NATIONAL
SENIOR CERTIFICATE
GRADE 11
NOVEMBER 2025
MATHEMATICAL LITERACY P2
MARKING GUIDELINES
MARKS: 100
Symbol Explanation
M Method
MA Method with accuracy
CA Consistent accuracy
A Accuracy
C Conversion
S Simplification
RT Reading from a table/graph/diagram
SF Correct substitution in a formula
O Opinion/Explanation/Reasoning
P Penalty, e.g. for no units, incorrect rounding off, etc.
R Rounding Off/Reason
NPR No penalty for correct rounding minimum two decimal places
AO Answer only
MCA Method with consistent accuracy
RCA Rounding with consistent accuracy
This marking guidelines consist of 10 pages.
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Mathematical Literacy Grade 11 P2 2025 Memo (Eng)
Grade 11 Mathematical Literacy Paper 2 memorandum from the Eastern Cape November Exam 2025 (English), 10 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 11 · Eastern Cape November Exam · 2025. Memorandum, 10 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 11
- Document type
- Memorandum
- Year
- 2025
- Exam period
- Eastern Cape November Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 10
- File size
- 736.9 KB
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Eastern Cape November Exam 2025
- Paper 1Question paper · EngMemo · Afr
- Paper 2AddendumMemo
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2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2025)
MARKING GUIDELINES
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.
• If the candidate presents any extra solution when reading from a graph, table, layout plan and
map, then penalize for every extra incorrect item presented.
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(EC/NOVEMBER 2025) MATHEMATICAL LITERACY P2 3
KEY TO TOPIC SYMBOL:
F = Finance; M = Measurement; MP = Maps, plans and other representations; P = Probability
QUESTION 1 [24 MARKS] ANSWER ONLY FULL MARKS
Ques. Solution Explanation Level
1.1.1 Total time = 1,5 minutes + 3 minutes + 15 minutes M 1M adding time M
= 19,5 minutes A 1A total time L1
(2)
1.1.2 Cream cheese =
10 ounces 1C dividing with M
C
35,274 35,274 L1
= 0,28349… 1A cream cheese in kg
≈ 0,28 kg A NPR
(2)
1.1.3 Salt = 1½ × 5 ml M 1M multiplying M
= 7,5 ml A 1A correct answer L1
(2)
1.1.4 5
℃ = 9 × (℉ - 32) M
5 1SF substitution L1
= 9 × (356⁰ - 32) SF 1A temperature in ℃
= 180 ℃ A (2)
1.2.1 Capacity refers to the maximum amount of liquid a 2A correct definition M
container can hold. ✓A (2) L1
1.2.2 Capacity = 500 ml ✓RT 2RT correct capacity M
(2) L1
1.3.1 Circumference of a circle = π × diameter 1SF substitution M
= 3,142 × 13SF 1A circumference L1
= 40,846 m A NPR
(2)
1.3.2 Z:X 1A correct ratio M
16 : 8A If ratio swopped = 1 mark 1S simplification L1
2 : 1 S (2)
1.4.1 Oribi Animal Clinic✓RT 2RT correct clinic MP
(2) L1
1.4.2 Southeast ✓A OR SE ✓A 2A correct direction MP
(2) L1
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4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2025)
1.4.3 Travelling distance = 21,1 km × 1 000 C 1C multiply by 1 000 MP
= 21 100 m A 1A distance in m L1
(2)
1.4.4 Arrival time = 10:15 1M adding time MP
+ :26 M 1A correct time L1
= 10:41 A (2)
[24]
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(EC/NOVEMBER 2025) MATHEMATICAL LITERACY P2 5
QUESTION 2 [27 MARKS]
Ques. Solution Explanation Level
2.1.1 Rober’s Grave ✓RT 2RT correct attraction MP
(2) L2
2.1.2 Probability = 0 ✓A OR None ✓A 2A correct answer P
(2) L2
2.1.3 Distance = 6,4 cm ✓A 2A correct distance MP
(Accept 6,2 cm – 6,6 cm) (2) L1
A CA from 2.1.3 MP
2.1.4 Scale line = 2,6cm (Accept 2,5 – 2,7 cm) 1A scale line in cm L3
2,6 cm 600 m 1M dividing correct
Scale = = M
2,6 2,6
values
= 1 cm : 230,7692308 m CA
1CA scale
Actual distance = 6,4 cm × 230,7692308 MCA
1 476,923077 m 1MCA multiplication
=
1 000
= 1,48 km CA
1CA distance in km
OR
OR
Scale line = 2,6 cm A
1A scale line in cm
Scale = 2,6 cm = 600 m
2,6 60 000 1M dividing correct
= : M
2,6 2,6 values
= 1 : 23 076,92308 CA
1CA scale
Actual distance = 6,4 × 23 076,92308 MCA
= 147 692,3077 cm 1MCA multiplication
147 692,3077
=
100 000
= 1,48 km CA
1CA distance in km
NPR
(5)
2.1.5 No of people =
1 721
M 1M dividing correct MP
25,40 values L1
= 67,7559… 1R number of people
≈ 68 people R
(Accept 67 people) (2)
2.2.1 Mozambique ✓RT OR Swaziland ✓RT 2RT any correct MP
country L1
(2)
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6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2025)
2.2.2 6 national roads ✓RT 2RT no of national MP
roads L2
(2)
2.2.3 Scale = 8,3 cm : 276 km MP
= 276 km × 100 000 C 1C multiplying by L2
= 27 600 000 cm 100 000
Scale =
8,3
:
27 600 000
M 1M dividing distances
8,3 8,3 1CA scale
= 1 : 3 325 301,205 CA NPR
(3)
2.2.4 Speed =
Distance MP
Time
L4
276 km 1SF substitution
80 km/h = SF
Time
Time =
276 𝑘𝑚 1M changing subject
M
80 𝑘𝑚/ℎ of formula
= 3,45 hoursCA 1CA time
= 0,45 × 60 = 27 minutes C 1C converting hrs to
Travel time = 3h27min min
Arrival time = 12:45
+ 03:27 M 1M adding time
= 16:12 CA
⸫ Siphokhazi’s claim is not validO 1CA arrival time
1O opinion
(7)
[27]
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(EC/NOVEMBER 2025) MATHEMATICAL LITERACY P2 7
QUESTION 3 [23 MARKS]
Ques. Solution Explanation Level
3.1.1 50
Radius = = 5 cm C 1C radius in cm M
10 L2
1SF substitution
Volume = 3,142 × 52 × 12 SF 1CA volume in cm3
= 942,6 cm3CA (3)
3.1.2 Width = 0,08 × 1 000 C 1C multiply by 1 000 M
= 80 mm A 1A width in mm L3
1C height in mm
Height = 14,5 × 10 1SF substitution
= 145 mm C
1S simplification
TSA = 2 (150 × 80) + 2 (150 × 145) + 2 (145 × 80)SF 1R rounding
= 24 000 mm2 + 43 500 mm2 + 23 200 mm2
= 90 700 mm2S
≈ 91 000 mm2R (6)
CA from 3.1.2 M
3.1.3 Amount of plastic needed =
91 000
C 1C divide by L3
1 000 000 1 000 000
= 0,091 m2 × 50 M 1M multiply by 50
= 4,55 m2
≈ 5 m2
1M multiply area with
Cost of plastic = 5 m2 × R27,55 M cost
= R137,75 CA 1CA total cost
(4)
3.1.4 1M calculating 7% of M
7 12 L2
Height = 100 × 12 cm = M
1CA answer
0,84 cm CA
1M subtracting 7%
⸫ 12cm – 0,84 cm M
from container height
= 11,16
≈ 11,2 cm
OR
OR
1M subtracting 7%
Height = 100% - 7% M
1CA beans height in
= 93% CA
%
93 1M calculating 93%
⸫ 100 × 12 cm M of 12
= 11,16 cm
≈ 11,2 cm (3)
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8 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2025)
3.1.5 Volume of cylindrical-shaped container = 942,6 cm3 M
L4
Rectangular shaped container: Length = 15 cm C 1C length in cm
Width = 8 cmC 1C width in cm
Height = 14,5 cm
Volume = L × W × H
= 15 cm × 8 cm × 14,5 cm SF 1SF substitution
= 1 740 cm3 CA 1CA volume in cm3
Her statement is incorrect / not valid O 1O opinion
(5)
3.1.6 To keep the coffee beans fresh ✓O 2O reason M
OR L4
To protect container against breakage ✓O
(Accept any other relevant explanation) (2)
[23]
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(EC/NOVEMBER 2025) MATHEMATICAL LITERACY P2 9
QUESTION 4 [26 MARKS]
Ques. Solution Explanation Level
4.1.1 7 1A correct fraction P
Probability = A × 100% M
11 1M multiply by 100% L2
1CA probability as a
= 63,64% CA %
(3)
4.1.2 Travelling time = 10:15 MP
- 08:30 M 1M subtracting time L4
= 01:45
⸫ 0,45 ÷ 60 = 0,75 h C 1C minutes to hours
Time = 1 h + 0,75 h = 1,75 hoursCA 1CA traveling time in
hours
Distance = Speed × Time
= 1,4 miles/h × 1,75 h SF 1SF substitution
= 2,45 miles 1C multiply with
⸫ 2,45 miles × 1,60934C conversion factor
= 3,942883 kmCA 1CA distance in km
Her statement is invalid O 1O opinion
OR OR
Travelling time = 10:15 1M subtracting time
- 08:30 M
= 01:45 1C minutes to hours
⸫ 0,45 ÷ 60 = 0,75 h C 1CA traveling time in
Time = 1 h + 0,75 h = 1,75 hoursCA hours
Speed = 1,4 miles/h × 1,60934 C 1C convert speed to
= 2,253076km/h km/h
Distance = Speed × Time 1SF substitution
= 2,253076km/h × 1,75 h SF 1CA distance in km
= 3,942883 kmCA 1O opinion
Her statement is invalid O NPR
(7)
4.2.1 Area of square = side × side 1SF substitution M
= 45 cm × 45 cm SF 1A area of box L2
= 2 025 cm2A
≈ 2 030 cm2 R 1R rounding
(3)
4.2.2 Value of D: 1M subtraction M
2D = 45 cm – 42 cm M 1M divide by 2 L2
= 3 cm 1CA value of D
3 𝑐𝑚 M
D=
2
= 1,5 cm CA (3)
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10 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2025)
4.2.3 Area of pizza (circle) = 3,142 × 212 A SF 1A radius M
2
= 1 385,622 cm 1SF substitution L3
Area of one slice =
1 385,622
M 1M dividing area by
8 8
= 173,20 cm2CA
1CA area of one
slice
(4)
4.3.1 Diameter = 160 cm + 35 cm + 35 cm M 1M adding correct M
= 230 cm values L2
230
Radius = M
2 1M divide diameter
= 115 cm CA by 2
1CA radius
(3)
4.3.2 Height of support pole = 78 cmM – (2,8 cm + 0,7 cm) M 1M subtraction M
= 74,5 cm CA 1M adding base and L2
thickness
1CA height in cm
(3)
[26]
TOTAL: 100
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