Downloaded from hlayiso.com
GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATION
2018
MARKING GUIDELINES
MATHEMATICAL LITERACY PAPER 2 (10602)
Symbol Explanation
M Method
MA Method with accuracy
CA Consistent accuracy
MCA Method with constant accuracy
A Accuracy
C Conversion
S Simplification
RT / RG / Reading from a table / graph / diagram
RD
SF Correct substitution in a formula
O Opinion / Example / Definition / Explanation
R Rounding off
NPR No penalty rounding or omitting units
AO Answer only, full marks
J Justify
KEY TO TOPIC SYMBOL:
F = Finance; M = Measure ment; MP = Maps, Plans and other
representations; DH = Data Handling; P = Probability
10 pages
You're offline
Skip to contentMemorandum 

%20June%202025%20Possible%20Answers--b592d0c3-9486-4357-be22-a8e995043cee/v1-cdd6cb1fcc7cfc379b4b/card.webp)


%20June%202024%20Possible%20Answers_hlayiso.com_--48d5582a-947d-4054-847b-d7bac4525b7d/v1-e67cb11e40db49b9008f/card.webp)


%20Preparatory%202017%20Possible%20Answers_hlayiso.com_--f2fdc5eb-8de9-4da5-b583-f3e5088fce88/v1-a361eac94c1f3222a6a9/card.webp)

%20Preparatory%202018%20Possible%20Answer_hlayiso.com_--b82b5337-d583-4576-8235-f8c6a3fe1432/v1-e0fc9a9293c5b8d6ade2/card.webp)

View all





Grade 12 NSC Math Lit P2 Preparatory 2018 Possible Answer hlayiso.com
Mathematical Literacy · Grade 12 · Gauteng Mock Exam · 2018. Memorandum, 10 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 12
- Document type
- Memorandum
- Year
- 2018
- Exam period
- Gauteng Mock Exam
- Paper
- 2
- Pages
- 10
- File size
- 371.8 KB
Loading document…
Loading document…
1 of 10
Document textSearch extracted text and jump to a page.
Downloaded from hlayiso.com
10602/18
GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATION
MATHEMATICAL LITERACY
(Paper 2)
MARKING GUIDELINES
Q1 ANSWER EXPLANATION LEVEL
1.1.1 Monthly taxable income 1 SF substituting
Monthly salary = R33 412,65
0,925
= R33 412,65 ÷ 0,925 1 A answer F3
1 MCA multiply by 12 [4]
= R36 121,78
1 CA answer
Annual salary = R36 121,78 x 12
= R433 461, 36
1.1.2 He is under 65 years old and earning an income of more 1 O explanation
than R75 750; therefore, he qualifies to pay tax. 1 J justification
F4
[2]
1.1.3 Taxable income before rebates 1 RT correct tax bracket
= R61 910 + 31% (R400 951,80 – R296 540) 1 SF substituting
= R94 277,66 R400 951,80
1 A answer
Medical credit = R303 x 2 x 12
= R7272 1 M multiplying by 2 and
12
Payable tax = R94 277,66 – R7272 – R13500 1 CA answer
= R73 505,66
1 M subtracting rebates
Monthly tax = R73 505,66 ÷ 12 and medical credits F4
= R6 125,47 1 CA answer [10]
1 MCA dividing by 12
Therefore, his claim is invalid. 1 CA answer
1 J justification
2
Downloaded from hlayiso.com
10602/18
Q1 ANSWER EXPLANATION LEVEL
1.2.1 Cost of wages = 30% of R321 021,56 1 M multiplying by 30% F2
(a) = R96 306,47 1 A answer [2]
(b) Cost of wages for each employee = R96 306,47 ÷ 4 1 MCA dividing by 4 F3
= R24 076,62 1 A answer [4]
Monthly cost = R24 076,62 ÷ 3 1 M dividing by 3
= R8025,54 1 A answer
OR
3 x 4 = 12 1 M multiplying 3 by 4
Cost of wages for each employee = R96 306,47 ÷ 12 1 A answer
= R8025,54 1 MCA dividing by 12
1 A answer
1.2.2 Cost of materials (VAT exclusive) 1 M subtracting F4
= R321 021,56 – R96 306,47 1 A answer [6]
= R224 715,09
Cost of materials (VAT inclusive) 2 M multiplying by 114%
= R224 715,09 × 114% 1 A correct values
= R256 175,20 1 J justification
His claim is valid
1.2.3 Institution A: 10% ÷ 2 = 5% 1 A for 5% F4
1 M multiplying by 105% [10]
Term 1: 105% of R210 050 = R220 552,50 1 A answer
1 CA answer
Term 2: 105% of R220 552,50 = R231 580,13
Institution B: 8% ÷ 4 = 2% 1 A for 2%
Term 1 : 102% of R210 050 = R214 251 1 A answer
Term 2 : 102% of R214 251 = R218 536,02 1 CA answer
Term 3: 102% of R218 536,02 = R222 906,74 1 CA answer
Term 4: 102% of R222 906,74 = R227 364,88 1 CA answer
True, he will repay less for a loan from institution B. 1 J opinion
[38]
3
Downloaded from hlayiso.com
10602/18
Q2 ANSWER EXPLANATION LEVEL
2.1. Bedrooms 3 and 4 1 A correct answer MP2
Because they are on the northern side of the house. 2 J justification [3]
2.2 M2
Area = + 1 SF substitution of [4]
= + correct values
= 21 + 12 1 A value of 3
= 33 1 A value of 4
1 CA answer
OR
Area = + 1 SF substitution of
= + correct values
= 24 +9 1 A value of 4
= 33 1 A value of 3
1 CA answer
OR
Area = (7m × 6m) – (3m × 3m)
= 42 -9
= 33
2.3 Number of boxes = CA from 2.2
= 16,6
1 R answer F3
17 boxes
1 M for multiplication [5]
1 A answer
Cost for tiling = R189,99 17
1 A answer for extra
= R3 229,83
boxes
1 CA final answer
Cost for two extra boxes of tiles
=R189,99 × 2
= R379,98
Total cost = R3 229,83 + R379,98
1 R for number of boxes
= R3 609,81
1 A for extra boxes
1 A answer
OR
1 M for multiplication
Number of boxes = 17 + 2 = 19
1 CA answer
Cost = 19 × R189, 99 = R3 609,81
2.4 3days : 8,40 1 M for the proportion M3
x : 73 [4]
1 M for multiplying
correctly
1 CA answer
26,0714285
1 R for rounding
≈ 4 weeks
[16]
4
Downloaded from hlayiso.com
10602/18
Q3 ANSWER EXPLANATION LEVEL
3.1.1 Surface Area = 2(ℓ× h) + 2(w × h) + 2(ℓ × w) 1 C for 0,86 M2
= 2(1,1 × 2) + 2 (0,86 × 2) + 2 (1,1 × 0,86) 1 SF for correct [3]
= 9,73m2 substitution
1 A answer
3.1.2 Radius = 0,65 1 A value of radius M4
2
Volume of a cylindrical geyser = πr h 1 SF for correct [6]
= 3,142 (0,65)2 × 2 substitution
= 2,65m3 1 A answer
Volume of a rectangular cylinder = ℓ x w x h 1 SF correct
=1,1 x 0,86 × 2 substitution
= 1,89m3 1 A answer
True, cylindrical geyser is bigger / has a greater capacity 1 J justification
NPR
3.1.3 Radius = 65 cm and height = 200 cm 1 C both r and h M3
[6]
Volume = 3,142 × (65 cm)2 × 200 cm 1 SF substitution of
= 265 4990 cm3 correct values
1 A answer
Therefore: 95% of 265 4990 cm3 = 252 2240,5 cm3
252 2240,5 cm3 ÷ 1000 = 2 522,2405ℓ 1 CA answer
≈ 2 522 ℓ 1 C conversion to
litres
1 R rounding of litres
3.2.1 Cost for 39 kℓ without levy F3
1 A for every 2 correct [6]
= (R0,00 × 6) + (R7,24 × 4) + (R11,84 × 5) + (R17,46 × 5) answers (3A)
+ (R23,34 × 10) + (R25,10 × 9) 1 CA answer
= R0,00 + R28,98 + R59,20 + R87,30 + R233,40 +
R225,90
= R634,78 1 CA answer
Water levy = 10% of R634,78
= R63,48 1 CA answer
Cost for 39 kℓ including levy
= R634,78 + R63,48
= R698,26
3.2.2 2 ÷ 5 of R698,26= R279,30 1 A for correct F4
fraction / proportion [4]
True, it is less than R300 1 M multiplying
correct values
1 A answer
1 J justification
5
Downloaded from hlayiso.com
10602/18
Q3 ANSWER EXPLANATION LEVEL
3.2.3 1 A for each correct P3
B branch (3A) [7]
B G
B
G
B G
B
Children B
G G
B
G
G
2 A for outcomes
Outcomes: 1 A probability
1 CA percentage
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG
The probability of having a boy as the first born and to have
a girl and boy in any order is 2 = 25%
8
[32]
6
Downloaded from hlayiso.com
10602/18
Q4 ANSWER EXPLANATION LEVEL
4.1.1 Free standing house 1 A correct answer D4
The average price of building a free standing house in RSA 2 O opinion [3]
is cheaper than building a townhouse.
4.1.2 Q 1 = (R4 698 + R4 609) ÷ 2 1 RG correct values D3
= R4 653,50 1 M adding values and [7]
dividing by 2
Q 3 = (R6 136 + R6 102) ÷ 2 1 A correct answer
= R6 119
1 A correct answer
IQR = Q 3 – Q1
= R6 119 – R4 653,50 1 A correct formula
= R1 465,50 1 SF correct values
1 CA answer
4.1.3 The type of landscape, dominated by rocky 1 O for reason / opinion MP4
lands. 1 O for reason / opinion [2]
Ocean, it is difficult to build in a sandy soil.
Tourist attraction.
More reinforcement is needed.
Accept any valid answer
4.1.4 An average is calculated by adding all the real values 2 J for reason D2
divided by the total number of plans. [2]
You cannot calculate an average by adding averages and
dividing by the number of provinces.
Accept any valid answer
7
Downloaded from hlayiso.com
10602/18
Q4 ANSWER EXPLANATION LEVEL
4.1.5 D2
1 A correct type [5]
Cost versus the price of houses 1 A for FS and NW
1 A for MP, LP and
EC
FS
1 A for NC and GP
1 A for WC and KZN
NW
MP
LP
EC
NC
GP
WC
KZN
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
COST
COSTFOR
FORTOWN HOUSES
TOWNHOUSES COST FOR FREE STANDING HOUSES
4.2.1 Number of full-time employees = 25% × 40 1 RT for 25% D2
= 10 employees 1 M for multiplying [3]
by 40
1 A correct answer
4.2.2 Number of part-time employees = 75% x 120 1 RT for 75% D2
= 90 employees 1 M for multiplying [3]
by 120
1 A correct answer
4.2.3 Median, because it shows salary paid to half of the 1 A correct answer D4
employees. 2 J explanation [3]
4.2.4 Range = Max value – Min value 1 A correct formula D2
= R14 000 – R6 000 1 SF correct values [3]
= R8 000 1 A answer
8
Downloaded from hlayiso.com
10602/18
Q4 ANSWER EXPLANATION LEVEL
4.2.5 It only focuses on outliers; few employees might fall in this 2 O explanation D2
category. [2]
It is only based on the difference between the highest salary
and the lowest salary.
4.2.6 50% of part time employees earn a salary of more than 2 A referring to D4
R11 000, whereas 50% of the full time employees earn a minimum, median or [4]
salary of more than R10 000. maximum
2 J conclusion
Therefore, part time employees are paid better salaries.
OR
The minimum and maximum salaries of full-time employees
are higher than those of part-time employees
Therefore Pule is correct
4.2.7 Full time employment 1 A correct choice D4
The least paid employee earns R7 000,00 2 O Explanation [3]
[40]
9
Downloaded from hlayiso.com
10602/18
Q5 ANSWER EXPLANATION LEVEL
5.1 Off-ramp to N12 and join N8 to Bloemfontein. 5 A for describing MP2
Off-ramp to N3 and join N5 after Harrismith, then join routes that pass [5]
N1 to Bloemfontein. through a national
road from
Johannesburg to
Bloemfontein.
5.2 Measured distance: 4,3 cm 1 A for 4,3 cm MP3
Bar: 2,2 cm (Accept: 4,1 - 4,5 cm) [7]
1 A for 2,2 cm
Actual distance = (4,3 cm ÷ 2,2 cm) × 300 km (Accept: 2,1 - 2,3 cm)
= 586 km
1 MA for division
Return distance = 586 km × 2 1 A for using 300 km
= 1172 km 1 A for answer
OR 1 MCA for
multiplying by 2
2,2 cm : 300 km 1 CA answer
2,2 cm : 30 000 000 km
1 : 13 636 364 Note: measure on
4,3 cm × 13 636 364 = 58 636 365 cm final copy
58 636 365 cm ÷ 100 000 = 586 km
586 km × 2 = 1172 km
5.3 Time taken = 15h15 min - 10h45 1 M for subtraction M3
= 4 hours and 30 min 1 A for answer [6]
= 4,5 hours 1 C for conversion
1 M Method
1 CA answer
= 130,22 km / h
1 R correct rounding
= 130 km / h
5.4 The general speed limits in terms of the South African 2 O Opinion MP2
National Road Traffic Act are: 60 km / h on a public road [2]
within an urban area; 100 km / h on a public road outside an Accept any sensible
urban area which is not a freeway; 120 km / h on freeways. explanation
5.5 It does not give enough detail. 4 O Opinion MP2
It does not have distances between cities and towns. [4]
Accept any sensible
explanation
[24]
TOTAL 150
10
Recommended for this subject
Published documents with matching subject and grade metadata.

Memorandum
MATHS LIT P1 GR12 MEMO JUNE 2025 AFRIKAANS FINAL

Memorandum
Maths LIT Grade 12 NSC P1 MEMO September 2025 Free State
%20June%202025%20Possible%20Answers--b592d0c3-9486-4357-be22-a8e995043cee/v1-cdd6cb1fcc7cfc379b4b/card.webp)
Memorandum
Mathematical Literacy P1 (English) June 2025 Possible Answers

Memorandum
Maths LIT Grade 12 NSC P1 MEMO September 2025 Limpopo

Memorandum
MATHS LIT P2 GR12 MEMO JUNE 2024 English hlayiso.com
%20June%202024%20Possible%20Answers_hlayiso.com_--48d5582a-947d-4054-847b-d7bac4525b7d/v1-e67cb11e40db49b9008f/card.webp)
Memorandum
Gr 12 Mathematical Literacy P1 (English) June 2024 Possible Answers hlayiso.com
Related documents
Matched using subject, grade, language, document type and exam metadata.

Memorandum
Grade 12 NSC Maths Literacy P1 Preparatory 2018 Possible Answer hlayiso.com

Memorandum
Mathematical Literacy P2 Feb March 2018 Memo Eng hlayiso.com
%20Preparatory%202017%20Possible%20Answers_hlayiso.com_--f2fdc5eb-8de9-4da5-b583-f3e5088fce88/v1-a361eac94c1f3222a6a9/card.webp)
Memorandum
Grade 12 NSC Mathematics Literacy P2 (Afrikaans) Preparatory 2017 Possible Answers hlayiso.com

Addendum
Grade 12 NSC Maths Literacy Addendum X5 Preparatory 2018 Possible Answer hlayiso.com
%20Preparatory%202018%20Possible%20Answer_hlayiso.com_--b82b5337-d583-4576-8235-f8c6a3fe1432/v1-e0fc9a9293c5b8d6ade2/card.webp)
Addendum
Grade 12 NSC Maths Literacy Addendum X5 (Afrikaans) Preparatory 2018 Possible Answer hlayiso.com

Question paper
Mathematical Literacy P2 Feb March 2018 Afr hlayiso.com
More from Grade 12 Mathematical Literacy
Explore more published documents in this catalogue.

Addendum
MATHS LIT P2 GR 12 SEPT 2025 ADDENDUM AFR

Question paper
Maths LIT Grade 12 NSC P1 QP September 2025 Free State

Addendum
Maths LIT Grade 12 NSC P1 ANSWER BOOK September 2025 KZN

Question paper
Maths LIT Grade 12 NSC P1 QP September 2025 Limpopo

Question paper
Maths LIT Grade 12 NSC P2 QP September 2025 Limpopo

Question paper