Antwoorde / Answers 1
5
1.1
P = R 1 200, i = ─── , n = 2 jaar / years
100
A = P(1 + in)
= R 1 200 (1 + 0,05 X 2)
= R 1 320
I = A ─ P
= R( 1 320 ─ 1 200)
= R 120
a = R 1 320
b = R 120
[ 1 ]
1.5
A = R 149 860, i = 0,18, n = 1,5 jaar / years
A 149 860
P = ───── = ──────────
1 + in 1 + 0,18 X 1,5
= R 118 000
I = A ─ P = R 149 860 ─ 118 000
= R 31 860
m = R 118 000
n = R 31 860
[ 1 ]
1.9
P = R 28 650, A = R 50 710,50,
n = 5,5 jaar / years
A = P(1 + in)
R 50 710,50 = R 28 650 (1 + i X 5,5)
R 50 710,50 = R 28 650 + R 157 5,5i
R 22 060,50 = R 157 575i
0,14 = i
r = 100 X i = 100 X 0,14
= 14 %
I = A ─ P = R 50 710,50 ─ 28 650
= R 22 060,50
w = 14 %
x = R 22 060,50
[ 1 ]
1.13
P = R 850 , i = 0,114 , A = R 1 189,15
A = P(1 + in) = P + Pin
A ─ P 1189,15 ─ 850
n = ────── = ───────────
Pi 850 X 0,114
= 3,5 jaar / years
I = A ─ P = R 1 189,15 ─ 850
= R 339,15
ee = 3,5 jaar / years
ff = R 339,15
[ 1 ]
1.14
gg = 2,25 jaar / years en / and
hh = R 1 350,35
[ 1 ]
1.16
pp = 8,75 jaar / years en / and
qq = R 18 843,75
[ 1 ]
1.2
c = R 28 6800 en / and d = R 6 300
[ 1 ]
1.3
e = R 31 099,75 en / and f = R 7 449,75
[ 1 ]
1.4
g = R 156 800 en / and h = R 76 800
[ 1 ]
1.6
p = R 655 840 en / and q = R 81 980
[ 1 ]
1.7
r = R 2 300 en / and s = R 368
[ 1 ]
1.8
t = R 16 000 en / and v = R 2 496
[ 1 ]
1.10
y = 8,6 % en / and z = R 74 304
[ 1 ]
1.11
aa = 12 % en / and bb = R 954 000
[ 1 ]
1.12
cc = 10,5 % en / and dd = R 1 161 342
[ 1 ]
OF / OR
I = A ─ P = R 1 189,15 ─ 850
= R 339,15
I = Pin
I 339,15
n = ─── = ────────────
Pi 850 X 339,15
= 3,5 jaar / years
ee = 3,5 jaar / years
ff = R 339,15
[ 1 ]
1.15
mm = 6 jaar / years en / and
nn = R 13 486,32
[ 1 ]
Antwoorde / Answers 2
6
2.1
P = R 250, r = 6 % , i = ───── = 0,06
100 × 1
tyd / time = 3 jaar / years; n = 3 X 1 = 3
A = P(1 + i)
n = R 250(1 + 0,06)
3
= R 297,75
a = 3 en / and b = R 297,75
[ 1 ]
2.6
A = R 26 702,71 , r = 15 %
15
i = ────── = 0,075
100 × 2
n = 2,5 × 2 = 5
A 26 702,71
P = ────── = ──────────
(1 + i)
n (1 + 0,075)
5
= R 18 600,00
p = R 18 600,00 en / and q = 5
[ 2 ]
2.10
[ 2 ]
2.2
c = 1 × 2 = 2 en / and d = R 1 323
[ 2 ]
2.3
e = 20 en / and f = R 19 985,99
[ 2 ]
2.4
g = 1,5 en / and h = R 24 811,08
[ 2 ]
2.5
m = 3 en / and n = R 178 130,70
[ 2 ]
2.7
r = R 125 800 en / and s = 13
[ 2 ]
2.8
t = R 640 820 en / and u = 15
[ 2 ]
2.9
v = R 8 520 en / and w = 2,5 jaar / years
[ 2 ]
2.11
aa = 6,4 % en / and bb = 24
[ 2 ]
2.12
cc = 112,5 % en / and
dd = 2,5 jaar / years
[ 2 ]
2.13
ee = 15,5 % en / and
ff = 1,25 jaar / years
[ 2 ]
P = R 116 850, n = 5 × 1 = 5 ⋯ inflasie word jaarlks saamgestel / inflation is compounded yearly
7,4
i = ────── = 0,074
100 × 1
A = P(1 + i)
n = R 116 850(1 + 0,074)
5 = R 116 974,49
Die kosprys van die masjien sal R 116 974 beloop ( korrek tot die naaste Rand) /
The cost price of the machine will be R 116 974 ( correct to the nearest Rand)
[ 4 ]
5. P = R 1 245; n = 3; i = 0,052
A = R 1245(1 + 0,052)
3 = R 1 494,49
Die DVD speler sal R 1 494,49 kos /
The DVD player will cost R 1 494,49
[ 5 ]
7.
Reguitlyn depresiasie /
Straight line depreciation
Gebruik / Use A = P(1 ─ in)
P = R 120 000; n = 5 jaar / years ;
i = 0,15
A = P(1 ─ in) = R 120 000(1 ─ 0,15 × 5)
= R 30 000
Die waarde van die motor is R30 000 /
The value of the car is R 30 000
[ 7 ]
9.
Verminderde balans metode /
Reducing-balance method
Gebruik / Use A = P(1 ─ i)
n
P = R 84 500; n = 3 jaar / years ;
i = 0,18
A = P(1 ─ i)
n = R 84 00(1 ─ 0,18)
3
= R 46 590,60
Die waarde is R46 590,60 /
The value is R 46 590,60
[ 9 ]
11.1
P = R 530 650; i = 0,08 ;
n = 5 jaar / years
A = R 530 650(1 ─ 0,08)
5
= R 349 741,61
[ 11 ]
11.3
A = R 744 650; i = 0,018 ; n = 20
A 744264
P = ────── = R ──────────
(1 + i)
n (1 + 0,018)
20
= R 520 920
[ 11 ]
12.
Gestel die huidige waarde is R P.
Die waarde na n jaar sal R 0,25P wees
en dus is A = R 0,25P en i = 0,12
0,25P = P(1 ─ 0,12)
n
0,25 = (1 ─ 0,12)
n OF 0,25 = 0,88
n
log 0,25 = n log 0,88
log 0,25
n = ──────
log 0,88
= 10,84
Na 10,84 jaar sal die waarde van
die masjien 'n kwart van sy
oorspronklike waarde wees.
[ 12 ]
13.
Gestel die waarde van die masjien
is Rx, dan sal die verminderde
2x
waarde R ─── wees en i = 0,12
3
2x
2
─── = x(1 ─ 0,12)
n OF
─── = 0,88
n
3
3
2
log ─── = n log 0,88
3
n = 3,17
Dit sal 3,17 jaar neem.
[ 12 ]
6. P = 650; n = 3; i = 0,021
A = 650(1 + 0,021)
3 = 691
Daar sal 691 voëls wees /
There will be 691 birds.
[ 6 ]
8.
Reguitlyn depresiasie /
Straight line depreciation
Gebruik / Use A = P(1 ─ in)
P = R 450 000; n = 4 jaar / years ;
i = 0,12
A = P(1 ─ in) = R 450 000(1 ─ 0,12 × 4)
= R 269 862,91
Die waarde van die trekker is R269 862,91 /
The value of the tractor is R 269 862,91
[ 8 ]
10.
Verminderde balans metode /
Reducing-balance method
Gebruik / Use A = P(1 ─ i)
n
P = R 215 700; n = 5 jaar / years ;
i = 0,112
A = P(1 ─ i)
n = R 215 700(1 ─ 0,112)
5
= R 119 100,88
Die waarde is R 119 100,88 /
The value is R 119 100,88
[ 10 ]
11.2
P = R 530 650; i = 0,07 ;
n = 5 jaar / years
A = R 530 650(1 + 0,07)
5
= R 744 264,08
= R 744 264 (tot naaste Rand)
= R 744 264 (to nearest Rand)
[ 11 ]
12.
Let the present value be R P.
After n years the value will be R 0,25P
and thus A = R 0,25P and i = 0,12
0,25P = P(1 ─ 0,12)
n
0,25 = (1 ─ 0,12)
n OF 0,25 = 0,88
n
log 0,25 = n log 0,88
log 0,25
n = ──────
log 0,88
= 10,84
After 10,84 years the machine will
have a value of one quarter of its
original value.
[ 12 ]
13.
Let the present value of the machine
be R x, then its diminished value
2x
will be R ─── and i = 0,12
3
2x
2
─── = x(1 ─ 0,12)
n OR
─── = 0,88
n
3
3
2
log ─── = n log 0,88
3
n = 3,17
It will take 3,17 years
[ 13 ]
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J0
J1
J2
J3
J4
J5
├──────────┼──────────┼──────────┼──────────┼──────────┤
40 000
├───────── 18 % ────────┼─────────────── 12 % ─────────────┤
maandeliks / monthly
kwartaalliks / quarterly
18
i (eerste 2 jaar / first two years) = ──────── = 0,015 en / and n = 2 × 12 = 24
100 × 12
12
i (laaste 3 jaar / final 3 years) = ──────── = 0,03 en / and n = 3 × 4 = 12
100 × 4
A = P(1 + i)
n
A (na 2 jaar / after 2 years) = R 40 000(1 + 0,015)
24
= R 57 180,11248
Hierdie bedrag form nou die kapitaal vir die laaste 3 jaar. /
This amount now becomes the principal for the last 3 years.
A (na laaste 3 jaar / after last 3 years) = R 57 180,11248(1 + 0,03)
12
= R 81 525,16788
Finale waarde / Final value = R81 525,17
OF / OR
Jy kan dit ook as een bewerking doen. Oppas vir die akies!!! / You can also do it as one calculation. Beware of the brackets!!!
A (finale waarde / final value) = R (40 000(1 + 0,015)
24)(1 + 0,03)
12
= R 81 525,17
Vraag / Question 14
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J0
J1
J2
J3
├──────────┼──────────┼──────────┤
P
23 682,66
├──── 8 % ───┼───────── 10 % ────────┤
kwartaalliks
half-jaarliks
quarterly
semi-annually
8
i (eerste jaar / first year) = ──────── = 0,02 en / and n = 1 × 4 = 4
100 × 4
10
i (laaste 2 jaar / final 2 years) = ──────── = 0,05 en / and n = 2 × 2 = 4
100 × 2
A = P(1 + i)
n
23 682,66 = (P(1 + 0,02)
4)(1 + 0,05)
4
23 682,66
P = ────────── = 18 000,00
1,02
4 × 1,05
4
Koos het R18 000 belê. /
Koos invested R18 000.
Vraag / Question 15
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J0
J1
J2
J3
J4
J5
├──────────
┼──────────
┼──────────
┼──────────
┼──────────
┤
30 000 ─ 8 000
─ 6 000
├───────────────────────── 18 % ───────────────────────────┤
maandeliks
monthly
18
i = ──────── = 0,015
100 × 12
A (na 2 jaar / after 2 years) = R 30 000(1 + 0,015)
24 = R 42 885,08436
Hy onttrek R8 000 sodat die nuwe kapitaal = R34 885,08436 / He withdraws R8 000 so that the new principal = R34 885,08436
A (na 3 jaar / after 3 years) = R 34 885,08436(1 + 0,015)
12 = R 41 709,24077
Hy onttrek nou R6 000 sodat die nuwe kapitaal = R35 709,24077 / He now withdraws R6 000 so that the new principal = R35 709,24077
A (finaal / final) = R 35 709,24077(1 + 0,015)
24 = R 51 046,46
Final waarde = R51 046,46 / Final value = R51 046,46
OF / OR
Bereken die waarde van die R30 000 belegging asof geen onttrekking plaasgevind het nie. Die R8 000 het nou vir 3 jaar rente verdien
wat afgetrek moet word. So ook die R6 000 wat vir 2 jaar rente verdien het. Doen dan die berekening in een stap.
Calculate the value of the R30 000 investment as if no withdrawals took place. The R8 000 has now received interest for 3 years and has
to be subtracted. The R6 000 is treated in the same way. Then do the calculation in one step.
A = R 30 000(1 + 0,015)
60 ─ 8 000(1 + 0,15)
36 ─ 6 000(1 + 0,15)
24
= R 51 046,46
Vraag / Question 16
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
M0
M4
M11
M24
├──────────┼──────────┼──────────┤
130 000 ─ 20 000 ─ 15 000
├────────────── 6 % ───────────────┤
maandeliks
monthly
6
i = ──────── = 0,005
100 × 12
A (na 4 maande / after 4 months) = R 130 000(1 + 0,005)
4 = R 132 619,5651
Hy onttrek R20 000 sodat die nuwe kapitaal = R112 619,5651 / He withdraws R20 000 so that the new principal = R112 619,5651
A (na 11 maande / after 11 months) = R 112 619,5651(1 + 0,015)
7 = R 116 620,8703
Hy onttrek nou R15 000 sodat die nuwe kapitaal = R101 620,8703 / He now withdraws R15 000 so that the new principal = R101 620,8703
A (finaal / final) = R 101 620,8703(1 + 0,015)
13 = R 108 428,07
Final waarde = R108 428,07 / Final value = R108 428,07
OF / OR
Bereken die waarde van die R130 000 belegging asof geen onttrekking plaasgevind het nie. Die R20 000 het nou vir 20 maande rente verdien
wat afgetrek moet word. So ook die R15 000 wat vir 13 maande rente verdien het. Doen dan die berekening in een stap.
Calculate the value of the R130 000 investment as if no withdrawals took place. The R20 000 has now received interest for 20 months and it has
to be subtracted. The R15 000 is treated in the same way. Then do the calculation in one step.
A = R 130 000(1 + 0,005)
24 ─ 20 000(1 + 0,005)
20 ─ 15 000(1 + 0,005)
13
= R 108 428,07
Vraag / Question 17
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J0
J1
J2
J3
├─────────┼────────┼─────────┼────────┼─────────┼────────┤
82 000
+ 8 000
+ 12 000
├─────────────────────────
8 % ───────────────────────────┤
half jaarliks
semi-annually
8
i = ──────── = 0,04
100 × 2
A (na 6 maande / after 6 months) = R 82 000(1 + 0,04)
1 = R 85 280
R8 000 word bygevoeg sodat die nuwe kapitaal = R93 280 / R8 000 is added so that the new principal = R93 280
A (na 18 maande / after 18 months) = R 93 280(1 + 0,04)
2 = R 100 891,648
'n Verdere R12 000 word belê sodat die P(nuut) = R112 891,648 / Another R12 000 is invested so that P(new) = R112 891,648
A (finaal / final) = R 112 891,648(1 + 0,04)
3 = R 126 987,75
Final waarde = R126 987,75 / Final value = R126 987,75
OF / OR
Bereken die waarde van die R82 000 belegging asof geen ander transaksie plaasgevind het nie. Die R8 000 het nou vir 2,5 jaar
of 5 periodes, rente verdien wat nou bygetel moet word. So ook die R12 000 wat vir 1,5, of 3 periodes, jaar rente verdien het.
Doen dan die berekening in een stap.
Calculate the value of the R82 000 investment as if no further transactions took place. The R8 000 has now received interest for
2,5 years, or 5 periods, and it has to be added. The R12 000 is treated in the same way, 3 periods.
Then do the calculation in one step.
A = R 82 000(1 + 0,04)
6 + 8 000(1 + 0,04)
5 + 12 000(1 + 0,04)
3
= R 126 987,75
Vraag / Question 18
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J/Y0
J/Y1
J/Y2
K/Q
0
1
2
3
4
5
6
7
8
├────────┼─────────┼────────┼────────┼────────┼────────┼────────┼─────────┤
65 000
+ 8 000
─ 12 000
├──────────────────────────────────
8 % ────────────────────────────────────┤
kwartaallliks
quarterly
8
i = ──────── = 0,02
100 × 4
A (na 4 maande / after 4 months) = R 65 000(1 + 0,02)
1 = R 66 300
R5 000 word bygevoeg sodat die nuwe kapitaal = R71 300 / R8 000 is added so that the new principal = R71 300
A (na 20 maande / after 20 months) = R 71 300(1 + 0,02)
4 = R 77 177,41301
R12 000 word onttrek sodat die P(nuut) = R65 177,41301 / R12 000 is withdrawn so that P(new) = R65 177,41301
A (finaal / final) = R 65 177,41301(1 + 0,02)
3 = R 69 166,79
Final waarde = R69 166,79 / Final value = R69 166,79
OF / OR
Bereken die waarde van die R50 000 belegging asof geen ander transaksie plaasgevind het nie. Die R5 000 het nou vir 7 kwartale
rente verdien wat nou bygetel moet word. So ook die R12 000 wat vir 3 kwartale, jaar rente verdien het.
Doen dan die berekening in een stap.
Calculate the value of the R65 000 investment as if no further transactions took place. The R8 000 has now received interest for
7 quarters, and it has to be added. The R12 000 is treated in the same way, 3 quarters.
Then do the calculation in one step.
A = R 65 000(1 + 0,02)
8 + 5 000(1 + 0,02)
7 ─ 12 000(1 + 0,02)
3
= R 69 166,79
Vraag / Question 19
Teken 'n tydlyn om jou te help. / Draw a time line to assist you.
J / Y
0
1
2
3
4
5
├───────────┼──────────┼───────────┼──────────┼───────────┤
55 600
+ 12 000 ─ 20 000
+ 8 300
├───────── 15 % ────────┼─────────────────
10 % ─────────────┤
maandeliks monthly kwartaalliks
quarterly
15
10
i (1ste 2 jaar / first 2 years) = ──────── = 0,0125
i (laasste 3 jaar / last 3 years) = ──────── = 0,025
100 × 12
100 × 12
A (na 12 maande / after 12 months) = R 55 600(1 + 0,0125)
12 = R 64 537,95119
R12 000 word bygevoeg sodat die nuwe kapitaal = R76 537,95119 / R12 000 is added so that the new principal = 76 537,95119
A (na 24 maande / after 24 months) = R 76 537,95119(1 + 0,04)
12 = R 88 841,77262
R20 000 word onttrek sodat die P(nuut) = R68 841,77262 / R20 000 is withdrawn so that P(new) = R68 841,77262
A (na 4 jaar / after 4 years) = R 68 841,77262(1 + 0,025)
8 = R 83 877,01522
R8 300 word belê sodat die P(nuut) = R92 177,01522 / A further R8 300 is invested so that P(new) = R92 177,01522
A (finaal / final) = R R92 177,01522(1 + 0,025)
4 = R 101 746,18
Final waarde = R101 746,18 / Final value = R101 746,18
OF / OR
Aangesien daar 'n rentekoers verandering is, is dit makliker om die bedrae vir die een koers by te tel by die bedrae van die tweede koers.
Bereken die waarde van die R55 600 belegging asof geen ander transaksie vir 2 jaar plaasgevind het nie. Die R12 000 het nou vir 1 jaar
of 12 periodes, rente verdien wat nou bygetel moet word. Die R20 000 word nou onttrek en die nuwe kapitaal verdien rente vir 3 jaar, of 12
kwartale, wat bygetel moet word. Die bykomende R8 300 verdien rente vir 1 jaar, of 4 kwartale. Doen dan die berekening in een stap.
Seeing that there is a change in interest rates, it is easier to add the amounts of the first interest rate to the amounts of the second
interest rate.
Calculate the value of the R55 600 investment as if for 2 years, 24 months, no further transactions took place. The R12 000 has now
received interest for 1 year, or 12 months, and it has to be added. The R20 000 is subtracted to form a new principal, which earns
interest for 3 years, 12 quarters, which has to be added. The additional R8 300 earns interest for 1 year or 4 quarters, and it must also
be added. Now do the calculation in one step.
A = R (55 600(1 + 0,0125)
24 + 12 000(1 + 0,0125)
12 ─ 20 000)(1 + 0,025)
12 + 8 300(1 + 0,025)
4
= R 101 746,18
Vraag / Question 20
Antwoord / Answer 24
6,25
i (nom) = ─────── = 0,015625
100 × 4
1 + i
eff = (1 + 0,015625)
4
The effective interest rate = 6,40 %
Vraag / Question 24
Antwoord / Answer 25
12
i (nom) = ─────── = 0,06
100 × 2
1 + i
eff = (1 + 0,06)
2
i
eff = 1,1236 ─ 1
= 0,1236
Die effektiewe rentekoers / = 12,36 %
The effective interest rate = 612,36 %
Vraag / Question 25
Antwoord / Answer 26
A = P(1 ─ i)
n
40 000 = 160 000(1 ─ i)
5
40 000
(1 ─ i)
5 = ──────── = 0,25
160 000
(1 ─ i) = 1,2421417167
i = 24,21 %
Vraag / Question 26