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NATIONAL
SENIOR CERTIFICATE
GRADE 11
NOVEMBER 2017
MATHEMATICAL LITERACY P2
MARKING GUIDELINE
MARKS: 100
Symbol Explanation
M Method
MA Method with accuracy
CA Consistent accuracy
A Accuracy
C Conversion
S Simplification
RT/RG/RM Reading from a table/Reading from a graph/Reading from a map
F Choosing the correct formula
SF Substitution in a formula
J Justification
P Penalty, e.g. for no units, incorrect rounding off etc.
R Rounding Off/Reason
AO Answer only
NPR No penalty for rounding
This marking guideline consists of 8 pages.
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Mathematical Literacy Grade 11 P2 2017 Memo (Eng)
Grade 11 Mathematical Literacy Paper 2 memorandum from the Eastern Cape November 2017 (English), 8 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 11 · Eastern Cape November · 2017 · English. Memorandum, 8 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 11
- Language
- English
- Document type
- Memorandum
- Year
- 2017
- Exam period
- Eastern Cape November
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 8
- File size
- 939.4 KB
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This exam
Eastern Cape November 2017
Questions in this paper
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2 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017)
QUESTION 1 [29]
Question Solution Explanation Topic
and
Level
1.1 Total number of cars and microbuses = (10 × 5) + (5 × 5) 1 M Calculated total P
= 50 + 25 number of cars and L2
= 75 microbuses
50 1 A Numerator
Probability =
75 1A Denominator
Accept 0,667 OR 66,7% (3)
1.2 Amount for cars washed Monday – Friday: 1M Multiplied by 10 F
10 × 30 = 300 × 5 and 5 L4
= R1 500 1CA Amount for cars
160 1M Increase by 60%
Amount for cars Saturday and Sunday = × 10
100 1CA Number of cars
= 16 1CA Amount for cars
= 16 × 30 × 2 on Saturday and
= R960 Sunday
Amount for microbuses (Monday – Friday)
= 5 × 40 1CA Number of
= R200 microbuses
200 × 5
= R1 000
1CA Number of
Amount for microbuses (Saturday and Sunday): microbuses
130
× 5 = 6,5 =7 1CA Amount for
100 microbuses
7 × 40 × 2
= R560
Total Amount = 1 500 + 1 000 + 960 + 560
1CA Total amount
= R4 020
1 O Claim valid
Claim is valid
(10)
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(EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 3
1.3 Water per car = 25 × 3 CA from 1.1
= 75 litres M
L3
Water for cars in 5 days = 10 × 5 × 75 1M Water for cars
= 3 750 litres Mondays to Fridays
Water for cars on Saturdays and Sundays = 16 × 2 × 75 1CA Water for cars on
= 2 400 litres Saturday and Sunday
Water per microbus = 25 × 4
= 100 litres
1CA Water for
Water per microbus in 5 days = 5 × 5 × 100 microbuses Mondays to
= 2 500 litres Fridays
Water per microbus on Saturdays and Sundays 1CA Microbuses on
= 7 × 2 × 100 Saturdays and Sundays
= 1 400 litres
Total number of litres = 3 750 + 2 400 + 2 500 + 1400 1M Addition
= 10 050 litres 1CA Answer
OR
Amount of water
= amount for cars + amount for microbuses
= (3 × 25 × 82) + (4 × 25 × 39)
= 6 150 litres + 3 900 litres
= 10 050 litres
(6)
1.4 Water for 7 days = 10 050 litres CA from 1.3 M&F
Water for a month = 10 050 × 4 L3
= 40 200 litres 1MA Amount of litres
40 200 l = 40,2 kilolitres 1C Amount of kilolitres
Cost of water from table:
1M First tariff
6 × 8.66 = R51,96 1M Second tariff
9 × 10.02 = R90,18 1M Third tariff
15 × 12,28 = R184,20 1M Fourth tariff
10,2 × 15.25 = R155,55
Total amount = R481,89 1M Calculated VAT
= R481,89 x 1,14 1CA Amount with
= R549,35 VAT
(8)
1.5 Most people do not work on Saturdays and Sundays. 2O Opinion DH
It is the only time they have available to wash their L4
cars.
Accept any other relevant answer. (2)
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4 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017)
QUESTION 2 [24]
2.1 2.1.1 R212 1RT Correct value from DH
table (2) L2
2.1.2 188−76 1M Used correct values F
× 100
76 1M Divided by correct L4
= 147,37% OR 147,4%
value
1CA %
Statement is valid
1J Valid
% increase is more than 100%
NPR (4)
2.1.3 Cost of burying a 5-year-old is less than that of F
burying a 15-year old. L4
Accept any other relevant answers. 2J Opinion (2)
2.2 2.2.1 Diameter = 150 cm 1M Subtraction M
Length of figure = 100 cm 1M Used 60 L4
A = 100 – 60 1CA Value of A
= 40 cm
B = 150 – 100
= 50 cm ÷ 2 1 Divided by 2
= 25 cm 1CA Answer
25 : 40 1M Ratio
=5:8 No penalty if not
simplified
Penalise 1 mark if ratio
is written with units
(6)
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(EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 5
2.2.2 Area of cross shaped glass = Area of longer rectangle
+ 2 × area of smaller rectangles M&F
= l × b + 2(l × b) 1M Substituted in L4
= 100 × 40 + 2(30 ×40) formula
= 4 000 + 2 400
= 6 400 cm2 1CA Simplification
OR
4(40 x 30) + (40 x 40)
= 6 400 cm2
OR
2(100 x 40) – 402
= 8 000 –1 600
= 6 400 cm2
6 400 1C Converted to square
= 0,64 m2
10 000 metres (m²)
Area wasted = Area of circle – Area of shape 1M Used formula
= πr2 – 0,64 1SF Substitution
= 3,142 × 0,75 × 0,75 – 0,64
= 1,767375 – 0,64 1 CA Unused area
= 1,127375m2
Amount lost = 1,127375 × 15 1CA Total amount
= R16,91 1M Added VAT
114
With VAT = × 16,91
100 1CA Answer
= R19,28
1O Opinion (10)
Statement is valid.
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6 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017)
QUESTION 3 [16]
3.1 Company A: DH
L3&4
29 000+25 000+24 000+15 000+15 000+8 000 1M Addition
Mean =
6
1M Divided by 6
116 000
= 6
= R19 333,33 1CA Mean
Company B:
31 000+29 000+17 0000+14 000+13 000+7 000
Mean = 6
111 000
= 6
=R 18 500 1CA Mean
Difference = 19 333,33 – 18 500
= R833,33 1CA Difference
Statement is not valid 1O Not valid (6)
3.2 No modal salary for company B. 2A DH
(2) L2
3.3 Modal value = 15 000 1RT Correct modal value
15 1M Calculated 15% of
× 15 000
100
modal value DH&F
= R2 250
1CA Value L3
105
Year 1 = 100 × 2 250
= R2 362,50 1M Increased by
105 5%
Year 2 = 100 × 2 362,50
1CA First year
= R2 480,63
1CA Final amount
(6)
3.4 0 OR 0% OR Impossible DH
2A Answer (2) L2
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(EC/NOVEMBER 2017) MATHEMATICAL LITERACY P2 7
QUESTION 4 [31]
4.1 4.1.1 A = 21 168 700 + 2 305 800 + 677 000 + 2 214 400 1 MA Added correct DH
= 26 365 900 values L2
1CA Total
22 165 000
B = 27 635 900 × 100
1 Correct values
= 80,2%
1CA % (4)
4.1.2 Indian / Asian 2A Answer DH
(2) L2
4.1.3 Two million two hundred and fourteen thousand four 2A Answer DH
hundred (2) L2
4.2 4.2.1 57 – 38 = 19 1M Correct number of M&P
seats on the left L2
1M Subtraction
1M Correct number of
seats on the right
1CA Answer (4)
4.2.2 4 1RM Correct values M&P
Percentage of seats for the handicapped 38 x 100
numerator and L3
denominator
10,526315789
1M Multiplied by 100
1CA Answer (3)
10,53%
4.3 4.3.1 Total distance = 35 + 57 + 11 + 21 + 59 + 41 1RM Correct values
M,
= 224 km
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
M&P
Speed = 𝑡𝑖𝑚𝑒 1CA Distance
L3&4
224
105 = 𝑡𝑖𝑚𝑒 1M Substituting in
224 formula
Time = 105
= 2,133 hours
1 S Change subject of
Minutes = 0,133 × 60
formula
= 7,98
1CA Time in hours
2 hours 8 minutes
1C Time in minutes
Time of arrival = 7:00 + 2 hours 8 minutes
1CA Arrival time
= 9:08 am
1O Claim valid (8)
Claim is valid
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8 MATHEMATICAL LITERACY P2 (EC/NOVEMBER 2017)
4.3.2 CA from 4.3.1 F
2,82 1M Divided by 1,068 L3
Amount before increase =1,068
1A Amount
= R2,64
I CA Distance
Total distance travelled = 224 x 2
= 448 km
1M Multiplied by
amount calculated
Amount = 448 x 2,64
1CA Amount
= R1 182,72
1R Nearest rand (6)
= R1 183
4.3.3 R61 1A R61 M&P
R349 1A R349 (2) L2
TOTAL: 100
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