasing power amount in 2014. What is the amount in 2014, beyond which the names of the Ontario's public servants should be published, assuming the recommended discount rate is 2.5% per year?

A) $ 152 162 B) $ 102 500 C) $ 65 720 D) $ 142 500

E) $ 100 000

140) S agheer received two offers for his property that he is selling. Offer 1 consists of $700 000 immediately and offer 2 consists of $500 000 now and $110,000 per annum for the next 2 years. If money earns 6% compounded semi-annually, which offer has a better economic advantage and by how much. Assume no inflation or discount?

A) O ffer 2 is better by ~$1600 B) O ffer 2 is better by ~$20 000 C) O ffer 1 is better by ~$5100 D) O ffer 1 is better by ~$1600 E) O ffer 2 is better by ~$5100

140) _____

141) F or a nominal interest rate of 9%, what is the compounding frequency if the periodic interest is 0.75%?

A) 1 2 B) 1 C) 2 4 D) 8 E) 6

141) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

142) A court ruled in favour of Professor Dumbledore. Professor Dumbledore successfully 142) ____________ claimed that that Professor Umbridge defaulted on two payments of 1000 galleons each.

One payment was due 18 months ago and the other 11 months ago. What is the appropriate amount for the court to order Umbridge to pay immediately if the court uses 6% compounded monthly for the interest rate money can earn?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

143) T wo payments of $10 000 each must be made one year and four years from now. If money can 143) _____ earn 9% compounded monthly, what single payment two years from now would be equivalent to the two scheduled payments?

A) $ 22 902 B) $ 19 296 C) $ 11 964 D) $ 8358 E) $ 10 938 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

144) Z ara borrowed $4000 from Duca financial at an interest rate of 7% compounded yearrespectively. semiannually. The loan is to be repaid in 3 equal payments in two years, three and five s, How much

will be each payment ?

____ ____ ____

144)

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

145) A five-year promissory note with a face value of $3500, bearing interest at 11% compounded 145) _____ semiannually, was sold 21 months after its issue date to yield the buyer 10% compounded quarterly. What amount was paid for the note?

A) $ 4336.93 B) $ 4251.71 C) $ 3500 D) $ 5978.51 E) $ 6021.50 146) H ow much will Ahsan's registered retirement savings deposit of $30 000.00 be worth in 21 years at 5% compounded quarterly?

A) $ 81 044.55 B) $ 38 941.88 C) $ 1 807 267.24 D) $ 83 578.88 E) $ 85 173.39

146) _____

147) C helsea's grand parents made a trust deposit of $5000.00 on November 30, 2002, at her birth.

This trust deposit can be withdrawn on Chelsea's twenty-fifth birthday. To what will the deposit amount on that date at 19.11% compounded semi-annually?

A) $ 531 830.45 B) $ 395 982.16 C) $ 48 952.52 D) $ 31 360 374.23 E) $ 479 269.91

147) _____

148) M eridian Credit Union expects an average annual growth rate of 3% for the next four years. If the assets of the credit union currently amount to $11.4 billion, what will the forecasted assets be in four years?

A) $ 1.45 billion B) $ 32.56 billion C) $ 1.43 billion D) $ 12.83 billion E) $ 12.85 billion

148) _____

149) G abriel defaulted on her tax worth $4189.00. CRA audited her and after 31 months sent her the tax bill, which also included an interest rate compounded annually at 3.75% p.a. What is the total tax due, if Gabriel pays it immediately on receiving the notice?

A) $ 4614.12 B) $ 4222.90 C) $ 13 114.19 D) $ 4592.83 E) $ 4606.94

149) _____

150) D erik bought a promissory note this morning. Find the maturity value of the promissory note for

150) _____

$1200.00 dated March 31, 2016, and due on August 31, 2026, if interest is 3.64% compounded quarterly.

A) $ 1750.28 B) $ 1724.05 C) $ 1318.75 D) $ 5323.00

E) $ 1741.51

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

151) A $25 641.00 investment matures in seven years, two months. Find the maturity value if 151) ____________ interest is 6.11% p.a. compounded quarterly.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

152) A n investment of $30 325 is accumulated at 9.45% compounded quarterly for nine and one-half years. At that time the interest rate is changed to 9.45% compounded monthly. How much is the investment worth two years after the change in interest rate?

A) $ 89 520.33 B) $ 73 646.89 C) $ 49 319.03 D) $ 88 902.62 E) $ 88 773.12

152) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

153) R BC offers three-year term deposits at 3.125% compounded annually while Meridian 153) ____________ credit union offers such deposits at 3.05% compounded quarterly. If you invest $23 000 at RBC and the same amount at Meridian, what is the maturity value of your deposit at each bank?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

154) T wo years after Michelle deposited $25 000 in a TD Bank savings account that earned interest at 154) _____ 2.25% compounded monthly, the rate of interest was changed to 5.75% compounded semi-annually. How much was in the account fifteen years after the deposit was made?

A) $ 54 641.30 B) $ 26 149.59 C) $ 58 510.72 D) $ 61 201.27 E) $ 54 629.87 155) S uman deposited $95 000.00 in an RRSP on April 1, 2016, at 8.22% compounded quarterly.

Subsequently the interest rate was changed to 8.5% compounded monthly on September 1, 2016, and to 8.8% compounded semi-annually on June 1, 2017. What was the value of the RRSP deposit on December 1, 2017, if no further changes in interest were made?

A) $ 112 975.44 B) $ 103 837.72 C) $ 97 446.55 D) $ 101 734.20 E) $ 108 406.58

155) _____

156) R omeo started an RRSP on March 1, 2016, with a deposit of $35 000.00. She added $19 000.00 on December 1, 2017, and $8500.00 on September 1, 2018. What is the accumulated value of her account on December 1, 2020, if interest is 6.6% compounded monthly?

A) $ 47 846.07 B) $ 39 272.78 C) $ 41 260.10 D) $ 80 850.58 E) $ 70 993.80

156) _____

157) R aman has a line of credit loan with the ICICI bank. The initial loan balance was $72 000.00.

Payments of and $25 000.00 were made after four months and ten months respectively. At the end of one year, he borrowed an additional $42 500.00. Seven months later, the line of credit loan was converted into a collateral mortgage loan. What was the amount of the mortgage if the line of credit interest was at prime (3%) + 1.25% compounded monthly?

A) $ 61 582.31 B) $ 19 560.45 C) $ 45 368.70 D) $ 77 002.58 E) $ 63 125.35

157) _____

158) A aron is planning to go on a vacation in 6 years. The vacation $7800.00. How much should he set-aside today in his savings account if the current interest rate is 7.6% p.a., compounded quarterly?

A) $ 4965 B) $ 7800 C) $ 5026 D) $ 4951 E) $ 12 254

158) _____

159) W hat is the discounted value of $9900.00 due in five years, seven months if money is worth 2.2% compounded quarterly.

A) $ 9031 B) $ 6876 C) $ 6855 D) $ 8757 E) $ 9118

159) _____

160) A $95 600.00, non-interest-bearing note due August 1, 2017, discounted on March 1, 2015, at 1.49% compounded monthly. Find the proceeds.

A) $ 92 106.54 B) $ 99 225.96 C) $ 92 243.34 D) $ 92 220.90 E) $ 92 335.41

160) _____

161) A $50 000.00, 5-year note bearing interest at 3.25% compounded quarterly, discounted three and a half years after the date of issue at 2.5% compounded monthly. Find the proceeds.

A) $ 58 783.81 B) $ 56 622.44 C) $ 40 965.02 D) $ 48 161.60 E) $ 42 528.72

161) _____

162) A ten-year promissory note dated April 1, 2011, with a face value of $700.00 bearing interest at

162) _____

7% compounded semi-annually, discounted six years later when money was worth 8.5% compounded monthly. What are the proceeds?

A) $ 498.84 B) $ 250.70 C) $ 992.58 D) $ 1392.85

E) $ 998.38

163) A lflah Islamic bank issued $9300.00 promissory note issued without interest for nine years on September 30, 2001, is discounted on July 31, 2006, at 5.5% compounded quarterly. Find the compound discount.

A) $ 7406.88 B) $ 15 205.27 C) $ 9300.00 D) $ 12 110.06 E) $ 1893.12

163) _____

164) A twenty-year note for $1000.00 bearing interest at 9% compounded monthly is discounted at 5% compounded quarterly four years and six months before maturity. Find the proceeds of the note.

A) $ 6009.15 B) $ 133.07 C) $ 4741.93 D) $ 4800.65 E) $ 4805.10

164) _____

165) C alculate the proceeds of $5500.00 due in 20 years, nine months discounted at 8.0% compounded semi-annually.

A) $ 1080.14 B) $ 1063.04 C) $ 1145.59 D) $ 1051.54 E) $ 1113.83

165) _____

166) A rjun has taken a loan and debt can be repaid by payments of $425.00 today, $377.00 in two years and $560.00 in five years. What single payment would settle the debt three years from now if money is worth 6.25% p.a. compounded quarterly?

A) $ 1407.70 B) $ 1546.98 C) $ 1341.13 D) $ 1387.92 E) $ 1362.00

166) _____

167) S aeed has taken a loan such that debts of $4000.00, $4500.00 and $5000.00 are due in one year, 167) _____ eighteen months and two years from now respectively. He won a lottery and would like to make a single payment now that would settle the debts if interest is 5.99% p.a. compounded quarterly.

E) $ 14 796.16 A) $ 12 324.59 B) $ 13 500.00 C) $ 13 079.58 D) $ 13 880.82 168) A de has to settle a debt, for which scheduled debt payments of $3800.00 was due seven months ago, $4700.00 was due two months ago, and $8800.00 due in five months. Debt has to be settled by two equal payments now and three months from now respectively. Determine the size of the equal replacement payments at 4.9% p.a. compounded monthly.

A) $ 8635.47 B) $ 8688.25 C) $ 17 270.93 D) $ 17 323.72 E) $ 8650.00

168) _____

169) D ebts of $2800.00 due three months from now, $3500 due 15 months and $4600.00 due twenty-one months from now are to be settled by two equal payments due in three months and nine months from now respectively. Determine the size of the equal replacement payments if interest is 5.5% p.a. compounded quarterly.

A) $ 5450.00 B) $ 10 900.00 C) $ 5246.71 D) $ 10 352.05 E) $ 5176.03

169) _____

170) T hree debt payments, the first in the amount of $1600.00 due today, the second in the amount of

170) _____

$7200.00 due in 15 months and the third in the amount of $5000 due in 24 months, are to be settled by two equal payments nine months from now and a final payment in 21 months.

Determine the size of the equal payments if the money is worth 12.12% p.a. compounded monthly.

A) $ 6415.26 B) $ 6801.60 C) $ 13 800.00 D) $ 12 830.51 171) H ow many number of compounding periods are there when compounding quarterly for 876

E) $ 6900.00 days?

171)

_____ A) 9 .0

B) 9 .6

C) 2 .4

D) 7 .2

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

172) F or a sum of money borrowed at 4.50% p.a. compounded semi-annually for 2.5 years, 172) ____________ state the numerical value of the compounding factor.

173) F or a sum of $2000 borrowed at 3.75% p.a. compounded monthly for 4.5 years, state the future value.

173) ____________

174) A sum of $10 000 earns 11.0% p.a. compounded quarterly for 3 years. After these three

174) ____________

years, new policies force the interest rate to be changed to 6.5% p.a. compounded biweekly. What is the value of the deposit one year after the change in interest rate?

175) F ind the numerical value of the discount factor if the interest rate is 8.0% compounded quarterly during a 3 year term.

175) ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

176) W hat principal amount will approximately result in the future value of $9189.41, given a 176) _____ nine-year term and an interest rate of 1.2% compounding quarterly.

A) $ 8250.45 B) $ 8250.00 C) $ 10 235.79 D) $ 9189.41 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

177) D etermine the proceeds on a non-interest bearing promissory note for $4750, discounted 177) ____________ 36 months before it's maturity date at 4% p.a. compounded semi-annually.

178) D etermine the proceeds for a $11 000 promissory note, compounding at 2% p.a. quarterly. The note is issued on October 1, 2017, due on April 1, 2020, and discounted on April 1, 2019 at 13% p.a. compounded biweekly.

178) ____________

179) A rjun wants to off some of his student debt through payments of $2200 six months from now, $1800 twelve months from now, $1400 eighteen months from now, and $1000 twenty-four months from now. Determine how much Arjun would have to pay if he chose to settle the debt now, if his money compounds 12% p.a. monthly.

179) ____________

180) Z ach has to make a tuition payment of $8200 today. However, he has the option to make

180) ____________

three equal payments two months, four months, and eight months from now. Determine the size of the equal payments if his money is compounded at 3.6% p.a. monthly.

1) U se PV = 5700.00; i = 0 .132/12 = 0.011; n = 84 FV = 5750(1 + 0 .011)84 = $114413.35 2) P V = 4100.00; i = 4%; n = 14 FV = 4100(1 + .04)14 = 4100(1.7316764) = $7099.87 3) P V = 3000; i =

= 2% = 0.02 ; n = 15(4) = 60; I/Y = 8; P/Y = C/Y = 4

FV = 3000(1 + 0.02)60 = 9843.09 Programmed solution:

Interest = 9843.09 - 3000.00 = $6843.09 4) P V = 11750.00; i = 8.44%/2 = 4.22%; n = 10*2 = 20 FV = 11750(1 + 0.0422)20 = 11750(2.285711) = 26857.11 I = 26857.11 - 11750.00 = 15107.11 5) P V = 10000.00; i = 6.0%/4 = 1.5%; n = 15 * 4 = 60 FV = 10000(1 + .015)60 = 10000(2.4432198) = 24 432.20 I = 24432.20 - 10000.00 = 14 432.20 6) P V = 3500.00; i =

= 0.0337

October 31, 2002 - July 31, 2016 contains 13 years, 9 months. n = 13 * 4 + 9/12 * 4 = 52 + 3 = 55 FV = 3500(1 + 0.0337)55 = 3500(6.1901104) = 21 665.39 7) P V = 7100.00 a) m = 1, i = 7.6%, n = 3 FV = 7100.00(1 + .076)3 = 7100.00(1.245767) = 8844.95 b) m = 2; i =

= 0.038; n = 3 * 2 = 6

FV = 7100.00(1 + .038)6 = 7100.00(1.2507892) = 8880.60 c) m = 4; i =

= 0.019; n = 3 * 4 = 12

FV = 7100.00(1 + .019)12 = 7100.00(1.2534015) = 8899.15 d) m = 12; i =

= 0.0063; n = 3 * 12 = 36

FV = 7100.00(1 + .0063333)36 = 7100.00(1.2551836) = 8911.80 8) P V = 4575.00; m = 12; i =

= .0102

a) n = 3.5 * 12 = 42 FV = 4575.00(1 + .0102)42 = 4575.00(1.5314728) = 7006.49 Interest = 7006.49 - 4575.00 = 2431.49 b) n = 7 * 12 = 84 FV = 4575.00(1 + .0102)84 = 4575.00(2.345409) = 10 730.25 Interest = 10730.25 - 4575.00 = 6155.25 c) n = 12.5 * 12 = 150 FV = 4575.00(1 + .0102)150 = 4575.00(4.5825227) = 20 965.04 Interest = 20965.04 - 4575.00 = 16390.04 9) P V = 16300000; m = 1; i = 16%; n = 4

FV = 1 + .16)4 = 16300000(1.81063936) = 29 513 421.57 1630Forecasted assets will amount to $29 513 421.57.

0000( 10) P V = 50 000 000; m = 1; i = 10%; n = 10 SFV = 50000000(1 + .10)10 = 50000000(2.59374246) = 129 687 123 Forecasted assets will amount to $129 687 123.

11) P V = 3875.00; i = 9.75%; n =

= 5.0833333

FV = 3875.00(1.0975)5.0833333 = 3875.00(1.6046846) = $6218.15 12) P V = 4320.00; i = 8.25% = .0825; n =

= 5.6666667

FV = 4320.00(1.0825)5.6666667 = 4320.00(1.5670811) = $6769.79 13) I nterest period 2016-02-29 to 2021-08-29 contains 6 years 6 months.

PV = 1400.00; i = .0191; n = 6 * 4 + 0.5 * 4 = 24 + 2 = 26 Maturity value = 1400.00(1.0191)26 = 1400.00(1.635448) = $2289.63 14) P V = 5725.00; i = .0229; n = 4 = 14.3333333 FV = 5725.00(1.0229)14.3333333 = 5725.00(1.3833823) = $7919.86 15) P V = 8000.00; i = .03; n = 4

= 21

FV = 8000.00(1.03)21 = 8000.00(1.860294572) = $14 882.36 16) B alance after 3.5 years FV 1 = 21700.00(1.0131)14 = 21700.00(1.1998651) = $26 037.07 Balance 2 years later FV 2 = 26037.07(1.0051)24 = 26037.07(1.1298546) = $29 418.10 n= 17) = 7.5833333 FV = 5400.00(1.0875)7.5833333 = 5400.00(1.8891014) = $10 201.15 18) n =

= 5.25

FV = 2000.00(1.1)5.25 = 2000.00(1.649345337) = $3298.69 19) a ) FV = 3000.00(1.09125)3 = 3000.00(1.299490) = $3898.47 b) FV = 3000.00(1.045)6 = 3000.00(1.302260) = $3906.78 20) a ) FV = 1000.00(1.10)5 = 1000.00(1.61051) = $1610.51 b) FV = 1000.00(1.024)20 = 1000.00(1.606938044) = 1606.94 21) a ) Bank: PV = 4400.00; m = 2; i =

= .03546; n = 2 * 2 = 4

FV 1 = 4400.00 (1 + .03564)4 = 4400.00(1.150364) = 5061.60 Trust: PV = 4400.00; m = 12; i =

= .0059167; n = 2 * 12 = 24

FV 2 = 4400.00(1 + .0059167)24 = 4400.00(1.1520945) = 5069.22 The Take Your Money Trust investment is preferable. b) Difference in interest = 669.22 - 661.60 = $7.62 22) a ) Boston: PV = 8100.00; m = 12; i = 1.2/12 = 0.001; n = 2 * 12 = 24 FV 1 = 8100.00 (1 + .001)24 = 8100.00(1.024278) = 8296.65 Albany: PV = 8100.00; m = 1; i = 1.236% = 0.01236; n = 2 * 1 = 2

F00(1 + .01236)2 = 8100.00(1.024873) = 8301.47 V2= The Albany Secure Savings bank investment is preferable.

8100.b) Difference in interest = 8301.47 - 8296.65 = $4.82 23) B alance after 2 years FV = 5000.00(1.005)24 = 5000.00(1.127159776) = 5635.80 Balance after a total of 15 years FV = 5635.80(1.032)26 = 5635.80(2.268151865) = $12 782.85 24) A t Sept. 1, 2002: FV = 17150(1.016)10 = 17150.00(1.1720256) = $20 100.24 Further: FV = 20100.24(1.0055)21 = 20100.24(1.1220794) = $22 554.07 (at June 1, 2014) FV = 22554.07(1.034)7 = 22554.07(1.2636994) = 28 501.56 (at Dec. 1, 2016 ) $28501.56 25) S tep 1 PV = 5000; i =

= 1.2% = 0.012;n =

(4) = 6; I/Y = 4.8; P/Y = C/Y = 4

FV = 5000(1 + 0.012)6 = 5370.97 Programmed solution:

Step 2 PV = 5370.97; i = n=

= 2.8% = 0.028;

(2) = 3 ; I/Y = 5.6; P/Y = C/Y = 2

FV = 5370.97(1 + 0.028)3 = $ 5834.88 Programmed solution:

26) S tep 1 PV = 2500.00; i =

= 2.3125% = 0.023125; n =

(4) = 6; I/Y = 9.25

FV = 2500(1 + 0.023125)6 = 2867.56 Programmed solution:

Step 2 PV = 2867.56; i =

= 4.875% = 0.04875; n =

FV = 2867.56(1 + 0.04875)3 = 3307.72 Programmed solution:

27) B alance after 3.5 years:

(2) = 3; I/Y = 9.75

PV = = .0083; n = 3.5 * 12 = 42 4300.FV = 4300.00(1 + .0083)42 = 4300.00(1.415045) = 6084.69 00; m Balance 6 years later: = 12; PV = 6084.69; m = 2; i = = .045; n = 6 * 2 = 12 i= FV = 6084.69(1 + .045)12 = 6084.69(1.695881) = 10318.91 The accumulated value 6 years after the change is $10318.91.

28) B alance after 2 years: PV = 5000.00; m = 12; i =

= .005; n = 2 * 12 = 24

FV = 5000.00(1 + .005)24 = 5000.00(1.127159776) = 5635.80 Balance 3.5 years later: PV = 5635.80; m = 2; i =

= .04; n = 3.5 * 2 = 7

FV = 5635.80(1 + .04)7 = 5635.80(1.315931) = 7416.33 The accumulated value 2.5 years after the change is $7416.33.

29) A fter 6 years: PV = 7110.00; i = .034; n = 12 FV = 7110.00(1 + .034)12 = 7110.00(1.4936418) = 10 619.79 4 years later (i.e., 10 years after the initial deposit): PV = 10619.79; i = .018; n = 16 FV = 10619.79(1 + .018)16 = 10619.79(1.3303455) = 14 127.99 30) B alance July 1, 2013: PV = 1720.00; m = 12; i =

= .007

March 1, 2011 - July 1, 2013 contains 2 years 4 months: n = 28 FV = 1720.00(1 + .007)28 = 1720.00(1.2156965) = 2091.00 Balance April 1, 2005: PV = 2091.00; m = 4; i =

= .022

July 1, 2003 - April 1, 2015 contains 1 year 9 months: n = 1.75 * 4 = 7 FV = 2091.00(1 + .022)7 = 2091.00(1.164545) = 2435.06 The balance on April 1, 2015 is $2435.06.

31) A fter 3 years: PV = 1700.00; i = .0103; n = 36 FV = 1700.00(1 + .0103)36 = 1700.00(1.4461789) = 2458.45 After 3 years: PV = 2458.45 - 800.00 = 1658.45; n = 24 After 5 years FV = 1658.45(1 + .0103)24 = 1658.45(1.2788172) = 2120.85 After 5 years: PV = 2120.85 - 950.00 = 1170.85; n = 48 FV = 1170.85(1 + .0103)48 = 1170.85(1.6353735) = 1914.78 32) m = 4; i =

= .0181

Balance on March 1, 2011 PV = 2470.00; n = 1.75 * 4 = 7 FV = 2470.00(1 + .0181)7 = 2470.00(1.1337911) = $2800.46 Balance on December 1, 2012 PV = 2800.46 + 1900.00 = $4700.46; n = 0.75 * 4 = 3 FV = 4700.46(1 + .0181)3 = 4700.46(1.05528876) = $4960.34 Balance on September 1, 2013 PV = 4960.34 + 1850.00 = 6810.34; n = 2.25 * 4 = 9 FV = 6810.34(1 + .0181)9 = 6810.34(1.1752058) = $8003.55

The accumulated value of the RRSP account on December 1, 2015 is $8003.55 33) P V = 1910.00; i = 0.0375; n = 14 FV = 1910.00(1.0375)14 = 1910.00(1.6743008) = $3197.91 Compound interest = 3197.91 - 1910.00 = $1287.91 34) P V = 5000.00; i = 0.016; n = 36 FV = 5000.00(1.016)36 = 5000.00(1.770816) = 8854.080995 Compound interest = 8854.08 - 5000.00 = $3854.08 35) B alance after 6 years FV = 4500.00(1.0039)72 = 4500.00(1.3234657) = $5955.60 Balance after a total of 13 years FV = 5955.60(1.032)14 = 5955.60(1.5542317) = $9256.38 36) m = 4; i =

= .015

Balance on March 1, 2007 PV = 2000.00; n = 1 * 4 = 4 FV = 2000.00(1 + .015)4 = 2000.00(1.061363551) = $2122.73 Balance on March 1, 2008 PV = 2122.73 + 1500.00 = $3622.73; n = 0.75 * 4 = 3 FV = 3622.73(1 + .015)3 = 3622.73(1.045678375) = $3788.21 Balance on December 1, 2008.

37) m = 12; i = = .0071 Balance after 4 months: PV = 7000.00; n = 4 FV = 7000.00(1 + .0071)4 = 7000.00(1.0287039) = 7200.93 Balance after 10 months: PV = 7200.93 - 3000.00 = 4200.93 FV = 4200.93(1 + .0071)6 = 4200.93(1.0433633) = 4383.10 Balance after 1 year: PV = 4383.10 - 2500.00 = 1883.10 FV = 1883.10(1 + .0071)2 = 1883.10(1.01425041) = 1909.93 Balance after 19 months: PV = 1909.93 + 4250.00 = 6159.93 FV = 6159.93(1 + .0071)7 = 6159.93(1.05077123) = 6472.68 The mortgage was for $6472.68 38) B alance after 6 months: PV = 20000.00; m = 2; i =

= 0.03; n = 1

FV = 20000.00(1 + .03)1 = 20000.00(1.03) = 20 600.00 Balance after 6 months: 20600-5000 = 15 600 Balance after 12 months PV = 15600.00; m = 2; i =

= .003; n = 1

FV = 15600.00(1 + .03)1 = 15600.00(1.03) = 16 068.00 Balance after 12 months: 16068 -10000 = 6068 Balance after 24 months PV = 6068.00; m = 12; i =

= 0.01; n = 12

FV = 6068.00(1 + .01)12 = 6068.00(1 + .01)12 = 6837.57 The final payment is $6837.57.

39) F V = 7800.00; i = 1.9%; n = 24 PV = 7800.00(1 + .019)-24 = 7800.00(.636531) = 4964.94 40) n = (4 +

)4 = 19.6666667

i = 0.052/4 = 0.013

PV = 18530.00(1.013)-19.6666667 = 18530.00(.77567699) = $14 373.29 41) n = (3 +

)4 = 13

i = 0.076/4 = 0.019

PV = 14000.00(1.019)-13 = 14000.00(.782953) = $10 961.34 42) P V = 6100.00(1.0413)-15 = 6100.00(.5449567) = $3324.24 Discount = 6100.00 - 3324.24 = $2775.76 43) n = (4 + )2 = 9.5 i = 0.08/2 = 0.04 PV = 5000.00(1.04)-9.5 = 5000.00(.6889429759) = $3444.71 44) F V = $12 500.00; n = (2 +

)(2) = 5.5; i =

= 3.9% = 0.039; I/Y = 7.8; P/Y = C/Y = 2

PV = FV(1 + i)-n = 12500(1 + 0.039)-5.5 = $10 128.02 Programmed solution:

45) F V = 6600.00; m = 4; i =

= 1.8% = 0.018; n = 7.25 * 4 = 29

PV = 6600.00(1.018)-29 = 6600.00(.5960936) = $3934.22 46) F V = 28000.00; i = 2.975%; n = 39 PV = 28000.00(1.02975)-39 = 28000.00(.318757) = $8925.20 47) F V = 10000.00; i = 0.2%; n = 325 PV = 10000.00(1.002)-325 = 10000.00(.522385) = 5223.85 48) F V = 23000.00; m = 12; i =

= .0081; n = 7 * 12 = 84

PV = 23000.00(1 + .00801)-84 = 23000.00(.5116269) = $11 767.42 49) F V = 37000.00; m = 1; i = 7.775% = .07775; n = 3 PV = 37000.00(1 + .07775)-3 = 37000.00(.7988144) = $29 556.13 50) n = (3 +

)4 = 14.6666667

P V = 5125.00(1.0206)-14.6666667 = 5125.00(.7415124) = $3800.25 51) n = (4 +

)2 =

= 8.8

PV = 7258.00(1.0329)-8.83 = 7258.00(.7513089) = $5453.00 52) n = (6 +

)4 = 26.3333333

PV = 114150.00(1.019)-26.3333333 = 114150.00(.6091812) = $69 538.03 53) n = (7 +

)2 = 15.8333333

PV = 8956.00(1.0375)-15.8333333 = 8956.00(.5582838) = $4999.99 54) D iscount period 2001-03-01 to 2003-08-01 contains 2 years 5 months FV = 41200.00; m = 12; i =

= .0061; n = 2 * 12 + 5 = 29

PV = 0.00(1 + .0061)-29 = 41200.00(.8383139) = $34 538.53 4120 55) D iscount period = 11 - 6 = 5 years FV = 71500.00; m = 4; i =

= .023; n = 5 * 4 = 20

PV = 71500.00(1 + .023)-20 = 71500.00(.6345814) = $45 372.57 56) D iscount period = 20-7 = 13 years FV = 40000.00; m = 4; i =

% = .025; n = 13 * 4 = 52

PV = 40000.00(1 + .025)-52 = 40000.00(.276923) = $11 076.92 57) M aturity value: PV = 5000.00; m = 4; i =

= .0206; n = 6 * 4 = 24

FV = 5000.00(1 + .0206)24 = 5000.00(1.6312989) = $8156.49 Proceeds: Discount period = 6 - 3.5 = 2.5 years FV = 8156.49; m = 12; i =

= .0055; n = 2.5 * 12 = 30

PV = 8156.49(1 + .0055)-30 = 8156.49(.8482771) = $6918.96 58) M aturity value: PV = 4700.00; m = 2; i =

= .036; n = 10 * 2 = 20

FV = 4700.00(1.036)20 = 4700.00(2.0285939) = $9534.39 Proceeds: Discount period 10 - 7 = 3 years FV = 9534.39; i =

= .0082667; m = 12; n = 3 * 12 = 36

PV = 9534.39(1.0082667)-36 = 9534.39(.7435064) = $7088.88 59) n = (3 +

)4 = 12.6666667

PV = 6532.00(1.0228)-12.6666667 = 6532.00(.7515956) = $4909.42 60) n = (6 +

)1 = 6.5833333

PV = 5640.00(1.0725)-6.5833333 = 5640.00(.6307895) = $3557.65 61) D ue date: 2010-09-30 Discount period: 2006-07-31 to 2010-09-30: 4 years 2 months n = (4 +

)4 = 16.6666667 i = 8.32%/4 = 0.0208

Proceeds = 4300.00(1.0208)-16.6666667 = 4300.00(.7095598) = 3051.11 Discount = 4300.00 - 3051.11 = $1248.89 62) M aturity value: PV = 8900.00; i = 0.0042; n = 84 FV = 8900.00(1.0042)84 = 8900.00(1.4219955) = 12 655.76 Proceeds: = (2 +

)2 = 5.1666667

= 6.5%/2 = 0.0325

P = 12655.76(1.0325)-5.1666667 = 12655.76(.8476854) = $10 728.10 63) D ue date: 2025-01-28 Maturity Value = 4000.00(1.018)40 = 4000.00(2.0413201) = 8165.28 Discount period n = (6 + 0.5)4 = 26

PV = 28(1.0205)-26 = 8165.28(.590013) = $4817.623389 8165.Discount = 8165.28 — 4817.62 = $3347.66 64) P V = 2220.00; i = 0.0055; n = 120 Maturity value = 2220.00(1.0055)120 = 2220.00(1.9312967) = 4287.48 = (4 +

)4 = 17.3333333

PV = 4287.48(1.0223)-17.333333 = 4287.48(.6822995) = $2925.35 65) P V = 10000.00; i = 0.01; n = 240 Maturity value = 10000.00(1.01)240 = 10000.00(10.89255365) = 108 925.54 = (4 +

)4 = 18

= 0.08/4 = 0.02

PV = 108925.54(1.02)-18 = 108925.54 (.700159375) = $76 265.24 66) D ue date: 2009-05-02 FV = 3100.00(1.02)28 = 3100.00(1.7410242) = 5397.18 n = (3 +

)2 = 7.3333333

Proceeds = PV = 5397.18(1.0425)-7.3333333 = 5397.18(.7369571) = $3977.49 67) I nterest period: May 1, 2011 to November 1, 2015 = 4.5 years.

Maturity value: PV = 5660.00; m = 2; i =

= .0325; n = 4.5 * 2 = 9

FV = 5660.00(1.0325)9 = 5660.00(1.3335538) = $7547.91 Discount period: Nov. 1, 2015 to May 1, 2013 FV = 7547.91; m = 12; i =

= .006; n = 2.5 * 12 = 30

PV = 7547.91(1.006)-30 = 7547.91(.8357196) = $6307.94 68) N = (5 +

)2 = 11.5

PV = 5000.00(1.04)-11.5 = 5000.00(.6369665) = $3184.83 69) L et the single payment be $x. Let the focal date be 3 years from now. i = 0.0988/4 = 0.0247 x = 4125.00(1.0247)12 + 3770.00(1.0247)4 + 5600.00(1.0247)-8 x = 4125.00(1.3401729) + 3770.00(1.025212) + 5600.00(.8226709) x = 5528.21 + 4156.50 + 4606.96 x = $14 291.67 70) L et the single payment be $x. Let the focal date be 4 years from now. i = 0.008. x = 1000.00(1.008)8 + 3000.00(1.008)4 x = 1000.00(1.065821) + 3000.00(1.032386) x = 1065.82 + 3097.16 x = $4162.98 71) L et the focal date be three and a half years from now. i = 0.108/12 = 0.009 x = 1600.00(1.009)42 + 8300.00(1.009)-6 + 12100.00(1.009)-42 x = 1600.00(1.4568974) + 8300.00(.947661) + 12100.00(.6863901) x = 2331.04 + 7865.59 + 8305.32 = $18501.95 The single payment 3.5 years from now is $18 501.95.

72) n =

= 14.6666667

PV = 7700.00(1.01875)-14.6666667 = 7700.00(.7615087) = $5863.62 73) L et the focal date be now .

For the $400.00 debt:

FV = Programmed solution: 400.0 0; n = 1(4) = 4; i For the $450.00 debt: = FV = 450.00; n = (4) = 6; I/Y = 8; P/Y = C/Y = 4 -6 = 2% P V = FV(1 + i)-n = 450( 1 + 0.02) = 399.59 = Programmed solution: 0.02; I/Y = 8; P/Y = C/Y For the $500.00 debt: =4 (4) and I/Y = 8 and P/Y = C/Y = 4 PV = FV = $500.00; n = FV(1 PV = FV(1 + i)-n = 500(1 + 0.02)-10 = 410.17 + i)-n = Programmed solution: 400(1 + 0.02) -4 = Payment = debts = 369.54 + 399.59 + 410.17 = $1179.30 369.5 4 74) L et the size of the equal payments be $x, and the focal point now.

For the $1500.00 debt: PV = 1500.00; n = 7; i = = 0.75% = 0.0075 ; I/Y = 9; P/Y = C/Y = 12 FV = PV(1 + i)n = 1500(1 + 0.0075)7 = 1580.54 Programmed solution:

For the 1200.00 debt the only difference is PV = 1200 and n = 2.

FV = PV(1 + i)n = 1200(1 + 0.0075)2 = 1218.07 Programmed solution: For the $1800.00 debt FV = $1800.00 n = 5 and I/Y = 9 and P/Y = C/Y = 12 PV = FV(1 + i)-n = 1800(1 + 0.0075)-5 = 1733.99 Programmed solution: For the first equal payment ($x) now, = $x For the second equal payment ($x) in three months FV = x and n = 3 .

PV = 833328 x FV(1 + i)-n Programmed solution: = x(1 + 0.007Payments = debts x + 0.977833328x = 1580.54 + 1218.07 + 1733.99 5)-3 = 0.977 1.977833328x = 4532.60 x = 2291.70 The size of the two equal payments is $2291.70.

75) L et the final payment be $x and the focal date at five years.

For the $5000.00 debt: PV = 5000.00; n = 5(2) = 10; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 5000(1 + 0.05)10 = 8144.47 Programmed solution:

For the $2000.00 payment: PV = 2000.00; n = 3(2) = 6; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 2000(1 + 0.05)6 = 2680.19 Programmed solution:

For the $2500.00 payment: PV = 2500.00; n = 2(2) = 4; i =

= 5% = 0.05; I/Y = 10; P/Y = C/Y = 2

FV = PV(1 + i)n = 2500(1 + 0.05)4 = 3038.77 Programmed solution:

Payments = debt 2680.19 + 3038.77 + x = 8144.47 x = $2425.51 76) L et the size of the equal payments be $x, and the focal point be at three months from now.

For the $2800.00 debt: 2800 For the 4600 debt: FV = 4600.00; n =

(4) = 6 ; i =

= 1.375% = 0.01375 ; I/Y = 5.5; P/Y = C/Y = 4

PV = FV(1 + i)-n = 4600(1 + 0.01375)-6 = 4238.11 Programmed solution:

$x For the second equal payment ($x) in nine months FV = x and n = -n -2 = x(1 + 0.01375) = 0.973056965x For PV = FV(1 + i) the first Programmed solution: equa l pay ment Payments = debts ($x) x + 0.973056965x = 2800.00 + 4238.11 at 3 1. 973056965x = 7038.11 mont x = 3567.11 hs, = The size of the two equal payments is $3567.11.

77) L et the focal day be in 21 months.

Payment of $7200: FV = 7200.00(1.024)5 = 7200.00(1.1258999) = 8106.48 At focal date: Let the final payment be $x. i = 12.12%/12 = 0.0101 1600.00(1.0101)21 + 8106.48(1.0101)6 = 3400.00(1.0101)12 + x 1600.00(1.2349569) + 8106.48(1.0621509) = 3400.00(1.1286456) + x 1975.93 + 8610.31 = 3835.76 + x $6750.48 = x Dorian's final payment will be $6750.48 78) L et the size of the payment be $x and the focal date be one year from now. x = 850.00(1.00625)6 + 700.00(1.00625)-4 + 1100.00(1.00625)-24 x = 850.00(1.0380908) + 700.00(.9753858) + 1100.00(.8611099) x = 882.38 + 682.77 + 947.22 x = $2512.37 79) I n 9 months: FV = 2000.00(1.03)3 = 2000.00(1.092727) = 2185.45 Balance: 2185.45 - 1500 = 685.45 Final payment: FV = 685.45(1.03)3 = 749.01 80) L et the size of the equal payments be $x and the focal date be now.

2500.00 = x(1.007)-12 + x(1.007)-24 + x(1.007)-48 2500.00 = x(.9197004) + x(.8458487) + x(.7154601) 2500.00 = 2.4810092x $1007.65 = x The size of the equal payments is $1007.65.

81) L et the size of the equal payments be $x.

Let the focal date be 2.5 years from now. i = 0.0074 5150.00 = x(1.0074)30 + x(1.0074)21 + x(1.0074)9 + x(1.0074)3 5150.00 = x(1.247551) + x(1.167457) + x(1.068606) + x(1.022365) 5150.00 = 4.505979x $1142.93 = x 82) C 83) B alance on Feb. 1, 2001 FV 1 = 2210.00(1.0261)4 = 2210.00(1.1085588) = 2449.91 2449.91 + 2000.00 = 4449.91 Balance on Feb. 1, 2004

(4) = 2.

FV 2 0261)12 = 4449.91(1.362311) = 6062.16 = 6062.16 + 1475.00 = 7537.16 4449.Balance on Aug. 1, 2005 91(1. FV 3 = 7537.16(1.0261)6 = 7537.16(1.167180775) = $8797.23 84) B alance on Jan. 1, 2009 FV 1 = 2000.00(1.03)4 = 2000.00(1.12550881) = 2251.02 2251.02 + 3000 = 5251.02 Balance on Jan. 1, 2010 FV 2 = 5251.02(1.03)4 = 5251.02(1.362311) = 5910.07 5910.07 + 2000 = 7910.07 Balance on July 1, 2010 FV 3 = 7910.07(1.03)2 = 7910.07(1.0609) = $8391.79 85) P V = 7500.00(1.012)-22 = 7500.00(.7691813) = $5768.86 Discount = 7500.00 - 5768.86 = $1731.14 86) F V = 175000.00; i = 0.04875; n = *2=9 PV = 17500.00(1.04875)-9 = 175000.00(.6515567) = $114 022.42 87) M aturity value = 12000.00(1.029)12 = 12000.00(1.4092385) = $16 910.86 Proceeds = 16910.86(1.0482)-2 = 16910.86(.91049473) = $15 391.37 88) L et the second payment be $x and the focal date be 16 months from now. = 7250.00(1.009)9 + x = 7250.00(1.083978) + x = 7858.84 + x 19299.47 = 7858.84 + x $11 440.63 = x 89) L et the single payment be $x and the focal date be 3 years from now. x = 1400.00(1.0201)12 + 5100.00(1.0201)5 + 1900.00(1.0201)-6 x = 1400.00(1.2697346) + 5100.00(1.1046221) + 1900.00(.8874492) x = 1777.63 + 5633.57 + 1686.15 x = $9097.35 90) L et the size of the equal payments be $x and the focal date be now.

8000.00(1.0061)24 = x + x(1.0061)-36 + x(1.0061)-72 8000.00(1.1571444) = x + .8033756x + .64541235x 9257.16 = 2.44878795x $3780.30 = x 91) M aturity value: PV = 17200.00; i = 11.6%/2 = 5.8%; n = 12 FV = 17200.00(1.058)12 = 17200.00(1.9671071) = $33 834.24 Proceeds: i = 0.09/4 = 0.0225 N = 2 × 4 = 8 PV = 33834.24(1.0225)-8 = 33834.24(.8369383) = $28 317.17 92) P V = 18875.00(1.002367)-93 = 18875.00(0.802647) = $15149.966740 Compound discount = 18875.00 — 15149.97 = $3725.03 93) B alance on Jan. 1, 2007 Maturity Value = 9000.00(1.031)28 = 9000.00(2.3509459) = $21 158.51 Discounted value on July 1, 2002 PV = 21158.51(1.0216)-9 = 21158.51(.8250344) = $17 456.50 94) L et the size of the three equal payments be $x and the focal date be now.

8125.+ x(1.017)-6 + x(1.017)-13 00 = 8125.00 = .9832842x + .903804x + .8032072x x(1.0 8125.00 = 2.6902954x 17)-1 $3020.11 = x 95) L et the size of the two equal payments be $x and the focal date be now.

4000.00 = x(1.02)-1 + x(1.02)-3 4000.00 = .9803921569x + .9423223345x 4000.00 = 1.922714491x $2080.39 = x 96) F ocal date is 24 months from now. Let the final payment be $x.

10200.00(1.023)12 + 13000.00(1.023)11 = 5800.00(1.023)8 + 10000.00(1.023)3 + x 10200.00(1.3137345) + 13000.00(1.2841979) = 5800.00(1.1995133) + 10000.00(1.0705992) + x 13400.09 + 16694.56 = 6957.18 + 10705.99 + x 30094.65 = 17663.17 + x $12 431.48 = x The size of the final payment is $7816.89.

97) C 98) E 99) B 100) C 101) B 102) C 103) D 104) E 105) B 106) E 107) E 108) D 109) C 110) E 111) A 112) C 113) E 114) C 115) E 116) D 117) E 118) B 119) C 120) E 121) A 122) E 123) D 124) C 125) i =

= 0.01

n=1×4=4 After 1st year: FV = PV(1 + i)n = $20000(1 + 0.01)4 = $20 812.08 After $5000 payment, PV after 1st year = $15 812.08 After 2nd year:

FV = $16 454.11 PV(1 After $5000 payment, PV after 2nd year = $11 454.11 + i)n After 3rd year: = $158FV = PV(1 + i)n = $11,454.11(1 + 0.01)4 = $11 919.20 12.08After $5000 payment, PV after 3rd year = $6919.20 (1 + 0.01) Payment required on the 4th anniversary to eliminate the debt = $6919.201(1 + 0.01)4 = $7200.14 4 = 126) i = = 0.005 n1 = 1 × 12 = 12 n2 = 1 × 6 = 6 Total payment during 3rd payment

= $3000 + $3000(1+ 0.005)12 + $3000(1+ 0.005)6 = $9276.17

127) C 128) i

= 0.0025

n=8 PV =

=

= $4673.70

Discount = $4768 - $4673.70 = $94.30 129) A 130) B 131) i = = 0.004083 For 3rd year payment made in 4th year PV = $49 000; n = 12 FV = PV(1 + i)n = $49000(1+ 0.004083)12 = $51 455.66 For 5th year payment made in 4th year FV = $49 000; n = 12 PV = FV(1 + i)-n = $49000(1+ 0.004083)-12 = $46 661.53 Single payment made in 4th year = $51 455.66 + $46 661.53 = $98 117.19 132) E 133) A 134) C 135) B 136) D 137) P V = $134 000 n = 17 i = 0.025 FV = $134 000 × 1.02517 = $203 897 138) P V = $100 000 in 1997 i = 0.025 n = 2013 -1997 = 16 FV in 2013 = $100 000 × 1.02516 = $148 451 139) A 140) A 141) A

142) i =

= 0.005

PV = 1000 galleons The amount today that is equivalent to the payment 18 months ago is FV 1 = PV(1 + i)n = 1000(1+ 0.005)18 = 1093.93 galleons The amount today that is equivalent to the payment 11 months ago is FV 1 = PV(1 + i)n = 1000(1+ 0.005)11 = 1056.40 galleons The appropriate amount for Umbridge to pay = 1093.93 + 1056.40 = 2150.33 galleons 143) B 144) S um of present value of the three payments = $4000 PV 1 + PV2 + PV3 = $4000 Since the payments are equal, represent FV of each payment by x PV 1 = x (1.035)-4 = 0.871442x 2 = x (1.035)-6 = 0.813501x PV PV3 = x (1.035)-10 = 0.708919x 0.871442x + 0.813501x + 0.708919x = $4000 ⇒ 2.393862x = $4000 ⇒ x = $1670.94 145) A 146) E 147) E 148) D 149) E 150) A 151) P V = 25641.00; i = .015275; n = 4[7 + 2/12] = 28.666667 FV = 25461.00(1.015275)28.666667 = 25461.00(1.544303) = $39 319.50 152) D 153) A t RBC, FV = 23000.00(1.03125)3 = 23000.00(1.0967102) = $25 224.33 At Meridian, FV = 23000.00(1.007625)12 = 23000.00(1.09543651) = $25 195.04 154) A 155) E 156) D 157) E 158) A 159) A 160) D 161) B 162) C 163) E 164) E 165) A 166) A 167) A 168) B 169) C 170) B

171) B 172) i = 0.045/2 = 0.0225 = .25%; n = 2.5 * 2 = 5 Compounding factor: (1 + 0.0225)^5 = 1.0225^5 ≈ 1.117678 173) P V = 2000; i = 0.0375/12 = 0.003125 = 0.3125%; n = 4.5 * 12 = 54 FV = 2000*(1 + 0.003125)^54 = 2000(1.183514) = $2367.03 174) F irst three years: PV = 10 000; i = 0.11/4 = 0.0275 = 2.75%; n = 3*4 = 12 FV = 10 000 * (1 + 0.0275)^12 = 10000 * (1.384784) = 13 847.8377547 Next year: PV = 13847.8377547; i = 0.065/26 = 0.0025 = 0.25%; n = 1 * 26 = 26 FV = 13847.837754*(1 + 0.0025)^26 = 13847.837754(1.067072) = $14 776.65 175) i = 0.08/4 = 0.02 = 2.0%; n = 3 * 4 = 12 Discount factor: (1 + 0.02)^-12 = 1/1.02^12 ≈ 0.788493 176) B 177) F V = 4750; i = 0.04/2 = 0.02 = 2%; n = (36/12)(2) = 6 PV = 4750/(1 + 0.02)^6 = 4750/1.126162 = $4217.86 178) C alculating future value PV = 11000; i = 0.02/4 = 0.005 = 0.5%; n = (2.5)(4) = 10 FV= 11000 * (1 + 0.005)^10 = 11000(1.051140) = $11 562.54 Calculating proceeds FV = 11562.54; i = 0.13/26 = 0.005 = 0.5%; n = 1 * 2 = 2 PV = 11562.54/(1 + 0.005)^2 = 11562.54/1.010025 = $11 447.78 179) i = 0.12/12 = 0.01 = 1%; n1 = 6; n2 = 12; n3 = 18; n4 = 24 PV1 = 2200/(1.01^6) = 2,072.499517559… = $2072.50 PV2 = 1800/(1.01^12) = 1,597.408605477… = $1597.41 PV3 = 1400/(1.01^18) = 1,170.4242398675 = $1170.42 PV4 = 1000/(1.01^24) = 787.56612742372 = $787.57 Total = 2072.50 + 1597.41 + 1179.42 + 787.57 = $5636.90 180) L et now be the focal date. Let x represent the amount of the equal payment. Let E1, E2 and E3 represent the dated values of the two payments. i = 0.036/12 = 0.003 = 0.3%; n1 = 2; n2 = 4; n3 = 8 E1 + E2 + E3 = 8200 x(1.003^-2) + x(1.003^-4) + x(1.003^-8) = 8200 x(0.99402689… + 0.98808946… + 0.9763207865…) = 8200 x(2.95843714176…) = 8200 x = $2771.73

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) A loan of $9000.00 was repaid together with interest of $3728.00. If interest was 9.4% 1) _____________ compounded quarterly, how long was the loan taken out? (Give answer in years and months to the nearest tenth.)

2) C alculate the number of years for money to triple at 3.6% p.a. compounded monthly.

2) _____________

3) I n how many days will $770.00 grow to $880.00 at 11.5% p.a. compounded monthly?

3) _____________

4) A loan of $4500.00 was repaid together with interest of $1164.00. If interest was 12 .4% compounded quarterly, for how many months was the loan taken out?

4) _____________

5) A 11-year $8000.00 promissory note, with interest at 8.4% compounded monthly, is discounted at 6.5% compounded semi-annually yielding proceeds of $14 631.15. How many months before the due date was the date of discount?

5) _____________

6) I n how many months will money double at 7.45% compounded semi-annually?

6) _____________

7) I n how many years will money double at 6.62% compounded semi-annually?

7) _____________

8) A five-year, $2000.00 note bearing interest at 10% compounded annually was discounted

8) _____________

at 12% compounded semi-annually yielding proceeds of $1900.00. How many months before the due date was the discount date?

9) I n how many years will money triple at 12% compounded semi-annually?

9) _____________

10) A promissory note for $3600.00 dated May 15, 2012, requires an interest payment of $370.00 at maturity. If interest is at 9.6% compounded monthly, determine the due date of the note.

10) _____________

11) A financial obligation requires the payment of $1000.00 in nine months, and $500.00 in twelve months. When can the obligation be discharged by a single payment of $1700.00 if interest is 12% compounded quarterly?

11) _____________

12) A financial obligation requires the payment of $250.00 in eighteen months, $350.00 in thirty months, and $300.00 in fifty-four months. When can the obligation be discharged by a single payment of $800.00 if interest is 5% compounded semi-annually?

12) _____________

13) A six-year, $1650.00 note bearing interest at 9.51% compounded annually was discounted at 11.2% compounded semi-annually yielding proceeds of $1916.75. How many months before the due date was the discount date?

13) _____________

14) J anica owes two debt payments–a payment of $6700 due in twelve months and a payment of $8750 due in twenty-one months. If Janice makes a payment of $6000 now, when should she make a second payment of $7900 if money is worth 11.5% compounded semi-annually?

14) _____________

15) A financial obligation requires payments of $2000.00 today, $2500.00 in three years, and

$3000.00 in five

years.

15) When can the obligatio n be discharg ed by a single payment equal to the sum of the required payment s if interest is 6%p.a. compou nded monthly ?

___ ___ ___ ___ _

16) A financial obligation requires the payment of $1500.00 in nine months, $700.00 in twenty-one months, and $1700.00 in 33 months. When can the obligation be discharged by a single payment of $3900.00 if interest is 8.44% compounded quarterly?

16) _____________

17) F ind the equated date at which two payments of $1600.00 due six months ago and $1850.00 due today could be settled by a payment of $4300.00 if interest is 9.48% compounded monthly.

17) _____________

18) L uciano sold a property and is to receive $14 200.00 in nine months, $14 000.00 in 42 months, and $15 500.00 in 57 months. The deal was renegotiated after six months at which time Luciano received a payment of ; he was to receive a further payment of $19 000.00 later. When should Luciano receive the second payment if money is worth 10% compounded quarterly?

18) _____________

19) A five-year, $4500.00 promissory note with interest at 7.5% compounded semi-annually was discounted at 8% compounded quarterly yielding proceeds of $6150.00. How many months before the due date was the discount date?

19) _____________

20) A t what nominal rate of interest compounded semi-annually will $11 800 earn $6800 interest in six years?

20) _____________

21) A principal of $4250.00 compounded monthly amounts to $4800.00 in 6.25 years. What is the nominal annual rate of interest?

21) _____________

22) A principal of $5000.00 compounded monthly amounts to $6000.00 in 7 years. What is the nominal annual rate of interest?

22) _____________

23) A t what nominal rate of interest compounded quarterly will $8100 earn $1700.00 interest in six years?

23) _____________

24) C alculate the nominal rate of interest compounded semi-annually that will double money in eight years.

24) _____________

25) A t what nominal rate of interest compounded monthly will $1700.00 earn $500.00 interest in three years?

25) _____________

26) A t what nominal rate of interest compounded quarterly will $2000.00 earn $400.00 interest in three years?

26) _____________

27) W hat is the nominal rate of interest at which money will triple itself in eight years and six months if compounded semi-annually?

27) _____________

28) I f $1400.00 accumulates to $2350.00 in five years, six months compounded semi-annually, what is the effective rate of interest?

28) _____________

29) I f the effective rate of interest on an investment is 6.52%, what is the nominal rate of interest compounded monthly?

29) _____________

30) I f the effective rate of interest on an investment is 7.34%, what is the nominal rate of interest compounded quarterly?

30) _____________

31) C alculate the nominal annual rate of interest compounded quarterly that is equivalent to 10% p.a. compounded semi-annually.

31) _____________

32) W hat is the nominal rate of interest compounded semi-annually which is equivalent to an effective rate of 5.89%?

32) _____________

33) T he treasurer of Lynn Lake Credit Union proposes changing the method of compounding interest on premium savings accounts to daily compounding. If the current rate is 4% compounded quarterly, what nominal rate should the treasurer suggest to the Board of Directors to maintain the same effective rate of interest?

33) _____________

34) T he accountant of Crystal Credit Union proposes changing the method of compounding interest on premium savings accounts to yearly compounding. If the current rate is 8% compounded quarterly, what nominal rate should the treasurer suggest to the Board of Directors to maintain the same effective rate of interest?

34) _____________

35) D etermine the effective rate of interest corresponding to 6% p.a. compounded monthly.

35) _____________

36) C alculate the nominal rate of interest compounded semi-annually that is equivalent to 8.4% compounded monthly.

36) _____________

37) A t what nominal rate of interest compounded semi-annually will $5900 earn $7400 interest in sixteen years?

37) _____________

38) A t what nominal rate of interest compounded annually will $5000 earn $2500 interest in seven years?

38) _____________

39) W hat is the nominal rate of interest compounded semi-annually that is equivalent to an effective rate of 8.25%?

39) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

40) I n how many years will money triple at 6.4% compounded monthly?

40) ______ A) 1 .48 years B) 2 .57 years C) 1 7.21 years D) 1 7.71 years E) 2 06.53 years 41) I n how many years will money double at 8% compounded yearly?

A) 0 .09 years B) 0 .9 years C) 2 years

41) ______ D) 9 0.1 years

E) 9 .01 years

42) H ow long, in compounding periods, will it take for $3327 to increase by $3799 if you are able to earn 6.2% compounded semi-annually?

A) 4 2.95 B) 4 .95 C) 2 4.95 D) 4 .56 E) 6 .45

42) ______

43) H ow long did it take for $1635 to increase to $2310 if the investment earned interest at a rate of 8.08% compounded quarterly (give your final answer in years and months, e.g., 3 years and 7.23 months)?

A) 3 years 4.79 months B) 7 years 7.69 months C) 8 years 1.69 months D) 4 years 3.84 months E) 4 years .64 months

43) ______

44) H ow many compounding periods does it take for a non-interest bearing bond of $420 150 to be reduced to $199 989 at a discount rate of 5.44% compounded monthly?

A) 1 84.12 B) 1 46.12 C) 1 54.12 D) 1 64.12 E) 1 21.12

44) ______

45) C alculate the nominal rate of interest p.a. compounded semi-annually if $5000.00 accumulates to $7002.64 in 78 months.

A) 2 .625% B) 0 .02625% C) 0 .0525% D) 0 .8656% E) 5 .25%

45) ______

46) A t what nominal rate of interest compounded quarterly will $13 000.00 earn $8407.27 in interest in six years? (nearest hundredth) A) 7 .33% B) 1 .83% C) 2 .10% D) 1 8.4% E) 8 .4%

46) ______

47) W hat nominal rate of interest-compounded quarterly will have to be earned on a savings account for it to grow from $1525 to $1955 over a period of 21 months?

A) 1 2.45% B) 1 4.45% C) 1 0.45% D) 3 .61%

47) ______ E) 1 5.45%

48) W hat is the quarterly interest rate that you will need to earn in order for an investment of $1635 to grow to be $1748.73 after 2.25 years?

A) 3 .00% B) 8 .93% C) 1 2.93% D) 1 3.93% E) 1 0.93%

48) ______

49) W hat is the monthly discount rate expressed as an nominal annual rate for a future value of $182 500 to be reduced to $178 663.58 over a period of 4.25 years?

A) 6 .88% B) 4 .88% C) 0 .50% D) 8 .04% E) 4 .06%

49) ______

50) C alculate the effective annual rate for 5% p.a. compounded semi-annually.

A) 0 .0255% B) 1 5.0625% C) 2 .55% D) 5 .0625%

50) ______

51) C alculate the effective annual rate for 10% p.a. compounded quarterly.

E) 0 .050625 51) ______

A) 0 .1038%

B) 1 0.38%

C) 1 .038%

D) 1 0%

52) Y ou have an interest rate of 8.44% compounded quarterly. What is the equivalent effective annual interest rate?

A) 8 .71% B) 7 .71% C) 8 .91% D) 7 .81%

E) 2 .5% 52) ______ E) 9 .71%

53) D eon has $4000.00 invested at 4.5% compounded semi-annually at her bank. In order to make a comparison with another financial institution she needs to know the effective rate of interest at her bank. What is the effective annual rate of interest?

A) 2 .28% B) 4 .5% C) 1 .12% D) 4 .55% E) 9 .10%

53) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

54) A ccording to the World Bank, the population in Georgia is decreasing by 0.83% per year.

54) _____________ According to the 2011 consensus, the population of Georgia is 4.5 million. Assuming the population growth rate remains the same; calculate the year in which the population of Georgia will be half the population in 2011?

55) H ow long will it take Whitby's population to grow from 141 000 (in 2019) to 200 000, if the annual growth rate is 3%?

55) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

56) H ow long will it take an investment to double in value if it earns 6% compounded quarterly?

56) ______ Include accrued interest and round the answer to the nearest month A) 1 1 years 9 months B) 1 1 years 8 months C) 3 years 10 months D) 1 1 years 11 months E) 3 years 11 months SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

57) B lackBerry took a loan contract which requires a payment of $40 million plus interest 57) _____________ two years after the contract's date of issue. The interest rate on the $40 million face value is 9.6% compounded quarterly. Before the maturity date, the original lender sold the contract to a pension fund for $43 million. The sale price was based on a discount rate of 8.5% compounded semi-annually from the date of sale. How many months before the maturity date did the sale take place?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

58) W hich is the most attractive of the following interest rates offered on a savings account?

58) ______ A) 2 .66% compounded quarterly B) 2 .68% compounded semiannually C) 2 .7% compounded annually D) 2 .64% compounded monthly E) A ll of them offer the same effective rate.

59) T he Brick store credit card quotes a rate of 1.75% per month on the unpaid balance. Calculate the effective rate of interest being charged.

A) 1 0.97% B) 9 .85% C) 2 3.14% D) 2 1% E) 5 1.64%

59) ______

60) C alculate the effective rate of interest, if $10 000 grew to $15 000 in 4 years with quarterly compounding.

A) 8 .45% B) 1 0.67% C) 1 2.28% D) 4 .06% E) 0 .85%

60) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

61) T he Brick store credit card quotes a rate of 1.75% per month on the unpaid balance.

61) _____________ When Leons took over Brick's management, they reduced the card's effective rate by 4%.

What will be the new periodic rate per month?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

62) F or a given interest rate of 10% compounded quarterly, what is the equivalent nominal rate of 62) ______ interest with monthly compounding?

A) 1 0.125% B) 5 .0625% C) 9 .918% D) 0 .8265% E) 1 0.381% 63) S ajid invested $20 000 in strip bonds earning him $2536.50 in one year. What is the nominal annual rate of interest compounded monthly?

A) 1 2% B) 1 2.68% C) 4 .19% D) 2 .05% E) 1 %

63) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

64) O ver a 15-year period, Tariq's investment in Cameco stock grew in value from $6000 to 64) _____________ $22 847. During the same period, the consumer price index rose from 88.31 to 118.91.

What was his real compound annual rate of interest on the stock during this period?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

65) A loan of $15 000.00 was repaid together with interest of $1089.00. If interest was 5% 65) ______ compounded quarterly, how long was the loan taken out? (Answers are rounded off to years and months) A) 5 years 8 months B) 5 2 years 10 months C) 1 year 11 month D) 3 years 7 months E) 1 year 5 months 66) M usa's parents deposited $20 000 in a long-term savings account as a wedding expenditure for their grand daughter at her birth, expecting it to triple by the time she gets married. Calculate the number of years for money to triple at 5% p.a. compounded monthly.

A) 2 2 years B) 2 2 years 6 months C) 1 8 years 7 months D) 3 0 years E) 2 5 years

66) ______

67) A n 11-year $5000.00 promissory note, with interest at 4.4% compounded monthly, is discounted at 3.5% compounded semi-annually yielding proceeds of $7651.15. How many months before the due date was the date of discount?

A) 4 7 months B) 2 0 months C) 1 16 months D) 1 32 months E) 7 6 months

67) ______

68) A promissory note for $7200.00 dated May 15, 2016, requires an interest payment of $740.00 at maturity. If interest is at 9.6% compounded monthly, determine the due date of the note.

A) 2 3 May 2017 B) 2 3 July 2016

68) ______

C) 2 4 Aug 2017 D) 2 1 Sep 2016 E) 1 1 Oct 2018 69) A financial obligation requires the payment of $25 000.00 in 12 months, $35 000.00 in 18 months, and in 30 months. When can the obligation be discharged by a single payment of $83 000.00 if interest is 6.5% compounded semi-annually?

A) 6 months B) 1 1 months C) 9 months D) 5 months E) 4 months

69) ______

70) M oriaty owes two debt payments–a payment of $6.7 million due in twelve months and a payment of $8.75 million due in twenty-one months. If Moriaty makes a payment of $7 million now, when should he make a second payment of $7.9 million if money is worth 11.5% compounded semi-annually?

A) 2 1 months B) 1 9 months C) 1 4 months D) 3 5 months E) 2 6 months

70) ______

71) M usa's parents deposited $20 000 in a long-term savings account as a wedding expenditure for their grand daughter at her birth, expecting to triple by the time she gets married at the age of 22. Calculate the rate of return compounded monthly for the savings account.

A) 0 .5% B) 0 .05% C) 4 .17% D) 0 .417% E) 5 %

71) ______

72) A t what nominal rate of interest compounded monthly will $40 000.00 earn $7200.00 interest in three years?

A) 4 .879% B) 6 % C) 4 .61% D) 5 9% E) 0 .461%

72) ______

73) W hat is the nominal rate of interest at which money will be 5 times itself in 20 years and six months if compounded semi-annually?

A) 4 .61% B) 4 % C) 0 .04% D) 8 % E) 4 8%

73) ______

74) W hat is the nominal rate of interest at which money will be 10 times itself in 30 years, if compounded monthly?

A) 0 .642% B) 6 .42% C) 7 .7% D) 4 .61%

74) ______ E) 0 .006%

75) C alculate the nominal annual rate of interest compounded quarterly that is equivalent to 15% p.a. compounded semi-annually.

A) 7 .5% B) 4 .61% C) 3 .682% D) 1 4.73% E) 7 .7%

75) ______

76) T he PC financial is considering changing the method of compounding interest on line of credit accounts to daily compounding. If the current rate is 5.75% compounded quarterly, what nominal rate should PC Financial propose to maintain the same effective rate of interest?

A) 0 .016% B) 4 .61% C) 0 .576% D) 1 .6% E) 5 .719%

76) ______

77) P WC recently proposed to BMO to change the method of compounding interest on line of credit accounts to monthly compounding, as part of the marketing campaign. If the current rate is 2.5% compounded quarterly, what nominal rate should the PWC suggest to BMO executives to maintain the same effective rate of interest?

A) 3 .773% B) 2 .5% C) 2 .75% D) 2 .495% E) 0 .208%

77) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

78) I n how many days would $7250 accumulate to $11 000 at 5.50% p.a. compounded 78) _____________ annually?

79) H ow many months would it take to triple an amount of money at 9% p.a. monthly?

79) _____________

80) H igh-profile movie star Rob Downer Sr. originally was contracted to receive four separate payments for his upcoming film: $1 million in 3 months, $2 million in 6 months, $4 million in a year and $8 million in two years. However, he was able to renegotiate terms of his contract and is now taking a single payment of $13.5 million. At a rate of 6% p.a. compounded monthly, when should Rob receive his payment?

80) _____________

81) D iana took out a loan of $600 today, and it must be repaid by payments of $200 in 3 months, $300 in 6 months, and then a payment of $125. If the interest rate is 7% p.a. compounded quarterly, when should the last payment be made?

81) _____________

82) W hat is the annually compounded rate if $1379 accumulates to $1519 in six years?

82) _____________

83) A t what nominal interest rate will money quadruple itself over the course of 22 years and 3 months if the interest rate is compounded quarterly?

83) _____________

84) S hayan took a student loan of $3400 from OSAP and wanted to pay it off within two years. If he repaid $3800, what was the nominal interest rate if it was compounded semi-annually?

84) _____________

85) N iyonella is trying to decide which local bank she should open a investment-linked savings account with. Uno Bank advertises a rate of 2.4% p.a. compounded monthly, while Bank de Deux promotes a rate of 2.6% compounded quarterly. Which bank would offer the higher rate of return on Niyonella's money?

85) _____________

86) W hat nominal annual rate compounded daily is equivalent to 5% p.a. compounded semi-annually?

86) _____________

87) I f the effective rate of interest on an investment is 5.25%, what is the nominal interest rate compounded biweekly?

87) _____________

1) P V = 9000; FV = 9000 + 3728 = 12 728; i =

n=

= 0.0235; P/Y = C/Y = 4

= 14.92069192 (quarters) ÷ 4 = 3.73017298 years = 3 years 8.8 months

Programmed solution:

2) P V = 1; FV = 3; i =

n=

= 0.003

= 366.7531281 (months) ÷ 12 = 30.56 years

Programmed solution:

3) P V = 770, FV = 880, i = 0.115 ÷ 12 = 0.0095833 n=

=

Time in days =

=

= 14.00037

= 1.166697509 × 365 = 425.85 = 426 days

4) P V = 4500, FV = 4500 + 1164 = 5664, i = 0.124 ÷ 4 = 0.031

n=

=

=

= 7.535504

Number of months = 7.535507 × 3 = 22 .61 = 23 months 5) M aturity value: PV = 8000.00; i = 0.084 ÷ 12 = 0.007, n = 11 × 12 = 132 FV = 8000.00(1 + 0.007)132 = 800.00(2 .5112509) = 20090.01 Discount:

n=

PV = 14631.15 i = 0.065 ÷ 2 = 0.0325

=

=

= 9.913701

The number of months = 6(9.913701) = 59.48 months 6) P V = 1, FV = 2, i = 0.0745 ÷ 2 = 0.03725

n=

=

= 18.95244

Number of months 18.95244 × 6 = 113.71 = 114 months = 7) P V = 1, FV = 2, i = 0.0662 ÷ 2 = 0.0331

n=

=

=

The number of years =

= 21.28569

= 10.64 years

8) P V = 2000, i = 0.10, n = 5 Maturity Value = 2000.00(1.10)5 = 2000.00(1.61051) = 3221.02 Discount: FV = 3221.02, PV = 1900, i = 0.12 ÷ 2 = 0.06

n=

=

=

= 9.058762388

The discount date is 9.058762388 × 6 = 54.35 months before the due date.

9) P V = 1, FV = 3, i = 0.12 ÷ 2 = 0.06

n=

=

=

= 18.85417668 sa

The number of years =

= 9.43 years

10) P V = 3600, FV = 3600 + 370 = 3970, i = 0.096 ÷ 12 = 0.008

n=

=

=

= 12.27788231 = 1 year + 0.27788231 × 365 days

0.27788231 × 365 = 101.438 ≈ 101 days May 15 + 101 days = 135 + 101 = 236 = August 24, 2012.

The due date is August 24, 2012.

11) L et the focal date be now; i = 0.12 ÷ 4 = 0.03; m = 4 Debts = 1000.00(1.03)-3 + 500.00(1.03)-4 = 1359.38 FV = 1700, PV = 1359.38

n=

=

=

= 7.56456693 quarters = 1.891141733 years

The discharge date is 1 year 325 days from now.

12) L et the focal date be now; i = 0.05 ÷ 2 = 0.025; m = 2 For the obligation: 250.00(1.025)-3 + 350.00(1.025)-5 + 300.00(1.025)-9

2payment: 800(1.025)-n 50.00 (.928 5994 11) + n = 350.0

=

=

0(.88 = 0.93611429 (semi-annual) ÷ 2 × 12 = 5.62 months 3854 Programmed solution: 288) + 300.0 0(.80 0728 362) 2 32.15 + 309.3 5+ 240.2 2= 781.7 2 For the 13) P V = 1650, i = 0.0951, n = 6 Maturity Value = 1650.00(1.0951)6 = 1650.00(1.7247362) = 2845.81 Discount: FV = 2845.81, PV = 1916.75, i = 0.112 ÷ 2 = 0.056

n=

=

=

= 7.253253

The discount date is 7.253253 × 6 = 43.52 months before the due date.

14) L et the focal date be now.

Today's balance 6700.00(1.0575)-2 + 8750.00(1.0575)-3.5 - 6000.00 6700.00(.8942094) + 8750.00(.822278) - 6000.00 5991.20 + 7194.93 - 6000.00 PV = 7186.13 owing today , FV = 7900 n = ln(7900/7186.13) / ln(1.0575) = ln(1.09933998)/ln(1.0575) n = 0.09470998/0.0559076 = 1.694045 The payment of $7900.00 should be made in 1.694045 × 6 = 10.164 months from now.

15) L et the focal date be now; i =

= 0.005; m = 12

For the obligation: 2000.00 + 2500.00(1.005)-36 + 3000.00(1.005)-60 2000.00 + 2500.00(.835644919) + 3000.00(.741372196) 2000.00 + 2089.11 + 2224.12 = 6313.23 For the payment: (2000 + 2500 + 3000) = 7500 FV = 7500, PV = 6313.23

N=

= 2 years 320 days The obligation can be discharged 2 years 320 days from now.

=

Programmed solution:

= 34.53 7174 01 mont hs ÷ 12 = 2.878 years 16) L et the focal date be now; i = 0.0844 ÷ 4 = 0.0211; m = 4 Debts = 1500.00(1.0211)-3 + 700.00(1.0211)-7 + 1700.00(1.0211)-11 = 3364.86 FV = 3900, PV = 3364.86 N=

= 7.06833429 quarters = 1.76708 years

= 1.767 years The discharge date is 1 year 280 days from now.

17) L et the focal date be today.

Payments: 1600.00(1.0079)6 + 1850.00 = = 1600.00(1.0483461) + 1850.00 = = 1677.35 + 1850.00 = 3527.35 due today PV = 3527.35, FV = 4300, i = 0.0948 ÷ 12 = 0.0079

n=

=

=

= 25.17 months from today

18) L et the focal date be at the time Luciano received $17000; i = 0.10 ÷ 4 = 0.025 He is still to receive: 14200.00(1.025)-1 + 14000.00(1.025)-12 + 15500.00(1.025) -17 - 17000.00 14200.00(.9756098) + 14000.00(.7435559) + 15500.00(.6571951) - 17000.00 13853.66 + 10409.78 + 10186.52 - 17000.00 34449.96 - 17000.00 = 17449.76 FV = 1900, PV = 17449.76

n=

=

=

= 3.4464414 quarters

= 3.4464414 × 3 = 10.34 months from the Focal day The second payment is to be received in 10.34 months after the first payment.

19) P V = 4500, i = 0.075 ÷ 2 = 0.0375 , n = 5 × 2 = 10 Maturity Value = 4500.00(1.0375)10 = 4500.00(1.4450439) = 6502.70 PV = 6150

n=

=

= 2.81606

It is due in 2.81606 × 3 = 8.448 = 8.448 months = 20) P V = 11 800; FV = 11 800 + 6800 = 18 600; n = 6(2) = 12; P/Y = C/Y = 2 i=

- 1 = 0.038650046 Rate = 0.038650046(2) (100%) = 7.73%

Programmed solution:

21) P V = 4250.00; FV = 4800.00; n = 6.25 × 12 = 75; m = 12 i=

- 1 = (1.1294118)1/75 - 1 = 1.0016239 - 1 = 0.001623942

The nominal annual rate is 0.001623943 × 12 × 100% = 1.94873% 22) P V = 5000.00; FV = 6000.00; n = 7 × 12 = 84; m = 12 i=

- 1 = (1.2)1/84 - 1 = 1.002172852 - 1 = 0.002172852

The nominal annual rate is 0.002172852 × 12 × 100% = 2.61% 23) P V = 8100, FV = 8100 + 1700 = 9800 n = 6 × 4 = 24 i=

- 1 = (1.2098765)1/24 - 1 = 1.0079699 - 1 = 0.0079698

The nominal annual rate = 0 .007969854 × 4 × 100% = 3.19% 24) P V = 1; FV = 2; n = 8(2) = 16; m = 2 2 = 1(1 + i)16 1 + i = 21/16 1 + i = 1.044273782 i = 0.044273782 i = 4.427% Nominal Rate = 4.427% × 2 = 8.85% p.a. compounded semi-annually Programmed solution :

25) P V = 1700, FV = 1700 + 500 = 2200, N = 3 × 12 = 36 i=

- 1 = (1.2941176)1/36 - 1 = 1.0071876 - 1 = 0.0071876

The nominal annual rate is 0.71876 × 12 × 100% = 8.62512%.

26) P V = 2000, FV = 2000 + 400 = 2400, n = 3 × 4 = 12

i=

(1.2)1/12 - 1 = 1.01530947 - 1 = 0.01530947 The nominal annual rate is 0.01530947 × 4 × 100% = 6.12%.

-1= 27) P V = 1, FV = 3 n = 8 × 2 +

i=

= 17

- 1 = (3)1/17 - 1 = 1.0667581 - 1 = 0.0667581

The nominal annual rate is 0.066758117 × 2 × 100% = 13.351623% 28) P V = 1400, FV = 2350 n = 5 × 2 + = 11

i=

- 1 = (1.6785714)1/11 - 1 = 1.0482118 - 1 = 0.0482118

f = (1+ .0482118)2 - 1 = 1.098748 - 1 = 0.098748 × 100% = 9.87% 29) i = (1+ 0.0652)1/12 - 1 = 1.005277 - 1 = 0.005277 The nominal annual rate is 0.005277 × 12 × 100% = 6.33% 30) i = (1+ 0.0734)1/4 - 1 = 1.017865508 - 1 = 0.01786551 The nominal annual rate is 0. 017865508 × 4 × 100% = 7.15% 31) i 1 = = 0.05 (1 + i)4 = (1 + 0.05)2 1 + i = (1.05)0.5 1 + i = 1.024695077 i = 0.024695077 × 4 × 100% = 9.878% i = 9.878% Programmed solution:

32) i = (1 + 0.0589)(1/2) - 1 = 1.0290287 - 1 = 0.0290287 The nominal rate = 2(0.0290287)(100%) = 5.8057336% 33) I nvest $1 for one year: /(1 + 0.01)4 = (1 + i)365 i = 1.01(4/365) - 1 = 1.0001091 - 1 = 0.0001091 Nominal rate = 0.0001091 × 365 = 0.0398215 = 3.98% compounded daily 34) I nvest $1 for one year: (1 + 0.02)4 = (1 + i)1 i = (1.02)4- 1 = 1.08243216 - 1 = 0.08243216 = 8.24% compounded yearly 35)

i=

f = (1 + i)m - 1 f = (1 + 0.005)12 - 1

f = 1.061677812 - 1 = 0.061677812 = 6.17% = 0.5% Programmed solution: = 0.005 ;m= 12 36) F or monthly; i =

= 0.007; n = 12

For semi-annually; n = 2 (1 + i)2 = (1 + 0.007)12 1 + i = (1.007)6 1 + i = 1.042741896 i = 0.042741896 = 4.2741%(2) = 8.55% Programmed solution:

37) P V = 5900, FV = 5900 + 7400 = 13300, n = 16 × 2 = 32 i = (13300/5900)^1/32 - 1

-1

= (2.2542373)1/32 - 1 = 1.0257257 - 1 = 0.0257257 The nominal annual rate is 0. 0257257 × 2 × 100% = 5.15%.

38) P V = 5000, FV = 5000 + 2500 = 7500, n = 7 i =

-1

= (1.5)1/7 - 1 = 1.059634023 - 1 = 0 .0596340227 The nominal annual rate is 0 .0596340227 × 100% = 5.693%.

39) i = (1+ 0.0825)0.5 - 1 = 1.0404326 - 1 = 0 .0404326 The nominal annual rate is 0 .0404326 × 2 × 100% = 8.08652% = 8.09%.

40) C 41) E 42) C 43) D 44) D 45) E 46) E 47) B 48) A 49) C 50) D 51) B 52) A

53) D 54) F V =

PV

i = -0.83% PV = FV (1 + i)-n ⇒ 2 = 0.9917-n ⇒ ln2 = -n ln 0.9917 ⇒ -n = -83.16 ⇒ n = 83 Year = 2011 + 83 = 2094 55) P V = 141 000 FV = 200 000 i = 3% per year n = ln(200,000/141,000) / ln(1.03) = 0.3495575/0.029559 = 11.83 years 56) B 57) T he selling price represents the present value of the loan's maturity value on the date of sale. Therefore, the solution requires two steps: (1) calculate the maturity value of the debt; (2) determine the length of time over which the maturity value was discounted to give a present value of $43 million. n = 4 × 2 = 8 and i =

= 2.4%

FV = $40 (1.024)8 = $48.357 million For discounting the maturity value, i =

n=

=

=

= 4.25%

= 2.82

Therefore, the date of sale was 2.82 × 6 = 17 months before the maturity date 58) C 59) C 60) B 61) E ffective rate for 1.75% per month = (1.0175)12 - 1 = 23.14% New effective rate = 23.14% - 4% = 19.14% 0.1914 = (1 + i)12 - 1 ⇒ ⇒ ⇒

(1 + i)12 = 1.1914 1 + i = (1.1914)1/12 i = (1.1914)1/12 - 1

⇒ i = 0.0147 = 1.47% 62) C 63) A 64) W ith the CPI up from 88.31 to 118.91, Tariq needed

times as many dollars at the end of the period to

purchase the same goods and services at the beginning.

The $3000 value of the stock at the beginning had to grow to $6000 × i=

- 1 = (22,847/8079.04)^(1/15) - 1 = 0.0717610 = 7.18%

= $8079.04

65) E 66) A 67) B 68) A 69) D 70) E 71) E 72) B 73) D 74) C 75) D 76) E 77) D 78) P eriodic rate of interest: 0.055/1 = 0.055 = 5.5% n = ln(11,000/7250) / ln(1.055) = 0.416893803931787 / 0.0535407669280298 n = 7.786474 years = 2842.06 days 79) P eriodic rate of interest: 0.09/12 = 0.0075 = 0.75% n = ln(3) / ln(1.0075) = 1.0986122886681 / 0.0074720148387 = 147.03 months 80) L et the focal date be now. Let n represent the number of compounding periods from the focal date to the equated date. Let E1, E2, E3, and E4 represent the equivalent values of the originally contracted payments. Let E5 represent the equivalent value of the single payment. Periodic rate of interest: 0.06/12 = 0.005 = 0.5% E5 = E1 + E2 + E3 + E4 13,500,000(1.005^-n) = 1,000,000(1.005^-3) + 2,000,000(1.005^-6) + 4,000,000(1.005^-12) + 8,000,000(1.005^-24) 13,500,000(1.005^-n) = 985,148.7593098 + 1,941,036.155939 + 3,767,621.358664 + 7,097,485.351129 13,500,000(1.005^-n) = 13,791,291.625042 n = ln(13,791,291.625042…/13,500,000) / ln(1.005) n = 0.02134766586 / 0.0049875415 = 4.28 months

81) L et the focal point be now. $200 in first quarter, $300 in second quarter.

Periodic interest rate: i = 0.07/4 = 0.0175 = 1.75% 600 = 200(1.0175^-1) + 300(1.0175^-2) + 125(1.0175^-n) 600 = 196.560197 + 289.769332 + 125(1.0175^-n) 113.670472 = 125(1.0175^-n) n = ln(125/113.670462) / ln(1.0175) n = 0.09501015922 / 0.01734863833 = 5.48 quarters 82) i = ((1519/1379)^(1/6)) - 1 i = (1.101522)^(1/6) - 1 i = 1.016246 - 1 i = 0.016246 = 1.62% 83) N umber of compounding terms: 22.25 years * 4 = 89 quarters i = 4^(1/89) - 1 i = 1.015698 - 1 i = 0.015698 = 1.569828% Nominal rate = 1.569828% * 4 = 6.279314 % 84) N umber of compounding terms: 2 years * 2 = 4 semi-annual terms. i = ((3800/3400)^(1/4)) - 1 i = 1.028196 - 1 i = 0.028196 = 2.819661% Nominal rate = 2.819661% * 2 = 5.639323% 85) U no Bank

Perio0.024/12 = 0.002 = 0.2%; m = 12 dic f = (1.002)^12 — 1 = 0.024266… = 2.426577…% interBank de Deux est Periodic rate of interest: i = 0.026/4 = 0.0065 = 0.65%; m = 4 rate: f = (1.0065)^4 — 1 = 0.026254… = 2.625460… % i= Therefore, Bank de Deux offers the better interest rate.

86) P eriodic rate of interest: i = 0.05/2 = 0.025 = 2.5% n=2 (1 + i)^365 = 1.025^2 i = 1.025^(2/365) - 1 i = 1.000135… - 1 i = 0.000135 = 0.013531… % 87) 0 .0525 = (1 + i)^26 - 1 1.0525 = (1 + i)^26 1.001970… = 1 + i i = 0.001970… = 0.196995… %

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) J ason deposited $100.00 at the end of each month for five years into an account paying 1) _____________ 6% compounded monthly. What will be the balance in the account at the end of the five-year term?

2) W hat will deposits of $470.00 made at the end of each month amount to after seven years if interest is 10.8% compounded monthly?

2) _____________

3) F ind the amount to which semi-annual deposits of $200.00 will grow in four years at 6.6% p.a. compounded semi-annually.

3) _____________

4) F ind the amount to which monthly deposits of $100.00 will grow in five years at 4.8% p.a. compounded monthly.

4) _____________

5) L eanne Simon made ordinary annuity payments of $81.00 per month for fourteen years earning 9% compounded monthly. How much interest is included in the future value of the annuity?

5) _____________

6) T he Olfert Contractors, Inc., are saving $958.00 every month in order to purchase a new paving machine in twelve years. Their savings certificates pay 6% p.a. compounded monthly. How much of the maturity value will be interest?

6) _____________

7) M r. and Mrs. Fox have each contributed $1825.00 per year for the last eight years into RRSP accounts earning 6.93% compounded annually. Suppose they leave their accumulated contributions for another five years in the RRSP at the same rate of interest. a) How much will Mr. and Mrs. Fox have in total in their RRSP accounts? b) How much did the Fox's contribute? c) How much will be interest?

7) _____________

8) M r. Hughes has contributed $4000.00 per year for the last ten years into a RRSP account earning 9.00% compounded annually. Suppose he leaves the accumulated contributions for another five years in the RRSP at the same rate of interest. a) How much will Mr. Hughes have in total in his RRSP account? b) How much did Mr. Hughes contribute? c) How much will be interest?

8) _____________

9) F or the last four years Joe Thiesman has made deposits of $300.00 at the end of every six months earning interest at 5.7% compounded semi-annually. If he leaves the accumulated balance for another ten years at 3.2% compounded quarterly, what will the balance be in Joe's account?

9) _____________

10) A man put aside $5710.00 at the end of every 3 months for seven years. How much will he have six years after the least deposit, if his account earned 5.6% p.a. compounded quarterly?

10) _____________

11) H ow much interest is included in the accumulated value of $4750 paid at the end of every six months for 4.5 years if the interest rate is 8.5% compounded semi-annually?

11) _____________

12) H ow much interest is included in the accumulated value of $250 paid at the end of every month for 5 years if the interest rate is 4.0% compounded monthly?

12) _____________

13) W hat is the discounted value of deposits of $150.00 made at the end of each month for fourteen years if interest is 4.5% compounded monthly?

13) _____________

14) F ind the present value for payments of $500.00 made at the end of each quarter for ten years, if interest is 8% compounded quarterly.

14) _____________

15) F ind the present value of ordinary semi-annual payments of $810.00 for five years at 5.45% p.a., compounded semi-annually.

15) _____________

16) A n installment contract for the purchase of a car requires payments of $568.60 at the end of each month for the next three years. Suppose interest is 12.0% p.a. compounded monthly. a) What is the amount financed? b) How much is the interest cost?

16) _____________

17) A n installment contract for the purchase of a computer requires payments of $30.00 at the end of each month for the next three years. Suppose interest is 18.00% p.a. compounded monthly. a) What is the amount financed? b) How much is the interest cost?

17) _____________

18) R andy bought an annuity to pay him $2700.00 at the end of every six months for twenty years. How much of the total annuity payments is interest, if interest is 6.5% p.a. compounded semi-annually?

18) _____________

19) A nnette bought a vacation property for $22 900.00 down and quarterly mortgage payments of $1224.51 at the end of each quarter for six years. Interest is 8.4% compounded quarterly. a) What was the purchase price of the property? b) How much interest will Annette pay?

19) _____________

20) I f a loan was repaid by ordinary monthly payments of $715.00 in seven years at 9.12%, compounded monthly, how much interest was paid?

20) _____________

21) I f a loan was repaid by ordinary yearly payments of $2000.00 in five years at 9.00%, compounded yearly, how much interest was paid?

21) _____________

22) J anice plans to retire in 10 years and would like to receive $3000.00 per month for fifteen years starting at the end of the first month after her retirement. Calculate the amount she must invest now if interest is 7.5% compounded monthly.

22) _____________

23) Y ou have inherited some money and you want to set some of that money aside for ten years. After ten years, you would like to receive $7600.00 at the end of each 6 months for nine years. If the interest is 6.5% compounded semi-annually, how much of your inheritance must you set aside?

23) _____________

24) P lanning for their child's college education, David and Carol Roberts opened an account paying 6.36% compounded monthly. If ordinary annuity payments of $700.00 per month are to be paid out of the account for three years starting seven years from now, how

muthe Roberts' ch deposit? did

24)

_____________ 25) P lanning for their child's college education, Sean and Jessica opened an account paying 8.00% compounded yearly. If ordinary annuity payments of $15 000.00 per year are to be paid out of the account for three years starting sixteen years from now, how much did the couple deposit?

25) _____________

26) A father wants to provide an annuity of $3350.00 payable at the end of each 6 months during the three years his son attends college. It will be six years before his son goes to college. What single deposit must he make today that will finance the annuity, if he can invest his money at 7.84% p.a. compounded semi-annually?

26) _____________

27) Y ankee Construction agreed to lease payments of $762.79 on construction equipment to be made at the end of each month for six years. Financing is at 15% compounded monthly. a) What is the value of the original lease contract? b) If, due to delays, the first eight payments were deferred, how much money would be needed after nine months to bring the lease payments up to date? c) How much money would be required to pay off the lease after nine months? d) If the lease were paid off after nine months, what would the total interest be? e) How much of the total interest would be due to deferring the first eight payments?

27) _____________

28) A loan was repaid in 7.25 years by end-of month monthly payments of $472. If interest was 6.12% compounded monthly, how much interest was paid?

28) _____________

29) A car was purchased for $3500.00 down and payments of $575.00 at the end of each month for 5 years. Interest is 9.72% compounded monthly. a) What was the purchase price of the car? b) How much will be the amount of interest paid?

29) _____________

30) A trust fund is set up to make payments of $1320.00 at the end of each month for seven years. Interest on the fund is 7.28% compounded monthly. a) How much money must be deposited into the fund? b) How much will be paid out of the fund? c) How much interest is earned by the fund?

30) _____________

31) I f a loan was repaid by quarterly payments of $715.00 over 7 years at 8.36% compounded quarterly, how much money had been borrowed?

31) _____________

32) D oris purchased a piano with $1300.00 down and monthly payments of $214.00 for two years at 9.72% compounded monthly. What was the purchase price of the piano?

32) _____________

33) K am expects to retire in 22 years. Beginning one month after his retirement he would like to receive $590.00 per month for 25 years. How much must he deposit into a fund today to be able to do so if the rate of interest on the deposit is 6.12% compounded monthly?

33) _____________

34) W hat is the size of end-of-month monthly deposits which will accumulate to $122 200.00 after six years if interest is 5.7% compounded monthly?

34) _____________

35) W hat payment is required at the end of each month for 3 years to repay a loan of $8000.00 at 6.4% compounded monthly?

35) _____________

36) W hat payment is required at the end of each quarter for 8 years to repay a loan of $20 000.00 at 6.0% compounded quarterly?

36) _ ____________

37) W hat payment made at the end of each month for 13 years will amount to $17 520.00 at 9.66% compounded monthly?

37) _ ____________

38) T he Chretiens bought a rental property valued at $125 000.00 by paying 25% down and mortgaging the balance over 21.5 years through equal payments at the end of each quarter at 8.6% compounded quarterly. What was the size of the quarterly payments?

38) _ ____________

39) T he Savoias bought an investment property valued at $160 000.00 by paying 25% down and mortgaging the balance over 25 years through equal monthly payments at 6% compounded monthly. What was the size of the monthly payments?

39) _ ____________

40) W hat is the size of semi-annual deposits that will accumulate to $113 200.00 after 9.5 years at 6.5% compounded semi-annually?

40) _ ____________

41) A $5000.00 loan requires payments at the end of each quarter for five years. If the interest rate on the loan is 8% compounded quarterly, calculate the size of each payment.

41) _ ____________

42) A $15 000.00 loan requires payments at the end of each month for five years. If the interest rate on the loan is 12% compounded monthly, calculate the size of each payment.

42) _ ____________

43) W hat deposit made at the end of each quarter will accumulate to $2510.00 in four years at 4% compounded quarterly?

43) _ ____________

44) W hat is the term of a mortgage of $235 000.00 repaid by monthly payments of $2475.00 if interest is 7.55% compounded monthly?

44) _ ____________

45) H ow many quarterly payments will it take for $400.00 deposited at the end of each quarter to amount to at 6% compounded quarterly?

45) _ ____________

46) H ow many semi-annually payments will it take for $500.00 deposited at the end of each half year to amount to $10 000.00 at 6% compounded semi-annually?

46) _ ____________

47) H ow many monthly payments will it take for $1000.00 deposited at the end of each month to amount to at 9% compounded monthly?

47) _ ____________

48) S uppose $726.56 is deposited at the end of every six months into an account earning 6.45% compounded semi-annually. If the balance in the account four years after the last deposit is to be $31 300.00, how many deposits are needed?

48) _ ____________

49) A loan of $19 850.00 is to be repaid in quarterly payments of $620.00. How many payments are required to repay the loan at 6.5% compounded quarterly?

49) _ ____________

50) H ow many yearly payments will it take for $25 000.00 deposited at the end of each year to amount to at 9% compounded yearly?

50) _ ____________

51) $ 1000.00 is deposited at the end of every month into an account earning 6.00% compounded monthly. If the balance in the account five years after the last deposit is to

be 000.00, how $50 many deposits

are needed?

51)

___ ___ ___ ___ _

52) I n what period of time could you pay back a loan of $13 900.00 by making monthly payments of $296.00 if interest is 6.5% compounded monthly?

52) _____________

53) W hat is the nominal rate of interest on a loan of $6800.00 repaid in semi-annual installments of $1175.00 in six years?

53) _____________

54) C ompute the nominal annual rate of interest (compounded monthly) at which $200.00 deposited at the end of each month for ten years will amount to $30 000.00.

54) _____________

55) N ote: The calculations for this question were done using Excel's RATE function.

What nominal rate of interest is paid on quarterly RRSP contributions of $573.00 made for 17.75 years if the balance just after the last contribution is $106 000.00?

55) _____________

56) W hat is the nominal rate of interest if a six-year loan of $71 000.00 is repaid by monthly payments of $1701.58?

56) _____________

57) N ote: The calculations for this question were done using Excel's RATE function.

A property worth $191 000.00 can be purchased for 15% down and quarterly mortgage payments of $1972.00 for 30 years. What effective rate of interest is charged?

57) _____________

58) B onny converted an RRSP balance of $176 875.67 into an RRIF that will pay her $2720.00 at the end of every month for 8 years. What is the nominal rate of interest?

58) _____________

59) W hat nominal rate of interest is earned by monthly deposits of $560.00 made for 6 years if the balance just after the last deposit is $52 700.00?

59) _____________

60) N ote: The calculations for this question were done using Excel's RATE function.

What is the effective annual rate of interest on a loan of $9200.00 repaid in semi-annual installments of $810.00 in nine years?

60) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

61) A n annuity with periodic payments made at the end of each payment period is called: 61) ______ A) g eneral B) o rdinary C) s imple D) a nnuity due E) n one of the above 62) W hen the conversion period is different from the payment period, an annuity is called: A) g eneral B) c ertain C) d eferred D) d ifferent 63) A n annuity due is an annuity for which: A) t he payments continue forever B) t he payments are made at the beginning of each payment period C) t he payment period is not the same as the conversion period D) t he payments are made at the end of each payment period E) t he payments are made to repay a loan

62) ______ E) o rdinary 63) ______

64) T he term of annuity is: A) t he length of time from the beginning of the first payment interval to the end of the last payment interval B) t he length of the conversion interval C) t he length of time between the successive payments D) t he length of the payment interval E) t he interval between the first and the last payments

64) ______

65) R egular contributions of $200 are made at the end of each month for five years into a savings account earning interest at 4% compounded quarterly. The annuity can be classified as: A) o rdinary general annuity B) p erpetuity C) o rdinary annuity due D) o rdinary simple annuity E) s imple general annuity

65) ______

66) C alculate the accumulated value of quarterly payments of $100.00 made at the end of each quarter for ten years just after the last payment has been made if interest is 8% p.a. compounded quarterly.

A) $ 2040.00 B) $ 4415.88 C) $ 4003.91 D) $ 6040.20 E) $ 2015.88

66) ______

67) C alculate the accumulated value of monthly payments of $100.00 made at the end of each month for ten years just after the last payment has been made if interest is 12% p.a. compounded monthly.

A) $ 330 003.87 B) $ 12 000 C) $ 33 003.87 D) $ 23 003.87 E) $ 1 046.22

67) ______

68) W hat is the future value of an annuity with monthly deposits of $277 for a period of 11.5 years at an interest rate of 7.44% compounded monthly? The deposits are made at the end of the month.

A) $ 60 611.68 B) $ 60 001.68 C) $ 60 616.68 D) $ 60 161.68 E) $ 60 186.68

68) ______

69) Y ou want to save $710 per quarter for 15.5 years towards the purchase of a motor home. You think that you can earn 6.56% compounded quarterly for this period of time. If your first deposit is in 3 months, what is the most expensive motor home that you will be able to purchase?

A) $ 79 999.79 B) $ 27 501.82 C) $ 75 399.79 D) $ 44 020.00 E) $ 75 999.79

69) ______

70) Y ou are offered payments of $475 at the end of each semi-annual period for 6.5 years. You think that the cost of money is 6.14% compounded semi-annually. What is the present cash value?

A) $ 5290.06 B) $ 5209.06 C) $ 5292.06 D) $ 5029.06 E) $ 5092.06

70) ______

71) C alculate the present value of five payments of $1500.00 made at the end of each of five consecutive years respectively if money is worth 6% compounded annually.

A) $ 7486.00 B) $ 7050.00 C) $ 6318.55 D) $ 7486.52

71) ______ E) $ 7950.00

72) Y ou want to be able to travel and withdraw $3750 at the end of every month during your yearlong trip. During your trip your travel fund will earn 6.24% compounded monthly. What is the size of your travel fund if you want to be able to start your trip in 4 years and 7 months?

A) $ 45 315.20 B) $ 32 715.58 C) $ 46 309.57 D) $ 45 555.20 E) $ 2849.74

72) ______

73) C alculate the present value of 120 payments of $50.00 made at the end of each of 120 consecutive months respectively if money is worth 12% compounded monthly.

A) $ 6000.00 B) $ 3485.02 C) $ 3485.03 D) $ 3000.00 E) $ 34 850.30

73) ______

74) Y ou currently have $4541 saved towards the purchase of a home theatre system. You want to be able to buy a system for $6225 plus GST in 4 years. Your money is earning 4.84% compounded monthly. What is the size of your monthly deposit?

A) $ 20.80 B) $ 13.55 C) $ 12.80 D) $ 22.80 E) $ 28.80

74) ______

75) Y ou currently have $4000 saved towards the purchase of a new car. You want to be able to buy a car for $15000 in 4 years. Your money is earning 5% compounded yearly. What is the size of your yearly deposit?

A) $ 2000.99 B) $ 2320.12 C) $ 2552.13 D) $ 928.05 E) $ 2352.13

75) ______

76) H ow long will it take for quarterly deposits of $425 to accumulate to be $16 440 at an interest rate of 8.48% compounded quarterly? (Final answer in years and months e.g., 7 years and 4.24 months) A) 7 years 1.64 months B) 7 years 6.14 months C) 7 years .136 months D) 2 8 years .55 months E) 2 8 years .64 months

76) ______

77) S ofia plans to save for next two years. She plans to put away $1000 every 6 months starting 6 months from now. What will be the value of her savings after two years if her savings accounts earns at a rate of 8% compounded semi-annually?

A) $ 2080 B) $ 6414.51 C) $ 4246.46 D) $ 4000 E) $ 6632.98

77) ______

78) C harter has been contributing $900 at the end of each month for the past 15 months to a savings plan that earns 4% compounded monthly. What amount will he have one year from now, if he continues with the plan?

A) $ 6898.05 B) $ 12 669.42 C) $ 13 500 D) $ 13 819.60 E) $ 25 382.84

78) ______

79) S alem plans to deposit $2200 every 6 months for 15 years to save for his son's higher education.

The rate of return will be 4% compounded semi-annually for the first 5 years and 8% compounded semi-annually for the subsequent 10 years. Calculate the future value of this simple annuity.

A) $ 106 237.02 B) $ 118 293.81

79) ______

C) $ 52 782.81 D) $ 24 089.39 E) $ 53 454.21 80) S alem plans to deposit $2200 every 6 months for 15 years to save for his son's higher education.

The rate of return will be 4% compounded semi-annually for the first 5 years and 8% compounded semi-annually for the subsequent 10 years. Calculate the future value of this simple annuity, if he stops payment after 5 years.

A) $ 53 454.21 B) $ 52 782.81 C) $ 118 293.81 D) $ 24 089.39 E) $ 106 237.02

80) ______

81) D on has just turned 43 and has accumulated $34 500 in his RRSP. He makes month-end contributions of $300 to the plan and intends to do so until he retires at the age of 60. The RRSP will be allowed to continue to accumulate until he reaches the age of 65. If the RRSP earn 8% compounded monthly for the next 22 years, what amount will his RRSP contain, when he reaches age 65?

81) ______

A) $ 129 539.17 B) $ 192 993.38 C) $ 199 361.27 D) $ 227 493.81 E) $ 392 354.65 82) N uclear power plants are required to have full decommissioning costs accumulated during the useful life of the plant. The new nuclear plant at Darlington is designed for 60 years of operational life starting in 2025. OPG plans to set aside $1 million per month during the operation of the plant. What is the value of this annuity in 2025, if OPG uses Ontario's recommended discount rate of 2.5% compounded monthly?

A) $ 373 million B) $ 67 million C) $ 720 million D) $ 1668 million E) $ 158 million

82) ______

83) N uclear power plants are required to have full decommissioning costs accumulated during the useful life of the plant. The new nuclear plant at Darlington is designed for 60 years of operational life starting in 2025. OPG plans to set aside $1 million per month in a sinking fund for the operational of the plant. What is the value of the value of this sinking fund at the end of the plant life, if OPG expects an average rate of return of 2.5% compounded monthly?

A) $ 373 million B) $ 720 million C) $ 67 million D) $ 1668 million E) $ 158 million

83) ______

84) H ow much will it cost Shaun to purchase a two-level retirement annuity that will pay $6000 at the end of each month for the first 10 years, and $9000 per month for the next 15 years? The payments represent a rate of return to Shaun of 6% compounded monthly.

A) $ 540 441 B) $ 2 193 174 C) $ 1 126 642 D) $ 1 066 532 E) $ 586 201

84) ______

85) A li bought Kia Forte 5 from Durham Kia for $20 078.00 on finance agreement with BMO at 5.99% compounded monthly. Calculate the monthly payment on the loan, if the financing period is 60 months.

A) $ 334.63 B) $ 388.06 C) $ 451.15 D) $ 336.30 E) $ 386.13

85) ______

86) P at won a lottery with two options: $1 million per year, at the beginning of the year, for 25 years or a single cash price of $18 million. If low-risk investments can earn 2.5% compounded annually, which option should Pat choose and what is the advantage in terms of current economic value?

A) 2 5 year annuity has an advantage of $424 000 B) L ump sum has an advantage of $424 000 C) L ump sum has an advantage of $885 000 D) 2 5 year annuity has an advantage of $885 000 E) B oth options are the same

86) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

87) O n his retirement, Louis received a lump sum of $800 000 from his employer. Taking 87) _____________ advantage of existing tax legislation, he invested his money in an annuity that provides for payments of $60 000 at the end of every 3 months. If interest is 6.25% compounded quarterly, how long will the annuity exist?

88) J ane borrowed $20 000 to help finance her BBA at UOIT with no interest until she finishes her education. She starts a job with a local bank immediately after finishing her BBA and contracts to pay her loan in monthly payments of $300 each. If the payments are due at the end of each month and the interest is compounded 4% monthly, how long will Jane have to make monthly payments?

88) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

89) A loan of $20 000 is paid over 5 years by monthly payments of $388. What is the nominal rate of 89) ______ interest on the loan?

A) 0 .089% B) 6 .14% C) 0 .625% D) 2 .47% E) 7 .5% 90) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, what will be the balance in the account at the end of the three-year term?

A) $ 17 390.19 B) $ 27 284.81 C) $ 18 600.50 D) $ 18 408.04 E) $ 18 603.37

90) ______

91) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, how much interest is included in the future value of the annuity?

A) $ 18 000.00 B) $ 18 603.37 C) $ 408.04 D) $ 18 408.04 E) $ 603.37

91) ______

92) J ordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost. If the production life of the plant is 60 years and the operator puts the money at the end of the month in a savings account, earning 7.25% compounded monthly, what will be the value of the decommissioning fund after 60 years of production?

92) ______

A) $ 10.87 billion B) $ 11.88 billion C) $ 720 million

D) $ 12.49 billion E) $ 917.65 million 93) J ordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded monthly. At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. What will be the value of the decommissioning fund in 2120?

93) ______

A) $ 74.57 billion B) $ 75.23 billion C) $ 60.90 billion D) $ 70.94 billion E) $ 12.49 billion 94) J ordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded monthly. At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. How much interest is included in the future value in 2120?

94) ______

A) $ 70.27 billion B) $ 74.57 billion C) $ 12.49 billion D) $ 11.77 billion E) $ 73.85 billion 95) W oods has contributed $3600.00 per year for the last ten years into a RRSP account earning 9% compounded annually. His wife Karen, on the other hand, contributed $300 at the end of every month for the same period into a RRSP account earning 9% compounded monthly. Who earns more interest at the end of 10 years and by how much?

A) K aren by $58 054 B) W ood by $54 695 C) K aren by $3360 D) W ood by $3360 E) K aren by $22 054

95) ______

96) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the end of each month, what will be the present value of the balance in the account at the end of the three-year term?

A) $ 17 390.19 B) $ 18 603.37 C) $ 27 284.81 D) $ 18 000.00 E) $ 18 408.04

96) ______

97) S am got a job at the Brick. He immediately finances a car paying $500 every month for 3 years.

The interest is 2.25% compounded monthly. How much is the interest cost of the financing?

A) $ 408.04 B) $ 9284.81 C) $ 17 390.19 D) $ 609.81 E) $ 603.37

97) ______

98) K eith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at

the end of

___ ___

each 98) month for 20 years.

Interest is 4.4% compou nded monthly.

What is the purchase price of the property ?

A) $ 195 214

B) $ 671 265

C) $ 469 885

D) $ 194 378

E) $ 278 878

99) S iri plans to retire now and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded monthly.

A) $ 539 042 B) $ 1 528 015 C) $ 633 665 D) $ 574 626 E) $ 594 425

99) ______

100) K nowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded monthly. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of each month for four years, how much did Roberta deposit?

A) $ 54 343 B) $ 65 533 C) $ 32 474 D) $ 32 723 E) $ 65 592

100) _____

101) S iri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now every month (at the end of the month) if interest is 4.9% compounded monthly.

A) $ 11 755 B) $ 1 528 015 C) $ 4420.74 D) $ 633 665 E) $ 10 660

101) _____

102) J amal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded monthly. What will be the size of his payments per month from the ordinary simple annuity for 20 years following his retirement?

A) $ 2000 B) $ 957 836 C) $ 1434 D) $ 8618 E) $ 333

102) _____

103) S iri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded monthly.

A) 3 6 years B) 1 8 years C) 2 16 years D) 1 .41 years E) 2 .41 year

103) _____

104) J amal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded monthly until retirement. The rate changes following the retirement. He wants $7000 per month paid to him for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel) A) 0 .86% B) 1 0.27% C) 1 0% D) 0 % E) 9 %

104) _____

105) W hat's the difference between a simple annuity and a general annuity?

A) p ayments are made at the end of each payment period for simple annuities B) p ayments are made at the beginning of each payment period for simple annuities C) s imple annuities are a type of general annuity D) t he compounding frequency and payment period are the same for simple annuities

105) _____

106) W hat type of annuity does not require a payment in the first payment period?

A) c ontingent annuity B) a nnuity certain C) g eneral annuity D) d eferred annuity

106) _____

107) L ease payments on equipment, installment payments on loans, and interest payments on bonds are all examples of what type of annuity?

B) P erpetuity A) A nnuity C) A nnuity due D) S imple annuity certain

107) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

108) A li deposits $25 of his biweekly paycheck into an RRSP that has an interest rate of 1.95% 108) ____________ p.a. compounded biweekly. If he continues to do this for the next 47 years, what will his total contribution be and what will be the balance of the RRSP? 109) A sifa invested $6000 in her TFSA at the end of every year for 4 years. Her TFSA is linked to an index mutual fund that returns around 4.8% p.a. compounded annually. After 4 years, she gave the deposit to her son to aid in the purchase of a house in the future. Her son left the deposit in a savings account that earned interest at 3.5% p.a. compounded annually. What is the balance of the savings account after 3 years?

109) ____________

110) F ind the present value for an annuity with payments of $100 paid at the end of every three months, with an interest rate of 5.2% p.a. compounded quarterly and a term of 8.75 years.

110) ____________

111) B uster and Dave purchase a cottage by making a down payment of $22 500 and paying $745 on a monthly basis for three years. If the interest rate is 4.2% p.a compounded monthly, what's the cash value of the cottage?

111) ____________

112) W hat payment is required at the end of every six months for 78 months to accumulate to $13 650 at 9.4% p.a. compounded semi-annually?

112) ____________

113) J ordan wants to buy a car selling for $31 500 by paying a down payment of $11 500 and

113) ____________

then the remaining balance in equal monthly payments for 5 years. Determine the monthly payments if the interest rate is 1.2% p.a. compounded monthly.

114) H ow long will it take $420 deposited at the end of each year to amount to $6830 at 2.5% p.a. compounded annually?

114) ____________

1) P MT = 100.00; n = 5(12) = 60; i = FV = 100

= 0.5% = 0.005; I/Y = 6; P/Y = C/Y = 12

= $6977.00

Programmed solution:

2) 3) 4) 5)

FV n = 470.00

= 470.00(124.7277964) = $58 622.06

FVn = 200.00

= 200.00(8.9875671) = $1797.51

FVn = 100.00

= 100.00(67.660180) = $6766.02

FVn = 81.00

= 81.00(334.5180794) = 27 095.96

Interest is 27095.96 - 81.00(168) = 27095.96 - 13608.00 = $13 487.96 6)

FVn = 958.00

= 958.00(210.1501631) = 201 323.86

Interest

= 201323.86 - 958.00(144) = 201323.86 - 137952.00 = $63 371.86

7) a) FVn = 3650.00

= 3650.00(10.2339355) = $37 354.00

FV = 37354.00(1.0693)5 = 37354.00(1.3979699) = $52 219.77 b) Contribution = 8(3650.00) = $29 200.00 c) Interest = 52219.77 - 29200.00 = $23 019.77 8) a) FVn = 4000.00

= 4000.00(15.19292972) = 60 771.72

FV = 60771.72(1.09)5 = 60771.72(1.538623955) = 93 504.82 b) Contribution = 10(4000.00) = $40 000.00 c) Interest = 93504.82 -40000.00 = $53 504.82 9) FVn = 300.00

= 300.00(8.8451439) = $2653.54

FV = 2653.54(1.008)40 = 2653.54(1.3753755) = $3649.61 10)

FVn = 5710.00

= 5710.00(33.9942668) = $194 107.26

FV = 194107.26(1.014)24 = 194107.26(1.396082) = $270 989.65 11)

FVn = 4750.00

= 4750.00(10.6918204) = $50 786.15

Total paid = 4750.00 × 9 = $42 750.00 Interest = 50786.15 - 42750.00 = $8036.15

12)

FVn = 250.00

= 250.00(66.298971) = 16 574.74

Total paid = 250.00 × 60 = $15 000.00 13)

Interest = 16574.74 - 15000 = $1574.74 PVn = $18 671.21 P MT==150.00 500.00; n = 10(4) = 40; I/Y ==8150.00(124.4747396) P/Y = C/Y = 4; i =

14)

= 2% = 0.02 PV = PMT

= 500

= $13 677.74

Programmed solution:

15)

PVn = 810.00 = 810.00(8.651179)

16)

= $7007.45 a) PVn = 568.60

= 568.60(30.10750504) = $17 119.13

b) Interest = 36(568.60) - 17228.86 = 20469.60 - 17119.13 = $3350.47 17)

a) PVn = 30.00

= 30.00(27.66068431) = 829.82

b) Interest = 36(30.00) - 829.82 = 1080.00 - 829.82 = $250.18 18)

PVn = 2700.00

= 2700.00(22.2084332) = $59 962.77

Interest = 2700.00(40) - 59962.71 = 108000.00 - 59962.77 = $48 037.23 19) a) PVn = 1224.51

= 1224.51(18.7014044)

= 22 900.06 Purchase price = 22 900.06 + 22900.00 = $45800.06 b) Interest = 24(1224.51) - 22900.06 = 29388.24 - 22900.06 = $6488.18 20)

PVn = 715

= 715.00(61.9193329) = $44 272.32

Interest: 715.00(84) - 44272.32 = 60060.00 - 44272.32 = $15 787.68 21)

PVn = 2000.00

= 2000.00(3.88965126) = $7779.30

Interest: 2000.00(5) -7779.30 = 10000.00 -7779.30 = $2220.70 22) S tep 1: PMT = 3000.00; n = 15(12) = 180; I/Y = 7.5; P/Y = C/Y = 12; i= PV = PMT

= 0.625% = 0.00625 = 3000

Step 2: FV = 323620.28; n = 10(12) = 120 PV = 323620.28(1.00625)-120 = $153 224.61

= $323 620.28

Prog ram med solu t ion: 23)

PVn = 7600.00

= 7600.00(13.4672608) = $102 351.18

PV = 102351.18(1.0325)-20 = 102351.18(.5274713) = $53 987.31 24) B alance in account 7 years from now.

PVn = 700.00

= 700.00(32.6954229) = $22 886.80

PV = 22886.80(1.0053)-84 = 22886.80(.6414498) = $14 680.73 25) B alance in account 15 years from now.

PVn = 15000.00

= 15000.00(2.577096987) = $38 656.45

PV = 38656.45(1.08)-15 = 38656.45(.315241705) = $12 186.13 26)

PVn = 3350.00

= 3350.00(5.255817) = $17 606.99

PV = 17606.99(1.0392)-12 = 17606.99(.6303915) = $11 099.30 27)

a) PVn = 762.79

= 762.79(47.2924743) = $36 074.23

b) FVn = 762.79

= 762.79(9.4633742) = $7218.57

c) PVn = 762.79

= 762.79(43.4234299)

d) Interest: 40341.53 - 36074.23 = $4267.30 e) Payment needed after 9 months = $7218.57 minus Normal payments = 9 × 762.7 = 6865.11 Additional interest = $ 353.46 28)

PVn = 472.00

= 472.00(70.1212093) = $33 097.21

Amount paid = 472.00 × 87 = $41 064.00 29)

Interest paid = 41064.00 - 33097.21 = $7966.79 a) PV

n = 575.00

= 575.00(47.3719563) = $27 238.87

Purchase price = 27238.87+ 3500.00 = $30 738.87 30)

b) Interest = 60(575.00) - 17764.48 = 34500.00 - 27238.87 = $7261.13 a) PVn = 1320.00

= 1320.00(65.6601227) = $86 671.36 b) Payout = 84(1320.00) = $110 880.00 c) Interest = 110880.00 - 86671.36 = $24 208.64 31) 32)

PVn = 715.00

= 715.00(21.0352112) = $15 040.18

PVn =

= 214.00(21.7316663) = $4650.58

214.0Purchase price = 4650.58 + 1300.00 = $5950.58 0 33) A mount needed at retirement PVn = 590.00 = 590.00(153.4550473) = $90 538.48 PV = 90538.48(1.0051)-264 = 90538.48(.261067) = $23 636.61 34)

122200.00 = PMT 122200.00 = PMT(85.6044029)

35)

$1427.50 = PMT The size of the monthly deposit is $1427.50 8000.00 = PMT 8000.00 = PMT(32.6760126)

36)

$244.83 20000.00==PMT PMT

20000.00 = PMT(25.26713874) 37)

$791.54 17520.00==PMT PMT

17520.00 = PMT(309.6943889) $56.57 = PMT 38) T he mortgaged balance is 125 000 × (1 - 0.25) = 93 750 93750.00 = PMT 93750.00 = PMT(39.0460809) $2401.35 = PMT 39) T he mortgaged balance is 160 000 × (1-0.25) = 120 000 120000.00 = PMT 120000.00 = PMT(155.206864) $773.16 = PMT 40) F V = 113 200, i = 0.065 ÷ 2 = 0.0325, n = 9.5 × 2 = 19 113200.00 = PMT 113200.00 = PMT(25.7280812 41)

$4399.86 = PMT P V = 5000.00; n = 5(4) = 20; i =

PV = PMT PV = PMT 5000 = PMT[16.35143334]

= 2% = 0.02 ; I/Y = 8; P/Y = C/Y = 4

PMT $305.78 = Programmed solution: = 42) P V = 15000.00; n = 5(12) = 60; i =

= 1% = 0.01

PV = PMT 15000 = PMT 15000 = PMT[44.95503841] PMT =

= $333.67

43) F V = 2510.00; n = 4(4) = 16; i =

= 1% = 0.01; I/Y = 4; P/Y = C/Y = 4

FV = PMT 2510 = PMT 2510 = PMT[17.25786449] PMT = $145.44 44) P V = 235000, PMT = 2475, i = 0.0755 ÷ 12 = 0.0062917 235000 = 2475

n=45) F V = 11453.40; PMT = 400.00; i =

=-

0.4295025 = 1.015n - 1 1.4295025 = 1.015n n ln 1.015 = ln 1.4295025 n = 24 (quarters) Programmed solution:

10000.00 = 500 0.6 = 1.03n - 1

= 145.056 = 145 ( months)

= 1.5% = 0.015 ; I/Y = 6; P/Y = C/Y = 4

11453.40 = 400

46) F V = 10000.00; PMT = 500.00; i =

=

= 3.0% = 0.03

1.6 = n ln 1.03 = ln 1.6 1.03n n = 15.9 (payments) ≈ 16 payments 47) F V = 50000.00; PMT = 1000.00; i =

= 0.75% = 0.0075

50000.00 = 1000 0.375 = 1.0075n - 1 1.375 = 1.0075n n ln 1.0075 = ln 1.375 n = 42.62 (months) ≈ 43 months 48) A t the moment of last deposit: FV = 31300, n = 4 × 2 = 8, i = 0.0645 ÷ 2 = 0.03225 PV = 31300.00(1.03225)-8 = 31300.00(0.7757484) = $24280.92 = FVg g = 24280.92, PMT = 726.56, i = 0.03225 Annuity: FV

n=

=

= 23.04 ≈

23 deposits

49) P V = 19 850, PMT 620, i = 0.065 ÷ 4 = 0.01625

n=-

=-

=

= 45.56 ≈ 46 payments

The required number of payments is 41 50) F V = 1 000 000.00; PMT = 25000.00; i =

= 9% = 0.09

1 000 000.00 = 25000 3.6 = 1.09n - 1 4.6 = 1.09n n ln 1.09 = ln 4.6 n = 17.71 (years) ≈ 18 years 51) T he balance at the moment of last deposit: FV = 50 000, i = 0.06 ÷ 12 = 0.005, n = 5 × 12 = 60 PV = 50000.00(1.005)-60 = 50000.00(0.7413722) = $37 068.61 For annuity: FV

n=

g = 37068.61, PMT = 1000, i = 0.005

=

= 34.09 ... ≈ 34 deposits

52) n=-

=-

53) P Vn = 6800.00; PMT = 1175.00; n = 12

=

= 54.33 months

The nominal annual rate = 2(0.135)(100%) = 27%

54) F V = 30 000.00; PMT = 200.00; n = 10(12) = 120; P/Y = C/Y = 12 Programmed solution:

55) N per 71 Pmt -573.00 FV 106000.00 Result 0.0242781 The nominal rate is 2.42781% × 4 = 9.71124% 56)

The nominal rate is 12(0.0166833)(100%) = 20.02% 57) N per 120 Pmt -1972.00 PV 191000.00 * (1 - .15) Result 0.0066885 f = (1 + 0.0066885)4 - 1 = 0.0270236 = 2.70236% 58) P V n = 176 875.67, PMT = 2 720, n = 8(12) = 96 Calculator answer: i = 0.0086532 The nominal rate = 12 (0.0086532)(100%) = 10.38384% 59) F V n = 52 700, PMT = 560, N = 6(12) = 72 Calculator answer: i = 0.007252 The nominal rate = 12(0.007252)(100%) = 8.7024% 60) N per 18 Pmt -810.00 PV 9200.00 Result 0.0537101 f = (1 + 0.0537101)2 - 1 = 11.0304975% 61) B 62) A 63) B 64) A 65) A

66) D 67) D 68) D 69) C 70) D 71) C 72) B 73) C 74) B 75) E 76) A 77) C 78) E 79) B 80) B 81) E 82) A 83) D 84) C 85) B 86) E 87) P V = $800 000; PMT = $60,000; i =

= 1.5625%

800000 = 60000

⇒ 1 - 1.015625-n = 0.2083 ⇒ 1.015625-n = 0.7917 ⇒ -n ln 1.015625 = ln 0.7917 ⇒ -n =

88)

⇒ n = 15 quarters ⇒ 3 years 9PMT months P Vn==$20,000; = $300; i = = 0.333% 20000 = 300 ⇒ 1 - 1.00333-n = 0.222 ⇒ 1.00333-n = 0.778 ⇒ -n ln 1.00333 = ln 0.778 ⇒ -n = ⇒ n = 76 months ⇒ n = 6 years 4 months

89) B 90) E 91) E 92) D 93) A 94) B 95) C

96) A 97) D 98) E 99) B 100) B 101) C 102) D 103) B 104) B 105) D 106) D 107) A 108) $ 25 * 26 bi-weeks/year * 47 years = $30 550 Periodic interest rate: i = 0.0195/26 = 0.00075 = 0.075% n = 26(47) = 1222 FV = 25

= 25

= $49 988.81

109) T FSA Return Periodic rate of interest: i = 0.048 n=4 FV1 = 6000

= 6000

= $25 783.96

Savings Account PV = FV1 = $25 783.96 Periodic rate of interest: i = 0.035 n=3 FV2 = 25783.96(1.035^3) = $28 587.14 110) P eriodic rate of interest: 0.052/4 = 0.013 = 1.3%; n = 8.75 * 4 = 35 PV = 100

= $2797.61

111) P eriodic rate of interest: 0.042/12 = 0.0035 = 0.35%; n = 3 * 12 = 36 PV = 745

= 745(33.768910) = $25 157.84

Cash Value = 22500 + 25157.84 = $47 567.84 112) P eriodic rate of interest: 0.094/2 = 0.047 = 4.7%; n = (78/12) * 2 = 13 PMT =

=

= $785.44

113) P eriodic rate of interest: 0.012/12 = 0.001 = 0.1%; PV = 31500 - 11500 = 20000; n = 5 * 12 = 60 PMT =

=

= $343.60

114) n=

=

= 13.815395 years = 14 years

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) C alculate the accumulated value after ten years of payments of $1000.00 made at the end 1) _____________ of each quarter if interest is 4% compounded semi-annually.

2) W hat is the accumulated value of deposits of $1120.00 made at the end of every six months for three years if interest is 8.48% compounded quarterly?

2) _____________

3) W hat is the accumulated value of deposits of $2000.00 made at the end of every year for three years if interest is 6.00% compounded quarterly?

3) _____________

4) W hat is the future value of payments of $230.00 made at the end of every three months in twenty-one years if interest is 7.65% compounded annually?

4) _____________

5) C onchita Martinez saves $17.25 at the end of each month and deposits the money in an account, paying 4.96% compounded quarterly. a) How much will she accumulate in 22 years? b) How much of the accumulated amount is interest?

5) _____________

6) A li saves $25.00 at the end of each month and deposits the money in a savings account, paying 3.00% compounded semi-annually. a) How much will she accumulate in 20 years? b) How much of the accumulated amount is interest?

6) _____________

7) M rs. Engleder has made deposits of $750.00 at the end of every six months for twenty-three years. If interest is 3.72% compounded monthly, how much will Mrs. Engleder have accumulated six years after the last deposit?

7) _____________

8) W hat is the accumulated value after 11.25 years of monthly deposits of $415.00 earning interest at 5.56% compounded semi-annually if the deposits are made at the end of each month?

8) _____________

9) W hat is the accumulated value after 15.50 years of deposits of $500.00 earning interest at

9) _____________

4.00% compounded semi-annually if the deposits are made at the end of each quarter?

10) A loan is repaid by making payments of $6000.00 at the end of every six months for twelve years. If interest on the loan is 8% compounded quarterly, what was the principal of the loan?

10) _____________

11) W hat is the discounted value of $4760.00 paid at the end of each year for 7 years if interest is 6.52% compounded quarterly?

11) _____________

12) W hat is the discounted value of $3000.00 paid at the end of every six months for 5 years if interest is 8.00% compounded yearly?

12) _____________

13) W hat is the principal from which $2279.00 can be withdrawn at the end of each month for 17.5 years if interest is 5.44% compounded quarterly?

13) _____________

14) A real estate developer bought land for $170 000.00 down and monthly payments of $10

450.00 for 5 years.

What is 14) the equivale nt cash price if money is worth 7.75% compou nded semi-ann ually?

15) A student bought a rental property for $30 000.00 down and monthly payments of

___ ___ ___ ___ _

15) _____________

$1100.00 for 5 years. What is the equivalent cash price if money is worth 5.25% compounded semi-annually?

16) I f a loan was repaid by monthly payments of $9230.00 in 7.5 years at 6.15% compounded annually, how much interest was paid?

16) _____________

17) I f a loan was repaid by semi-annual payments of $4720.00 in 6.5 years at 8.24% compounded quarterly, how much interest was paid?

17) _____________

18) I f a student loan was repaid by monthly payments of $800.00 in 5.0 years at 8.00% compounded yearly, how much interest was paid?

18) _____________

19) A 21-year mortgage is amortized by making payments of $3052.61 at the end of every three months. If interest is 8.45% compounded annually, what was the original mortgage balance?

19) _____________

20) F or her daughter's university education, Carla Hackl has invested an inheritance in a fund paying 9.2% compounded quarterly. If ordinary annuity payments of $4750.00 per month are to be made out of the fund for 5 years and the annuity begins 7.75 years from now, how much was the inheritance?

20) _____________

21) Y ou won $150 000.00 in a lottery and you want to set some of that sum of money aside for 6 years. After 6 years you would like to receive $2710.00 at the end of every 6 months for 8 years. If interest is 6% compounded quarterly, how much of your winnings must you set aside?

21) _____________

22) Y ou won $250 000.00 in a lottery and you want to set some of that sum of money aside for 11.5 years. After 11.5 years, you would like to receive $4500.00 at the end of each month for 4.75 years. If the interest is 6.79% compounded annually, how much of your winnings must you set aside?

22) _____________

23) W hat is the principal invested at 6.76% compounded semi-annually from which monthly withdrawals of $450.00 can be made at the end of each month for 22 .75 years?

23) _____________

24) A 22-year mortgage is amortized by payments of $1761.50 made twice a year. If interest is 9.65% compounded semi-annually, what is the mortgage principal?

24) _____________

25) A loan was repaid in 6.5 years by quarterly payments of $515.00 at 9.85% compounded

semi-annually.

What is 25) the value of the loan?

26) C alculate the amount of money that must be deposited at the end of every three months

___ ___ ___ ___ _ 26) _ ____________

into an account paying 6% compounded monthly to accumulate to $12 500.00 in ten years?

27) A $ 120 000.00 mortgage is amortized over 25 years. If interest on the mortgage is 8.5% compounded semi-annually, calculate the size of monthly payments made at the end of each month.

27) _____________

28) H ow much must be deposited at the end of each quarter for 5 years to accumulate to $27 000.00 at 6.84% compounded biweekly?

28) _____________

29) H ow much must be deposited at the end of each month for 10 years to accumulate to $100 000.00 at 6.00% compounded yearly?

29) _____________

30) A loan of $91 450.00 is to be repaid by equal semi-annual payments for 11 years. What is the size of each semi-annual payment if the interest is 9.24% compounded annually?

30) _____________

31) A loan of $40 000.00 is to be repaid by equal quarterly payments for 5 years. What is the size of each semi-annual payment if the interest is 5.00% compounded annually?

31) _____________

32) K elsey bought a car priced at $19 700.00 for 10% down and equal monthly payments for 4.5 years. If interest is 8.22% compounded semi-annually, what is the size of the monthly payment?

32) _____________

33) E dward Martin borrowed $15 810.00 from his uncle to finance his graduate studies. The loan agreement calls for equal payments at the end of each month for 6 years. The payments are deferred for 3 years and interest is 8.12% compounded semi-annually.

What is the size of the monthly payments?

33) _____________

34) A mortgage of $198 000.00 is to be amortized by payments over 22.5 years. If the

34) _____________

payments are made at the end of each month and interest is 8.75% compounded semi-annually, what is the size of the monthly payments?

35) W hat sum of money can be withdrawn from a fund of $46 950.00 invested at 6.5% compounded semi-annually at the end of every three months for twelve years?

35) _____________

36) S ean Paul borrowed $20 000.00 from his father to finance his new business. The loan agreement calls for equal payments at the end of each month for 10 years. The payments are deferred for 4 years and interest is 4.00% compounded semi-annually. What is the size of the monthly payments?

36) _____________

37) H ow much must be contributed into an RRSP at the end of each year for 37 years to accumulate to if interest is 8.04% compounded quarterly?

37) _____________

38) H ow many deposits of $6100.00 made at the end of every 6 months are needed to accumulate to $172 000.00 at 5.24% compounded quarterly?

38) _____________

39) F or how long will Harold have to make payments of $550.00 at the end of every three months to repay a loan of $16 900.00 if interest is 9% compounded monthly?

39) _____________

40) F or how long will Jessica have to make payments of $1200.00 at the end of every three months to repay a loan of $10 000.00 if interest is 12% compounded semi-annually?

40) _____________

41) M r. Lavergne accumulated $320 000.00 in an RRSP. He converted the RRSP into an RRIF and started to withdraw $7500.00 at the end of every three months from the fund. If interest is 4.68% compounded monthly, how many withdrawals can Mr. Lavergne make?

41) _____________

42) M r. Amer accumulated $102 800.00 in an RRSP. He converted the RRSP into an RRIF and started to withdraw $2510.00 at the end of every month from the fund. If interest is 4.5% compounded yearly, for how long can Mr. Sepaba make withdrawals?

42) _____________

43) D eon makes monthly contributions of $250.00 at the end of every month into an RRSP. Calculate, to the nearest month, how long it will take to grow to $100 000.00 if interest is 8% compounded annually.

43) _____________

44) A loan of $54 000.00 is repaid by making payments of $3000.00 at the end of every three months. If the interest is 9% compounded monthly, calculate how long the payments have to be made.

44) _____________

45) F or how long must $179.23 be deposited at the end of each month to accumulate to $9700.00 at 6.18% compounded quarterly?

45) _____________

46) A $360 000.00 mortgage is amortized by making monthly payments of $2600. If interest is 7.5% compounded semi-annually, what is the term of the mortgage?

46) _____________

47) S abrina deposits $300.00 into a savings account at the end of each month for five years. If the accumulated value of the deposits is $20 000.00 and interest is compounded semi-annually, calculate the nominal rate of interest

47) _____________

48) C alculate the nominal interest rate of interest compounded quarterly if a loan of 43 000.00 is repaid in seven years by payments of $4000.00 made at the end of every six months.

48) _____________

49) N ote: The calculations for this question were done using Excel's RATE function.

Compute the nominal annual rate of interest on a loan of $67 000.00 repaid in semi-annual installments of $6540.00 in 11.5 years.

49) _____________

50) N ote: The calculations for this question were done using Excel's RATE function.

What nominal annual rate of interest compounded semi-annually is earned by quarterly deposits of $7327.00 made for six years if the balance just after the last deposit is $289 000.00?

50) _____________

51) N ote: The calculations for this question were done using Excel's RATE function.

What is the rate of interest compounded quarterly if a loan of $31 500.00 is repaid in seven years by payments of $2712.00 made at the end of every six months?

51) _____________

52) N ote: The calculations for this question were done using Excel's RATE function.

A $345 000.00

mortgag52) e is repaid in 19 years by making monthly payment s of $2486.44.

What is the nominal annual rate of interest compou nded semi-ann ually?

___ ___ ___ ___ _

53) N ote: The calculations for this question were done using Excel's RATE function.

A property worth $15 000.00 is purchased for 10% down and semi-annual payments of $1750.00 for twelve years. What is the effective rate of interest if interest is compounded semi-annually?

53) _____________

54) N ote: The calculations for this question were done using Excel's RATE function.

A $160 000.00 mortgage with a 20-year term is repaid by making monthly payments of $1361.00. What is the rate of interest compounded semi-annually on the mortgage?

54) _____________

55) F or their daughter's future education David and Bonnie set a fund in which they invest at the end of every six months. The first investment is $215 and they plan to increase their investments at a constant rate of 0.5%. If the fund earns 2.4% compounded semi-annually, how much money will be in the fund in 18 years when their daughter goes to college?

55) _____________

56) H ow much should you invest today into account paying 3.2% compounded quarterly if

56) _____________

you want to be able to withdraw at the end of every 3 month the amounts of money growing at a constant rate of 3% for the period of 5 years , if the first withdrawal is to be $100 five years three months from now?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

57) C alculate the accumulated value after ten years of payments of $1000.00 made at the end of each 57) ______ year if interest is 6% compounded monthly.

A) $ 13 285.11 B) $ 13 228.03 C) $ 10 600.00 D) $ 10 228.56 E) $ 10 228.03 58) C alculate the accumulated value after twenty years of payments of $12 000.00 made at the end of each year if interest is 6% compounded semi-annually.

58) ______

A) $ 441 427.09 B) $ 322 444.73 C) $ 445 721.73 D) $ 240 000.00 E) $ 642 766.07 59) W hat is the final amount of an investment where you are making quarterly deposits of $672.31 at the end of the payment period for 7 years and 9 months? The interest rate is 6.16% compounded semi-annually.

A) $ 28 611.16 B) $ 26 407.57 C) $ 28 811.16 D) $ 26 811.16 E) $ 26 611.57

59) ______

60) Y ou miss the 11th to 13th payments of a loan. The loan payments are $519.27 each month and the

60) ______

interest rate is 8.45% compounded annually. How much are you behind in your payments on the day that you miss the 13th payment?

A) $ 1586.40 B) $ 1508.40 C) $ 1458.40 D) $ 1658.40 E) $ 1568.40 61) Y ou are to receive $16 700 from your trust fund in 7.5 years. You negotiate a deal where you can withdraw $350 per month starting 2.75 years before you can withdraw the remainder of the funds. You are able earn 4.18% compounded annually on your money. How much is your trust fund reduced to?

A) $ 4195.40 B) $ 4159.40 C) $ 4155.40 D) $ 4459.40 E) $ 4495.40

61) ______

62) A mortgage requires payments of $4000.00 at the end of every six months for twelve years. If interest is 8% compounded quarterly, calculate the principal of the loan.

A) $ 88 320.00 B) $ 84 948.26 C) $ 75 655.70 D) $ 60 738.85 E) n one of the above

62) ______

63) A mortgage requires payments of $1000.00 at the end of every month for twenty-five years. If interest is 6% compounded semi-annually, calculate the principal of the loan.

A) $ 46 188.41 B) $ 156 297.23 C) $ 33 328.64 D) $ 155 206.86 E) $ 300 000.00

63) ______

64) T he loan will have a term of 54 months and monthly payments of $385.13. The interest rate on the loan is 8.84% compounded quarterly. What is the amount of money that was borrowed?

A) $ 17 130.71 B) $ 17 430.71 C) $ 17 030.71 D) $ 17 310.71 E) $ 20 368.27

64) ______

65) Y ou lease a car for monthly payments of $610.23 at the end of each month for 54 months. There is a no security deposit sale going on. The interest rate on the lease is 3.9% compounded

anny. The uallbuy-out

is $19 65) 552 at the end of the lease.

How expensiv e is the vehicle?

___ ___

A) $ 46 863.26 B) $ 48 633.26 C) $ 48 863.26 D) $ 46 683.26 E) $ 48 683.26 66) W hat is the payment size of a loan that has semi-annual payments and an interest rate of 5.52% compounded monthly? The loan principal is $1641.12 and the loan is for 48 months.

A) $ 123.74 B) $ 274.31 C) $ 321.74 D) $ 221.31 E) $ 231.74

66) ______

67) C alculate the amount of money that must be deposited at the end of every three months into an account paying 6% compounded monthly to accumulate to $12 500.00 in ten years.

A) $ 304.94 B) $ 293.75 C) $ 80.77 D) $ 229.97 E) n one of the above

67) ______

68) C alculate the amount of money that must be deposited at the end of every month into an

68) ______

account paying 12% compounded annually to accumulate to $100 000.00 in ten years.

A) $ 416.67 B) $ 434.71 C) $ 833.33 D) $ 45.06 E) $ 450.59 69) Y ou buy a motorcycle for $52 523 plus freight of $724 and other delivery charges of $561. You have saved 25% of the total purchase price towards a down payment. The bank is willing to finance the purchase of the motorcycle at 9.86% compounded quarterly. What is the size of your monthly payment if the loan is for 4.5 years?

A) $ 262.80 B) $ 962.80 C) $ 996.80 D) $ 296.80 E) $ 926.80

69) ______

70) Y ou can buy a piece of equipment for $8600. The equipment is expected to last for 9 years and you want to have it completely paid off 3.5 years before it becomes unusable. Current loan rates are 9.5% compounded annually. What is the size of your semi-annual payment?

A) $ 1515.98 B) $ 1055.98 C) $ 1105.98 D) $ 1015.98 E) $ 1115.98

70) ______

71) M ark deposits $900 at the end of each month in a savings account earning 4.5% quarterly. How soon can he buy his first car assuming that he needs $8 000 for a decent car.

A) 8 .76 months B) 8 .86 months C) 8 .52 months D) 9 .06 months E) 9 years

71) ______

72) A mortgage of $326 000 is to be repaid by making payments of $2500 at the end of each month. If interest is 4.75% compounded semi-annually, what is the term of the mortgage?

72) ______

A) 1 84 months B) 1 06 months C) 1 83 months D) 1 05 months E) 1 82 months 73) Y ou make 6 quarterly deposits starting at $135 and increasing at a constant rate of 4.5%. The interest rate is 8.4% compounded quarterly. What is the size of the last deposit?

A) $ 186.23 B) $ 668.23 C) $ 168.23 D) $ 135.00 E) $ 128.63

73) ______

74) Y ou make 4 semi-annual deposits starting at $350 and increasing at a constant rate of 2.14%. The interest rate is 5.12% compounded annually. How much has been deposited in total?

A) $ 1459.59 B) $ 1445.59 C) $ 1400.00 D) $ 1454.59 E) $ 1444.59

74) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

75) S aba gets a year-end bonus in December every year before the Christmas. She plans to 75) _____________ invest $1000 from the bonus at the end of every year into mutual funds RRSP, which are expected to grow at the rate of 8% compounded semi-annually. What will be the total value of her periodic investments after 25 years?

76) B hamani is contributing to a RESP they have set up for their grandchildren. What amount will they have in the RESP after 13 years of contributing $185 at the end of every month if the plan earns 5.5% compounded quarterly?

76) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

77) A series of payments of $900 are being made at the end of each year for the next 8 years. What is 77) ______ the future value 8 years from now, if money can earn 9% compounded semi-annually?

A) $ 9999 B) $ 7200 C) $ 14 346 D) $ 77 200 E) $ 30 220 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

78) D on turned 60 years old today and has retired from Hydro One. Current economic value 78) _____________ of his pension is his significant asset. He will be receiving pension payments of $7000 at the end of each month. Based on a 22-year life expectancy and money worth 8% compounded semi-annually, calculate the current economic value of Don's pension.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

79) S ilva is contemplating to buy a 20 year annuity paying $2500 at the end of each month. What 79) ______ amount will be required to purchase the annuity, if the annuity provides a return of 6.75% compounded annually?

A) $ 600 000 B) $ 37 037 C) $ 567 436 D) $ 47 246 E) $ 334 000 80) A dealer sold a car to Kristy for $2000 down and monthly payments of $259.50 for 3.5 years, including interest at 7.5% compounded annually. What was the selling price of the car?

A) $ 12 899 B) $ 9600 C) $ 11 600 D) $ 10 899 E) $ 6012.50

80) ______

81) S am took a $5000 loan to build a deck in the backyard. The loan requires payments at the end of each quarter or 4 years. If the interest on the loan is 9% compounded monthly, what is the size of each payment Sam is required to make?

A) $ 1321.64 B) $ 353.98 C) $ 332.80 D) $ 376.08 E) $ 447.83

81) ______

82) I n order to purchase a truck Shandex Transport (ST) obtained a $50 000 loan for 5 years at 7.8% compounded semi-annually. The loan is structured to reduce the balance owing to $10 000 at the

end the 5 of year

___ ___

period.

82) How much are ST's payment s at the end of each month?

A) $ 868.84

B) $ 804.87

C) $ 1006.08

D) $ 834.79

E) $ 1070.11

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

83) D ion is planning to buy a brand new BMW. He plans to keep the car for 3 years and then 83) _____________ trade it in. He borrows $37 000 at 7.2% compounded semi-annually to purchase the car.

What monthly payment will reduce the balance on the loan after 3 years to the expected trade-in value of $16 500?

84) O ne month from now, Shiva will make his first monthly contribution of $250 to a Tax-Free Savings account (TFSA). Over the long run, he expects to earn 8% compounded annually. How long will it take for the contributions and accrued earnings to reach $100 000?

84) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

85) M omin plans to buy a house on cash instead of paying mortgage. He targets spending $400 000 85) ______ on his dream house. He plans to set aside $1200 every month to reach his target. He puts his savings in a Tax Free Savings Account (TFSA) and invests them in a high risk mutual fund, which has traditionally earned 12.2% annually. How long will it take Momin to reach his target?

A) 1 5 years 3 months B) 2 4 years 9 months C) 1 3 years 9 months D) 1 2 years 6 months E) 2 7 years 10 months SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

86) Note: This problem is solved using CalcPro for iPad 86) _____________ Lahore Gym allows members to pay the members to pay the annual membership fee by a single payment of $240 at the beginning of the year, or by payments of $22 at the beginning of each month. What effective rate of interest is paid by members, who choose to pay by month?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

87) W hat annually compounded rate of return must Thiva earn in her TFSA account in order for 87) ______ month end contributions of $950 to accumulate $1 million after 25 years?

A) 0 .716% B) 7 .16% C) 0 % D) 8 .97% E) 8 .62% SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

88) M omin plans to buy a house on cash instead of paying mortgage. He plans to set aside 88) _____________ $14 400 at the end of each year for 15 years. He puts his savings in a Tax Free Savings Account (TFSA) and invests them in a high risk mutual fund, which has traditionally earned 12.2% annually. Money decreases in value by 2.5% per annum as well. How long will it take Momin to reach his target?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

89) D on turned 60 years old today and has retired from Hydro One. Current economic value of his 89) ______ pension is his significant asset. He will be receiving pension payments of $84 000 every year.

Based on a 22-year life expectancy and money worth 8% compounded annually and money depreciating at 2% annually, calculate the current economic value of Don's pension.

A) $ 1 001 884

B) $ 832 256

C) $ 1 234 888

D) $ 740 932

E) $ 877 326

90) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, what will be the balance in the account at the end of the three-year term?

A) $ 17 390.19 B) $ 27 284.81 C) $ 18 408.04 D) $ 18 600.50 E) $ 18 603.37

90) ______

91) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, how much interest is included in the future value of the annuity?

A) $ 408.04 B) $ 600.50 C) $ 18 603.37 D) $ 603.37 E) $ 18 600.50

91) ______

92) J ordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost. If the production life of the plant is 60 years and the operator puts the money at the end of the month in a savings account, earning 7.25% compounded semi-annually, what will be the value of the decommissioning fund after 60 years of production?

A) $ 917.65 million B) $ 11.88 billion C) $ 10.87 billion D) $ 12.49 billion E) $ 720 million

92) ______

93) J ordan is considering building a nuclear power plant, which will be ready for production in 2030. The country's governing body is also considering a decommissioning liability law for the operator to put aside $1 million every month towards decommissioning cost, when the plant is producing electricity. During the 60 years of production life of the plant, the operator will put $1 million at the end of the month in a savings account, earning 7.25% compounded semi-annually.

At the end of the production life of the plant, there are no more contributions and the money is expected to grow at the rate of 6% compounded quarterly for the next 30 years. What will be the value of the decommissioning fund in 2120?

93) ______

A) $ 74.57 billion B) $ 11.88 billion C) $ 60.90 billion D) $ 70.94 billion E) $ 12.49 billion 94) S ilva saves $100 at the end of each month and deposits the money in an account, paying 7.49% compounded semi-annually. How much of the accumulated amount is interest after 40 years?

A) $ 253 521 B) $ 243 872 C) $ 291 872 D) $ 299 531 E) $ 251 531

94) ______

95) W oods has contributed $3600.00 per year for the last ten years into a RRSP account earning 9%

has saving bigat the ger end of

compounded semi-annually. His wife Karen, on the other hand, contributed $300 at the end of every month for the same period into a RRSP account earning 9% compounded quarterly. Who

10 years and by how much?

95)

___ ___

A) W ood by $2580 B) K aren by $3360 C) K aren by $2580 D) W ood by $3360 E) K aren by $2262 96) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the end of each month, what will be the present value of the balance in the account at the end of the three-year term?

A) $ 18 603.37 B) $ 17 392.96 C) $ 18 600.50 D) $ 18 408.04 E) $ 17 390.19

96) ______

97) S am got a job at the Brick. He immediately finances a car paying $500 every month for 3 years.

The interest is 2.25% compounded semi-annually. How much is the interest cost of the financing?

A) $ 607.04 B) $ 17 392.96 C) $ 609.81 D) $ 408.04 E) $ 603.37

97) ______

98) K eith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the end of each month for 20 years. Interest is 4.4% compounded semi-annually. What is the purchase price of the property?

A) $ 279 831 B) $ 194 378 C) $ 254 646 D) $ 195 881 E) $ 278 878

98) ______

99) S iri plans to retire now and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded quarterly.

A) $ 1 528 015 B) $ 633 665 C) $ 539 042 D) $ 1 530 569 E) $ 793 072

99) ______

100) K nowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded semi-annually. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of each month for four years, how much did Roberta deposit?

A) $ 65 533 B) $ 32 474 C) $ 32 723 D) $ 54 343 E) $ 65 592

100) _____

101) S iri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now every month (at the end of the month) if interest is 4.9% compounded semi-annually.

A) $ 4420.74 B) $ 11 755 C) $ 1 534 362 D) $ 1 528 015 E) $ 4461.50

101) _____

102) J amal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded semi-annually. What will be the size of his payments per month from the ordinary annuity for 20 years following his retirement?

A) $ 8721 B) $ 8618 C) $ 8372 D) $ 941 547 E) $ 957 836

102) _____

103) S iri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded semi-annually.

A) 2 17.35 years

103) _____

B) 1 8 years C) 2 16 years D) 3 6 years E) 1 8.11 years 104) J amal plans to retire in 17 years. He is saving $2000 every month in a retirement savings account paying him a long-term interest of 9% compounded semi-annually until retirement. The rate changes following the retirement. He wants $7000 per month paid to him for 20 years after the retirement. At what rate (semi-annually) should his savings account pay him to fulfill his dream?

(Use the Rate function in Excel.)

A) 0 .86% B) 6 .55% C) 0 .54% D) 3 .28% E) 1 0.27%

104) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

105) Y u Yi receives annuity payments on a quarterly basis and then puts the money into an 105) ____________ account earning 3.0% p.a. monthly. What's the equivalent rate of interest per payment period?

106) E very 3 months, Carolyn gets an allowance of $80 that she puts into a savings account with a 1.2% p.a. return compounded semi-annually. After 6 years, what is the balance of her savings account?

106) ____________

107) W hile planning for his retirement, Alishan purchased an annuity that paid him $3200

107) ____________

every year for 13 years at a rate of 3.8% compounding semi-annually. What was Alishan's initial investment into the annuity?

108) W hat is the loan principal for a loan in which payments of $460 must be made at the end of every two weeks for 1.5 years, and the interest rate is 4.4% p.a. compounded quarterly?

108) ____________

109) R aahim, a new business owner, puts a certain amount of money aside into a mutual fund for future use. The fund returns 4.6% compounding interest quarterly. He hopes that after 5.5 years, he'll have $500 000 in the account. How much money on a monthly basis should Raahim put into the fund?

109) ____________

110) A ntonio bought a big boat for $17 400 and paid $1100 as a down payment. The remainder of the balance must be paid in equal installments at the end of each year for six years. If interest compounds on a semi-annual basis at 8.2% p.a., what is the size of the yearly payment?

110) ____________

111) A start-up known as Spec-lights recently decided to lease new equipment valued at $33 500. The company has to make payments of $6500 at the end of each year, with an interest rate of 4.5% p.a. compounded quarterly. How long will payments have to be made?

111) ____________

112) H ow many monthly payments of $100 would it take for an RRSP to have a balance of $12,000 at an interest rate of 3.4% p.a. compounding semi-annually?

112) ____________

113) W ithdrawals from Mo's RRSP are made semi-annually, starting with $3000 and increasing by 3% per term for 8 years. The interest rate on the money is 6.8% p.a. compounding semi-annually.

113) ____________

a) What is the size of the 13th withdrawal? b) Compute the compounding factor for the constant-growth annuity.

114) ____________ 114) P etra has started to make contributions towards her child's RESP, starting with a contribution of $4500 and decreasing by 2% at the end of every year when she makes the next contribution. If she plans on doing this for 18 years with an interest rate of 3.0% p.a. compounded annually, what will be the balance of the RESP at the end of the 18 years?

1) P MT = 1000.00; n = 10(4) = 40; i =

= 2% = 0.02; c = 2/4 = 0.5

p = (1 + 0.02)^0.5 - 1 = 0.00995049 FV = 1000

= $48 836.51

2) P MT = 1120, i = 0.0848 ÷ 4 = 0.0212, n = 3 × 2 = 6, c =

=2

p = (1.0212)2 - 1 = 0.0428494 FVg = 1120.00 = 1120.00(6.6804902) = $7482.34 P MT = 2000, i = 0.06 ÷ 4 = 0.015, n = 3, c = 3)

=4 p = (1.015)4 - 1 = 0.0613636 FVg = 2000.00

= 2000.00(3.1878563) = $6375.71

4) P MT = 230, i = 0.0765 , n = 21 × 4 = 84, c = p=

- 1 = 0.0185996

FVg = 230.00 5)

= 230.00(199.0432233) = $45 779.94

P MT = 17.25, i = 0.0496 ÷ 4 = 0.0124, n = 22 × 12 = 264, c = = a) p = (1 + 0.0124)1/3 - 1 = 0.00411637 FVg = 17.25

= 17.25(475.6458279) = $8204.89

b) Interest = 8204.89 - 264(17.25) = 8204.89 - 4554.00 = $3650.89 6) P MT = 25, i = 0.03 ÷ 2 = 0.015, n = 20 × 12 = 240, c = = a) p = (1 + 0.015)1/6 - 1 = 0.00248452 FVg = 25.00

= 25.00(327.636659) = $8190.92

b) Interest = 8190.92 - 240(25.00) = 8190.92 - 6000.00 = $2190.92 7) P MT = 750, i = 0.0372 ÷ 12 = 0.0031, n = 23 × 2 = 46, c = =6 Amount after 23 years. p = (1 + 0.0031)6 - 1 = 0.0187447 FVg = 750.00

= 750.00(72.0026939) = $54 002.02

Amount after another 6 years 8)

FV = 54002.02(1.0031)72 = $67 483.03 P MT = 415, i = 0.0556 ÷ 2 = 0.0278, n = 11.25 × 12 = 135, c =

=

p = (1 + 0.0278)1/6 - 1 = 0.0045806 FVg = 415.00 9)

= 415.00(186.2861243) = $77 308.74

PMT 0.02, n = 15.5 × 4 = 62, c = = 0.5 = 500, i p = (1 + 0.02)0.5 - 1 = 0.0099505 = 0.04 FVg = 500.00 =500.00(85.18052518) = 42 590.30 ÷2= 10) P MT = 6000.00; n = 12(2) = 24; i =

= 2% = 0.02; c =

= 2; I/Y = 8; P/Y = 2; C/Y = 4

p = (1 + 0.02)2 - 1 = 1.0404 - 1 = 0.0404 PV g = PMT

= 6000

= $91 108.28

11) P MT = 4760, i = 0.0652 ÷ 4 = 0.0163, n = 7, c =

=4

p = (1 + 0.0163)4 - 1 = 0.0668115 PVg = 4760.00

= 4760.00(5.4497186) = $25 940.66

12) P MT = 3000, i = 0.08, n = 5 × 2 = 10, c = p = (1 + 0.08)1/2 - 1 = 0.0392305 PVg = 3000.00 = 3000.00(8.142055396) = 24 426.17 P MT = 2279, i = 0.0544 ÷ 4 = 0.0136, n = 17.5 × 12 = 210, c = 13)

= p = (1 + 0.0136)1/3 - 1 = 0.0045129 PVg = 2279.00 = 2279.00(135.5103403) = $308 828.07 P MT = 10450, i = 0.0775 ÷ 2 = 0.03875, n = 5 × 12 = 60, c =

14)

= p = (1 + 0.03875)1/6 - 1 = 0.0063565 PVg = 10450.00

= 10450.00(49.7543772) = $519 933.24

Cash price = 170000.00 + 519933.24 = $689 933.24 15) P MT = 1100, i = 0.0525 ÷ 2 = 0.02625, n = 5 × 12 = 60, c =

=

p = (1 + 0.02625)^(1/6) - 1 = 0.004327902 PVg = 1100.00

= 1100.00(52.74261) = $58 016.87

Cash price = 30000 + 58016.87 = $88 016.87 16) P MT = 9 230, i = 0.0615, n = 7.5 × 12 = 90, c = p = (1 + 0.0615)1/12 - 1 = 0.004986 PVg = 9230.00

= 9230.00(72.3737824) = $668 010.79

Interest paid = 90(9230.00) - 66801.01 = 830700 - 668010.01 = $162 689.99 17) P MT = 4720, i = 0.0824 ÷ 4 = 0.0206, n = 6.5 × 2 = 13, c = p = (1 + 0.0206)2 - 1 = 0.0416244

=2

PVg =

= 4720.00(9.8857467) = $46 660.74

4720.Interest paid = 13(4720.00) - 46660.72 00 = 61360.00 - 46660.72 = $14 699.28 18) P MT = 800, i = 0.08, n = 5 × 12 = 60, c = p = (1 + 0.08)1/12 - 1 = 0.0064340 PVg = 800.00

= 800.00(49.6449458) = $39 715.92

Interest paid = 60(800.00) - 39715.96 = 48000.00 -39715.96 = $8284.04 19) P MT = 3052 .61, i = 0.0845, n = 21 × 4 = 84, c = p = (1 + 0.0845)1/4 - 1 = 0.0204868

20)

PVg = 3052.61 = 3052.61(39.925945) = $121 178.36 P MT = 4750, i = 0.092 ÷ 4 = 0.023, n = 5 × 12 = 60, c = = p = (1 + 0.023)1/3 - 1 = 0.0076083 The amount in the fund 7.75 years from now (annuity begins): PV g = 4750.00 Today's balance:

= 4750.00(48.0273124) = $228 129.73

PV = 228129.73 (1+ 0.023)-7.75 × 4 = $112 729.41 P MT = 2710, i = 0.06 ÷ 4 = 0.015, n = 8 × 2 = 16, c = 21)

=2 Balance at 6 years (annuity starts): p = (1 + 0.015)2 - 1 = 0.030225 PVg = 2710.00

= 2710.00(12.539523) = $33 982.11

Balance now: PV = 33982.11(1 + 0.015)-6 × 4 = $23 771.98 P MT = 4500, i = 0.0679, n = 4.75 × 12 = 57, c = 22) Balance after 11.5 years (annuity starts) p = (1 + 0.0679)1/12 - 1 = 0.0054895 PVg = 4500.00

23)

= 4500.00(48.8299465) = $219 734.76

Balance today PV = 219734.76(1.0679)-11.5 = $103 227.88 P MT = 450, i = 0.0676 ÷ 2 = 0.0338, n = 22 .75 × 12 = 273, c = = p = (1 + 0.0338)1/6 - 1 = 0.0055556 PVg = 450.00

= 450.00(140.3336252) = $63 150.13

24) P MT = 1761.50, i = 0.0965 ÷ 2 = 0.04825, n = 22 × 12 = 264, c = 2/2 = 1 p = 1.04825 - 1 = 0.04825

PVg 50 = 1761.

= 1761.50(20.725307) = $36 507.63

25) P MT = 515, i = 0.0985 ÷ 2 = 0.04925, n = 6.5 × 4 = 26, c =

= 0.5

p = (1 + 0.04925)0.5 - 1 = 0.024329

26)

PVg = 515.00 = 515.00(19.108482) = $9837.45 F V = 12 500.00; n = 10(4) = 40; i = = 0.5% = 0.005; I/Y = 6; P/Y = 4; C/Y = 12; c =

=3

p = (1 + i)c - 1 = (1 + 0.005)3 - 1 = 0.015075125 FV = PMT 12500 = PMT 12500 = PMT[54.35422486] PMT = $229.97 Programmed solution:

27) P V = 120 000.00; n = 25(12) = 300; i =

= 4.25% = 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

p = (1 + i)c - 1 = (1 + 0.0425)1/6 - 1 = 0.006961062 120000 = PMT 120000 = PMT[125.7286733] PMT =

= $954.44

Programmed solution:

28) F V g = 27000, i = 0.0684 ÷ 26 = 0.002631, n = 5 × 4 = 20, c = 26/4 = 6.5 p = (1 + 0.002631)6.5 - 1 = 0.0171977 27000.00 = PMT 27000.00 = PMT(23.637339) 29)

$1142.26 = PMT FV g = 100 000, i = 0.06, n = 10 × 12 = 120, c = p = (1 + 0.06)1/12 - 1 = 0.0048675506

=

100 000 = PMT 100 000.00 = PMT(162.4734413)

$615.49 = PMT 30) P V g = 91 450, i = 0.0924, n = 11 × 2 = 22, c = p = (1 + 0.0924)1/2 - 1 = 0.0451794 91450.00 = PMT 91450.00 = PMT(13.7613682) 31)

$6645.41 = PMT PV g = 40 000, i = 0.05, n = 5 × 4 = 20, c = p = (1 + 0.05)1/4 - 1 = 0.0122722344 40000.00 = PMT 40000.00 = PMT(17.63931702)

$2267.66 = PMT 32) T he mortgaged amount is 19700(1-0.1) = 17730 = PV i = 0.0822 ÷ 2 = 0.0411, n = 4.5 × 12 = 54, c =

g

=

p = (1 + 0.0411)1/6 - 1 = 0.0067356 17730.00 = PMT 17730.00 = PMT(45.1433762) $392.75 = PMT 33) B alance in 3 years: i = 0.0812 ÷ 2 = 0.0406, N = 3 × 2 = 6 FV = 15810.00(1.0406)6 = 15810.00(1.2694802) = $20 074.04 For annuity: PV

g = 20074.04, i = 0.0406, n = 6 × 12 = 72, c =

=

p = (1 + 0.0406)1/6 - 1 = 0.006655 20074.04 = PMT

34)

20074.04 = PMT(57.05678) $351.83 = PMT PV

g = 198000, n = 22 .5 × 12 = 270, i = 0.0875 ÷ 2 = 0.04375, c =

p = (1 + 0.04375)1/6 - 1 = 0.0071622 198000.00 = PMT 198000.00 = PMT(119.2931506) 35)

$1659.78 = PMT PV g = 46 950, i = 0.065 ÷ 2 = 0.0325, n = 12 × 4 = 48, c = p = (1 + 0.0325)0.5 - 1 = 0.0161201 46950.00 = PMT

= 0.5

=

4695= PMT(33.24246) 0.00 $1412.35 = PMT 36) B alance in 4 years: i = 0.04 ÷ 2 = 0.02, n = 4 × 2 = 8 FV = 20000(1.02)8 = $23433.19 = PV g For annuity: i = 0.02, n = 10 × 12 = 120, c =

=

p = (1 + 0.02)1/6 - 1 = 0.0033058903 23433.19 = PMT 23433.19 = PMT(98.9230334) 37)

$236.88 = PMT FV g = 213000, i = 0.0804 ÷ 4 = 0.0201, c =

=4

p = (1 + 0.0201)4 - 1 = 0.0828567 213000.00 = PMT 213000.00 = PMT(217.4395489) 38)

$979.58 = PMT FV g = 172 000, PMT = 6100, i = 0.0524 ÷ 4 = 0.0131, c =

=2

p = (1 + 0.0131)2 - 1 = 0.0263716

N=

=

=

= 21.36

Twenty-one deposits of $6100.00 each and a final deposit smaller than $6100.00 are needed.

39) P Vg = 16900, PMT = 550, i = 0.09 ÷ 12 = 0.0075, c = =3 p = (1.0075)3 - 1 = 0.022669172

n=-

= 53.2 ≈ 53 quarters

40) P Vg = 10 000, PMT = 1200, i = 0.12 ÷ 2 = 0.06, c =

=

p = (1.06)1/2 -1 = 0.029563014

n=

= 9.708 ≈ 10 periods.

41) P Vg = 320 000, PMT = 7500, i = 0.0468 ÷ 12 = 0.0039, c =

=3

p = (1 + 0.0039)3 - 1 = 0.01174569

n=

=

=

= 59.56 (withdrawals)

42) P V g = 102 800, PMT = 2510, i = 0.045, c = p = (1 + 0.045)1/12 - 1 = 0.0036748094

n=

= 44.47 months = 3.71 years

43) F V = 100 000.00; PMT = 250.00; i =

= 8% = 0.08; I/Y = 8; P/Y = 12; C/Y = 1; c =

p = (1 + 0.08)1/12 -1 = 0.00643403 100000.00 = 250

2.573612044 = (1.00643403)n - 1 3.573612044 = (1.00643403)n n ln 1.00643403 = ln 3.573612044 n(0.00641342) = 1.273576861 n = 198.58 ≈ 199 months Programmed solution:

44) P V = 54 000.00; PMT = 3000.00; i = !

= 0.0075; I/Y = 9; P/Y = 4; C/Y = 12; c =

=3

p = (1 + 0.075)3 -1 = 0.022669172 54000.00 = 3000 0.408045094 = 1 - (1.022669172)-n 0.591954906 = (1.022669172)-n -n ln 1.022669172 = ln 0.591954906 -n(0.022416045) = 0.524324819 n = 23.39060393 quarters ÷ 4 ≈ 6 years

Programmed solution:

45) F Vg = 9700, PMT = 179.23, i = 0.0618 ÷ 4 = 0.01545, c =

=

p = (1 + 0.01545)1/3 - 1 = 0.0051237

n=

=

= 47.88968 ≈ 48 months

=

46) P V g = 360 000, PMT = 2564.19, i = 0.075 ÷ 2 = 0.0375, c =

=

p = 0.0061545 (1 + 0.037 5)1/6 n =

=

=

-1= 47) F V = 20 000.00; PMT = 300; n = 5(12) = 60; P/Y = 12; C/Y = 2; c = Programmed solution:

The rate is 4.26% p.a. compounded semi-annually.

48) P V = 43 000.00; PMT = 4000.00; n = 7(2) = 14; P/Y = 2; C/Y = 4 Programmed solution:

The rate is 7.40% p.a. compounded quarterly.

49) N per 23 Pmt -6540.00 PV 67000.00 Result 0.0815144 The nominal annual rate is 8.15144% × 2 = 16.30288% 50) N per 24 Pmt -7327.00 FV 289000.00 Result 0.0407147 p = (1 + 0.0407147)2 - 1 = 0.0830871 2(0.0830871) = 16.61742% per year comp. semi-annually 51) N per 14 Pmt -2712.00 PV 31500.00 Result 0.0259414 f = (1 + 0.0259414).5 - 1 = .0128877 The nominal annual rate is 1.28877 * 4 = 5.15506% 52) N per 228 Pmt -2486.44 PV 345000.00 Result 0.0047739 1.0047739 = (1 + i)1/6 1 + i = 1.00477396 i = 0.0289874 The nominal annual interest rate is 0.0289874 × 2 × 100% = 5.79748% 53) N per 24 Pmt -1750.00 PV 15000.00 * (1 - 0.10) Result 13500 Effective annual rate f = 1.1213292 - 1 = 25.5737873% 54) N per 240

= 311.5625381 ≈ 312 months =

Pmt 160000.00 -1361Result 0.0068559 .00 f = (1 + 0.0068559)12/2 - 1 = 0.0418469 PV The interest rate is 4.18469% × 2 = 8.36938% 55) P MT = 215, k = 0.005, i = 0.024 ÷ 2 = 0.012, n = 18 × 2 = 36 FV = 215

= $10433.61

56) T he amount needed to be in the account five years from now: PMT = 100, i = 0.032 ÷ 4 = 0.008, k = 0.03, n = 5 × 4 = 20 PV = 100

= $2454.76

The amount to be invested today: PV = 2454.76(1 + 0.008)-20 = $2093.14 57) A 58) C 59) B 60) E 61) E 62) D 63) B 64) A 65) D 66) E 67) D 68) E 69) E 70) D 71) A 72) C 73) C 74) B 75) P MT = $1000;

n = 1 × 25 = 25; i =

c=

= 4%

=2

p = (1 + i)c - 1 = (1+ 0.04)2 - 1 = 0.0816 per year FV = PMT 76) P MT = $185;

= $1000 n = 12 × 13 = 156;

c=

= $74 836.81 i=

= 1.375%

= 0.3333

p = (1 + i)c - 1 = (1+ 0.01375)0.3333 - 1 = 0.0045625 FV = PMT

= $185

77) A 78) P MT = $7000;

n = 12 × 22 = 264;

= $41 937

i=

= 4%

c=

p = (1 + i)c - 1 = (1+ 0.04)0.1667 - 1 = 0.0065582

PV = PMT = $7000 = $877 325.83 = 0.166 7 79) E 80) C 81) D 82) A 83) $ 37000 = (PV of the loan payment annuity) + (PV of the $16500 balance) For PV annuity n = 36; i = 3.6% c= p = 1.0362/12 - 1 = 0.0059 PV annuity = PMT

= 32.34PMT

For PV balance n = 6; i = 3.6% V balance = FV(1 + i)

-n

-6 = $16500 (1+ 0.036) = $13 345.21

32.34PMT = $37000 - $13,345.21 = $23 654.79 ⇒ PMT = $731.44 84) F V = $100 000; PMT = $250;

i = 8%

c= p = 1.081/12 -1 = 0.006434

n=

=

= 198.58

⇒ n = 199 months = 16 years 7 months 85) D 86) T he first monthly payment is due on the same date as the full annual fee. In effect, Lahore Gym lends $240 - $22 = $218 to a member choosing the monthly payment option. The member then repays this loan in 11 month-end payments of $22.

PV = $218; PMT = $22; P/Y = 12; C/Y = 1; n = 11 Using above values in CalcPro, i = 23.62% compounded annually 87) D 88) P MT = $14 400; k = -2.5%; i = 12.2%; n = 15 years FV = PMT FV = $483 723 89) D 90) D

= $14400

= $14400 × 33.59

91) B 92) B 93) D 94) B 95) C 96) B 97) A 98) A 99) D 100) E 101) E 102) C 103) E 104) B 105) c = 12/4 = 3; i = 0.03/12 = 0.0025; p = 1.0025^3 - 1 = 0.751877% 106) c = 2/4 = 0.5; i = 0.012/2 = 0.006 p = 1.006^0.5 - 1 = 0.002996 n = 6 * 2 = 12 FV = 80

= 80(12.199691) = $975.98

107) c = 6/12 = 0.5; i = 0.038/2 = 0.019 = 1.9% p = 1.019^0.5 — 1 = 0.009455 n = 13 * 2 = 26 PV = 3200

= 3200

= 3200(22.955082) = $73.456.26

108) c = 4/26 = 2/13; i = 0.044/4 = 0.011 = 1.1% p = 1.011^(2/13) — 1 = 0.001684 n = 1.5 * 26 = 39 PV = 460

= 460

= 460(37.715822) = $17349.28

109) c = 4/12 = 1/3; i = 0.046/4 = 0.0115 = 1.15% p = 1.0115^(1/3) — 1 = 0.003819 n = 12 * 5.5 = 66 PMT =

=

= $6675.51

110) c = 2/1 = 2 i = 0.082/2 = 0.041 = 4.1% p = 1.041^2 — 1 = 0.083681 n=6 PMT = =

= $3565.41

111) c = 4/1 = 4; i = 0.045/4 = 0.01125 = 1.125% p = 1.01125^4 — 1 = 0.045765

n=

=

112) c = 2/12 = 1/6; i = 0.034/2 = 0.017 = 1.7% p = 1.017^(1/6) - 1 = 0.002813…

= 6.011617 years

n=

=

= 103.53700 payments

Therefore it would take 103 monthly payments of $100 and a smaller final payment to result in a $12 000 balance in the RRSP. 113) a ) 13th payment = 3000(1.03)^12 = 4227.28 b) i = 0.068/2 = 0.034 = 3.4%; n = 8 * 2 = 16 compounding factor = =

=

= 25.666847

114) k = -2% = -0.02, i = 3% = 0.03, n = 18 FV = 4500

= 4500

= 4500 (20.145955) = $90 656.80

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) H ector makes payments of $2000.00 into an RRSP at the beginning of every six months 1) _____________ starting today for 33 years. If the interest rate is 8% compounded semi-annually, calculate the accumulated value.

2) A contract valued at 60 000.00 requires payments of $3000.00 at the beginning of each quarter. If interest is 6% compounded quarterly, calculate the term of the contract.

2) _____________

3) C alculate the present value of payments of $100.00 made at the beginning of each quarter for ten years if the interest is 6% compounded quarterly.

3) _____________

4) A lease agreement valued at $15 000.00 requires payment of $900.00 at the beginning of

4) _____________

every quarter for five years. What is the nominal rate of interest charged (compounded quarterly)?

5) W hat semi-annual payment must be made into a fund at the beginning of every six months to accumulate to $4800.00 in ten years at 7% compounded semi-annually?

5) _____________

6) A shlyn just graduated and plans to buy a house in 15 years worth $1 million. How much does she have to save at the beginning of every three months for 15 years to accumulate $1 million dollars if interest is 4% compounded quarterly?

6) _____________

7) P ayments of $500.00 are made at the beginning of each month for four years. The interest rate is 4.5% compounded monthly. If no further deposits are made a) Calculate the accumulated value twelve years after the first deposit. b) Calculate the amount deposited. c) Calculate the interest.

7) _____________

8) B ala bought mutual funds within TFSA as part of his retirement planning in 2019.

Payments of $500.00 will be made into the fund at the beginning of every month for 20 years. If the fund is expected to earn interest at 8.4% compounded monthly, what will be the balance in the fund after 20 years?

8) _____________

9) D etermine the accumulated value after 13.5 years of deposits of $470.00 made at the beginning of every three months and earning interest at 7.2% compounded quarterly.

9) _____________

10) P ayments of $1000.00 are made into a fund at the beginning of every month for 10 years.

If the fund earns interest at 12% compounded monthly, what will the balance in the fund at the end of the term?

10) _____________

11) D etermine the accumulated value after 15 years of deposits of $9000.00 made at the beginning of every three months and earning interest at 8% compounded quarterly.

11) _____________

12) W hat is the discounted value of deposits of $570.00 made at the beginning of every three months for 8.25 years if money is worth 8.4% compounded quarterly?

12) _____________

13) W hat is the discounted value of deposits of $250.00 made at the beginning of every three months for 10.5 years if money is worth 6% compounded quarterly?

13) _____________

14) L inda contributes $3000.00 at the beginning of every six months into an RRSP paying interest at 6% compounded semi-annually. a) How much will her RRSP deposits amount to in 15 years? b) How much of the amount will be interest?

14) _____________

15) S teve deposited $600.00 in a trust account on the day of his son's birth and every three months thereafter in order to plan and pay for his son's higher education. If interest paid is 7.2% compounded quarterly, what will the balance in the trust account be before the

15) _____________

deposit is made on the son's 18th birthday?

16) T he monthly rent payment on a storage warehouse is $3971.00 payable in advance. What yearly payment in advance would satisfy the lease if interest is 6.84% compounded monthly?

16) _____________

17) J udy paid for a living room suite over time, signing an installment contract that requires semi-annual payments of $2560.00 for 8.5 years. The first payment is made on the signing date and interest is 14.5% compounded semi-annually. What was the cash price?

17) _____________

18) J essica bought a bedroom suite on credit, signing an installment contract with a finance company that requires monthly payments of $100.00 for 5 years. The first payment is made on the date of signing and interest is 24% compounded monthly. a) What was the cash price? b) How much will Tricia pay in total? c) How much of what she pays will be interest?

18) _____________

19) R oxanne contributes $167.00 at the beginning of every six months into an RRSP paying interest at 8.9% compounded semi-annually. a) How much will her RRSP deposits amount to in 31.5 years? b) How much of the amount will be interest?

19) _____________

20) T he monthly rent payment on computer equipment is $30.00 payable in advance. What yearly payment in advance would satisfy the lease if interest is 12% compounded monthly?

20) _____________

21) V icki paid for a kitchen set over time, signing an installment contract that requires semi-annual payments of $500.00 for 5 years. The first payment is made on the signing date and interest is 18% compounded semi-annually. What was the cash price?

21) _____________

22) G ibran bought a boat valued at $110 194.00 on the installment plan. He made equal semi-annual payments for 6 years. If the first payment is due on the date of purchase and interest is 7.5% compounded semi-annually, what is the size of the semi-annual payments?

22) _____________

23) D etermine the future value and the present value of monthly payments of $200.00 each for five years at 12% compounded monthly if a) the payments form an annuity due; b) the payments form an ordinary annuity.

23) _____________

24) T ricia bought a living-room suite on credit, signing an installment contract with a finance company that requires monthly payments of $167.25 for 4.5 years. The first payment is made on the date of signing and interest is 28.8% compounded monthly.

a) was the cash Whprice? at b) How much

will Tricia 24) pay in total? c) How much of what she pays will be interest?

___ ___ ___ ___ _

25) D etermine the future value and the present value of monthly payments of $173.00 each for six years at 6.48% compounded monthly if a) the payments form an annuity due; b) the payments form an ordinary annuity.

25) _ ____________

26) J ack Chen makes deposits of $415.00 at the beginning of every three months. Interest earned by the deposits is 4.12% compounded quarterly. a) What will the balance in Jack's account be after eight years? b) How much of the balance will Jack have contributed? c) How much of the balance is interest?

26) _ ____________

27) S ean Paul makes deposits of $500.00 at the beginning of every three months. Interest earned by the deposits is 8% compounded quarterly. a) What will the balance in Sean's account be after ten years? b) How much of the balance will Sean have contributed? c) How much of the balance is interest?

27) _ ____________

28) H ow many months will it take to accumulate $17 897.20 at 5.87% compounded semi-annually, if $1140.00 is deposited in an account at the beginning of every 6 months?

28) _ ____________

29) H ow much does a depositor have to save at the beginning of every 6 months for twenty years to accumulate $1 000 000.00 if interest is 6.00% compounded semi-annually?

29) _ ____________

30) A sum of money is deposited at the beginning of each year for 3 years at 12% compounded annually. After the last deposit interest for the account is to be 8.24% compounded quarterly and the account is to be paid out by payments of $370.00, made at the beginning of each quarter for nine years. What is the size of the annual deposit?

30) _ ____________

31) K elly is saving $1600.00 at the beginning of each month. How soon can he retire if he wants to have a retirement fund of $345 000.00 and interest is 11.04% compounded monthly?

31) _ ____________

32) H ow much does a depositor have to save at the beginning of every three months for 13.25 years to accumulate $41 600.00 if interest is 8% compounded quarterly?

32) _ ____________

33) A contract valued at $81 500.00 requires payment of $3350.00 at the beginning of every six months. If interest is 6.5% compounded semi-annually, what is the term of the contract?

33) _ ____________

34) R obert Bernard deposited $1000.00 in a trust account on the day of his son's birth and every three months thereafter. If interest paid is 8% compounded quarterly, what will

the ance in the trust balaccount be

before 34) the deposit is made on the son's twenty-fi rst birthday ?

___ ___ ___ ___ _

35) H ow much does a depositor have to save at the beginning of every 6 months for eleven years to accumulate $17 448.00 if interest is 6.25% compounded semi-annually?

35) _ ____________

36) E qual sums of money are withdrawn monthly from a fund of $33 000.00 for 15.25 years.

If interest is 9% compounded monthly, what is the size of each withdrawal if the withdrawal is made at the beginning of each month?

36) _ ____________

37) E qual sums of money are withdrawn monthly from a fund of $60 000.00 for 15 years. If interest is 12% compounded monthly, what is the size of each withdrawal if the withdrawal is made at the beginning of each month?

37) _ ____________

38) W hat is the accumulated value after 15 years of monthly deposits of $500.00 earning interest at 8% compounded semi-annually if the deposits are made a) at the end of each month? b) at the beginning of each month?

38) _ ____________

39) A lease requires payments of $1200.00 at the beginning of every quarter for ten years. If interest is 6% compounded monthly, calculate the value of the lease.

39) _ ____________

40) W hat deposit made at the beginning of each month will accumulate to $36 000.00 at 5% compounded quarterly at the end of eight years?

40) _ ____________

41) A vacation property was bought by making semi-annual payments of $7500.00 for seven years. If the first payment is due on the date of purchase and interest is 6% p.a. compounded quarterly, calculate the purchase price of the property.

41) _ ____________

42) E llora wants to accumulate $150 000.00 in an RRSP by making annual contributions of $5000.00 at the beginning of each year. If interest is 5.5% compounded quarterly, calculate how long she has to make contributions.

42) _ ____________

43) C ompute the nominal annual rate of interest compounded monthly at which $1500.00 deposited at the beginning of every three months for ten years will amount to $90 000.00.

43) _ ____________

44) D etermine the accumulated value after 14.5 years of deposits of $172.00 made at the beginning of every 6 months and earning interest at 4.45% compounded monthly?

44) _ ____________

45) D etermine the accumulated value after 15 years of deposits of $500.00 made at the beginning of every 6 months and earning interest at 12% compounded monthly?

45) _ ____________

46) S ean contributes $545.00 at the beginning of each month into TFSA paying interest at 6.25% compounded semi-annually. What will be the accumulated balance in the TFSA at

the of 10.25 years? end

46)

_____________ 47) A property was purchased for quarterly payments of $1070.00 for 9 years. If the first payment was made on the date of purchase and interest is 5.57% compounded annually, what was the purchase price of the property?

47) _____________

48) A property was purchased for quarterly payments of $3000.00 for 10 years. If the first payment was made on the date of purchase and interest is 6% compounded annually, what was the purchase price of the property?

48) _____________

49) S lick Sean contributes $500.00 at the beginning of each month into RRSP paying interest at 6.00% compounded semi-annually. What will be the accumulated balance in the RRSP at the end of 15 years?

49) _____________

50) A ubrey made quarterly deposits of $750.00 at the beginning of every 3 months for 11.25 years in a non-registered savings account. If the interest is prime+0.5 compounded monthly, what amount can be withdrawn 6 years after the last deposit? Assume a fixed prime rate of 2.25%.

50) _____________

51) W hat is the accumulated value after 14 years of monthly deposits of $415.00 earning interest at 5.8% compounded semi-annually if the deposits are made a) at the end of each month? b) at the beginning of each month?

51) _____________

52) Q uarterly deposits of $2000.00 were made at the beginning of every 3 months for 10 years. If the interest is 6% compounded monthly, what amount can be withdrawn 7 years after the last deposit?

52) _____________

53) G ary and Sandra Duncan bought a property by making semi-annual payments of $4210.00 for 8.5 years. If the first payment is due on the date of purchase and interest is 8.53% compounded quarterly, what is the purchase price of the property?

53) _____________

54) B ow Valley Credit Union entered a lease contract valued at $3100.00. The contract provides for payments at the beginning of each month for three years. If interest is 8.8% compounded quarterly, what is the size of the monthly payment?

54) _____________

55) A lease requires monthly payments of $1150.00 due in advance. If interest is 16% compounded quarterly and the term of the lease is 4 years, what is the cash value of the lease?

55) _____________

56) A t the beginning of every 6 months, $1280.00 is deposited for 8 years at 6.6% compounded monthly. After 8 years interest for the account is to be 9.66% compounded quarterly and the account is to be paid out by equal end-of-quarter quarterly payments over 6 years. What is the size of the quarterly payment?

56) _____________

57) A my contributed $1011.00 at the end of every six months for 21.5 years into an RRSP earning interest at 6.65% compounded semi-annually. Seven years after the last contribution, Amy converted the RRSP into an RRIF that is to pay her equal quarterly amounts for 17.25 years. If the first payment is due three months after the conversion into the RRIF and interest on the RRIF is 7.66% compounded quarterly, how much will Amy receive every three months?

57) _____________

58) S tarting three months after her grandchild Crystal's birth, Mrs. Robinson made deposits of $63.25 into a trust fund every three months until Crystal was 19 years old. The trust fund provides for equal withdrawals at the end of each quarter for 5 years beginning three months after the last deposit. If interest is 8.24% compounded quarterly, how much will Crystal receive every three months?

58) _____________

59) K aren Peters bought their neighbor's farm for $450 000.00 down and payments of $67100.00 at the end of every six months for 7 years. What is the purchase price of the farm if the semi-annual payments are deferred for four years and interest is 5.5% compounded semi-annually?

59) _____________

60) M ike acquired his sister's share of their business by agreeing to make payments of $4711.00 at the end of each year for 8 years. If the payments are deferred for three years and money is worth 5.96% compounded quarterly, what is the cash value of the sister's share of the business?

60) _____________

61) T he Saskatchewan Junior Achievers need to borrow to finance a project. Repayment of the loan involves payments of $8711.00 at the end of every three months for six years.

No payments are to be made during the development period of three years. Interest is 9.24% compounded quarterly. a) How much did the Achievers borrow? b) What amount will be repaid? c) How much of that amount will be interest?

61) _____________

62) E qual payments are to be made at the end of each month for 12 years with interest at 9.66% compounded monthly. After the last payment, the fund is to be invested for 10 years at 6.82% compounded quarterly and have a maturity value of $27 600.00. What is the size of the monthly payment?

62) _____________

63) F lin Flon Aqua Club bought swimming equipment on a contract requiring monthly payments of $815.50 for 4 years beginning 19 months after the date of purchase. What was the cash value of the equipment if interest is 7.8% compounded monthly?

63) _____________

64) A sum of $15 700.00 is invested at 6.48% compounded semi-annually for eight years.

After the eight years, the balance in the fund is converted into an annuity paying equal payments at the end of every 6 months for 5.5 years. If interest on the annuity is 7.95% compounded monthly, what is the size of the equal payments?

64) _____________

65) O akville Aqua Club bought swimming equipment on a contract requiring monthly payments of $1000 for 5 years beginning 6 months after the date of purchase. What was the cash value of the equipment if interest is 12% compounded monthly?

65) _____________

66) A sum of money is deposited at the end of every month for 9 years at 8.5% compounded monthly. After the last deposit, interest for the account is to be 8.1% compounded quarterly and the account is to be paid out by end-of-quarter quarterly payments of $1800.00 over 7 years. What is the size of the monthly deposit?

66) _____________

67) T he local high school soccer club need to borrow to finance a new soccer field.

Repayment of the loan involves payments of $10 000.00 at the end of every three months for eight years. No payments are to be made during the development period of five years. Interest is 8% compounded quarterly. a) How much did the Achievers borrow?

b) nt will be Whrepaid? at c) How much of amthat amount ouwill be interest?

67)

_____________ 68) P aul received an inheritance payment of $38 000.00 at age 50. He invested that sum of money at 8.5% compounded semi-annually until he was 65. At that time he converted the existing balance into an ordinary annuity paying $9150.00 every 6 months, with interest at 5% compounded semi-annually. How many payments will be made?

68) _____________

69) A sum of money is deposited at the end of every 6 months for 9.5 years at 8% compounded semi-annually. After the last deposit, interest for the account is to be 4.5% compounded quarterly and the account is to be paid out by making payments of $4360.00 at the end of each month over 7.75 years. What is the size of the semi-annual deposit?

69) _____________

70) T he amount of $75 400.00 is invested at 9.65% compounded quarterly for 3 years. After 3 years the balance in the fund is converted into an annuity. If interest on the annuity is 6.75% compounded semi-annually and payments are made at the end of every six months for 8 years, what is the size of the payments?

70) _____________

71) M onica would like to receive $3760.00 at the end of every six months for 8 years after her retirement. If she retires ten years from now and interest is 6.5% compounded semi-annually, how much must she deposit into an account every six months starting now?

71) _____________

72) J ack took a loan of $140 000.00 to rebuild his cottage. The loan is to be repaid in installments due at the end of each month for 10 years. If the payments are deferred for a year and interest is 7.28% compounded quarterly, what is the size of the monthly payments?

72) _____________

73) M ark contributed $270.00 at the end of every 3 months for 5.75 years into an RRSP earning 12.6% interest compounded quarterly. Four years after the last contribution he converted the RRSP into a RRIF that is to pay him quarterly payments for 29.5 years. If the first payment is due three months after the conversion into the RRIF and interest on the RRIF is 9.5% compounded quarterly, how much will Mark receive every 3 months?

73) _____________

74) $ 6420.00 was invested at a fixed rate of 6.5% compounded semi-annually for 5 years.

After 5 years, the fund was converted into an ordinary annuity paying $550.00 per month. If interest on the annuity was 6% compounded monthly, what was the term of the annuity?

74) _____________

75) M r. Caldwell received a retirement gratuity of $481 700.00, which he deposited in an RRSP. He intends to leave the money for 6 years, then transfer the balance into an RRIF and make equal withdrawals at the end of every six months for 25 years. If interest is 8.5% compounded semi-annually, what will be the size of each withdrawal?

75) _____________

76) M rs. Lavergne deposits $562.71 at the beginning of every three months. Starting three months after the last deposit, she intends to withdraw $587.00 every three months for 16 years. If interest is 8.14% compounded quarterly, for how long must Mrs. Lavergne make deposits?

76) _____________

77) T J invested a retirement gratuity of $43 000.00 in an income averaging annuity paying 7% compounded monthly. He withdraws the money in equal monthly amounts over five years. If the first withdrawal is made nine months after the deposit, what is the size

of withdrawal? eac h

77)

_____________ 78) A nne-Marie received $445 000.00 from her mother's estate. She wants to set aside part of her inheritance for her retirement 12 years from now. At that time she would like to receive a pension supplement of $2600.00 at the end of each month for 15 years. If the first payment is due one month after her retirement and interest is 3.6% compounded monthly, how much must Anne-Marie set aside?

78) _____________

79) M rs. Martin contributes $1830.00 at the beginning of every three months to an RRSP. Interest on the account is 7.44% compounded quarterly. a) What will the balance in the account be after 6 years? b) How much of the balance will be interest? c) If Mrs. Martin converts the balance after 6 years into an RRIF paying 5.41% compounded quarterly and makes equal quarterly withdrawals for 11 years starting three months after the conversion into the RRIF, what is the size of the quarterly withdrawal? d) What is the combined interest earned by the RRSP and the RRIF?

79) _____________

80) B everley intends to retire in 7.25 years and would like to receive $2700.00 every month for 12.25 years starting on the date of her retirement. How much must she deposit in an account today if interest is 11.36% compounded annually?

80) _____________

81) J ennifer intends to retire in 15 years and would like to receive $5000.00 every month for 20 years starting on the date of her retirement. How much must she deposit in an account today if interest is 9% compounded annually?

81) _____________

82) S heila intends to retire in 14 years and would like to receive $2472.00 every six months for 9 years starting on the date of her retirement. How much must Sheila deposit in an account today if interest is 6.65% compounded semi-annually?

82) _____________

83) S andra intends to retire in 15 years and would like to receive $3000.00 every six months for 19 years starting on the date of her retirement. How much must Sandra deposit in an account today if interest is 8% compounded semi-annually?

83) _____________

84) S undeep wants to withdraw $1000.00 at the beginning of each quarter for twelve years.

If the withdrawals are deferred to begin ten years from now and interest is 4.5% compounded monthly calculate the amount that must be invested today to be able to make the withdrawals.

84) _____________

85) T he sale of property provides for payments of $4500.00 due at the beginning of every three months for five years. If the payments are deferred for two years and interest is 6% compounded monthly, what is the cash value of the property?

85) _____________

86) T he sale of property provides for payments of $10 000.00 due at the beginning of every three months for ten years. If the payments are deferred for five years and interest is 12% compounded monthly, what is the cash value of the property?

86) _____________

87) E rica intends to retire in 37.5 years and would like to receive $6600.00 every 6 months for 14 years, starting on the date of her retirement. How much must she deposit in an account today if interest is 6.25% compounded semi-annually?

87) _____________

88) P aul intends to retire in 15 years and would like to receive $8000.00 every 6 months for

20 years, starting

on the 88) date of his retireme nt. How much must he deposit in an account today if interest is 5% compou nded semi-ann ually?

___ ___ ___ ___ _

89) A n investment in a lease offers returns of $3713 per month due at the beginning of each month for 4.5 years. What investment is justified if the returns are deferred for two years and the interest required is 6.12% compounded monthly?

89) _____________

90) A n investment in a lease offers returns of $4000 per month due at the beginning of each month for 5 years. What investment is justified if the returns are deferred for three years and the interest required is 12% compounded monthly?

90) _____________

91) W hat is the principal invested at 6.5% compounded semi-annually from which monthly withdrawals of $840.00 can be made at the beginning of each month for 15.5 years but deferred for 2 years?

91) _____________

92) W hat is the principal invested at 8% compounded semi-annually from which monthly withdrawals of $1000.00 can be made at the beginning of each month for 10 years but deferred for 3 years?

92) _____________

93) A lease requires semi-annual payments of $6100.00 for 7.5 years. If the first payment is due in 3 years and interest is 9.72% compounded monthly, what is the cash value of the lease?

93) _____________

94) P ayments on a six-year lease valued at $12 975.00 are to be made at the beginning of each month during the last 3 years of the lease. If interest is 9% compounded monthly, what is the size of the monthly payments?

94) _____________

95) J oshua deposited $3560.00 at the beginning of every 6 months for 11 years into a fund

95) _____________

paying 3.5% compounded semi-annually. Fourteen years after the first deposit he converted the then existing balance into an annuity due paying him equal annual payments for 22 years. If interest for the annuity due is 5% compounded annually, what is the size of the annual payment?

96) M r. Dawson wants to receive payments of $1230.00 at the beginning of every three months for 19 years starting on the date of his retirement. If he retires in 21 years, how much must he deposit in an account at the beginning of every three months if interest on the account is 6.84% compounded quarterly?

96) _____________

97) A n annuity provides payments of $4580.00 at the end of every three months. The annuity is bought for $53 500.00 and payments are deferred for 10 years. If interest is 9.72% compounded monthly, for how long will payments be received?

97) _____________

98) I f you save $150.00 at the beginning of each quarter for 20 years, for how long can you withdraw $800.00 at the beginning of each quarter, starting 20 years from now, and assuming that interest is 9% compounded quarterly?

98) _____________

99) F raser deposits $790.00 at the beginning of every three months. He wants to build up his account so that he can withdraw $1550.00 every three months starting three months after the last deposit. If he wants to make the withdrawals for 16 years and interest is 6.64% compounded quarterly, for how long must Fraser make the quarterly deposits?

99) _____________

100) M rs. Muirhead paid $27 700.00 into a retirement fund paying interest at 9.32% compounded semi-annually. If she retires in 13.5 years, for how long can Mrs. Muirhead withdraw $5000.00 from the fund every six months? Assume that the first withdrawal is on the date of retirement.

100) ____________

101) T he sum of $59 760.00 is invested at 9.5% compounded monthly for four years. After the

101) ____________

four years, the balance in the fund is converted into an annuity due paying annually $15 240.00. If interest on the annuity is 9.57% compounded annually, what is the term of the annuity?

102) A lease valued at $47 600.00 requires payments of $4500.00 every three months. If the first payment is due 2 years after the lease was signed and interest is 6.88% compounded quarterly, what is the term of the lease?

102) ____________

103) A retirement gratuity of $123 900.00 is invested in annuity deferred for 10 years. The annuity provides payments of $9600.00 due at the beginning of every six months. If interest is 7.41% compounded annually, for how long will annuity payments be made?

103) ____________

104) A debt of $23 900.00 is repaid by making payments of $3617.00 at the beginning of every six months deferred for three years. If interest is 9.6% compounded monthly, for how long will payments have to be made?

104) ____________

105) C arla Craig invested a retirement gratuity of $17 570.00 in an RRSP paying 6.5% compounded semi-annually for 11.5 years. At the end of 11.5 years, she rolled the RRSP balance over into an RRIF paying $1500.00 at the beginning of each month starting with the date of rollover. If interest on the RRIF is 9.5% compounded monthly, for how long will Carla receive monthly payments?

105) ____________

106) N ote: The calculations for this question were done using Excel's RATE function.

A vacation property valued at $62 000.00 was bought for 40 payments of $2150.00 due at the beginning of every six months. What nominal rate of interest was charged?

106) ____________

107) Y ou bought a camper priced at $15 660.00 on February 1. You agreed to make monthly payments of $575.00 beginning December 1 of the same year. For how long will you have to make these payments if interest is 8.5% compounded monthly?

107) ____________

108) S haun's maternal grandmother Abbey gifted a handsome amount of money to Shaun on his birthday. The money is held in a trust fund providing a scholarship of $1000.00,

payaat the ble beginning of

each 3 108) months, if interest is 8.4% compou nded quarterly ?

____ ____ ____

109) A rental property provides a monthly income of $1472.00 due at the beginning of every month. What is the cash value of the property if money is worth 7.32% compounded monthly?

109) ____________

110) T riad is considering a medical process that is expected to reduce annual costs by $43 000.00. What is the maximum amount of money that could be invested in the process to be economically feasible if interest is 7.44% compounded quarterly?

110) ____________

111) E nermax offers to acquire a right-of-way from a property owner who receives annual lease payments of $3725.00 due in advance. What is a fair offer if money is worth 6.5% compounded quarterly?

111) ____________

112) A aron Perry bought an income property for $41 321.00 three years ago. He has held the property for the three years without renting it. If he rents the property now, what should be the size of the monthly payment due in advance be if money is worth 6.24% compounded monthly?

112) ____________

113) W hat monthly lease payment due in advance should be charged for a tract of land

113) ____________

valued at $135 000 if the agreed interest is 8.15% compounded semi-annually?

114) G eomax pays $8710.00 at the beginning of each year for using a tract of land. What should the company offer the property owner as a purchase price if interest is 10.18% compounded annually?

114) ____________

115) A fund to provide an annual scholarship of $5700.00 is to be set up. If the first payment is due in three years and interest is 6.4% compounded quarterly, what sum of money must be deposited in the scholarship fund today?

115) ____________

116) W hat sum of money must be deposited in a trust fund to provide a scholarship of $630.00 payable at the end of each month if interest is 7.8% compounded monthly?

116) ____________

117) C alculate the amount of money that needs to be invested today at 4.8% compounded semi-annually to provide monthly payments of $500.00 in perpetuity starting: a) one month from today b) one year from today

117) ____________

118) C alculate the amount of money that needs to be invested today at 7% compounded semi-annually to provide monthly payments of $2000.00 in perpetuity starting: a) one month from today b) one year from today

118) ____________

119) C arla plans to invest in a property that after 6.25 years will yield $1457.00 at the end of each month indefinitely. How much should Carla be willing to pay if an alternative

invent yields stme12.96%

compou nded monthly ?

119)

____ ____ ____

120) N ote: The calculations for this question were done using Excel's RATE function.

What is the effective rate of interest charged on a lease valued at $17 040.00 if payments of $1180.00 are made at the beginning of each 3 months for four years?

120) ____________

121) N ote: The calculations for this question were done using Excel's RATE function.

At what nominal rate of interest compounded semi-annually will $2200.00 deposited at the beginning of every six months accumulate to $68 000.00 in 11 years?

121) ____________

122) N ote: The calculations for this question were done using Excel's RATE function.

An insurance policy provides for a lump sum benefit of $450 000.00 fifteen years from now. Alternatively, payments of $25700.00 may be received at the beginning of each of the next fifteen years. What is the effective rate of interest if interest is compounded quarterly?

122) ____________

123) N ote: The calculations for this question were done using Excel's RATE function.

What is the nominal annual rate of interest charged quarterly on a lease valued at $2340.00 if payments of $240.00 are made at the beginning of every three months for 3.25 years?

123) ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

124) Y ou make six monthly deposits of $125 starting now. The interest rate is 6.36% compounded 124) _____ monthly. How much interest has been earned in total?

A) $ 12.10 B) $ 13.10 C) $ 11.01 D) $ 9.01 E) $ 14.04 125) Y ou start to save for a major purchase. You can invest $329 every three months starting today for 5 years and 9 months. You are able to earn 5.87% compounded semi-annually. What is the amount of interest that you earn during the entire term?

A) $ 2229.49 B) $ 2029.49 C) $ 1475.72 D) $ 2092.49 E) $ 2292.49

125) _____

126) Y ou start to save for a major purchase. You can invest $500 every three months for 10 years. You are able to earn 8% compounded semi-annually. What is the amount of interest that you earn during the entire term?

A) $ 15 072.94 B) $ 30 072.94 C) $ 20 072.94 D) $ 10 072.94 E) $ 25 072.94

126) _____

127) S tarting in 5 years you want to be able to withdraw $6220 every six months for 7.5 years. You want to deposit a single amount immediately and then let it grow at a rate of 4.88% compounded quarterly. How much do you have deposit?

A) $ 61 126.62 B) $ 62 216.62 C) $ 62 116.62 D) $ 62 226.62 E) $ 61 216.62

127) _____

128) S tarting in 5 years and 6 months you want to be able to withdraw $950 every three months for 3 years and 3 months. You want to deposit a single amount immediately and then let it grow at a rate of 7.24% compounded quarterly. How much interest did you earn during the entire period of time?

A) $ 4589.47

B) $ 2090.47

C) $ 4888.47

D) $ 4859.47

128) _____

E) $ 2099.47

129) H ow much extra interest do you gain by starting today rather than waiting one year if you want to deposit $1710 per year for 28 years? You are able to earn 7.07% compounded annually.

A) $ 9896.63 B) $ 9889.63 C) $ 9999.63 D) $ 9689.63 E) $ 9869.63

129) _____

130) Y our child is currently six years old. When he turns 19 you want him to be able to withdraw $14700 per year in order to either travel or go to school for four years. If you are able to earn 6.62% compounded semi-annually for the entire time period, what is the single amount of the deposit three years from now that will ensure that your child is able to do this?

A) $ 28 877.49 B) $ 27 777.49 C) $ 28 777.49 D) $ 29 877.49 E) $ 27 877.49

130) _____

131) A trust fund will start to pay you $15 200 per year at the beginning of each year once you turn 18. The prevailing interest rate is 8.79% compounded annually. For how long can you receive payments if you are celebrating your 14th birthday today and the current value of your trust fund is $111 345?

A) 2 4.97 years B) 2 9.77 years C) 2 0.07 years D) 2 9.07 years E) 2 0.97 years

131) _____

132) I f you want to be able to set up a scholarship fund that will last forever and you feel that you can

132) _____

earn 4.8% compounded annually. How much money do you have to deposit today in order to be able to provide 6 scholarships of $3000 each?

A) $ 735 000.00 B) $ 375 000.00 C) $ 357 000.00 D) $ 333 000.00 E) $ 37 500.00 133) J awad makes semi-annual contributions of $2000 starting on his 27th birthday. How much will he accumulate in his RRSP by age 60, if RRSP earns 8% compounded semi-annually and no contribution is made on his sixtieth birthday?

A) $ 669 842 B) $ 587 937 C) $ 640 156 D) $ 615 535 E) $ 565 324

133) _____

134) J awad makes semi-annual contributions of $2000 starting on his 27th birthday. How much will he accumulate in his RRSP by age 60, if RRSP earns 8% compounded annually and no contribution is made on his sixtieth birthday?

A) $ 565 324 B) $ 595 254 C) $ 640 156 D) $ 615 535 E) $ 618 606

134) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

135) J enna intends to contribute $2500 to her RRSP at the beginning of every six months, contions for 20 starting today. The financial adviser has advised Jenna that if she continues her

ributyears, the

RRSP 135) would earn 8% compou nded semi-ann ually for the first 7 years and 7% compou nded semi-ann ually, thereafte r. What amount will Jenna have in the plan after 20 years?

____ ____ ____

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

136) H eavy Mechanical Complex (HMC) acquired a giant lathe machine under a capital lease 136) _____ agreement. HMC pays the lessor $2400 at the beginning of every 3 months for 5 years. If HMC can obtain 5-year financing at 10% compounding quarterly, what long-term lease liability will HMC initially record?

A) $ 38 349 B) $ 62 840 C) $ 61 307 D) $ 37 414 E) $ 48 000 137) H eavy Mechanical Complex (HMC) acquired a giant lathe machine under a capital lease agreement. HMC pays the lessor $2400 at the beginning of every 3 months for 5 years. If HMC can obtain 5-year financing at 10% compounding quarterly, what liability will be reported 2 years later?

A) $ 33 937 B) $ 33 109 C) $ 37 414 D) $ 25 234 E) $ 38 349

137) _____

138) D ell Canada advertises a computer bundle including printer, router and home network attached storage for $2995. The same system bundle may be leased for 24 months at $130 per month, at the beginning of each month. At the end of the lease, the bundle may be purchased for 10% of its retail price. What is the economic cost of the lease or purchase in present value, if a two year loan can be obtained at 12% compounded annually to purchase the computer?

A) L ease is better by $109.38 B) B oth options are the same.

C) P urchase is better by $109.38 D) L ease is better by $48.64 E) P urchase is better by $48.64

138) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

139) S ue won a "Millionaire Life" lottery. As a winner of the Grand Prize, she can choose low stments, either $1 million per year for 25 years or a single cash payment of $17 million. What risk money can option should be chosen if the payments are made at the beginning of each year and on inveearn 3.2%

compou nded annually ?

____ ____ ____

139)

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

140) F ord Oshawa sells Ford Fusion special edition for a price of $27 900. During a month long 140) _____ promotion, the manufacturer offers an interest rate of 1.8% compounded monthly on a 3-year lease. If the residual value is $14 500, what will be the lease payment at the start of each month with a $2500 down payment?

A) $ 311 B) $ 350 C) $ 332.50 D) $ 403.82 E) $ 724.30 141) O livia plans to have $1 000 000 in her RRSP at her retirement at the age of 60. How long will it take her to accumulate $1 000 000 in an RRSP if the first quarterly contribution of $2000 is made today and RRSP is expected to earn 8% compounded quarterly?

A) 3 0 years 6 months B) 2 8 years 9 months C) 2 9 years D) 3 0 years and 3 months E) 4 7 years and 3 months

141) _____

142) J eff is retired and has an investment fund worth $210 000 in an RRSP, which earns 9% compounded semi-annually. He plans to withdraw only $2000 at the beginning of each month starting today. When will the fund become depleted?

A) 1 7 years B) 1 6 years 10 months C) 1 6 years and 6 months D) 8 years and 9 months E) 9 years

142) _____

143) C ameron is 35 years old and intends to retire at the age of 60. He is planning to contribute $3000

143) _____

at the beginning of each six-month period to an RRSP. What semi-annually compounded rate of interest must his RRSP earn in order to reach $600 000 after 25 years?

A) 9 .72% B) 9 .52% C) 9 .43% D) 4 .86%

E) 4 .72%

144) L i buys a $100 000 life insurance policy, which requires an annual premium of $420 or a monthly premium of $37. In either case, the premium is payable at the beginning of the period of coverage. What is the effective rate of interest Li pays, if he chooses monthly payment plan?

A) 1 0.39% B) 1 3.04% C) 1 .09% D) 1 0.89% E) 5 .7%

144) _____

145) W hitby Honda advertised Honda Pilot for sale at $33 610. The same vehicle could also be leased for 5 years at $379 per month, based on $4310 down payment and a residual value of $10 711.

What monthly compounded interest rate was built into the lease?

A) 0 .38% B) 0 .9% C) 4 .13% D) 7 .45% E) 4 .05%

145) _____

146) S agheer is setting up a fund to help his grandson's university education. He wants his grandson to be able to withdraw $3000 every 3 months for 3 years after he starts university. His first withdrawal will be 5.5 years from now. If the fund can earn 7.2% compounded quarterly, what single amount contributed today will provide for the withdrawal?

A) $ 42 041 B) $ 36 000 C) $ 22 083 D) $ 52 077 E) $ 32 119

146) _____

147) B asim purchased a deferred annuity from an insurance company for $10 971. The money used to

purchase

the 147) annuity will earn 6% quarterly . The annuity will provide 16 quarterly payment s of $1000. If the first payment is to be received on October 1, 2017, when did Basim purchase the deferred annuity?

____ _

A) O ctober 1, 2013 B) O ctober 1, 2017 C) J une 1, 2014 D) A pril 1, 2013 E) A pril 1, 2011 148) L eons has a promotion on a refrigerator selling for $1750. Buyers will pay "no money down and no payments for six months." The first of 12 equal monthly payments is required 6 months from the purchase date. What should the monthly payments be if Leons is to earn 15% compounded monthly on its account receivable during both the deferred period and repayment period?

A) $ 182.24 B) $ 170.17 C) $ 1885.42 D) $ 1862.14 E) $ 168.07

148) _____

149) A sif received a bonus of $10 000, which he immediately invested in a fund earning 9.25% compounded semi-annually through his Tax Free Savings Account. Five years later, Asif plans to make the first semi-annual withdrawal of $1000. After how many withdrawals, the fund will be completely depleted?

A) 2 0 B) 2 9 C) 2 7 D) 1 0 E) 2 8

149) _____

150) N ational Research Council is considering the establishment in perpetuity of a Visiting Professor chair at UOIT. The ongoing cost will be $12 500 at the end of each month. If the money can earn 8% compounded monthly in perpetuity, what endowment is required to fund the position?

A) $ 1 875 000 B) $ 937 500 C) $ 2 000 000 D) $ 156 250 E) $ 312 500

150) _____

151) N ational Research Council is considering the establishment of $2 million endowment held in

perpetuity

____ _

of a151) Visiting Professor chair at UOIT. The ongoing cost will be $12 500 at the end of each month.

What monthly compou nded nominal rate of return must be earned to fully fund the position?

A) 6 %

B) 7 .5%

C) 6 .25%

D) 0 .075%

E) 0 .625%

152) G oggle Inc. is planning to pay a quarterly dividend of $0.50 on its perpetual preferred share. The market requires a dividend yield of 5% compounded annually on preferred shares of similar risks. What is the fair market value of its perpetual preferred share just after payment of a dividend at the end of the quarter?

A) $ 40 B) $ 20 C) $ 20.25 D) $ 40.74 E) $ 10

152) _____

153) A fter retirement, Andy plans to buy bonds in a perpetual fund. What amount must Andy place in a perpetual fund today, if it earns 4.8% compounded semi-annually and the first monthly payment of $500 in perpetuity will be made one year from today?

A) $ 120 873 B) $ 126 244 C) $ 125 000 D) $ 119 629 E) $ 120 396

153) _____

154) A t the age of 75, Mrs. Walker decided to set up a trust fund for her grandson, which can take care of some of his expenses for the rest of his life. She made an initial contribution of $150 000 to set up the trust fund, which would make the first semi-annual payment of perpetuity to her grandson, 5 years from now. If the funds in trust earn 5% compounded annually, what is the maximum payment the trust can make to Mrs. Walker's grandson in perpetuity?

A) $ 4727.68 B) $ 5746.53 C) $ 4613.74 D) $ 6033.85 E) $ 4844.43

154) _____

155) S am received a lump sum of $135 000 as an inheritance from his great grandmother. He plans to invest them for 20 years in a long-term investment fund with an expected rate of return of 9% compounded monthly. After 20 years, what payment can he withdraw at the beginning of each month for 10 years?

A) $ 4161 B) $ 66 985 C) $ 7245 D) $ 10 200 E) $ 2955

155) _____

156) W hen Nancy died, she left $20 000 for her 13 year old grandson, Max, to be paid to him as an

annual

____ _

scholarsh156) ip of $8000 at the beginnin g of each year, if he ends up in the universit y. The inheritan ce was invested in a scholarsh ip fund for 5 years at 9% compou nded annually.

How long will the scholarsh ip be paid, once Max enters the universit y?

A) 3 years

B) 7 years

C) 6 years

D) 4 years

E) 5 years

157) O PG has set up a fund to help pay the expenses for the monitoring of Pickering nuclear power plant once it is in safe storage. The balance in the fund today is $700 million, and it will earn interest at 7% compounded annually. Starting 10 years from now, $100 million will be withdrawn from the fund at the beginning of every year for 30 years. What is the annually compounded interest rate for the fund during the final 30 years?

A) 4 .33% B) 0 .72% C) 7 % D) 6 .61% E) 0 .67%

157) _____

158) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the beginning of each month, what will be the balance in the account at the end of the three-year term?

A) $ 18 408.04 B) $ 18 638.25 C) $ 17 390.19 D) $ 18 603.37 E) $ 18 600.50

158) _____

159) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account

at the

begin ning

____ _

of each159) month, what will be the present value of the balance in the account at the end of the threeyea r term?

A) $ 18 408.04

B) $ 17 422.79

C) $ 18 603.37

D) $ 18 638.25

E) $ 17 390.19

160) K eith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the beginning of each month for 20 years. Interest is 4.4% compounded monthly. What is the purchase price of the property?

A) $ 195 930 B) $ 195 214 C) $ 194 378 D) $ 278 878 E) $ 279 900

160) _____

161) J amal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded monthly. What will be the size of his payments at the start of each month from the ordinary simple annuity for 20 years following his retirement?

A) $ 957 836 B) $ 8682 C) $ 2000 D) $ 8618 E) $ 965 020

161) _____

162) S iri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the start of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded monthly.

A) 1 .41 years B) 3 6 years C) 2 16 years D) 1 8 years E) 2 .41 year

162) _____

163) J amal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded monthly until retirement.

The rate changes following the retirement. He wants $7000 per month paid to him at the start of the month for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel) C) A) 9% 0% B) 1 0.27% D) 6 .22% E) 1 0%

163) _____

164) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded semi-annually. If he puts the money in his savings account at the beginning of each month, what will be the balance in the account at the end of the three-year term?

A) $ 18 603.37 B) $ 18 635.22 C) $ 18 638.25 D) $ 18 600.50 E) $ 17 390.19

164) _____

165) S am got a job at the Brick. He plans to save $500 every month for 3 years to buy a car. The savings account earns 2.25% compounded monthly. If he puts the money in his savings account at the beginning of each month, what will be the present value of the balance in the account at the end of the three-year term?

A) $ 17 390.19 B) $ 18 603.37 C) $ 17,425.42 D) $ 18 638.25 E) $ 17 422.79

165) _____

166) K eith bought a lakeside cottage for 30% down and monthly mortgage payments of $1224.51 at the beginning of each month for 20 years. Interest is 4.4% compounded semi-annually. What is the purchase price of the property?

A) $ 279 900 B) $ 195 930 C) $ 196 593 D) $ 280 847 E) $ 194 378

166) _____

167) J amal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded quarterly. What will be the size of his payments at the start of each month from the annuity for 20 years following his retirement?

A) $ 8682 B) $ 8618 C) $ 965 020 D) $ 8517 E) $ 958 266

167) _____

168) S iri plans to retire when her simple annuity savings account has enough money to receive $10 000.00 per month for twenty years starting at the start of the first month after her retirement. She starts saving $4420 per month. Calculate when should Siri retire from today if her savings account pays 4.9% compounded quarterly.

A) 2 16 years B) 1 .41 years C) 3 6 years D) 1 8 years E) 1 8 years 6 months

168) _____

169) J amal plans to retire in 17 years. He is saving $2000 every start of the month in a retirement savings account paying him a long-term interest of 9% compounded quarterly until retirement.

The rate changes following the retirement. He wants $7000 per month paid to him at the start of the month for 20 years after the retirement. At what rate should his savings account pay him to fulfill his dream? (Use the Rate function in Excel.)

A) 6 .34% B) 6 .22% C) 1 0% D) 9 % E) 1 0.27%

169) _____

170) K nowing that the education costs for her kids would be huge, Roberta puts a lump sum of

170) _____

money from her inheritance in a savings account paying an interest of 4.69% compounded monthly. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of the month for four years starting fifteen years from now, how much did Roberta deposit?

A) $ 32 723 B) $ 65 592 C) $ 32 474 D) $ 54 343

E) $ 65 533

171) K nowing that the education costs for her kids would be huge, Roberta puts lump sum money from her inheritance in a savings account paying an interest of 4.69% compounded semi-annually. If ordinary annuity payments of $1500.00 per month are to be paid out of the account at the end of the month for four years starting fifteen years from now, how much did Roberta deposit?

A) $ 32 474 B) $ 65 592 C) $ 32 723 D) $ 65 533 E) $ 54 343

171) _____

172) S iri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded monthly.

A) $ 633 665 B) $ 539 042 C) $ 594 425 D) $ 1 528 015 E) $ 574 626

172) _____

173) S iri plans to retire in 18 years and would like to receive $10 000.00 per month for twenty years starting at the end of the first month after her retirement. Calculate the amount he must invest now if interest is 4.9% compounded quarterly.

A) $ 1 528 015 B) $ 636 990 C) $ 1 530 569 D) $ 633 665 E) $ 539 042

173) _____

174) R ob bought a yacht on a contract requiring monthly payments of $2500 for 5 years beginning on the 6th month after the date of purchase. What was the cash value of the yacht, if interest is 6.9% compounded monthly?

A) $ 136 979 B) $ 126 556 C) $ 122 277 D) $ 135 267 E) $ 122 980

174) _____

175) A be deposited a sum of money at the end of every month for 15 years at 8.5% compounded monthly. After the last deposit, interest for the account is to be 3.5% compounded quarterly and the account is to be paid out by end-of-quarter quarterly payments of $5905.00 over 17 years.

What is the size of the monthly deposit?

A) $ 560 735 B) $ 756 C) $ 1466 D) $ 1160 E) $ 1550

175) _____

176) W itzke retired from Ontario Hydro and received a payment of $980 000.00 at age 56. He invested that sum of money at 5.5% compounded semi-annually until he was 65. At that time he converted the existing balance into an ordinary annuity paying $10 250.00 every month, with interest at 5% compounded monthly. How many payments will be made?

A) 1 83 B) 2 1 C) 2 3 D) 2 71 E) 2 52

176) _____

177) R ousan contributed $1900.00 at the beginning of every month to an RRSP. Interest on the account is 4.95% compounded monthly. If she converted the balance after 13 years into an RRIF paying 7.1% compounded quarterly and makes equal quarterly withdrawals for 20 years starting three months after the conversion into the RRIF, what is the size of the quarterly withdrawal?

A) 3 831.66 B) $ 9789.48 C) 1 071.47 D) $ 11 463.51 E) $ 8232.21

177) _____

178) B abar wants to withdraw $27 000.00 at the beginning of each quarter for twelve years. If the withdrawals are deferred to begin ten years from now and interest is 5.6% compounded monthly calculate the amount that must be invested today to be able to make the withdrawals.

A) $ 950 943 B) $ 1 859 180 C) $ 546 034 D) $ 543 895 E) $ 1 063 365

178) _____

179) T om started his career with Oracle with an intention to retire in 25 years. When he retired, he started receiving $66 000.00 every 6 months for 15 years, starting on the date of his retirement.

How much did Tom deposit deposit in an account at the start of his career, if interest is 9.25% compounded semi-annually?

A) $ 147 332 B) $ 79 360 C) $ 115 594 D) $ 1 449 334 E) $ 1 108 432

179) _____

180) P ayments on a six-year lease value of a car at $27 175.00 are to be made at the beginning of each

180) _____

month during the last 3 years of the lease. If interest is 7.25% compounded monthly, what is the size of the monthly payments?

A) $ 937.65 B) $ 1039.85 C) $ 754.86 D) $ 33 755 E) $ 1029.42 181) W hat sum of money must be deposited in a savings scholarship account to provide a scholarship of $3320.00, payable at the beginning of each 3 months, if interest is 3.59% compounded quarterly?

A) $ 373 236 B) $ 92 799 C) $ 92 479 D) $ 369 916 E) $ 558 195

181) _____

182) W hat monthly lease payment due in advance should be charged for a house valued at $835 000 if the agreed interest is 4.4% compounded semi-annually?

A) $ 3050.48 B) $ 18 136.54 C) $ 3022.99 D) $ 17 974.56 E) $ 3039.40

182) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

183) C ote heard about the Tesla truck being launched in 3 years and plans to buy the truck in for 3 s to cash for $78 000. How much does he have to save at the beginning of every three months yearaccumulate $78

000 183) dollars if interest is 4% compou nded quarterly ?

184) R ick wants to celebrate his 25th wedding anniversary in style and has a budget of $50

____ ____ ____

184) ____________

000. How much does he have to save at the beginning of every three months for 3 years to accumulate $50 000 dollars if interest is 6% compounded monthly?

185) L auren bought mutual funds within TFSA as part of her retirement planning in 2019, immediately after graduation at the age of 22. Payments of $500.00 will be made into the fund at the beginning of every month for 33 years. If the fund is expected to earn interest at 8.4% compounded monthly, what will be the balance in the fund after 33 years? How much interest did she earn in 33 years?

185) ____________

186) I n the previous question, what is the discounted value of deposits of $500.00 made at the beginning of every month for 33 years if money is worth 8.4% compounded quarterly?

186) ____________

187) T he monthly rent payment on a domestic water heater is $16.99 payable in advance.

What yearly payment in advance would satisfy the lease if interest is 6.84% compounded monthly?

187) ____________

188) S teve wants to save for a dream vacation with his girlfriend. What deposit made at the beginning of each month will accumulate to $16 000.00 at 5% compounded quarterly at the end of eighteen months?

188) ____________

189) R yan has bought a vacation property in Florida by making semi-annual payments of $9000.00 for 10 years. If the first payment is due on the date of purchase in 2019 and interest is 6% p.a. compounded quarterly, calculate the total amount Ryan paid for the property.

189) ____________

190) A rjun is 25 years old and planning his retirement. The financial advisor has advised him to accumulate a minimum of $2 150 000.00 in an RRSP to live comfortably after retirement. He can make annual contributions of $15 000.00 at the beginning of each year. If interest is 5.5% compounded quarterly, at what age can he retire.

190) ____________

191) A sim bought a cottage property near lake Huron for quarterly payments of $2470.00 for 18 years. If the first payment was made on the date of purchase and interest is 3.57% compounded annually, what was the purchase price of the property?

191) ____________

192) W hat is the difference in accumulated value after 15 years of monthly deposits of $995.00 earning interest at 5.8% compounded semi-annually, between the deposits made at the end of each month vs. at the beginning of each month?

192) ____________

193) M ohsin contributed $9100.00 at the end of every six months for 25 years into an RRSP earning interest at 6.65% compounded semi-annually. Two years after the last contribution, Mohsin converted the RRSP into an RRIF that is to pay her equal monthly amounts for 20 years. If the first payment is due a month after the conversion into the

RRIest on the RRIF is 7.66% F and compounded intermonthly, how

much will Mohsin receive every month?

193)

____ ____ ____

194) A month after his grandchild Mobin's birth, Mr. Akbar started a savings accounts to help in Mobin's higher education when he reaches 18 years of age. Mr. Akbar made deposits of $150 into the savings account every month until Mobin was 18 years old. At the age of 18, the savings account was converted into a trust fund providing for equal withdrawals at the end of each month for 4 years beginning one month after the last deposit. If interest is 7.2% compounded quarterly, how much will Mobin receive every month?

194) ____________

195) S uncor bought an oil field in Libya for $4.5 billion down and payments of $671 million at the end of every six months for 7 years. What is the purchase price of the oil field if the semi-annual payments are deferred for four years and interest is 5.5% compounded semi-annually?

195) ____________

196) D elawri acquired his partner's share in Dehli Hydrogenated Oil by agreeing to make payments of $471 100.00 at the end of each year for 5 years. If the payments are deferred for three years and money is worth 5.96% compounded quarterly, what is the cash value of the partner's share of the business?

196) ____________

197) T he Sask Power needs to borrow to finance the carbon capture project for their coal power plants. Repayment of the loan involves payments of $871 100.00 at the end of every three months for six years. No payments are to be made during the design, development and installation period of three years. Interest is 9.24% compounded quarterly. a) How much did the Sask Power borrow? b) What amount will be repaid? c) How much of that amount will be interest?

197) ____________

198) D ave would like to receive $22 500.00 at the end of every six months for 15 years after his retirement. If he retires 12 years from now and interest is 7% compounded semi-annually, how much must he deposit into an account every six months starting now?

198) ____________

199) A farm provides a monthly income of $7972.00 due at the beginning of every month.

What is the cash value of the farm if money is worth 5.32% compounded monthly?

199) ____________

200) G opal invested $40 000 two years ago to pay for his son's tuition fee at the Ivey School of Business. His son dropped out and started his own business. Gopal converted that amount into a trust fund, such that it provides a monthly scholarship to his son. What will be the amount of scholarship if money is worth 6.24% compounded monthly?

200) ____________

201) K nowing he has cancer, Dier has created a fund to provide an annual scholarship to his children of $9500.00. If the first payment is due in three years and interest is 5.0% compounded quarterly, what sum of money must be deposited in the scholarship fund today?

201) ____________

1) P MT = 2000.00; n = 33(2) = 66; i = FV(due) = 2000

= 0.04; I/Y = 8; P/Y = C/Y = 2 = $640 155.60

Programmed solution:

2) P MT = 3000.00; PV = 60 000; i =

= 0.015; I/Y = 6; P/Y = C/Y = 4

60 000 = 3000(1.015) 0.295566502 = 1 - 1.015-n -n ln 1.015 = ln 0.704433498 -n(0.014888612) = -0.350361349 n = 23.53216923 quarters ÷ 4 = 5.88 ≈ 6 years Programmed solution:

3) P MT = 100.00; n = 10(4) = 40; i = PV(due) = 100

= 0.015; I/Y = 6; P/Y = C/Y = 4 = $3036.46

Programmed solution:

4) P V = 15 000.00; n = 5(4) = 20; PMT = 900.00; P/Y = C/Y = 4 Programmed solution:

The nominal rate of interest is 8.03% (approx.) p.a. compounded quarterly.

5) F V(due) = 4800; n = 10(2) = 20; i = = 0.035; I/Y = 7; P/Y = C/Y = 2 FV(due) = PMT(1 + i) 4800 = PMT(1 + 0.035) 4800 = PMT(1 + 0.035)[28.27968181] 4800 = PMT(29.26947068)

PMT $163.99 = Programmed solution:

PMT = 6) F V n (due) = 1000000 1000000.00 = PMT

(1.01)

1000000.00 = PMT(81.66966986)(1.01) $12 123.22 = PMT 7) a ) Step 1: PMT = 500; n = 4(12) = 48; i =

= $26 340.32

FV(due) = 500 Step 2:

PV = 26340.32; n = 8(12) = 96 FV = 26340.32(1.00375)96 = $37 728.94

Programmed solution:

b) Amount deposited is 48(500.00) = $24 000.00 c) Interest is 37 728.94 - 24 000.00 = $13 728.94 8)

FV (due) = 500.00

(1.007)

= 500.00(619.1778116)(1.007) 9)

= $311 756.03 FV (due) = 470.00

(1.018)

= 470.00(90.026386)(1.018) 10)

= $43 074.02 FV (due) = 1000.00

(1.01)

= 1000.00(230.0386895)(1.01) 11)

= $232 339.08 FV (due) = 9000.00

(1.02)

= 9000.00(114.05153394)(1.02) 12)

= $1 046 993.13 PV (due) = 570

= 0.00375; I/Y = 4.5; P/Y = C/Y = 12

(1.021)

= 570(2=3 $.6134 755147.45)8( 1 .0 21) PV (due) = 250 13) (1.015) = 250(30.99405004)(1.015) 14)

= $7864.74 a) FV (due) = 3000

(1.03)

= 3000(47.57541571)(1.03) = $147008.03 b) Total deposits = (3000) × 30 = $90 000.00 Interest = 147008.03 - 90000 = $57 008.03 15)

FVn (due) = 600

(1.018)

= 600(145.1537512)(1.018) = $88 659.91 16)

PV (due) = 3971

(1.0057)

= 3971(11.5669781)(1.0057) 17)

= $46 194.29 PV (due) = 2560

(1.0725)

= 2560(9.5964039)(1.0725) 18)

= $26 347.89 a) PV (due) = 100

(1.02)

= 100(34.76088668)(1.02) = $3545.61 b) Total paid = 60(100) = $6000 c) Interest = 6000 - 3545.61 = $2454.39 19)

a) FV (due) = 167

(1.0445)

= 167(326.5655028)(1.0445) = $56963.31 b) Total deposits = (167) × 63 = $10 521 Interest = 56963.31 - 10521 = $46 442.31 20)

PV (due) = 30

(1.01)

= 30(11.25507747)(1.01) 21)

= $341.03 PV (due) = 500

(1.09)

= 500(6.4176577)(1.09) = $3497.62 22) P V n (due) = 110 194 110 194 = PMT

(1.0375)

110 194.00 = PMT(9.5226939)(1.0375) $11 153.47 = PMT 23) i = 0.12 ÷ 12 = 0.01

a) ) = 200(81.66966986)(1.01) = $16 497.27 FVn (1.01) (due) PVn (due) = 200 = = = 200(44.95503841)(1.01) $9080.92 200 b) FVn = 200(81.66966986) = $16 333.93 (1.01 24)

PVn = 200(44.95503841) = $8991.01 a) PV (due) = 167.25

(1.024)

= 167.25(30.0897982)(1.024) = $5153.30 b) Total paid = 54(167.25) = $9031.50 c) Interest = 9031.50 - 5153.30 = $3878.20 25) i = 0.0648 ÷ 12 = 0.0054 a) FVn (due) = 173.00

(1.0054 ) = 173.00(87.7161525)(1.0054) = $15 256.84

PVn (due) = 173.00

(1.0054)

= 173.00(59.522361)(1.0054) = $10352.97 b) FVn = 173.00(87.7161525) = $15 174.89 PVn = 173.00(59.522361) = $10 297.37 26)

a) FVn (due) = 415.00

(1.0103)

= 415.00(37.676677)(1.0103) = $15 796.87 b) Total deposit = 32(415.00) = $13 280.00 c) Interest = 15796.87 - 13280.00 = $2516.87 27)

a) FVn (due) = 500

(1.02)

= 500(60.40198318)(1.02) = $30 805.01 b) Total deposit = 40(500.00) = $20 000.00 c) Interest = 30805.01 - 20000.00 = $10 805.01 28) FVn (due) = 17897.20 = 1140.00

(1.02935)

15.251662 = 0.4476363 = 1.02935n - 1 ln 1.447363 = n ln 1.02935 =n n = 12 .78 ≈ 13 semi-annual periods 29) F V n (due) = 1 000 000 1 000 000.00 = PMT

30)

(1.03)

1 000 000.00 = PMT(75.40125973)(1.03) $12876.10 = PMT

PVn 06) (due) = 370.00(25.2449471)(1.0206) = = $9533.05 = FVn (due) 370.0 0 9533.05 = PMT (1.12) 9533.05 = PMT(3.3744)(1.12) (1.02 31)

$2522.42 = PMT 345 000.00 = 1600.00

(1.0092)

213.6593341 = 1.9656659 = 1.0092n - 1 2.9656659 = 1.0092n ln 2 .9656659 = n ln 1.0092 1.0871016 = n(.0091579) n = 118.71 ≈ 119 months 32) F V n (due) = 41 600 41 600.00 = PMT

(1.02)

41 600.00 = PMT(92.816737)(1.02) $439.41 = PMT 33) P V n (due) = 81 500, PMT = 3350 81 500 = 3350

(1.0325)

23.5625745 = 0.7657837 = 1 - 1.0325-n -n ln 1.0325 = ln 0.2342163 -n(0.031983) = -1.4515102 n = 45.383804 Answer is: 45 full payments and one less than $3350 34)

Sn (due) = 1000

(1.02) = 1000(213.8666068)(1.02) = $218 143.94

35) F V n (due) = 17 448 17 448 = PMT

(1.03125)

17 448.00 = PMT(30.9733986)(1.03125) $546.25 = PMT 36) P V n (due) = 33 000 33 000 = PMT

(1.0075)

33 000 = PMT(99.3634773)(1.0075) $329.64 = PMT 37) P V n (due) = 60 000 60 000 = PMT

(1.01)

60 000 = PMT(83.32166399)(1.01)

$712.97 = PMT 38) a) p=

- 1 = 0.006558197

FVg (due) = 500

= 500(342.0753463) = $171 037.67

b) FVg = 500(342.0753463)(1.006558197) = $172 159.37 39) P MT = 1200; n = 10(4) = 40; i = = 0.005; I/Y = 6; P/Y = 4; C/Y = 12; c =

=3

p = (1.005)3 -1 = 0.015075125 PV(due) = 1200(1.015075125)

= $36 390.27

Programmed solution:

40) F V(due) = 36 000.00; n = 8(12) = 96; i = c=

= 0.0125; I/Y = 5; P/Y = 12; C/Y = 4

=

p = (1 + i)c - 1 = (1 + 0.0125)1/3 - 1 = 0.004149425 FV(due) = PMT(1 + p) 36 000 = PMT(1 + 0.004149425) 36 000 = PMT(1.004149425)[117.6381043] 36 000 = PMT(118.1262348) PMT =

= $304.76

Programmed solution:

41) P MT = 7500.00; n = 7(2) = 14; i =

= 0.015; c =

= 2; I/Y = 6; P/Y = 2; C/Y = 4

p = (1.015)2 - 1 = 0.030225 PV(due) = 7500

= $87 147.51

Programmed solution:

42) P MT = 5000.00; FV(due) = 150 000; i = p = (1.01375)4 - 1 = 0.056144809

= 0.01375; c =

= 4; I/Y = 5.5; P/Y = 1; C/Y = 4

150 09n -1 n ln 1.056144809 = ln 2.594804293 000 = n(0.054625306) = 0.953511097 5000 n = 17.455483

Programmed solution: 1.594 8042 93 = 1.056 1448 n = 18 years (approx.).The last payment will be less than $5000.00.

43) F V(due) = 90 000.00; n = 10(4) = 40; PMT = 1500.00; P/Y = 4; C/Y = 12 Programmed solution:

The nominal rate of interest is 7.48% (approx.) p.a. compounded monthly.

44)

p=

- 1 = 0.0224573

FV g (due) = 172.00

(1.0224573)

= 172.00(40.26262)(1.0224573) 45)

= $7080.69 p=

- 1 = 0.061520151

FVg (due) = 500

(1.061520151)

= 500(81.20594577)(1.061520151) = $43 100.87 46)

p=

- 1 = 0.0051418

FVg (due) = 545

(1.0051418)

= 545(170.9815136)(1.0051418) = $93 664.06 47)

p=

- 1 = 0.013643245

PVg (due) = 1070

(1.013643245)

= 1070(28.2958162)(1.013643245) = $30 689.59 48)

p=

- 1 = 0.014673846

PVg (due) = 3000

(1.014673846)

= 3000(30.09471567)(1.014673846) = $91 608.96 49)

p=

938622 (1.004938622)

FVg (due) = 500 -1= 0.004 50)

= 500(289.0001444)(1.004938622) = $145 213.70 p=

- 1 = 0.006891

FVg (due) = 750

(1.006891)

= 750(52.5455)(1.006891) = $39 680.69 FV = 39680.69(1.006891)24 = 39680.69(1.179177) = $46 790.56 51)

a)

p=

- 1 = 0.004775944

FVg = 415 b)

= 415(256.8161929) = $106 578.72

FVg (due) = 415(256.8161929)(1.004775944) = $107 087.73

52) p =

- 1 = 0.015075125

FVg (due) = 2000

(1.015075125)

= 2000(54.35422486)(1.015075125) = $110 347.24 FV = 110347.24 (1.015075125)28 = 110347.24(1.520369636) = $167 768.59 53)

p=

- 1 = 0.0431048

PVg (due) = 4210

(1.0431048)

= 421(11.8778937)(1.0431048) = $52 161.43 54) P V n (due) = 3100 p=

- 1 = 0.0072802

3100.00 = PMT

55)

(1.0072802)

3100.00 = PMT(31.5685288)(1.0072802) $97.49 = PMT p= - 1 = 0.013159404 PVn (due) = 1150

(1.013159404)

= 1150(35.41891644)(1.013159404) = $41 267.76 56)

p= FVg(due) = 1280.00

- 1 = 0.03345709 (1.03345709) = 27403.23

1517.84 = PMT 27403.23 = PMT The size of the quarterly payment is $1517.84 27403.23 = PMT(18.0540866) 57)

FVn = 1011 = 1011(92.680366) = $93 699.85 FV = 93699.85(1.03325)14 = 93699.85(1.5807959) = $148 120.34 148120.34 = PMT

58)

148120.34 = PMT(38.1133915) $3886.31 = PMT FVn = 63.25

= 63.25(180.0981175) = $11391.21 PVn = 11391.21 = PMT 11391.21 = PMT(16.2571498) 59)

$700.69 = PMT PVn = 67100.00 = 67100.00(11.4910081) = $771046.64 PV = 771046.64(1.0275)-8 = 771046.64(0.8049064) = 620620.34 Purchase price = 620620.34 + 450000 = $1 070 620.34

60)

p=

- 1 = 0.0609453

PVn = 4711.00 = 4711.00(6.1866279) = $29 145.20 PV = 29145.20(1.0609453)-3 = 29145.20(0.837377) = $24 405.52 61)

a) PV n = 8711.00 = 8711.00(18.2662014) = $159116.88 PV = 159116.88(1.0231)-12 = 159116.88(.7602965) = $120 976.01

b) Total repaid = 24(8711.00) = $209 064.00 c) Interest = 209064.00 - 120976.01 = $88 087.99 62) B alance after 15 years: PV = 27600.00(1.01705)-40 = 27600.00(.50851976) = $14 035.15 Deposits: FVn = 14035.15 = PMT 14035.15 = PMT(269.891221) $52.00 = PMT 63)

PVn = 815.50

= 815.50(41.119857) = $33 533.24

PV = 33533.24(1.0065)-18 = $29 841.97 64) F V = 15700.00(1.0324)16 = 15700.00(1.6655893) = $26 149.75

p=

4142 26149.75 = PMT

- 1 = 26149.75 = PMT(8.7409393) 0.040$2991.64 = PMT 65) PVn = 1000 = 1000(44.95503841) = $44 955.04 PV = 44955.04 (1.01)-5 = 44955.04(0.951465687) = $42 773.18 66) PVn = 1800 = 1800(21.2125066) = $38182.51 = FVn 38182.51 = PMT 38182.51 = PMT(161.3936213) $236.58 = PMT The monthly deposit is $236.58.

67) a) PVn = 1000.00 = 10000.00(23.46833482) = $234683.35 PV = 234683.35(1.02)-20 = 234683.35(.6729713331) = $157 935.17 b) Total repaid = 32(10000) = $320 000 c) Interest = 320000 - 157935.17 = $162 064.83 68) F V = 38000.00(1.0425)30 = 38000.00(3.485635) = $132 254.13 = PVn n= n= 69)

=

p=

= 18.16 - 1 = 0.003736025

PVn = 4360.00 = 4360.00(78.44041381) = $342000.20 = FVn 342000.20 = PMT 342 000.20 = PMT(27.6712294) $12 359.42 = PMT 70) F V = 75400.00(1.024125)12 = 75400.00(1.3311764) = $100 370.70 = PVn 100370.70 = PMT

71)

100370.70 = PMT(12.2084378) $8221.42 = PMT PVn = 3760.00

= 3760.00(12.3243576) = $46339.58 = FVn (due)

4633 9.58 =

(1.0325)

46339.58 = PMT(27.5642438)(1.0325) PMT $1628.23 = PMT 72) F V = 140000.00(1.0182)4 = 140000.00(1.074812) = $150 473.63 = PV n p=

- 1 = 0.0060302

150473.63 = PMT

73)

150473.63 = PMT(85.2299222) $1765.50 = PMT FVn = 270.00

270.00(33.0400611) = $8920.82 FV = 8920.82(1.0315)16 = 8920.82(1.6425088) = $14 652.53 = PVn 14652.53 = PMT 14652.53 = PMT(39.4661717) $371.27 = PMT 74) F V = 6420(1.0325)10 = 6420(1.3768943) = $8839.66 = PVn

n=

n= n=

= 16.7966

The annuity term is 17 months.

75) F V = 481700.00(1.0425)12 = 481700.00(1.6478314) = $793 760.39 = PVn 793760.39 = PMT

76)

793760.39 = PMT(20.5930613) $38 545.04 = PMT PVn = 587.00

= 587.00(35.6040979) = $20899.61 = FVn (due) 20899.61 = 562.71 36.4002511 = 0.7407451 = 1.02035n - 1

(1.02035)

1.74035n 7451 ln 1.7407451 = n ln 1.02035 = 0.5543132 = n(0.0201457)

1.020n = 27.51521 ≈ 28 quarters 77) F V = 43000(1.0058333)8 = 43000(1.0476304) = $45048.12 = PV n 45048.12 = PMT

78)

45048.12 = PMT(50.5020416) $892.01 = PMT PVn = 2600

= 2600(138.9268243) = $361209.74 =FV PV = 361209.74(1.003)-144 = 361209.74(.6496294) = $234 652.45 79)

a) FVn (due) = 1830

(1.0186)

= 1830.00(29.9076016)(1.0186) = $55 748.91 b) 55748.91 - 24(1830.00) = 55748.91 - 43920.00 = $11 828.91 c) 55748.91 = PMT 55748.91 = PMT(32.9973043) $1689.50 = PMT d) Total interest = 44(1689.50) - 24(1830.00) = 74338.00 - 43920.00 = $30 418.00 80)

p=

- 1 = 0.00900682

PVg (due) = 2700

(1.0090068)

= 2700(81.3107476)(1.00900682) = $221516.02 PV = 221516.02(1.00900682)-87 = 221516.27(.4583676) = $101 535.59 81)

p=

- 1 = 0.0072073233

PVg (due) = 5000

× 1.0072073233

= 5000.00(114.362113)(1.0072073233) = $575 931.79 PV = 575931.79 (1.0072073233 )-180 = 575931.79(0.274536) = $158 115.19 82)

PV (due) = 2472

(1.03325)

= 2472(13.383091)(1.03325) = $34183.01 PV = 34183.01(1.03325)-28 = 34183.01(.4001736) = $13 679.14 83)

PV(due) = 3000

(1.04)

= 3000(12.65929697)(1.04)

3949=7 $.0319(419.074.0)-13 0 = 39497.01(.308318668) = $12 177.67 PV = 84) S tep 1: PMT = 1000.00; n = 12(4) = 48; i =

= 0.00375; c =

= 3; I/Y = 4.5; P/Y = 4; C/Y = 12

p = (1.00375)3 - 1 = 0.01129224 PV(due) = 1000

= $37 314.82

Step 2: FV = 37314.82; n = 10(4) = 40

-40 PV = 37314.82(1 - 1.01129224) = $23 813.01 Programmed solution:

85)

- 1 = 0.015075125

p= PVg (due) = 4500

(1.015075125)

= 4500(17.15593096)(1.0150751) = $78 365.51 PV = 78365.51(1.015075125)-8 = 78365.51(0.887185668) = $69 524.76 86)

p=

- 1 = 0.030301

PVg (due) = 10000

(1.030301)

= 10000(23.00271345)(1.030301) = $236997.19 PV = 236997.19(1.030301)-20 = 236997.19(0.5504496159) = $130 455.01 87)

PV (due) = 6600

(1.03125)

= 6600(18.480549)(1.03125) = $125783.24 PV537-543 = 125783.24(1.03125)-75 = 125783.24(.0994725) = $12 511.97 88)

PV (due) = 8000

(1.025)

= 8000(25.10277505)(1.025) = $205 824.76 PV = 205824.76(1.025)-30 = 205824.76(0.4767426852) = $98 125.45 89) i = 0.0612 ÷ 12 = 0.0051

PVn (due) = 3713

(1.0051) = 3713(47.0980497)(1.0051) = $175 766.92 PV = 175766.92(1.0051)-24 = 175766.92(.8850696) = $155 565.96

90) i = 0.12 ÷ 12 = 0.01 PVn (due) = 4000

(1.01)

= 4000(44.95503841)(1.01) = $181 618.36 PV = 181618.36 (1.01)-36 = 181618.36 (.6989249496) = $126 937.60 91)

p=

- 1 = 0.00534474

PVg (due) = 840

(1.00534474)

= 840(117.6803282)(1.00534474) = $99 379.81 PV = 99379.81(1.00534474)-24 = 99380.12(0.879913049) = $87 445.59 92)

p=

- 1 = 0.006558197

PVg (due) = 1000

(1.006558197)

= 1000(82.89062668)(1.006558197) = $83 434.24 PV = 83,434.24 (1.006558197)-36 = 83333.65 (.790314523) = $65 939.29 93)

p=

- 1 = 0.04959484

PVg (due) = 6100

(1.04959484) = 6100(10.40814249)(1.04959484) = $66 638.43

PV = 66638.43(1.04959484)-6 = 66638.43(0.747945354) = $49 841.90 94) F V = 12975.00(1.0075)36 = 12975.00(1.3086454) = $16979.67 = PVn (due) 16979.67 = PMT

95)

(1.0075)

16979.67 = PMT(31.4468053)(1.0075) $535.93 FVn(due)==PMT 3560 = $96193.53

(1.0175) = 3560.00(26.5559262)(1.0175)

FV = 96193.53(1.0175)6 = 96193.53(1.1097024) = $106 746.19

FVn (due) = 106746.19 = PMT

(1.05)

106746.19 = PMT(13.1630026)(1.05) $7723.39 = PMT 96)

PVn (due) = 1230

(1.0171)

= 1230(42.3594904)(1.0171) = $52993.12 = FVn (due) 52993.12 = PMT

(1.0171)

52993.12 = PMT(184.4894598)(1.0171)

$282.41 = PMT 97) p=

- 1 = 0.0244974

FV = 53500.00(1.0081)120 = 53500.00(2.6328964) = $140859.96 = PVn

n

=

=

=

98) P MT = 150.00; n = 20(4) = 80; i =

= 57.8502657 ≈ 58 quarters = 2.25% = 0.0225; I/Y = 9; P/Y = 4; C/Y = 4

FV = 150.00(1 + 0.0225)

= 150(1.0225)(219.1175688) = 33607.16

PV = 33607.16; PMT = 800.00; 33607.16 = 800(1.0225) 0.924402323-1 = -(1.0225)-n -n ln 1.0225 = -ln 0.075597677 -n = n = 116.0565865 (quarters) ÷ 4 ≈ 29 years Programmed solution:

99)

PVn (due) = 1550.00

(1.0166)

= 1550.00(39.237778)(1.0166) = $61 828.14 = FVn (due) 61828.14 = 790.00

(1.0166)

76.986256 = 1.2779719 = 1.0166n - 1 ln 2.2779719 = n ln 1.0166 0.8232855 = n(0.0164637) 50.01 deposits = n Approximately 12.5 years.

100) F V = 27700(1.0466)27 = 27700.00(3.4204222) = $94 745.70 = PVn (due)

9474 66) 5.70 18.1054271 0.843712902== 1 - 1.0466-n = -n ln 1.0466 = ln.156287097 5000 -n(0.045546814) = -1.856060596 n = 40.75 semi-annual periods = 20.4 years

(1.04 101) F V = 59760.00(1.0079167)48 = 59760.00(1.4601006) = $87255.47 = PVn (due) 87255.47 = 15240.00

(1.0957)

5.225357799 = 0.500066741 = 1 - 1.0957-n -n ln 1.0957 = ln .499933258 -n(0.0913934) = -0.693280672 n ≈ 7.6 years 102) F V = 47600.00(1.0172)8 = 47600.00(1.1461747) = $54 557.91 = PVn (due) 54557.91 = 4500

(1.0172)

11.91897365 = .205006346 = 1 - 1.0172-n -n ln 1.0172 = ln 0.794993653 -n(0.01705375) = -0.229421147 n = 13.4528 ≈ 13 quarters 103)

p=

- 1 = 0.03638796

FV = 123900.00(1.03638796)20 = 123900.00(2.043841277) = $253 231.93 = PVg (due) 253231.93 = 9600

(1.03638796)

25.45217342 = 0.926152623 = 1 - 1.03638796-n -n ln 1.03638796 = ln 0.073847376 -n(0.0357416) = -2.605754795 n = 72.9054766 semi-annual periods ≈ 36.5 years 104)

p=

- 1 = 0.0489703

FV = 23900.00(1 + .008)36 = 23900(1.3322298) = $31840.29 = PVg (due) 31840.29 = 3617.00

(1.0489703)

8.391.4109586 = 1 - 1.0489703-n 9969 -n ln 1.0489703 = 0.5890414 = -n(0.047809) = -0.5292588 n = 11.0702755 ≈ 11 semi-annual periods

105) F V = 17570(1 + .0325)23 = 17570(2.0867546) = $36664.28 = PVn (due) 36664.28 = 1500

(1.007916667)

24.2508673 = 0.191986032 = 1 - 1.0079167-n -n ln 1.007916667 = ln 0.808013967 -n(0.007885494) = -0.213175934 n = 27.0333934 ≈ 27 months 106) N per 40 Pmt -2150.00 PV 62000.00 Type 1 Result 0.0180267 The nominal rate is 1.80267% * 2 = 3.60534% 107) F V = 15660.00(1.0070833)9 = 15660.00(1.0655861) = $16687.08 = PVn

n=

=

=

Payments are required for 33 months.

108) P V (due) = 1000.00 + = 1000.00 + 47619.05 = $48 619.05 109) P V(due) = 1472.00 + 110)

111)

p=

= 1472.00 + 241311.48 = $242 783.48 - 1 = 0.0765016

PV =

= $562 079.75

p=

- 1 = 0.0666016

PV (due) = 3725.00 +

= 3725.00 + 55929.59 = $59 654.59

112) F V = 41321.00(1.0052)36 = 41321.00(1.2052837) = $49803.53 = PV(due) 49803.53 = PMT + 49803.53 = PMT + 192.3076923 PMT 49803.53 = 193.3076923 PMT

= 32.60312148

$257.64 = PMT 113) p=

- 1 = 0.006679141

135000 = PMT + 135000 = PMT + 149.7198483 PMT 135000 = (150.7198483)PMT $895.70 = PMT 114) P V(due) = 8710.00 + = 8710.00 + 85559.92 = $94 269.92 115)

p=

- 1 = 0.06555245

PV 1 =

= $86 953.27

PV 2 = 86953.27(1.06555245)-3 = 86953.27(.82656213) = $71 872.28 116) P V = = $96 923.08 117) a ) PMT = 500; i =

= 0.024; c =

=

p = (1.024)1/6 - 1 = 0.003960577 PV = b)

PV = 500 +

= $126244.24 = 500 + 126244.24 = 126744.24

FV = 126244.24; i = 0.024; n = 1(12) = 12 PV = 126744.24(1.003960577)-12 = $12 0872.73 118) a ) PMT = 2000; i = = 0.035; c = = p = (1.035)1/6 - 1 = 0.005750039 PV = b) PV = 2000 +

= $347823.70 = 2000 + 347823.70 = 349823.70

FV = 349823.70; i = 0.035; n = 1(12) = 12 PV = 349823.70(1.005750039)-12 = $326 564.17 119) P V = = $134907.41 PV 2 = 134907.41(1.0108)-75 = 134907.41(0.4467942) = $60 275.85 120) N per 16 Pmt -1180.00 PV 17040.00 Type 1 Result 0.0140451 f = (1 + 0.0140451)4 - 1 = 0.057375 Effective rate = 5.7375% 121) N per 22 PMT -2200.00

FV Type 1 6800 Result .0285814 0.00 The nominal rate is 2.85814% × 2 = 5.711628% 122) N per 15 Pmt = -25700.00 FV = 450000.00 Type 1 Result 0.019101047 The effective annual rate is 1.9101047% 123) N per 13 Pmt -240.00 PV 2340.00 Type 1 Result 0 .0522876 Nominal annual rate = 5.22876% × 4 = 20.91503% 124) E 125) C 126) D 127) C 128) D 129) E 130) E 131) E 132) B 133) C 134) E 135) F or the first 7 years, i =

= 4% and n = 2 × 7 = 14

FV after first 7 years of contribution = ⇒

FV 1 = $2500

= PMT

(1 + i)

(1.04) = $47 559

The future value of $47 559 after an additional 13 years at i =

= 3.5% and n = 2 × 13 = 26 is:

FV = $47559 × 1.03526 = $116 327 The FV for the last 26 payments will be FV

2 = $2500

= $106 898

The total amount in the RRSP after 20 years will be $116327 + $106898 = $223 225 136) A 137) D 138) C 139) P MT = $1 000 000; n = 25; i = 3.2% PV(due)

= PMT = $1 000 000

(1 + i) 1.032

= $17 576 379 ∴ 25 year annuity has $576 379 more economic value than the lump-sum payment.

140) C 141) D 142) C 143) C 144) B 145) C 146) C 147) D 148) E 149) C 150) A 151) C 152) D 153) A 154) C 155) D 156) E 157) D 158) E 159) B 160) E 161) B 162) D 163) D 164) B 165) C 166) D 167) D 168) E 169) A 170) C 171) C 172) A 173) B 174) E 175) E 176) E 177) B 178) D 179) C 180) B 181) A 182) C 183) F V n (due) = 78000 78000.00 = PMT

(1.01)

78000.00 = PMT(16.09689554)(1.01) $4797.68 = PMT 184) F V n (due) = 50000 50000.00 = PMT

(1.005)

5000= PMT(39.33610496)(1.005) 0.00 $1264.77 = PMT 185) FV (due) = 500.00

(1.007)

= 500.00(2119.558001)(1.007) = $1 067 197.45 Interest = 1067197.45 — (500 × 12 × 33) = $869 197.45 186)

PV (due) = 500

(1.007)

= 500 × 134.7734782 187)

= $67 386.34 PV (due) = 16.99

(1.0057)

= 16.99(11.5669781)(1.0057) = $197.64 188) F V(due) = 16000.00; n = 18; i = c=

= 0.0125; I/Y = 5; P/Y = 12; C/Y = 4

=

p = (1 + i)c - 1 = (1 + 0.0125)1/3 - 1 = 0.004149425 FV(due) = PMT(1 + p) 16000 = PMT(1 + 0.004149425) 16000 = PMT(1.004149425)[18.64913287] 16000 = PMT(18.72651605) PMT = $854.40 189) P MT = 9000.00; n = 10(2) = 20; i = = 0.015; I/Y = 6; P/Y = 2; C/Y = 4; c =

=2

p = (1.015)2 - 1 = 0.030225 PV(due) = 9000

= $137 657.80

190) P MT = 15000.00; FV(due) = 2150000; i =

= 0.01375= 4; I/Y = 5.5; P/Y = 1; C/Y = 4 ; c =

p = (1.01375)4 - 1 = 0.056144809 2150000 = 15000 7.619620486 = 1.056144809n - 1 n ln 1.056144809 = ln 8.619620486 n(0.054625306) = 2.154041057 n = 39.433 n = 40 years (approx.).The last payment will be less than $15000.00.n Arjun can retire at the age of 65 191)

p=

- 1 = 0.00880795

PVg (due) = 2470

(1.00880795)

= 2470(52.71499667)(1.00880795) = $131 352.89

192)

p=

- 1 = 0.004775944

FVg = 995

= 995(284.2477768) = $282 826.54

FVg (due) = 995(284.2477768)(1.004775944) = $284 177.30 Difference = 284177.30 — 282826.54 = $1350.76 193)

FVn = 9100 = 9100(124.265) = $1 130 811.57 FV = 1130811.57(1.03325)4 = 1130811.57(1.139782) = $1 288 878.26 0.00638333 240 1288878.26 = PMT 1288878.26 = PMT(122.6385)

194)

$10 FVn =509.57 150 = PMT

= 150(440.0872454) = $66 013.09 PVn = 66013.09 = PMT 66013.09 = PMT(41.598819) 195)

$1586.90 = PMT PVn = 671 = 671(11.4910081) = $7710.47 million PV = 7710.47(1.0275)-8 = 7710.47(0.8049064) = $ 6206.20 million Purchase price = 6206.20 + 4500 = $10 706.20 million or $10.71 billion

196)

p=

- 1 = 0.0609453

PVn = 471100.00 = 471100.00(4.2015533) = $1 979 351.77 PV = 1979351.77(1.0609453)-3 = 1979351.77(0.837377) = $1 657 463.65 197)

a) PV n = 871100.00 = 871100.00(18.2662014) = $15 911 688.04 PV = 15911688.04(1.0231)-12 = 15911688.04(.7602965) = $12 097 600.73 b) Total repaid = 24(871100.00) = $20 906 400 c) Interest = 20906400 - 12097600.73 = $8 808 799.27

198) PVn = 22500.00

= 2= FV n (due) 2250 0 (1.035) (18.3413821.02 = PMT 92) = 413821.02 ==PMT(32.9878333)(1.035) 120.44 PMT + $413 $12 P V(due) = 7972.00 821.0 199)

= 7972.00 + 1798195.49 = $1 806 167.49

200) F V = 40000.00(1.0052)24 = 40000.00(1.13256) = $45302.22 = PV(due) 45302.22 = PMT + 45302.22 = PMT + 192.3076923 PMT 45302.22 = 193.3076923 PMT $234.35 = PMT 201) p = - 1 = 0.050945 PV =

= $186 474.38 = FV

-3 PV 2 = 186474.38(1.050945) = $160 649.44

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) A debt of $10 000.00 with interest at 8% compounded quarterly is to be repaid by equal 1) _____________ payments at the end of every three months for two years. a) Calculate the size of the monthly payments. b) Construct an amortization table. c) Calculate the outstanding balance after three payments.

2) A contractor's price for a new building was $85 100.00. Stampede Inc., the buyers of the building, paid down and financed the balance by making equal payments at the end of every six months for 14 years. Interest is 9.8% compounded semi-annually. a) What is the size of the semi-annual payment? b) How much will Stampede Inc. owe after 6 years? c) What is the total cost of the building for Stampede Inc.? d) What is the total interest included in the payments?

2) _____________

3) V icki receives payments of $4170.00 at the beginning of each month from a pension fund of $208 500.00. Interest earned by the fund is 6% compounded monthly. What is the number of payments Vicki will receive?

3) _____________

4) A investor's price for a townhouse was $160 000.00. Sepaba Investments., the buyers of the rental unit, paid $40 000.00 down and financed the balance by making equal payments at the end of every six months for 25 years. Interest is 6% compounded semi-annually. a) What is the size of the semi-annual payment? b) How much will Sepaba Investments. owe after 20 years? c) What is the total cost of the building for Sepaba Investments? d) What is the total interest included in the payments?

4) _____________

5) A loan of $14 100.00 is amortized over 11 years by equal monthly payments at 5.4% compounded monthly. Construct an amortization schedule showing details of the first three payments, the fortieth payment, the last three payments, and totals.

5) _____________

6) A loan of $8380.00 is repaid by equal payments made at the end of every three months for 3 years. If interest is 7% compounded quarterly, find the size of the quarterly payments and construct an amortization schedule showing the total paid and the total cost of the loan.

6) _____________

7) R ola Inc. borrowed $42 000.00 at 7% compounded semi-annually. The loan is repaid by payments of $4700.00 due at the end of every six months. a) How many payments are needed? b) How much of the principal will be repaid by the 6th payment? c) Prepare a partial amortization schedule showing the details of the last three payments and totals.

7) _____________

8) A debt of $12 500.00 with interest at 5.5% compounded semi-annually is amortized by making payments of $1000.00 at the end of every six months. Calculate the outstanding balance after seven years.

8) _____________

9) O lfert Inc. is repaying a loan of $52 500.00 by making payments of $4700.00 at the end of every six months. If interest is 7.5% compounded semi-annually, construct an amortization schedule showing the total paid and the total cost of the loan.

9) _____________

10) B arbara borrowed $12 000.00 from the bank at 9% compounded monthly. The loan is amortized with end-of-month payments over five years. a) Calculate the interest included in the 20th payment. b) Calculate the principal repaid in the 36th payment. c) Construct a partial amortization schedule showing the details of the first two payments, the 20th payment, the 36th payment, and the last two payments. d) Calculate the totals of amount paid, interest paid, and the principal repaid.

10) _____________

11) M r. Lamb borrowed $8321.00 at 11.12% compounded monthly. He agreed to repay the loan in equal monthly payments over five years. a) What is the size of the monthly payment? b) How much of the 24th payment is interest? c) What is the principal repaid in the 37th payment? d) Prepare a partial amortization schedule showing details of the first three payments, the last three payments, and totals.

11) _____________

12) A loan of $19 000.00 is repaid by quarterly payments of $900.00 each at 8% compounded quarterly. What is the principal repaid by the 21st payment?

12) _____________

13) A loan of $12 000.00 is repaid by quarterly payments of $1200.00 each at 12% compounded quarterly. What is the principal repaid by the 11th payment?

13) _____________

14) C ody borrowed $9100.00 from his credit union. He agreed to repay the loan by making equal monthly payments for 4 years. Interest is 9% compounded monthly. a) What is the size of the monthly payments? b) How much will the loan cost him? c) How much will Cody owe after 16 months? d) How much interest will he pay in his 26th payment? e) How much of the principal will be repaid in the 18th payment? f) Prepare a partial amortization schedule showing details of the first three payments, Payments 24, 25, 26, the last three payments, and totals.

14) _____________

15) D uguid and Partners bought a property valued at $87 300.00 for $17 000.00 down and a mortgage amortized over 17 years. The firm makes equal payments due at the end of every three months. Interest on the mortgage is 6.85% compounded annually and the mortgage is renewable after five years. a) What is the size of each quarterly payment? b) What is the outstanding principal at the end of the five-year term? c) What is the cost of the mortgage for the first five years? d) If the mortgage is renewed for a further five years at 7.17% compounded semi-annually, what will be the size of each quarterly payment?

15) _____________

16) S ean and friends bought a property valued at $50 000 for $5000.00 down and a mortgage

c) t of the Whmortgage for at the first five is years? the d) If the cosmortgage is

amortized over 10 years. The group makes equal payments due at the end of every three months. Interest on the mortgage is 6.00% compounded annually and the mortgage is renewable after five years. a) What is the size of each quarterly payment? b) What is the outstanding principal at the end of the five-year term?

renewed 16) for a further five years at 9% compou nded semi-ann ually, what will be the size of each quarterly payment ?

___ ___ ___ ___ _

17) T he owner of the Pink Flamingo Motel borrowed $19 500.00 at 8.11% compounded semi-annually and agreed to repay the loan by making payments of $1110.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest?

17) _____________

18) A debt of $12 970.00 with interest at 7.23% compounded semi-annually is repaid by payments of $1880.00 made at the end of every three months. Construct an amortization schedule showing the total paid and the total cost of the debt.

18) _____________

19) T he Taylors agreed to monthly payments rounded up to the nearest $100.00 on a mortgage of $136 000.00 amortized over 15 years. Interest for the first five years was 8.5% compounded semi-annually. After 30 months, as permitted by the mortgage agreement, the Taylors increased the rounded monthly payment by 10%. a) Determine the mortgage balance at the end of the five-year term. b) If the interest rate remains unchanged over the remaining term, how many more of the increased payments will amortize the mortgage balance? c) How much did the Taylors save by exercising the increase-in-payment option?

19) _____________

20) A $30 000.00 mortgage is amortized by monthly payments over twenty years and is renewable after five years. a) If the interest rate is 8.5% compounded semi-annually, calculate the outstanding balance at the end of the five-year term. b) If the mortgage is renewed for a further three-year term at 8% compounded semi-annually, calculate the size of the new monthly payment. c) Calculate the payout figure at the end of the three-year term.

20) _____________

21) A debt of $43 100.00 is repaid by payments of $3350.00 made at the end of every six months. Interest is 8.4% compounded quarterly. a) What is the number of payments needed to retire the debt? b) What is the cost of the debt for the first five years? c) What is the interest paid in the 11th payment? d) Construct a partial amortization schedule showing details of the first three payments,

the nts, and totals. last thr ee pay me

21)

_____________ 22) A $180 000.00 mortgage is to be amortized by making monthly payments for 22.5 years.

Interest is 7.2% compounded semi-annually for a four-year term. a) Compute the size of the monthly payment. b) Determine the balance at the end of the four-year term. c) If the mortgage is renewed for a five-year term at 8.66% compounded semi-annually, what is the size of the monthly payment for the renewal term?

22) _____________

23) A $248 000.00 mortgage amortized by monthly payments over 35 years is renewable after five years. Interest is 8.12% compounded semi-annually. a) What is the size of the monthly payments? b) How much interest is paid during the first year? c) How much of the principal is repaid during the first five-year term? d) If the mortgage is renewed for a further five-year term at 7.16% compounded semi-annually, what will be the size of the monthly payments?

23) _____________

24) A debt of $42 500.00 is repaid by payments of $4850.00 made at the end of each year.

Interest is 7% compounded semi-annually. a) How many payments are needed to repay the debt? b) What is the cost of the debt for the first three years? c) What is the principal repaid in the 3rd year? d) Construct an amortization schedule showing details of the first three payments, the last three payments, and totals.

24) _____________

25) K aren receives pension payments of $2604.90 at the end of every six months from a retirement fund of . The fund earns 8.06% compounded semi-annually. a) How many payments will Karen receive? b) What is the size of the final pension payment?

25) _____________

26) T om receives pension payments of $6000.00 at the end of every six months from a retirement fund of . The fund earns 8.00% compounded semi-annually. a) How many payments will Tom receive? b) What is the size of the final pension payment?

26) _____________

27) T he owner of the Guelph Inc. borrowed $20 000.00 at 1.2% compounded semi-annually and agreed to repay the loan by making payments of $1000.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years?

27) _____________

28) A debt of $41 000.00 is repaid in monthly installments of $660.00. If interest is 8.4% compounded quarterly, what is the size of the final payment?

28) _____________

29) F ar East Imports Inc. owes $64 000.00 to be repaid by monthly payments of $2475.00.

Interest is 6.12% compounded monthly. a) How many payments will Far East Imports Inc. have to make? b) How much interest is included in the 15th payment? c) How much of the principal will be repaid by the 17th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals.

29) _____________

30) A debt of $42 000.00 is repaid in monthly installments of $525.00. If interest is 8.0% compounded quarterly, what is the size of the final payment?

30) _____________

31) A factory valued at $50 000.00 is purchased for a down payment of 28% and payments of $2000.00 at the end of every three months. If interest is 9% compounded monthly, calculate the size of the final payment.

31) _____________

32) P ayments of $1000.00 deferred for nine years are received at the end of each month from a fund of $20 000.00 deposited today at 6% compounded monthly. Calculate the size of the final payment.

32) _____________

33) A ndre has saved $152 000.00. If he withdraws $1750.00 at the beginning of every month and interest is 7.5% compounded monthly, what is the size of the last withdrawal?

33) _____________

34) L isa saved $750 000.00. If she withdraws $12 000.00 at the beginning of every month and interest is 10.8% compounded monthly, what is the size of the last withdrawal?

34) _____________

35) A ngelina has agreed to purchase her partner's share in the business by making payments of $1720.00 every three months. The agreed transfer value is $19 500.00 and interest is 9.11% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment?

35) _____________

36) J ohn has agreed to purchase his partner's share in real estate by making payments of $5650.00 every three months. The agreed transfer value is $45 200.00 and interest is 10% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment?

36) _____________

37) A contract valued at $47 500.00 requires payments of $6951.00 every six months. The first payment is due in four years and interest is 6.6% compounded semi-annually. a) How many payments are required? b) What is the size of the last payment? c) How much will be paid in total? d) How much of what is paid is interest?

37) _____________

38) A mortgage balance of $137 960.70 is to be repaid over a eleven-year term by equal monthly payments at 8.1% compounded semi-annually. At the request of the mortgagor, the monthly payments were set at $1140.00. a) How many payments will the mortgagor have to make? b) What is the size of last payment? c) Determine the difference between the total actual amount paid and the total amount required to amortize the mortgage by the contractual monthly payments.

38) _____________

39) C atrina receives payments of $1120.00 at the beginning of each month from a pension fund of $79 500.00. Interest earned by the fund is 11.36% compounded monthly. a) What is the number of payments Catrina will receive? b) What is the size of the final payment?

39) _____________

40) A contract worth $59 000.00 provides benefits of $21 000.00 at the end of each year. The benefits are deferred for six years and interest is 8.95% compounded quarterly. a) How many payments are to be made under the contract? b) What is the size of the last benefit payment?

40) _____________

41) A $176 000.00 mortgage is to be repaid over a 21-year period by monthly payments rounded up to the next higher $100.00. Interest is 9.44% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment.

41) _____________

42) T he Acme Parts Company is repaying a debt of $15 000.00 by payments of $1410.00 made at the end of every three months. Interest is 7.2% compounded monthly. a) How many payments are needed to repay the debt? b) What is the size of the final payment?

42) _____________

43) A debt of $17 000.00 is repaid by quarterly payments of $1430.00. If interest is 6.12% compounded quarterly, what is the size of the final payment?

43) _____________

44) A $151 000.00 mortgage is to be repaid over a 17-year period by monthly payments rounded up to the next higher $50.00. Interest is 8.2% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. c) Calculate the amount of interest saved by rounding the payments up to the next higher $50.00.

44) _____________

45) A demand (variable rate) mortgage of $137 000.00 is amortized over 20 years by equal monthly payments. After 21 months the original interest rate of 6% p.a was raised to 6.6% p.a. Three years after the mortgage was taken out, it was renewed at the request for the mortgagor for a five-year term at a fixed rate of 7.25% p.a. a) Calculate the mortgage balance after 21 months. b) Compute the size of the new monthly payment at the 6.6% rate of interest. c) Determine the mortgage balance at the end of the five-year term.

45) _____________

46) A bank has an interest rate of 5% compounded semi annually for a five year closed mortgage. If the mortgage has monthly payments, calculate the effective rate of interest per each term.

46) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

47) H ow much interest is paid in total on a 3-year loan for $28 600 (rounded to the nearest dollar)?

47) ______ The interest rate is 8.7% compounded monthly and the payments are monthly.

A) $ 3999 B) $ 3997 C) $ 3779 D) $ 4000 E) $ 3977 48) H ow much interest is paid in total on a 1-year loan for $5000? The interest rate is 12 % compounded monthly and the payments are monthly.

A) $ 330.88 B) $ 300.88 C) $ 333.00 D) $ 333.88

48) ______ E) $ 330.00

49) H ow much principal is repaid in the 74th payment interval on a $142 300 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 7.44% compounded monthly.

A) $ 572.16 B) $ 257.16 C) $ 527.16 D) $ 275.16 E) $ 574.16

49) ______

50) H ow much total principal is repaid between the 1st and 7th payment interval of a 4.5-year loan for $4887 at an interest rate of 7.4% compounded monthly and the payments are also monthly.

Round your answer to the nearest dollar.

A) $ 546 B) $ 445 C) $ 456 D) $ 423 E) $ 556

50) ______

51) W hat is the outstanding balance after 23 payments on a 20-year mortgage that has monthly

payments

of 51) $1062.32 and an interest rate of 6.6% compou nded monthly ?

___ ___

A) $ 143 402.54 B) $ 134 322.54 C) $ 144 302.54 D) $ 134 302.54 E) $ 134 402.54 52) W hat is the monthly payment size of a 21-year mortgage for $169 600 and an interest rate of 7.2% compounded semi-annually?

A) $ 1296.11 B) $ 1926.11 C) $ 1211.96 D) $ 1222.96 E) $ 1169.11

52) ______

53) W hat is the monthly payment size of a 25-year mortgage for $100 000 and an interest rate of 6% compounded semi-annually?

A) $ 600.81 B) 6 33.81 C) $ 630.81 D) $ 1639.81 E) $ 639.81

53) ______

54) H ow much interest is paid in the 53rd monthly payment interval of a loan for $43 200? The loan is amortized at a rate of 9.17% compounded annually over 7 years.

A) $ 400.11 B) $ 144.11 C) $ 444.11 D) $ 411.11 E) $ 111.11

54) ______

55) H ow much total interest is paid between the 6th and 13th payments on a loan that has monthly payments for 3 years and an original principal of $14 750? The loan rate is 6.6% compounded quarterly.

A) $ 504 B) $ 450 C) $ 545 D) $ 405 E) $ 540

55) ______

56) H ow much interest is paid after the first payment interval on a $100 000 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 6% compounded semi-annually. Round your answer to the nearest dollar.

A) $ 494 B) $ 527 C) $ 401 D) $ 146 E) $ 640

56) ______

57) W hat is the outstanding balance after the 23rd payment interval of a 16-year loan for $14 734 with semi-annual payments and an interest rate of 5.95% compounded quarterly?

A) $ 5566.21 B) $ 5562.21 C) $ 5625.21 D) $ 5526.21 E) $ 5652.21

57) ______

58) W hat is the size of the final unequal payment of a loan that has 69.47 payments and the equal payment size is $215.64? The interest rate is 3.1% per compounding period.

A) $ 99.10 B) $ 120.17 C) $ 102.17 D) $ 99.17 E) $ 210.17

58) ______

59) W hat is the size of the final unequal payment of a loan for $2723.44 if it has quarterly payments of $500? The interest rate is 8.4% compounded quarterly.

A) $ 432.14 B) $ 424.14 C) $ 414.14 D) $ 441.14 E) $ 423.14

59) ______

60) M unawar is a first time buyer and bought a brand new townhouse in Oshawa for $212 000. He has saved enough to make a 5% down payment and will have to pay mortgage loan insurance at

2.75of the % mortga

ge 60) balance.

Munawa r wants the insuranc e premium added to the maximu m allowabl e loan balance.

Determi ne his maximu m initial mort gag e balance.

___ ___

A) $ 212 000 B) $ 201 400 C) $ 5538.50 D) $ 206 938.50 E) $ 212 477 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

61) _ ____________ 61) L i bought a house in Whitby for $305 000 5 years ago. He made a down payment of $46 000 and financed the house through Scotia Bank with a mortgage amortized over 25 years. Interest for the first 5 years was 4.4% compounded semi-annually and payments were made monthly. At the end of the 5 year term, Li has renewed the mortgage for another 5 years, at 3.7% compounded semi-annually. What is the monthly payment for the second term?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

62) ______ 62) S hiva transferred his mortgage of $145 981 from CIBC to BMO for the remaining amortization period at 3.59% compounded semi-annually. He has agreed to pay rounded up payment of $2000 every month. How long will it take Shiva to pay-off his mortgage?

A) 8 years B) 6 years 11 months C) 6 years 6 months D) 1 0 years E) 6 years 1 months SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

63) J ohn makes a monthly mortgage payment of $1940 at the end of every month. On mont on December nth31? The October 31, his loan balance was $145 000 after making a payment of $1940. His financial ly effective institute allows him to make one additional payment equal to or less than his monthly paymonthly fixed payment, once a month. Taking advantage of that, John made a payment of $1500 on merate is 0.306%.

November 15 and $1900 on December 20. What is his year-end balance after making

63)

_____________ 64) _____________

64) P eter worked for GM for 31 years. On his retirement, he opted a pension buy-out and received $1 467 921. He invested his money in an annuity that provided for payments of $9000 at the end of every month. If interest is 4.4% compounded monthly, determine the size of the final payment.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

65) T he Korean community is planning to buy a place for their community in Markham valued at $1 65) ______ 697 000. They made a down payment of 10% and payments of $400 000 are required at the end of every 3 months. If interest is 9% compounded monthly, what is the size of the final payment A) $ 15 685 B) $ 355 547 C) $ 131 373 D) $ 205 512 E) $ 48 381 66) ______

66) K irk had accumulated $975 000 in his RRSP at the age of 55. He converted the RRSP into an RRIF at that time and started withdrawing $5000 at the beginning of each month. If interest is 4.5% compounded monthly, what is the size of his final withdrawal?

A) $ 4947.31 B) $ 1409.28 C) $ 1739.27 D) $ 412.07

E) $ 33.82

67) H onda is selling a 2014 Accord Coupe Ex M for $30 056.90 including freight, PDI and all applicable fees. The lease payment of $382 is due at the beginning of each month. If the interest rate compounded annually is 3.99% and the residual value is $13 200, determine the size of the final lease payment.

A) $ 285.95 B) $ 5.92 C) $ 214.13 D) $ 150.28 E) $ 240.51

67) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

68) Z ara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% 68) _____________ compounded annually and amortized by payments of $378 at the end of every month.

She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made?

69) R ebecca bought a house in Oshawa and took a $259 000 mortgage amortized by monthly payments over 25 years that is renewable after 5 years. If interest is 4.4% compounded semi-annually, what is the outstanding balance at the end of the five year term?

If the mortgage is renewed for a further 5 year term at 3.7% compounded semi-annually, what is the balance after the second 5 year term?

69) _____________

70) A manda borrowed $57 800 from BMO to buy a 2014 E-Class Mercedes Sedan at 3.9% compounded monthly. Payments of $2056 are required at the end of every month.

Construct a complete amortization schedule in MS Excel showing the details of all payments and the total.

70) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

71) R ebecca borrowed $20 400 from her local credit union to pay for the upgrades in her kitchen. She 71) ______ wants to pay the loan with monthly payments over the period of 3 years. What is the interest included in the tenth payment, if the rate charged by the credit union is 6% monthly?

A) $ 75.48 B) $ 15 638.09 C) $ 5698.49 D) $ 21 336.56

E) $ 78.19

72) S amuel paid $850 000.00 for a new house. He paid $140 000.00 down and financed the balance by making equal payments at the end of every month for 25 years. Interest is 5.1% compounded monthly.

What is the total cost of the building for Samuel?

A) $ 1 404 198 B) $ 1 257 618 C) $ 1 397 618 D) $ 1 400 254

E) $ 1 412 058

72) ______

73) K eith receives payments of $14 500.00 at the beginning of each month from a pension fund of $1 500 73) ______ 000.00. Interest earned by the fund is 7.6% compounded monthly. What is the number of payments Keith will receive? B) 1 69 A) 1 04 C) 1 40 D) 1 67 E) 1 56 74) A car loan of $46 700.00 is amortized over 8 years by equal monthly payments at 2.9% compounded monthly. What is the monthly payment and outstanding principal balance after 40 payments.

A) $ 545.66 and $18 153.12 B) $ 545.66 and $28 546.88 C) $ 486.46 and $27 241.67 D) $ 544.34 and $28,478.06

74) ______

75) Z am Zam Inc. borrowed $142 000.00 at 4.4% compounded semi-annually. The loan is repaid by payments of $9700.00 due at the end of every six months. How many payments are needed and what will be the outstanding balance after 8 payments?

A) 1 8 and $85 158 B) 1 8 and $56 842 C) 1 8 and $81 520 D) 1 8 and $60 480 E) 1 5 and $64 800

75) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

76) _ ____________ 76) B ig Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded quarterly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan?

77) B ig Printing Inc. is repaying a loan of $37 300.00 by making payments of $2300.00 at the end of every three months. If interest is 3.5% compounded monthly, construct an amortization schedule showing the total paid and the total cost of the loan. What is the total interest paid and cost of the loan?

77) _ ____________

78) T ariq bought a commercial property valued at $71 000 for $26 000.00 down and a mortgage amortized over 10 years. He makes equal payments due at the end of every quarter. Interest on the mortgage is 6% compounded annually. a) What will be the payment amount needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest?

78) _ ____________

79) T he owner of the Pink Flamingo Motel borrowed $19 500.00 at 8.11% compounded semi-annually and agreed to repay the loan by making payments of $1110.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years? c) How much of the payments made at the end of five years will be interest?

79) _ ____________

80) T he Taylors agreed to monthly payments rounded up to the nearest $100.00 on a mortgage of $136 000.00 amortized over 15 years. Interest for the first five years was 8.5% compounded semi-annually. After 30 months, as permitted by the mortgage agreement, the Taylors increased the rounded monthly payment by 10%. a) Determine the mortgage balance at the end of the five-year term. b) If the interest rate remains unchanged over the remaining term, how many more of

the nts will incramortize the easmortgage ed balance? payc) How much medid the Taylors

save by 80) exercisin g the increasein-paym ent option?

___ ___ ___ ___ _

81) A $30 000.00 mortgage is amortized by monthly payments over twenty years and is renewable after five years. a) If the interest rate is 8.5% compounded semi-annually, calculate the outstanding balance at the end of the five-year term. b) If the mortgage is renewed for a further three-year term at 8% compounded semi-annually, calculate the size of the new monthly payment. c) Calculate the payout figure at the end of the three-year term.

81) _____________

82) A debt of $43 100.00 is repaid by payments of $3350.00 made at the end of every six months. Interest is 8.4% compounded quarterly. a) What is the number of payments needed to retire the debt? b) What is the cost of the debt for the first five years? c) What is the interest paid in the 11th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals.

82) _____________

83) A $180 000.00 mortgage is to be amortized by making monthly payments for 22.5 years.

Interest is 7.2% compounded semi-annually for a four-year term. a) Compute the size of the monthly payment. b) Determine the balance at the end of the four-year term. c) If the mortgage is renewed for a five-year term at 8.66% compounded semi-annually, what is the size of the monthly payment for the renewal term?

83) _____________

84) A $248 000.00 mortgage amortized by monthly payments over 35 years is renewable after five years. Interest is 8.12% compounded semi-annually. a) What is the size of the monthly payments? b) How much interest is paid during the first year? c) How much of the principal is repaid during the first five-year term? d) If the mortgage is renewed for a further five-year term at 7.16% compounded semi-annually, what will be the size of the monthly payments?

84) _____________

85) A debt of $42 500.00 is repaid by payments of $4850.00 made at the end of each year.

Interest is 7% compounded semi-annually. a) How many payments are needed to repay the debt? b) What is the cost of the debt for the first three years? c) What is the principal repaid in the 3rd year? d) Construct an amortization schedule showing details of the first three payments, the last three payments, and totals.

85) _____________

86) K aren receives pension payments of $4570.00 at the end of every six months from a retirement fund of . The fund earns 8.06% compounded semi-annually. a) How many payments will Karen receive? b) What is the size of the final pension payment?

86) _____________

87) T om receives pension payments of $6000.00 at the end of every six months from a . The fund earns 8.00% compounded semi-annually. retirement fund of a) How many payments will Tom receive? b) What is the size of the final pension payment?

87) _____________

88) T he owner of the Guelph Inc. borrowed $20 000.00 at 1.2% compounded semi-annually and agreed to repay the loan by making payments of $1000.00 at the end of every three months. a) How many payments will be needed to repay the loan? b) How much will be owed at the end of five years?

88) _____________

89) A debt of $41 000.00 is repaid in monthly installments of $660.00. If interest is 8.4% compounded quarterly, what is the size of the final payment?

89) _____________

90) F ar East Imports Inc. owes $64 000.00 to be repaid by monthly payments of $2475.00.

Interest is 6.12% compounded monthly. a) How many payments will Far East Imports Inc. have to make? b) How much interest is included in the 15th payment? c) How much of the principal will be repaid by the 17th payment? d) Construct a partial amortization schedule showing details of the first three payments, the last three payments, and totals.

90) _____________

91) A debt of $40 000.00 is repaid in monthly installments of $500.00. If interest is 8.0% compounded quarterly, what is the size of the final payment?

91) _____________

92) A factory valued at $100 000.00 is purchased for a down payment of 28% and payments

92) _____________

of $4000.00 at the end of every three months. If interest is 9% compounded monthly, calculate the size of the final payment.

93) P ayments of $1000.00 deferred for nine years are received at the end of each month from a fund of $20 000.00 deposited today at 6% compounded monthly. Calculate the size of the final payment.

93) _____________

94) A ndre has saved $152 000.00. If he withdraws $1750.00 at the beginning of every month and interest is 7.5% compounded monthly, what is the size of the last withdrawal?

94) _____________

95) L isa saved $750 000.00. If she withdraws $12 000.00 at the beginning of every month and interest is 12% compounded monthly, what is the size of the last withdrawal?

95) _____________

96) A ngelina has agreed to purchase her partner's share in the business by making payments of $1720.00 every three months. The agreed transfer value is $19 500.00 and interest is 9.11% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment?

96) _____________

97) J ohn has agreed to purchase his partner's share in real estate by making payments of $5000.00 every three months. The agreed transfer value is $40 000.00 and interest is 10% compounded annually. If the first payment is due at the date of the agreement, what is the size of the final payment?

97) _____________

98) A contract valued at $47 500.00 requires payments of $6951.00 every six months. The first payment is due in four years and interest is 6.6% compounded semi-annually. a) How many payments are required?

b) is the size of the Whlast payment? at c) How much

will be paid in total? d) How much of what is paid is interest?

98)

___ ___ ___ ___ _

99) A mortgage balance of $137 960.70 is to be repaid over an eleven-year term by equal monthly payments at 8.1% compounded semi-annually. At the request of the mortgagor, the monthly payments were set at $1140.00. a) How many payments will the mortgagor have to make? b) What is the size of last payment? c) Determine the difference between the total actual amount paid and the total amount required to amortize the mortgage by the contractual monthly payments.

99) _____________

100) C atrina receives payments of $1120.00 at the beginning of each month from a pension

100) ____________

fund of $79 500.00. Interest earned by the fund is 11.36% compounded monthly. a) What is the number of payments Catrina will receive? b) What is the size of the final payment?

101) A contract worth $59 000.00 provides benefits of $21 000.00 at the end of each year. The benefits are deferred for six years and interest is 8.95% compounded quarterly. a) How many payments are to be made under the contract? b) What is the size of the last benefit payment?

101) ____________

102) A $176 000.00 mortgage is to be repaid over a 21-year period by monthly payments rounded up to the next higher $100.00. Interest is 9.44% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment.

102) ____________

103) T he Acme Parts Company is repaying a debt of $15 000.00 by payments of $1410.00

103) ____________

made at the end of every three months. Interest is 7.2% compounded monthly. a) How many payments are needed to repay the debt? b) What is the size of the final payment?

104) A debt of $17 000.00 is repaid by quarterly payments of $1430.00. If interest is 6.12% compounded quarterly, what is the size of the final payment?

104) ____________

105) A $151 000.00 mortgage is to be repaid over a 17-year period by monthly payments rounded up to the next higher $50.00. Interest is 8.2% compounded semi-annually. a) Determine the number of rounded payments required to repay the mortgage. b) Determine the size of the last payment. c) Calculate the amount of interest saved by rounding the payments up to the next higher $50.00.

105) ____________

106) A demand (variable rate) mortgage of $137 000.00 is amortized over 20 years by equal monthly payments. After 21 months the original interest rate of 6% p.a was raised to 6.6% p.a. Three years after the mortgage was taken out, it was renewed at the request for the mortgagor for a five-year term at a fixed rate of 7.25% p.a. a) Calculate the mortgage balance after 21 months.

b) of the new Commonthly pute payment at the the 6.6% rate of size interest.

____ ____ ____

c) 106) Determi ne the mort gag e balance at the end of the five-year term.

107) ____________

107) A bank has an interest rate of 5% compounded semi annually for a five year closed mortgage. If the mortgage has monthly payments, calculate the effective rate of interest per each term.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

108) H ow much interest is paid in total on a 3-year loan for $28 600 (rounded to the nearest dollar)?

108) _____ The interest rate is 8.7% compounded monthly and the payments are monthly.

A) $ 3977 B) $ 3779 C) $ 3999 D) $ 4000 E) $ 3997 109) H ow much interest is paid in total on a 1-year loan for $5000? The interest rate is 12 % compounded monthly and the payments are monthly.

A) $ 333.00 B) $ 330.00 C) $ 300.88 D) $ 333.88

109) _____ E) $ 330.88

110) H ow much principal is repaid in the 74th payment interval on a $142 300 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 7.44% compounded monthly.

A) $ 527.16 B) $ 574.16 C) $ 275.16 D) $ 257.16 E) $ 572.16

110) _____

111) H ow much total principal is repaid between the 1st and 7th payment interval of a 4.5-year loan for $4887 at an interest rate of 7.4% compounded monthly and the payments are also monthly.

Round your answer to the nearest dollar.

A) $ 445 B) $ 546 C) $ 556 D) $ 423 E) $ 456

111) _____

112) W hat is the outstanding balance after 23 payments on a 20-year mortgage that has monthly payments of $1062.32 and an interest rate of 6.6% compounded monthly?

A) $ 134 402.54 B) $ 134 322.54 C) $ 144 302.54 D) $ 143 402.54 E) $ 134 302.54

112) _____

113) W hat is the monthly payment size of a 21-year mortgage for $169 600 and an interest rate of 7.2% compounded semi-annually?

A) $ 1169.11 B) $ 1296.11 C) $ 1222.96 D) $ 1211.96 E) $ 1926.11

113) _____

114) W hat is the monthly payment size of a 25-year mortgage for $100 000 and an interest rate of 6% compounded semi-annually?

A) 6 33.81 B) $ 600.81 C) $ 639.81 D) $ 630.81 E) $ 1639.81

114) _____

115) H ow much interest is paid in the 53rd monthly payment interval of a loan for $43 200? The loan is amortized at a rate of 9.17% compounded annually over 7 years.

115) _____

A) $ 144.11

B) $ 411.11

C) $ 400.11

D) $ 111.11

E) $ 444.11

116) H ow much total interest is paid between the 6th and 13th payments on a loan that has monthly payments for 3 years and an original principal of $14 750? The loan rate is 6.6% compounded quarterly.

A) $ 540 B) $ 545 C) $ 450 D) $ 405 E) $ 504

116) _____

117) H ow much interest is paid after the first payment interval on a $100 000 mortgage? The mortgage is amortized over 25 years and the payments are monthly. The interest rate is 6% compounded semi-annually. Round your answer to the nearest dollar.

A) $ 146 B) $ 494 C) $ 527 D) $ 640 E) $ 401

117) _____

118) W hat is the outstanding balance after the 23rd payment interval of a 16-year loan for $14 734 with semi-annual payments and an interest rate of 5.95% compounded quarterly?

A) $ 5652.21 B) $ 5625.21 C) $ 5562.21 D) $ 5566.21 E) $ 5526.21

118) _____

119) W hat is the size of the final unequal payment of a loan that has 69.47 payments and the equal payment size is $215.64? The interest rate is 3.1% per compounding period.

A) $ 102.17 B) $ 99.10 C) $ 210.17 D) $ 99.17 E) $ 120.17

119) _____

120) W hat is the size of the final unequal payment of a loan for $2723.44 if it has quarterly payments of $500? The interest rate is 8.4% compounded quarterly.

A) $ 414.14 B) $ 424.14 C) $ 432.14 D) $ 423.14 E) $ 441.14

120) _____

121) M unawar is a first time buyer and bought a brand new townhouse in Oshawa for $212 000. He has saved enough to make a 5% down payment and will have to pay mortgage loan insurance at 2.75% of the mortgage balance. Munawar wants the insurance premium added to the maximum allowable loan balance. Determine his maximum initial mortgage balance.

A) $ 212 477 B) $ 206 938.50 C) $ 201 400 D) $ 5538.50 E) $ 212 000

121) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

122) L i bought a house in Whitby for $305 000 5 years ago. He financed the house through 122) ____________ Scotia Bank with a mortgage of $259 000 amortized over 25 years. Interest for the first 5 years was 4.4% compounded semi-annually and payments were made monthly. At the end of the 5 year term, Li has renewed the mortgage for another 5 years, at 3.7% compounded semi-annually. What is the monthly payment for the second term?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

123) S hiva transferred his mortgage of $145 981 from CIBC to BMO for the remaining amortization 123) _____ period at 3.59% compounded semi-annually. He has agreed to pay rounded up payment of $2000 every month. How long will it take Shiva to pay-off his mortgage?

A) 6 years 1 months B) 6 years 6 months C) 1 0 years D) 6 years 11 months E) 8 years SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

124) J ohn makes a monthly mortgage payment of $1940 at the end of every month. On October 31, his loan balance was $145 000 after making a payment of $1940. His financial institute allows him to make one additional payment equal to or less than his monthly payment, once a month. Taking advantage of that, John made a payment of $1500 on November 15 and $1900 on December 20. What is his year-end balance after making monthly payment on December 31? The effective monthly fixed rate is 0.306%.

124) ____________

125) P eter worked for GM for 31 years. On his retirement, he opted for a pension buy-out and received $1 467 921. He invested his money in an annuity that provided for payments of $9000 at the end of every month. If interest is 4.4% compounded monthly, determine the size of the final payment.

125) ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

126) T he Korean community is planning to buy a place for their community in Markham valued at $1 126) _____ 697 000. They made a down payment of 10% and payments of $400 000 are required at the end of every 3 months. If interest is 9% compounded monthly, what is the size of the final payment?

A) $ 131 373 B) $ 48 381 C) $ 205 512 D) $ 15 685 E) $ 355 547 127) K irk had accumulated $975 000 in his RRSP at the age of 55. He converted the RRSP into an RRIF at that time and started withdrawing $5000 at the beginning of each month. If interest is 4.5% compounded monthly, what is the size of his final withdrawal?

A) $ 4947.31 B) $ 412.07 C) $ 33.82 D) $ 1409.28 E) $ 1739.27

127) _____

128) H onda is selling a 2014 Accord Coupe Ex M for $30 056.90 including freight, PDI and all applicable fees. The lease payment of $382 is due at the beginning of each month. If the interest rate compounded annually is 3.99% and the residual value is $13 200, determine the size of the final lease payment.

A) $ 240.51 B) $ 285.95 C) $ 214.13 D) $ 150.28 E) $ 5.92

128) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

129) Z ara bought a 2013 Kia Forte by taking a $19 500 loan from BMO with interest at 5.99% 129) ____________ compounded annually and amortized by payments of $378 at the end of every month.

She receives an inheritance after making 36 payments and plans to pay the balance in full. What is the payout figure just after the last regular payment made?

130) R ebecca bought a house in Oshawa and took a $259 000 mortgage amortized by monthly payments over 25 years that is renewable after 5 years. If interest is 4.4% compounded semi-annually, what is the outstanding balance at the end of the five year term?

130) ____________

If the mortgage is renewed for a further 5 year term at 3.7% compounded semi-annually, what is the balance after the second 5 year term?

131) A manda borrowed $57 800 from BMO to buy a 2014 E-Class Mercedes Sedan at 3.9% compounded monthly. Payments of $2056 are required at the end of every month.

Construct a complete amortization schedule in MS Excel showing the details of all payments and the total.

131) ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

132) R ebecca borrowed $20 400 from her local credit union to pay for the upgrades in her kitchen. She 132) _____ wants to pay the loan with monthly payments over a period of 3 years. What is the interest included in the tenth payment, if the rate charged by the credit union is 6% monthly?

A) $ 15 638.09 B) $ 75.48 C) $ 78.19 D) $ 21 336.56 E) $ 5698.49

133) H owden is interested in purchasing a jet ski for $15 834. He chose to finance this purchase across a 5-year term at an interest rate of 3.6% compounded quarterly. For the first, second and third payment, how much of the payment is interest?

A) $ 153.98, $147.42, $145.88 B) $ 726.13, $728.67, $735.23 C) $ 864.64, $864.64, $864.64 D) $ 142.51, $135.97, $129.41 E) $ 142.51, $142.51, $142.51

133) _____

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

134) J acob took out a loan of $11 256 to furnish his new studio apartment that charged an 134) ____________ interest rate of 10.8% compounded monthly over 3 and a half years. After the third payment, how much of the loan remains to be paid?

135) H uzaifa received a loan of $43 610 from the bank at 9.1% compounded semi-annually to help start his business. As per the terms of the loan, Huzaifa must pay off the loan in 8 years and 6 months. What's the loan balance after 4 years?

135) ____________

136) T he Allenoyin Association needed to cover some of their off-season costs and took a loan of $24 755 from the bank at 6.24% compounded bi-weekly. Payment of $680 were expected to be made every other week. How many payments are necessary?

136) ____________

137) M irsu Materials acquired a loan of $68 900 at 5.4% compounded quarterly to begin operations. They must make equal monthly instalments for the next 11 years to pay off this loan. After the first payment, how much does Mirsu Materials still owe?

137) ____________

138) I n the contract for actor Dwayne "The Large Pebble" Johnson's new movie, he has agreed to receive payments of $15 000 every month for the next 6.5 years. At the time the contract was established, he interest rate was 8.6% compounded semi-annually. Using the prospective method, determine the present value of all these payments.

138) ____________

139) A t the end of every year, The Train Trust must make a payment of $675 to the bank towards their loan of . The loan agreement has an interest rate of 2.3% compounded annually. What's the size of the final payment?

139) ____________

140) I vey Inc has decided to lease a new building for operations. The lease obligation is $10 900 with lease payments of $315 made every three months. If the interest rate is 4.4% compounded quarterly, what's the size of the last payment?

140) ____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

141) _ _______ is an affordability measure used to determine qualification for mortgage default 141) _____ insurance. It takes the monthly housing costs such as property tax as a percentage of the gross household monthly income, and to qualify, this measure must be no more than ________.

A) T otal debt service; 35% B) G ross debt service; 35% C) V ariable equity mortgage ratio; 25% D) T otal debt service; 25% E) G ross debt service; 25%

1) a ) PV = 10 000.00; n = 2(4) = 8; i =

= 2% = 0.02; I/Y = 8; P/Y = C/Y = 4;

10000 = PMT 10000 = PMT[7.32548144] PMT = PMT = $1365.10 Programmed solution:

b) Amortization Table (done using Excel):

c) PMT = 1365.10; n = 5(number of payments left); i = 0.02 PV = 1365.10

= $6434.34

Programmed solution:

The difference from the chart is due to rounding.

2) a) 68850.00 = PMT 68850.00 = PMT(15.06140866) $4571.29 = PMT b) PVn = 4571.29

= 4571.29(8.913452135) = $40745.97

c)

= 16250.00 + 127996.12 = $144246.12 Total d) Total interest = 144246.12 - 85100.00 = $59 146.12 paid = 16250.00 + 28(4571.29) 3) 208500 = 4170 (1.005) 49.75124378 = 0.2487562189 = 1 - 1.005-n -n ln 1.005 = ln 0.7512437811 -n(0.0049875415) = -0.2860250712 n = 57.3479079 payments 4)

a) 120000.00 = PMT 120000.00 = PMT(25.72976401) $4663.86 = PMT b) An = 4663.86

= 4663.86(8.530202837) = $39 783.67

c)

5)

Total paid = 40000.00 + 50(4663.86) = 40000.00 + 233193.00 = $273 193.00 d) Total interest = 273193.00 - 160000.00 = $113 193.00 14100 = PMT

14100 = PMT(99.3665357) $141.90 = PMT Partial Amortization Table (rounded to the nearest cent):

PVn = 141.90

= 141.9 0(75.

6)

8552917) = 10763.87 PVn = 141.90 = 141.90(2 .97320114) = $421.90 8380.00 = PMT 8380.00 = PMT(10.7395497) $780.29 = PMT Amortization Schedule (done using Excel):

7) a)

n=

=

The number of payments is 11. b)

PVn = 4700 = 4700.00(5.2509011) = $24 679.24 Interest = 24679.24(.035) = $863.77 Principal repaid = 4700.00 - 863.77 = $3836.23

c) Partial Amortization (rounded the nearest cent):

=

= 10.90304697

PVn = 4700 = 4700(2.71554292) = $12 763.05

8) P V = 12 500.00; n = 7(2) = 14; i =

% = 0.0275; I/Y = 5.5; P/Y = C/Y = 2

The accumulated value of the original principal after five years FV = 12500(1.0275)14 = $18 274.93 The accumulated value of the first ten payments FV = 1000

= $16 799.79

The outstanding balance after five years is $18 274.93 - $16 799.79 = $1475.14 9) A mortization Schedule (done using Excel)

10) a ) PV = 12 000.00; n = 5(12) = 60; i = 12000 = PMT 12000 = PMT[48.17337352] PMT =

= 0.75% = 0.0075; I/Y = 9; P/Y = C/Y = 12;

paid in the 20th payment is 8764.10(0.0075) = $65.73

PMT = b) The outstanding balance after the 35th payment: $249.

PV = 249.10 = $5659.24 10 Prog ram Programmed solution: med solut ion:

The interest included in Payment 36 is 5659.24(0.0075) = $42.44 The principal repaid by Payment 36 is 249.10 - 42.44 = $206.66 c) The outstanding balance after Payment 58: The PV = 249.10 = $492.65 outst andi ng bala nce Programmed solution: after the 19th pay mentThe total amount paid is 249.10(59) + 249.09 = $14 945.99 The principal repaid is $12 000.00.

The total interest paid is 14945.99 - 12000.00 = $2945.99 : PV = 249.1Partial Amortization Schedule: (rounded to the nearest cent): 0

= $87 6 4.10 Prog ram med solu t ion:

The inter est 11)

a) 8321.00 = PMT 8321.00 = PMT(45.86668688)

$ 181.4 2=

PVn = 181.42 = 181.42(21.42999316) = $3887.83 181.42 - 3887.83(.009266667) = 181.42 - 36.03 = $145.39

PMT b)

d) Partial amortization table (done by hand, rounded to the nearest cent).

PVn = 181.4 2

= 181.42(31.2027738) = $5660.81 5 660.8 1(0.0 0926 6667) = $52.4 6 c) 12) n=

=

=

= 27.7016753

PVn = 900

= 900(7.07263077) = $6365.37 Interest = 6365.37(0.02) = $127.31 Principal repaid = 900.00 - 127.31 = $772.69 13) n=

=

=

PVn = 1200 = 1200(10) = $12000 Interest = 12000(0.03) = $360 Principal repaid = 12000 - 360 = $11 640 14)

a) 9100.00 = PMT 9100.00 = PMT(40.1847819) $226.45 = PMT b) Total paid = 226.45(48) = $10 869.60 Amount borrowed = 9100.00 Cost of financing = $ 1769.60

= 12.06662371

c)

5

= 226.45(21.0533147) = $4767.52

Interest in 26th payment = 4767.52(.0075) = $35.76 Bala nce e) Balance after 17th payment: n = 48 - 17 = 31 after An = 226.45 = 226.45(27.5683178) = $6242.85 16 mont Interest in 18th payment = 6242.85(.0075) = $46.82 hs: n = 48 Principal repaid = 226.45 - 46.82 = $179.63 16 = 32 f) Balance after 23rd payment: n = 48 - 23 = 25 An = 226.4

An = 226.45

= 226.45(22.7187555) = $5144.66

A n = 226.45

= 226.45(2.9555562) = $669.29

5 Partial Amortization Table (each step rounded to the nearest cent). = 226.4 5(28.

3556 505) = $642 1.14 d) Bala nce after 25th pay ment :n= 48 25 = 23 An = 226.4 15)

a)

p=

- 1 = 0.01670189

70300 = PMT 70300 = PMT(40.4615867) $1737.45 = PMT b) PVg = 1737.45

$3=4 1774397.0.405 ( 32.8376079) = $57053.70 Principal repaid = 70300.00 - 57053.70 = $13 246.30 Interest cost = 34749.00 - 13246.30 = $21 502.70

c) Total paid = 1737.

45(20 ) = 16)

d) p =

- 1 = 0.01776716

57053.70 = PMT 57053.70 = PMT(32.1146991) $1776.56 = PMT p= a)

- 1 = 0.01467385 45000 = PMT 45000 = PMT(30.0947153) $1495.28 = PMT

b) PVg = 1,495.28 = 1519.67(17.2239653) = $25754.65 c)

Total paid = 1495.28(20)

= $29 905.60

Principal repaid = 45000.00 - 25754.65 = $19 245.35 Interest cost = 29905.60 -19245.35 = $10 660.25 d) p =

- 1 = 0.022252415

25754.65 = PMT

17)

25754.65 = PMT(16.00151134) $1609.51 = PMT p= a)

- 1 = 0.02007353

n=

=

=

= 21.879938

The number of payments is 22.

18)

b) PVg = 1110

= 1110(1.826974901) = $2027.94

c)

= $22 200 = $17 472.06 = $4727.94

p=

Total paid = 1110(20) Principal repaid = 19500.00 - 2027.94 Interest = 22200.00 - 17472.06 - 1 = 0.01791454

Loan Amortization Schedule (done in Excel):

19)

a) p =

- 1 = 0.00696106

136000 = PMT 136000 = PMT(102 .442161) $1327.58 = PMT Payment rounded to $1400.00 FV = 136000(1.00696106)60 = 136000.00(1.51621447) = $206 205.17 FVn = 1400

= 1400(74.1574287) = $103 820.40

Balance = 206205.17 - 103820.40 = $102 384.77

b) n =

=

=

= 89.5752224 payments

c) n =

=

=

= 102.561916 payments

PVn = 1400.00 = 1400.00(0.58865597) = $782.43 782.43(1.00696106) = $787.88 PVn = 1540.00 = 1540.00(0.57208741) = $881.01 881.01(1.00696106) = $887.15 Rounded payments = 102(1400.00) + 787.88 = 142800.00 + 787.88 = $143 587.88 Increased rounded payments = 89(1540.00) + 887.15 = 137,060.00 + 887.15 = $137 947.15 Savings = 143 587.88 - 137 947.15 = $5640.73 20) a ) PV = 30 000.00; n = 20(12) = 240; i = = 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

=

p ) = 180 = (1 + PV = 257.57 i)c = (1 + PV = $26 386.10 0.042 5)1/6 Programmed solution: -1= 0.006 9610 62 3 0000 = PMT

3 0000 = PMT [116.

b) The outstanding balance of $26 386.10 is to be amortized the remaining 15 years.

PV = 26 386.10; n = 15(12) = 180; i = p=

=

= 0.04; I/Y = 8; P/Y = 12; C/Y = 2; c =

- 1 = 0.006558197

26386.10 = PMT

4742 048]

26386.10 = PMT[105.4682161] PMT = $250.18

P MT = $257.Programmed solution: 57

T he num ber of c) At the end of the 3 year term the outstanding payments is 144. outst PV = 250.18 andi ng PV = $23 265.45 pay ment Programmed solution: s after five years is 15(12 21)

a)

p=

n=

- 1 = 0.042441

=

=

= 18.9998078

=

18 3350.00(7.3531857) = $24 633.17 full Total paid = 10(3350.00) = $33 500.00 pay Principal repaid = 43100.00 - 24633.17 = $18 466.83 ment Cost = 33500.00 - 18466.83 = $15 033.17 s are needc) $24633.17(.042441) = $1045.46 ed and d) Partial Amortization Schedule (rounded to 2 decimal places): one small er pay ment . b) PVg = 3350.

00

= 22)

a)

p=

- 1 = 0.00591193

180000.00 = PMT 180000.00 = PMT(134.708006) $1336.22 = PMT b) PVn = 1336.22 = 1336.22(123.444897) = $164 949.54 c)

p=

- 1 = 0.00708981

164949.54 = PMT

23)

164949.54 = PMT(111.6558184) $1477.30 = PMT p=

- 1 = 0.00665496

a) 248000 = PMT 248000 = PMT(140.995881) $1758.92 = PMT

b)

$ 21 107.04 Principal repaid = 248000.00 - 246,649.91 = 1350.09 Interest paid = $ 19756.95

PVn = 1758. c) 92

PVn = 1758.92 = 1758.92(136.465612) = $240 032.09 248 000.00 - 240 032.09 = $7967.91 principal repaid = 1758.92(140.2280408) = $246,649.91

Total paid = 12(17 58.92 ) = 24)

d) p =

- 1 = 0.0058796

240032.09 = PMT 240032.09 PMT(149.469597) $1605.89 ==PMT a)

p=

- 1 = 0.071225

n= =

=

b) PVn = 4850.00 Total paid = 3(4850.00) Principal repaid 42500.00 - 36632.42 Total cost c)

= 14.22222256 payments

= 4850.00(7.5530728) = $36 632.42 = $14 550.00 = 5867.58 $ 8682.42

PVn = 4850.00 = 4850.00(7.9843889) = $38 724.29 38724.29 - 36632.42 = $2091.87 repaid in year 3.

d) (Rounded to the nearest cent)

PVn = 4850 = 4850(1.99059863) = $9654.40

25) a)

n=

=

=

b) PVn = 2604.90

= 32.5019768

= 2604.90(0.48727778) = $1269.31

Final payment = 1269.31(1.0403) = $1320.46 26) a)

n=

=

=

b) PVn = 6000.00

= 23.36241894

= 6000(0.352844567) = $2117.07

Final payment = 2117.07 (1.04) = $2201.75 27)

a)

P=

- 1 = 0.0029955134

n=

=

=

= 20.6550349

The number of payments is 21. b) PVg = 1000 28)

= 1000(0.653415465) = 653.42

p=

- 1 = 0.006951564

n=

=

Present value of final payment:

=

= 81.6095827

PVg = 660(0594-601.6069114) = $400.09 = 660 Final payment = 400.09(1.006951564) = $402.87 29) a ) n=

=

=

= 27.8008341

28 payments are required. b) PVn = 2475.00 = 2475.00(13.29365829) = $32 901.80 32901.80(.0051) = $167.80 c)

PVn = 2475.00 = 2475.00(11.4244994) = $28 275.64 2475.00 - 28,275.64(0.0051) = 2475.00 - 144.21 = $2330.79

d) Partial Amortization Table (rounded to the nearest cent)

30)

p=

- 1 = 0.00662271

n=

=

=

= 114.323133

Present value of final payment: PVg = 525

= 525(0.321724458) = $168.91

Final payment = 168.91(1.00662271) = $170.02 31)

PV = Y = 12; c = =3 50 000(0 .72) = 36 000.0 0;PM T = p = (1 + 0.0075)3 - 1 = 0.022669172 36000 = 2000 4000.n = 23.39060393 quarters 00; i = -n 0.408045094 = 1 - 1.022669172 -n ln 1.022669172 = ln 0.591954906

Present Value of final payment: PMT = 2000.00; p = 0.022669172; n = 0.39060393 PV = 2000 = 0.007PV = $769.12 5; I/Y = 9; Interest is 769.12(0.022669172) = $17.44 P/Y = Final Payment is 769.12 + 17.44 = $786.56 4;C/ 32) P V = 20 000.00; n = 9(12) = 108; i = = 0.005;I/Y = 6; P/Y = C/Y = 12 FV = 20000 (1 + 0.005)108 = $34 273.99 34273.99 = 1000 0.17136995 = 1 - 1.005-n 1.005-n = 0.82863005 -n ln 1.005 = ln 0.82863005 n = 37.6902094 months PV = 1000 PV = $687.31 Programmed solution:

The size of the final payment is 687.31(1.005) = $690.75 33)

152000.00 = 1750.00

(1.00625)

86.316 = 1 - 1.00625-n 7657 -n ln 1.00625 = ln 0.46051464 5 = n(0.00623055) = -0.77541063 n = 124.4530037

PVn(due) = 1750.00

(1.00625) = 1750.00(.45095723)(1.00625) = $794.11

0.539The final payment is $794.11.

4853 34) 750000 = 12000

(1.009)

61.9425 = 0.55748 = 1 - 1.01-n -n = ln(1 - 0.55748)/ln(1.01) n = 81.934488 (1.009) = $11 217.14

PVn(due) = 12000 The final payment is $11 217.14 35)

p=

- 1 = 0.02203587

19500 = 1720

(1.02203587)

11.0927704 = 0.24443886 = 1 1.0220359-n -n ln 1.0220359 = ln 0.75556114 -n(0.021796590) = -0.280294578 n = 12.85956074 PVn(due) = 1720

(1.02203587)

= 1720(0.84231184)(1.02203587) = $1480.70 The final payment is $1480.70.

36)

p=

- 1 = 0.0241136891

45200 = 5650

(1.0241136891)

7.811632717 = 0.18836728271 = 1 - 1.0241136891-n -n ln 1.0241136891 = ln 0.8116327173 -n(0.023827545) = -0.2087073597 n = 8.75907946 PVn(due) = 5650

(1.0241136891) = 4301.08

The final payment is $4301.08 37) a ) FV = 47500.00(1.033)8 = 47500.00(1.2965897) = $61 588.01 61588.01 = 6951.00

(1.033)

81.033 = ln 0.716950422 .5772 -n(0.032467190) = -0.332748578 n = 10.24876426 5993 The number of payments is 11.

7= b) Present value of last payment: PVn (due) = 6951.00 (1.033) = 6951.00(0.237620547)(1.033) 0 .2830 = $1750.30 4957 The final payment is $1750.30.

8=1 c) Total paid = 6951.00(10) + 1750.30 1.033 = 69510.00 + 1750.30 = $71 260.30 -n d) Interest = 71260.30 - 47500.00 = $23 760.30 n ln 38)

a)

p=

- 1 = 0.006638834

n=

=

=

246 payments b) PVn = 1140.00

= 1140.00(.8293897) = $949.40

949.40(1.006638834) = $955.70 c) 137960.70 = PMT

39)

a)

137960.70 = PMT(87.73874749) $1572.40 = PMT = $207 556.80 Contractual = 1572.40(132) Actual = 1140.00(245) + 955.70 = 280 255.70 = $72 698.90 Difference in payments 79500.00 = 1120 (1.0094667) 70.31648017 = 0.665662679 = 1 - 1.0094667-n -n ln 1.0094667 = ln 0.334337321 -n(0.009422139) = -1.095604853 n = 116.2798491 payments

b) PVn = 1120

(1.009466667)

= 1120(0.2789165895)(1.009466667) = $314.50 40)

a)

p=

- 1 = 0.092548902

FV = 59000.00(1.092548902)6 = 59000.00(1.700768902) = $100 345.37

= 245.8378879

=

n

=

= 6.595730861 payments

= b) PVn = 21000 = 21000(0.554993627) = $11 654.87 Final payment = 11654.87(1.092548902) = $12 733.52 41)

a)

p=

- 1 = 0.007716274

176000.00 = PMT 176000.00 = PMT(110.914783) $1586.77 = PMT The rounded payment is $1600.00.

n= =

b)

=

= 245.7618375 payments.

PVn = 1600 = 1600.00(0.756695498) = $1210.71 Final payment = 1210.05(1.007716274) = $1220.05

42)

a) p =

- 1 = 0.018108216

N=

=

=

= 11.92375947

The number of payments is 12. b) PVg = 1410

= 1410.00(0.907947708) = $1280.21

1280.21(1.018108216) = $1303.39 43) n=

=

n = 13.22144230 PVn = 1430 = 1430.00(.219396293) = $313.74 = Balance Final Payment = 313.74(1.0153) = $318.54 44)

a)

p=

- 1 = 0.00671944

=

1 5100 0.00 = PMT

$1362.07 = PMT Rounded payments $1400.00

n=

=

=

= 192.6301891

192 equal payments plus one less than $1400 1 b) PV g = 1400 = 1400.00(0.626757778) = $877.46 5100 0.00 Last Payment = 877.46(1.00671944) = $883.36 = PMT c) Unrounded 204(1362.07) 862.28 192(1400.00) + 883.36 ==$277 269 683.36 $ 8178.92 (110.

Rounded Savings 8605 = 388) 45) a) p = - 1 = 0.00486755 137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431) = $130 336.56 b) p =

- 1 = 0.005340319

130335.56 = PMT 130335.56 = PMT(128.9285035) $1010.91 = PMT c)

PVn = 1010.91 1010.91(124.0775295) = $125431.22 p=

- 1 = 0.005849741

125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62 46) i =

= 0.025; c =

=

p = 1.0251/6 - 1 = 0.004123915 = 0.4124% Programmed solution:

= 1054.62(97.14104632) = $102 446.89

p = 0.4124%

47) B 48) A 49) B 50) A 51) E 52) A 53) E 54) B 55) A 56) A 57) C 58) C 59) E 60) D 61) F irst calculate the balance at the end of the first 5 year term. This requires calculating the FV of the payments made in first 5 years and the future value of the original loan.

PV = 305000 - 46000 = $259 000;

∴p=

n = 12 × 25 = 300;

i=

= 2.2%; c =

- 1 = 0.0036335

PMT for original term =

= $1419.13

∴FV after 1st term = 1419.13

= $94 950

FV of the original loan at the end of 5 year term = 259000 × 1.003633560 = $321 965 Balance = $321 965 - $94 950 = $227 015 Balance is the PV for the second term For second term PV = $227,015;

∴p=

n = 12 × 20 = 240;

i=

= 1.85%; c =

- 1 = 0.00306

PMT for second term =

= $1336.72

62) B 63) F or p = 0.306%, equivalent simple annual rate is 3.67% Using MS-Excel,, mortgage statement is created as follows Payment Date

Number of Amount Days Paid

Interest Paid Principal

31-Oct 15-Nov 30-Nov 20-Dec

15 15 20

$218.69 $216.76 $285.55

$1500 $1940 $1900

$1281.31 $1723.24 $1614.45

Balance Repaid $145 000.00 $143 718.69 $141 995.45 $140 381.00

31-Dec 11 $1940 $155.27 Year-end balance = $138 596.26 64) P V = $1 467 921; PMT = $9000; P/Y = 12 i=

$1784.73

$138 596.26

C/Y = 12;

I/Y = 4.4%

= 0.367%

1467921 = 9000 ⇒ -n ln 1.00367 = ln 0.402 ⇒ n = 248.7953157 months For last month's payment: PMT = $9000; i = 0.367%; PV = $9000 × 1.00367 ×

n = 0.7953157 = $7160.52

65) A 66) C 67) C 68) P V = $19 500; n = 3; i = 5.99% FV = 19500 × 1.05993 = $23 218.24 Accumulate the value of 36 payments: PMT = $378; n = 36; P/Y = 12; C/Y = 1; I/Y = 5.99%; c =

p=

; i = 5.99%

- 1 = 0.00485965

FV = 378

= $14 831.65

Outstanding balance after 36 months = $23 218.24 - $14 831.65 = $8386.59 69) P V = $259 000; n = 25 × 12 = 300; P/Y = 12; C/Y = 2; I/Y = 4.4%; c = ;i=

p=

= 2.2%

- 1 = 0.0036335

259000 = PMT ⇒ PMT = $1419.13 FV = 259000 × 1.003633560 = $321 965.04 FV 60 = 1419.13

= $94 950.38

Balance at the end of 5 year term = $321 965.04 - $94 950.38 = $227 014.66 For second 5 year term: PV = $227 014.66; n = 20 × 12 = 240; P/Y = 12; C/Y = 2; I/Y = 4.4%; c =

p=

-1 = 0.003059831

;i=

= 1.85%

PMT = $1336.72 2270 14.66 FV = 227014.66 × 1.00305983160 = $272 686.88 = PMT FV 60 = 1336.72 = $87 890.34 Balance at the end of second 5 year term = $272686.88 - $87890.34 = $184 796.54 ⇒

70) Payment Number 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Totals

Principal Amount Paid Interest Paid Repaid $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $1,089 $60,71 3

$188 $182 $176 $170 $163 $157 $151 $145 $139 $132 $126 $120 $114 $107 $101 $95 $88 $82 $75 $69 $63 $56 $50 $43 $37 $30 $23 $17 $10 $4 $2,913

71) E 72) C 73) D 74) B 75) A 76) A mortization Schedule (done using Excel)

$1,868 $1,874 $1,880 $1,886 $1,893 $1,899 $1,905 $1,911 $1,917 $1,924 $1,930 $1,936 $1,942 $1,949 $1,955 $1,961 $1,968 $1,974 $1,981 $1,987 $1,993 $2,000 $2,006 $2,013 $2,019 $2,026 $2,033 $2,039 $2,046 $1,085 $57,80 0

Outstanding Principal balance $57,800 $55,932 $54,058 $52,177 $50,291 $48,398 $46,500 $44,595 $42,684 $40,766 $38,843 $36,913 $34,977 $33,035 $31,086 $29,131 $27,170 $25,202 $23,228 $21,248 $19,261 $17,267 $15,267 $13,261 $11,248 $9,229 $7,203 $5,170 $3,131 $1,085 $0

9 $2,300.00 $183.92 $2,116.08 $18,902.82 $2,300.00 $165.40 $2,13O4.u6t0standin$1g6 ,768.22 1 $146.P7r2incipal $2,153.P2r8incipal$ 14,614.94 $2,300.00 0 Am12ount Pai$d2 ,30I0n.0te0rest Paid $127.8R8epaid $2,172.1B2alance $12,442.82 1 1 0 13 $2,300.00 $108.87 $2,191.13$37,30$01.000,251.70 $1,973$.623,210.30$35,326$.388,041.40 1 14$2,300$.020,300. $326.38$89.7 $1,990$.829,229.64$33,335$.458,811.76 2 0 00 $2,008$.321,249.15$31,327$.137,562.62 3 15$2,300$.020,300. $309.11$70.3 $2,025$.829,268.83$29,301$.218,293.79 4 6 00 $2,043$.611,293.79$27,257.67$0.00 5 $291.69$50.8 16$2,300$.020,300.

6 5 00 $2,300.00 $2,061.50 $25,196.1 $274.11$31.1 17$2,300$.020,300. $2,079.53 7 Total7 Interest $p2a,3id0 0=. 0$03105.11 7 00 $23,116.6 Total8 cost of t$h2e, 3lo0a0n.0 0= $37300 +$ $230120.257.11 = $40$ 420,059.171.7 3 $256.39$11.3 18$2,300$.010,305.

4 77) p = 0.88% 2 Excel): $238.50 11 $21,018.9 Loan Amortization Schedule (done in $220.47 1

Amount Paid 01 23 45 67 89 10 11 12 13 14 15 16 17 18

$2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $2,300.00 $1,315.13

Interest Paid $327.33 $310.02 $292.55 $274.94 $257.17 $239.24 $221.15 $202.91 $184.51 $165.94 $147.22 $128.32 $109.27 $90.04 $70.65 $51.08 $31.35 $11.44

Principal Repaid

$1,972.6 7 $1,989.9 8 $2,007.4 5 $2,025.0 6 $2,042.8 3 $2,060.7 6 $2,078.8 5 $2,097.0 9 $2,115.4 9 $2,134.0 6 Total Interest paid = $3115.13 $2,152.7 Total cost of the loan = $37300 + $3115.13 = $40 415.13 8 78) $2,171.6 a) p = - 1 = 0.01467385 8 $2,190.7 45000 = PMT 3 $2,209.9 45000 = PMT(30.0947153) 6 $1495.28 = PMT $2,229.3 5 b) PVg = 1,495.28 $2,248.9 2 $2,268.6 5 $1,303.6 9

Outstanding Principal Balance $37,300.00 $35,327.33 $33,337.34 $31,329.90 $29,304.83 $27,262.00 $25,201.24 $23,122.39 $21,025.31 $18,909.81 $16,775.76 $14,622.97 $12,451.30 $10,260.57 $8,050.61 $5,821.26 $3,572.34 $1,303.69 $0.00

= = $25754.65 1519.

67(17c) Total paid = 1495.28(20) = $29 905.60 .2239 Principal repaid = 45000.00 - 25754.65 = $19 245.35 653) Interest cost = 29905.60 -19245.35 = $10 660.25 79) a) p = - 1 = 0.02007353

n=

=

=

= 21.879938

The number of payments is 22.

80)

b) PVg = 1110

= 1110(1.826974901) = $2027.94

c)

= $22 200 = $17 472.06 = $4727.94

Total paid = 1110(20) Principal repaid = 19500.00 - 2027.94 Interest = 22200.00 - 17472.06

a) p =

- 1 = 0.00696106

136000 = PMT 136000 = PMT(102 .442161) $1327.58 = PMT Payment rounded to $1400.00 FV = 136000(1.00696106)60 = 136000.00(1.51621447) = $206 205.17 FVn = 1400

= 1400(74.1574287) = $103 820.40

Balance = 206205.17 - 103820.40 = $102 384.77

b) n =

=

=

= 89.5752224 payments

c) n =

=

=

= 102.561916 payments

PVn = 1400.00 = 1400.00(0.58865597) = $782.43 782.43(1.00696106) = $787.88 PVn = 1540.00 = 1540.00(0.57208741) = $881.01 881.01(1.00696106) = $887.15 Rounded payments = 102(1400.00) + 787.88 = 142800.00 + 787.88 = $143 587.88 Increased rounded payments = 89(1540.00) + 887.15

Savings = 143 5=8 173.878 0 -6 01.3070 9+4 878.175.1 =5 $=5 $614307.7 934 7.15 81) a ) PV = 30 000.00; n = 20(12) = 240; i = = 0.0425; I/Y = 8.5; P/Y = 12; C/Y = 2; c =

=

p = (1 + i)c = (1 + 0.0425)1/6 - 1 = 0.006961062 30000 = PMT 30000 = PMT[116.4742048] PMT = $257.57 The number of outstanding payments after five years is 15(12) = 180 PV = 257.57 PV = $26 386.10 Programmed solution:

b) The outstanding balance of $26 386.10 is to be amortized the remaining 15 years.

PV = 26 386.10; n = 15(12) = 180; i =

= 0.04; I/Y = 8; P/Y = 12; C/Y = 2; c =

p = (1 + i)c - 1 = 0.006558197 26386.10 = PMT 26386.10 = PMT[105.4682161] PMT = $250.18 Programmed solution:

c) At the end of the 3 year term the outstanding payments is 144.

PV = 250.18 PV = $23 265.45 Programmed solution:

=

82)

- 1 = 0.042441

a) p =

n=

=

=

= 18.9998078

18 full payments are needed and one smaller payment. b) PVg = 3350.00

= 3350.00(7.3531857) = $24 633.17

Total paid = 10(3350.00) Principal repaid = 43100.00 - 24633.17 Cost = 33500.00 - 18466.83

= $33 500.00 = $18 466.83 = $15 033.17

c) $24633.17(.042441) = $1045.46 d) Partial Amortization Schedule (rounded to 2 decimal places):

83)

a)

p=

- 1 = 0.00591193

180000.00 = PMT 180000.00 = PMT(134.708006) $1336.22 = PMT b)

PV n = 1336.22 = 1336.22(123.444897) = $164 949.54

c)

p=

- 1 = 0.00708981

1649 49.54 = PMT 84)

164949.54 PMT(111.6558184) $1477.30 ==PMT

p=

- 1 = 0.00665496

a) 248000 = PMT 248000 = PMT(140.995881) $1758.92 = PMT b) PVn = 1758.92 = 1758.92(140.2280408) = $246,649.91 Total paid = 12(1758.92) = $ 21 107.04 1350.09 Principal repaid = 248000.00 - 246,649.91 = 19756.95 Interest paid =$ c) PVn = 1758.92 = 1758.92(136.465612) = $240 032.09 248 000.00 - 240 032.09 = $7967.91 principal repaid d) p =

- 1 = 0.0058796

240032.09 = PMT

85)

240032.09 = PMT(149.469597) $1605.89 = PMT a) p =

- 1 = 0.071225

= =

=

= 14.22222256 payments

b) PVn = 4850.00 Total paid = 3(4850.00) Principal repaid 42500.00 - 36632.42 Total cost

= 4850.00(7.5530728) = $36 632.42 = $14 550.00 = 5867.58 $ 8682.42

c) PVn = 4850.00 = 4850.00(7.9843889) = $38 724.29 38724.29 - 36632.42 = $2091.87 repaid in year 3.

d) (Rou nded to the near est cent)

PVn = 4850 = 4850(1.99059863) = $9654.40 86) a)

n=

=

b)

PVn = 4570.00

=

= 32.5019768

= 4570.00(0.48727778) = $2226.86

Final payment = 2226.86(1.0403) = $2316 .60 87) a)

n=

=

b)

PVn = 6000.00

=

= 23.36241894

= 6000(0.352844567) = $2117.07

Final payment = 2117.07 (1.04) = $2201.75 88)

a)

P=

- 1 = 0.0029955134

n=

=

=

= 20.6550349

The number of payments is 21. b) 89)

PVg = 1000

= 1000(0.653415465) = 653.42

p=

- 1 = 0.006951564

n=

=

=

= 81.6095827

e of final payment: Prese PVg = 660 = 660(0594 - 601.6069114) = $400.09 nt valu Final payment = 400.09(1.006951564) = $402.87 90) a) n =

=

=

= 27.8008341

28 payments are required. b) PVn = 2475.00 = 2475.00(13.29365829) = $32 901.80 32901.80(.0051) = $167.80 c) PVn = 2475.00 = 2475.00(11.4244994) = $28 275.64 2475.00 - 28275.64(0.0051) = 2475.00 - 144.21 = $2330.79 d) Partial Amortization Table (rounded to the nearest cent)

91)

p=

- 1 = 0.00662271

n=

=

=

= 114.323133

Present value of final payment: PVg = 500

= 500(0.321724458) = $160.86

Final payment = 160.86(1.00662271) = $161.93 92) P V = 100 000(0.72) = 72 000.00;PMT = 4000.00; i =

= 0.0075; I/Y = 9; P/Y = 4;C/Y = 12; c =

=3

p = 72 = ln 0.591954906 (1 + n = 23.39060393 quarters 0.007 5)3 - Programmed solution: 1= 0.022 6691 72 7200 Present Value of final payment: 4000.00; p = 0.022669172; n = 0.39060393 0 = PMT PV = =4000 4000 PV = $1538.23 0.408 Programmed solution: 0450 94 = 11.022 6691 72-n Interest is 1538.23(0.022669172) = $34.87 -n ln Final Payment is 1538.23 + 34.87 = $1573.10 1.022 6691 93) P V = 20 000.00; n = 9(12) = 108; i = = 0.005; I/Y = 6; P/Y = C/Y = 12 FV = 20000 34273.99 = 1000 0.17136995 = 1 - 1.005-n 1.005-n = 0.82863005 -n ln 1.005 = ln 0.82863005 n = 37.6902094 months PV = 1000 PV = $687.31 Programmed solution:

= $34 273.99

The size of the final payment is 687.31(1.005) = $690.75

94)

152000.00 = 1750.00

(1.00625)

86.3176575 = 0.53948536 = 1 1.00625-n -n ln 1.00625 = ln 0.46051464 -n(0.00623055) = -0.77541063 n = 124.4530037 PVn(due) = 1750.00

(1.00625) = 1750.00(.45095723)(1.00625) = $794.11

The final payment is $794.11.

95)

750000 = 12000

(1.01)

61.88 = 0.6188 = 1 - 1.01-n ln 0.38118811 = -nln 1.01 (0.9644622765) = -n(-0.0099503309) n = 96.927659 PVn(due) = 12000

(1.01)

= 12000(0.9303234212) = $11 135.91 The final payment is $11 135.91 96)

p=

- 1 = 0.02203587

19500 = 1720

(1.02203587)

11.0927704 = 0.24443886 = 1 1.0220359-n -n ln 1.0220359 = ln 0.75556114 -n(0.021796590) = -0.280294578 n = 12.85956074 PVn(due) = 1720

(1.02203587)

= 1720(0.84231184)(1.02203587) = $1480.70

The final payment is $1480.70.

97) p= - 1 = 0.0241136891 40000 = 5000

(1.0241136891)

7.811632717 = 0.18836728271 = 1 - 1.0241136891-n -n ln 1.0241136891 = ln 0.8116327173 -n(0.023827545) = -0.2087073597 n = 8.75907946 PVn(due) = 5000

(1.0241136891) = 3806.27

The final payment is $3806.27 98) a ) FV = 47500.00(1.033)8 = 47500.00(1.2965897) = $61 588.01 61588.01 = 6951.00

(1.033)

8.577259937 = 0.283049578 = 1 - 1.033-n -n ln 1.033 = ln 0.716950422 -n(0.032467190) = -0.332748578 n = 10.24876426 The number of payments is 11. b) Present value of last payment: PVn (due) = 6951.00

(1.033) = 6951.00(0.237620547)(1.033)

= $1750.30 The final payment is $1750.30. c)

Total paid = 6951.00(10) + 1750.30 = 69510.00 + 1750.30 = $71 260.30 d) Interest = 71260.30 - 47500.00 = $23 760.30 99)

a) p =

- 1 = 0.0066638834

n=

=

=

= 245.8378879

246 payments b) PVn = 1140.00 949.40(1.006638834) = $955.70 c) 137960.70 = PMT 137960.70 = PMT(87.73874749)

= 1140.00(.8293897) = $949.40

$1 Contractual = 1572.40(132) 572.4 Actual = 1140.00(245) + 955.70 0= Difference in payments PMT 100) a) 79500.00 = 1120

= $207 556.80 = 280 255.70 = $72 698.90 (1.0094667)

70.31648017 = 0.665662679 = 1 - 1.0094667-n -n ln 1.0094667 = ln 0.334337321 -n(0.009422139) = -1.095604853 n = 116.2798491 payments b) PVn = 1120

(1.009466667)

= 1120(0.2789165895)(1.009466667) = $314.50 101)

a)

p=

- 1 = 0.092548902

FV = 59000.00(1.092548902)6 = 59000.00(1.700768902) = $100 345.37

n=

=

=

= 6.595730861 payments

b) PVn = 21000 = 21000(0.554993627) = $11 654.87 Final payment = 11654.87(1.092548902) = $12 733.52 102)

a)

p=

- 1 = 0.007716274

176000.00 = PMT 176000.00 = PMT(110.914783) $1586.77 = PMT The rounded payment is $1600.00. n= = 245.7618375 payments. b)

PVn = 1600 = 1600.00(0.756695498) = $1210.71 Final payment = 1210.05(1.007716274) = $1220.05

103)

a) N=

p=

=

=

= 11.92375947

The number of payments is 12.

-1= b) 0.018 1082

PVg = 1410 1280.21(1.018108216) = $1303.39

= 1410.00(0.907947708) = $1280.21

16 104) n=

=

=

n = 13.22144230 PVn = 1430 = 1430.00(.219396293) = $313.74 = Balance Final Payment = 313.74(1.0153) = $318.54 105)

a) p =

- 1 = 0.00671944

151000.00 = PMT 151000.00 = PMT(110.8605388) $1362.07 = PMT Rounded payments $1400.00

n=

=

=

= 192.6301891

192 equal payments plus one less than $1400 b) PVg = 1400

= 1400.00(0.626757778) = $877.46

Last Payment = 877.46(1.00671944) = $883.36 c) Unrounded 204(1362.07) Rounded 192(1400.00) + 883.36 Savings 106)

a) p =

- 1 = 0.00486755

137000 = PMT 137000 = PMT(141.3843090) $968.99 = PMT PVn = 968.99 = 968.99(134.5076431) = $130 336.56

= $277 862.28 = 269 683.36 = $ 8178.92

b) p =

130335.56 = PMT(128.9285035) $1010.91 = PMT

- 1 = c) PV n = 1010.91 0.005 3403 1010.91(124.0775295) = $125431.22 19 p= - 1 = 0.005849741 13 0335.

56 = PMT

125431.22 = PMT 125431.22 = PMT(118.9350584) $1054.62 = PMT PVn = 1054.62

107) i =

= 0.025; c =

= 1054.62(97.14104632) = $102 446.89 =

p = 1.0251/6 - 1 = 0.004123915 = 0.4124% Programmed solution:

p = 0.4124% 108) E 109) E 110) D 111) B 112) A 113) B 114) C 115) A 116) E 117) B 118) B 119) A 120) D 121) B 122) F irst calculate the balance at the end of the first 5 year term. This requires calculating the FV of the payments made in first 5 years and the future value of the original loan.

PV = $259 000; n = 12 × 25 = 300;

∴p=

i=

= 2.2%;

- 1 = 0.0036335

PMT for original term =

= $1419.13

∴FV after 1st term = 1419.13

= $94 950

c=

FV of the = 259000 origi nal× 1.003633560 = $321 965 loan atBalance = $321 965 - $94 950 = $227 015 the Balance is the PV for the second term end of 5 For second term year PMT for second term = term PV = $227015; n = 12 × 20 = 240; i=

∴p=

= 1.85%; c =

- 1 = 0.00306 = $1336.72

123) D 124) F or p = 0.306%, equivalent simple annual rate is 3.67% Using MS-Excel,, mortgage statement is created as follows Payment Date 31-Oct 15-Nov 30-Nov 20-Dec 31-Dec

Number of Amount Days Paid

Interest Paid

Principal

15 15 20 11

$218.69 $216.76 $285.55 $155.27

$1281.31 $1723.24 $1614.45 $1784.73

Balance Repaid $145,000.00 $143,718.69 $141,995.45 $140,381.00 $138,596.26

P/Y = 12

C/Y = 12;

I/Y = 4.4%

$1500 $1940 $1900 $1940

Year-end balance = $138 596.26 125) P V = $1 467 921; PMT = $9000; i=

= 0.367%

1467921 = 9000 ⇒ -n ln 1.00367 = ln 0.402 ⇒ n = 248.7953157 months For last month's payment: PMT = $9000; i = 0.367%; PV = $9000 × 1.00367 ×

n = 0.7953157 = $7160.52

126) D 127) E 128) C 129) P V = $19 500; n = 3; i = 5.99% FV = 19500 × 1.05993 = $23 218.24 Accumulate the value of 36 payments: PMT = $378; n = 36; P/Y = 12; C/Y = 1; I/Y = 5.99%; c =

p= FV = 378

; i = 5.99%

-1 = 0.00485965 = $14 831.65

Outstanding balance after 36 months = $23 218.24 - $14 831.65 = $8386.59 130)

PV = 6335 $259 000; n = 259000 = PMT 25 × ⇒ PMT = $1419.13 12 = FV = 259000 × 1.003633560 = $321 965.04 300; FV60 = 1419.13 P/Y = = $94 950.38 12; C/Y ForBalance second at 5 year term: the end of 5 year term = $321965.04 - $94950.38 = $227 014.66 = 2; PV = $227 014.66; n = 20 × 12 = 240; P/Y = 12; C/Y = 2; I/Y = 4.4%; c= p= -1 = 0.003059831 227014.66 = PMT ;i=

I/Y = 4.4%; c =

;i=

⇒ PMT = $1336.72 =

FV = 227014.66 × 1.00305983160 = $272 686.88

2.2% p = FV60 = 1336.72

= $87 890.34

-1 = Balance at the end of second 5 year term = $272686.88 - $87890.34 = $184 796.54 0.003 131) Payment Number 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Amount Paid

Interest Paid

Principal Repaid

$2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056 $2,056

$188 $182 $176 $170 $163 $157 $151 $145 $139 $132 $126 $120 $114 $107 $101 $95 $88 $82 $75 $69 $63

$1,868 $1,874 $1,880 $1,886 $1,893 $1,899 $1,905 $1,911 $1,917 $1,924 $1,930 $1,936 $1,942 $1,949 $1,955 $1,961 $1,968 $1,974 $1,981 $1,987 $1,993

Outstanding Principal balance $57,800 $55,932 $54,058 $52,177 $50,291 $48,398 $46,500 $44,595 $42,684 $40,766 $38,843 $36,913 $34,977 $33,035 $31,086 $29,131 $27,170 $25,202 $23,228 $21,248 $19,261 $17,267

= 1.85%

22 23 24 25 26 27

28 29 30 Totals

$2,056$ 2,056 $2,056$ 2,056 $2,056$ 1,089 $2,056$ 60,713 $2,056 $2,056

$56 $50 $43 $37 $30 $23

$17 $10 $4 $2,913

$2,000$ 2,039 $2,006$ 2,046 $2,013$ 1,085 $2,019$ 57,800 $2,026 $2,033

132) C 133) D 134) P V = 11 256; n = 3.5(12) = 42; i = 0.108/12 = 0.009;

$15,26$73 ,131 $13,26$11 ,085 $11,24$80 $9,229 $7,203 $5,170

11 256 = PMT PMT = 323.03 First period interest = 11 256(0.009) = 101.30 First period principal payment = 323.03 - 101.30 = 221.73 Owed after first period = 11 256 - 221.73 = 11034.27 Second period interest = 11034.27(0.009) = 99.31 First period principal payment = 323.03 - 99.31 = 223.72 Owed after second period = 11 034.27 - 223.72 = 10810.55 Third period interest = 10810.55(0.009) = 97.29 Third period principal payment = 323.03 - 97.29 = 225.74 Owed after third period: 10810.55 - 225.74 = 10584.81 135) P V = 43 610; n = 8.5(2) = 17; i = 0.091/2 = 0.0455; 43 610 = PMT 3739.25 = PMT Original Debt: FV = 43610(1.0455)8 = $62255.59 4 years of payments: FV = 3739.25

= $35136.88

Balance: Outstanding balance = $62255.59 - $35136.88 = $27 118.71 136) P V = $24 755; PMT = $680; i = 0.0624/26 = 0.0024 24 755 = 680 0.087370588 = 1 - 1.0024-n 1.0024-n = 0.9126294118 -n = n = 38.13960377 Therefore, 38 payments of $680 and one smaller payment will be necessary to cover the loan.

137) P V = $68 900; n = 11(12) = 132; c = 4/12 = 1/3; i = 0.054/4 = 0.0135 p=

- 1 = 0.004479901

68 900 = PMT 692.56 = PMT First period interest = 68900(0. 004479901) = 308.67 First period principal payment = 692.56 - 308.67 = 383.89

Owed after first period = 68900 - 383.89 = 68 516.11 138) P MT = 15 000; n = 6.5(12) = 78; c = 2/12 = 1/6; i = 0.086/2 = 0.043 p=

- 1 = 0.007041539

PV = 15000

= $897 884.38

139) P V = $8400; PMT = $675; i = 0.073 8400 = 675 0.28622222 = 1 - 1.023-n 1.023-n = 0.71377777 -n = n = 14.828109 Fractional value of n for last payment period: 14.828109 - 14 = 0.828109 PV = 675 PV = $547.47 140) P V = $10 900; PMT = $315; i = 0.044/4 = 0.011; 10

900

=

315

0.389635 = 1.011-n 1.011-n = 0.619365 -n = n = 43.7900498 Fraction value of n for last payment period: 0.7900498 PV = 315 141) B

= $246.44

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) A $5000 bond that pays 6% semi-annually is redeemable at par in 14 years. Calculate the 1) _____________ purchase price if it is sold to yield 8% compounded semi-annually.

2) A $25 000, 6% bond redeemable at par is purchased 11 years before maturity to yield 6.9% compounded semi-annually. If the bond interest is payable semi-annually, what is the purchase price of the bond?

2) _____________

3) A $50 000 bond bearing interest at 6.5% bond payable semi-annually matures in 10 years. If it is bought to yield 5.7% compounded semi-annually, what is the purchase price of the bond?

3) _____________

4) A $200 000 bond is redeemable at 108 in 14 years. If interest on the bond is 5.5% payable semi-annually, what is the purchase price to yield 4% compounded semi-annually?

4) _____________

5) W hat is the purchase price of a $10 000, 3.5% bond with semi-annual coupons redeemable at 104 in seven years if the bond is bought to yield 2.5% compounded semi-annually?

5) _____________

6) A $100 000 bond is redeemable at 105 in 20 years. If interest on the bond is 5% payable

6) _____________

semi-annually, what is the purchase price to yield 6% compounded semi-annually?

7) A $100 000, 4% bond with semi-annual coupons is redeemable at 108. What is the purchase price to yield 5.5% compounded semi-annually seven years before maturity?

7) _____________

8) A $5000, 6.5% bond with semi-annual coupons redeemable at 108 is bought two years before maturity to yield 6% compounded semi-annually. What is the purchase price?

8) _____________

9) A $200 000.00, 6% bond with semi-annual coupons is redeemable at par. What is the purchase price of the bond six years before maturity to yield 8% compounded semi-annually?

9) _____________

10) A $100 000, 7.5% bond with semi-annual coupons redeemable at 106 is bought three years before maturity to yield 8% compounded semi-annually. What is the purchase price?

10) _____________

11) A $150 000 bond bearing interest at 6% payable semi-annually is bought eight years before maturity to yield 4.5% compounded annually. If the bond is redeemable at par, what is the purchase price?

11) _____________

12) A $100 000 bond bearing interest at 8% payable semi-annually is bought five years before maturity to yield 6% compounded annually. If the bond is redeemable at par, what is the purchase price?

12) _____________

13) S ix $1500 bonds with 4.5% coupons payable semi-annually are bought to yield 5% compounded monthly. If the bonds are redeemable at par in eight years, what is the purchase price?

13) _____________

14) A $150 000 bond redeemable at par on October 1, 2026, is purchased on January 15, 2014.

Interest is 6% payable semi-annually and the yield is 7.5% compounded semi-annually. a) What is the market price of the bond? b) How much interest has accrued? c) What is the cash price?

14) _____________

15) B onds in denominations of $10 000 redeemable at par in six years and four months are offered for sale. If the coupon rate is 6.5% payable quarterly and the expected yield is 8% compounded quarterly, determine a) the market price; b) the accrued interest; c) the cash price.

15) _____________

16) B onds in denominations of $100 000 redeemable at 104 are offered for sale. If the bonds mature in ten years and six months and the coupon rate is 5.5% payable quarterly, what is the market price of the bonds to yield 7.2% compounded quarterly?

16) _____________

17) A $100 000, 8.75% bond with interest payable annually is redeemable at 103 in four years. What is the purchase price to yield 7% compounded quarterly?

17) _____________

18) A $1 000 000, 9% bond with interest payable annually is redeemable at 104 in 5 years.

What is the purchase price to yield 8% compounded quarterly?

18) _____________

19) A $50 000 bond bearing interest at 5.5% payable semi-annually is redeemable at par on August 10, 2033. The bond is sold on the primary market on December 10, 2013, to yield 5% compounded semi-annually. Determine a) calculate the market price; b) the accrued interest; c) the cash price.

19) _____________

20) A $50 000 bond bears interest at 4.5% payable semi-annually and is redeemable at par on November 4, 2022. The bond is sold on March 20, 2013, to yield 5.5% compounded semi-annually. What is the cash price?

20) _____________

21) A $250 000, 7% bond redeemable at par with interest payable semi-annually is bought 4 years before maturity. Determine the premium or discount and the purchase price if the bond is purchased to yield: a) 6.5% compounded semi-annually b) 8% compounded semi-annually

21) _____________

22) A $10 000, 7.2% bond with semi-annual coupons is purchased 3 years before maturity.

Calculate the discount or premium if the bond is sold to yield 6% compounded semi-annually.

22) _____________

23) A $50 000, 4% bond with semi-annual coupons is purchased three years before maturity.

Calculate the discount or premium if the bond is sold to yield 6% compounded semi-annually.

23) _____________

24) A $1000, 7% bond redeemable at par in eight years bears coupons payable annually.

b) 8.5% 7.5 % c)

Compute the premium or discount and the purchase price if the yield, compounded annually, is: a) 6.5%

24)

_____________ 25) A $1000, 6.5% bond redeemable at par in four years bears coupons payable annually.

Compute the premium or discount and the purchase price if the yield, compounded annually, is 7.5%.

25) _____________

26) B onds with a face value of $130 000 redeemable at par bearing 4.5% coupons payable quarterly are sold six years before maturity to yield 5.25% quarterly. Determine a) the premium or discount; b) the purchase price?

26) _____________

27) N ine $10 000, 4% bonds with interest payable semi-annually and redeemable at par are purchased 8.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 6%.

27) _____________

28) A $50 000.00, 8% bond with quarterly coupons redeemable at par is purchased five years before maturity to yield 4% compounded semi-annually. Determine the premium or discount.

28) _____________

29) T en $10 000, 4% bonds with interest payable semi-annually and redeemable at par are purchased 18.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 7%.

29) _____________

30) T wenty $5 000 bonds redeemable at par bearing 6% coupons payable quarterly are sold eight years before maturity to yield 5.5% compounded annually. What is the purchase price of the bonds?

30) _____________

31) A $125 000 bond, redeemable at par in three years with 7.5% coupons payable quarterly, is bought to yield 6% compounded quarterly. (i) Compute the premium or discount and the purchase price. (ii) Construct a schedule for amortization of premium.

31) _____________

32) A $100 000 bond, redeemable at 110 in seven years with 6.75% coupons payable annually, is bought to yield 7.25% compounded annually. (i) Determine the discount and the purchase price. (ii) Construct a schedule of accumulation of discount.

32) _____________

33) A $5000.00 6.5% bond redeemable at 109 with interest payable annually is purchased six years before maturity to yield 7% compounded annually. Construct the appropriate bond schedule.

33) _____________

34) A $5000, 6.25% bond with interest payable annually redeemable at par in eight years is

34) _____________

purchased to yield 7.5% compounded annually. Find the premium or discount and the purchase price and construct the appropriate bond schedule.

35) A $1000, 6% bond redeemable at par with semi-annual coupons was purchased 10 years before maturity to yield 5% compounded semi-annually. The bond was sold 3 years later at 102. Calculate the gain or loss on the sale of the bond.

35) _____________

36) F ind the gain or loss on the sale without constructing a bond schedule for six $5000, 9.5% bonds with interest payable semi-annually redeemable at par bought twenty-one years before maturity to yield 10.5% compounded semi-annually. The bonds were sold three

year at 103.625. rs late

36)

_____________ 37) F ind the gain or loss on the sale without constructing a bond schedule for a $100 000, 8% bond with semi-annual coupons redeemable at par purchased eleven-and-a-half years before maturity to yield 9% compounded semi-annually. The bond was sold five years later at 99.125.

37) _____________

38) F ind the gain or loss on the sale without constructing a bond schedule for a $50 000, 6% bond with semi-annual coupons redeemable at par purchased twelve years before maturity to yield 10% compounded semi-annually. The bond was sold five years later at 99.375.

38) _____________

39) T hree $25 000, 6% bonds with semi-annual coupons redeemable at par were bought eight years before maturity to yield 7% compounded semi-annually. Determine the gain or loss if the bonds are sold at 89.625 four years later.

39) _____________

40) A $40 000.00, 5% bond with semi-annual coupons redeemable at par in 14 years is purchased to yield 7% compounded semi-annually. What is the gain or loss if the bond is sold three years before maturity at 99.75?

40) _____________

41) A $50 000.00, 6% bond with semi-annual coupons redeemable at par in 14 years is purchased to yield 8% compounded semi-annually. What is the gain or loss if the bond is sold four years before maturity at 99.25?

41) _____________

42) F ind the gain or loss on the sale without constructing a bond schedule for three $100 000, 5.5% bonds with quarterly coupons redeemable at par on September 1, 2017, bought on May 1, 2011, to yield 6% compounded quarterly. The bonds were sold on January 21, 2014, at 93.5.

42) _____________

43) A 4.5% annuity bond of $500 000 with interest payable quarterly is to be redeemable at par in seven years. a) What is the purchase price to yield 6% compounded quarterly? b) What is the book value after 6 years? c) What is the gain or loss if the bond is sold six years after the date of purchase at

43) _____________

44) A $25 000, 8% bond with semi-annual coupons, redeemable at par in 12 years, is purchased to yield 6% compounded semi-annually. Determine the gain or loss if the bond is sold two years later at 107.25.

44) _____________

45) A $25 000, 7% bond with semi-annual coupons redeemable at par in twenty-two years is

45) _____________

purchased to yield 6% compounded semi-annually. Determine the gain or loss if the bond is sold seven years after the date of purchase at 98.25.

46) A $50 000, 6% bond with semi-annual coupons redeemable at par on April 25, 2018, was purchased on June 25, 2009, at 94.378. What was the approximate yield rate?

47) A $100 000, 8% bond with semi-annual coupons redeemable at par on April 25, 2018,

46) _____________

47) _____________

was purchased on June 25, 2009, at 94.125. What was the approximate yield rate?

48) A $40 000 bond with semi-annual coupon payments at 5.5% compounded semi-annually is redeemable at par in 12 years. Calculate the yield rate if the bond is purchased at 102.5.

48) _____________

49) A $100 000, 6.0% bond with semi-annual coupons redeemable at par is bought 17.5 years before maturity at 74.25. What was the approximate yield rate?

49) _____________

50) A $10 000.00, 5% bond with semi-annual coupons redeemable at 105 in 23 years is purchased at 102.5. What is the approximate yield rate?

50) _____________

51) A $50 000 bond that pays 5% semi-annually is redeemable at par on July 15, 2025.It is quoted at 97.5 on December 2, 2013. Determine the yield rate.

51) _____________

52) A $250 000, 6.5% bond with semi-annual coupons redeemable at par is bought 17.5 years before maturity at 78.25. What was the approximate yield rate?

52) _____________

53) A $100 000, 9.0% bond with semi-annual coupons redeemable at par on March 1, 2015, was purchased on September 22, 2007, to yield 10.00% compounded semi-annually.

What was the purchase price?

53) _____________

54) A $10 000, 8.5% bond with semi-annual coupons redeemable at par on March 1, 2005, was purchased on September 22, 1997, to yield 10.06% compounded semi-annually.

What was the purchase price?

54) _____________

55) A $10 000 bond bearing interest at 6% payable semi-annually redeemable at par on

55) _____________

March 1, 2002, was purchased on September 30, 1996, to yield 5% compounded semi-annually. Determine the purchase price.

56) A $200 000, 6% bond with semi-annual coupons redeemable at par March 1, 2009, was purchased on November 15, 1999, to yield 5% compounded semi-annually. What was the purchase price of the bond?

56) _____________

57) A $100 000.00, 7% bond with semi-annual coupons redeemable at 105 on November 1, 2001, is purchased on July 23, 1999, to yield 8% compounded semi-annually. Determine the quoted price.

57) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

58) W hat is the purchase price of a bond that has 4 years and 6 months until it matures? The face 58) ______ value of the bond is $3000 and the coupon rate is 5.2% compounded semi-annually. The yield rate is 7.5% compounded semi-annually.

A) $ 2547.53 B) $ 2770.53 C) $ 2740.53 D) $ 2447.53 E) $ 2440.53 59) W hat is the purchase price of a bond that is issued on May 1, 2002 for 22 years at a coupon rate of 9.4% compounded semi-annually? The purchase date of the bond is May 11, 2007 and the yield rate is 8.5% compounded semi-annually. The bond has a face value of $5000 and is redeemable at par.

A) $ 4531.30 B) $ 5413.30 C) $ 4533.30 D) $ 5431.30 E) $ 4513.30

59) ______

60) W hat is the amount of premium/discount for a bond that has 4 years till it matures? The face

60) ______

value of the bond is $100 000 and it has a bond rate of 6.4% compounded semi-annually. The bond is sold to yield 7.4% compounded semi-annually.

A) $ 3084.47 B) $ 3408.47 C) $ 3840.47 D) $ 3184.47

E) $ 3048.47

61) W hat is the amount of premium/discount amortized or accumulated in the first payment interval for a bond that has a face value of $6000 and it sold for $5700? The coupon rate is 9%

comnded pousemi-an

___ ___

nually and 61) the market rate is 10.1% compou nded semi-ann ually.

A) $ 18.75

B) $ 11.85

C) $ 17.85

D) $ 15.75

E) $ 14.75

62) W hat is the amount of premium/discount amortized or accumulated for the second payment interval of a $1500 bond that sold for $1627.32? The bond rate is 6.95% compounded semi-annually and the yield rate is 6% compounded semi-annually.

A) $ 4.30 B) $ 3.40 C) $ 1.40 D) $ 5.30 E) $ 2.30

62) ______

63) W hat is the quoted price of a bond that has 4.5 years until maturity? The face value of the bond is $5000 and it has a coupon rate of 8.1% compounded semi-annually and a yield rate of 8.6% compounded semi-annually.

A) $ 4898.32 B) $ 4908.32 C) $ 4998.32 D) $ 4809.32 E) $ 4888.32

63) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

64) A $1000 bond, with interest at 8% payable semi-annually on January 1 and July 1, was 64) _____________ purchased on October 8 at 104 plus accrued interest. What is the purchase payment for the bond?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

65) C larington issued 15 year bonds in the amount of $500 000. Interest on the bonds is 2.2% payable 65) ______ annually. What is the promised payment at the end of each year?

A) $ 13 925 B) $ 360 750 C) $ 693 000 D) $ 374 675 E) $ 11 100 66) C larington issued 15 year bonds in the amount of $500 000. Interest on the bonds is 2.2% payable annually. What is the issue price of the bonds, if the bonds are sold to yield 2.8% compounded quarterly?

A) $ 330 425 B) $ 329 004 C) $ 132 950 D) $ 343 749 E) $ 461 954

66) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

67) A $1000 bond, with interest at 8% payable semi-annually on January 1 and July 1, was 67) _____________ purchased on February 8, 2012. The bond matures on July 1, 2015 and yields 2.2% compounded semi-annually. What is the purchase payment for the bond?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

68) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 68) ______ 3.5% compounded quarterly. Determine the premium or discount.

A) $ 3829 premium B) $ 1347 discount C) $ 2481 premium D) $ 2481 discount E) $ 1347 premium 69) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Determine the premium or discount.

69) ______

A) $ 1073.95 premium B) $ 1073.95 discount C) $ 4698.55 premium D) $ 3624.60 discount E) $ 4698.55 discount 70) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Calculate the shortage of the actual interest received compared to the required interest based on the yield rate.

A) $ 1350 B) $ 437.50 C) $ 100 D) $ 337.50 E) $ 1750

70) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

71) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, 71) _____________ to yield 3.5% compounded quarterly. Construct a schedule of amortization of premium.

72) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 7% compounded quarterly. Construct a schedule of accumulation of discount.

72) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

73) N ick buys a $25 000, 5.4% bond with quarterly interest coupons, 3 years before maturity, to yield 73) ______ 7% compounded quarterly. The bond was sold a year later at 102. What is the gain or loss on the sale of the bond?

A) $ 1240.50 loss B) $ 740.51 loss C) $ 740.51 gain D) $ 500 gain E) $ 1240.50 gain 74) F rom his retirement fund, Jack buys a $50 000, 5.4% bond with quarterly interest coupons at 104.1, redeemable at par in 15 years. What is the approximate yield rate?

A) 4 .92% B) 1 .256% C) 5 .024% D) 1 .23% E) 5 .19%

74) ______

75) A $30 000, 3% bond redeemable at par is purchased 7 years before maturity to yield 5.5% compounded quarterly. If the bond interest is payable quarterly, what is the purchase price of the bond?

A) $ 5199.75 B) $ 225.00 C) $ 34 719.51 D) $ 20 467.13 E) $ 25 666.88

75) ______

76) A $63 000 bond bearing interest at 9.1% bond payable semi-annually matures in 20 years. If it is bought to yield 11.1% compounded semi-annually, what is the purchase price of the bond?

A) $ 7261.24 B) $ 52 956.98 C) $ 45 695.74 D) $ 2866.50 E) $ 74 510.72

76) ______

77) A $20 000, 3.6% bond with semi-annual coupons is purchased 5 years before maturity. Calculate the discount or premium if the bond is sold to yield 3% compounded semi-annually.

A) $ 553.33 premium B) $ 544.64 premium C) $ 554.64 discount D) $ 553.33 discount E) $ 360 premium

77) ______

78) N ine $2000, 9% bonds with interest payable semi-annually and redeemable at par are purchased 7 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 4%.

78) ______

A) $ 4600.27 premium B) $ 5447.81 discount C) $ 5447.81 premium D) $ 4600.27 discount E) $ 11.14 discount 79) S ix $20 000, 2% bonds with interest payable semi-annually and redeemable at par are purchased 18.5 years before maturity. Find the premium or discount and the purchase price if the bonds are bought to yield 5.9%.

A) $ 8711.47 discount B) $ 56 268.82 discount C) $ 56 268.82 premium D) $ 12 011.81 premium E) $ 12 011.81 discount

79) ______

80) T wenty $1 000 bonds redeemable at par bearing 6% coupons payable quarterly are sold eight years before maturity to yield 5.5% compounded annually. What is the purchase price of the bonds?

A) $ 20 633.46 B) $ 19 861.49 C) $ 17 969.05 D) $ 13 031.98 E) $ 20 229.82

80) ______

81) A $1000, 6% bond redeemable at par with semi-annual coupons was purchased 10 years before maturity to yield 5% compounded semi-annually. The bond was sold 3 years later at 102.

Calculate the gain or loss on the sale of the bond.

A) $ 76.48 Gain B) $ 58.45 Loss C) $ 76.48 Loss D) $ 38.45 Gain E) $ 38.45 Loss

81) ______

82) F ind the gain or loss on the sale without constructing a bond schedule for six $3500, 3.5% bonds with interest payable semi-annually redeemable at par bought twenty-one years before maturity to yield 5.5% compounded semi-annually. The bonds were sold three years later at 102.

A) $ 5180.68 Gain B) $ 5153.98 Loss C) $ 5612.65 Gain D) $ 5180.68 Loss E) $ 5612.65 Loss

82) ______

83) A 4.5% annuity bond of $50 000 with interest payable quarterly is to be redeemable at par in seven years with a yield 6% compounded quarterly. What is the gain or loss if the bond is sold six years after the date of purchase at 99.625?

A) $ 535.20 Gain B) $ 916.87 Loss C) $ 535.20 Loss D) $ 4073.76 Gain E) $ 4073.76 Loss

83) ______

84) A 6.3% annuity bond of $12 000 with interest payable monthly is to be redeemable at par in nine years with a yield 7.9% compounded monthly. What is the gain or loss if the bond is sold eight years after the date of purchase at 101.325?

A) $ 343.03 Loss B) $ 184.03 Loss

84) ______

C) $ 343.03 Gain D) $ 336.94 Gain E) $ 184.03 Gain 85) T hree 9.1% annuity bond of $15 000 with interest payable quarterly are to be redeemable at par in seventeen years with a yield 7.1% compounded quarterly. What is the gain or loss if the bond is sold six years after the date of purchase at 102?

A) $ 6831.18 Loss B) $ 5931.18 Loss C) $ 5931.18 Gain D) $ 5815.32 Loss E) $ 6831.18 Gain

85) ______

86) A $15 000, 4.4% bond with semi-annual coupons redeemable at par on April 25, 2025, was purchased on June 25, 2016, at 93.725. What was the approximate yield rate?

A) 2 .55% B) 2 .63% C) 5 .10% D) 5 .26% E) 4 .54%

86) ______

87) A bond for which no security is offered by the issuer is called a: A) U nsecure bond B) I ndex Fund C) P remium D) M utual Fund E) D ebenture

87) ______

88) W hen the market rate is greater than the bond rate, the bond is described as: A) P ar B) E xpensive C) D ebenture D) P remium

88) ______ E) D iscount

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

89) J ack purchases a $4300 bond with an interest rate of 2.8% p.a. payable semi-annually and 89) _____________ a maturity date of 9 years from now. If the yield rate is 4.2% p.a. compounded semi-annually, what is the purchase price of Jack's bond?

90) T he Government of Canada begins to sell $10 000 bonds that have an interest rate of 3.8% p.a. payable semi-annually. What is the price of the bonds if they mature in 5 years and are sold to yield 3.6% p.a. monthly?

90) _____________

91) W endy decided to redeem her $9100 bond 6 years and 3 months early. The bond has an interest rate of 4.4% payable semi-annually and the yield rate is 5.0% compounded semi-annually. What is the cash price?

91) _____________

92) A $100 provincial government bond has a bond rate of 9.4% p.a. payable semi-annually and a yield rate of 6.8% p.a. compounded semi-annually. The bond is redeemable in 2 years. Determine the premium or discount.

92) _____________

1) P MT =

= $150

PP = 5000(1 + 0.04)-28 + 150 PP = 1667.39 + 2499.46 PP = $4166.85 Programmed solution:

The purchase price is 1667.39 + 2499.46 = $4166.85 2) F V = $5000; n = 11(2) = 22; P/Y = C/Y = 2; I/Y = 6.9; i =

= 0.0345; PMT = 25000(0.06) = $750.00

PP = 25000(1.0345)-22 + 750 = 25000.00(.4741646) + 750.00(15.2416064) = 11854.11 + 11431.20 = $23 285.32 3) F V = $50 000; n = 10(2) = 20; P/Y = C/Y = 2; PMT = 50000(0.065) = $1625; i = PP = 50000(1.0285)-20 + 1625

= 0.0285

= 28502.57 + 24514.61 = $53 017.18

4) PP = 200000.00(1.08)(1.02)-28 + 5500.00 = 200000.00(1.08)(.5743746) + 5500.00(21.2812724) = 124064.90 + 117047.00 = $241 111.90 5) F V = $10000(1.04) = $10400; P/Y = C/Y = 2; PMT = 10000(0.035/2) = $175.00; n = 7(2) = 14.

PP = 10400.00(1.0125)-14 + 175.00 = 10400.00(.8403681) + 175.00(12.7705528) = 87398.28143 + 2234.84673 = $89 633.13 6) F V = 100000(1.05) = $105000; PMT = 100000(0.05) = $2500; n = 20(2) = 40; P/Y = C/Y = 2; i =

= 0.03

PP = 105000(1.03)-40 + 2500 = 105000(.3065568408) + 2500(23.11477197) = 32188.47 + 57786.93 = $89 975.40 7) F V = $100000(1.08) = $108 000; PMT = 100000(0.04/2) = $2000; P/Y = C/Y = 2; i = PP = 108000(1.0275)-14 + 2000 = 73871.70584 + 22982.01627 = $96 853.72

= 0.0275; n = 7(2) = 14.

8) F V = $5000(1.08) = $5400; P/Y = C/Y = 2; PMT = $5000

= $162.50; i =

= 0.03

PP = 5400(1.03)-4 + 162.50 = 5400.00(.888487) + 162.50(3.7170984) = 4797.83 + 604.03 = $5401.86 9) F V = $200 000; P/Y = C/Y = 2; PMT = 200000(0.06/2) = $6000; i =

= 0.04; n = 6(2) = 12.

PP = 200000(1.04)-12 + 6000 = 124919.4099 + 56310.44256 = $181 229.85 10) F V = $100000(1.06) = $106000; P/Y = C/Y = 2; PMT = $100000

= $3 750; n = 3(2) = 6

PP = 106000.00(1.04)-6 + 3750.00 = 106000.00(.7903145257) + 3750.00(5.242136857) = 83773.34 + 19658.01 = $103 431.35 11) F V = $150 000; PMT = 150000(0.06)0.5 = $4500; n = 8(2) = 16; P/Y = 2; C/Y = 1; c =

= 0.5; i =

= 0.045

p = (1 + 0.054)1/2 - 1 = 0.0222524 PP

= 150000(1.0222524)-16 + 4500

= 150000(.7031853) + 4500(13.3385481) = 105477.77 + 60023.46 = $165 501.23 12) F V = $100000; PMT = $100000(0.08)0.5 = $4000; P/Y = 2; C/Y = 1; c = i=

= 0.06; p =

; n = 5(2) = 10;

- 1 = .029563014

PP = 100000(1.029563014)-10 + 4000 = 100000(.7472581729) + 4000(8.549257745) = 74725.82 + 34197.03 = $108 922.85 13) F V = $1500(6) = $9 000.00; PMT = 9000(0.05)0.5 = $225.00; n = 8(2) = 16; P/Y = 2; C/Y = 12; c=

= 6; p =

- 1 = 0.02526187

PP

= 9000 (1.02526187)-16 + 225

= 6037.90 + 2931.40 = $8969.30 14) F V = $150 000; PMT = $4500; P/Y = C/Y = 2; The interest date preceding the date of purchase is October 1, 2013. The time period from October 1, 2013 to the date of maturity is 13 years. n = 13(2) = 26; i =

= 0.0375

a) PP on October 1, 2013 = 150000(1.0375)-26 + 4500 = 57597.09 + 73922.33 = $131 519.42 Market price = $131 519.42

b) est period Oct 1, 2013-Apr 1, 2014: 182 days; Days Days between Oct 1, 2013 and Jan 15, 2014: 106 in Accrued interest = 150000(.03) = $2620.88 the interc) Cash price = 131519.42 + 2620.88 = $134 140.30 15) F V = $10 000; PMT = 10 000(0.065)(0.25) = $162.50; a) The interest date preceding the purchase date is 6.5 years before maturity. n = 6.5(4) = 26; P/Y = 4; C/Y = 4; I/Y = 8; i =

= 0.02

PP = 10000(1.02)-26 + 162.50 = 5975.79 + 3269.67 = $9245.46 Market price = $9245.46 b) The accrued interest two months later PV = $10 000; r =

= 0.01625; t =

= 0.5

Interest = 10 000(0.01625)0.5 = $81.25 c) Cash price = $9245.46 + 81.25 = $9326.71 16) F V = 100 000(1.04) = $104 000; PMT = 100000(0.055)0.25 = $1375.00; n = 10.5(4) = 42; P/Y = 4; C/Y = 4; I/Y = 7.2; = 0.018; The purchase price on the interest date preceding the date of purchase PP = 104000(1.018)-42 + 1375 = 49161.67 + 40279.22 = $89 440.89 17) F V = $100 000; P/Y = 1; C/Y = 4; PMT = 100000(0.0875/1) = $8750.00; c = p=

= 4; n = 4(1) = 4.

- 1 = 0.071859031

PP = 103000.00(1.071859031)-4 + 8750 = 103000.00(.7576164) + 8750.00(3.373044266) = 78034.49 + 29514.14 = $107 548.63 18) F V = $1 000 000(1.04) = $1040 000; P/Y = 1; C/Y = 4; PMT = 1 000 000(0.09/1) = $90 000; c = p=

- 1 = .08243216

PP = 1040000(1.08243216)-5 + 90000 = 1040000.00(.6729713331) + 90000.00(3.96724612) = 699890.19 + 357052.15 = $1 056 942.34 19) F V = $50 000; P/Y = C/Y = 2; PMT = 50000 = $1 375.00; a) The interest date preceding the purchase is August 10, 2013.

The time period until maturity is 20 years. n = 20(2) = 40; i =

= 0.025;

PP = 50000(1 + 0.025)-40 + 1375

= 4;

i=

P 5 P = The purchase price is $53 137.85 1862b) The number of days from August 10, 2013 to February 10, 2014 is 184 days.

1.531 The number of days from August 10, 2013 to December 10, 2013 is 122 days.

18 +PV = $53 137.85; r = = 0.0275; t = 3451 6.315 Interest = 53137.85(0.0275) = $968.90 7= $53 The accrued interest is $968.90 c) The cash price is $53 137.85 + 968.90 = $5106.75 137.8 20) F V = 50 000; PMT = 5000

= $1125.00; P/Y = C/Y = 2; i =

= 0.0275

Since the maturity date is Nov.4, the semi-annual interest dates are May 4 and Nov. 4. The interest date preceding the date of purchase is November 4, 2012. The time period from Nov. 4, 2012 to November 4, 2022 is 10 years. n = 10(2) = 20 The purchase price on the date preceding the date of purchase PP(Nov. 4, 2012) = 50000(1.0275)-20 + 1125 = 29062.52832 + 17130.65865 = $46 193.19 The purchase price is $46 193.19 The number of days from Nov.4, 2012 to May 4, 2013 is 181 days.

The number of days from Nov.4, 2012 to March 20, 2013 is 136 days.

PV = $46 193.19; r = i = 0.0225; t = FV = $46193.19

= $46 974.14

The cash price on March 20, 2013 is $46 974.14 21) F V = $250 000; P/Y = 2; C/Y = 2; b = = 0.035; PMT = 250 000(0.035) = $8750.00 a) i =

= 0.0325; n = 4(2) = 8.

Since b > i, the bond will sell at a premium.

PP = 250 000(1.0325)-8 + 8750

=

= 193 561.744 + 60 779.661 = $254 341.41 The premium is $254 341.41-$250 000 = $4341.41 b) i =

= 0.04; n = 4(2) = 8.

Since b < i, the bond will sell at a discount.

PP = 250 000(1.04)-8 + 8750

=

= 182 675.551 + 58 911.518 = $241 587.07 The discount is $250 000 - $241 587.07 = $8412.93 22) F V = $10 000; P/Y = 2; C/Y = 2; n = 3(2) = 6; b =

= 0.036; i =

= 0.03.

Since the bond rate is greater than the yield rate (b > i), the bond is sold at a premium.

PMT = 10 000

= $360.00

PP = 1000 0(1.0 = 8374.84 + 1950.189 = $10 325.03 The purchase price is $10 325.03 3)-6 The premium is 10 325.03 - 10 000 = $325.03 + 360 23) F V = $50 000; P/Y = C/Y = 2; PMT = 50 000

= $1 000; n = 3(2) = 6; i =

= 0.03.

Since the bond rate is less than the yield rate (b < 1), the bond is sold at a discount.

PP = 50000(1 + 0.03)-6 + 1000 PP = 41 874.21283 + 5 417.19144 = $47 291.40 The purchase price is $47 291.40 The discount is 50 000 - 47 291.40 = $2708.60 24) F V = $1000; P/Y = C/Y = 1; b = = 0.07; n = 8(1) = 8 a) i =

= 0.065; Since b > i the bond will sell at a premium.

Premium = [1000(0.07) - 1000(0.065)] = 5(6.088751) = $30.44 PP = 1000.00 + 30.44 = $1034.44 b) i =

= 0.075; Since b < i the bond will sell at a discount.

Discount = [1000(0.07) - 1000(0.075)] = -5(5.8573036) = -$29.29 PP = 1000 - 29.29 = $970.71 c) i =

= 0.085; Since b < i the bond will sell at a discount.

Discount = [1000(0.07) - 1000(0.085)] = -15.00(5.639183) = -$84.59 PP = 1000 - 84.59 = $915.41 25) F V = $1000; P/Y = C/Y = 1; n = 4(1) = 4; b = 0.065; i = 0.075; Since b < i, the bond will sell at a discount Discount = [1000(0.065) - 1000(0.075] = -10(3.349326) = $33.49 PP = 1000 - 33.49 = $966.51 26) F V = $130 000; P/Y = C/Y = 4; n = 6(4) = 24; b =

= 0.01125; i =

= 0.013125

a) Since b < i , the bond will sell at a discount Discount = [130 000(0.01125) - 130 000(0 .013125)]

27)

= -243.75(20.473452) = - $4990.40 b) = 130 000) 000 -=4990.40 $125 009.60 F VPP = 9($10 $90 000;= P/Y = C/Y = 2; b = = 0.02; i = Since b < i, the bonds are sold at a discount.

= 0.03; n = 8.5(2) = 17.

Disc ount 0(0.02) - 90000(0.03)] = = -900(13.1661185) = $11 849.51 [9000 = 90000 FV = $50PP 000; P/Y = -4;11849.51 C/Y = 2; n==$78 5(4)150.49 = 20; c = 28)

= 0.5; p ==

- 1 = .00995049

Since b > i adjusted, the bond sells at a premium Premium = [50000(0.02) - 50 000(0.00995049)] 502.48(18.054551) = $9072.05 29) F V = 10($10 000) = $100 000; P/Y = C/Y = 2; b =

= 0.02; i =

= 0.035; n = 18.5(2) = 37.

Since b < i, the bonds are sold at a discount.

Discount = [100000(0.02) - 100000(0.035)]

30)

= -1500(20.570525) = -$30 855.79 PP 30 855.79 = $69 F V == 100000.00 20(5 000) =-$100 000; P/Y = 4;144.21 C/Y = 1; c = = 0.25; n = 8(4) = 32; b = p=

- 1 = 0.01347517. Since b > i, the bonds are sold at a premium.

Premium = [100000(0.015 ) - 100 000(0.01347517] = 152.482556(25.855038) = $3942.44 PP = 100 000.00 + 3 942.44 = $103 942.44 31) F V = $125 000; P/Y = C/Y = 4, b = = 0.01875; i =

= 0.015; n = 3(4) = 12;

PMT = 125 000(0.01875) = $2 343.75. Since b > i, bond sells at a premium. (i) Premium = [125000(0.01875) - 125000(0.015)] = 468.75(10.907505) = $5 112.89 PP = 125 000.00 + 5112.89 = $130 112.89 (ii)

32)

= 0.015;

FV = = 0.0725; $100 PMT = 100 000(0.0675) = $6750.00. Since b < i, the bond sells at a discount.

000; P/Y = (i) Discount = [100000(0.0675) -110000(0.0725)] C/Y = -1225(5.342633) = -$6544.73 = 1; n PP = 110000.00 - 6544.73 = $103 455.27 (ii) = 7(1) = 7; b =

= 0.06 7 5; i= 33) F V = $5 000(1.09) = $5450.00; P/Y = C/Y = 1; b =

= 0.065; i =

PMT = $5000(0.065) = $325 Discount = [5000(0.065) - 5450(0.07)] = -56.50(4.7665397) = -$269.31 PP = 5450.00 - 269.31 = $5180.69

34) F V = $5000; P/Y = C/Y = 1; b = 0.0625; i = 0.075; n = 8(1) = 8 Discount = [5000.00 * .0625 - 5000.00 * .075] = -$62.50(5.8573036) = -$366.08 PP = 5000.00 - 366.08 = $4633.92

= 0.07;

35) F V = 1000; P/Y = 2; C/Y = 2; n = (10 - 3)(2) = 14; b = 3%; i = 2.5% Since b > i, the bond sells at a premium Premium = [0.03(1000) - 0.025(1000)] = 5(11.690912) = $58.45 Programmed solution: PMT = (0.03 × 1000 - 0.025 × 1000) = 5

The purchase price is 1000 + 58.45 = $1058.45 The proceeds from the bond are 1000(1.02) = $1020.00 The loss on the sale is 1058.45 - 1020 = $38.45 36) F V = 6(5000) = $30 000; P/Y = C/Y = 2; n = (21-3)2 = 36; b = Since b < i, the bonds are sold at a discount.

= 0.0475; i =

Discount = [30 000(0.0475) - 30 000(0.0525)]

37)

= -150(16.0287454) = -$2404.31 Book value 30000.00 - 2404.31= $27 595.69 Proceeds = 30000(1.03625 ) = $31 087.50 Gain F V = $100 000; P/Y = C/Y = 2; b = $ 3491.81 = 0.035; i =

= 0.04; n = (11.5 - 5)2 = 13

Since b < i, the bonds are sold at a discount.

Discount = [100000(0.035) - 100 000(0.04)]

38)

= -500(9.985648) = -$4992.83 Book value = 100 000 - 4992.83 = $95 007.17 Proceeds = 100000(0.99125 = $99 125.00 Gain $ 4117.83 F V = $50 000; P/Y = C/Y = 2; b = = 0.03; i =

= 0.05; n = (12 - 5)2 = 14;

Since b < i, the bonds are sold at a discount.

Discount = [50000(0.03) - 50 000(0.05)]

39)

= -1000(9.898641) = -$9898.64 Book value = 50 000 - 9898.64 = $40 101.36 Proceeds = 50000(0.99375) = $49 687.50 Gain 49 687.50 40 101.36 = $9586.14 F V = =3($25 000) =- $75 000; P/Y = C/Y = 2; b = 0.03; i = Discount = [75000(.03 - .035)]

= -375.00(6.873956) = -$2577.73

Book value = 75 000.00 - 2577.73 = $72 422.27 Proceeds = 75000.00 * .89625 = $67 218.75 Loss on sale = $5203.52 40)

= 0.035; n = 4(2) = 8.

= 0.0525.

Disc ount 40000.00 * .035] = = -400.00(5.328553) = -$2131.42 [4000 Book value = 40000.00 - 2131.42 = $37868.58 0.00 * Proceeds = 40000.00 * .9975 = = $2031.42 39900.00 .025 - Gain 41) F V = $50 000; P/Y = C/Y = 2; n = 4(2) = 8; Discount = [50000.00 * .03 - 50000.00 * .04]

42)

= -500.00(6.732744875) = -$3366.37 Book value = 50000.00 - 3366.37 = $46633.63 Proceeds = 50000.00 * .9925 = 49625.00 Gain $2991.37 F V = $3(100 000) = $300 000; P/Y = =C/Y = 4; b = = 0.01375; i =

= 0.015;

The interest date preceding the purchase date is December 1, 2013, 3.75 years before maturity n = (3.75)4 = 15 Since b < i, the bonds are sold at a discount.

Discount = [300000(0.01375) - 300 000(0.015)] = -375(13.343233) = -$5003.71 Book value = 300 000 -5003.71 = $294 996.29 on December 1, 2013 The number of days from Dec.1, 2013 to Jan.21,2014 is 51 days.

The number of days from Dec.1,2013 to March 1, 2014 is 90 Accumulated price on Jan. 21, 2014 294 996.29

= $297 503.76

Accrued interest to Jan. 21, 2004 Interest = 300000(0.01375)

= $2337.50

Proceeds = 300000(0.935)+ 2337.50 = $282 837.50 43)

Loss on sale==[500000.00 297 503.76*-.01125 282 837.50 = $14 666.26 a) Discount - 500000.00 * .015]

= -1875.00(22.7267167) = -$42612.59 PP = 500000.00 - 42612.59 = $457 387.41 b) Discount = 500000.00(-.00375) -1875.00(3.8543847) = -$7226.97 Book value = 500000.00 - 7226.91 = $492 773.09 = $498 125.00 c) Proceeds: 500000.00 * .99625 $5351.91 Gain 44) F V = $25 000; P/Y = C/Y = 2; b =

= 0.04; i =

Since b > i, the bond sells at a premium.

Premium = [25 000(0.04 ) - 25 000(0.03)] = 250(14.877475) = $3719.37 Book value: 25 000 + 3719.37 = $28 719.37 Proceeds: 25 000(1.0725) = $26 812.50 Loss = $1906.87

= 0.03; n = 10(2) = 20

45) F V = $25 000; P/Y = C/Y = 2; b =

= 0.035; i =

= 0.03; n = (22 - 7)2 = 30;

Premium = [25000(0.035) - 25 000(0.03)] = 125(19.6004414) = $1274.03 Book value: 27000.00 + 1274.03 = $28 274.03 = $24 562.50 Proceeds: 25000(0.9825) = $3711.53 Loss 46) Q uoted price = 50000(0.94375) = $47 187.50 Redemption price = $50 000.00 Average book value = .5(50 000 + 47 187.50) = $48 593.75 the number of interest payments to maturity is 9(2) = 18 Total interest = 18(50000)(.03) = $27 000.00 Bond discount = 50 000.00 - 47 187.50 = $2 812.50 Average income per payment interval = Approximate value of i =

(27 000 + 2 812.50) = $1656.25 = .0340836

Approximate yield = .0340836 * 2 = 6.81672% 47) Q uoted price = 100000(0.94125) = $94 125.00 Redemption price = $100000.00 Average book value = .5(100000.00 + 94125.00) = $97 062.5 Total interest = 18(100000.00)(.04) = $72 000.00 Bond discount = 100000.00 - 94125.00 = $5 875.00 Average income per payment interval = Approximate value of i =

(72000.00 + 5875.00 ) = $4326.39

= .044573239

Approximate yield = .044573239 * 2 = 8.1946% 48) T he quoted price (initial book value) is 40 000(1.025) = $41 000 Redemption Value is 40 000 Average book value =

(41000 + 40000) = $40 500

The semi-annual interest payment is 40 000

= $1100.00

The number of interest payments to maturity is 12(2) = 24 The total interest payments 24(1100) = $26 400 The premium is 41000 - 40 000 = $1000 Average income per interest payment interval is Approximate value of i =

= 0.026131605 = 2.61%

The yield rate is 2(2.61) = 5.22% 49) Q uoted price: 100000.00 ∗ .7425 = $74 250.00 Average book value: .5(100000.00 + 74250.00) = $87 125.00 Semi-annual interest: 100000.00 * .03 = $3000.00 Total interest: 35(3000.00) = $105000.00 Discount: 100000.00 - 74250.00 = $25750.00 Average income:

(3000.00 + 105000.00) = $3085.71

= 1058.33

Appr e of i is oxim ate valu

= .0354170445

Annual rate: 2(.0354170445) = 7.0834%

50) I nitial book value: 100000.00 * 1.025 = $102 500.00 Redemption price: 100000.00 * 1.05 = $105 000.00 Average book value: .5(102500.00 + 105000.00) = $103 750.00 Total interest: 46(100000(0.025) = $115 000.00 Discount = 102500.00 - 105000.00 = $2500.00 Average income per interest payment interval = Approximate value of i:

= $2554.35

= .0246202

Annual rate: 2(.0246202) = 4.92405% 51) T he interest dates are January 15 and July 15. The nearest interest date is January 15, 2004 (11.5 years to maturity) The quoted price is 50 000(0.975) = $48 750 The redemption value is 50 000 The average book value is

(48750 + 50000) = $49375.00

The semi-annual interest is 50000

= $1250

The number of interest payments to maturity is 11.5(2) = 23 The total interest payments are 23(1250) = $28 750.00 Discount is 50 000 - 48 750 = $1250 = $1304.35

Average income per interest interval is Approximate value of i is

= 0.026417215 = 2.64%

The yield rate is 2(2.64) = 5.28% 52) Q uoted price: 250000(0.7825) = $195 625.00 Average book value: .5(250 000 + 195 625) = $222 812.50 Semi-annual interest: 250 000(0.0325) = $8125.00 Total interest: 35(8125.00) = $284 375.00 Discount: 250000.00 - 195625.00 = $54 375.00 Average income:

(8125.00 + 284375.00) = $8357.14

Approximate value:

= .0375075

Annual rate: 2(.0375075) = 7.5015% 53) 100000.00(1.05)-15 + 4500.00 100000.00(.4810170981) + 4500.00(10.37965804) 48101.71 + 46708.46 = $94 810.17 PP = 94810.17 = 94810.17(1.005769231) = $95 357.15 54) 10000.00(1.0503)-15 + 425.00 10000.00(.4789603) + 425.00(10.3586421) 4789.60 + 4402.42 = $9192.02

PP = 9192 . = 9192.02(1.005803846) = $9245.37 0255) [10000.00 * .03 - 10000.00 * .025] 50.00(9.5142087) = $475.71 Premium PP1 = 10000.00 + 475.71 = $10 475.71 PP2 = 10475.71 = 10475.71(1.0040055) = $10 517.67 56) PP1 = 200000.00(1.025)-19 + 6000.00 = 200000.00(.6255277) + 6000.00(14.9788913) = 125105.54 + 89873.35 = $214 978.89 PP2 = 214978.89 = 214978.89(1.0103591) = $217 205.88 57) 105000.00(1.04)-5 + 3500.00 105000.00(.8219271) + 3500.00(4.4518223) 86302.35 + 15581.38 = $101883.73 PP = 101883.73 = 101883.73(1.0180435) = $103 722.07 Accrued interest = 100000.00 * .035 *

= $1578.80

Quoted price = 103722.07 - 1578.80 = $102 143.27 58) C 59) B 60) B 61) C 62) B 63) B 64) D ays of accrued interest = 99 Market Value = $1000 × 104% = $1040 Accrued interest = $1000 × 0.08 ×

= $22

Purchase payment for bond = $1040 + $22 = $1062 65) E 66) E 67) F V = $1000; PMT = $1000 × = $40 n = 3.5 × 2 = 7;

P/Y = 2;

C/Y = 2;

I/Y = 2.2%;

i=

Purchase price of the bond on January 1, 2012 = $1,000 × 1.011-7 + $40 Number of days from January 1, 2012 and February 8, 2012 = 39 Number of days from January 1, 2012 and July 1, 2012 = 183 Interest = $1,000 × 0.04 ×

= $8.53

= 1.1% = $1194.35

Purc hase price of the bond = $1202.87 68) E 69) B 70) C 71) F V = $25 000; n = 3 × 4 = 12; b =

= 1.35%; i =

= 0.875% = $1347.16

Premium = $25000 × (0.0135 - 0.00875) Purchase price = $25000 + $1347.16 = $26 347.16 E nd of

Bond Interest Interest on Book value at Amount of Interest Received (Coupon) yield rate Premium Payment i = 0.875% Amortized b = 1.35% Interval 012345 $230.54 $106.96 6 7 8 9 10 $337.50 $337.50 $229.60 $107.90 11 12 $337.50 $228.66 $108.84 $337.50 $227.71 $109.79 $337.50 $226.74 $110.76 $337.50 $225.78 $111.72 $337.50 $224.80 $112.70 $337.50 $223.81 $113.69 $337.50 $222.82 $114.68 $337.50 $221.81 $115.69 $337.50 $220.80 $116.70 $337.50 $219.78 $117.72 72) F V = $25000;

n = 3 × 4 = 12;

b=

= 1.35%; i =

Discount = $25000 × (0.0175 - 0.0135)

Book Value of Remaining Bond Premium $26,347.16 $26,240.20 $26,132.30 $26,023.46 $25,913.66 $25,802.91 $25,691.18 $25,578.48 $25,464.79 $25,350.11 $25,234.42 $25,117.72 $25,000.00

$1,347.16 $1,240.20 $1,132.30 $1,023.46 $913.66 $802.91 $691.18 $578.48 $464.79 $350.11 $234.42 $117.72 $0.00

= 1.75%

= $1073.95

Purchase price = $25000 - $1073.95 = $23 926.05 End of Interest Payment Interval

Bond Interest Interest on Received Book value at (Coupon) yield rate i = b = 1.35% 1.75%

Amount of Discount Accumulated

Book Value of Remaining Bond Discount

0 1 2 3 4 5 6 7 8 9 10 11

$337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50 $337.50

$81.21 $82.63 $84.07 $85.54 $87.04 $88.56 $90.11 $91.69 $93.30 $94.93 $96.59

$23,926.05 $24,007.26 $24,089.88 $24,173.96 $24,259.50 $24,346.54 $24,435.11 $24,525.22 $24,616.91 $24,710.21 $24,805.14 $24,901.73

$418.71 $420.13 $421.57 $423.04 $424.54 $426.06 $427.61 $429.19 $430.80 $432.43 $434.09

$1,073.95 $992.74 $910.12 $826.04 $740.50 $653.46 $564.89 $474.78 $383.09 $289.79 $194.86 $98.27

12 $337.50 $435.77 $98.27 73) E 74) C 75) E 76) B 77) A 78) C 79) B 80) E 81) E 82) A 83) A 84) C 85) B 86) D 87) E 88) E 89) i = 0.042/2 = 0.021 = 2.1%; n = 9 * 2 = 18 Periodic interest payment = 4300 * 0.028/2 = $60.20 Present Value of Principal PV = 4300(1.021)-18 = $2958.05

$25,000.00

$0.00

Present Value of Interest Payments PV = 60.20

= 60.20(14.861050) = $894.64

Purchase Price = 894.64 + 2958.05 = $3852.69 90) i = 0.036/12 = 0.003 = 0.3%; c = 12/2 = 6; p = 1.003^6 - 1 = 0.018136; n = 5 * 2 = 10 Periodic interest payment = 10 000(0.038/2) = $190 Present Value of Principal PV = 10000(1.018136)-10 = $8534.95 Present Value of Interest Payments PV = 190

= 190(9.070845) = $1723.46

Purchase Price = 8354.95 + 1723.46 = $10 078.41 91) i = 0.05/2 = 0.025 The last interest date prior to her redemption is 6.5 years before maturity: n = 6.5 * 2 = 13 Periodic interest payment = 9100(0.044/2) = 200.20 Present Value of Principal PV = 9100(1.025^-13) = $6601.33 Present Value of Interest Payments PV = 200.20

= 200.20(10.983185) = $2198.83

Purchase Price on interest date preceding purchase date Purchase Price = 6601.33 + 2198.83 = $8800.16 Accrued interest and Accumulated Price n = 3/6 = 0.5 Cash Price = 8800.16(1.025)0.5 = 8909.48 92) S ince the bond rate b is greater than the yield rate i, the bond will sell at a premium b = 0.094/2 = 0.047; i = 0.068/2 = 0.034; n = 2 * 2 = 4 Periodic Interest Payment: 0.047(100) = $4.70 Required Interest: 0.034(100) = $3.40 Excess actual interest: 4.70-3.40 = $1.30

Prem ium = 1.30

= 1.30( 3.681 816) = $4.79

Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

1) A n obligation can be settled by making a payment of $7500.00 now and a final payment 1) _____________ of $10 000.00 in five years. Alternatively the obligation can be settled by payments of $750.00 at the end of every three months for five years. If interest rate is 10% compounded quarterly, determine the preferred alternative.

2) Y ou win a lottery and have a choice of taking $200 000.00 immediately or taking payments of $7311.15 at the end of every three months for ten years. Which offer is preferable if interest is 8% compounded quarterly?

2) _____________

3) A n obligation can be settled by making a payment of $12 000 now and a final payment of $32 000 in 4 years. Alternatively, the obligation can be settled by payments of $2700 at the end of every three months for five years. Interest is 10% compounded quarterly.

3) _____________

Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion.

4) A n obligation can be settled by making a payment of $16 000 now and a final payment of $30 000 in 3 years. Alternatively, the obligation can be settled by payments of $2500 at the end of every three months for four years. Interest is 12% compounded quarterly.

4) _____________

Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion.

5) A n expenditure may be met by outlays of $1700 now and $2210 at the end of every six months for 6 years or by making monthly payments of $500 in advance for seven years.

Interest is 11% compounded annually.

5) _____________

Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion.

6) A n expenditure may be met by outlays of $3000 now and $1000 at the end of every six months for 5 years or by making monthly payments of $250 in advance for three years.

Interest is 12% compounded annually.

6) _____________

Compute the present value of each alternative and determine the preferred alternative according to the discounted cash flow criterion.

7) A selection has to be made between two investment alternatives. The first alternative offers a net return of $42 430.00 after three years, $30 000.00 after five years and $26 000.00 after seven years. The second alternative provides a net return of $13 000.00 per year for seven years. Determine the preferred alternative according to the discounted cash flow criterion if money is worth 11%.

7) _____________

8) D onna and Keith want to sell their business. They have received two offers. If they accept Offer A they will receive $61 000 immediately and $20 000 in three years. If the accept Offer B they will receive $37000 now and $3000 at the end of every six months for 5 years. If interest is 6.67%, which offer is preferable?

8) _____________

9) S ean and Jessica want to sell their interest in a small business. They have received two offers. If they accept Offer A they will receive $40 000 immediately and $30 000 in two years. If the accept Offer B they will receive $50 000 now and $2000 at the end of every six months for 5 years. If interest is 8%, which offer is preferable?

9) _____________

10) A piece of property may be acquired by making an immediate payment of $125 000 and payments of $37 500 and $50 000 three and five years from now respectively.

Alternatively, the property may be purchased by making quarterly payments of $11 150 in advance for five years. Which alternative is preferable if money is worth 12.2% compounded semi-annually?

10) _____________

11) A car costs $29 700. Alternatively, the car can be leased for three years by making payments of $540 at the beginning of each month and can be bought at the end of the lease for $14 750. If interest is 9.2% compounded bi-weekly, which alternative is preferable?

11) _____________

12) A new car costs $21 000. Alternatively, the car can be leased for three years by making payments of $360 at the beginning of each month and can be bought at the end of the lease for $10 000. If interest is 8% compounded semi-annually, which alternative is preferable?

12) _____________

13) A wireless telephone system with a disposable value of $5 000 after five years can be purchased for $15 000. Alternatively, a leasing agreement is available that requires an immediate payment of $2000 plus payments of $100.00 at the beginning of each month for five years. If money is worth 6% compounded monthly, should the telephone system be leased or purchased?

13) _____________

14) A telephone system with a disposable value of $4200 after five years can be purchased for $16 600. Alternatively, a leasing agreement is available that requires an immediate payment of $1900 plus payments of $150.00 at the beginning of each month for five years. If money is worth 9.6% compounded monthly, should the telephone system be leased or purchased?

14) _____________

15) A contract is estimated to yield net returns of $7000.00 quarterly for seven years. To secure the contract, an immediate outlay of $80 000.00 and a further outlay of $60 000.00 three years from now are required. If interest is 6% compounded quarterly, determine if the investment should be accepted or rejected.

15) _____________

16) A n investment project requires an initial expenditure of $160 000.00 with a salvage value of $30 000.00 after ten years. It is estimated that it will have annual returns of $23 149.00 for ten years. Should the company undertake this project if it wants to achieve a 9% rate of return?

16) _____________

17) T amia Industries plans to replace the outdated equipment that will cost the company $100 000.00 now and $60 000.00 six years from now. This replacement will result in revenues of $6000.00 at the end of each quarter for twelve years. At an interest rate of 9% compounded annually and using the Net Present Value criterion, should the company replace this equipment or not?

17) _____________

18) R eplacing old equipment at an immediate cost of $75 000 and an additional outlay of $10 000 six years from now will result in savings of $3120 per quarter for 11 years. The

reqed rate of return uiris 11.4%

compou 18) nded annually.

Use the net present value method to determin e whether the company should replace old equipme nt or not.

___ ___ ___ ___ _

19) R eplacing old equipment at an immediate cost of $75 000 and an additional outlay of $10 000 six years from now will result in savings of $3120 per quarter for 11 years. The required rate of return is 8% compounded annually. Use the net present value method to determine whether the company should replace old equipment or not.

19) _____________

20) A company has two investment choices. Alternative A requires an immediate outlay of $4000.00 and offers a return of $14 000.00 after seven years. Alternative B requires an immediate outlay of $3600.00 in return for which $500.00 will be received at the end of every six months for the next seven years. If the rate of return is 6% compounded semi-annually, determine which alternative is preferable.

20) _____________

21) P roject A requires an immediate investment of $18 000 and another $16 000 in three years. Net returns are $4500 after two years, $13 000 after four years, and $8900 after six years. Project B requires an immediate investment of $4000, another $6000 after two years, and $4000 after four years. Net returns are $3375 per year for 8 years. Determine the net present value at 11%. Which project is preferable according to the net present value criterion?

21) _____________

22) S uppose you are offered two investment alternatives. If you choose Alternative 1, you will have to make an immediate outlay of $26 000. In return, you will receive $1500 at the end of every three months for the next ten years. If you choose Alternative 2, you will have to make an outlay of $14 000 now and $8000 in two years. In return, you will receive $60 000 ten years from now. Interest is 8.22% compounded semi-annually.

Compute the net present value for each alternative and determine which investment should be accepted or rejected according to the net present value criterion.

22) _____________

23) T he SHREK Company has to make a decision about expanding its production facilities.

Research indicates that the desired expansion would require an immediate outlay of $160 000 and an outlay of a further $60 000 in five years. Net returns are estimated to be $25 000 per year for the first five years and $20 000 per year for the following 9 years.

Find the net present value of the project. Should the expansion project be undertaken if the required rate of return is 10% compounded annually?

23) _____________

24) A company is considering a project that will require a cost outlay of $17 200 per year for 24) _____________ 3 years. At the end of the project the salvage value will be $15 000. The project will yield returns of $60 000 in Year 4 and in Year 5. There are no returns after Year 5.

Alternative investments are available that will yield a return of 14.2%. Should the company undertake the project?

25) A company is considering a project that will require a cost outlay of $25 000 per year for

25) _____________

3 years. At the end of the project the salvage value will be $15 000. The project will yield annual net returns of $17 500 for 5 years. Alternative investments are available that will yield a return of 15%. Should the company undertake the project?

26) B ig Sky Company expects a demand of 30 000 units per year for a special purpose component for six years. Net return per unit is $4.10. To produce the component, the company must buy a machine costing $245 000 with a life of six years and a salvage value of $47 000 after six years. The company estimates that repair costs will be $18 000 per year during Years 2 to 6. If Big Sky company requires a return on investment of 15.5%, should it market the component?

26) _____________

27) L D Winery expects a demand of 10 000 bottles per year for a special purpose wine for six

27) _____________

years. Net return per unit is $6.10. To produce the wine, LD must buy equipment costing $200 000 with a life of six years and a salvage value of $10 000 after six years. The company estimates that repair costs will be $8000 per year during Years 2 to 6. Should LD invest in the equipment if it requires a return of 16% on its investment?

28) R eplacing old equipment at an immediate cost of $5000 and $7000 six years from now

28) _____________

will result in a savings of $3000 semi-annually for ten years. At 11% compounded annually, should the old equipment be replaced?

29) R eplacing old office equipment at an immediate cost of $3000 and $4000 four years from now will result in savings of $400 semi-annually for ten years. Should the old equipment be replaced a) at 8% compounded annually? b) at 12% compounded annually?

29) _____________

30) A company is considering a project that will require a cost outlay of $30 000 per year for

30) _____________

three years. At the end of the project, the company expects to salvage the physical assets for $50 000. The project is estimated to yield net returns of $60 000 in Year 4, $40 000 in Year 5, and $15 000 for each of the following five years. Alternative investments are available yielding a rate of return of 14.5%. Compute the net present value of the project.

31) A new venture that requires outlays of $127 000 for each of the first two years will yield net returns of in each year for years 3 to 6 and $70 000 for each of the following four years. A residual value of can be recovered at the end of the last income period. Should the venture be undertaken if a yield of 11.56% is required?

31) _____________

32) A special chemical development project requires an immediate outlay of $110 000 and $50 000 at the end of each year for 3 years. Net returns are nil for the first 3 years and $60 000 per year thereafter for fourteen years. What is the net present value of the project at 17%?

32) _____________

33) A commitment on the project requires an initial outlay of $10 000.00 and a further outlay

of $5000.00 after

eleven 33) years.

Net returns are $5 000.00 per year for five years.

What is the net present value of the project at 16%?

___ ___ ___ ___ _

34) A project requires an initial outlay of $350 000 and a further outlay of $100 000 after one year. Net returns are $105 000 per year for five years. What is the net present value of the project at 9.9%?

34) _____________

35) A once in a lifetime project requires an immediate outlay of $100 000 and $25 000 at the end of each year for 3 years. Net returns are nil for the first 3 years and $30 000 per year thereafter for fourteen years. What is the net present value of the project at 16%?

35) _____________

36) A ssume that the net present value of a project is $ 3870 at 10%, and -$1853 at 12%. Use linear interpolation to compute the rate of return correct to the nearest tenth of a percent.

36) _____________

37) A ssume that the net present value of a project is $ 220 at 22%, and -$305 at 24%. Use linear interpolation to compute the rate of return correct to the nearest tenth of a percent.

37) _____________

38) A n investment of $180 000 yields annual net returns of $27 700 for seven years. What is the rate of return on the investment (correct to the nearest tenth of a percent) if part of the initial investment, in the amount of, $80 000 is recovered at the end of 7 years? a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR.

38) _____________

39) T he introduction of a new product requires an immediate outlay of $145 000 and has a residual value of after 10 years. The anticipated net returns from the marketing of the product are expected to be per year for ten years. What is the rate of return on the investment (correct to the nearest tenth of a percent)? a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR.

39) _____________

40) A restaurant may be purchased for $250 000. Annual net income from the operation of the restaurant is expected to be $61 000 for each of the first 4 years and $30 000 for each of the next three years. After 7 years, the restaurant can be sold for $315 000. Determine the rate of return on the investment correct to the nearest tenth of a percent. a) Use linear interpolation to find the approximate value of the rate of return. b) Find the answer using Cash Flow and IRR.

40) _____________

41) A project requiring an immediate investment of $150 000 and a further outlay of $60 000 after four years has a residual value of $50 000 after nine years. The project yields a negative net return of $10 000 in Year 1, a zero net return in Year 2, $60 000 per year for the following four years, and $70 000 per year for the last three years. Find the rate of return (correct to the nearest tenth of a percent).

41) _____________

42) T he Radium Hot Springs plans to install a swimming pool. Construction of the lift is estimated to require an immediate outlay of $420 000. The life of the pool is estimated to be 20 years with a salvage value of $20 000. Cost of preparing the area is expected to be $30 000 for each of the first 2 years of operation. Net cash inflows from the pool are expected to be $49 000 for each of the first five years and $90 000 for each of the following 15 years. Find the rate of return (correct to the nearest tenth of a percent).

42) _____________

Use Excel's IRR function to answer the question.

43) T he introduction of a new product requires an initial outlay of $610 000. The anticipated net returns from the marketing of the product are expected to be $92 300 per year for 12 years. Find the rate of return (correct to the nearest tenth of a percent).

43) _____________

44) T he owner of a music store is considering remodelling the store in order to carry a larger

44) _____________

inventory. The cost of remodelling and additional inventory is $43 200. The expected increase in net profit is $7000 per year for the next 3 years and $10 000 each year for the following 7 years. After ten years, the owner plans to retire and sell the business. She expects to recover the additional $40 000 invested in inventory but not the $43 200 invested in remodeling. Compute the rate of return.

45) A project requires an initial outlay of $100 000 and promises net returns of $18 500 per year over a twelve-year period. If the project has a residual value of $4000 after twelve years, what is the rate of return?

45) _____________

46) B illy Bob wants to convert his farm into a ski resort. He asked you to determine his rate of return based on the following estimates.

- Development cost for each of the first 2 years, $87 000.

- Construction of a chalet in Year 3, $1 240 000.

- Upon his retirement in fifteen years, improvements in the property will yield him $270 000.

- Net returns from the operation of the golf course will be nil for the first three years and $200 000 per year afterwards until his retirement.

46) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

47) A company buys equipment for $15 000 today and has annual net cash inflows of $6000 for 3 47) ______ years. The discount rate is 12% compounded annually. What is the Net Present Value (NPV)?

A) - $652.90 B) - $525.90 C) $ 552.90 D) $ 652.90 E) $ 992.50 48) A company needs to spend $15 000 for each of the next three years. Net returns beginning in Year 4 are estimated at $2500 per year for ten years. The required rate of return is 9.75% compounded annually. The NPV is?

A) - $27 521.41 B) $ 25 217.41 C) - $27 721.41 D) - $25 721.41 E) $ 27 521.41

48) ______

49) A company has the following net cash inflows. Today -$7000, Year 1 -$15 000, Year 2 -$9000, Year 3 + $12 000, Year 4 -$3000, Year 5 + $19 000. Compute the net present value if the required rate of return is 7.22% compounded annually.

A) $ 4709.89 B) - $7009.89 C) - $3000.00 D) - $4709.89 E) - $7409.89

49) ______

50) A company is looking to invest in a very risky project. They have a required rate of return of 27% compounded annually. The project has the following cash inflows: Year 1 $17500, Year 2 $15000, Year 3 $27500. It also has the following cash outflows: Immediately -$10 000, Year 1 -$15 000, Year 3 -$9500. What is the NPV? A) $ 9718.06 B) - $9718.06 C) $ 7918.06 D) $ 7819.06 E) $ 9918.06

50) ______

51) A company has cash outflows of $21 275 starting today for each of the next 5 years. After that they will have cash inflows of $16 000 for each of the following 7 years. The discount rate is 9% compounded annually. What is the NPV? A) - $30 415.15 B) - $41.51 C) - $34 015.15 D) - $341.51 E) - $4015.15

51) ______

52) A company spends $117 500 today and has positive cash inflows of $34 000 for each of the next 6 years. What is the Internal Rate of Return (IRR)?

A) 1 4.87% B) 1 5.87% C) 1 7.48% D) 1 6.84% E) 1 8.47%

52) ______

53) A company has the following pattern of cash flows. Today -$47 000, Year 1 -$15 500, Year 2 + $16 000, Year 3 + $35 000, Year 4 + $57 000. What is the IRR? A) 4 9.301% B) 3 9.301% C) 9 .301% D) 1 9.301% E) 2 9.301%

53) ______

54) A company is planning on investing the following monies. They spend Today -$26 000, Year 1 -$7000, Year 4 -$11 000, and Year 7 -$10 100. Their cash inflows are Years 1-3 inclusive + $12 000, Years 4-9 inclusive + $16 000. What is their IRR? A) 1 5.18% B) 2 5.18% C) 5 .18% D) 1 0.18% E) 3 5.18%

54) ______

55) A company has an immediate cash outlay of -$325 000 and additional subsequent cash outlays of

55) ______

-$6000 per year for the life of the project. They will receive cash inflows in year 3 for + $77 000 and increasing by $3000 per year until the project ends 7 years later (total of 10 years). What is the IRR? A) 0 .79% B) 7 .79% C) 1 0.91% D) 3 .79%

E) 9 .79%

56) W hat is the IRR for the following net annual cash flows today -$45 000, Year 1 + $12 000, Year 2 + $37 000, Year 3 + $12 000, Year 4 + $17 000?

A) 4 6.722% B) 1 6.722% C) 2 6.722% D) 3 6.722% E) 6 .722%

56) ______

57) N ick has a choice to pay $1499 for a TV now or pay $1599 3 years from now. What should be the

57) ______

minimum interest rate at which Nick should invest $1499 to make a decision to pay 3 years from now?

A) 2 .176% B) 3 % C) 6 .67% D) 0 .725% E) 6 .528% SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

58) M olly needs to decide whether to buy a water heater for $2200 cash and enter a service a years lease contract requiring a payment of $30 at the end of every 3 months for 10 years or to enter 10 requiring a

payment 58) of $90 at the beginnin g of every 3 months.

If the leased water heater can be bought after 10 years for $80, should Molly buy or lease the water heater, if money is compou nded quarterly at 2.5%?

59) K evin needs to decide whether to buy a Honda Civic for $19 900 with a salvage price of $6500 after 5 years or lease the car for 5 years making monthly payments of $269 at the beginning of each month. If money is worth 5% compounded annually, should Kevin buy or lease?

___ ___ ___ ___ _

59) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

60) A house is on sale in Markham. Marlene has an option to pay $575 000 lump sum or pay $6000 at 60) ______ the end of every month for the next 10 years. If money earns 5% compounded monthly, which option has a better economic advantage?

A) P aying lump sum has an advantage of $9312 in PV B) P aying monthly has an advantage of $9312 in PV C) P aying lump sum has an advantage of $85 129 in PV D) P aying monthly has an advantage of $85 129 in PV E) B oth options are the same. Choose any.

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

61) I f Ontario decides to build two new units of nuclear power plants, the overnight cost of 61) _____________ construction is $15 billion. The annual net rate of return from the two units is $1 billion every year for the next 60 years, at which time the plants should be decommissioned and have a residual value of $0.96 billion. Should Ontario decide to build the two units of nuclear power plants if the required rate of return is 7%?

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

62) A local community church is contemplating installment of solar cells on its roof and sell the access

___ ___

electricit 62) y to the Ontario grid via a power purchase agreeme nt. The project requires an initial investme nt of $75 000 with a residual value of $2000 after 10 years. It is estimate d to yield annual net returns of $15 000 for 10 years.

What is the NPV of the project given a rate of return of 7%?

A) $ 32 387

B) $ 31 370

C) - $31 370

D) - $32 387

E) $ 1017

63) P fizer is planning to embark on developing a new medicine for the cure of common cold. If undertaken, the project requires $10 million dollar on R&D, every year for 5 years. The medicine will be marketed in Year 6 and net returns are estimated to be $10 million for the next 5 years at the end of which Pfizer plans to sell the formula to its competitor for $1 million. If corporation requires a ROI of 15%, what is the NPV for the project?

63) ______

A) - $16.85 million B) $ 16.6 million C) - $16.6 million D) $ 0.123 million E) $ 16.85 million 64) A firm invests $200 000 in machinery that yields net after-tax cash flows of $90 000 at the end of

each of the

next 64) three years.

The opport u nity cost of capital is 12%.

What is the net present value of the project (to the nearest thousand dollars)?

A) $ 16 000

___ ___

B) - $16 000

C) - $8000

D) $ 8000

E) $ 0

65) S aint Mary's is offered a contract, which offers a net present value of $15 000 if the required rate

65) ______

of return is 12%. Alternatively, net present value is -$12 000 if the required rate of return is 18%.

At what rate of return, would the NPV be zero?

A) 1 3% B) 1 4% C) 1 3.7% D) 1 4.5% E) 1 5% 66) S aint Mary's is offered a contract, which requires an immediate investment of $25 million. The estimated returns are $5 million per year for 20 years. Compute the rate of return.

A) 1 3.79% B) 7 .75% C) 9 9.9% D) 1 9.42% E) 1 2.45%

66) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

67) K orea Nuclear tritium removal facility (TRF) project requires an immediate investment 67) _____________ of $33 million with a residual value of $7 million at the end of the project. It is expected to yield a net return of $1 million in Year 1 from the sale of immobilized tritium to ITER, $5 million dollars in Year 2, $8 million per year for the following six years, and $7 million per year for the remaining four years. Find the rate of return using Excel's IRR function.

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

68) S uleiman must make a choice between two investment alternatives. Alternative 1 will provide 68) ______ him returns of $300 000 at the end of second year and $700 000 at the end of fifth year.

Alternative 2 will provide him returns $180 000 at the end of each of the next six years.

Alternative 3 will provide him returns $370 000 at the end of second, fourth and sixth years.

Which alternative is preferable if money is worth 9.4%?

A) A lternative 1 B) A lternative 2 C) A lternative 3 D) A lternative 2 or Alternative 3, as both are essentially same E) A lternative 1 or Alternative 3, as both are essentially same 69) S uleiman must make a choice between two investment alternatives. Alternative 1 will provide him returns of $300 000 at the end of second year and $700 000 at the end of fifth year.

Alternative 2 will provide him returns $180 000 at the end of each of the next six years.

Alte3 will rnatprovide ive him

returns 69) $370 000 at the end of second, fourth and sixth years.

Which alternati ve is preferabl e if money is worth 5.5556%?

A) A lternative 1 B) A lternative 2 C) A lternative 3 D) A lternative 2 or Alternative 3, as both are essentially the same E) A lternative 1 or Alternative 3, as both are essentially same 70) J eremy has a number of outstanding credit cards debts. He seeks help from BDO to consolidate his debts and make it easy for him to make the debt payments. BDO offers him two alternatives.

In the first alternative, the debt can be settled by making a payment of $30 000 now and a final payment of $42 000 in 3 years. Alternatively, the obligation can be settled by payments of $8400 at the end of every three months for three years. Which alternative is preferred if the interest is 8.5% compounded quarterly.

___ ___

70) ______

A) A lternative 1 B) A lternative 2 C) N either Alternative D) D on't take offer from the BDO E) A lternative 1 or Alternative 2, as both are essentially same 71) J eremy has a number of outstanding credit cards debts. He seeks help from BDO to consolidate his debts and make it easy for him to make the debt payments. BDO offers him two alternatives.

In the first alternative, the debt can be settled by making a payment of $30 000 now and a final payment of $42 000 in 3 years. Alternatively, the obligation can be settled by payments of $8400 at the end of every three months for three years. Which alternative is preferred and by how much if the interest is 13.5% compounded quarterly.

71) ______

A) A lternative 1 B) A lternative 2 C) N either Alternative D) D on't take offer from the BDO E) A lternative 1 or Alternative 2, as both are essentially same 72) J asmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 6.3% compounded semi-annually. What is Jasmine's most preferred option?

72) ______

A) A lternative 1 B) A lternative 2 C) A lternative 3 D) A lternative 3 or Alternative 2, as both are essentially same E) A lternative 1 or Alternative 2, as both are essentially same 73) J asmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 7.4% compounded semi-annually. What is Jasmine's most preferred option?

73) ______

A) A lternative 1 B) A lternative 2 C) A lternative 3 D) A lternative 3 or Alternative 2, as both are essentially same E) A lternative 1 or Alternative 2, as both are essentially same 74) J asmine has two investment choices. Alternative 1 requires an immediate outlay of $150 000 and offers a return of $417 000 after seven years. Alternative 2 requires an immediate outlay of $180 000 in return for which $25 000 will be received at the end of every six months for the next seven years. Alternative 3 requires an immediate outlay of $200 000 in return for which $60 000 will be received at the end of every year for the next seven years. The required rate of return on investment is 6.58% compounded semi-annually. What is Jasmine's most preferred option?

A) A lternative 1 B) A lternative 2 C) A lternative 3 D) A lternative 3 or Alternative 2, as both are essentially same E) A lternative 1 or Alternative 2, as both are essentially same

74) ______

75) C GI Inc. is developing a new software platform, which requires an immediate investment of $780 000 and another $260 000 in three years. Net returns expected are $450 000 after two years, $313 000 after four years, and $889 000 after six years. However, if they simply upgrade their current platform, it requires an immediate investment of $400 000, another $260 000 after two years, and $340 000 after four years. Net returns are $187 500 per year for 8 years. Determine the net present value at 5.9%. Which project is preferable according to the net present value criterion?

75) ______

A) U pgrade by $53 309 B) N ew Software platform by $76 599 C) N ew Software platform by $53 309 D) N ew Software platform by $70 203 E) U pgrade by $277 277 76) S imion Inc is considering to build a new office in Clagary, which requires an immediate investment of $78 million and another $35 million in one year. Being near to the customers, it is expected a revenue of $2 million in Year one, $10 million in year 2 and then a steady stream of $15 million revenue each year in the subsequent 12 years. However, if they simply expand their current offices in Toronto and build a satellite office in Calgary, it requires an immediate investment of $30 million, another $15 million after one year, and $10 million after two years.

Net returns are $expected to be $10 million per year from year 2 to year 14. Determine the net present value at 7.4%. Which option is preferable according to the net present value criterion?

A) B uild a new office with NPV of $21 million

76) ______

B) E xpanding current office is $30 million better in NPV compared to building a new office C) B uild a new office with NPV of -$21 million D) E xpand current office with NPV of $9 million E) E ither B or D 77) L G expects a demand of 10 000 headphones per year for a special purpose headphone for six years. Net return per unit is $6.10. To produce the headphones, LG must buy equipment costing $200 000 with a life of six years and a salvage value of $10 000 after six years. The company estimates that repair costs will be $8000 per year during Years 2 to 6. Should LG invest in the equipment if it requires a return of 16% on its investment? Calculate the NPV. A) N o, NPV = -$2679 B) N o, NPV = -$5530 C) Y es, NPV = $2679 D) Y es, NPV = $6783 E) N o, NPV = -$6783

77) ______

78) A ssume that the net present value of a project is $74.3 million at 6%, and -$21.2 million at 9%.

Use linear interpolation to compute the rate of return correct to the nearest hundredth of a percent.

A) 1 0.2% B) 2 .33% C) 8 .33% D) 6 .67% E) 4 .2%

78) ______

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

79) M atthew purchased a motorcycle and is offered two alternatives for payment. The first 79) _____________ option requires a $5550.00 now and a final payment of $12 325.00 in four years. The second alternative would require Matthew to make payments of $1085.00 at the end of every three months for four years. If interest rate is 8.2% compounded quarterly, determine the preferred alternative.

80) A ki meets with his grandparents, who offer him either $85 right now or $100 on his birthday 11 months from now. Which option should Aki choose to maximize the amount of money he earns? Suppose the interest rate is 12.5% compounded annually.

80) _____________

81) A afia is overseeing a project that will establish a new Faculty of Business building at her university. While analysts estimate that the building would result in $325 500 in annual net returns over the first four years, the project would require $1 100 500 in initial investment costs. If there is a 14.5% required rate of return, and the building will leave a residual value of $200 000 after four years, should Aafia pursue the development of the building?

81) _____________

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

82) A rjun is interested in opening a new branch for Habitat for Humanity, which will take $102 325 82) ______ in initial startup costs but will yield $24 775 in annual net returns every year for 6 years. Assume that after 6 years, there are no further benefits nor any residual value left on Arjun's investment.

If the required rate of return is 10%, should he start work on the project? What if the required rate of return is 17%?

A) Y es if the rate of return is 10%, No if the rate of return is 17%.

B) Y es for both rates of return.

C) Y es if the rate of return is 17%, No if the rate of return is 10%.

D) N o for both rates of return.

E) M ore information is required.

83) I f the present value of inflows is greater than the present value of outflows:

83) ______

A) N PV > 0, Profitability Index > 1, Rate of Return on Investment = i B) N PV = 0, Profitability Index > 1, Rate of Return on Investment > i C) N PV > 0, Profitability Index > 1, Rate of Return on Investment > i D) N PV < 0, Profitability Index < 1, Rate of Return on Investment > i E) N PV > 0, Profitability Index = 1, Rate of Return on Investment < i 84) T he annual net return on a project is $160.30 for 5 years, and the initial investment is $635. The required rate of return is 8.3%, and there is no residual value after 5 years. Which of the following are true?

A) N PV = 0, Profitability Index = 1, Rate of Return on Investment = i B) N PV < 0, Profitability Index > 1, Rate of Return on Investment > i C) N PV = 0, Profitability Index < 1, Rate of Return on Investment > i D) N PV > 0, Profitability Index = 1, Rate of Return on Investment < i E) N PV > 0, Profitability Index > 1, Rate of Return on Investment = i

84) ______

1) P V of 10 000: n = 5(4) = 20; i = PV = 10000(1.025)-20 = 6102.71

= 0.025; FV = 10 000; I/Y = 10;P/Y = C/Y = 4

ALT. 1 = 7500 + 6102.71 = 13 602.71 = $13 603 Programmed solution:

For ALT. 2: PMT = 750; n = 5(4) = 20; i = 0.025; I/Y = 10; P/Y = C/Y = 4 PV = 750

= 11 691.87 = $11 692

Programmed solution:

ALT. 2 is better (less to pay) 2) P MT = 7311.15; n = 10(4) = 40; i = 0.02; I/Y = 8; P/Y = C/Y = 4 PV = 7311.15

= $200 000.01 = $200 000

Either option is preferable.

3) ALT. 1 = 12000.00 + 32000.00(1.025)-16 = 12000.00 + 32000.00(.6736249) = 12000.00 + 21556.00 = $33 556 ALT. 2 = 2700.00

= 2700.00(15.5891623) = $42090.74 = $42 091

Since the present value of Alternative 1 is less than the present value of Alternative 2, Alternative 1 is preferable.

4) A LT. 1 = 16000.00 + 30000.00(1.03)-12 = 16000.00 + 30000.00(.701378802) = 16000.00 + 21041.40 = $37041.40 = $37 041.

ALT. 2 = 2500.00

= 2500.00(12.56110203) = $31402.76 = $31 403

At 12% compounded quarterly Alternative 2 is preferable.

5)

ALT. 1 p =

- 1 = .05356538

= 1700.00 + 2210.00 = 1700.00 + 2210.00(8.6876858) = 1700.00 + 19199.79 = $20899.79 = $20 900. BEST ALT. 2 p = = 500.00

- 1 = .0087346 (1.0087346)

= 500.00(59.3435117)(1.0087346) = $29930.93 = $29 931 6)

ALT. 1 p =

- 1 = .05833005

= 3000.00 + 1000.00 = 3000.00 + 1000.00(7.419712748) = 3000.00 + 7419.71 = $10419.71 = $10 420 - 1 = .00948879

ALT. 2 p = = 250.00

(1.00948879)

= 250.00(30.3747539)(1.009488793) = $7665.74 = $7666 BEST 7) A LT. 1 = 42430.00(1.11)-3 + 30000.00(1.11)-5 + 26000.00(1.11)-7 = 42430.00(.7311914) + 30000.00(.5934513) + 26000.00(.4816584) = 31024.45 + 17803.54 + 12523.12 = $64692.66 = $61 351.11.

ALT. 2 = 13000.00

8)

= 13000.00(4.7121963) = $61258.55 = $61 259

Since at 11% the present value of Alternative 1 is greater than the present value of Alternative 2, Alternative 1 is preferable. p= -1 = .0328117 ALT. A = 61000.00 + 20000.00(1.0667)-3 = 61000.00 + 20000.00(.8238974) = 61000.00 + 16477.95 = $77477.95 = $77 478 BEST ALT. B = 37000.00 + 3000.00 = 37000.00 + 3000.00(8.4090961) = 37000.00 + 25227.29 = $62227.29 = $62 227

9)

p=

-1 = .0392305 = 40000.00 + 30000.00(1.08)-2

ALT. A

= 40000.00 + 30000.00(.857339) = 40000.00 + 25720.16 = $65720.16 = $65 720 = 50000.00 + 2000.00

ALT. B

= 50000.00 + 2000.00(8.142056025) = 50000.00 + 16284.11 = $66284.11 = $66 284 BEST 10) A LT. 1 = 125000.00 + 37500.00(1.061)-6 + 50000.00(1.061)-10 = 125000.00 + 37500.00(.7009833) + 50000.00(.5531541) = 125000.00 + 26286.87 + 27657.71 = $178944.58 = $178 945.

ALT. 2:

I/Y = 12.2%; P/Y = 4; C/Y = 2; c =

PV of ALT.2 = 11 150

= 0.5; n = 4(5) = 20; p =

-1 = .03004854

(1.03004854)

= 11150.00(14.87080004)(1.03004854) = $170791.75 = $170 792 Since the present value of the Alternative 2 is smaller than the present value of the Alternative 1, Alternative 2 is preferable.

11) F V = $14 750; PMT = $540; c = 26/12 = 13/6 ; n = 12(3) = 36; i = 0.092/26 = 0.00353846 p=

- 1 = 0.007682495

ALT. CASH = $29 700.00

ALT. LEA SE

= 540(31.34623955)(1.007682495) + 14750(.7591827) = $28254.96 Since the present value of the lease is smaller than the present value of the cash payment, the lease is preferable (1.007682495) + 14750(1.007682495)-36 = 540

12) F V = $10 000; PMT = $360; I/Y = 8%; P/Y = 12; C/Y = 2; c = p=

; n = 3(12) = 36;

- 1 = .0065582.

ALT. CASH = $21 000.00 ALT. LEASE = 360.00

(1.0065581969) + 10000.00(1.0065581969)-36

= 360.00(31.97303733)(1.0065581969) + 10000.00(.7903145268) = 11585.78 + 7903.15 = $19488.93 = $19 489 BEST 13) P /Y = C/Y = 12; I/Y = 6%; i = 0.005; n = 5(12) = 60 PV of Purchase: = 15000.00 - 5000.00(1.005)-60 = 15000.00 - 5000.00(.7413721962) = 15000.00 - 3706.86 = $11 293.14 = $11 293 PV of Lease: = 2000 + 100

(1.005)

= 2000.00 + 100(51.72556075)(1.005) = 2000 + 5198.42 = $7198.42 At 6% monthly the present value of the decision to lease is smaller than the present value of the decision to buy.

The wireless telephone system should be leased.

14) P MT = $150; P/Y = C/Y = 12; n = 5(12) = 60; I/Y = 9.6%; i = 0.008 PV of Purchase: = 16 600 - 4200(1.008)-60 = 16 600 - 4200(.6199663) = 16600.00 -2603.86 = $13 996 PV of Lease: = 1900 + 150

(1.008)

= 1900 + 150(47.5042142)(1.008) = 1900 + 7182.64 = $9083 At 9.6% monthly the present value of the decision to lease is smaller than the present value of the decision to buy.

The telephone system should be leased.

15) P V of 60 000: n = 3(4) = 12; i = = 0.015; FV = 60 000; I/Y = 6; P/Y = C/Y = 4 PV = 60000(1.015)-12 = 50183.25 PVout = 80 000 + 50 183.25 = 130 183.25 = $130 183

Programmed solution:

For PVIN: PMT = 7000; n = 7(4) = 28; i = 0.015; I/Y = 6; P/Y = C/Y = 4 PVIN = 7000

= 159087.02 = $159 087

Programmed solution:

NPV = 159 087 - 130 183 = $28 904, accept the project.

16) F V = $30 000; n = 10(1) = 10; i =

= 0.09; PMT = 21 000;

PVout = 160 000 - 30000(1.09)-10 = $147 328 PVin = 23149

= 148562.35 = $148 562

NPV = 148 562 - 147 328 = $1234 > 0, accept the project.

17) P V of $60 000 Outlay is: FV = 60 000; n = 6(4) = 24; i = 0.09; I/Y = 9; P/Y = 4; C/Y = 1; c =

;

PV = 60000(1.09)-6 = 35 776.04 PVOUT = 100 000 + 35 776.04 = $135 776.04 = $135 776 Inflow is: PMT = 6000; n = 12(4) = 48 p = (1.09)1/4 - 1 = 0.0217782 PV = 6000

= 177 553.47

PVIn = $177 553 Programmed solution:

NPV = $177 553 - 135 776 = $41 777 > 0; Replace the equipment.

18) O utlays: FV = $10 000; I/Y = 11.4%; P/Y = C/Y = 1; n = 6(1) = 6; PV = 10 000(1.114)-6 = $5232.25 PVOUT = 75 000 + 5232.25 = $80 232 Inflows: PMT = $3 120; P/Y = 4; C/Y = 1; c = p=

= 0.25; n = 4(11) = 44;

- 1 = .0273568

PV = 3120

= $79 266.63 = $79 267

PVIn = $79 267 NPV = 79 267 - 80 232 = -$965 < 0, The equipment shouldn't be replaced at the current rate of return.

19) O utlays: FV = 10 000; I/Y = 8%; P/Y = C/Y = 1; n = 6(1) = 6 PV = 10 000 (1.08)-6 = $6301.70 PVOUT = 75 000 + 6301.70 = $81 302 Inflows: PMT = $3120; P/Y = 4; C/Y = 1; c =

= 0.25; n = 11(4) = 44;

p = 1.081/4 - 1 = .0194265; PVIN = 3120

= 91 724.25 = 91 724

NPV = 91 724 - 81 302 = $10 422 Since NPV > 0, the old equipment should be replaced.

20) A lterative A: PV of 14 000: n = 7(2) = 14; i =

= 0.03; FV = 14 000; I/Y = 6; P/Y = C/Y = 2

PV = 14000(1.03)-14 = $9255.65 Programmed solution:

NPVA = 9255.65 - 4000.00 = $5255.65 = $5256 Alternative B: PMT = 500; n = 7(2) = 14; i = 0.03; I/Y = 6; P/Y = C/Y = 2 PV = 500

= $5648.04

Programmed solution:

NPVb = 5648.04 - 3600.00 = $2048.04 = $2048 Since NPVA > NPVB, alternative A is preferred.

21) A LT.A: PVIN = 4500

+ 13 000

+ 8900

= $16 974.

PVOUT = 18 000 + 16 000

= 3652.30 + 8563.50 + 4758.30

= 18 000 + 11 699.06 = $29 699

NPVA = 16 974 - 29699 = -$12 725 ALT. B: PVIN= 3375

= $17 368

PVOUT = 4000 + 6000

+ 4000

= 4000 + 4869.73 + 2634.92 = $11 505

NPVB = 17 368 - 11 505 = $5863 Since NPVB > NPVA , alternative B is preferable.

22) A LT.1: PMT = $1500; I/Y = 8.22%; P/Y = 4; C/Y = 2; n = 10(4) = 40; c = p= NPV1

- 1 = .02034308 = -26000 + 1500 = -26000.00 + 1500.00(27.19160583) = -26000.00 + 40787.41 = $14787.41 = $14 787

ALT. 2: Outlays $14 000 now and $8000 in two years PVOUT = 14 000 + 8000(1.0411)-4 = $20 809.58 = $20 810 PVIN = 60000(1.0411)-20 = $26 810.34 = $26 810 NPV2 = 26 810 - 20 810 = $6000 Since NPV1 > NPV2, alternative 1 is preferred.

= 0.5; i =

= 0.0411;

23) P V OUT = 160000 + 60000(1.10)-5 = $197 255 PVIN = 25000

(1.10)-5

+ 20000

= 25000(3.7907868) + 20000(5.7590238)(.6209213) = $166 287.68 = $166 288 NPV = 166 288 - 197 255 = -$30 967 Since NPV < 0, the project shouldn't be undertaken.

24) O utlays: PMT = $17 200; P/Y = C/Y = 1; I/Y = 14.2%; i = 0.0142 (1.142) -15000(1.142)-5 = 45449.8213 - 7722.5502 = 37 727.27

PVOUT = 17200

PVOUT = $37 727 PVIN = 60000(1.142)-4 + 20000(1.142)-5 = 35276.6093 + 10296.7336 = 45 573.34 PVIN = $45 573 NPV = 45 573 - 37 727 = $7 846 The company should undertake the project.

25) O utlays: PMT = 25 000; P/Y = C/Y = 1; I/Y = 15%; i = 0.015 PVOUT = 25 000

(1.15) - 15 000

= 65642.7221 - 7457.6510 = 58 185.07

PVOUT = $58 185 PVIN = 17 500

= 58 662.71

PVIN = $58 663 NPV = 58 663 - 58 185 = $478. The net present value is slightly positive, which means that the net present value method does not provide a clear signal as to whether to undertake the project or not.

26) I nflows: PMT = 30 000(4.10) = $123 000; I/Y = 15.5%; n = 6(1) = 6; P/Y = C/Y = 1 PVIN = 123000.00

= 459 290.62 = $459 291

Outlays: FV = $47 000; PMT = 19 600 - 47000.00(1.155)-6

PVOUT = 245 000 + 18 000

= 245 000 + 59 631.3170 - 19 797.2995 = $284 834 NPV = 459 291 - 284 834 = $174 457 Since the net present value is positive, the rate of return on investment is greater than 15.5%. Big Sky Company should market the component.

27) I nflows: PMT = $10 000(6.10) = $61 000; P/Y = C/Y = 1; I/Y = 16%; n = 6 PVIN = 61000

= 224 769

PVOUT = 200 000 + 8 000

- 10 000

= 200 000 + 26 194.3492 - 4104.4225 = 222 089.9267 = $222 090 NPV = 224 769 - 222 090 = $2679 Since NPV > 0, the return on the investment is greater than 16%. LD Winery should invest in the equipment.

28)

PMT = $3 000; P/Y = 2; C/Y = 1; I/Y = 11; c = PVIN = 3000 PVOUT = 5000 + 7000 NPV = 36 282 - 8742 = $27 540

= 0.5; n = 10(2) = 20; p =

= 36 281.7688 = $36 282 = 8742.48 = $8742

- 1 = .053565

Since NPV > 0, the old equipment should be replaced.

29) a) PMT = 400; P/Y = 2; C/Y = 1; I/Y = 8%; n = 10(2) = 20; c = PVIn = 400

= 0.5; p =

- 1 = 0.039230

= 5473.36 = $5473

PVOUT = 3000 + 4000

= $5940 NPV = 5473 - 5940 = -$467 Since NPV is slightly negative, the net present value method doesn't provide a clear signal as to whether to replace the equipment or not. b) PMT = 400; P/Y = 2; C/Y = 1; I/Y = 8%; n = 10(2) = 20; c = = 0.5; p = - 1 = 0.058301 PVIn = 400

= 4651.94 = $4652

PVOUT = 3000 + 4000

= 3000 + 2542.07 = $5542 NPV = 5542 - 4 652 = $890 Since NPV is positive, the return on the investment is greater than 12%. The old equipment should be replaced.

30) O utlays: PMT = 30 000; P/Y = C/Y = 1; I/Y = 14.5% (1.145) - 50 000(1.145)-10 = 79 084 -12 909.6699 = $66 174

PVOUT = 30 000 Inflows:

PVIN = 60000(1.145)-4 + 40000 (1.145)-5 + 15000

(1.145)-5

= 34 908.3486 + 20 325.0938 + 25 855.2358 = $81 089 NPV = 81 089- 66 174 = $14 915 Since NPV is positive, the rate of return on the project is greater than 14.5%. The project should be undertaken.

31) O utlays: PVOUT = 127 000 + 127 000 - 130 000 = 127 000 + 113 840.09 - 43 536.98 = $197 303 Inflows: (1.1156)-2 + 70000

PVIN = 85000

(1.1156)-6

= 209 379.10 + 111 321.34 = $320 700 NPV = 320 700 - 197 303 = $123 397 Since NPV > 0, the rate of return on the venture is greater than 11.56%. The venture should be undertaken.

32) I nflows: PMT = 60 000; P/Y = C/Y = 1; I/Y = 17%; n = 14(1) = 14 (1.17)-3 = $195 901 PVIN = 60000 Outlays: PMT = 50 000; n = 3(1) PVOUT = 110 000 + 50000

= 110 000 + 110 479.2481 = $220 479

NPV = 195 901 - 220 479 = -$24 578 The net present value of the project at 17% is -$24 578.

33) P V OUT = 10 000 + 5000 PVIN = 5000

= $10 977.08 = $16 371

NPV = 16 371 - 10 977 = $5394 The net present value of the project at 16% is $5394.

34) P V OUT = 350 000 + 100 000

= $440 992

PVIN = 105 000

= $399 052

NPV = 399 052 - 440 992 = -$41 940 Since NPV < 0, the rate of return on the project is less than 9.9%.

35)

(1.16)-3 = $105 084

PVIN = 30 000 PVOUT = 100 000 + 25000

= 100 000 + 56 147.2385 = $156 147

NPV = 105 084 - 156 147 = -$51 063 36) T he ROI is contained between 10% < ROI < 12% Rate

NPV

10% 12% x%

$3870 -$1853 0

d=

= 1.35%

The rate of discount for which the NPV = 0 is approximately 10% + 1.35% = 11.4% 37) T he ROI is contained between 22% < ROI < 24% Rate

NPV

22% 24% x%

$220 -$305 0

d=

= 0.84%

The rate of discount for which the NPV = 0 is approximately 22% + 0.84% = 22.8% 38) P V OUT = 180 000 - 80 000 (1 + i)-7; PVIN = 27 700 NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4 rounded.

Calculations below show the attempts in approximating the ROI i = 12% i = 10% i = 8% PVOUT 143812 138947 133321 PVIN NPV Index Inc/Decr Rate

126416 -17396 88 -3 ROI < 12%

134855 -4092 97 -1 ROI < 10%

144216 10896 108 2 ROI > 8%

Hence, 8% < ROI < 10%. Use linear approximation to find the approximate value of ROI. d=

= 1.45%

The correct to the nearest tenth of a percent, is 9.5% rate b) CF0 = 180 000± of C01 = 27 700 F01 = 6 retur C02 = 107 700 F02 = 1 n, Compute IRR = 9.429578 = 9.4% 39) P V OUT = 145 000 - 30 000(1 + i)-10; PVIN = 25 500 NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4 rounded

Calculations below show the attempts in approximating the ROI Attempts i = 16% i = 10% i = 14% PVOUT 138 199 133 434 136 908 PVIN NPV Index Inc/Decr Rate

123 247 -14 952 89 -3 ROI < 16%

156 686 232 534 117 4 ROI > 10%

133 011 -38973897 97 -1 ROI < 14%

i = 12% 135 341 144 081 8740 106 2 ROI > 12%

Hence, 12% < ROI < 14%. Use linear approximation to find the approximate value of ROI. d=

= 1.38%

The rate of return, correct to the nearest tenth of a percent, is 13.4% b) CF0 = 145 000 ± C01 = 25 500 F01 = 9 C02 = 55 500 F02 = 1 Compute IRR = 13.353832 = 13.4% 40) P V OUT = 250 000 - 315 000(1 + i)-7; PVIN = 61 000

+ 30 000

(1 + i)-4;

NPV = PVIN - PVOUT Index =

× 100%; Increase/Decrease in Rate = (Index - 100%)/4

Calculations below show the attempts in approximating the ROI Attempts i = 16% i = 28% i = 24% PVOUT 138 544 194 045 180 119 PVIN NPV Index Inc/Decr Rate

207 901 69 357 150 12 ROI > 16%

157 580 -364 644 81 -5 ROI < 28%

171 802 -8317 95 -1 ROI < 24%

i = 22% 171 695 179 768 8073 105 1 ROI > 22%

Hence, 22% < ROI < 24%. Use linear approximation to find the approximate value of ROI. d=

= 0.985%

The rate of return, correct to the nearest tenth of a percent, is 22 + 0.985 = 23.0% b) CF0 = 250 000±

C01 = 61 000

F01 = 4 C02 = 30 000 C03 = 345 000

F02 = 2 F03 = 1 Compute IRR = 22.958989 = 23.0% 41) C F0 = 150 000 ± C01 = 10 000 ± F01 = 1 C02 = 0 C03 = F02 = 1 60 000 C04 = 0 F03 = 1 C05 = 60 000 F04 = 1 C06 = 70 000 F05 = 2 C07 = 120 000 F06 = 2 F07 = 1 Compute IRR = 17.261192 = 17.3% 42) C F0 = 420 000 ± C01 = 19 000 F01 = 2 C02 = 49 000 F02 = 3 C03 = 90 000 F03 = 14 C04 = 110 000 F04 = 1 Compute IRR = 13.596638% = 13.6% 43)

44)

45)

46)

47) B 48) D 49) E 50) C 51) A 52) E 53) D 54) E 55) C 56) C 57) A 58) P resent value of cash payment for the water heater = $2200 For PMT = $30;

n = 40; P/Y = 4;

C/Y = 4;

I/Y = 2.5%;

i = 0.625%

Present value of the service contract involves an ordinary annuity = $30

= $1058.85

Present value of decision to buy the water heater = $3258.85 For PMT = $100;

n = 40;

i = 0.625%

Present value of the lease payment annuity due = $90 × 1.00625 × Present value of the purchase price after 10 years = $80 × 1.00625-40 = $62.35 Present value for the decision to lease = $3258.74

= $3196.39

Both options result into the same economic value, within the rounding error.

59) P resent value of the salvage price = $6500 × 1.05-5 = $5092.92 Net present value if Kevin decides to buy = $19 900 - $5092.92 = $14 807.08 For monthly lease: PMT = $269;

p=

P/Y = 12;

C/Y = 1;

I/Y = 5%;

c=

;

n = 60;

i = 5%

-1 = 0.004074

PV = $269 × 1.004074 ×

= $14 351.23

PV for lease is smaller than PV to buy. Therefore Kevin should lease the Honda Civic.

60) B 61) P MT = $1 billion; n = 60; i = 7% PV IN = $1 = $14.04 OUT = $15 billion - $0.96 billionbillion PV

= $14.9834329 billion

NPV = $14.04 - $14.9834329 billion = -$0.94 billion Since NPV is negative, Ontario should not build the two units of nuclear power plants.

62) B 63) C 64) A 65) C 66) D 67) I nitial Cash Flow Net Cash Flow (in $million) (in $million) -$33 $1 New Cash Flow Period 1 $5 New Cash Flow Period 2 $8 New Cash Flow Period 3 $8 New Cash Flow Period 4 $8 New Cash Flow Period 5 $8 New Cash Flow Period 6 $8 New Cash Flow Period 7 $8 New Cash Flow Period 8 $7 New Cash Flow Period 9 $7 New Cash Flow Period 10 $7 New Cash Flow Period 11 $14 New Cash Flow Period 12 16.38% Internal Rate of Return 68) B 69) D 70) B 71) A 72) C 73) B 74) D 75) A

76) D 77) C 78) C 79) n = 4(4) = 16; i = 8.2%/4 = 0.0205; PMT = 1085

Alternative 1 PV = 12325(1.0205)-16 = 8907.97 Total Payment = 5550 + 8907.97 = 14 459.97 Alternative 2 PV = 1085

= 14 673.63

Therefore, Alternative 1 is the better option as there is less to pay.

80) F V = $100; i = 0.125; n = 11/12 PV = 100(1.125)-11/12 = 89.77 Since the PV of the $100 payment is larger than the $85 alternative, Aki should choose to receive $100 on his birthday.

81)

PV in = 325500

PV out = 1100500 - 200000

= 938773.85

= 984138.83

NPV = 938773.85 - 984138.83 = -45364.98 Since the NPV is negative, Aafia should not pursue development of the building.

82) A 83) C 84) A