•••

••
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 III
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 . ••
••
(A) 168,000 (B) 182,900 (C) 184,425 (D) 200,000 (E) 201,675
••
•••
2. Which of the following are true? ••
•••
;hCi) :v-2 1J"m)) ~ v-(";') m~(i) ~ 1 +,
••
I.

(A) I and ITonly

IT.

(B) I and ill only (C) ITand ill onl~) (D) I, ITand ill
••
(E) The correct answer is not given by (A), (B)vEC) or (D) •••
~ ( •••
3.
f-:5
A fund starts with a zero balance at time zero. The fund accumulates with a varying force of
••
-fJ-, •••
interest Ot ==
t + 1 for t > O. A deposit 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 (B) 1.00 (C) 1.50 (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) Fund Y accumulates at a rate ofinterestj, compounded semiannually.
(iii) At the end of eight years, Fund X is 1.05 times as large as Fund Y.

Calculate j.

(A) .022 (B) .023 (C) .042 (D) .044 (E) .046

5. 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 29

(C) :L
1=10
I? (D) :L 1f
1=]0

7. Gloria borrows 100,000 to be repaid over 30 years. You are given:

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

(A) 5505 (B) 5555 (C) 5605 (D) 5655 (E) 5705

8. You are given a perpetuity with annual payments as follows:

(i) Payments of 1 at the end of the first year and every three years thereafter.
(ii) Payments of2 at the end of the second year and every three years thereafter.
(iii) Payments of 3 at the end of the third year and every three years thereafter.
The interest rate is 5% convertible semiannually. Calculate the present value of this
perpetuity.

(A) 24 (B) 29 (C) 34

\ "---'/ (E) 47
r-~ 39

9. 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 (B) 19 (D) 21 (E) 22

10. 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 38

/ CE)
 ---- .••4,-
-.
ll-28

11. A loan is to be amortized by r. re- •.e: ::'::'.l2.1 payments of X, where n > 5, You are giver::
•••
It

(i) The amount of i:l:e:-es: ::: :~e first payment is 604.00,

••
(ii) The amount ofin:e:-es:::: ::le third payment is 593,75 . ••
(iii) The amount of :r::e:-es: :~ tr.e fifth payment is 582.45 . •••
Calculate X ••
(~A~ (B) 739 (C) 1163 (D) 1198 (E) --
''':':' - •••
"-/ 704
••
...

••
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 (B) 94 (C) 99 (0) 104

'~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 ~ 1600 (D) 1650 (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% com-
pounde~ semiannually. Fiona earns the same amount of profit as Henry. Calculate X

CA) 500,000 (B) 502,000 \9 505,000 (D) 571,000 (E) 574,000

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

1/1/99 511/99 9/1/99 1/1/00

Account Value 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

(B) 6.4% ,A/A.. 9.8% (D) 12.3% (E) 15.8%

~2.3% !~
 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
(B) At least 550, but less than 600 tE) At
\~) At least
least 650,
700 but less than 700
(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.

(C) 1- k (E) (1+il-J

(A) 1 (B) k
 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 ••
III
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,
(2.3230573)-1/8J = .40.
ANSWER D ,.
••
2. 1.
-,
v -,yes . •••
d .
dd(z) _- d (
([(l r=a d ) -
_ (1 - (1
d) -- d)2
d( - 1) = (1 - I d)2 _
••
n. :i (i(m)) = :i [m(1+i)I/m - 1] = (1+i)l/m -J = v(m-I)/m, no.

m. -isCi) = 1s(eO - 1) = eO = 1+ i,yes.

I and ill are true, ANSWERB

3.
The fund accu~ulates from t = 2 to t = s, where exp [is Ot dt] = exp [15 t2~ 1 dt] = 2.

t -J1-+
J2 t 1 dt = In(t2+ 1)1~ =: In(s2+ 1) - In 5 = In (~). Then

exp [In s2t1 ] =~ = 2, so that s = 3. The fund accumulates from time 2 to time 3, a
period of 1.0 year, ANSWER B

4.
After eight years Fund
. X has e80 = eA, and Fund Y has ( I+ ~')16 . We are given that
ANSWER
eA = (1.05) ( 1+ ~')16 ' so that j = 2 [( 1~05
A )1/16
- 1] = .04438,
D

5. We view the withdrawals at times t = 4 and t = 5 as being of amount K(1.05) to include the
penalty. The withdrawals at times t = 6 and t = 7 are of amount K only, since there is no
penalty. Thus we have the equation of value

10,000(1.04)]0 - K(1.05)(1.04)6 - K(1.05)(1.04)5 - K(1.04)4 - K(1.04)3 = 10,000.

I.
So vmg for K we have K = -----~---~----10,000[(1.04)]0 - I] = 979.92
(1.05)(1.04)6 + (1.05)(1.04)5 + (1.04)4 + (1.04)3 '
ANSWERC.
 ill-.::~-

6. The current value at t =2 of a.-. a.-: . .1: :y-due is (1 +i? . a201 if the interest rate were i in eve::.
year. More generally, the CU!7e:::·:::':-.1e 1S

if the rate in year t is it· Here we have it = 8 it, so that 1 + it = ~ t~,

and (1+i:)(l-i:) = ¥
(l-i2) = M
11
TT
1 _ 11
l+i3 - 12
1 11
(1+i3)(1 +i4) = 13

1 11
(1+i3)(1+i4) .. ·(1+i!9) = 28

28

Finally, X = L
t=9
\1, ANSWER B

7. The loan is the present value of the repayments, so

100,000 = X· v -+ (X +100)v2 + (X+200)v3 + ... + (X + 1900)v2o + (X +1900)v20 • a-;Qj

= X· a301 + 1900v20 • aTOJ + 100v . (I a )T9!'

where all functions are calculated at 5%.

ThenX 100,000 - 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:

2 3 1 3 1 2 3 1 2 3
•
••
t
o 2
t t
3
t
4
,

5
J

6
t
7 8
t t
9 10
t
11
t I
12 ••
••
Consider the value of each set of three consecutive payments. At t =0 the value of the first ••
•••
= (1.025
1 )2 + (1.025)4 + Cl . _~ =
three payments is X
=
2
= 4, =
_. f. 5.350614. This value X also exists
=
••
at t 3 (for the payments at t 5, 6), at t 6 (for the payments at t 7, 8, 9), and so on .
••
The overall present value is
ANSWERD
••
PV = X[1 + I + (1.025)12
1 + ...] = I 5.350614
] = 3886 •••
(1.025)6 .025
_ (_1_)6 .,
••
? •••
9. PV _ 1 1.07 (1.07)-
- 1.12 + (1.12i + (1.12)3 + ... ~
ANSWERC

10. The sinking fund payment is 1000

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

SF B4 = lODO
101·04. 8'4104
. ' so the interest earned in the fifth year is

.04 X SFB4 _- 1000 [(1.04) 4 -

8-101·04 1] -_ 121000 8-
.nt)~1 (.169 6) - 14.15.

The direct interest payment is .05(1000) = 50, so the requested difference IS

50 - 14.15 = 3S.85, ANSWERD

(Calculator comments: Following is an alternate solution based on the relationship between the
sinking fund method with i = .04 and if = .OS, and the amortization method with i = .04, and
making use of the BA-35's ability to split payments into principal and interest. The net interest
in the Sth year under the sinking fund method is the same as the interest in the 5th payment under
the amortization method at i = .04, plus the quantity (if - i)L = (.05 - .04)(1000) = 10. To
obtain the interest in the amortization method, hit IAC/ONI, enter 10, hit fED, enter 4, hit I %i I,
enter 1000, hit /PVI, and then hit ICPTIIPMTI to determine the annual payment of 123.29. To
obtain the interest in the 5th payment, enter 5, and hit II/PI, resulting in 2S.85. Then the net
interest in the sinking fund method is 25.85 + 10 = 35.85.)
 ill-37

11. lfthe loan is L the payment is X = a!:....'

nli
and the interest in the tth payment is X(1 - vn-t+1).

Thus we have (a) X(1 - vn) = 604.00, (b) X(l - vn-2) = 593.75, and
(c) X(1 - vn-4) = 582.45. Subtracting (a) - (b), we have X(vn-2 - vn) = 10.25.
Subtracting (b) - (c), we have X(t,n-4 - vn-2) = X(1+i)2(vn-2 - vn) = 11.30.. Taking the

ratio of ~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. We

lmow that the ratio ~ ~~ = (l +N, so we have Xx-}~tJ5 = 1.10.244, from which we find
X - 593.75 = (1. 10244) (X - 60.4), leading to X = (1.10.244)\6gi1- 593.75 = 704.0.6,
ANSWER A.

12. We are given 10.0. = p. aTOT.04+2p· vlO. a201.04'so that

P _ 10.0. _ 10.0 _
- _ + 2.
101
vlO . a-
201 - (R, 1109) -+ 2U;7556)(13.5903) - 3.77743. Under the new arrange-

ment the equation of value becomes 10.0. = 3.77743aTOT,045 +X . V.b~5' so that

X = (lo.o. - 3.77743awr)(1.0.45)10 = 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 = 10.8.88.)

13. We are given the following, where j = 2- 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 X: .40.77 = (.0.7 - j)a2nlj; Y: .2718 = (.0.6 - j)a2nlF Taking the

ratio we find :g~ = ~ = :in~,which solves for j = .0.4 and i= 2j = .0.8. Then we have

.40.77 = .03a2nl.04, so a2nl.04 = 13.59. Inspecting a 4% interest table we find azol = 13.59, so
2n = 20.. Finally the price of Bond X at i = .0.7is
P = 100.0.+ 100.0.(.0.7- .0.35)a201.035= 100.0.+ (35)(14.2124) = 1497.43, ANSWER A

(Calculator comment: \
The annuity keys make it possible to see that 2n = 20 without interest
tables. 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 =
P.02 = 1,000,000 [1 + (.03 - .02)a!oi.02]

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

ForFiona,po2 = X[1+(.03-.02)a201.02]
=
= 1,089,826, and

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

= 1.163514X,and
= ,
•
•
= =
.......... ~- ~
-.' -- - •
P.OI X[1+(.03-.01)a201.Dl] 1.360912X.
=
Her profit is POI - P02
~~;~OOOQ= 504,564,
=
•
which is the same as Henry's. Thus we have X
•
(Calculator comment: P02 - P.Ol can be readily obtained using the BA-35's annuity ke:.~ .-...
•
••
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 = 1,089,825.80. Now enter 1, hit I%il, and ther, :-.:: :?-=-
~
IPV! to obtain POI = 1,189,426.10.)
••
••
15. •••
. h te:d 1 + '/.
T·Ime welg . = (75,000)
50, 000 (90,000)
90, 000 (67,000)
65, 000 =. 15462 , so 'l.tw
. = ..5462
~
C = Let = -10,000 .
Dollar weighted: C1/3 = 15,000, C2/3 = -25,000, and
The equation of value is 67,000 = 50,000 + C + I = 50,000 - 10,000 + I, from whlcr. 'xe
••
•••
find 1=27,000
difference is
and
itw - idw
idw

= .5462
= 50,000
-
+ 15,000d~
.5226 = .0236,
1~~~_
ANSWER A
"H' MM' "I'" = .5226. Tt;::-. ::.e .-
~
•••
16. The initial annual payment is gooo
101·07
= 1139.02. Let the new annual payment be P. 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 (1+i)"-l (1+i)k-l
d = d(1+i) = i
ANSWERD