SENIOR CERTIFICATE EXAMINATIONS/
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
SENIORSERTIFIKAAT-EKSAMEN/
NASIONALE SENIORSERTIFIKAAT-EKSAMEN
MATHEMATICAL LITERACY P1/
WISKUNDIGE GELETTERDHEID V1
MAY/JUNE/MEI/JUNIE 2024
MARKING GUIDELINES/NASIENRIGLYNE
MARKS/PUNTE: 150
Symbol/Kode Explanation/Verduideliking
MA Method with accuracy/Metode met akkuraatheid
CA Consistent accuracy/Volgehoue akkuraatheid
A Accuracy/Akkuraatheid
C Conversion/Herleiding
S Simplification/Vereenvoudiging
RT Reading from a table/graph/document/diagram/Lees vanaf tabel/grafiek/dokument/diagram
SF Correct substitution in a formula/Korrekte vervanging in 'n formule
O Opinion/Explanation/Opinie/Verduideliking
P Penalty, e.g. for no units, incorrect rounding off, etc./Penalisasie, bv. vir geen eenhede,
verkeerde afronding, ens.
R Rounding off/Afronding
NPR No penalty for rounding/Geen penalisasie vir afronding nie
NPU No penalty for omitting correct unit/Geen penalisasie vir die uitlos van die korrekte eenheid nie.
AO Answer only/Slegs antwoord
MCA Method with consistent accuracy/Metode met volgehoue akkuraatheid
RCA Rounding consistent with accuracy/ Afronding met volgehoue akkuraatheid
These marking guidelines consist of 19 pages.
Hierdie nasienriglyne bestaan uit 19 bladsye.
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Mathematical Literacy Grade 12 P1 NSC June 2024 Memo (Afr)
Grade 12 Mathematical Literacy Paper 1 memorandum from the NSC June Exam 2024 (Afrikaans), 19 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 12 · NSC June Exam · 2024 · Afrikaans. Memorandum, 19 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 12
- Language
- Afrikaans
- Document type
- Memorandum
- Year
- 2024
- Exam period
- NSC June Exam
- Paper
- 1
- Curriculum
- CAPS
- Pages
- 19
- File size
- 530.0 KB
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This exam
NSC June Exam 2024
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Mathematical Literacy/P1/Wiskundige Geletterdheid/V1 2 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out (cancelled) an attempt to a question and NOT redone the solution,
mark the crossed out (cancelled) version.
• Consistent accuracy (CA) applies in ALL aspects of the marking guidelines; however it stops at the
second calculation error or break-down.
• If the candidate presents any extra solution when reading from a graph, table, layout plan and map, then
penalise for every extra item presented.
• Rounding is an independent mark.
• General principle of marking, if the candidate makes one mistake one mark is deducted.
• A conclusion mark can only be given if relevant calculations precedes it (at least 1 mark before
conclusion).
• No penalty for rounding (NPR) if the first decimal is correct, except questions involving money.
LET WEL:
• As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
• As 'n kandidaat 'n antwoord van 'n vraag doodtrek (kanselleer) en nie oordoen nie, sien die doodgetrekte
(gekanselleerde) poging na.
• Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyne toegepas; dit hou egter op
by die tweede berekeningsfout of ‘break-down’.
• Wanneer 'n kandidaat aflesings vanaf 'n grafiek, tabel, uitlegplan en kaart geneem en ekstra antwoorde
gee, penaliseer vir elke ekstra item.
• Afronding tel as 'n afsonderlike punt.
• Die algemene beginsel van merk as 'n leerder een fout maak, word een punt afgetrek.
• 'n Gevolgtrekkingspunt kan slegs gegee word indien relevante berekeninge dit voorgaan (ten minste een
punt voor die gevolgtrekking).
• Geen penalisering vir ronding (NPR) as die eerste desimaal korrek is nie, behalwe as vrae geld insluit.
QUESTION/VRAAG 1 [32 MARKS/PUNTE] ANSWER ONLY FULL MARKS
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
1.1.1 Kenya / Kenia ✓✓A 2A correct country L1
(2) E
D
1.1.2 Zambia / Zambië ✓✓A 2A first correct country L1
Malawi / Malawi ✓A 1A second correct country E
(3)
D
1.1.3 Malawi / Malawi ✓✓A 2A correct country L1
(2) M
* D
1.1.4 People for Studying Purposes / Mense vir Studiedoeleindes 1RT correct values L1
✓RT 1MA adding four correct M
= 83 + 98 + 475 + 83 ✓MA values
= 739 tourists ✓A 1A simplification
(3)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
F
1.2.1 Price of 1 teabag / Prys van 1 teesakkie L1
E
= R50,00 ÷ 40 ✓MA 1MA R50 divided by 40
= R1,25 ✓A 1A simplification
OR / OF OR / OF
✓A
= R1,25 × 40 ✓MA 1A R1,25
= R50 1MA R1,25 multiplied by 40
(2)
* ✓RT F
1.2.2 185 : 100 ✓MA 1RT correct values L1
1MA values in correct order M
37 : 20 OR / OF 1 : 0,54 OR / OF 1,85 : 1 ✓A 1A simplification
(3)
F
1.2.3 Total of purchase / Totaal van aankope L1
✓RT 1RT correct values E
= R185,00 + R100,00 + R16,00 ✓MA 1MA adding 3 correct values
= R301,00 ✓A 1A simplification
(3)
* F
1.3.1 Namibia / Namibië ✓✓A 2A correct country L1
(2) E
* F
1.3.2 Zambian Kwacha ✓✓A L1
E
OR / OF 2A correct currency
Zambia / ZMW ✓✓A
(2)
F
1.3.3 1 ✓MA 1MA dividing correct values L1
Malawian Kwacha = 1RT reading correct value E
56, 211355 ✓RT
= 0,017790
OR / OF
OR / OF
1MA dividing correct values
1 ✓MA
Malawian Kwacha =
0, 017790 1RT calculating correct
= 56,211355 ✓RT value
NPR
(2)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
1.4.1 One million one hundred and five thousand six hundred and L1
eighty five / Een miljoen een honderd en vyf duisend ses E
honderd vyf en tagtig. ✓✓A 2A answer in words
(2)
D
1.4.2 P = 131 693 + 254 139 ✓MA L1
E
= 385 832 ✓A
OR / OF 1MA adding correct values
P = 64 943 + 66 694 + 56 +136 510 + 114 436 + 3 193 ✓MA 1A simplification
= 385 832 ✓A
(2)
* D
1.4.3 Increase / Verhoging L1
E
= 359 686 – 131 693 ✓MA 1MA subtracting correct
values
= 227 993 ✓A 1A simplification
(2)
✓A ✓A 1A correct month D
1.4.4 February / Februarie 2022 OR / OF Feb '22 OR / OF 1A correct year L1
02/2022 (2) E
[32]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 2 [31 MARKS/PUNTE] NPU FOR QUESTION 2.2
Q/V Solution/Oplossing Explanation/Verduideliking T&L
F
2.1.1 R2 000,00 ✓✓A 2A correct amount L1
Accept (– R2 000,00) E
NPU
(2)
* F
2.1.2 For security reasons / Vir veiligheidsredes ✓✓A 2A reason L4
(2) E
1RT 2 correct values F
2.1.3 Available money / Geld beskikbaar 1MA adding correct values L2
✓MA ✓RT 1MA subtracting from M
=R20 000 – (R5 656,22 + R6 020,00) ✓MA R20 000
= R20 000 – R11 676,22
= R8 323,78 OR / OF
OR / OF 1RT 2 correct values
✓MA ✓RT 1MA subtracting correct
= R20 000 – R5 656,22 – R6 020,00 ✓MA values
= R8 323,78 1MA subtracting from
R20 000
OR / OF OR / OF
✓MA ✓RT 1RT 2 correct values
= R5 656,22 + R6 020,00 + R8 323,78 1MA adding correct values
= R20 000 ✓MA 1MA getting to R20 000
(3)
* F
2.1.4 Price per litre / Prys per liter L2
✓RT 1RT correct rand value M
R1 376,15
= ✓MA 1MA dividing by litres
54,1365 litres
= R25,42 ✓CA 1CA simplification
OR / OF OR / OF
✓MA
= R25,42/ℓ × 54,1365 ℓ ✓RT 1RT correct rand value
1MA multiplying
= R1 376,15 ✓CA 1CA simplification
AO
(3)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
F
2.1.5 100% – 17,5% = 82,5% ✓MA 1MA calculate discounted L2
percentage D
Original price / Oorspronklike prys
1RT correct value
100 ✓RT 1MA percentage calculation
= R3 299,99 ✓MA
82,5
= R3 999,99 ✓CA 1CA simplification
OR / OF OR / OF
✓MA
1MA calculate discounted
100% – 17,5% = 82,5% = 0,825 percentage
✓RT 1RT correct value
R3 299,99 1MA divide by 0,825
= ✓MA
0,825
= R3 999,99 ✓CA 1CA simplification
NPR
(4)
F
2.2.1 Bar sales and Functions / Kroegverkope en Funksies ✓✓RT 2RT correct item L1
Accept: Bar sales E
(2)
F
2.2.2 Projected amount / Geprojekteerde bedrag L2
D
47 49 1MA concept of ratio
= ✓MA
23 500 B 1RT 23 500
✓RT
1MA divide by 47
B = 1 151 500 47 ✓MA
= 24 500 ✓CA
1CA simplification
OR / OF
OR / OF
47 49
: ✓MA 1MA concept of ratio
23 500 B
✓RT 1RT 23 500
23 500
B= × 49
47 ✓MA 1MA divide by 47
= 500 × 49
= R24 500 ✓CA 1CA simplification
OR / OF OR / OF
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
2.2.2 Total ratio / Totale verhouding = 47 + 49
= 96
✓RT 96 1RT 23 500
Total Income / Totale Inkomste = 23500 × 47 ✓MA 1MA concept of ratio
= 48 000
1MA subtracting values
B = 48 000 – 23 500 ✓MA
1CA simplification
= 24 500 ✓CA
(4)
F
2.2.3 Difference in income / Verskil in inkomste L3
✓RT 1RT correct values M
= £(455 869 – 396 453) 1CA difference
= £59 416 ✓CA
1MA % calculation
59 416 1MA correct denominator
% change / verandering = 100%✓MA
396 453 ✓MA ✓CA
= 14,9868… % = 14,99 OR 15% 1CA simplification
OR / OF
OR / OF
% change / verandering
new price/new prys - old price/ou prys
= × 100% 1RT correct values
old price/ ou prys
1MA % calculation
✓RT 1MA correct denominator
455 869 - 396 453
= 100% ✓MA
396 453 ✓MA 1CA correct difference
✓CA
59 416 1CA simplification
= 100
396 453
= 14,9868 ... %= 14,99% OR/OF 15% ✓CA OR / OF
OR / OF
Current percentage / Huidige persentasie 1RT correct values
✓RT 1MA % calculation
455 869
= 100% ✓MA
396 453 1A correct percentage
= 114,9937067% ✓A
Difference in % income / Verskil in % inkomste
1MCA correct difference
= 114,9937067% – 100% ✓MCA 1CA simplification
= 14,99% OR/OF 15% ✓CA (5)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
F
2.2.4 Year 1's inflation / Jaar 1 se inflasie: L4
1RT correct value D
✓RT
1,82
= × £257 460
100
= £4 685,772
After year 1's inflation / Na jaar 1 se inflasie:
= £257 460 + £4 685,772 ✓MCA 1MCA adding correct values
= £262 145,77 ✓CA 1CA simplification
Year 2's inflation / Jaar 2 se inflasie:
2
= × £262 145,77
100
= £5 242£,9154
After year 2's inflation / Na jaar 2 se inflasie:
= £262 145,77 + £5 242£,9154
= £267 388,69 ✓CA 1CA amount year 2
Difference / Verskil
= £284 000 – £267 388,69 ✓MCA 1MCA subtracting values
= £16 611,31
His statement is VALID / Sy bewering is GELDIG. ✓O 1O conclusion
OR / OF OR / OF
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
2.2.4 After year 1's inflation / Na jaar 1 se inflasie:
✓MA
✓RT = × 1,0182
101,82 1RT correct value
£257 460 1MA percentage increase
100
= £262 145,77 ✓CA 1CA simplification
After year 2's inflation / Na jaar 2 se inflasie:
102
£262 145,77
100
= £267 388,69 ✓CA 1CA amount year 2
Difference / Verskil
= £284 000 – £267 388,69 ✓MCA 1MCA subtracting values
= £16 611,31
His statement is VALID / Sy bewering is GELDIG. ✓O 1O conclusion
(6)
[31]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 3 [30 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
3.1.1 29 ✓✓A 2A correct number L1
(2) E
D
3.1.2 30 ✓✓A 2A mode L2
(2) E
✓RT D
3.1.3 3 1RT correct fraction L2
D = × 40 ✓MA 1MA multiplying with 40 M
5
= 24 periods ✓CA 1CA simplification
OR / OF OR / OF
3 + D + 26 + 30 + 32 ...
33 =
29
933 + D
33 = ✓RT 1RT correct fraction
29
33 × 29 = 933 + D ✓MA 1MA changing the subject of
the formula
D = 957 – 933
D = 24 ✓CA 1CA simplification
OR / OF OR / OF
✓RT
D = 0,6 × 40 ✓MA 1RT correct fraction
1MA multiplying with 40
D = 24 periods ✓CA
1CA simplification
AO
(3)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
3.1.4 Median / Mediaan = 35 ✓✓A 2A median L4
E
The median has half the staff above and half the staff
below. ✓O
Die mediaan toon die helfte van die personeel bo en helfte
van die personeel onder.
OR / OF
The outliers affect the calculation of the mean, hence it is
not a reliable average.
Die uitskieters affekteer die berekenining van die
gemiddels, daarom is dit nie 'n betroubare punt nie. ✓O 1O reason
OR / OF
More than 58,62% of staff has 35 periods or more. ✓O
Meer as 58,62% van die personeel het 35 periodes of meer.
(3)
P
3.1.5 Probability / Waarskynlikheid L2
E
16 ✓A 1A numerator
=
21 ✓A 1A denominator
(2)
* D
3.2.1 Scatter plot / Spreidingsdiagram ✓✓A 2A correct graph L1
(a) (2) E
D
3.2.1 Range / Omvang L2
(b) ✓RT ✓RT 1RT highest E
= 81 − 15 1RT lowest
= 66 ✓CA 1CA simplification
AO
(3)
* D
3.2.1 Learner / Leerder H ✓✓A 2A correct learner L4
(c) D
The marks for both Task 1 and Task 2 are much lower
compared to the other learners / Beide Taak 1 en Taak 2 se
punte is heelwat laer in vergelyking met die ander leerders.
✓✓O 2O correct reason
The learner failed while all the other learners passed / Die
leerder het gedruip terwyl al die ander leerders geslaag
het.
(4)
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
3.2.2 53 + 69 + 53 + 49 + 50 + 47 + 61 + 15 + 47 + 81 ✓MA 1MA correct concept of mean L4
Mean = M
10
525 ✓CA 1CA correct adding of values
=
10
= 52,5 ✓CA 1CA mean
Difference / Verskil
= 66,7 – 52,5
= 14,2 ✓CA 1CA simplification
1O conclusion
VALID / GELDIG ✓O
(5)
* D
3.2.3 90 L2
M
80
70
✓A A
60 L
50
Task 2
40 ✓A A
K
30
20
10
0
0 20 40 60 80 100
Task 1
2A first point plotted (68;40)
2A second point plotted (88;64) (4)
[30]
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 4 [26 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T&L
* F
4.1.1 Total cost before discount /Totale koste voor afslag L3
✓MA ✓MA 1MA multiply correct values M
= (R149,95 16,7) + (R99,99 13) 1MA multiply correct values
= R2 504,165 + R1 299,87 ✓MCA 1MCA adding cost
= R3 804,04 ✓CA 1CA simplification
Discount amount / Afslag bedrag
15
= R3 804,04 ✓MCA 1MCA calculating 15%
100
= R570,61 ✓CA
1CA simplification
Total amount /Totale koste
85
= R3 804,04 – R570,61 = R3 804,04
= R3 233,43 ✓CA 100 1CA simplification
OR / OF OR / OF
Discounted chops / Afslag tjops
✓MA ✓MCA
85 1MA multiply correct values
= R149,95 16,7 = R2 128,54 ✓CA 1MCA calculating 85%
100 1CA simplification
Discounted boerewors / Afslag boerewors
✓MA ✓MCA
85 1MA multiply correct values
= R 99,99 13 = R1 104,89 ✓CA 1MCA calculating 85%
100
1CA simplification
Total amount / Totale bedrag
= R2 128,54 + R1 104,89
= R3 233,43 ✓CA 1CA simplification
OR / OF OR / OF
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SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
*
4.1.1 Total cost before discount /Totale koste voor afslag
✓MA ✓MA 1MA multiply correct values
= (R149,95 16,7) + (R99,99 13) 1MA multiply correct values
= R2 504,165 + R1 299,87 ✓MCA 1MCA adding cost
= R3 804,04 ✓CA 1CA simplification
Total amount /Totale koste
✓MCA 1MCA calculating 15%
= R3 804 ,04 – (15% × R3 804,04)
= R3 804,04 – R570,606 ✓MCA 1MCA subtracting correct
values
= R3 233, 43 ✓CA 1CA simplification
(7)
CA from Question 4.1.1 F
4.1.2 Total cost / Totale koste 1MCA adding value from L2
✓MCA Q 4.1.1 M
= R3 233,43 + R850 ✓A 1A adding R850
= R4 083,43
Cost of 1 plate / Koste vir 1 bord
= R4 083,43 200 ✓MCA 1MCA correct value 200
= R20,42 ✓CA 1CA simplification
Accept: R20,40
OR / OF
OR / OF
Cost of meat for 1 plate / Koste van vleis vir 1 bord
✓MCA 1MCA dividing value from
R3 233,43 Q 4.1.1
=
200 ✓MCA 1MCA correct value 200
= R16,16715
Cost of salad, relish etc for 1 plate / Koste van slaai, sous ens
vir een bord
R850
=
200
= R4,25 ✓A 1A R4,25
Total cost / Totale koste
= R16,16715 + R4,25
= R20,42 ✓CA 1CA simplification
Accept: R20,40
(4)
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Q/V Solution/Oplossing Explanation/Verduideliking T&L
CA from Question 4.1.2 F
4.1.3 Profit per plate / Wins per bord L3
D
= R35,00 – R20,42 ✓MA 1MA subtracting values
= R14,58 ✓CA 1CA simplification
Number of plates / Getal borde
= R2 850,00 ÷ R14,58 ✓MCA 1MCA dividing by profit
= 195,47 ✓CA 1CA simplification
Total number of plates / Totale getal borde
= 200 + 195,47
= 395,47 ✓CA 1CA total number of plates
= 400 ✓R 1R correct rounding
OR / OF OR / OF
Profit per plate / Wins per bord
✓MA
(200 R35) − R4 083, 43 1MA subtracting values
=
200
R2916,57
=
200
= R14,58 ✓CA
1CA simplification
Number of plates / Aantal borde
R2 850
= ✓MCA
R14,58 1MCA dividing by profit
=195,47 ✓CA
1CA simplification
Total number of plates / Totale aantal borde
= 200 + 195,47
= 395,47 ✓CA 1CA total number of plates
= 400 ✓R 1R correct rounding
(6)
P
4.2.1 15% ✓✓RT 2RT correct value L1
Accept: 14,5 – 15,2 E
(2)
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Q/V Solution/Oplossing Explanation/Verduideliking T&L
* P
4.2.2 Probability / Waarskynlikheid L2
M
4 052 572 ✓A 1A numerator
=
43 378 959 ✓A 1A denominator
= 0,0934… ✓CA 1CA simplification
= 0,093 OR 9,342% ✓R 1R rounding
(4)
D
4.2.3 25 – 34 age group / ouderdomsgroep ✓✓A 2A correct group L4
M
The largest population is in this group / Die grootste
bevolking is in die groep. ✓O 1O reason
(3)
[26]
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QUESTION/VRAAG 5 [31 MARKS/PUNTE]
Q/V Solution/Oplossing Explanation/Verduideliking T&L
* F
5.1.1 2 / TWO / TWEE ✓✓A 2A number of rebates L1
(2) E
CA from Question 5.1.1 F
5.1.2 Tax before rebates / Belasting voor kortings L3
✓RT 1RT correct bracket M
251 258 + 41% of taxable income above 857 900
✓SF
= R251 258 + 41% (R981 500 − R857 900) 1SF substitute R981 500
= R251 258 + 41% (R123 600)
= R251 258 + R50 676
= R301 934 ✓CA 1CA amount before rebates
Tax after rebates / Belasting na kortings
= R301 934 − R17 235 − R9 444 ✓MCA 1MCA subtracting rebates
= R275 255 ✓CA 1CA simplification
Monthly Tax / Maandelikse belasting
R275 255
=
12 ✓MCA 1MCA dividing by 12
= R22 937,92 ✓CA 1CA simplification
(7)
F
5.2.1 Interest rate is the percentage of the total value you have to L1
pay extra for taking the loan. E
Rentekoers is die persentasie van die totale waarde wat jy
ekstra moet betaal vir die uitneem van die lening.
OR / OF ✓✓A 2A correct definition
Interest rate is a percentage charged on the loan taken to buy
a motor vehicle / Rentekoers is 'n persentasie gehef op die
lening uitgeneem om die voertuig te koop.
(2)
F
5.2.2 Difference / Verskil L1
E
= R6 115,47 – R5 498,19 ✓MA 1MA subtracting correct
values
= R617,28 ✓CA 1CA simplification
AO
(2)
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Mathematical Literacy/P1/Wiskundige Geletterdheid/V1 18 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
✓SF 1SF substitution in bracket F
5.2.3 X = (R6 115,47 × 6 × 12) – R300 000 ✓A 1A subtracting R300 000 L2
E
= R440 313,84 – R300 000
= R140 313,84 ✓CA 1CA simplification
AO
(3)
* CA from Question 5.2.3 – F
5.2.4 Balloon payment / Ballonpaaiement R334 000 only L3
✓RT 1RT correct values M
= 20% × R300 000 1A simplification
= R60 000 ✓A
✓MA 1MA correct value × 72
Y = (R5 498,19 × 72) + R60 000 ✓MCA 1MCA adding balloon
payment
= R455 869,68 ✓CA 1CA simplification
(5)
* F
5.2.5 The vehicle serves as security for the loan / L4
✓✓O 2O reason
Die voertuig dien as sekuriteit vir die lening. E
(2)
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Mathematical Literacy/P1/Wiskundige Geletterdheid/V1 19 DBE/May/June/Mei/Junie 2024
SC/NSC/SS/NSS – Marking Guidelines/Nasienriglyne
Q/V Solution/Oplossing Explanation/Verduideliking T&L
D
5.3.1 24 ✓✓RT 2RT correct answer L2
NPU E
(2)
CA from Question 5.3.1 D
5.3.2 Interquartile Range / Interkwartielomvang L4
D
Electrical / Elektries
1RT using correct values
✓RT (28,8 – 29,2)
= 29% − 24% ✓MA
1MA concept of IQR
= 5% ✓CA
1CA simplification(4,8 – 5,2)
Plug-in hybrid / Inprophibried
= 24% − 6%
= 18% ✓CA 1CA simplification
Quarter of Plug-in / Kwart van 'n inprop
1 ✓CA 1CA simplification
= 18% = 4,5% ≠ 5% OR
4 Electric × 4
= 5% × 4 = 20% ≠ 18%
OR / OF
5 1 1
= = ≠ ✓CA
18 3,6 4
OR / OF
5
= 100 = 27,8% ✓CA
18
INVALID / NIE GELDIG NIE ✓O 1O conclusion
(6)
[31]
TOTAL/TOTAAL: 150
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