WISKUNDIGE GELETTERDHEID
GRAAD 10
NOG OEFENINGE
Finansiële documente, tariewe : antwoorde
GRAAD 10
NOG OEFENINGE
Finansiële documente, tariewe : antwoorde
MATHEMATICAL LITERACY
GRADE 10
MORE EXERCISES
Financial documents, tariffs : answers
GRADE 10
MORE EXERCISES
Financial documents, tariffs : answers
1.1.1 Van 0 tot 6 kl: gratis en
van 7 tot 8 kl R2/kl
[ V 1.1 ]
1.1.2 Van 0 tot 6 kl: gratis en
van 7 tot 8 kl R2/kl
d.i. koste = R(6 x 0 + 2 x 2) = R4,00
[ V 1.1 ]
1.1.1 From 0 to 6 kl: free and
from 7 to 8 kl R2/kl
[ Q 1.1 ]
1.1.2 From 0 to 6 kl: free and
from 7 to 8 kl R2/kl
i.e. cost = R(6 x 0 + 2 x 2) = R4,00
[ Q 1.1 ]
1.2 49,12 kl word benader tot / is rounded up to 50 kl.
Bedrag betaalbaar / Amount due = R(6 x 0 + (29 - 6) x 2 + (49 - 30) x 2,40 + 2 x 3,50)
= R98,60 [ V/Q 1.2 ]
1.3 Bedrag betaalbaar / Amount due = R(6 x 0 + (29 - 6) x 2 + (49 - 30) x 2,40 + 7 x 3,50)
= R116,10
Die rekenig is dus verkeerd - dit is te groot / The account is thus wrong - it is in excess.
Bedrag te groot Amount in excess = R135,65 ─ R116,10
= R 19,55 [ V/Q 1.3 ]
2.1 Totale uitgawe = R8 520,00
Nee, die uitgawe is
R(8 520 - 7 600) = R920
meer as die inkomste.
[ V 2.1 ]
2.2 Nog nie. Die uitgawe is nog
R420 meer as die inkomste.
Gert kan afskaal op vermaak en
klere. Spandeer R200 minder op
klere en R420 op vermaak.
Dan klop die begroting.
[ V 2.2 ]
2.3 Spandeer R300 op klere,
vermaak inkort tot R200 en
motor onderhoud inkort met R20.
[ V 2.3 ]
2.1 Total expenses = R8 520,00
No, the expenses are
R(8 520 - 7 600) = R920
more than the income.
[ Q 2.1 ]
2.2 Not yet. The expenses are still
in excess by R420 over the income.
Jack can spend less on entertainment
and clothes. Spend R200 less on
clothes and R420 on entertainment.
Then the budget balances.
[ Q 2.2 ]
2.3 Spend R300 on clothes,
decrease entertainment by R200
and motor maintenance by R20.
[ Q 2.3 ]
3.1 A = R17 000 en B = R18 300
[ V 3.1 ]
3.1 Koos kan nie al sy uitgawes
bekostig nie, want die uitgawes
oorskry die inkomste met R1 300.
[ V 3.2 ]
3.3 Die tekort is R1 200. Besnoei
R500 op klere en R500 op
vermaak en R200 op motor
onderhoud - ry minder,
spaar op brandstof.
[ V 3.3 ]
3.1 A = R17 000 and B = R18 300
[ Q 3.1 ]
3.2 Peter cannot afford all his expenses,
because the expenses exceed
the income by R1 300.
[ Q 3.2 ]
3.3 The shortage is R1 200. Cut down
expenses by spending R500
less on clothes, R500 on
entertainment and R200 on
on fuel, use the car less.
[ Q 3.3 ]
4.1 Die foonnommer is 001 968 0001.
[ V 4.1 ]
4.2 Die posadres is
Bus 34, Komhier 001908
[ V 4.2 ]
4.3 Die datum van die faktuur / rekening
28 / 03/ 2016.
[ V 4.3 ]
4.4 Laaste datum vir betaling
is 14/04/2016.
[ V 4.4 ]
4.5 Die rekening is betaal en dus
moet daardie bedrag afgetrek
word sodat die uitstaande
bedrag R0,00 is.
[ V 4.5 ]
4.6 Water verbruik = (1241 - 1206) kl
= 35 kl
en elektrisiteit = (8324 - 80136)
= 188 eenhede
[ V 4.6 ]
4.7 Bedrag = R(6 x 0 + (35 - 6) x 1,10)
= R31,90
BTW = R31,90 x 0,14 = R4,47
en dus is
Totale koste = R31,90 + R4,47
= R36,37
Die rekening is dus heel verkeerd.
[ V 4.7 ]
4.8 Elektrisiteit verbruik is 188 eenhede
en nie 288 nie. Die koste is dus
verkeerd.
Koste = R188 x 1,45 = R272,60
sodat BTW = R272,60 x 0,14
= R38,16 en
Totaal = R310,76.
Die bedrag betaalbaar is dus
verkeerd. Die bedrae moet wees :
Bedrag = R424,50; BTW = R59,43
en Bedrag = R894,05
[ V 4.8 ]
4.9 Nee. Teen 1,6% is die belasting
R350 000 x 0,016 = R5 600,00.
Die belasting van R4 921,44 word
bereken teen (4921,44 ÷ 350000) x 100
= 1,406%
[ V 4.9 ]
4.1 The phone number is 001 968 0001.
[ Q 4.1 ]
4.2 The postal addres is
P. O. Box 34, Hereweare 001908
[ Q 4.2 ]
4.3 The invoice is dated
28 / 03 / 2016.
[ Q 4.3 ]
4.4 Last date for payment is
14/04/2016.
[ Q 4.4 ]
4.5 The account is settled and thus
that amount must be deducted
from the account balance so that
the outstanding amount is R0,00.
[ Q 4.5 ]
4.6 Water consumed = (1241 - 1206) kl
= 35 kl
and electricity = (8324 - 80136)
= 188 units
[ Q 4.6 ]
4.7 Amount = R(6 x 0 + (35 - 6) x 1,10)
= R31,90
VAT = R31,90 x 0,14 = R4,47
so that
Total cost = R31,90 + R4,47
= R36,37
The account is completely in error.
[ Q 4.7 ]
4.8 Electricity consumption is
188 units and not 288. The cost
is therefore erroneous.
Cost = R188 x 1,45 = R272,60
so that VAT = R272,60 x 0,14
= R38,16 and
Total = R310,76.
The amount due is therefore wrong.
The amounts should be :
Amount = R424,50; VAT = R59,43
and Amount = R894,05
[ Q 4.8 ]
4.9 No. At 1,6% the levy is
R350 000 x 0,016 = R5 600,00.
The levy of R4 921,44 is calculated
at (4921,44 ÷ 350000) x 100
= 1,406%
[ Q 4.9 ]
5.1 Die datum is 2 Mei 2005 en die
nommer van die staat is 51.
[ V 5.1 ]
5.2 Die kommas word gebruik as
honderde en duisende skeiers en
die punte word gebruik as
die desimale skeiers.
[ V 5.2 ]
5.3 Die rentekoers is 1% per jaar
of 0,083333% per maand.
[ V 5.3 ]
5.4 Die rente verdien = Pi
= R1 986.32 x 0.00083333
= R1.66
[ V 5.4 ]
5.5 Dit beteken dat die rente verdien
by die Saldo, geld in die rekening,
of kapitaal gevoeg word.
[ V 5.5 ]
5.6 Saldo = R1,985.32 + 1.66
= R1,987.98
[ V 5.6 ]
5.7 Die bedrae in die Debiete kolom
is die bedrae wat sekere dinge
betaal of geld wat uit die
rekening onttrek word.
Die Saldo word dus met
daardie bedrag verminder.
[ V 5.7 ]
5.8 Die rente verdien = I
= R1,168.78 x 0.00083333
= R0.97
sodat die Saldo = R1,169.75
[ V 5.8 ]
5.1 The date is May 2, 2005 and the
number of the statement is 51.
[ Q 5.1 ]
5.2 The commas are used to separate
hundreds and thousands and the
fullstops are used to separate
the integers from the decimals.
[ Q 5.2 ]
5.3 The interest rate is 1% per annum
or 0,083333% per month.
[ Q 5.3 ]
5.4 The interest earned = Pi
= R1 986.32 x 0.00083333
= R1.66
[ Q 5.4 ]
5.5 It means that the interest earned
is added to the Balance to form
the new Capital.
[ Q 5.5 ]
5.6 Balance = R1,985.32 + 1.66
= R1,987.98
[ Q 5.6 ]
5.7 The amounts in the Debit column
are the amounts that are
withdrawn from the Balance to
pay other accounts or fees.
The Balance is therefore
decreased by that amount.
[ Q 5.7 ]
5.8 The interest earned = I
= R1,168.78 x 0.00083333
= R0.97
so that the Balance = R1,169.75
[ Q 5.8 ]
6.1 Die datum is 2 Mei 2005 en die
nommer van die staat is 251.
[ V 6.1 ]
6.2 Die Saldo oorgebring is 'n bedrag
wat Mej. Marais aan die Bank skuld
- haar rekening is oortrokke -
sy leen geld by die Bank.
[ V 6.2 ]
6.3 Die bedrae in die Debiete kolom
is die bedrae wat sekere dinge
betaal of geld wat uit die rekening
onttrek word. Geld word uit die
rekening verwyder en daarom
moet die Saldo dus met
daardie bedrag verminder.
[ V 6.3 ]
6.4 Saldo = R3,814.21- + R7,420.00
= R3,605.79
Die teken is positief wat beteken
dat die rekening nie meer
oortrokke is nie, daar is weer
fondse in die rekening.
[ V 6.4 ]
6.5 Saldo = R20.96-
Die rekening is weer met
R20.96 oortrokke.
[ V 6.5 ]
6.6 Saldo = R2,451.21-
[ V 6.6 ]
6.7 Saldo = R2,4763.21-
Daar is nou R800 in die
rekening in betaal wat die
oortrokke rekening verminder.
[ V 6.7 ]
6.8 Saldo = R4,423.76-
[ V 6.8 ]
6.9 P = R4,423.76- + 99.00-
= R4,522.76- en
i = 13,5 ÷ (100 x 12)
= 0.01125
I = Pi = R4,522.76 x 0.01125
= R50.88
[ V 6.9 ]
6.10 Saldo = R4,522.76- + (R50.88-)
= R4,573.64-
[ V 6.10 ]
6.1 The date is May 2, 2005 and
the number of the statement is 251
[ Q 6.1 ]
6.2 The Balance brought forward is an
amount that Miss Marais owes
the Bank - her account is overdrawn
- she "borrows" money from the Bank.
[ Q 6.2 ]
6.3 The amounts in the Debits column
are the amounts that were used to
pay certain other debts or cash
withdrawals. Money is withdrawn
from the account and therefore the
Balance must be decreased by
that amount.
[ Q 6.3 ]
6.4 Balance = R3,814.21- + R7,420.00
= R3,605.79
The sign is positive which means
that the account is no longer
overdrawn, there is "money in
the Bank", there are funds available.
[ Q 6.4 ]
6.5 Balance = R20.96-
The account is once more
overdrawn by R20.96.
[ Q 6.5 ]
6.6 Balance = R2,451.21-
[ Q 6.6 ]
6.7 Balance = R2,4763.21-
An amount of R800 is deposited
into the account which decreases
the amount overdrawn.
[ Q 6.7 ]
6.8 Balance = R4,423.76-
[ Q 6.8 ]
6.9 P = R4,423.76- + 99.00-
= R4,522.76- and
i = 13,5 ÷ (100 x 12)
= 0.01125
I = Pi = R4,522.76 x 0.01125
= R50.88
[ Q 6.9 ]
6.10 Balance = R4,522.76- + (R50.88-)
= R4,573.64-
[ Q 6.10 ]
7.1 20 items is gekoop.
[ V 7.1 ]
7.2 4 plastiese drasakke is gekoop.
[ V 7.2 ]
7.3 24 liter. [ V 7.3 ]
7.4 6 items is BTW vry. [ V 7.4 ]
7.5 R109,70 [ V 7.5 ]
7.6 R146,45 [ V 7.6 ]
7.7 R10,99 [ V 7.7 ]
7.1 20 items were bought.
[ Q 7.1 ]
7.2 4 plastic carry bags were bought.
[ Q 7.2 ]
7.3 24 liter [ Q 7.3 ]
7.4 6 items are VAT free. [ Q 7.4 ]
7.5 R109.70 [ Q 7.5 ]
7.6 R146.45 [ Q 7.6 ]
7.7 R10,99 [ Q 7.7 ]
7.8 0,96 kg kos / costs R34,56
R34,56
1 kg kos / costs ──────
0,96
= R36,00 [ V/Q 7.8 ]
7.9 18 eiers / eggs kos / costs R29,99
R29,99
1 eier / egg kos / costs ──────
18
d.i. / i.e. R1,6661111
6 eiers / eggs kos / costs R9,99
R9,99
1 eier / egg kos / costs ──────
6
d.i. / i.e. R1,665186
Die 6 eier verpakking is dus effens goedkoper. / The 6 egg pack is thus a little cheaper.
[ V/Q 7.9 ]
7.10 0,75 kg wortels kos / carrots costs R11,24
R11,24
1 kg wortels / carrots kos / costs ──────
0,75
d.i. / i.e. R14,98666667
3 kg wortels / carrots kos / costs 3 X R 14,98666667
= R44,96
[ V/Q 7.10 ]
8.1
| Tyd / Time (minute / minutes) | 0 | 10 | 50 | 200 | 250 | 300 | 400 |
| Koste / Cost (R) | 0 | 5 | 25 | 100 | 125 | 150 | 200 |
[ V 8.1 ]
Koste
8.2 Ja, want die verhouding ──────
tyd
bly konstant
[ V 8.2 ]
Cost
8.2 Yes, because the ratio ──────
time
remains constant / the same
[ Q 8.2 ]
8.3 Sien die grafiek langsaan.
[ V 8.3 ]
8.3 See the graph in 8.3
[ V 8.3 ]
8.4.1 A; R50 [ V 8.4.1 ]
8.4.2 B; 200 minute / minutes [ V 8.4.2 ]
8.4.3 C; R75 [ V 8.4.3 ]
8.4.4 D; 250 minute / minutes [ V 8.4.4 ]
8.5 Die lyn sal opwaarts skuif met
die helfte van die oorspronklike
y-waarde (Koste-waarde),
bv. by 100 minute sal die punt
met die helfte van R50,
d.w.s. R25, opskuif sodat
die nuwe Koste nou R75 is.
Sien grafiek in 8.5
[ V/Q 8.5 ]
8.5 The line will shift upwards by half
of the original y-value
(Cost-value),
e.g. at 100 minutes the point
will shift upwards by half of R50,
i.e. R25, so that the new Cost
is now R75.
See the graph in 8.5
[ V/Q 8.5 ]
8.2 Ja, want die verhouding ──────
tyd
bly konstant
[ V 8.2 ]
Cost
8.2 Yes, because the ratio ──────
time
remains constant / the same
[ Q 8.2 ]
8.3 Sien die grafiek langsaan.
[ V 8.3 ]
8.3 See the graph in 8.3
[ V 8.3 ]
8.4.1 A; R50 [ V 8.4.1 ]
8.4.2 B; 200 minute / minutes [ V 8.4.2 ]
8.4.3 C; R75 [ V 8.4.3 ]
8.4.4 D; 250 minute / minutes [ V 8.4.4 ]
8.5 Die lyn sal opwaarts skuif met
die helfte van die oorspronklike
y-waarde (Koste-waarde),
bv. by 100 minute sal die punt
met die helfte van R50,
d.w.s. R25, opskuif sodat
die nuwe Koste nou R75 is.
Sien grafiek in 8.5
[ V/Q 8.5 ]
8.5 The line will shift upwards by half
of the original y-value
(Cost-value),
e.g. at 100 minutes the point
will shift upwards by half of R50,
i.e. R25, so that the new Cost
is now R75.
See the graph in 8.5
[ V/Q 8.5 ]
9.1
Die grafiek begin in die
punt (0 ; 50). Daar is nou
'n vaste R50 koste wat
onafhanklik van die tyd is.
[ V 9.1 ]
9.2.1 R90,00 [ V 9.2.1 ]
9.2.2 120 minute [ V 9.2.2 ]
9.2.3 R120,00 [ V 9.2.3 ]
9.2.4 310 minute [ V 9.2.4 ]
punt (0 ; 50). Daar is nou
'n vaste R50 koste wat
onafhanklik van die tyd is.
[ V 9.1 ]
9.2.1 R90,00 [ V 9.2.1 ]
9.2.2 120 minute [ V 9.2.2 ]
9.2.3 R120,00 [ V 9.2.3 ]
9.2.4 310 minute [ V 9.2.4 ]
9.1
The graph starts at the point
(0 ; 50). There is now a fixed
cost that is independent of
the time.
[ Q 9.1 ]
9.2.1 R90,00 [ Q 9.2.1 ]
9.2.2 120 minute [ Q 9.2.2 ]
9.2.3 R120,00 [ Q 9.2.3 ]
9.2.4 310 minute [ Q 9.2.4 ]
(0 ; 50). There is now a fixed
cost that is independent of
the time.
[ Q 9.1 ]
9.2.1 R90,00 [ Q 9.2.1 ]
9.2.2 120 minute [ Q 9.2.2 ]
9.2.3 R120,00 [ Q 9.2.3 ]
9.2.4 310 minute [ Q 9.2.4 ]
9.3
Die vaste fooi van R50 moet afgetrek word van die R150 om die koste vir 250 minute te kry.
The fixed cost of R50 must be deducted from thr R150 to calculate the cost for 250 minutes.
R100
Koste per minuut / Cost per minute = ———————— = R0,40
250 minute(s) [ V/Q 9.3 ]
9.4 9.2.1 Koste / Cost = R50 + (0,40 x 100) = R(50 + 40) = R90,00
R50
9.2.2 Tyd / Time = ────── = 125 minute(s)
R0,40
9.2.3 Koste / Cost = R50 + (0,40 x 150) = R(50 + 60) = R110,00
9.2.4 Koste vir minute gepraat = R(178 - 50) = R128
R128
Tyd / Time = ────── = 320 minute(s)
R0,40 [ V/Q 9.4 ]
The fixed cost of R50 must be deducted from thr R150 to calculate the cost for 250 minutes.
R100
Koste per minuut / Cost per minute = ———————— = R0,40
250 minute(s) [ V/Q 9.3 ]
9.4 9.2.1 Koste / Cost = R50 + (0,40 x 100) = R(50 + 40) = R90,00
R50
9.2.2 Tyd / Time = ────── = 125 minute(s)
R0,40
9.2.3 Koste / Cost = R50 + (0,40 x 150) = R(50 + 60) = R110,00
9.2.4 Koste vir minute gepraat = R(178 - 50) = R128
R128
Tyd / Time = ────── = 320 minute(s)
R0,40 [ V/Q 9.4 ]
10.1
Die grafiek begin by
punt (0 ; 50). Daar is nou
'n vaste R50 koste wat
onafhanklik van die tyd is.
[ V 10.1 ]
10.2 Van A na B is die grafiek
ewewydig aan die horisontale
tyd-as sodat daar geen
verandering in waarde op
die vertikale koste-as is nie
alhoewel die tyd toeneem,
d.w.s. van A na B bly die
koste konstant op R50
alhoewel die tyd toeneem
van 0 tot 100 minute.
[ V 10.2 ]
10.3 Van B na C is die grafiek
skuins sodat die waardes op
die vertikale koste-as toeneem
soos wat die waardes op die
horisontale tyd-as toeneem
sodat daar 'n styging in koste
is soos wat die tyd toeneem.
[ V 10.3 ]
10.4.1 R50,00 [ V 10.4.1 ]
10.4.2 150 minute [ V 10.4.2 ]
10.4.3 R125,00 [ V 10.4.3 ]
10.4.4 350 minute [ V 10.4.4 ]
punt (0 ; 50). Daar is nou
'n vaste R50 koste wat
onafhanklik van die tyd is.
[ V 10.1 ]
10.2 Van A na B is die grafiek
ewewydig aan die horisontale
tyd-as sodat daar geen
verandering in waarde op
die vertikale koste-as is nie
alhoewel die tyd toeneem,
d.w.s. van A na B bly die
koste konstant op R50
alhoewel die tyd toeneem
van 0 tot 100 minute.
[ V 10.2 ]
10.3 Van B na C is die grafiek
skuins sodat die waardes op
die vertikale koste-as toeneem
soos wat die waardes op die
horisontale tyd-as toeneem
sodat daar 'n styging in koste
is soos wat die tyd toeneem.
[ V 10.3 ]
10.4.1 R50,00 [ V 10.4.1 ]
10.4.2 150 minute [ V 10.4.2 ]
10.4.3 R125,00 [ V 10.4.3 ]
10.4.4 350 minute [ V 10.4.4 ]
10.1
The graph starts at the point
(0 ; 50). There is now a fixed
cost of R50 which does not
depend on the time.
[ Q 10.1 ]
10.2 From A to B the graph is
parallel to the horizontal
time-axis so that there is no
change in the value on the
vertical cost-axis although
the time increases,
i.e. from A to B the cost
remains constant on R50
although the time increases
from 0 to 100 minutes.
[ Q 10.2 ]
10.3 From B to C the graph slopes
upward so that the values on
the vertical cost-axis increase
as the values on the horizontal
time-axis increase causing an
increase in the cost as
the time increases.
[ Q 10.3 ]
10.4.1 R50,00 [ Q 10.4.1 ]
10.4.2 150 minutes [ Q 10.4.2 ]
10.4.3 R125,00 [ Q 10.4.3 ]
10.4.4 350 minutes [ Q 10.4.4 ]
(0 ; 50). There is now a fixed
cost of R50 which does not
depend on the time.
[ Q 10.1 ]
10.2 From A to B the graph is
parallel to the horizontal
time-axis so that there is no
change in the value on the
vertical cost-axis although
the time increases,
i.e. from A to B the cost
remains constant on R50
although the time increases
from 0 to 100 minutes.
[ Q 10.2 ]
10.3 From B to C the graph slopes
upward so that the values on
the vertical cost-axis increase
as the values on the horizontal
time-axis increase causing an
increase in the cost as
the time increases.
[ Q 10.3 ]
10.4.1 R50,00 [ Q 10.4.1 ]
10.4.2 150 minutes [ Q 10.4.2 ]
10.4.3 R125,00 [ Q 10.4.3 ]
10.4.4 350 minutes [ Q 10.4.4 ]
10.5
Die vaste fooi van R50 moet afgetrek word van die R200 om die koste vir 300 minute te kry.
The fixed cost of R50 must be deducted from thr R200 to calculate the cost for 300 minutes.
Koste vir die minute gebruik / Cost for the minutes used = R200 - R50 = R150
R150
Koste per minuut / Cost per minute = ———————— = R0,50
300 minute(s) [ V/Q 10.5 ]
10.6 10.4.1 Koste / Cost = R50 + (0,40 x 100) = R(50 + 40) = R90,00
10.4.2 Koste vir minute gebruik / Cost for minutes used = R(80 - 50) = R30
R30
Tyd / Time = ────── = 60 minute(s)
R0,50
10.4.3 Koste / Cost = R50 + (0,50 x (250 - 100)) = R(50 + 75) = R125,00
10.4.4 Koste vir minute gebruik / Cost for minutes used = R(175 - 50) = R125
R125
Tyd / Time = ────── = 250 minute(s)
R0,50
Tyd / Time = 250 + 100 minute(s) = 350 minute(s) [ V/Q 10.6 ]
The fixed cost of R50 must be deducted from thr R200 to calculate the cost for 300 minutes.
Koste vir die minute gebruik / Cost for the minutes used = R200 - R50 = R150
R150
Koste per minuut / Cost per minute = ———————— = R0,50
300 minute(s) [ V/Q 10.5 ]
10.6 10.4.1 Koste / Cost = R50 + (0,40 x 100) = R(50 + 40) = R90,00
10.4.2 Koste vir minute gebruik / Cost for minutes used = R(80 - 50) = R30
R30
Tyd / Time = ────── = 60 minute(s)
R0,50
10.4.3 Koste / Cost = R50 + (0,50 x (250 - 100)) = R(50 + 75) = R125,00
10.4.4 Koste vir minute gebruik / Cost for minutes used = R(175 - 50) = R125
R125
Tyd / Time = ────── = 250 minute(s)
R0,50
Tyd / Time = 250 + 100 minute(s) = 350 minute(s) [ V/Q 10.6 ]
11.1
Die grafiek begin nie by 0
nie, want daar is 'n vaste
koste van R50
onafhanklik van die tyd is.
[ V 11.1 ]
11.2 kontrak A R90
kontrak B R60 [ V 11.2 ]
11.3 kontrak A 375 minute
kontrak B 333.3 minute
[ V 11.3 ]
11.4 R is die punt (250 ; 150)
[ V 11.4 ]
nie, want daar is 'n vaste
koste van R50
onafhanklik van die tyd is.
[ V 11.1 ]
11.2 kontrak A R90
kontrak B R60 [ V 11.2 ]
11.3 kontrak A 375 minute
kontrak B 333.3 minute
[ V 11.3 ]
11.4 R is die punt (250 ; 150)
[ V 11.4 ]
11.1
The graph does not start at
0 because there is a fixed
cost of R50 which does not
depend on the time.
[ Q 11.1 ]
11.2 contract A R90
contract B R60 [ Q 11.2 ]
11.3 contract A 375 minutes
contract B 333,3 minutes
[ Q 11.3 ]
11.4 R is the point (250 ; 150)
[ Q 11.4 ]
0 because there is a fixed
cost of R50 which does not
depend on the time.
[ Q 11.1 ]
11.2 contract A R90
contract B R60 [ Q 11.2 ]
11.3 contract A 375 minutes
contract B 333,3 minutes
[ Q 11.3 ]
11.4 R is the point (250 ; 150)
[ Q 11.4 ]
11.5
By punt R is die koste R150 vir 250 minute.
At point R the cost is R150 for 250 minutes.
R150
Koste per minuut / Cost per minute = ———————— = R0,60 = 60 c
250 minute(s) [ V/Q 10.5 ]
11.6 By punt R is die koste R150 vir 250 minute.
At point R the cost is R150 for 250 minutes.
Die vaste fooi van R50 moet afgetrek word van die R150 om die koste vir 250 minute te kry.
The fixed cost of R50 must be deducted from thr R150 to calculate the cost for 250 minutes.
Koste vir die minute gebruik / Cost for the minutes used = R150 - R50 = R100
R100
Koste per minuut / Cost per minute = ———————— = R0,40
250 minute(s) [ V/Q 11.6 ]
11.7.1 Kontrak B omdat die koste vir 200 en minder minute op kontrak B minder is as die op
kontrak A. By 200 minute is die koste op kontrak B R120 terwyl dit op kontrak A R130 is.
Die koste is dus minder. [ V/Q 11.7.1 ]
Contract B because the cost for 200 minutes and less on contract B is less than that on
contract B. At 200 minutes the cost on contract B is R120 while on contract A it is R130.
The cost is thus smaller. [ V/Q 11.7.1 ]
11.7.2 Kontrak A omdat die koste vir 260 en meer minute op kontrak A minder is as die op
kontrak B. By 260 minute is die koste op kontrak A R154 terwyl dit op kontrak B R156 is.
Die koste is dus minder. [ V/Q 11.7.2 ]
Contract A because the cost for 260 minutes and more on contract A is less than that on
contract B. At 260 minutes the cost on contract A is R154 while on contract B it is R156.
The cost is thus smaller. [ V/Q 11.7.2 ]
At point R the cost is R150 for 250 minutes.
R150
Koste per minuut / Cost per minute = ———————— = R0,60 = 60 c
250 minute(s) [ V/Q 10.5 ]
11.6 By punt R is die koste R150 vir 250 minute.
At point R the cost is R150 for 250 minutes.
Die vaste fooi van R50 moet afgetrek word van die R150 om die koste vir 250 minute te kry.
The fixed cost of R50 must be deducted from thr R150 to calculate the cost for 250 minutes.
Koste vir die minute gebruik / Cost for the minutes used = R150 - R50 = R100
R100
Koste per minuut / Cost per minute = ———————— = R0,40
250 minute(s) [ V/Q 11.6 ]
11.7.1 Kontrak B omdat die koste vir 200 en minder minute op kontrak B minder is as die op
kontrak A. By 200 minute is die koste op kontrak B R120 terwyl dit op kontrak A R130 is.
Die koste is dus minder. [ V/Q 11.7.1 ]
Contract B because the cost for 200 minutes and less on contract B is less than that on
contract B. At 200 minutes the cost on contract B is R120 while on contract A it is R130.
The cost is thus smaller. [ V/Q 11.7.1 ]
11.7.2 Kontrak A omdat die koste vir 260 en meer minute op kontrak A minder is as die op
kontrak B. By 260 minute is die koste op kontrak A R154 terwyl dit op kontrak B R156 is.
Die koste is dus minder. [ V/Q 11.7.2 ]
Contract A because the cost for 260 minutes and more on contract A is less than that on
contract B. At 260 minutes the cost on contract A is R154 while on contract B it is R156.
The cost is thus smaller. [ V/Q 11.7.2 ]
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