Solutions to Interest Theory Sample Questions - Exam FMactuarialbrew.com/data/documents/AB_FM_Sample_Exam_Solutions.pdfSolutions to Interest Theory Sample Questions Solution 1 C Chapter

SOA FM Sample Exam Solutions

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Solutions to Interest Theory Sample Questions

Solution 1

C Chapter 4, Interest Rate Conversion

After 7.25 years, the value of each account is the same:

7.25 27.25

7.25

0.04100 1 1002

1.3326ln(1.3326) 7.25

e

e

0.0396

Solution 2

E Chapter 7, Level Annuities

Let j be the effective interest rate for an interval of 4 years:

4(1 ) 1j i

There are 10 4-year intervals in 40 years, and there are 5 4-year intervals in 20 years. Therefore, the accumulated value at the end of 10 intervals is equal to 5 times the accumulated value at the end of 5 intervals:

10 510 5

10

5

5

5

100 5 100

(1 ) 1 (1 ) 15/ ( 1) / ( 1)

(1 ) 1 5(1 ) 1

(1 ) 1 5

(1 ) 40.3195

j js s

j jj j j j

jj

j

jj

The accumulated amount at the end of 40 years is:

10

10 0.31951.3195 1100 100 1.3195

0.3195X s

6,194.72

The BA-II Plus can be used as follows:

4 [yx] 0.2 [=] [] 1 [=] [] 100 [=] [I/Y]

[2nd] [BGN] [2nd] [SET] [2nd] [QUIT]

10 [N] 100 [PMT] [CPT] [FV]

Result is 6,194.72. Solution is 6,194.72.


Page 2 © ActuarialBrew.com 2015

Solution 3

C Chapter 3, Simple Interest

Eric and Mike earn the same amount of interest during the last 6 months of the 8th year:

15

15

100 1 2002 2 2

1 22

i i i

i

i

0.946

Solution 4

A Chapter 12, Sinking Fund

The deposits to the sinking fund are equal to 1,627.45 minus the interest on the loan:

1,627.45 10,000 0.10 627.45

The accumulated value of the deposits is:

10

10 0.141.14 1627.45 627.45 627.45 19.3373 12,133.1858

0.14s

After repaying the loan, the balance is:

12,133.1858 10,000 2,133.19

The BA-II Plus can be used to answer this question:

10 [N] 14 [I/Y] 627.45 [PMT] [CPT] [FV]

[+/-] [] 10,000 [=]

Solution is 2,133.19.

Solution 5

E Chapter 13, Dollar-weighted Rate of Return

The income is the withdrawals minus the deposits, treating the initial balance as a deposit and the final balance as a withdrawal:

Income Withdrawals Deposits 60 5 25 80 35 75 10 12 10

The fund exposure is the average amount in the fund:

Fund exposure (Net deposit)(Time deposit is in the fund)

11 10 1 0 10 6 2.5 275 10 5 25 80 3512 12 12 12 12 12 12 1211 12 47075 10 90.83332 12 12



The simple interest approximation for the dollar-weighted rate of return is the income divided by the fund exposure:

Income 10Fund exposure 90.8333

i 0.1101

Solution 6

C Chapter 8, Varying Annuities

The equation of value at the end of 1 year can be used to solve for na :

77.1 1.105 ( )

77.1 1.105

77.1 0.1051.105

8.0955

nn

n nn

n

n

nIa vi

i a nv nv

a

a

The BA-II Plus can be used to solve for n:

10.5 [I/Y] 8.0955 [PV] 1 [PMT] [CPT] [N]

Result is 19.00.

Solution 7

C Chapter 8, Varying Annuities

The accumulated value is:

9 8

9 8

1,000(0.06) 100 1.09 900(0.06) 100 1.09 100(0.06) 100

160 1.09 154 1.09 106

Using the PIn method, we have:

1 160 6 10P I n

The present value is:

10

100 1

6 1 (1.09) 6 10160 (1.09)0.09 0.09 0.09

93.3333 6.4177 281.6072 880.5886

nn

I InPV P a vi i

The accumulated value at the end of 10 years is:

10880.5886 1.09 2,084.67

Using the BA II Plus, we have:

10 [N] 9 [I/Y] 160 [] 6 [] 0.09 [=] [PMT]

6 [] 10 [] 0.09 [=] [FV]

[CPT] [PV]



PV = 880.5886

0 [PMT]

[CPT] [FV] Answer = 2,084.67

Question 8 has been deleted from the set of sample questions.

Solution 9

D Chapter 12, Loans

The first 10 payments pay the principal down at a rate that is equal to 50% of the interest rate. Since the interest rate is 10%, the portion of the principal that is paid down by each of the first 10 payments is:

(1.50 1.00) 0.10 0.05

After 10 years, the original principal has been reduced by 5% for 10 years. The equation of value at the end of 10 years is:

1010 0.10

10

1,000 0.95

1 1.10598.74690.10

Xa

X

X

97.44


1,000 [] 0.95 [yx] 10 [=] [PV]

10 [N] 10 [I/Y] [CPT] [PMT]

Result is 97.4417. Solution is 97.44.

Solution 10

B Chapter 15, Bonds

The book value at the end of 6 years is:

6 4 0.06 410,0000.08 10,000 800 3.4651 7,920.93661.06

10,693.0211

BV a

The interest (which is also known as the investment income) portion of the 7th payment is:

7 6 10,693.0211 0.06InvInc BV y 641.58


4 [N] 6 [I/Y] 0.08 [] 10,000 [=] [PMT] 10,000 [FV]

[CPT] [PV]

Result is 10,693.0211.

[] 0.06 [=]



Result is 641.5813. Answer is 641.58.

Solution 11

A Chapter 11, Geometric Varying Annuities

At the end of 5 years, the value of the perpetuity’s remaining payments is equal to the value of the 25-year annuity-immediate:

2 2 3 24 25

2 3 252 24

100 1.08 1.08 1.080.08

100 1 1 1 11.08 1.08 1.080.08 1.08 1.08 1.08 1.08

100 250.08 1.08

100 1.080.08 25

X v v v v

X

X

X

X

54

Solution 12

C Chapter 5, Accumulated Value

The $10 initial deposit accumulates over the first 10 years (40 quarters) at a nominal discount rate of d compounded quarterly and then over the next 20 years (40 half years) at a nominal interest rate of 6% compounded semiannually. The $20 deposit at time 15 years accumulates for 15 years (30 half years) at a nominal interest rate of 6% compounded semiannually.

The equation of value at the end of 30 years can be solved for d:

40 3040

10 (1.03) 20(1.03) 100

14d

d

0.0453

Solution 13

E Chapter 5, Varying Force of Interest

At time 3, before the deposit of X, the value of the fund is:

3323 27300

0 100 0 300100 100 100t

t dte e e

A deposit of X is made at time 3, and the interest earned from time 3 to time 6 is equal to X:



3 36 327 300 300300

27300 0.63

100 1

100 1

96.0259 0.1224

e X e X

e X e X

XX

784.59

Solution 14

A Chapter 11, Geometric Progression Annuities

Problems involving geometric annuities can be solved using the formula for a geometric series:

2 1 (1 )1

First term Term that would come next1 Ratio

nn

na rS a ar ar ar

r

The present value of the perpetuity-immediate is 167.50:

2 3 4 5 6 2 7

6

5

5 6

6

1(1 )

10 10 10 10 10 10 (1 ) (1 ) 167.50

(1 ) 010 10 167.501 (1 )

1 1.092 (1 )10 10 167.500.092 1 (1 )

(1 )10 3.8696 10 167.501 (1 )

1 21.8407

0.0400K

v v v v v K v K v

K vaK v

K vK v

K vK v

v

K

The question referred to K% instead of K, so we multiply the value of K found above by 100:

0.0400 100 4.00

Solution 15

B Chapter 12, Loans

The amount of the equal annual payments under option (i) is:

10 0.0807

2,000 2,000 2,990.00736.6889a



Alternatively, the amount of the equal annual payments under option (i) can be found using the BA-II Plus calculator:

10 [N] 8.07 [I/Y] 2,000 [+/-][PV] [CPT] [PMT]

Result is 299.0007.

The sum of the payments under option (i) is:

299.0007 10 2,990.0073

Since the payments of 200 under option (ii) are over and above the payment of the interest, the balance of the loan decreases by 200 year. Therefore, the interest payments decline each year. The sum of the payments under option (ii) is:

200 10 (2,000 1,800 200) 2,000 200 (10 9 1)

10 112,000 200 2,000 11,0002

i i

i i

Setting the sum of the payments under option (i) equal to the sum of the payments under option (ii) allows us to solve for i:

2,990.0073 2,000 11,000990.007311,000

i

i

i

0.0900

Solution 16

B Chapter 11, Geometric Progression Annuities

There are 60 monthly payments. We use one month as the unit of time:

(12)

1

0.09 0.007512 12

1.0075 0.99256

i

v

The 60 payments are described below:

Time Payment 1 1,000 2 1,000 0.98 3 21,000 0.98 40 391,000 0.98 41 401,000 0.98 60 591,000 0.98



The outstanding balance after the 40th payment is the present value of the payments after the 40th payment:

40 41 2 59 2040

40 60 21

1,000 0.98 1,000 0.98 1,000 0.98

0.98 0.981,0001 0.98

PV v v v

v vv

6,889.11

Solution 17

C Chapter 6, Level Annuities

The equation of value at the end of 3n years can be used to find i:

222 2

8,000 98 (1 ) 196

8,000 98 (1 ) 1 (1 ) 196 (1 ) 1

8,000 98 2 1 4 196 4 1

nn n

n n n

s i s

i i i i

i

i

0.1225

Solution 18

B Chapter 8, Varying Annuities

We use one month as the unit of time. The monthly effective interest rate is:

1/30.091 1 0.00744

4

Using the PIn method, we have:

1 2 2 60 0.00744P I n i


0 1

60602 1 (1.00744) 2 602 (1.00744)

0.00744 0.00744 0.00744270.6568 48.2485 10,329.5813

nn

I InPV P a vi i

2,729.21


0.09 [] 4 [] 1 [=] [yx] 3 [1/x] [=] [] 1 [=] [STO] 1

60 [N] [RCL] 1 [] 100 [=] [I/Y] 2 [+] 2 [] [RCL] 1 [=] [PMT]

2 [] 60 [] [RCL] 1 [=] [+/-] [FV]

[CPT] [PV]

Result is PV = 2,728.11. Answer is 2,729.21



Solution 19

C Chapter 13, Dollar-Weighted and Time-Weighted Return

We can use the simple interest approximation to find the dollar-weighted rate of return for account K. The income and fund exposure are:

Income Withdrawals Deposits 125 100 2 25


100 0.5 0.25(2 ) 100

X X X

X X

The dollar-weighted rate of return is approximately:

Income 25Fund exposure 100

Xi

We set the dollar-weighted return for account K equal to the time-weighted return for account L:

2

25 125 105.8 1100 100 125

125 105.825 10012513,225125125

(125 ) 13,225125 115

10 or 240

XX

XX

XX

XX

X X

There are two possibilities for the value of i:

25 25 10 25 240 2150.15 or 2.15100 100 100 100

X

We use the positive value of 15%.

Solution 20

A Chapter 3, Present Value

The equation of value at the outset is:

2 10

2 10

10

100 200 300 600

100 200 0.76 300 0.76 600

0.7088

n nv v v

v

vi

0.03502



Solution 21

A Chapter 10, Continuously Payable Annuities

The equation of value at time 10 is:

10 110 (8 )

020,000 (8 )

bst

t

br ds

b ta

s ds

AV Pmt e dt

k t e dt

We begin by evaluating the integral in the exponent:

10 101 18(8 ) ln(8 ) ln(18) ln(8 ) ln8tt

s ds s tt

The equation of value can now be used to solve for k:

10

0

10

0

1820,000 (8 )8

20,000 18

20,000 (180 0)

k t dtt

kdt

kk

111.11

Solution 22

D Chapter 15, Bonds

The price of the bond is:

22

2

2 2

11,000 381.50

1,000 1.03125(1 ) 381.50 1,031.25(1 0.5889 ) 381.50

nn

ni

n

vP Coup a Rv ri

v

1,055.11

Solution 23

D Chapter 13, Net Present Value

The net present value of Project P is:

0 0 0 22,000 4,000(Cash Inflows) (Cash Outflows) 4,0001.10 1.10

1,123.9669

NPV PV PV

We can set this equal to the next present value of Project Q:

24,0001,123.9669 2,0001.10 1.10

X

X

5,460



Solution 24

E Chapter 12, Sinking Fund

We can use the BA-II Plus calculator to find the amount of the annual payment:

20 [N] 6.5 [I/Y] 20,000 [PV] [CPT] [PMT]

Result is 1,815.1279.

We reduce the payment amount by the interest that is paid to the lender in order to obtain the sinking fund payment that must accumulate to 20,000 at the end of 20 years. We can use the calculator to find the interest rate that the sinking fund must earn. Continuing from the calculation above, the keystrokes are:

[+] 0.08 [] 20,000 [=] [PMT]

Result is 215.1279, indicating that the sinking fund payment is 215.1279.

0 [PV] 20,000 [FV] [CPT] [I/Y]

Result is 14.1792.

The solution is 14.18%.

Solution 25

D Chapter 6, Perpetuities

Brian’s share of the present value of the perpetuity is 40%:

0.4

1 0.4

0.6

n

n

n

XXai

v

v

The charity’s share of the present value of the perpetuity is K:

2

20.6

nX Xv Ki i

KK

0.36

Solution 26

D Chapter 12, Loans

The interest paid by Seth is:

105,000 1.06 1 3,954.2385

The interest paid by Janice is:

5,000 10 0.06 3,000



The interest paid by Lori is the sum of the 10 payments minus the total principal paid. The total principal paid is equal to the initial principal of 5,000:

10 0.06

5,000 5,00010 5,000 10 5,000 679.3398 10 5,0007.3601

1,793.3979

a

The total amount of interest paid on all 3 loans is:

3,954.2385 3,000 1,793.3979 8,747.64

Solution 27

E Chapter 3, Accumulated Value

Since the amount of interest earned in Bruce’s account during the 11th year is equal to the amount of interest earned in Robbie’s account during the 17th year, the amount in Bruce’s account at the end of the 10th year must be equal to the amount in Robbie’s account at the end of the 16th year:

10 16

6100(1 ) 50(1 )

2 (1 )0.1225

i i

ii

The interest earned in Bruce’s account in the 11th year is:

10 10100(1 ) 100(1.1225) 0.1225X i i 38.88

Solution 28

D Chapter 12, Loans

The outstanding principal at the end of ( 1)t years is:

( 1) 1n t n ta a

The outstanding principal at the end of t years is:

n ta

The sum of the interest paid in year t and the principal paid in year ( 1)t is:

1 1

1 (1 ) 1 1 (1 ) 1

1 ( 1) 1

n t n t n t n tn t n t

n t n t

X a i a i v v v v

v v v d

The answer is Choice D.



Solution 29

B Chapter 7, Perpetuities

The present value of the first annuity can be used to find the effective 3-year interest rate:

(1/3)

(1/3)

1/3

1/3

10 32

0.3125

i

i

The 4-month effective interest rate is found below:

( ) ( )

3(3)1/3

(3) 1/31/3

1 1

1.3125 13

1.3125 1 0.030683

m pm pi im p

i

i

The present value of the second perpetuity is:

10.03068

X 32.60

Solution 30

D Chapter 17, Asset-Liability Matching

On 12/31/2017, the company will receive the par value of 822,703, leaving the following net liability:

1,000,000 822,703 177,297

Under Scenario A, the profit is:

4

4 0.0451.045 10.05 822,703 177,297 0.05 822,703 177,297

0.0451,312.97

s

Under Scenario B, the profit is:

4

4 0.0551.055 10.05 822,703 177,297 0.05 822,703 177,297

0.0551,322.78

s

Choice D best describes the insurance company’s profit.



Solution 31

D Chapter 11, Geometric Progression Annuities


Time Payment 1 5,000 1.07 2 25,000 1.07 3 35,000 1.07 20 205,000 1.07

The present value of the payments is:

2 2 20 20

215,000 1.07 5,000 1.07 5,000 1.07

1.07 1.071.05 1.055,000

1.0711.05

v v v

122,633.60

Solution 32

C Chapter 13, Net Present Value

The 60,000 received at the end of 3 years is lent for 1 year at 4%. The net present value is:

460,000(1.04) 60,000 100,000

1.05NPV

698.78

Solution 33

B Chapter 18, Spot Rates

The value of the bond is:

2 360 60 1,060

1.07 1.08 1.09 926.03

Solution 34

E Chapter 18, Spot Rates

We can use the BA-II Plus calculator to answer this question:

60 [] 1.07 [+] 60 [] 1.08 [x2] [+] 1,060 [] 1.09 [yx] 3 [=]

(Result is 926.0296)

[+/-] [PV] 3 [N] 60 [PMT] 1,000 [FV]

[CPT] [I/Y]

(Result is 8.9180)



Answer is 8.92%.

Solution 35

C Chapter 16, Duration

The modified duration can be expressed in terms of the derivative of the bond’s price or in terms of the Macaulay duration:

( )

( )( )

'

1

700100 1 0.08

m

mm

P y MacDurModDuryP ym

MacDur

MacDur

7.56

Solution 36


The price of the stock and the derivative of its price are:

1

2

( )

'( )

DivP y Div yy

P y Div y

The modified duration is:

2

1' 1 1 10

0.10P y Div yModDurP y yDiv y

The Macaulay duration is:

(1 ) 10 1.10MacDur ModDur y 11

Solution 37

B Chapter 16, Duration

The price of the stock and the derivative of its price are:

1

2

( ) ( )

'( ) ( )

DivP y Div y gy g

P y Div y g


2

1' ( ) 1 1 1

0.05 0.02 0.03( )33.3333

P y Div y gModDurP y y gDiv y g




(1 ) 33.3333 1.05MacDur ModDur y 35

Questions 38-44 have been deleted from the set of sample questions.

Solution 45

A Chapter 13, Project Appraisal

We can use the fact that the time-weighted rate of return is zero to solve for X:

12(1 0)10 12

1012 12120 10 12

60

XX

XX

X XX

The investment income and the exposure of the fund to interest are:

Income Withdrawals Deposits 10 10

Fund exposure (Net deposit)(Time deposit is in the fund) 10 0.5

10 0.5 60 40

X XX

The simple interest approximation for the dollar-weighted rate of return is:

Income 10Fund exposure 40

i 0.25

Solution 46

A Chapter 12, Loans

The final payment is the loan balance at the end of year 3 accrued with interest:

559.12 1.08 603.8496Pmt

The initial loan balance is:

4

41 1.08603.8496 2,000.0265

0.08Pmt a

The first principal payment is equal to the level payment minus the interest on the initial loan balance:

603.8496 2,000.0265 0.08 443.8475

Solution 47

B Chapter 15, Bonds



Since the yield is equal to the coupon rate, the price of the bond is 1,000. Since the investment in the bond results in a yield of 7%, we have the following equation of value:

0.510

20(1 ) 11,000(1.07) 30 1,000is

We can use the BA-II Plus calculator to answer this question:

1,000 [] 1.07 [yx] 10 [] 1,000 [=] [FV]

20 [N] 30 [+/-] [PMT] [CPT] [I/Y]

Result is 4.7597.

[] 100 + 1 [=] [x2] [] 1 [=]

Solution is 0.09746.

Solution 48

A Chapter 7, Level Annuities

The monthly effective interest rate is:

0.08 0.0066712

To have 3,000 of monthly income beginning on his 65th birthday, the man needs the following lump sum on his 65th birthday:

3,000 1,000 310,880.839.65

His contributions must accumulate to 310,880.83:

25 120.00667300

310,880.83

1.00667 1 1.00667 310,880.830.00667

957.3666 310,880.83

Xs

X

XX

324.72

We can use the BA II Plus to answer this question:


3,000 [] 9.65 [] 1,000 [=] [FV]

25 [] 12 [=] [N] 8 [] 12 [=] [I/Y]

[CPT] [PMT]

Answer is 324.72



Solution 49

D Chapter 6, Level Annuities

The parents make 17 contributions of X. On the daughter’s 18th birthday, the equation of value is:

17 16 2 3

172 3

1

1.05 1.05 1.05 50,000 1

1.05 50,000 1k

k

X v v v

X v v v

Only Choices A and D show 17 contributions, so the correct answer must be Choice A or Choice D.

The right side of the equation in Choice A shows the value of the withdrawals at time 17, but the left side of Choice A shows the value of the contributions at time 0, so the equation of value is not correct.

The right side of Choice D shows the value of the withdrawals at time 18, and the left side of Choice D shows the value of the contributions at time 18, so the equation of value is valid, and Choice D is the correct answer.


Solution 51

D Chapter 17, Dedication

The quantity of Bond II to purchase is the quantity that produces a cash flow of $1,000 at time 1:

1,000 0.975611,025IIQ

The quantity of Bond I to purchase is the quantity that produces the net liability remaining after the payment from Bond II is received:

1,000 25 0.97561 0.938091,040IQ

Choice D is the correct answer.

Solution 52

C Chapter 17, Dedication

The liability of 1,000 due in two years is met by arranging a payment of 1,000 in two years from Mortgage II. Since Mortgage II makes an equal-sized payment at time 1, it also pays 1,000 at time 1. That leaves 1,000 of net liability at time 1 to be met by Mortgage I. The amount lent is:

21,000 1,000 1,0001.06 1.07 1.07

X Y 2,751.41



Solution 53

A Chapter 18, Forward Rates

Bond II must produce a cash flow at time 2 that is sufficient to grow to 2,000 at a 6.5% interest rate:

2,000 1,877.93431.065

The combined prices of Bond I and Bond II are:

21,000 1,877.93431.06 1.07

2,583.66

Solution 54

C Chapter 15, Callable Bonds

Since the coupon rate is greater than the yield-to-worst, the bond is a premium bond. The yield-to-worst can therefore be found by identifying each interval defined by level redemption prices and considering the possibility that the bond is called at the beginning of each interval. In this case, there is only one such interval, and it runs from time 15 years until maturity. The yield-to-worst is based on the beginning of the interval, which occurs at the time 15 years. The equation of value is:

30 0.03 301,722.25 0.041.03

XXa


30 [N] 3 [I/Y] 0.04 [PMT] 1 [FV]

[CPT] [PV]

Result is 1.1960.

[1/x] [] 1,722.25 [=]

Result is 1,440.0030. Answer is 1,440.00.

Solution 55

B Chapter 15, Callable Bonds

We observe that the coupon of 40 is greater than the yield-to-worst times the final redemption value:

0.03 1,100 33YTW R

Therefore, the bond is a premium bond, and the earliest possible redemption within each interval of level redemption prices should be considered. The price-to-worst is the minimum of the two resulting prices:

300.03 40 0.0330 401,200 1,10040 , 401.03 1.03

Min a a




30 [N] 3 [I/Y] 40 [PMT] 1,200 [FV]

[CPT] [PV]

Result is 1,278.4018.

[STO] 1

40 [N] 1,100 [FV] [CPT] [PV]

Result is 1,261.8034.

Since 1,261.8034 is less than 1,278.4018, the price-to-worst is 1,261.8034.

Solution 56

E Chapter 15, Callable Bonds

The coupon is less than the product of the yield-to-worst and the final redemption value, so the bond is a discount bond:

0.02 0.03 ktX X Coup YTW R

Therefore, the yield-to-worst is based on the latest possible redemption, which occurs at the end of 10 years. The equation of value is:

200.03 201,021.50 0.021.03

XXa


20 [N] 3 [I/Y] 0.02 [PMT] 1 [FV]

[CPT] [PV]

Result is 0.8512.

[1/x] [] 1,021.50 [=]


Solution 57

B Chapter 15, Callable Bonds

Since the price is less than the redemption value of 1,100, the bond is a discount bond. Therefore, the yield-to-worst is based on the latest possible redemption, which occurs at the end of 10 years. The equation of value is:

20 201,1001,021.50 (0.02 1,100)

(1 )yay


20 [N] 1,021.50 [+/-] [PV] 0.02 [] 1,100 [=] [PMT] 1,100 [FV]

[CPT] [I/Y]

Result is 2.4559.



[] 2 [=]

Result is 4.9117.

Answer is 4.91%.


Solution 59

C Chapter 17, Asset-Liability Management

The Macaulay duration of the liability is:

15150

0 15

0 150

1515,000( ) 66.07210.06215,000 9.5866 9.5866

6.8921

tt

tt

a vt PV CFIa

MacDuraPV CF

Let X be the percentage of the present value that is invested in the 5-year bonds:

6.8921 5 (1 )100.6216

X XX

The amount invested in the 5-year bonds is:

1535,000 0.6216 35,000 9.5866X a 208,556.21

The BA II Plus can be used to answer this question:

15 [N] 6.2 [I/Y] 1[PMT] [CPT] [PV] [+/-]

Result is 9.5865.

[STO] 1 [] 1.062 [] 15 [] 1.062 [yx] 15 [=] [] 0.062 [=]

Result is 66.0721.

[] [RCL] 1 [=]

Result is 6.8921.

[] 10 [=] [] 5 [+/-] [=]

Result is 0.6216.

[] [RCL] 1 [] 35,000 [=]

Solution is 208,556.21.



Solution 60

A Chapter 11, Geometric Progression Annuities


Time Payment 1 2,000 2 2,000 1.03 3 22,000 1.03 8 72,000 1.03 9 72,000 1.03 0.97 10 7 22,000 1.03 0.97 16 7 82,000 1.03 0.97

The present value of the payments is:

2 8 7

7 8 2 2 8 8

8 9 9 97 8

2,000 2,000 1.03 2,000 1.03

2,000 1.03 0.97 0.97 0.97

1.03 0.97 0.972,000 2,000 1.031 1.03 1 0.97

2,000 6.5682 1,431.5956 5.275413,136.4140 7,552.2193

L v v v

v v v v

v v v vvv v

20,688.

63

Solution 61

E Chapter 5, Varying Force of Interest

The equation of value is:

0

2100

3150

2100

3150

0 3

0 3

500 2,000

exp 4

ln4

ts

s

s

s

s

ds

t

t

e

ds

ds

To evaluate the integral, let’s use the following substitution:

3

2

3150

50

su

sdu ds



The integral is:

23100

3150

33

10 1500 03 0

150150

0.5 0.5ln 0.5ln 3

30.5 ln 3 ln(3) ln

3

s

s

s tt s t s t sss s

tt

ds u du u

We can now solve for t:

3

15034

3

t

t

18.9

Solution 62

E Chapter 15, Bonds

The rate of growth of the accumulation of discount is equal to the yield:

5

(1 )

306.69 (1 )194.82

0.09500

kt k

t

DA yDA

y

y

We use the discount accumulated in the 15th year to find the difference between the yield times the redemption value and the coupon:

1

40 15 1

( )

194.821.09500

2,062.4201

n ttDA Ry Coup v

Ry Coup

Ry Coup

The discount at the time of purchase is:

40 0.09500Discount ( ) 2,062.4201

2,062.4201 10.2475nyRy Coup a a

21,134.59

Solution 63

A Chapter 12, Loans

We begin by finding the level payment amount:

1

8 5 1699.681.0475

842.3946

n ttPrn v Pmt

Pmt

Pmt



The total amount of interest paid on the loan is equal to the total amount of the payments minus the initial loan balance:

8

81 1.04758 842.3946 842.3946 6,739.1570 842.3946

0.04756,739.1570 842.3946 6.5290 6,739.1570 5,500.0251

a

1,239.13

The BA II Plus can be used to answer this question:

699.68 [] 1.0475 [yx] 4 [=]

Result is 842.3946.

[PMT] 8 [N] 4.75 [I/Y] [CPT] [PV]

Result is 5,500.0251.

[+] 8 [] [RCL] [PMT] [=]

Answer is 1,239.13.

Solution 64

D Chapter 7, Level Annuities

The monthly interest rate for the first 18 months is:

0.084 0.00712

The accumulated value of the loan after 18 months minus the accumulated value of the payments is:

18

18180.007

1.007 122,000(1.007) 450.30 24,943.2564 450.300.007

24,943.2564 450.30 19.1121 16,337.0983

s

The new interest rate is to refinance the loan is:

0.048 0.00412

The equation of value for the refinanced loan after 18 months is:

24 0.00424

16,337.0983

1 1.00416,337.09830.004

16,337.0983 22.8405

Xa

X

XX

715.27


18 [N] 0.7 [I/Y] 22,000 [PV] 450.30 [+/-] [PMT] [CPT] [FV]

(Result is 16,337.0983)

[PV] 24 [N] 0.4 [I/Y] 0 [FV] [CPT] [PMT]



Answer is 715.27.

Solution 65


Since the bond is priced at par, its Macaulay duration is:

14

( ) (2)7

1 1.038 1.0380.076

mn

MacDur a a

5.56

Solution 66

A Chapter 16, Duration


( )7.959 7.42441.072

1m

MacDurModDurym

The estimated percentage change in the price is:

( )% 7.4244 (0.08 0.072) 0.05940mP ModDur y

The estimate for the new price is:

1,000(1 % ) 1,000(1 0.05940)P 940.60

Solution 67

E Chapter 18, Forward Rates

The wording of this question is a little ambiguous, as it seems that we are being asked for either:

2f , the forward rate that applies from time 2 to time 3, or

3f , the forward rate that applies from time 3 to time 4

Since we are not given enough information to find 3f , we conclude that we are being asked for 2f :

1 11

2

2

(1 )1(1 )

10.858921 10.90703

tt

t tt

sfs

f

f

0.05601



Solution 68


Since we are not told otherwise, we assume that the yield remains constant at 5%.

After the first coupon is paid, the bond is still priced at par, so the remaining 7-year bond has a price of 5,000.

The duration of an immediate payment of 250 is 0, and the duration of the 7-year bond is 2d . A portfolio consisting of an immediate payment of 250 and the 7-year bond has a duration of 1d .

1 21

2

250 5,0000250 5,000 250 5,000

0.9524

k

Port j jj

d MacDur w MacDur d

d

The ratio is:

1 2

2 2

0.9524d dd d

0.9524

Solution 69

A Chapter 17, Dedication

The quantity of Bond C that is purchased is:

100105CQ

The quantity of Bond A that is purchased is the amount needed to cover the remaining liability after the cash flow from Bond C is received:

10099 5105

107AQ

0.8807

Solution 70

B Chapter 17, Redington Immunization

A is true, because unless the portfolio is cash-flow-matched, the duration of the assets can change at a different rate from the duration of the liabilities.

B is false because Redington immunization requires frequent rebalancing.

C is true. Full immunization protects against large changes in the interest rate, but Redington immunization only protects against small changes.

D is true, because Redington immunization is based on an assumption that the yield curve is flat.

E is true, because Redington immunization is based on an assumption that any shifts to the yield curve are parallel shifts.



Solution 71

D Chapter 17, Full Immunization

Since the liability is equidistant from the 2 asset cash flows, the weights (or market values) of the 2 asset cash flows must be equal. This implies that the value of A at time 4 is 3,000 and the value of B is also 3,000:

2

2

(1.05) 3,000 2,721.0884

3,000 3,307.50001.05

A AB B

The absolute value of the difference is:

2,721.0884 3,307.5000A B 586.41

Solution 72

A Chapter 17, Full Immunization

The present value of the assets is equal to the present value of the liability:

5 8 8

8

5,000 12,0001.03 1.03 1.03

5,159.86691.03

b

b

B

B

The duration of the assets is equal to the duration of the liability:

5 8 8

5 8

5 5,000 (8 ) 8 12,0001.03 1.03 1.03

25,000 96,000(8 ) 5,159.86691.03 1.03

8 10.50762.5076

bb B

b

bb

We can now solve for B:

10.5076 5,159.86691.03

7,039.2687

B

B

The ratio is:

7,039.26872.5076

Bb

2,807.12

Solution 73

D Chapter 17, Immunization

The first sentence of the question could be more clearly stated as, “Trevor has asset cash flows at time 2 of A and at time 9 of B.”



The duration of the asset portfolio must be equal to duration of the liability. Let X be the percentage of the asset portfolio that is invested in the asset that pays at time 2:

2 9(1 ) 57 4

47

X XX

X

Since the present value of the asset portfolio is equal to the present value of the liability, the present values of the asset cash flows at time 0 are:

4737

A L

B L

PV PV

PV PV

The amounts of the cash flows are found by accumulating their present values:

22

9 79

4 1.041.04 4731.04 3 1.041.047

LA

B L

PVPVAB PV PV

1.0132

Solution 74

D Chapter 15, Bonds

The price of Bond A exceeds the price of Bond B by 5,341.12:

20 /2 20 /22 22 2

20 /2

10,000 10,00010,000 0.02 10,000 0.022 21 1

5,341.1210,000 0.04 5,341.12

i ii i

i

i ia a

a


5,341.12 [] 10,000 [] 0.04 [=] [PV]

20 [N] 1 [+/-] [PMT] [CPT] [I/Y]

[] 2 [=]




Solution 75

D Chapter 12, Loans

The initial loan payment is:

18015 12 0.0075

400,000 400,000 400,000 4,057.066398.59341 1.0075

0.0075

Pmta

The balance after the 36th payment can be found using the prospective method. At the new interest rate, the smaller payments pay off the balance in 12 years:

144 0.0075 144 /12

144 /12

144 /12

4,057.0663 (4,057.0663 409.88)

4,057.0663 87.8711 3,647.1863

356,498.8491 3,647.1863

j

j

j

a a

a

a

The easiest way to find j is to use the BA II Plus calculator. Let’s use the calculator from the beginning of this question:

180 [N] 9 [] 12 [=] [I/Y] 400,000 [+/-] [PV] [CPT] [PMT]

Result is 4,057.0663.

144 [N] [CPT] [PV]

Result is 356,498.8491.

[RCL] [PMT] [] 409.88 [=] [PMT] [CPT] [I/Y] [] 12 [=]

Result is 6.9000. Answer is 6.90%.

Solution 76

D Chapter 15, Bonds

The equation of value that equates the prices of the two bonds is:

60 0.025 60 /260 60

2

1,200 80025 251.025 1

j ja a

The easiest way to find j is to use the BA II Plus calculator:

60 [N] 2.5 [I/Y] 25 [PMT] 1,200 [FV] [CPT] [PV]

Result is 1,045.4567.

800 [FV] [CPT] [I/Y] [] 2 [=]

Result is 4.3985. Answer is 4.40%.



Solution 77

E Chapter 3, Interest Rate Conversions

Interest is credited only at the end of each interest conversion period, so Lucas receives interest only every 6 months. Therefore, the moment at which Lucas’s account is at least double the amount in Danielle’s account will occur on some multiple of 6 months.

Let t be the number of 6-month periods until Lucas’s account is at least double the amount in Danielle’s account. We need to find the minimum integer value of t that satisfies the following:

60.06 0.031 2 12 12

ln(1.03) ln(2) 6 ln(1.0025)47.5490

t t

t tt

The smallest integer that satisfies the equation above is t = 48. The number of months in 48 6-month intervals is:

48 6 288

Solution 78

B Chapter 13, Time-Weighted Rate of Return

The balance at the end of the year is:

0.55,000 1.09 2,600 1.09 8,164.4797

The time-weighted rate of return is:

5,200 8,164.4797 15,000 5,200 2,600

0.08860

Solution 79

A Chapter 5, Varying Rates

The interest accumulation factors must be the same for Bill and Joe over the course of 4 years. The interest accumulation factor for Joe is:

44

0 0

4

1 ln( 0.25 )exp exp exp 4 ln( 1) 4 ln( )0.25 0.25

1 1exp 4 ln

K tdt K KK t

K KK K



The interest accumulation factor for Joe is equal to the interest accumulation factor for Bill:

4 4

2

2

1 125

1 125

125

1255

K KK

K KK

KK K

K

K

Joe’s accumulated value at the end of 4 years is:

4 41 610 10

5K

K

20.736

Solution 80

C Chapter 12, Sinking Funds

Since the student pays the interest on the loan each year, the amount needed to pay off the loan at the end of 5 years is the original amount of 1,000. Therefore, the sinking fund payments must accumulate to 1,000.

50.85

5

1,000 2 1,000

(1 0.8 ) 12,000 1,0000.8

(1 0.8 ) 1 0.40.8 0.06961

ii s

iii

ii

i

0.08701

Solution 81

D Chapter 12, Loans

The principal repaid in year 26 is the payment of 2,500 minus the interest on the outstanding balance at the end of 25 years:

252,500X iL

The interest paid in the first year is the interest rate times the initial loan balance:

25 25 250 25 2525

25 25 2525 25

25

2,500 2,500(1 )

2,500 2,500 2,500 2,500

2,500

i L i a v L v iv L

v iv L v iL

Xv

The final expression above matches Choice D.



Solution 82

A Chapter 13, Dollar-Weighted Weight of Return

The formula describes the simple interest approximation for the internal rate of return. The smaller the net deposits (i.e., the cash flows) between time 0 and time 1, relative to the initial deposit, the more exact the estimate becomes. This matches Choice A.

Solution 83

E Chapter 13, Time-Weighted Weight of Return

The time-weighted rate of return is based on the product of the time intervals’ corresponding accumulation factors:

120,000 130,000 100,0001100,000 120,000 30,000 130,000 50,000

i

i

0.30

Solution 84


At the end of 20 years, the value in the fund is:

20

200.08161.0816 11,000 1,000 1.0816 50,382.1558

0.0816s

The effective 6-month interest rate is:

0.51.0816 1 0.04

Let n be the number of 6-month periods that the fund can support withdrawals of 3,000:

0.043,000 50,382.1558

1 1.04 1.04 16.79410.0426.4719

nn

a

n

The 26th payment of 3,000 is made at the beginning of the 26th 6-month period, which is the same as the end of the 25th period, so the 26th payment is made at the end of 12.5 years.

Six months after the 26th payment of 3,000 is made, the balance in the fund is:

2626 0.0426

50,382.1558(1.04) 3,000

1.04 1139,683.0046 3,000 1.040.04

139,683.0046 3,000 46.0842

s

1,430.36





20 [N] 8.16 [I/Y] 1,000 [PMT] [CPT] [FV]

(Result is 50,382.1558)

[PV] 4 [I/Y] 3,000 [PMT] 0 [FV] [CPT] [N]

(Result is 26.4719)

26 [N] [CPT] [FV]

Answer is 1,430.36.

Solution 85

D Chapter 7, Perpetuities

We can use the price of the first perpetuity to find the value of i:

217.21 1

(1 ) 10.0775

ii

The annual effective interest rate used to value the second annuity is:

0.01 0.0775 0.01 0.0875i

The second perpetuity begins at the end of 1 year, so it can be valued as a perpetuity immediate accumulated for two years:

2

37.21 1.08751.0875 1

R

R

1.7446

Solution 86

E Chapter 8, Varying Annuities

At the end of 5 years, the balance of the loan is:

5 55

510,000(1.05) 100( ) 12,762.82 100

0.055.8019 512,762.82 100 12,762.82 100 16.0383

0.0511,158.99

sIs



The loan is paid off with 15 level payments of X:

1515

11,158.99

1 11,158.990.05

Xa

vX

X

1,075.08


5 [N] 5 [I/Y] 1 [PMT] [CPT] [FV]

[+/-] [] 1.05 [] 5 [=] [] 0.05 [=] [] 100 [+/-] [=]

[+] 10,000 [] 1.05 [yx] 5 [=] [PV]

[15] [N] 0 [FV] [CPT] [PMT]


Solution 87


The balance of the fund at the end of 10 years is:

55,000 3,917.63081.05

The 10 level payments of X accumulate to the balance at the end of 10 years:

100.0610

3,917.6308

1.06 1 3,917.630820.06

Xs

X

X

297.22


5,000 [] 1.05 [yx] 5 [=] [FV]

10 [N] 6 [I/Y] [CPT] [PMT]


Solution 88

E Chapter 12, Loans

The monthly effective rate at which the loan is originally made is:

0.08 0.00666712

The original payment amount is:

180 0.006667

65,000 621.1739a



After the 12th payment, the remaining balance can be found using the prospective method:

12 168 0.006667621.1739 62,661.3994L a

The new payment amount is the amount needed to pay off the remaining balance at the new interest rate of 6% compounded monthly:

0.0612 1680.005180 12

62,661.3994 62,661.3994a a

552.19

Alternatively, we can use the BA II Plus to answer this question:

180 [N] 8 [] 12 [=] [I/Y] 65,000 [+/-] [PV] [CPT] [PMT]

Result is 621.1739.

168 [N] [CPT] [PV]

Result is 62,661.3994.

6 [] 12 [=] [I/Y] [CPT] [PMT]

Answer is 552.19.

Solution 89

E Chapter 6, Level Annuities

The first tuition payment is due at the beginning of the 18th year, which is at the end of 17 years. The payment of X is made at the end of 18 years. The equation of value at the end of 17 years is:

17 1818

17 1818

18

750 6,000 1.05 1.05

750 6,000 1.05 1.05

1.07 1750 6,000 4.54120.07 1.07

750 33.9990 27,247.17101.07

25,499.2744 27,247.17101.07

s X v v

s Xv v

X

X

X

X

1,870.25


18 [N] 7 [I/Y] 750 [PMT] [CPT] [FV]

1.05 [yx] 17 [+] 1.05 [yx] 18 [] 1.07 [=] [] 6,000

[+] [RCL] [FV] [=] [] 1.07 [=]

Answer is 1,870.25.



Solution 90

B Chapter 15, Bonds

The price of the bond is:

40 0.05 401,20045 45 17.1591 170.45481.05

nnyP Coup a Rv a

942.61


40 [N] 5 [I/Y] 45 [PMT] 1,200 [FV] [CPT] [PV]


Solution 91

A Chapter 15, Callable Bonds

The bond is a discount bond, because the coupon is less than the yield-to-worst times the redemption value:

0.05 1,000 0.06 1,000 ktCoup YTW R

Since the bond is a discount bond, its price-to-worst is calculated based on the latest possible redemption:

2020 0.0650 1,000 50 11.4699 311.8047P a v 885.30


20 [N] 6 [I/Y] 50 [PMT] 1,000 [FV] [CPT] [PV]


Solution 92

C Chapter 18, Forward Rates

The question is asking for the rate that applies from time 4 to time 5.

1 11

54 4

4

(1 )1(1 )

1.09511.09

tt

t tt

sfs

f

f

0.1152

Solution 93

D Chapter 11, Geometric Progression Annuities


0.06 0.00512



The annual effective interest rate is:

121.005 1 0.06168

The present value of the first year’s payments is:

12

12 0.0051 1.0052,000 2,000 23,237.8641

0.005a

The payments in the 24 subsequent years increase by 2% per year:

25

24

1.0211.0616823,237.8641 1 1.02 (1.02 ) 23,237.8641

1.0211.06168

23,237.8641 16.1135 374,443.38

P v v

The difference between the lump sum and P is:

374,500 374,500 374,443.38 56.62P

Choice D is the correct answer.

Solution 94

A Chapter 3, Accumulated Value


0.072 0.00612

The equation of value that equates the value of the deposits with $6,500 five years from today is:

60 60 2

2 60

2

1,700 1.006 3,400 1.006 6,500

1,700 1.006 3,400 1.006 6,500 1.006

3,400 1,700 4,539.7769 0 where: 1.006

n n

n n

nX X X

We can use the quadratic formula to solve for X:

21,700 1,700 4(3,400)( 4,539.7769)2 3,400

1.4323 or 0.93226

X

X X

Using the positive value of X, we have the maximum possible value of n:

1.006

0.9323 1.00611.7264

n

nX

n

Since n must be less than or equal to 11.7264, the maximum integral value of n is 11.



Solution 95

C Chapter 3, Accumulated Value

We can use the ratio of S to T to find d:

4

44

24

4

4

2

3938

1,000 1 39381,000 1

1 39381

39 19.5 38 9.51 10

0.1

d

d

d

d

ST

d dd

d

The value of d convertible semiannually is equivalent to an annual effective interest rate of i:

2

2

1 12

0.95 1

d i

ii

0.1080

Solution 96

C Chapter 7, Level Annuity


0.042 0.003512

May 1, of the year (y + 10) is 124 months after January 1, of the year y. During these 124 months, there are 41 quarterly payments of 100. The equation of value at time 0 is:

3 6 123 124

41

3 1241

100 100 100 1.91.0035 1.0035 1.0035 1.0035

100 1.91.0035 1.0035k

k

XX

XX

The answer is Choice C.



Solution 97

D Chapter 7, Level Annuity

Let’s work in 2-year periods. For the first six years, the 2-year effective interest rate is:

21.04 1 0.0816

The accumulated value at time six years of three $100 payments made at the beginning of each 2-year period is:

3

38.16%1.0816 1100 100 351.6778

0.0816 /1.0816s

For the last four years, the 2-year effective interest rate is:

21.05 1 0.1025

The accumulated value at time ten years of the time-6 accumulated value and of the last two $100 payments made at the beginning of each of the remaining 2-year periods is:

2 210.25%2

351.6778 1.1025 100

1.1025 1351.6778 1.2155 1000.1025 / 1.1025

427.4665 231.8006 659.2671

s

We use the BA II Plus to obtain the annual effective yield:


5 [N] 100 [PMT] 659.2671 [FV] [CPT] [I/Y]

(Result is 9.3636)

[] 100 [+] 1 [=] [yx] 0.5 [=] 1 [=]

Answer is 0.0458.

Solution 98

C Chapter 7, Level Annuity


1/121.08 1 0.006434

The withdrawals of $25,000 are made at times 15, 16, 17, and 18.



The equation of value at time 0 can be used to find X:

18 12 0.006434 40.0815

216 4

25,0001.08

1 1.006434 1 1.081.06434 7,881.0426 1.080.06434 0.08

117.2787 7,881.0426 3.5771

Xa a

X

XX

240.3782



1.08 [yx] 12 [1/x] [=] [] 1 [=] [] 100 [=] [I/Y]

18 [] 12 [=] [N] 1 [PMT] [CPT] [PV]

(Result is 117.2787) [STO] 1

4 [N] 8 [I/Y] 25,000 [PMT] [CPT] [PV]

(Result is 89,427.4247) [] 1.08 [yx] 15 [=]

[] [RCL] 1 [=]

Answer is 240.3782.

Solution 99

B Chapter 6, Level Annuity

The equation of value at time 0 can be used to find X:

110.10 100.10 150.0810 10

11 10 15

0.10 0.10 10 0.081.10 1.10 1.08

1 15,00015,0000.081.10 1.10

1 1.10 1 1.10 1 1 1.0815,000 72,289.371.10

15,000 7.1446 72,289.37 6.7590 3.5641

Xa Xa a

X

X

X

17,384.

14



10 [N] 10 [I/Y] 1 [PMT] [CPT] [PV]

(Result is 6.7590) [+/-] [STO] 1

15 [N] 8 [I/Y] 1 [PMT] [CPT] [PV] [] 1.10 [yx] 10 [=]

(Result is 3.5641) [+/-] [STO] 2

11 [N] 10 [I/Y] 15,000 [PMT] [CPT] [PV] [+/-]

[+] 15,000 [] 0.08 [] 1.10 [yx] 10 [=]

(Result is 179,457.8734) [] [(] [RCL] 1 [+] [RCL] 2 [)] [=]

Answer is 17,384.14.



Solution 100

A Chapter 8, Varying Annuity

The 6-month effective interest rate is:

0.06 0.032

Since the last coupon payment was 22.50, the next coupon payment is 22.50X . The present value of the bond is 1,050.50. The time 0 equation of

value can be used to solve for X:

2 14 14

140.03 14 0.03 14

141414

22.50 22.50 2 22.50 14 300 1,050.501.03 1.03 1.03 1.03

30022.50 ( ) 1,050.501.0314(1.03)1 1.0322.50 198.3353 1,050.50

0.03 0.0311.6350 14(1.03)22.50 11.2960

X X X

a X Ia

aX

X

14198.3353 1,050.50

0.03254.1616 79.3102 198.3353 1,050.50XX

7.5401


14 [N] 3 [I/Y] 1 [PMT] [CPT] [PV]

[] 1.03 [=] [+/-] [] 14 [] 1.03 [yx] 14 [=] [] 0.03 [=] [STO] 1

22.50 [PMT] 300 [FV] [CPT] [PV] + 1,050.50 [=]

[] [RCL] 1 [=]

Answer is 7.54001.

Solution 101

D Chapter 8, Varying Annuity

Let’s use the PIn method find the value of the annuity at the end of 9 years. We have:

1 1,000 500 30P I n


3030

9 1500 1 (1.05) 500 301,000 (1.05)0.05 0.05 0.05

11,000 15.3725 69,413.2346 99,683.7267

nn

I InPV P a vi i

To find the present value at time zero, we discount for 9 years:

999,683.7267

1.05 64,257.02




30 [N] 5 [I/Y] 1,000 [+] 500 [] 0.05 [=] [PMT]

500 [] 30 [] 0.05 [=] [+/-] [FV]

[CPT] [PV]

[] 1.05 [yx] 9 [=]


Solution 102

C Chapter 11, Geometric Progression Annuity

The equation of value after 30 years can be used to solve for i:

29 28 29

29

303029

303029

29

5,000 (1 ) (1 ) (1.03) 1.03

50,000 1 1.03 (1.03 )

1.031.03 1(1 ) 115,000 50,0001.03 1.031 11 1

1.03 1.03(1 ) 10 11 1

1.03(1 ) 11

i i

v v

i ii

i i

ii i

i

30 30

29

1.0310 11

(1 ) 100.08264

i i

ii

The account balance after

the final deposit is:

30302929 1.031.03 (1.08264)(1 )

1.0826415,000 5,0001.03 1.031 11 1.08264

ii

i

797,836.82

Solution 103

D Chapter 14, Dividend Discount Model

Although this does not refer to the perpetuity as payments from a share of common stock, we can treat the payments as dividends and use the dividend discount model to find the present value of the payments.

The rate of growth of the quarterly payments is:

2,010 1 0.0052,000



The present value of the payments can be used to solve for the quarterly effective interest rate:

(4)

4(4)

2,010100,000 2,0000.005

0.025514

i

i


4(4)

41 1 1.02551 14

i

0.1060

Solution 104

A Chapter 10, Continuously Payable Annuity

Let’s break the perpetuity into two parts.

The first part consists of the payments made in the first 10 years. The present value of the first part is:

10

101 1 1.06 7.5787

ln(1.06)

nvar

The second part consists of the payments made after 10 years. The formula for the present value of a continuously payable annuity is:

t

sab

r dsa t

a

PV Pmt e dt

The present value at time 10 of the payments made after 10 years is:

10 ln(1.06)10 10 ln(1.06)( 10)10

10 1010

10 (10 )

10 10

10 10

1.031.06 10

1.03 1.03

1.06 1.031.03 1.061.03 1.06

1.06 1.03 1 1.06 1.0301.03 1.06 1.03 1ln

t dst t t

tt t

t

PV e dt e dt

dt dt

10

1.031.06

1.031.06

1.06 ln

1 34.8309ln

The

present value at time 0 of the payments made after 10 years is the present value at time 10, discounted for 10 years:

10 10

0 101 1 34.8309 19.4494

1.06 1.06PV PV



The present value of the perpetuity is equal to the sum of the present values of the two parts of the perpetuity:

7.5787 19.4494 27.0282

Solution 105

C Chapter 5, Varying Force of Interest

The present value at time 5 of the 75,000 payment is:

1010 10155 1 5ln( 1)

5 10

ln(6) ln(11)

75,000 75,000675,000 75,00011

s s dsr ds sPV AV e e e

e

The equation of value at time 0 can be used to solve for X:

3 56 110,000 75,000111.06 1.06

X

X

24,498.79

Solution 106

D Chapter 12, Drop Payments

If we accumulate the initial loan balance to time 2, then we can treat the loan as a loan with the first payment occurring one year later:

215,000,000 16,796,842.19(1 0.055)


0.055 0.058201 0.055

We can solve the time-0 equation of value for n:

16,796,842.19 1,200,000

1 1.0582016,796,842.19 1,200,0000.05820

0.1853 1.0582029.7960

nn

n

a

n

Therefore, there are 29 payments of 1,200,000 and a final drop payment at time 30:

3029

2930

16,796,842.19 1,200,000 0.945

1 1.0582016,796,842.19 1,200,000 0.9450.05820

a DropPmt

DropPmt

DropPmt

960,723.59




15,000,000 [] 0.945 [x2] [=] [+/-] [PV]

0.055 [] 0.945 [] 100 [=] [I/Y] 1,200,000 [PMT]

[CPT] [N]

Result is 29.7960.

29 [N] [CPT] [FV] [] 0.945 [=]

Answer is 960,723.59.

The closest answer choice is Choice D.

It appears that when the answer choices were prepared for this question, there was an unusually large discrepancy introduced by internal rounding.

Solution 107

C Chapter 12, Loans

We make use of the following formula for the interest portion of a payment:

1(1 )n ttInt v Pmt

The formula can be used to find the following expressions:

21

43

1

(1 )

(1 )

(1 )

n

nn

Int v Pmt

Int v Pmt

Int v Pmt

The interest portion of the payment at time (n 1) is equal to 0.5250 of the interest portion of the payment at time (n 3), which allows us to solve for v:

2 4

2(1 ) 0.5250(1 )

1 0.5250(1 )0.9512

v Pmt v Pmt

vv

The interest portion of the payment at time (n 1) is equal to 0.1427 of the interest portion of the first payment, which allows us to solve for n:

2(1 ) 0.1427(1 )

0.09524 0.1427(1 )

0.3326

n

n

n

v Pmt v Pmt

v

vn

22.00



Solution 108

B Chapter 12, Sinking Funds

The sinking fund payment is:

110.047

LSFPs

The amount owed to the lender is remains constant at L until the loan is paid off, so the equation of value at the end of 7 years is:

7110.047

7

11

6,241

1.047 11 6,2411.047 1

14,749.3182

LL ss

L

L

The sinking fund payment is:

110.047

14,749.318213.9861

LSFPs

1,054.57

Solution 109

C Chapter 12, Drop Payments

The level payments satisfy the following time-0 equation of value:

360 0.005200,000

200,000 166.79161,199.1011

Pmt a

PmtPmt

When we subtract the present value of the extra payments, the new equation of value is:

12 0.00551.005 1

50.06168

200,000 10,000 1,199.1011

1 1.005200,000 10,000 1,199.10110.005

1 1.005200,000 10,000 4.1932 1,199.10110.005

1 1.005158,067.9347 1,199.10110.005

158,067.9347 1,199.1

nn

n

n

a a

a

1 1.005011

0.0051.005 0.3409

215.7768

n

n

n



The final payment therefore occurs after 216 months, which is 18 years:

216 1812

The loan originated on January 1, 2003, and the final payment is made 18 years later. Adding 18 years to January 1, 2003 brings us to January 1, 2021. The beginning of January 1, 2021 is the same as the end of December 31, 2020.


200,000 [PV] 360 [N] 0.5 [I/Y] [CPT] [PMT]

Result is 1,199.1011.

[STO] 1

5 [N] 1.005 [yx] 12 [] 1 [=] [] 100 [=] [I/Y] 10,000 [PMT] [CPT] [PV]

[+] 200,000 [=]

Result is 158,067.9347

[PV] 0.5 [I/Y] [RCL] 1 [PMT] [CPT] [N]

Result is 215.7768, so the final payment occurs after 216 months.

216 [] 12 [=]

Result is 18.

[+] 2003 [=]

Result is 2021.

The final payment is made at the beginning of January 1, 2021, which is equivalent to the end of December 31, 2020.

Solution 110

D Chapter 12, Loans


0.08 0.086961 0.08

i

We can solve for the amount of the 5 level payments:

5 0.08696500,000

500,000 3.9206127,532.7207

Pmt a

PmtPmt

If the first 4 payments were instead 128,000, then the balance at the end of 4 years would be:

4 0.086964 4500,000 500,000128,000 128,000 4.55260.92 0.92

115,202.7473

s



The final payment is the balance at the end of 4 years, accumulated for one additional year of interest:

115,202.74730.92

125,220.38


0.08 [] 0.92 [] 100 [=] [I/Y]

5 [N] 500,000 [+/-] [PV] [CPT] [PMT]

Result is 127,532.7207.

128,000 [PMT] 4 [N] [CPT] [FV]

[] 0.92 [=]

Answer is 125,220.38.

Solution 111

B Chapter 18, Spot Rates

The price of a zero-coupon bond as a percentage of its redemption value is equal to the inverse of the accumulation factor achieved by investing in the bond.

Therefore investing X in the 6-month bond, for example, results in an accumulated value of:

0.94

X

Investing X in each of the bonds results in an accumulated value of:

0.94 0.95 0.96 0.97 0.98 0.991 1 1 1 1 1 6.2196

0.94 0.95 0.96 0.97 0.98 0.99

X X X X X X

X X

Setting this accumulated value equal to 100,000 allows us to solve for X:

6.2196 100,000XX

16,078.29

Solution 112

D Chapter 12, Loans

We use the following formulas:

1

1

(1 )n tt

n tt

Int v Pmt

Prn v Pmt

Using the information provided in the question, we have:

10

15

6

(1 ) 3,600

4,871

Int v Pmt

Prn v Pmt



Dividing the first equation by the second equation allows us to solve for v:

10

5

10 5

10 5

(1 ) 3,6004,871

3,60014,871

3,600 1 04,871

v Pmtv Pmt

v v

v v

We can use the quadratic formula to solve for 5v :

23,600 3,600

4,871 4,8715

5 5

4 1 ( 1)

20.6966 or 1.4356

v

v v

We use the positive value of 5v to solve for i:

5 0.6966

0.07500vi

Since the interest paid in the first year is 3,600, we have:

3,600

0.07500 3,600XiXX

48,000.18

Solution 113

A Chapter 15, Bonds

The equation of value can be solved for R:

0.5501.0705 1 25

25

0.5 25

25

10,000 0.0351.0705

1 1.0705 110,000 0.0351.0705 1 1.0705

110,000 0.035 23.60451.0705

nnyP Coup a Rv

RR a

R

R

R

9,917.99


1.0705 [yx] 0.5 [] 1 [=] [] 100 [=] [I/Y]

50 [N] 0.035 [PMT] 1 [FV] [CPT] [PV]

Result is 1.008269.

[1/x] [] 10,000 [=]




Solution 114

B Chapter 15, Bonds

At the end of each month, the net cash flow to Jeff is the coupon payment from the bond minus the interest on the loan:

0.09 0.0810,000 2,000 61.666712 12

At the end of 10 years, Jeff receives 10,000 from the bond and pays back the 2,000 loan, giving him a net cash flow of:

10,000 2,000 8,000

Since the cost of entering this position is 8,000, the time-0 equation of value is:

(12)

(12)12120120

12

8,0008,000 61.66671

ii

a


120 [N] 8,000 [+/-] [PV]

10,000 [] 0.09 [] 12 [] 2,000 [] 0.08 [] 12 [=] [PMT]

8,000 [FV] [CPT] [I/Y]

Result is 0.7708.

[] 100 [+] 1 [=] [yx] 12 [] 1 [=]

Answer is 0.0965.

Solution 115

B Chapter 15, Bonds

The first equation of value below sets the value of the first bond equal to the value of the second bond, and the second equation sets the value of the second bond equal to the value of the third bond:

0.0528 1,000 1,000 0.0440 1,100 1,100

0.0440 1,100 1,100 1,320 1,320

n nn n

n nn n

a v a v

a v r a v

The two equations above can be simplified as follows:

4.4 100

(48.4 1,320 ) 220

nn

nn

a v

r a v



Dividing the first equation into the second equation allows us to solve for r:

(48.4 1,320 ) 2204.4 100

48.4 1,320 2204.4 100

48.4 1,320 9.68

nn

nn

r a va vr

rr

0.02933

Solution 116

A Chapter 15, Bonds

The time-0 equation of value can be used to solve for c:

2 12

6

13

250 1.02 1.02 1.02582.531.04 1.04 1.041.04

1.02 1.021.04 1.04384.9514 1.021

1.04

c

c

c

32.04

Solution 117

E Chapter 15, Bonds

We can use the book values provided to find the coupon amount:

4 3 10.06

(1 )

(1.06)

1,277.38 1,254.87(1.06) 152.7822

kt k t k yBV BV y Coup s

BV BV Coup s

CoupCoup



We can now use the prospective formula to find n:

33 30.06

( 3)3

3 3

3

3

52.7822 1,890

1 1.061,254.87 52.7822 1,8900.06

1,254.87 879.7033 1 1,890

375.1667 1,010.2967

0.3713 1.063 17.0010

n tt n t y

nn

nn

n n

n

n

BV Coup a Rv

BV a v

v

v v

v

nn

20.00


1,254.87 [] 1.06 [] 1,277.38 [=]

Result is 52.7822.

[PMT] 6 [I/Y] 1,254.87 [+/-] [PV] 1,890 [FV]

[CPT] [N]

Result is 17.0010.

[+] 3 [=]

Answer is 20.00.

Solution 118

A Chapter 15, Bonds

The discount accrued with the fourth coupon is 8.44:

1

4 3 4

4 4 3

4 8.44

t t tBV BV DABV BV DADA BV BVDA

The discount accrued can also be written as:

1( ) n ttDA Ry Coup v

We can now solve for n:

4 1

38.44 (2,500 0.04 2,500 0.035)

0.6752 1.043 10.0137

13.0137

n

nv

nn



The n that was found above is the number of coupons paid, but the question asks for the number of years until the bond matures. Therefore, we must divide the n above by 2:

13.01372

6.51

Solution 119

C Chapter 18, Forward Rates

The 3-year accumulation factor calculated with the spot rate is the same as the 3-year accumulation factor calculated with the forward rates:

0 1 13

3 0 1 23

3

3

(1 ) (1 )(1 ) (1 )

(1 ) (1 )(1 )(1 )

(1 ) 1.04 1.06 1.08

tt ts f f f

s f f f

ss

0.0599

Solution 120

D Chapter 13, Time-Weighted & Dollar-Weighted ROR

The income and fund exposure are:

Income Withdrawals Deposits 55,000 10,000 50,000 8,0007,000


850,000 8,000 10,000(1 )12

45,333.3333 10,000

t

t

The dollar-weighted rate of return is equal to the time-weighted rate of return:

7,000 52,000 62,000 55,000 145,333.3333 10,000 50,000 60,000 52,000

7,000 0.136745,333.3333 10,00045,333.3333 10,000 51,219.5122

t

tt

t

0.5886

Solution 121


The formula for Macaulay Duration is:

00

00

tt

tt

t PV CFMacDur

PV CF



We can use the duration of the 3-year annuity to solve for v:

2

2

2 2

2

2

0 20.931

0.93 0.93 0.93 2

1.07 0.07 0.93 0

0.07 0.07 4(1.07)( 0.93)2 1.07

0.9002 or 0.9656

v vv vv v v v

v v

v

v v

We use the positive discount factor to find the duration of Annuity B:

2 3 2 3

2 3 2 30 2 3 0.9002 2 0.9002 3 0.9002 4.7088

3.43981 1 0.9002 0.9002 0.9002v v vv v v

1.3689

Solution 122

D Chapter 16, Duration


2 3 4

2 3 4

40 2 25 3 100 400 1.07 1.07 1.07

40 25 1000

1.07 1.07 1.070

436.2555131.6345

tt

tt

t PV CFMacDur

PV CF

3.314

Solution 123


As of 10/6/2015, the Exam FM Sample Solutions available on the Society of Actuaries’ web site incorrectly showed the answer as being Choice C.

The Macaulay duration of Bond A is:

3

31 1.10 1.10 2.7355

0.10AMacDur a

The modified duration of Bond A is:

( )2.7355 2.48691.10

1

AA m

MacDurModDurym

The modified duration of Bond B is equal to the modified duration of Bond A:

2.4869BModDur



The Macaulay duration of Bond B is found below:

( )1

2.4869 0.1012

2.4869 1.05

BB m

B

B

B

MacDurModDurym

MacDur

MacDurMacDur

2.6112

Solution 124


The price and the derivative of the price can be used to obtain an expression for the modified duration:

2

2

1 2

11

'

' 1 1(1 )1

Py

P y

P yModDurP y yy y y

The relationship between the modified duration and the Macaulay duration can be used to find the yield:

11 30

(1 ) (1 )1 30

130

MacDurModDury

y y y

y

y


1 1

30 30

1 1(1 ) 1

ModDury y

29.03



Solution 125

D Chapter 14, Common Stock

Let JDiv be the next dividend of Stock J. The value of Stock F is twice the value of Stock J:

Value of Stock F 2 (Value of Stock J)0.5 2

0.088 0.0882(0.088 ) 0.5(0.088 )4(0.088 ) 0.0883 0.088 5

J JDiv Divg g

g gg g

gg

0.0528

Solution 126

B Chapter 17, Immunization

Statement I is false because an inverted yield curve does not reduce the need for immunization.

Statement II is true, because the modified duration is equal to the Macaulay duration divided by a one-period accumulation factor. Therefore, if either the modified or Macaulay durations are equal, then the other will be equal as well.

Statement III is false, because exact matching is accomplished only when the cash flows themselves are matched.

Solution 127

A Chapter 17, Immunization

The present value, Macaulay duration, and Macaulay convexity of the liabilities are:

4

0 40

00

2 2 20 40

00

500 1,000 1,137.55891.10 1.10

500 1,0001 41.10 1.10 2.8013

1,137.5589

500 1,0001 41.10 1.10 10.0063

1,137.5589

L

tt

Lt

t

tt

Lt

t

PV

t PV CFMacDur

PV CF

t PV CFMacC

PV CF

Since the duration of the assets is equal to the duration of the liabilities, the average duration of the assets is 2.8013. Let the weight of the cash flow of X be w:

0 (1 ) 3 2.80130.06625

w ww



Since the cash flow of X occurs at time 0, the value of X is its weight times the present value of the liabilities:

0.06625 1,137.5589X 75.36

The Macaulay convexity of the assets is:

20

2 20

00

0 (1 ) 3

0.06625 0 (1 0.06625) 9 8.4038

tt

At

t

t PV CFMacC w w

PV CF

Since the Macaulay convexity of the assets is less than the Macaulay convexity of the liabilities, the conditions of Redington immunization are not satisfied. Choice A is the correct answer.

Solution 128

D Chapter 17, Full Immunization

Let w be the weight of the first asset cash flow:

6

8

300,0001.04 0.324481,000,0001.04

w

The Macaulay duration of the assets must be equal to the Macaulay duration of the liability:

6 (1 ) 86(0.32448) (1 0.32448) 8

8.9607

w y wy

y

The present value of the assets must be equal to the present value of the liabilities:

6 8.9607 8300,000 1,000,0001.04 1.04 1.04

X

X

701,458.26



Solution 129


The present value, Macaulay duration, and Macaulay convexity of the liabilities are:

2 5

0 2 50

00

2 2 20 2 50

00

573 701 1,000.28371.07 1.07

573 7012 51.07 1.07 3.4990

1,000.2837

573 7012 51.07 1.07 14.4929

1,000.2837

L

tt

Lt

t

tt

Lt

t

PV

t PV CFMacDur

PV CF

t PV CFMacC

PV CF

The present value, Macaulay duration, and Macaulay convexity of Choice A are:

2 2

500 500 1,0000.5 1 0.5 6 3.5

0.5 1 0.5 6 18.5

A

A

A

PVMacDur

MacC

Since the present value and the Macaulay duration of Choice A are equal (within rounding tolerance) to the present value and Macaulay duration of the liabilities, and the Macaulay convexity of Choice A exceeds the Macaulay convexity of the liabilities, Choice A is the correct answer.

Choice B’s Macaulay duration can be found as the weighted average of the timing of its cash flows:

0 0.572 1 0.428 6 3.14BMacDur

Since 3.14 is not equal to the duration of the liabilities, Choice B is not correct.

Choices C and D have present values in excess of the present value of the liabilities, so neither is correct.

Choice E has a single cash flow, so its Macaulay convexity is equal to the square of the time of the cash flow:

23.5 12.25EMacC

Since 12.25 is less than the Macaulay convexity of the liabilities, Choice E is not correct.



Solution 130

D Chapter 17, Immunization

The Macaulay duration of the liabilities is:

0 2 30

00

402.11 402.11 402.111 2 31.10 1.10 1.10 1.9365

1,000

tt

Lt

t

t PV CFMacDur

PV CF

The Macaulay duration of the assets must be equal to the Macaulay duration of the liabilities. Let w be weight of the one-year bond:

3(1 ) 1.93650.53173

w ww

The amounts to invest in the one-year bond and the 3-year bond are:

One-year bond: 1,000 1,000 0.53173Two-year bond: 1,000(1 ) 1,000(1 0.53173)

ww

531.73468.26

Solution 131


All of the answer choices have the same present value as the present value of the liabilities, so we’ll focus on the requirements that the Macaulay duration of the assets be equal to that of the liabilities and that the Macaulay convexity of the assets be greater than that of the liabilities.

For Choice A, we have:

2 2

3,077 6,6205 20 15.24039,697 9,6973,077 6,6205 20 281.00709,697 9,697

A

A

MacD

MacC

Since the Macaulay duration of the assets is equal (within rounding tolerance) to the Macaulay duration of the liabilities, and the Macaulay convexity of the assets is greater than the Macaulay convexity of the liabilities, Choice A is the correct answer.

For Choice B, the Macaulay duration is not equal to the Macaulay duration of the liabilities:

6,620 3,0775 20 9.75979,697 9,697BMacD

For Choice C, the Macaulay duration is not equal to the Macaulay duration of the liabilities:

465 9,23215 20 19.76029,697 9,697CMacD



For Choice D, the Macaulay duration is not equal to the Macaulay duration of the liabilities:

4,156 5,54115 20 17.85719,697 9,697DMacD

For Choice E, the Macaulay convexity is not greater than the Macaulay convexity of the liabilities:

2 2

9,232 46515 20 15.23989,697 9,6979,232 46515 20 233.39189,697 9,697

E

E

MacD

MacC

Solution 132

E Chapter 17, Asset-Liability Management

The liability cash flows occur at time 1 and time 2 in the following amounts:

Time 1: 20,000 1.10 0.5 11,000Time 2: 20,000 1.10 0.5 1.10 12,100

The strategy that costs the least to implement produces the highest profit. To answer this question, we:

1. Determine the cost of each strategy.

2. Find the lowest cost strategy that provides the necessary cash flows.

The cost of each strategy is listed below:

A: 9,091 8,264 2,145 19,500B: 10,000 10,000 20,000C: 10,000 9,821 19,821D: 8,910 731 10,000 19,641E: 8,821 10,804 19,625

Choice A has the lowest cost, so we check to see if it provides the necessary cash flows:

2Time 1: 9,091 1.10 2,145 0.12 10,257.50

Time 2: 8,264 1.11 2,145 1.12 12,584.4744

Since Choice A does not provide at least 11,000 in year 1, it is not correct.

Since Choice E has the next lowest cost, we check to see if provides the necessary cash flows:

Time 1: 8,821 1.10 10,804 0.12 10,999.58Time 2: 10,804 1.12 12,100.48

Choice E provides (within rounding tolerance) the necessary cash flows to make the liability payments, so Choice E is the correct answer.



Solution 133

C Chapter 17, Dedication

We do not use the 4.50% bond, because:

1. It has the lowest yield, and

2. It produces cash flow at time 1, reducing our ability the use the highest-yielding bond, which is the 6.75% bond.

The smallest cost of assets that provide the necessary cash flows is:

220,000 25,000

1.06751.05 41,559.79

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Solutions to Interest Theory Sample Questions - Exam FMactuarialbrew.com/data/documents/AB_FM_Sample_Exam_Solutions.pdfSolutions to Interest Theory Sample Questions Solution 1 C Chapter