Scan 0007

•••

••

., ••

IT-26

PRACTICE MULTIPLE CHOICE TEST 7 1.

Investment X for 100,000 is invested at a nominal rate of interest, j, convertible semiannually. After four years it accumulates to 214,358.88. Investment Y for 100,000 is invested at a nominal rate of discount, k, convertible quarterly. After two years it accumulates to 232,305.73. Investment"Z for 100,000 is invested at an annual effective rate of interest equal to j in year one and .m annual effective rate of discount equal to k in year two. Calculate the value ofinvestmentZ at the end of two years .

2.

(D) 200,000

(C) 184,425

(B) 182,900

(A) 168,000

(E) 201,675

Which of the following are true?

I.

;hCi) :v-2

(A) I and ITonly

IT.

1J"m)) ~ v-(";')

(B) I and ill only

m~(i)

(C) ITand ill onl~)

~ 1 +, (D) I, ITand ill

(E) The correct answer is not given by (A), (B)vEC) or (D) ( ~ A fund starts with a zero balance at time zero. The fund accumulates with a varying force of interest

Ot ==

-fJ-, t + 1 for t > O. A deposit

III

•• •• ••

•• ••• •• ••• ••

•• •••

•••

f-:5 3.

•• •• ••

of 100,000 is made at time 2. Calculate the number

of years from the time of deposit for the fund to double.

••

••• •••

I

•.. ,

(A) .50

(C) 1.50

(B) 1.00

(D) 2.00

(E) 2.50

~ ~ 4.

Two funds, X and Y, start with the same amount.

You are given:

-e

(if

Fund X accumulates at a force of interest of 5%.

(ii) (iii)

Fund Y accumulates at a rate ofinterestj, compounded semiannually. At the end of eight years, Fund X is 1.05 times as large as Fund Y.

Calculate j. (A) .022

5.

(B) .023

(C) .042

(D) .044

(E) .046

Carl puts 10,000 into a bank account that pays an annual effective interest rate of 4% for ten years. If a withdrawal is made during the first five and one-half years, a penalty of 5% of the withdrawal amount is made. Carl withdraws K at the end of each of years 4, 5, 6 and 7. The balance in the account at the end of year 10 is 10,000. Calculate K. (A) 929

(B) 958

(C) 980

(D) 1005

(E) 1031

II-27

6.

You are given: (i)

Xis the current val1.1ea: :he end of year two ofa 20-year annuity-due of 1 per annum.

(ii)

The annual effective :n:e:-es: rate for year t is 8 ~ t·

Calculate X 29

(C)

:L

29

I?

(D)

1=10

7.

:L 1f

1=]0

Gloria borrows 100,000 to be repaid over 30 years. You are given: (i) (ii) (iii)

Her first payment is X at the end of year 1. Her payments increase at the rate of 100 per year for the next 19 years and remain level for the following 10 years. The effective annual rate of interest is 5%.

Calculate X (A) 5505

8.

(B) 5555

(E) 5705

Payments of 1 at the end of the first year and every three years thereafter. Payments of2 at the end of the second year and every three years thereafter. Payments of 3 at the end of the third year and every three years thereafter.

The interest rate is 5% convertible perpetuity. (A) 24

(B) 29

semiannually.

(C) 34

Calculate the present

\ "---'/ r-~

39

value of this

(E) 47

You are given a perpetual annuity-immediate with annual payments increasing in geometric progression, with a common ratio of 1.07. The annual effective interest rate is 12%. The first payment is 1. Calculate the present value of this annuity. (A) 18

10.

(D) 5655

You are given a perpetuity with annual payments as follows: (i) (ii) (iii)

9.

(C) 5605

(D) 21

(B) 19

(E) 22

A loan of 1000 is taken out at an annual effective interest rate of 5%. Level annual interest payments are made at the end of each year for 10 years, and the principal amount is repaid at the end of 10 years. At the end of each year, the borrower makes level annual payments to a sinking fund that earns interest at an annual effective rate of 4%. At the end of 10 years, the sinking fund accumulates to the loan principal. Calculate the difference between the interest payment on the loan and the interest earned by the sinking fund in the fifth year. CA)

23

(B) 26

(C) 33

~

/

36

CE)

38

---ll-28

11. A loan is to be amortized by r. (i) (ii) (iii)

re- •.e: ::'::'.l2.1

payments of X, where n

>

5, You are giver::

The amount of i:l:e:-es: ::: :~e first payment is 604.00, The amount ofin:e:-es:::: ::le third payment is 593,75 . The amount of :r::e:-es: :~ tr.e fifth payment is 582.45 .

Calculate X

(~A~ "-/ 704

(B) 739

(C)

(D) 1198

1163

(E)

--

''':':' ...

12. Tom borrows 100 at an annual effective interest rate of 4% and agrees to repay it with 30 annual installments. The amount of each payment in the last 20 years is set at twice that in the first 10 years. At the end of 10 years, Tom has the option to repay the entire loan with a final payment X; in addition to the regular payment. This will yield the lender an annual effective rate of 4.5% over the 1O-year period. Calculate X (A) 89

(0) 104

(C) 99

(B) 94

'~109

13. You are given two n-year 1000 par value bonds. Bond X has 14% semiannual coupons and a price of 1407.70, to yield i, compounded semiannually. Bond Y has 12% semiannual coupons and a price of 1271.80, to yield the same rate i, compounded semiannually. Calculate the price of Bond X to yield i-I %. / (B) 1550

~

(D) 1650

1600

(E) 1700

14. Henry has a five-year 1,000,000 bond with coupons at 6% convertible semiannually. Fiona buys a 10-year bond with face amount X and coupons at 6% convertible semiannually. Both bonds are redeemable at par. Henry and Fiona both buy their bonds to yield 4% compounded semiannually and immediately sell them to an investor to yield 2% compounde~ semiannually. Fiona earns the same amount of profit as Henry. Calculate X CA) 500,000

(B) 502,000

\9

(D) 571,000

505,000

(E) 574,000

15. You are given the following information on an investment account:

Account Value

1/1/99

511/99

9/1/99

1/1/00

50,000

75,000

90,000

67,000

(before deposit or withdrawal) Deposit

15,000

Withdrawal

25,000

Calculate the absolute value of the difference in the yield rates by the dollar-weighted and time-weighted methods . •.-A7

~2.3%

(B) 6.4%

,A/A.. !~

9.8%

(D) 12.3%

(E) 15.8%

.••4,-

-. It

•••

•• •• ••• •• ••• •• ••

•• •••

II-29

16.

Paul lends 8000 to Peter. Peter agrees to pay it back in 10 annual installments at 7% with the fIrst payment due in one year. After making 4 payments, Peter renegotiates to payoff the debt with 4 additional annual payments. The new payments are calculated so that Paul will get a 6.5% annual yield over the entire 8-year period. Determine how much money Peter saved by renegotiating. r (A) Less than 550 \~) tE) At At least least 650, 700 but less than 700 (B) At least 550, but less than 600 (C) At least 600, but less than 650

17.

You are given an n-year annuity-due of 1 per year plus a fInal payment at time n + k - 1, for 0 < k < 1. The present value of the payments can be simplified to J _~n+k. Calculate the final payment.

(A) 1

(B) k

(C) 1-

k

(E) (1+il-J

••

ill-34

SOLUTIONS TO PRACTICE MULTIPLE CHOICE TEST 7 1.

= •••

We are given the following information: X: 100,000 (1

+}')8

•••

=: 214,358.88

Y: 100,000 (1 _ ~) -8 =: 232,305.73 Solving we have j =: 2[(2.1435888)1/8 1] =: .20, and k = 4(1 Then the value of Investment Z is 100,000(1.20)(.60)-1 = 200,000,

2.

1.

d . dd(z)

n.

:i

m.

-isCi)

I and

_-

d

:i

=

(i(m))

=

r=a

d ( ([(l

1s(eO

-

= .40. ANSWER D

(2.3230573)-1/8J

(1 - (1 d) -- d)2 d( - 1)

1]

-

eO

=

=

1+

=

(1+i)l/m

(1 - I d)2

=

-J

t -J1-+ t

1 dt

= In(t2+

exp [In s2t1 ]

=~

= 2 to t = s,

v(m-I)/m, no.

i,yes.

where exp

1)1~ =: In(s2+ 1) - In 5

= 2,

so that

s = 3.

[is

Ot

= In (~).

=

dt]

exp

[15 t2~

=

=

2.

The fund accumulates from time 2 to time 3, a ANSWER B

e80

=

eA,

(1.05) ( 1+ ~')16 ' so that j

=

2 [( 1~05 A )1/16

After eight years Fund . X has eA

1 dt]

Then

period of 1.0 year,

and Fund Y has ( -

1]

I+ ~')16 . We are given that =

ANSWER

.04438,

D 5.

t = 4 and t = 5 as being of amount K(1.05) to include the The withdrawals at times t = 6 and t = 7 are of amount K only, since there is no

We view the withdrawals at times penalty.

penalty. Thus we have the equation of value

10,000(1.04)]0 - K(1.05)(1.04)6

I.

So vmg for K we have K

-

K(1.05)(1.04)5

-

K(1.04)4

-

K(1.04)3

=

10,000[(1.04)]0 - I] = -----~---~---= (1.05)(1.04)6 + (1.05)(1.04)5 + (1.04)4 + (1.04)3

,. •• ••

ANSWERB

from t

III

•••

v -,yes .

_

ill are true,

The fund accu~ulates J2

_

=

1)

•• -,

-

)

[m(1+i)I/m

3.

4.

~

10,000. 979.92

' ANSWERC.

ill-.::~6.

The current value at

=2

t

of a.-. a.-: . .1: :y-due is (1 +i?

year. More generally, the CU!7e:::·:::':-.1e

if the rate in year and

t is it·

Here we have it

(1+i:)(l-i:) =

¥

=

M

(l-i2)

. a201

if the interest rate were

i in eve::.

1S

=

8 it, so that 1 + it

t~,

= ~

11 TT 1

l+i3 1 (1+i3)(1 +i4)

1

(1+i3)(1+i4)

.. ·(1+i!9)

_

11

-

12 11

= 13

11

=

28

28

Finally, X

7.

=

L t=9

\1,

ANSWER B

The loan is the present value of the repayments, so 100,000

=

X·

=

X·

+ (X+200)v3

v -+ (X +100)v2 a301

+ 1900v20

•

aTOJ

+ 100v

+ ... + (X + . (I a

1900)v2o

+ (X

+1900)v20

•

a-;Qj

)T9!'

where all functions are calculated at 5%.

100,000 -

ThenX -

1900v20.

aTOl -

100v'

(I a)T9i

a301

100,000 - (1900)(.37689)(7.7217) - (100)(.95238)(103.4139) 15.3725

=

5504.74,

ANSWER A

--

•• • ••

ill-36

8.

The payments are shown on the follo\\ing diagram:

t

o

1

2

3

1

2

3

J

t

t

t

t

t

6

7

8

I

9

10

11

3

2

3

1

t

t

t

,

2

3

4

5

At t

Consider the value of each set of three consecutive payments. three payments is X at t

=

2 = (1.025 1 )2 + (1.025)4 + Cl . _~

3 (for the payments at t

= 4,

5, 6), at t

=

=

_. f.

12

=0

the value of the first

5.350614. This value X also exists

6 (for the payments at t

=

7, 8, 9), and so on .

The overall present value is

PV = X[1

+

I + (1.025)12 1 + ...] (1.025)6

=

I

=

]

5.350614 .025 _ (_1_)6

ANSWERD 3886 .,

PV _

1 1.12

-

1.07

+ (1.12i +

(1.07)(1.12)3

The sinking fund payment is

SF B4 .04

X

+ ...

The

direct

50 - 14.15

1000 [(1.04) 4 8101·04

interest

payment

earned in the fifth year is

1] -_ is

8121000 .nt)~1 (.169 6) -

.05(1000)

= 50,

so

14.15. the

requested

difference

= 3S.85,

(Calculator comments:

Following is an alternate solution based on the relationship between the

i = .04 and

if

=

.OS,

Sth

year under the sinking fund method is the same as the interest in the

the amortization

method at

i = .04, and

and the amortization method with

making use of the BA-35's ability to split payments into principal and interest. in the

IS

ANSWERD

sinking fund method with

•••

~

1000 . After four years the sinking fund balance is 10).04

= lODO 101·04. 8'4104 . ' so the interest SFB4 _-

•• •• ••

••• ANSWERC

10.

•••

••

?

9.

•• •• •• ••

i = .04, plus the quantity

(if - i)L

=

The net interest

5th

payment under

(.05 - .04)(1000)

obtain the interest in the amortization method, hit IAC/ONI, enter 10, hit

fED,

=

10. To

enter 4, hit I %i I,

enter 1000, hit /PVI, and then hit ICPTIIPMTI to determine the annual payment of 123.29. obtain the interest in the 5th payment, enter 5, and hit II/PI, resulting in 2S.85. interest in the sinking fund method is 25.85

+ 10 =

35.85.)

To

Then the net

ill-37

11.

=

lfthe loan is L the payment is X Thus we have (a) X(1 - vn) (c)

X(1

-

vn-4)

=

nli

and the interest in the

= 604.00, (b) X(l -

vn-2)

X(t,n-4

-

vn-2)

payment is X(1 -

tth

= 593.75,

582.45. Subtracting (a) - (b), we have

Subtracting (b) - (c), we have ratio of

a!:....'

vn-t+1).

and

X(vn-2

= X(1+i)2(vn-2

-

vn)

-

vn)

= 10.25. = 11.30.. Taking

~j= ~ j, we find (1 +ii = i6:~~= 1.10.244. Next we note that the principal

repaid in

the first payment is X - 60.4, and the principal repaid in the third payment is X - 593.75. lmow that the ratio ~ ~~

= (l +N,

= (1. 10244) (X

X - 593.75

so we have

- 60.4), leading to X

=

Xx-}~tJ5

=

the

We

1.10.244, from which we find

(1.10.244)\6gi1-

593.75

= 704.0.6, ANSWER A.

12.

We are given 10.0.

P _

-

= p.

10.0.

_ + 2. 101

aTOT.04+2p·

_ 201 -

a201.04'so that

vlO.

10.0

vlO . a-

ment the equation of value becomes 10.0. = X

=

(lo.o. -

_

(R, 1109) -+ 2U;7556)(13.5903)

=

3.77743awr)(1.0.45)10

-

3.77743aTOT,045

3.77743.

+X

Under the new arrange-

. V.b~5'

so that

10.8.88,

ANSWER E

(Calculator comment: The last equation for X can be easily handled with the annuity keys of the BA-35: hit lAC/aNI, enter la, hit [H], enter 4.5, hit I %il, enter 3.77743, hit IPMTI, enter lOa, hit IPVI, and then hit ICPTIIFV/ to obtain X

13.

We are given the following, where j

=

2-

= 10.8.88.) is the effective semiannual yield:

X: 140.7.70. = 100.0.+ 100.0.(.0.7- j)a2nlj Y: 1271.80.

=

100.0.+ 100.0.(.0.6- j)a2nlj,

Simplifying we have

:g~ = ~ =

ratio we find .40.77 =

2n P

X:

.03a2nl.04,

so

.40.77

=

(.0.7 - j)a2nlj; Y: .2718

:in~,which

a2nl.04

solves for j

=

.0.4 and

=

(.0.6 - j)a2nlF

i=

2j

\

tables.

.0.8. Then we have

= 13.59. Inspecting a 4% interest table we find

= 20.. Finally the price of Bond X at i = .0.7is = 100.0.+ 100.0.(.0.7- .0.35)a201.035= 100.0.+ (35)(14.2124) =

(Calculator comment:

=

1497.43,

The annuity keys make it possible to see that 2n

Taking the

= 20

azol

=

13.59, so

ANSWER A

without interest

Hit iAC/ONI, enter 4, hit I %il, enter 1, hit IPMTI, enter 13.59, hit IPVI, and then hit

ICPTIlli] to obtain 2n

= 20..)

,,

•

ill-38

14.

For Henry, P.OI

=

=

=

=

1,000,000 [1 + (.03 - .02)a!oi.02]

=

1,000,000[1+(.03-.01)a!oi.oI]

ForFiona,po2 P.OI

=

P.02

=

X[1+(.03-.01)a201.Dl]

(Calculator comment: P02 -

1.163514X,and Her profit is POI - P02

1.360912X.

=

Thus we have X

which is the same as Henry's.

=

1,189,426, for a profit ofP.ol - P02

=

X[1+(.03-.02)a201.02]

,

=

..........

~- ~ -.' --

-

~~;~OOOQ= 504,564,

can be readily obtained using the BA-35's annuity ke:.~

P.Ol

• •

1,089,826, and

.-...

IAC/ONI, enter 10, hit [Rj, enter 2, hit I %il, enter 30,000, hit IPMTI. enter 1,000,000, :-.:: ::". and then hit ICPTIIPVI, to obtain P02 IPV! to obtain

POI

=

= 1,089,825.80.

Now enter 1, hit I%il, and ther, :-.::

:?-=-

•

•

• •

•• ~

1,189,426.10.)

••

•• 15.

T·Ime welg . h te:d 1 + '/. . = (75,000) 50, 000

(90,000) 90, 000

(67,000) 65, 000

•••

. = ..5462 =. 15462 , so 'l.tw

C = Let = -10,000 . The equation of value is 67,000 = 50,000 + C + I = 50,000 - 10,000 + I, from whlcr. 'xe Dollar weighted:

find 1=27,000 difference is

C1/3

and

itw - idw

=

idw

15,000,

=

= .5462

C2/3

= -25,000,

+ 15,000d~ .5226 = .0236,

and

50,000

1~~~_

-

ANSWER A

"H' MM'

"I'" =

.5226.

Tt;::-. ::.e

~

•• •••

.~

••• 16.

The initial annual payment is gooo 101·07

=

Let the new annual payment be P.

1139.02.

TI'.e:-.

to

give an annual yield of 6.5% to the lender, we have 8000

= 1139.02a4j.065 + P

. a4j.065. v~65'

which solves for P = 1538.87. Then Peter pays 4(1139.02 + 1538.87) = 10,771.57 under the new arrangement, versus 10(1139.02) = 11,390.20 under the original arrangement. The savings is 11,390.20 - 10,771.57 = 678.63, A.~SWER D

17.

The present value is

a~ + p.

vn+k-1

=

d I_vn+k

Therefore

p

(l_~n+k _ l-;r) (1 + i)n+k-J

=

(1+i)H_(1+W1 d

=

(1+i)"-l d(1+i)

=

(1+i)k-l

i

ANSWERD