WISKUNDE
GRAAD 10
NOG OEFENINGE
Rente, saamgestelde groei, appresiasie,
depresiasie, inflasie : antwoorde.
GRAAD 10
NOG OEFENINGE
Rente, saamgestelde groei, appresiasie,
depresiasie, inflasie : antwoorde.
MATHEMATICS
GRADE 10
MORE EXERCISES
Interest, compound growth, appreciation,
depreciation, inflation : answers.
GRADE 10
MORE EXERCISES
Interest, compound growth, appreciation,
depreciation, inflation : answers.
1.1 I = Prn
6,5
I = R4 500 x ——— x 4
100
= R1 170
A = P + I
= R4 500,00 + 1 170,00
= R5 670,00
Vr. / Qu. 1.
1.5 P = A − I
= R2 497 − 279
= R2 200
100 × I
r = —————
P × t
100 x 279
r = ——————
2 200 x 3
= 4,5%
Vr. / Qu. 1.
1.9 P = A − I
= R68 965,44 − R23 365,44
= R45 600
100I
n = ———
Pr
100 × 23 365,44
= —————————
45 600 × 6,5
= 4,2 jaar / years
Vr. / Qu. 1.
1.13 I = A − P
= R25 755,65 − R18 650,00
= R7 105,65
100I
n = ———
Pr
100 x 7 105,65
n = ————————
18 650 x 15,24
= 3,25 jaar / years
Vr. / Qu. 1.
6,5
I = R4 500 x ——— x 4
100
= R1 170
A = P + I
= R4 500,00 + 1 170,00
= R5 670,00
Vr. / Qu. 1.
1.5 P = A − I
= R2 497 − 279
= R2 200
100 × I
r = —————
P × t
100 x 279
r = ——————
2 200 x 3
= 4,5%
Vr. / Qu. 1.
1.9 P = A − I
= R68 965,44 − R23 365,44
= R45 600
100I
n = ———
Pr
100 × 23 365,44
= —————————
45 600 × 6,5
= 4,2 jaar / years
Vr. / Qu. 1.
1.13 I = A − P
= R25 755,65 − R18 650,00
= R7 105,65
100I
n = ———
Pr
100 x 7 105,65
n = ————————
18 650 x 15,24
= 3,25 jaar / years
Vr. / Qu. 1.
1.2 c = R3 854 d = R22 654 Vr. / Qu. 1.
1.3 e = R618,75 f = R6 118,75 Vr. / Qu. 1.
1.4 g = R55 215 h = R300 615 Vr. / Qu. 1.
1.6 p = R15 400 q = 5,7% Vr. / Qu. 1.
1.7 r = R26 700 s = 7,4% Vr. / Qu. 1.
1.8 t = R112 000 u = 8,2% Vr. / Qu. 1.
1.10 y = R55 600;
z = 2,5 jaar / years Vr. / Qu. 1.
1.11 aa = R112 650;
bb = 4 jaar / years Vr. / Qu. 1.
1.12 cc = R250 850;
bb = 3,25 jaar / years Vr. / Qu. 1.
1.14 mm = 3 jaar / years;
nn = R24 057,60 Vr. / Qu. 1.
1.15 pp = 8,5 jaar / years;
qq = R98 489,00 Vr. / Qu. 1.
1.16 ss = R98 489;
rr = 3,5 jaar / years Vr. / Qu. 1.
1.3 e = R618,75 f = R6 118,75 Vr. / Qu. 1.
1.4 g = R55 215 h = R300 615 Vr. / Qu. 1.
1.6 p = R15 400 q = 5,7% Vr. / Qu. 1.
1.7 r = R26 700 s = 7,4% Vr. / Qu. 1.
1.8 t = R112 000 u = 8,2% Vr. / Qu. 1.
1.10 y = R55 600;
z = 2,5 jaar / years Vr. / Qu. 1.
1.11 aa = R112 650;
bb = 4 jaar / years Vr. / Qu. 1.
1.12 cc = R250 850;
bb = 3,25 jaar / years Vr. / Qu. 1.
1.14 mm = 3 jaar / years;
nn = R24 057,60 Vr. / Qu. 1.
1.15 pp = 8,5 jaar / years;
qq = R98 489,00 Vr. / Qu. 1.
1.16 ss = R98 489;
rr = 3,5 jaar / years Vr. / Qu. 1.
Let op : Please note
A = P(1 + i)n waar / where
A = die eindbedrag / final amount
P = hoofsom of oorspronkilike bedrag belê /
principal or capital or sum initially invested
i = rentekoers per periode uitgedruk as 'n desimale breuk /
interest rate per period expressed as a dcimal fraction
n = aantal rentedraende periodes /
number of interest-bearing periods
m = aantal rentedraende periodes per jaar /
number of interest-bearing periods per year
As die rente nie jaarliks saamgestel word nie
maar
kwartaalliks deel i dan deur 4 en
; vermenigvuldig termyn (aantal jare) met 4
halfjaarliks deel i dan deur 2 en
vermenigvuldig termyn met 2.
maandeliks deel i dan deur 12 en
vermenigvuldig termyn met 12.
maar
kwartaalliks deel i dan deur 4 en
; vermenigvuldig termyn (aantal jare) met 4
halfjaarliks deel i dan deur 2 en
vermenigvuldig termyn met 2.
maandeliks deel i dan deur 12 en
vermenigvuldig termyn met 12.
If interest is not compounded yearly but
quarterly divide i by 4 and multiply
number of years by 4.
half-yearly or semi-annually divide i by 2
and multiply number of years by 2.
monthly divide i by 12 and multiply
number of years by 12.
quarterly divide i by 4 and multiply
number of years by 4.
half-yearly or semi-annually divide i by 2
and multiply number of years by 2.
monthly divide i by 12 and multiply
number of years by 12.
2.1 P = R12 500; r = 7,8%; n = 5 jaar / years; m = 1
r
i = —————
100 × m
7,8
i = —————
100 × 1
= 0,078
A = P(1 + i)n = R12 500 (1 + 0,078)5
= R18 197,17 Vr. / Qu. 2.
2.5 A = R80 374,54; r = 7,2%;
n = 36 periodes / periods; m = 12
n = 36; m = 12 sodat / so that aantal jare /
number of years = 3
7,2
i = ————— = 0,006
100 × 12
A
A = P (1 + i)n : P = —————
(1 + i)n
80 371,54
P = ——————— = 64 800
(1,006)36
2.9 P = R350 000; A = R464 894,71;
n = 36 periodes / periods; m = 12
n = 36; m = 12 sodat / so that
x = 3 jaar / years
Vr. / Qu. 2.
r
i = —————
100 × m
7,8
i = —————
100 × 1
= 0,078
A = P(1 + i)n = R12 500 (1 + 0,078)5
= R18 197,17 Vr. / Qu. 2.
2.5 A = R80 374,54; r = 7,2%;
n = 36 periodes / periods; m = 12
n = 36; m = 12 sodat / so that aantal jare /
number of years = 3
7,2
i = ————— = 0,006
100 × 12
A
A = P (1 + i)n : P = —————
(1 + i)n
80 371,54
P = ——————— = 64 800
(1,006)36
2.9 P = R350 000; A = R464 894,71;
n = 36 periodes / periods; m = 12
n = 36; m = 12 sodat / so that
x = 3 jaar / years
Vr. / Qu. 2.
2.2 c = 5; d = R133 299,53
Vr. / Qu. 2.
2.3 e = 12; f = R319 695,98 Vr. / Qu. 2.
2.4 g = 30; h = R100 409,83 Vr. / Qu. 2.
2.6 p = R184 400; q = 4 Vr. / Qu. 2.
2.7 r = R121 000; s = 11 Vr. / Qu. 2.
2.8 t = R546 000; u = 2,5 Vr. / Qu. 2.
2.10 y = 7,5% z = 5 Vr. / Qu. 2.
2.11 aa = 8,2% bb = 12 Vr. / Qu. 2.
2.12 cc = 12,2% dd = 10 Vr. / Qu. 2.
2.13 ee = 7,6% ff = 3 Vr. / Qu. 2.
2.3 e = 12; f = R319 695,98 Vr. / Qu. 2.
2.4 g = 30; h = R100 409,83 Vr. / Qu. 2.
2.6 p = R184 400; q = 4 Vr. / Qu. 2.
2.7 r = R121 000; s = 11 Vr. / Qu. 2.
2.8 t = R546 000; u = 2,5 Vr. / Qu. 2.
2.10 y = 7,5% z = 5 Vr. / Qu. 2.
2.11 aa = 8,2% bb = 12 Vr. / Qu. 2.
2.12 cc = 12,2% dd = 10 Vr. / Qu. 2.
2.13 ee = 7,6% ff = 3 Vr. / Qu. 2.
Moontlikheid / Possibility 1 : r = 7,5% m = 2
r
i = —————
100 × m
7,5
i = ————— = 0,0375
100 × 2
n = 3 × 2 = 6
A = P(1 + i)n
= R20 000 (1 + 0,0375)6
= R24 943,57
r
i = —————
100 × m
7,5
i = ————— = 0,0375
100 × 2
n = 3 × 2 = 6
A = P(1 + i)n
= R20 000 (1 + 0,0375)6
= R24 943,57
Moontlikheid / Possibility 2 : r = 8% m = 1
r
i = —————
100 × m
8
i = ————— = 0,08
100 × 1
n = 3 × 1 = 3
A = P(1 + i)n
= R20 000 (1 + 0,08)3
= R25 194,24
r
i = —————
100 × m
8
i = ————— = 0,08
100 × 1
n = 3 × 1 = 3
A = P(1 + i)n
= R20 000 (1 + 0,08)3
= R25 194,24
Moontlikheid 2 is dus die beste omdat
dit meer rente verdien.
dit meer rente verdien.
Possibilty 2 is the better option because
it earns the most interest.
it earns the most interest.
7,5
i = ————— = 0,075
100 × 1
Waarde / Value = P(1 − i)n
= R885 000 (1 − 0,075)5
= R599 310,57
12
i = ————— = 0,12
100 × 1
Waarde / Value = P(1 − i)n
= R275 560 (1 − 0,12)5
= R145 469,30
6,5
i = ————— = 0,075
100 × 1
Waarde / Value = P(1 + i)n
= 3 105 560 (1 + 0,065)4
= 3 995 198
6,8
i = ————— = 0,068
100 × 1
Waarde / Value = P(1 + i)n
= R45 800 (1 + 0,068)4
= R59 586,86
8,2
i = ————— = 0,006833...
100 × 12
Waarde / Value = P(1 + i)n
= R33 000 (1 + 0,006833...)36
= R42 168,39
i = ————— = 0,006833...
100 × 12
Waarde / Value = P(1 + i)n
= R33 000 (1 + 0,006833...)36
= R42 168,39
8,2
i = ————— = 0,082
100 × 1
Waarde / Value = P(1 + i)n
= R48 350 (1 + 0,82)3
= R61 246,07
8,2
9.1 I = Prn = R1 200 × ——— × 3
100
= R270
A = R(1 200 + 270) = R1 470
Saldo / Amount = R1 470
6,85
9.3 i = —————
100 × 1
= 0,0685
Koste oor 3 jaar is R1 463,88
9.1 I = Prn = R1 200 × ——— × 3
100
= R270
A = R(1 200 + 270) = R1 470
Saldo / Amount = R1 470
6,85
9.3 i = —————
100 × 1
= 0,0685
Koste oor 3 jaar is R1 463,88
9.2 Deposito / Deposit = 10% van / of R1 150
= R115
Koste / Cost = deposito / deposit + 24 paaiemente / instalments
= R115 + 24 × R55
= R1 435
Waarde / Value = R1 200 (1 + 0,0685)3
= R1 463,88
The player will cost R1 463,88 3 years from now.
11.1 Persentasie deposito / Percenage of deposit
deposit(o) 100
= ———————— × ——— %
koopsom / price 1
675 100
= ———— × ——— %
4 500 1
= 15 %
11.3 Totale bedrag betaal / Total amount paid
= deposit(o) + som van paaiemente /
sum of instalments
= R(675 + 24 × 199,22) = R5 456,25
Bedrag ekstra betaal / Amount paid extra
= R(5 45625 − 4 500)
= R956,25
deposit(o) 100
= ———————— × ——— %
koopsom / price 1
675 100
= ———— × ——— %
4 500 1
= 15 %
11.3 Totale bedrag betaal / Total amount paid
= deposit(o) + som van paaiemente /
sum of instalments
= R(675 + 24 × 199,22) = R5 456,25
Bedrag ekstra betaal / Amount paid extra
= R(5 45625 − 4 500)
= R956,25
11.2 Uitstaande bedrag / Amount outstanding
= R(4 500 − 675) = R3 825
12,5
i = ———— = 0,125
100 × 1
Bedrag verskuldig / Amount due
= P(1 + in) = R3 825(1 + 0,125 × 2)
= R4 781,25
Paaiement / Instalment
Bedrag verskuldig / Amount due
= ————————————————————
Aantal paaiemente / Number of instalments
4 781,25
= R ————— = R199,22
24
= R(4 500 − 675) = R3 825
12,5
i = ———— = 0,125
100 × 1
Bedrag verskuldig / Amount due
= P(1 + in) = R3 825(1 + 0,125 × 2)
= R4 781,25
Paaiement / Instalment
Bedrag verskuldig / Amount due
= ————————————————————
Aantal paaiemente / Number of instalments
4 781,25
= R ————— = R199,22
24
12.1 Deposito / Deposit = 10% van koopsom /
of purchase-price
10 2 490
= R ——— × ————
100 1
= R249
of purchase-price
10 2 490
= R ——— × ————
100 1
= R249
12.2 Totale uitstaande bedrag / Total amount due
= Aantal paaiemente / Number of instalments ×
bedrag / amount
= R 30 × 137 = R4 110
Uitstaande bedrag voor rente /
Amount due before interest
= Koopprys / Purchase price − deposit(o)
= R2 490 − R249 = R2 241
Rente gehef / Interest imposed = A − P
= R4 110 − 2 241
= R1 869
100 i 100 × 1 869
i = Prn : r = ——— = —————— %
Pn 2 241 × 2,5
= 33,36 %
= Aantal paaiemente / Number of instalments ×
bedrag / amount
= R 30 × 137 = R4 110
Uitstaande bedrag voor rente /
Amount due before interest
= Koopprys / Purchase price − deposit(o)
= R2 490 − R249 = R2 241
Rente gehef / Interest imposed = A − P
= R4 110 − 2 241
= R1 869
100 i 100 × 1 869
i = Prn : r = ——— = —————— %
Pn 2 241 × 2,5
= 33,36 %
12.3 Ekstra koste / Extra cost
= Totale bedrag betaal / Total amount
paid − kontantprys / cash price
= Deposit(o) + som van paaiemente / sum
of instalments − R2 490
= R(249 + 30 × 137) − R2 490
= R1 869
= Totale bedrag betaal / Total amount
paid − kontantprys / cash price
= Deposit(o) + som van paaiemente / sum
of instalments − R2 490
= R(249 + 30 × 137) − R2 490
= R1 869
8,25
12.4 i = ————— = 0,006875
100 × 12
A = R2 490,00
A 2 490
P = ———— = —————————
(1 + i)n (1 + 0,006875)30
= R2 027,37
12.4 i = ————— = 0,006875
100 × 12
A = R2 490,00
A 2 490
P = ———— = —————————
(1 + i)n (1 + 0,006875)30
= R2 027,37
7,25
13.1 i = ————
100 × 1
= 0,0725
A = R 18 504,75
13.1 i = ————
100 × 1
= 0,0725
A = R 18 504,75
5,4
13.2 i = ———— = 0,0045; n = 36; A = R18 504,75
100 × 12
A 18 504,75
P = ———— = ——————— = R15 742,92
(1 + i)n (1 + 0,0045)36
Hy moet meer belê as wat die stel nou kos!!
He must invest more then the cash price!!
13.2 i = ———— = 0,0045; n = 36; A = R18 504,75
100 × 12
A 18 504,75
P = ———— = ——————— = R15 742,92
(1 + i)n (1 + 0,0045)36
Hy moet meer belê as wat die stel nou kos!!
He must invest more then the cash price!!
14.1 Uitstaande bedrag / Amount due
= Koopprys / Purchase-price − deposit(o)
= R6 999 − 700 = R6 299
12
i = ———— = 0,12
100 × 1
Rente / Interest = Pin = R6 299 × 0,12 × 2,5
= R1 889,70
Bedrag plus rente / Amount plus interest
= R8 188,70
Totale bedrag verskuldig / Total amount due
= uitstaande bedrag + rente + versekering /
outstanding amount + interest + insurance
= R 6 999 + 1 889,70 + 30 × 20
= R8 788,70
8 788,70
Paaiement / Instalment = R ————— = R292,96
30
= Koopprys / Purchase-price − deposit(o)
= R6 999 − 700 = R6 299
12
i = ———— = 0,12
100 × 1
Rente / Interest = Pin = R6 299 × 0,12 × 2,5
= R1 889,70
Bedrag plus rente / Amount plus interest
= R8 188,70
Totale bedrag verskuldig / Total amount due
= uitstaande bedrag + rente + versekering /
outstanding amount + interest + insurance
= R 6 999 + 1 889,70 + 30 × 20
= R8 788,70
8 788,70
Paaiement / Instalment = R ————— = R292,96
30
14.2 Totale bedrag betaal / Total amount paid
= R(30 × 292,96 + 700)
= R 9 488,70
14.3 Ekstra betaal / Paid extra
= totale bedrag betaal / total amount paid −
koopsom / purchase-price
= R 9 488,70 − 6 999,00
= R 2 489,70
= R(30 × 292,96 + 700)
= R 9 488,70
14.3 Ekstra betaal / Paid extra
= totale bedrag betaal / total amount paid −
koopsom / purchase-price
= R 9 488,70 − 6 999,00
= R 2 489,70
Maak gebruik van tydlyne in hierdie soort vrae. / Use time-line graphs to help solve problems like these.
Bereken i (eerste 4 jaar) /
Calculate i (first 4 years).
6,5
i = ————— = 0,065
100 × 1
1 + i1 = 1,065
6,5
i = ————— = 0,065
100 × 1
1 + i1 = 1,065
Bereken i (laaste 4 jaar) /
Calculate i (last 4 years)
7,2
i = ————— = 0,072
100 × 1
1 + i1 = 1,072
7,2
i = ————— = 0,072
100 × 1
1 + i1 = 1,072
Trek nou die tydlyn met al die inligting daarop.
No draw the time line containing all the information.
A(na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4
= R13 314,92673 (P vir die volgende berekening. / P for the
next calclation.)
A(na 5 jaar / after 5 years) = P(1 + i)n = R13 314,92673(1 + 0,072)1
= R14 273,60145 (P volgende berekening. / P next calclation.)
Hy onttrek nou R6 300 sodat nuwe / He now withdraws R6 300 so that new
P = A − 6 300 = R14 273,60145 − 6 300 = R7 973,60145
A(finaal / final) = P(1 + i)n = R7 973,60145(1 + 0,072)3
= R9 822,880944
Bedrag in die rekening / Balance in account = R9 822,88
OF / OR
Hierdie is 'n samevatting van die boonste berekening. / This is a condensed version of the calculation above.
A4(A na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4 = P1(volgende P / next P)
A5 = P1(1 + i)n = R(10 350(1 + 0,065)4)(1 + 0,072)4 − 6 300 = P2
Afinaal / final = A8 = P2(1 + i)n = R((10 350(1 + 0,065)4)(1 + 0,072)4 − 6 300)(1 + 0,072)3
Skryf dit nou as een uitdrukking soos hieronder. Let op na die hakies - hulle is belangrik as jy
die uitdrukking op die sakrekenaar intik.
Write it as one expression as below. Make sure that you enter the brackets correctly - they are
important when you type the expression on your calculator.
A(finaal / final) = R(((10 350(1,065)4)(1,072)1 − 6 300)(1,072)3
= R9 822,88
OF / OR
A4(A na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4 = P1(volgende P / next P)
A8(A na volgende 4 jaar / after next 4 years) = P1(1 + i)n = R((10 350(1 + 0,065)4)(1 + 0,072)4)) = P2
Let op : R6 300 word nog nie afgetrek nie, maar bedra rente. / Please note that the R6 300 is not
subtracted yet but it earns interest.
Trek nou die R6 300 plus die rente verdien af om die finale saldo te verkry. /
Subtract the R6 300 plus interest earned to calculate the final balance.
Afinaal / final = R((10 350(1 + 0,065)4)(1 + 0,072)4) − (6 300(1 + 0,072)3)
A(finaal / final) = R((10 350(1,065)4)(1,072)4) − (6 300(1,072)3)
= R9 822,88
Vraag / Question 15
A(na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4
= R13 314,92673 (P vir die volgende berekening. / P for the
next calclation.)
A(na 5 jaar / after 5 years) = P(1 + i)n = R13 314,92673(1 + 0,072)1
= R14 273,60145 (P volgende berekening. / P next calclation.)
Hy onttrek nou R6 300 sodat nuwe / He now withdraws R6 300 so that new
P = A − 6 300 = R14 273,60145 − 6 300 = R7 973,60145
A(finaal / final) = P(1 + i)n = R7 973,60145(1 + 0,072)3
= R9 822,880944
Bedrag in die rekening / Balance in account = R9 822,88
Hierdie is 'n samevatting van die boonste berekening. / This is a condensed version of the calculation above.
A4(A na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4 = P1(volgende P / next P)
A5 = P1(1 + i)n = R(10 350(1 + 0,065)4)(1 + 0,072)4 − 6 300 = P2
Afinaal / final = A8 = P2(1 + i)n = R((10 350(1 + 0,065)4)(1 + 0,072)4 − 6 300)(1 + 0,072)3
Skryf dit nou as een uitdrukking soos hieronder. Let op na die hakies - hulle is belangrik as jy
die uitdrukking op die sakrekenaar intik.
Write it as one expression as below. Make sure that you enter the brackets correctly - they are
important when you type the expression on your calculator.
A(finaal / final) = R(((10 350(1,065)4)(1,072)1 − 6 300)(1,072)3
= R9 822,88
A4(A na 4 jaar / after 4 years) = P(1 + i)n = R10 350(1 + 0,065)4 = P1(volgende P / next P)
A8(A na volgende 4 jaar / after next 4 years) = P1(1 + i)n = R((10 350(1 + 0,065)4)(1 + 0,072)4)) = P2
Let op : R6 300 word nog nie afgetrek nie, maar bedra rente. / Please note that the R6 300 is not
subtracted yet but it earns interest.
Trek nou die R6 300 plus die rente verdien af om die finale saldo te verkry. /
Subtract the R6 300 plus interest earned to calculate the final balance.
Afinaal / final = R((10 350(1 + 0,065)4)(1 + 0,072)4) − (6 300(1 + 0,072)3)
A(finaal / final) = R((10 350(1,065)4)(1,072)4) − (6 300(1,072)3)
= R9 822,88
Maak gebruik van tydlyne in hierdie soort vrae. / Use time-line graphs to help solve problems like these.
Bereken i (eerste 3 jaar) /
Calculate i (first 3 years).
6,2
i = ————— = 0,062
100 × 1
1 + i1 = 1,062
6,2
i = ————— = 0,062
100 × 1
1 + i1 = 1,062
Bereken i (laaste 4 jaar) /
Calculate i (last 4 years)
5,8
i = ————— = 0,058
100 × 1
1 + i1 = 1,058
5,8
i = ————— = 0,058
100 × 1
1 + i1 = 1,058
Trek nou die tydlyn met al die inligting daarop.
No draw the time line containing all the information.
A2 = A(na 2 jaar / after 2 years) = P(1 + i)n = R13 400(1 + 0,062)2 = P2(volgende P / next P)
= R15 113,1096
Hy belê R3 800 sodat / He invests R3 800 so that
P2 = R(15 113,1096 + 3 800) = R18 913,1096
A3 = P2(1 + 0,062)1 = R(15 113,1096 + 3 800)(1,062)1 = R20 085,7224 = P3
A5 = P3(1 + 0,058)2 = R(20 085,7224)(1,058)2 = R22 483,23456
Hy belê nog R2 500 sodat / He invests another R2 500 so that
P4 = R(22 483,23456 + 2 500) = R24 983,23456
Afinaal / final = P4(1 + 0,058)2 = R(24 983,23456(1,058)2) = R27 965,33
OF / OR
Kapitaal / Principal (P0) = R13 400; i(eerste 3 jr / first 3 years) = i1 = 0,062 en / and i2 = 0,058
A2 = P0(1 + i)n = R13 400(1 + 0,062)2 = P2(P vir volgende berekening / for next calculation)
= R13 400(1,062)2 = P2
Hy belê R3 800 sodat / He invests R3 800 so that
P2 = R(13 400(1,062)2 + 3 800)
A3 = P2(1 + i)n = R(13 400(1,062)2 + 3 800) × 1,062 = P5
A3 i verander na / changes to 0,058
A5 = P5(1 + i)n = R((13 400(1,062)2 + 3 800) × 1,062) × 1,0582 = P7
Hy belê R2 500 sodat / He invests R2 500 so that
P7 = R(((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500
A7 = P7(1 + i)n = R((((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500) × 1,0582
= R27 965,33
Vir jou antwoord tik jy onderstaande uitdrukking in - maak net baie seker van al die hakies!!
As your answer type in the expression below - just make very sure about the brackets!!
Afinaal / final = R((((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500) × 1,0582
= R27 965,33
OF / OR
Bereken die rene op die R13 400 vir 3 jaar. / Calculate the interest on the R13 400 investment for 3 years.
A1 = P0(1 + i)n = R13 400(1 + 0,062)3 = P2(P vir volgende berekening / for next calculation)
Bereken nou die rene op die bykomende R3 800 vir 1 jaar en tel dit by die eerste bedrag. /
Now calculate the interest on the additional R3 800 investment for 1 year and add that to the first amount.
A2 = R(13 400 × 1,0623) + (3 800 × 1,062)
Bereken nou die rene op hierdie bedrag vir die laaste 4 jaar teen die nuwe rentekoers. /
Now calculate the interest on this amount for the last 4 years at the new interest rate.
A3 = R((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584
Bereken nou die rente op die bykomende belegging van R2 500 vir 2 jaar en tel dit by. /
Now calculate the interest on the additional amount of R2 500 for the last 2 years and add it.
A3 = R(((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584) + (2 500 × 1,0582)
Vir jou antwoord tik jy onderstaande uitdrukking in - maak net baie seker van al die hakies!!
As your answer type in the expression below - just make very sure about the brackets!!
Afinaal / final = R(((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584) + (2 500 × 1,0582)
= R27 965,33
Vraag / Question 16
A2 = A(na 2 jaar / after 2 years) = P(1 + i)n = R13 400(1 + 0,062)2 = P2(volgende P / next P)
= R15 113,1096
Hy belê R3 800 sodat / He invests R3 800 so that
P2 = R(15 113,1096 + 3 800) = R18 913,1096
A3 = P2(1 + 0,062)1 = R(15 113,1096 + 3 800)(1,062)1 = R20 085,7224 = P3
A5 = P3(1 + 0,058)2 = R(20 085,7224)(1,058)2 = R22 483,23456
Hy belê nog R2 500 sodat / He invests another R2 500 so that
P4 = R(22 483,23456 + 2 500) = R24 983,23456
Afinaal / final = P4(1 + 0,058)2 = R(24 983,23456(1,058)2) = R27 965,33
Kapitaal / Principal (P0) = R13 400; i(eerste 3 jr / first 3 years) = i1 = 0,062 en / and i2 = 0,058
A2 = P0(1 + i)n = R13 400(1 + 0,062)2 = P2(P vir volgende berekening / for next calculation)
= R13 400(1,062)2 = P2
Hy belê R3 800 sodat / He invests R3 800 so that
P2 = R(13 400(1,062)2 + 3 800)
A3 = P2(1 + i)n = R(13 400(1,062)2 + 3 800) × 1,062 = P5
A3 i verander na / changes to 0,058
A5 = P5(1 + i)n = R((13 400(1,062)2 + 3 800) × 1,062) × 1,0582 = P7
Hy belê R2 500 sodat / He invests R2 500 so that
P7 = R(((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500
A7 = P7(1 + i)n = R((((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500) × 1,0582
= R27 965,33
Vir jou antwoord tik jy onderstaande uitdrukking in - maak net baie seker van al die hakies!!
As your answer type in the expression below - just make very sure about the brackets!!
Afinaal / final = R((((13 400(1,062)2 + 3 800) × 1,062) × 1,0582) + 2 500) × 1,0582
= R27 965,33
Bereken die rene op die R13 400 vir 3 jaar. / Calculate the interest on the R13 400 investment for 3 years.
A1 = P0(1 + i)n = R13 400(1 + 0,062)3 = P2(P vir volgende berekening / for next calculation)
Bereken nou die rene op die bykomende R3 800 vir 1 jaar en tel dit by die eerste bedrag. /
Now calculate the interest on the additional R3 800 investment for 1 year and add that to the first amount.
A2 = R(13 400 × 1,0623) + (3 800 × 1,062)
Bereken nou die rene op hierdie bedrag vir die laaste 4 jaar teen die nuwe rentekoers. /
Now calculate the interest on this amount for the last 4 years at the new interest rate.
A3 = R((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584
Bereken nou die rente op die bykomende belegging van R2 500 vir 2 jaar en tel dit by. /
Now calculate the interest on the additional amount of R2 500 for the last 2 years and add it.
A3 = R(((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584) + (2 500 × 1,0582)
Vir jou antwoord tik jy onderstaande uitdrukking in - maak net baie seker van al die hakies!!
As your answer type in the expression below - just make very sure about the brackets!!
Afinaal / final = R(((13 400 × 1,0623) + (3 800 × 1,062)) × 1,0584) + (2 500 × 1,0582)
= R27 965,33
Maak gebruik van tydlyne in hierdie soort vrae. / Use time-line graphs to help solve problems like these.
Bereken i vir die eerste 30 maande, d.i. 2,5 jaar.
Calculate i for the first 30 months, i.e. 2,5 years.
7,8
i = ————— = 0,039
100 × 2
1 + i1 = 1,039
Calculate i for the first 30 months, i.e. 2,5 years.
7,8
i = ————— = 0,039
100 × 2
1 + i1 = 1,039
Bereken i vir die laaste 18 maande /
Calculate i for the last 18 months.
6,6
i = ————— = 0,0055
100 × 12
1 + i1 = 1,0055
Calculate i for the last 18 months.
6,6
i = ————— = 0,0055
100 × 12
1 + i1 = 1,0055
Trek nou die tydlyn met al die inligting daarop.
No draw the time line containing all the information.
A1 = A(na 30 maande / after 30 months) = P(1 + i)n = R7 300(1 + 0,039)2,5 × 2 = P2(volgende P / next P)
A2 = A(na 36 maande / after 36 months) = R(7 300 × 1,0392,5 × 2) × 1,00556
Sy onttrek nou R4 800. / She now withdraws R4 800
Afinaal final = R(((7 300 × 1,0392,5 × 2) × 1,00556) − 4 800) × 1,005512
= R4 629,58
OF / OR
Afinaal final = R(((7 300 × 1,0392,5 × 2) × 1,00556) − 4 800) × × 1,005512
= R4 629,58
OF / OR
Afinaal final = R((7 300 × 1,0392,5 × 2) × 1,005518) − (4 800 × × 1,005512)
= R4 629,58
Vraag / Question 17
A1 = A(na 30 maande / after 30 months) = P(1 + i)n = R7 300(1 + 0,039)2,5 × 2 = P2(volgende P / next P)
A2 = A(na 36 maande / after 36 months) = R(7 300 × 1,0392,5 × 2) × 1,00556
Sy onttrek nou R4 800. / She now withdraws R4 800
Afinaal final = R(((7 300 × 1,0392,5 × 2) × 1,00556) − 4 800) × 1,005512
= R4 629,58
Afinaal final = R(((7 300 × 1,0392,5 × 2) × 1,00556) − 4 800) × × 1,005512
= R4 629,58
Afinaal final = R((7 300 × 1,0392,5 × 2) × 1,005518) − (4 800 × × 1,005512)
= R4 629,58