· 23. Identification: Annuity u64 10 4 40 40 0,1225 2500000 1 4 0,1225 2500000 1 1 0,1225 4 3 4...
Transcript of · 23. Identification: Annuity u64 10 4 40 40 0,1225 2500000 1 4 0,1225 2500000 1 1 0,1225 4 3 4...
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DSC1630
MAY/JUNE 2009
SOLUTIONS
()lIESTlO'" I
S :..:::1'(1 of I'll .. ...J 500::: 4317.26(1 + 0.151) .. 1 = 103 Jays. Fmm till.' 1.lhk", 5 \la~ is IhI..' 1251h
d;t~ or III•..· .\I..';lf.: 103 + '25 ::: 22Xlll day of Ih~ y car \\ hie-h is 16 (\ugllst. 121
I·=.\'(\~d/). :.35()O='''{1-0.I3X~}- :.S= 10925.23 151
()IIESTlO'" .1
d ,. 0.1858r = --. :. d = -- = ,- D.IXD027= IX% III
1-"1 1'1+ 1 O.185Xx ~-, + 1
365
(I+~r =r""" : 1+± =r"'2.<71' :.i=12.77% HI
PV = ±..f{) 000. FV = 56000. N:::: 5. C()~1P I x 2. 1-'1
I'V=±IO(I. N=4. 1= I~. COMI'I'V. N=52. COMI'I '21114
()IIESTIO ..••7
)IV = ±140 000. \I:::: 10. I = 2X. CO\IP FV. rv after 10 Yl:ars= R 1652 82R.27
PV:::: ±1652 X2X.27. N = Q. I:::: 5. CO\W FV ISI
/
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DSC1630 May/June 2009 Solutions
()I'ESTlO:\ 'J
100L~ 6 000o I :! 3 ~ :;
NO\\
13.75 .
la) P\'=±10000. ,'\!=IO.I=~. CO\IPI"V. rV=R1C)~-C!.9In n(l)
Ih) F\'=6000. N=6. 1= 13.75. CO\IPPV.PV=R ..t()26.1(,L.-nnn-(1)2
( 1) + (2) = R23 4(,9.10 151
Qn:STIO:\ III
90tlO 23-169.10
(I
Raphael P;)ys Pinky 4 yt.'ars from now
QlJESTlO:\ II
PV = ±] 64 1100. N = II. I = 9,9. CO~IP FV 14)
QlIESTlO:\ 12
~
PV = 500000(1 + 0.224 t' + 250 OOO(]+ 0,224) <, + 150 000(1 + IU24 r' + 700 OOO(]+ 0.224) ,
= R440401 ill
QIJESTIO:\ U
~20()O
(I
;!Jl!lli14 15
·1 000
II>
4w)Q
17lJlliQ
25
x = (I + i)2000(l,~11-t 4000aw1, (1 +i) 15
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.DSC1630 !-'I_ay/June 2009 Solutions
()UESTION I~
-' OOQ."lirndilll: J
6J!JlP" QJl!lO
IJ
Page 3
QJlliQ
19
I he first timdillc shmys payments made at the end of each month.]1) the s",'CU!llj timclinc the first
paymcll! is Illw ..h: al the beginning oflhe 131h month and the last paymcnt is t11<Kh.'at the hcginlling
(if till' 20lh I1wllth.
The I'Y of the raIments made at the end of each month ~ 3 ooo(] + ~) 12 ~ ]U2 ~52.37.. 12
The [)V al the heginning or the 131" month or al1 the payments mauL' at the heginning of each
month = (I + ~)6 OOOal( l (J I%!P = R45 385.16. \Ve now determine the present valuc of this12 . ~
-I'
a1l1(!unt al the beginning of the term. Thus PV = R45 38S.16( (1 + 0./;6) - = R37 366.28
lIence py o!'thc loan = R32 452.37+RJ7 366.28 = R69 818.65 121
QUESTION IS
Interest rate per month = 6036.40 x 100 = 1.2% = 14.4% per annllm503 033.57
QUESTION 16
A=49) ]61.53-(]5908.46-5917.94)=R483 ]70.99
QUESTION 17
I'M f ~ ± 15'iOS.46. N ~ 2~0. ] ~ J.12. COMI'I'Y. 131]2
QUESTION IS
( ] 7(0) .1005962= x+-- ')~lIl 0.1
(LI -
QUESTION ]<J
20x] 700
0.]'. x = 6 son approximately 121
7.5"1~1 I = 90.80770 - J2.09XXX = 64.70X82
PV 64J,OS82. Pr\1T-cc±7.5, N'-'-IS.(CO:--'WJ)x2 121
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DSC16JO May/June 2009 Solutions
(H TSTIO.'\ .211
PI= ~PV+~.ill\l.'s(Il1l.'~ iL' I ")...J.17.::,=195000+lnilialill\('stm~Initial inn's[lt1l.'nt ... - .• Initial inn:sllllt'1l1
Initial ill\'l.'SIIllL'111 = XOO oon I~I
()IIESTlO:\ 21
Page 4
"'9 -'larch ~OOI.) t) "'la, 21XN 29 ScrHcmbcr 2009 29 SCDtcml~r 20 15
() 12
( l'1'(29Sep~()09)= (J.Julll ~75 1100 1+ ~;~ RI1J.9JJS\tYlI.Sim.'elhcfirstl'lIlJpol1datc
and tbe sc:ltlcmcilt oale arc more than 10 days apart. we add tht: value ()f the WUptll1ltl this price.
J kIK'C prkc R 120.23381. L\lso the numhcr of days betwcen <) ~1ay 2009 and 29 Sl."ptcmhcr1-0
2(1)1) is I...J.J. I h:ncc all-in-price = 120.1338{] + .t.7S rli~ = R 115.974 7...J.%HI100
()lJESTION 22
(. = PV I (I + ~IIRR)" :. _c_ = (1 + \IIRR)". T~lkin1! thl.' natural log on hoth sides wc gel•••• PV""I ~
101-1'- = /lln( I + t\I/RR).Solving for t1 givesPVn,,!
( (' J
In
n- PV"UI 121
11l(1+~IIRR)
()IIF:STIO:\ 2J
121
()IIESTlO:\ 24
10, lx, yl. I 500. Ie,,"
15. (x, .1'). 1~(). LN"
RCIe h ISI
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DSC1630 May/June 2009 Solutions
Rei r III
~
(a) 1':'-.11 ± 7.35. N:= ::!9.13.5
CO\lP I-V. PV = 92.50885------.. (I)
(h) FV -'= ] 00. N = 19. I = 1J.5 .. CO\IP PV. PV = 15.042~\9_hUU~h (2)1
:. P(24 JUIlC 2(09)=(1)+ (2):= RI07.5517..J
Sill'-'c thl.' sclth:mcnt date and the first (,.'OUP0!ldale an: more than I() days "part. add the
valuL' ol"lh(" (;'oupon to the aome price. lienee all-in-price on 24 June 2009 is R 114.90174%.
Clean Price = all-in-price - accrued interest
= 114.90174 - 4.J09J2 = R 110.59242% 151
QIIESTlO:\ 27L-
'\;0\\:~4
X
8
QIIESTlOI'O 2H
Loan = 0.&0 x I -150000 = J 160000
P\IJ=±1527·t76. 1=12.95. N=2·t CO\I!'P\'. 1'\'=130775312
IIcncl.' real cost ofloan = I J07753-1160000=RI47753 121
QII":STIO:\ 2')
FV = X50 000. N = 32. 1= 12J) .. COMP PMT 1.'14
QI'ISIIO:\ .111
't'earh cost = \Ionthl\" interest )(S + quarh:rI\' pa\ll1('Il(S )(-t.. 12 ..
~ ~ x R50000 + 15 nO.ORx 4 ~ R 188880100
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MAY/JUNE 2010
1 Identification: Simple interest
Pr
29125000 0.0845
365
1684.21 (1)
SI T
R option
u u
2 Identification: Discount
5(1 )
748000(1 0.119 )12
44668 (1)
P dt
R option
�
� u
3 Identificatio: Discount
(1 ) ] (1 )
1
1
(5)
If S P rt and P s dt
dt d
r
d
rt option
d
� �
§ · � y¨ ¸© ¹§ ·�¨ ¸© ¹
4 Identification:Compounding interest
5 mar
291 days
21 Dec
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9152
52
(1 )
0.296117500 1
52
29338.66 (3)
nS P i
R option
u
�
§ · �¨ ¸© ¹
5. Identification: Continuous compounding
ln 1
0.1942ln 1
12
19.24% (5)
jmC m
m
option
§ · �¨ ¸© ¹§ · �¨ ¸© ¹
6. Identification: Normal and effective interest rates
� �1 1 (5)m
jm m jeff optionª º � �¬ ¼
7. Identication: Annuities
� �
9 3
1 1
0.11321 1
32200
0.1132
66173 (1)
ni
S Ri
R option
u
ª º� � « »
« »¬ ¼ª º§ ·� �« »¨ ¸© ¹« » « »« »¬ ¼
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8. Identification:Deferred annuity
5 2
: 1
0.18695000 5 20.186 2 1
2
601708.66 (3)
tmvm
PV Ranm
a
R option
�
� u
§ · �¨ ¸© ¹
§ · u y �¨ ¸© ¹
9. Identification: Differed annuity
� �4 2
1
0.186601708.66 1
2
601708.66 (3)
nPV S i
R option
�
� u
�
§ · �¨ ¸© ¹
10. Identification: Continuous compounding
(0.0847 8)53580
27 209.76 (2)
ct
ct
S Pe
P Se
e
R option
�
� u
11. Identification: Evaluation of cash flows
1 2 3 4 5 6 7 8 9
p1 p2 p3 p4 p5
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330000 5
250000
26,4% (5)
Average after tax incomeAPR
Investment level
option
�
y
12. Identification: band
Settlement date of the stock is 11 August 2010 as the next coupon date of stock 538 is
28/09/2010.
Option (3)
13. Identification: Perpetuity
3500
1,112
12
375000 (4)
RP
L
R option
§ ·¨ ¸© ¹
14. Identification: Evaluation of cash flows
� � � �� � � �
3 5
7 9
25000 1 0,0864 50000 1 0,0864
75000 1 0,0864 30000 1 0,0864
19497,13 33038,57 41 988,75 14230,28
108755 (1)
PVout
R option
� �
� �
u � � u � �
u � � u �
� � �
15. Identification: Evaluation of cash flows
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� �� �
� �
1
1
108755 1 0,18472
500024
500000 2
i
n
n
CMIRR
PVout
C PVout i
R
R option
§ · �¨ ¸© ¹
�
�
16. Identification: Sinking funds
� �
� �
Pr 350000
8486,23 20 12 0,105 12
850000
850000 350000
120000 4
iceof the house R PV of theloan
PV a
R
Total price R
R option
�
u y
�
17. Identification: Sinking funds
� �
.
20 12 8486,23 850000
2036695 850000
1186695,20
1186695 3
I no of payments payments theloan
R
R
R option
u � u u � �
18. Identification: Bond
Pr
90,61202 5,31616%
85229586%
Financial calculator
BOND
Clean ice All in price Accrued Interest
R
� �
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19. Identification: Amortisation
� �
intint
707,25
24522,83 8926,30
707,25
15596,53
4,5346%
4,5% 1
erest dueApplicable erest rate
accumulated amount in the fund
option
�
20. Identification: Amortisation
� �
Pr
8151,38 7444,64
1481,67 1
Interest rate incipal paid payment
R
R option
� �
21. Identification: Amortisation
� �
� �
7444,64 45000 4,5%
7 444,64 2025
5419,64 5
Valueof B additionto fund
Additionto fund
R option
o
� u
�
22. Identification: Bond
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� �� �� �� �
� � � �� �� �
:
9,6
100
17 05 2010
14 12 2033
/ 2 2
1/ 10,4
92,
Financial calculator Clear memory and press BONDbutton
COUPON PMT ENT
REDEMPT FV ENT
Settlement date ENT
Maturity date ENT
CPN Y N always ENT
YIELD Y ENT
PRICE PV PRESS COMP BUTTON
o �
o �
� � �
� � �
o �
o �
o o
� �
� �
99634150
4,061538462
Pr
92,99634150 4,061538462
97,05788% 3
ACCU INT PRESS COMP BUTTON
All in price ice PV ACCU INT
R option
o o
? �
�
23. Identification: Annuity
� �
� �
10 4
406 4
40
0,12252500000 1
4
0,12252500000 1 1 0,1225 4 3
4
0,12252500000 1 5,06261
4
8355393,89
5,06261
1650412,32
3 4951236,95 4
Obligations
Payments
R
R option
u
u
§ ·� ¨ ¸© ¹
§ ·� u � y u u¨ ¸© ¹
§ ·� u¨ ¸© ¹
u
u
uu
now 4 10 years
p1
x 3x
p2
x
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24. Identification: Simple interest Compound interest
1612
12
int
1
375000
271 0,0791
365
372818,55
int
1
0,0791372818,55 1
12
335617,209
int
1
355617
tm
Simple erest for odd periods
SP
rt
R
R
Coumpound erest for full terms
jmP S
m
R
Simple erest for odd periods
SP
rt
�
� u
�
� u
§ · �¨ ¸© ¹
§ · �¨ ¸© ¹
�
� �
, 21
1 0,0791 23 365
333952,66 1R option
� u y
25. Identification: Regression line
� �92,308 5
Financial calculator
Slope for the angle y a bx
rcl r option
�
5 may 2009
16 months
24 Oct 2010
27 days 23 days
1 June 2009 1 Oct 2010
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26. Identification: Regression line
� �0,97,15 2
Financial calculator
rcl r option �
27. Identification: Annuity due
� �
� �
� �
52 4
1
0,134209000 1
52
122366,87
122366,87 54 2 0,134 52
758,77
758,77 1
nPV FV i
R
R
R R
weekly deposits R option
�
� u
�
§ · �¨ ¸© ¹
u y
28. Identification: Annuity
1 2 3 4 5 6 7 8 910
P1=12,65%
c/y=2
n=7
15 500now
3
P2=35 000
C/y=12
9,5%
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� �
2 7
12 3
1
0,126515500 1
2
36578,61
0,0952 35000 1
12
26350,053
1 2
62928,66 5
Payment
S
R
P S
R
Total amount Frank has to pay back
P P
R option
u
� u
§ · �¨ ¸© ¹
§ · �¨ ¸© ¹
�
29. Identification: Annuity
� �
4 5 2 4
8 20
0,109 0,10915000 1 62928,66 1
4 4
0,109 0,10962928,66 1 15000 1
4 4
52907,43 5
Payments Obligations
x
R option
u u
§ · § ·� � u �¨ ¸ ¨ ¸© ¹ © ¹
§ · § · � � �¨ ¸ ¨ ¸© ¹ © ¹
30. Identification: Annuity due and ordinary annuity
4 7
5yrs9
10
35 000now
15 000 Amount received
here
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� � � �
� �
� �� �
9
12
912
12
1 :
1
15 0,1641 0,164 12 4000 12
12 12
54655,23
2 :
1 1
1
0,6141 1
127000
0,614 0,6141
12 12
n
n
st set of payments annuity due
P i R a n i
a
PV R
nd set of payments ordinary annuity
iP R
i i
u
�
§ · � y u¨ ¸© ¹
§ ·� � ¨ ¸
¨ ¸�© ¹§ ·
§ ·¨ � �¨ ¸¨ © ¹ ¨¨ § ·�¨ ¸¨ © ¹© ¹
� �
58902,19
1 2
54655,23 58902,19
113557,42
113557 4
R
The amount thatVicky owes Cindy
P P
R
R option
¸¸¸¸¸
� �
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MAY/JUNE 2011 SOLUTION
Question 1
Identification: Simple interest
� �
� �� �
7,5%
545
365
1
54525000 1 0,075
365
25000 0,075 1,4935
27799,66 1
Interest rate per year
time
S P rt
Option
? �
§ · � u¨ ¸© ¹
� u
Question 2
19.4cm
25 000
P
S545 days
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� �� �� �� �
� � � �
� �
� �� �
� �
1
1
1 1
1 1
11
1
11
1
1 1
1
1 1
1
1
1
1
1
1
1
1 / 1
S P rt
P S dt
S S dt rt
Sdt rt
S
rtdt
rtdt
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OCTOBER/NOVEMNER 2011
1. Identification: Simple interest
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4 Jan2011
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4.
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9 months
13 days 20 days
1 Mar16 Feb
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13. Identification: Deferred annuity
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16. Identification: Perpetuity
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20. Identification: Correlation coefficient
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29.
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MAY/JUNE 2012 SOLUTION
Question 1
Identification: Simple interest
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12620000 1 0,12
12
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Question 4
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09/04/2012
Settlements
Date
16/07/2012
Next
Coupon date
09/10/2012
Maturity
Date
09/10/2038
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erest rate is R
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BASIC SOLUTIONS FOR DSC1630 EXAMINATION QUES-
TION PAPER OCT/NOV 2012
1. Identification: Simple interest rate
2 months 10 months now
11,5%
R9 000 R10 000
Savings worth at month 10 – move R9 000 from 2 month’s ago to 10 month’s in future:
S = P (1 + rt)
= 9 000
!
1 + 0,115×12
12
"
= 9 000(1,115 . . .)
= R10 035,00.
Money Short:
R10 035,00− R10 500 = −465,00
She shorts R465,00.
Option [3]
2. Identification: Simple discount rate
31 Aug
R5 000
2 Nov
0,18 d =
t = day number 306 - day number 243 = 63 days
P = S(1− dt)
5 000 = S
!
1− 0,18×63
365
"
5 000#
1− 0,18× 63365
$ = S
S = R5 160,32.
Option [2]
1
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3. Identification: Compound interest
jm = 12%
m = 12
P = P
S = 2P
t = ?
S = P
!
1 +jmm
"tm
2P = P (1 +0,12
12)t12
2P
P=
!
1 +0,12
12
"12t
ln 2 = 12t ln
!
1 +0,12
12
"
ln 2
ln#
1 + 0,1212
$ = 12t
t =
%
ln 2ln(1+ 0,12
12)
&
12t = 5,80506 years
t ≈ 5,81.
Option [1]
4. Identification: Odd periodes – method given
R375 000
7 March 1 April 1 May 1 June 1 July 1 Aug 1 Sept 1 Oct 1 Nov 28 Nov
25 days odd
10,45%/12 Full=7 months 27 days odd
S = P (1 + rt)(1 + r)t(1 + rt)
= 375 000
!
1 + 0,1045×25
365
"!
1 +0,1045
12
"7
12×
12
1
!
1 +27
365× 0,1045
"
= 404 419,5870 . . .
≈ R404 419,59.
Option [4]
2
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5. Identification: Fractional compounding
S = P
!
1 +jmm
"tm
= 375 000
!
1 +0,1045
12
"( 7
12+ 25
365+ 27
365)× 12
1
= R404 415,85.
Option [4]
6. Identification: Compound interest
5 years
R50 000 ? 0,1388/52
S = P (1 + jm/m)tm
50 000 = P
!
1 +0,1388
52
"5×52
P = R25 001,79.
Option [2]
7. Identification: Continuous compounding rate
?
57 weeks
R32 412,87
10,15% c =
S = Pect
P = S/ect
=32 412,87
e(0,1015×57
52)
= R29 000,00.
Option [1]
3
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8. Identification: Equivalent compound interest rate
jn = n
'
!
1 +jmm
"m
n
− 1
(
j52 = 52
'
!
1 +0,149
4
"4
52
− 1
(
= 0,14650
≈ 14,65%.
Option [1]
9. Identification: Effective interest rate
jeff = 100
!!
1 +jmm
"m
− 1
"
= 100
'
!
1 +0,165
6
"6
− 1
(
= 17,67684
≈ 17,677%.
Option [4]
10. Identification: Re-scheduling of debt – equation of value.
now 4 years
10 years
X 3X
2 500 000 0,1225/4
2 500 000
!
1 +0,1225
4
"10×4
= X
!
1 +0,1225
4
"6×4
+ 3X
2 500 000
!
1 +0,1225
4
"10×4
= X
)
!
1 +0,1225
4
"24
+ 3
*
2 500 000
!
1 +0,1225
4
"10×4
= 5,06261X
2 500 000
!
1 +0,1225
4
"+
5,06261 = X
X = R1 650 412,32
3X = R4 951 236,95.
4
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Option [4]
11. Identification: Re-scheduling of debt – time value of money.
150 000
6 years - now 3 years
X
6 years 9 years
R250 000
0,155
12
0,164
2
X = 150 000
!
1 +0,155
12
"9×12
+ 250 000
!
1 +0,164
2
"
−(2×3)
= 599 863,8759 + 155 803,23
= R755 667,10.
Option [4]
12. Identification: Payments paid indefinitely – Perpetuity
PMT = 3 500 i = 0,112/12
P = R/i
= 3 500
+
(0,112/12)
= R375 000.
Option [4]
13. Identification: Equal payments in equal time intervals plus compound interest rate – annuity
but time intervals of payments not equal to compounding periods thus change compound
interest rate from quarterly to monthly.
5
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jn = n
'
!
1 +jmm
"m
n
− 1
(
= 12
'
!
1 +0,0775
4
"4
12
− 1
(
= 0,07700.
Thus
S = Rs n i
= 1 200s 10×12 0,077
12
= R215 899,01.
Option [2]
14. Identification: Maths manipulation of equation.
S = Pect
S ÷ P = ect
ln(S ÷ P ) = ln ect
ln(S ÷ P ) = ct ln e
ln(S ÷ P )
t= c
Option [5]
15. Identification: Payments that are made on equal time periods but payments increase each
time period with a constant amount – increasing annuity.
S =
!
R +Q
i
"
s n i −nQ
i
=
!
3 600 +360
0,10
"
s 20 0,10 −20(360)
0,1
= R340 379,99
≈ R340 380.
Option [2]
16. Identification: Payment being postponed – deffered annuity.
R25 000 R25 000
5 years 6 years
0,169/6
R25 000
A now
6
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A = Ra n i
= 25 000a 6×6 0,169/6
= R561 047,91.
Option [4]
17. Identification: Moving money back in time – time value of money and compound interest.
P = S/(1 + jm/m)tm
=R561 047,91
(1 + 0,1696
)5×6
= R243 834,05.
Option [4]
18. Identification: Equal amount’s deposited in equal time periods + payments made immedi-
ately − annuity due.
R350 000 5 000 5 000
0,124/12 now 4 years
S = (1 + i)Rs n i
= (1 +0,124
12)5 000s 4×12 0,124
12
= R311 882,75.
Amount needed still:
= 350 000− 311 882,75
= R38 117,25.
Option [1]
19. Identification: Sinking fund.
5 years
R275 000
0,16/4
sinking fund 0,14
2
7
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S = Rs n i
275 000 = Rs 5×2 0,14
2
R = R19 903,81
R ≈ R19 904.
Option [3]
20. Identification: Amortisation schedule.
A = payment
Interest + principal repaid = 3081,86 + 1119,21
= R4 201,07.
Option [5]
21. Identification: Internal rate of return.
Using your calculator:
IRR = 15,23893%
≈ 15,24%.
Option [5]
22. Identification: Equal payments in equal time intervals – annuity or armotisation
A = Ra n i
= 5 311,69a 20×12 0,0975
12
= 559 999,54
≈ R560 000.
Option [3]
23. Identification: Percentage calculation.
560 000 = 80%
? = 100%560 000
1×
100
80= R700 000.
Option [4]
8
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24. Identification: Total real cost
Total real cost = PV of annuity using inflation rate − PV of annuity
PV = 5 311,69a 20×12 0,0467
12
= 827 543,12.
Real cost = 827 543,12− 560 000
= R267 543,13.
Option [2]
25. Identification: Correlation coefficient.
Using your calculator:
r = −0,98185.
Option [2]
26. Identification: Slope of regression line.
Using your calculator:
intercept a = 1 021,13
slope = b = −207,59
Option [1]
27. Identification: Bond
7 Aug 12 15 Nov 12 7 Feb 13 7 Aug 13 7 Feb 14 7 Aug 14 7 Feb 35 7 Aug 2035
11,25% YTM
2 = 9,75% c =
n = 22,5 years
= 22× 2 + 1
= 44 + 1 = 45
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P = da n z + 100(1 + z)−n
=9,75
2a 45 0,1125/2 + 100
!
1 +0,1125
2
"
−45
= 87,80282
Add coupon as number of days > 10 days between settlement date and next coupon date.
87,80282 + 4,8785
= 92,67782%.
Option [5]
28. Identification: Bond
107,55174 = da 29 0,135 + 100
!
1 +0,135
2
"
−2
107,55174− 100
!
1 +0,135
2
"
−29
= da 29 0,135
92,50885 = da 29 0,135
This looks like an annuity formule. Thus use your calculator’s financial mode with
92,50885 = PV ; N = 29; P/Y = 2; I/Y = 13,5 and solve for your payment which is d.
c/2 = d = 7,35%.
Option [2]
29. Identification: Profitability index
Typing error in answers – Question ignored in October/November 2012 exams.
PI =NPV + initial
initial
1,0514 =25 700 + x
x1,0514x = 25 700 + x
0,0514x = 25 700
x = 500 000.
30. Identification: NPV ; PI and IRR
Investment A Investment B
NPV < 0 reject NPV > 0 accept
PI < 1 reject PI > 1 accept
IRR < K reject IRR > K accept
Accept Invest B Option [2]
10
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DSC 1630: MAY/JUNE 2013
QUESTION 1
� �
int
1
129000 1 0.115
12
10035
10500 10035 465
3
simple erest
S P rt
option
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QUESTION 2
� �int
1
9000
1 8%
4
?
5.12
2
n
compound erest
FV PV i
FV
Y
C Y
N
N yrs calculator
option
�
?
QUESTION 3
� �1
830000 1 0.165
12
26700
1
simple discount
P S dt
option
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� �38
612
int
1
0.113513300 1
6
18988.31
5
n
compound erest
S P i
option
u
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QUESTION 5
25000
1 7.5%
6
6
721181.68
3
PMT R
Y
P Y
N
PV
option
QUESTION 6
sin
870000 8 120.092 12
870000
8 120.092 12
6166.12
2
king funds
S Rs ni
Rs
RS
option
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u u
QUESTION 7
40000
/ 2
/ 2
1/ 9.56
7
?
401574.30
4
PMT
C Y
P Y
Y
N years
PV
PV
option
?
QUESTION 8
19cm
7 years
513
1
PMT
40 000
01 2 3 4 5
5 years
401 574.30
401574.30
1/ 9.56
/ 2
5
?
251757.24
4
FV
Y
C Y
N
PV
PV
option
?
QUESTION 9
189002279396
0.0995
12
2
Q
i
option
QUESTION 10
6.6 tan
3
calculator s dard deviation
option
QUESTION 11
5;500
15;900
19;1500
7;2000
0.16428
2
correlation coefficient
DATA
DATA
DATA
DATA
RCL r
r
option
QUESTION 12
5500
1/ 8.9%
/ 12
/ 10
10
?
173000
4
PMT
Y
C Y
P Y
N
FV
FV
option
?
QUESTION 13
� �� �
6
3
45000 1 0.1159 86.876
90000 1 0.1159 125059.97
115000
326950
5
TOTAL
option
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�
QUESTION 14
1
1
9
1
3269501
95000
14.72%
1
n
out
CMIRR
PV
option
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2
5000
/ 4
1/ 9.5
/ 4
10
?
131242.59
4
calcultor nd F FV BGN
PMT
P Y
Y
C Y
N yrs
PV
PV
option
o
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?
QUESTION 16
Pr
3081.86 1119.21
4201.07
2
payment Interest due incipal repaid
option
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QUESTION 17
Pr
4201.07 1617.22
2583.85
3
incipal repaid Payment Interest due
option
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QUESTION 18
sin
1200 20 12007500 0.12
0.12 0.12
160917.74 200000
1060917.74
4
Increa g annuity
Q nQS ni R
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S n
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QUESTION 19
3 4
100000 1/ 9.71% / 4
4 ?
146779.42
0.971146779.42 1 195732.99
4
12000
/ 4
/ 4
1/ 9.71
3
164869.85
360602.85
3
PV Y C Y
N FV
FV
PMT
P Y
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14 days3 months28 days
QUESTION 22
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1600000
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QUESTION 24
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QUESTION 28
34 x 2 = 68 Maturity date
8/10/2047
183 days
132 days
Next coupon date
29/05/20138/04/2013
Prev coupo Settlement
8/10/2013
H
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QUESTION 7
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QUESTION 12
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QUESTION 16
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QUESTION 17
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QUESTION 18
Maturity date
14/12/2036
182 days
28 days
Next coupon date
17/05/201314/12/2012
Prev coupo Settlement
8/10/2013
H
14 /12 / 2036 14 / 06 / 2013
23
Years �
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QUESTION 20
sin
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20
360 20 3603600 0.10
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412380 72000
340380
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QUESTION 21
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QUESTION 22
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4
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2
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QUESTION 24
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QUESTION 25
int
2080.54 2014.27
66.27
3
Monthly payment principal repaid
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option
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QUESTION 26
Nov
End of May Aug 15000 Oct 15 000 Dec 15 0002 months
4 months
415000 1 0.10 15500
12
215000 1 0.10 15250
12
15000
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QUESTION 27
0.90828
4
calculator r
option
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QUESTION 28
Nov 10 months 28 months 32 months
3 000 18 months 4 months25 000
28 18 46 6 6
12 12 12
14 9 2
9 2
14
0.1475 0.1475 0.14751 3000 1 25000 1
6 6 6
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QUESTION 29
1
1
8.04
Positive NPV
OPTION
PI greater than
IRR greater than
QUESTION 30
17287.42 2310.26
40497.68
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