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NATIONAL
SENIOR CERTIFICATE
GRADE 12
SEPTEMBER 2020
MATHEMATICAL LITERACY P2
MARKING GUIDELINE
MARKS: 150
Symbol Explanation
M Method
M/A Method with accuracy
MCA Method with consistent accuracy
CA Consistent accuracy
A Accuracy
C Conversion
S Simplification
RT/RG/RM Reading from a table OR Reading from a graph OR Read from 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 OR Reason
AO Answer only
NPR No penalty for rounding
This marking guideline consists of 11 pages.
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Mathematical Literacy Grade 12 P2 2020 Memo (Eng)
Grade 12 Mathematical Literacy Paper 2 memorandum from the Eastern Cape Mock Exam 2020 (English), 11 pages. Read online or download the free PDF.
Mathematical Literacy · Grade 12 · Eastern Cape Mock Exam · 2020 · English. Memorandum, 11 pages. Read online or download the PDF.
- Subject
- Mathematical Literacy
- Grade
- Grade 12
- Language
- English
- Document type
- Memorandum
- Year
- 2020
- Exam period
- Eastern Cape Mock Exam
- Paper
- 2
- Curriculum
- CAPS
- Pages
- 11
- File size
- 475.0 KB
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2 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020)
QUESTION 1 [37]
Ques. Solution Explanation Level
1.1.1 Amount used for 3 batches = 3 × 125 MA 1MA Multiplying M
= 375 mℓ correct values L2
375 1MA Dividing by
Number of cups =
250
MA
250
= 1,5 cups OR 1½ cups
A 1A Number of cups
(3)
14,99
1.1.2 Price of 3 eggs = M 1M Dividing by 12 F
12
1M Multiply by 3 L3
= R1,249166667 × 3 M
1S Simplification
= R3,7475 S
1CA Cost
≈ R3,75 CA
(4)
1.1.3 °Fahrenheit = 1,8 × °Celsius + 32° 1SF Substitute M
= 1,8 × 180° + 32°SF correct °C L2
= 356°F – 330°F M 1M Subtract correct
= 26°F values
(2)
1.1.4 Time taken for 9 batches = 25 min + 45 min 1MA Total time for M
MA preparation and L4
= 70 min × 9 baking
= 630 min CA 1CA Total time for
Time in hours and minutes = 10 hours 30 minC 9 batches
Time completed = 09 hours 15 min + 10 hours 30 min M 1C Convert to hours
= 19:45 CA and minutes
Not valid O 1M Adding time
1CA Time
completed
1O Not valid
(6)
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(EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 3
1.1.5 3 1C mℓ to gram F&M
Cake flour = × 250 = 150 grams
5
21,99 M 1M Dividing L4
Cost of cake flour in grams = 1C kg to g
2500 C MCA
= 0,008796 × 150 × 9 1MCA Multiply by
= R11,87 CA 150 and 9
35,99
Cost of oil in millilitres = 1CA Cost of flour
2000
= 0,017995 × 125 × 9 1CA Cost of oil
= R20,24 CA
14,99 1CA Cost of eggs
Cost of eggs = CA from 1.1.2
12
= 1,249166667 × 3 × 9 1CA Total cost
= R33,73CA
Total cost = R11,87 + R20,24 + R33,73 1O Invalid
= R65,84 CA
Invalid O
OR
1C mℓ to gram
1M Dividing
3
Cake flour = × 250 = 150 grams C 1C g to kg
5
21,99
M 1M Multiply by
Cost for cake flour in kg =
2,5 C 0,15 and 9
= 8,796 × 0,15 × 9 M 1CA Cost of flour
= R11,87 CA
35,99
Cost of oil in litre = 1CA Cost of oil
2
= 17,995 × 0,125 × 9
= R20,24 CA
14,99
Cost of eggs = 1CA Cost of eggs
12
= 1,249166667 × 3 × 9 CA from 1.1.2
= R33,73 CA 1CA Total cost
Total cost = R11,87 + R20,24 + R33,73 1O Invalid
= R65,84 CA
Invalid O (9)
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4 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020)
1.2.1 7 feet 8 inches = (7 × 0,3048) + (8 × 0,0254) C 1C cm to m M
= 2,1336 m + 0,2032 m S 1M Multiplying and L3
= 2,3368 m CA adding
1S Simplification
6 feet 6 inches = (6 × 0,3048) + (6 × 0,0254) 1CA Answer
= 1,8288 m + 0,1524 m
= 1,9812 m CA 1CA Answer
Length = 2,3368 m + 1,9812 m 1CA Length
= 4,318 m CA
OR 1M Adding
Total feet in metres = 7 feet + 6 feet M 1M Multiply
= 13 feet × 0,3048 M 1CA Answer
= 3,9624 m CA
Total inches in metres = 8 inches + 6 inches
= 14 inches × 0,0254 C 1C cm to m
= 0,3556 m CA 1CA Answer
Length = 3,9624 m + 0,3556 m 1CA Length
= 4,318 m CA (6)
1.2.2 Top view A 1A View M&
All features clearly visible R 2A Reason P
(3) L4
1.2.3 People preparing meals in the kitchen. A 2A First reason M&
People helping to prepare meals. A 2A Second reason P
People in and out of the kitchen. A L4
It is frequently visited by all. A
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(EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 5
QUESTION 2 [39]
Ques. Solution Explanation Level
2.1.1 Simple Interest A 1SF Substitution F
Interest = 280 000 × 0,0975 × 2,916666667 SF 1A Number of years L2
= R79 625 CA 1CA Interest
(3)
2.1.2 Compound Interest F
First year = 280 000 + (280 000 × 0,0825) M 1M Adding and L3(6)
= R303 100 CA Multiplying L4(1)
Second year = 303 100 + (303100 × 0,0825) 1CA Amt 1st year
= R328 105,75 CA A 1CA Amt 2nd year
11 months = R328 105,75 + (R328 105,75 × 0,075625) 1A Interest rate for
= R352 918,75 CA 11 months
OR 1CA Total amount
Amount after 35 months
= 280 000 × 1,0825 × 1,0825 × 1,075625
= R352 918,75
1CA Interest
Interest = R352 918,75 – R280 000
= R72 918, 75 CA
1O Option 2
She will pay less interest on Option 2 (compound interest) CA from 2.1.1
(7)
2.2.1 Factory 1 M D
IQR = 75 000 – 40 000 RD 1RD Q1 and Q3 L3(5)
= 35 000 CA 1M Concept of IQR L4(3)
Median = 60 000 1CA IQR
Factory 2
IQR = 80 000 – 35 000 RD 1RD Q1 and Q3
= 50 000 CA 1CA IQR
Median = 50 000
A O
Factory 2 performed worse because their IQR is greater 1A Factory 2
and their median is lower O 1O Comparing
median
1O Comparing
IQR’s
(8)
2.2.2 No, factory 2 is still new A CA from 2.2.1 D
2A No with reason L4
(2)
94 MA
2.3.1 Percentage = × 100% M 1MA Dividing M
150
correct values L2
= 62,7%
1M Multiply by 100
(2)
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6 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020)
2.3.2 310 𝑚𝑚 M 1A Correct M
Over the length =
78 𝑚𝑚 A
diameter L4
= 3,97…
1M Divide correct
≈ 3 coffee mugs A
220 𝑚𝑚 values
Over the width = 1A Number of
78 𝑚𝑚
= 2,8… mugs (length)
≈ 2 coffee mugs CA 1CA Number of
150 𝑚𝑚
Over the height = mugs (width)
94 𝑚𝑚
= 1,6… 1CA Number of
≈ 1 coffee mug CA mugs (height)
Number of mugs in 1 box = 3 × 2 × 1
= 6 coffee mugs CA 1CA Total
66
Number of boxes = 1CA Number of
6
= 11 boxes CA boxes
Statement invalid O 1O Invalid
(8)
2.4.1 Northeast A 2A Northeast M&P
North A 2A North (4) L2
Probability = A
2.4.2 1 1A Numerator P
4 A
1A Denominator L2
(2)
2.4.3 Turn left on the N6 in East London to Queenstown A 1A Left on N6 M&P
Turn right on the R56 to Kokstad A 1A Right on R56 L4
Turn left on the N2 A 1A Left on N2 (3)
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(EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 7
QUESTION 3 [36]
Ques. Solution A Explanation Level
3.1.1 Initial tax = 0,18 × 195A
850 1A Use correct % F
= R35 253 1A Use correct L2
amount
(2)
3.1.2 7,5 1MA 7,5% of F
Pension = × 37 537,75 MA
100
correct value L3
= R2 815,33125 × 12 MA
1MA Multiply by
= R33 783,98CA
12
1CA Pension
Donation = 575 × 12
amount
= R6 900 A
1A Donation
amount
Total = R33 783,98 + R6 900
1CA Total amount
= R40 683,98 CA
(5)
MA MCA
3.1.3 Taxable income = (37 537,75 × 12) – R40 683,98 CA from 3.1.2 F
= R450 453 – R40 683,98 1M Multiply by 12 L2
= R409 769,02 CA 1MCA Subtract
pension and
donations
1CA Taxable
income
(3)
A
3.1.4 Tax payable = 63 853 + 31% of taxable income above 305 850 CA from 3.1.3 F
MCA 1A Correct tax L3&4
= 63 853 + 0,31 × (409 769,02 – 305 850) bracket
= 63 853 + 0,31 × 103 919,02 1MCA Amount
= 63 853 + 32 214,8962 above
S M 1S Simplification
= R96 067,8962 – R14 220 1M Subtract rebate
81 847,8962
= M 1M Divide by 12
12
1CA Monthly tax
= R6 820,66 CA
1O Invalid or Less
Invalid OR less than R6 850 O
than
(7)
3.1.5 They receive 3 rebates A 2A Explanation F
OR L4
Their rebate is higher A (2)
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8 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020)
3.1.6 A
37 537,75 1A Correct gross F
Gross monthly salary in 2018/2019 =
1,064 MA
salary V2
= R35 279,84A 1MA Divide by
1,064
OR 1A Gross salary
A CA from 3.1.3
450 453
Gross monthly salary in 2018 = 1A Correct gross
1,064 MA
= R35 279,84A salary
1MA Divide by
1,064
1A Gross salary
(3)
3.2.1 Value of A = 90 – (8 + 13 + 30 + 15 + 10) 1MA Subtract from D
= 90 – 76 MA 90 L2
14 1M Divide by 2
= M
2
1CA Value of A
= 7 CA
(3)
D
3.2.2 16 years A 2A Age (2) L2
3.2.3 Average age MCA CA from 3.2.1 D
(13 × 8) + (14 × 7)+ (15 × 13)+(16 × 30)+(17 × 15)+(18 × 7)+(19× 10) 1MCA Adding L3
=
90 correct values
104 + 98 +195 + 480 + 255 + 126 + 190
= 1M Dividing
90
1 448 M 1CA Average age
=
90
1O Not valid
= 16,088… years CA
(4)
Statement invalid O
3.2.4 CA from 3.2.1 P
Number of boys = 13 + 30 + 15 + 7 1MCA Numerator L2
1CA Denominator
65 MCA 1R 3 decimal
Probability =
90 A
places
= 0,722 R
(3)
D
3.2.5 The weight of the boys should also be taken into account.A 2A Reason (2) L2
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(EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 9
QUESTION 4 [38]
Ques. Solution Explanation Level
4.1.1 14,202957 A R 1A Correct value D
The lower the value, the stronger the Rand 1R Reason L2&4
(2)
4.1.2 Amount after exchange fee = 40 830 – (40 830 × 0,045) F
= 40 830 – 1837,35 MA 1MA Subtract 4,5% L3&4
= R38 992,65 CA 1CA Value
38 992,65 1MCA Dividing
At 14,983385 = MCA
14,983385
correct values
= $2 602,39 CA 1CA Dollar value
38 992,65
At 14,398064 =
14,398064
= $2 708,18 CA 1CA Dollar value
Difference = $2 708,18 – $2 602,39
= $105,79 CA 1MCA Difference
Statement invalid O 1O Invalid
OR
40 830 1MA Dividing
14,983385 MA
At 14,983385 =
CA correct values
= $2 725,018412 – (2 725,018412 × 0,045) 1CA Value
= 2 725,018412 – 122,6258285 MCA 1MCA Subtract
= $2 602,39 MCA 4,5%
40 830
1CA Dollar Value
At 14,398064 =
14,398064
= $2 835,797924 – (2 835,797924 × 0,045)
= 2 835,797924 – (127,6109066)
= $2 708,18 CA 1CA Dollar value
Difference = $2 708,18 – $2 602,39
= $105,79MCA 1MCA Difference
Statement invalid O 1O Invalid (7)
4.1.3 Political factors A 2A 1st reason D
OR 2A 2nd reason L4
Economic factors A
OR
Supply and demand of countries A
OR
Inflation rate A
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10 MATHEMATICAL LITERACY P2 (EC/SEPTEMBER 2020)
Accept any other relevant responses
(4)
4.2.1 Enlargement: M&P
Northern Line (A) = 7,2 cm A (Accept 7,2 cm – 7,4 cm) 1A Measure A L4
Scale is 4,2A
cm = 300 km (Accept 4,1 cm – 4,4cm) 1A Measure scale
7,2 𝑐𝑚 1M Ratio
Distance = × 300 km M
4,2 𝑐𝑚
1CA Kilometres
= 514,286 km CA
Map:
Northern Line (A) = 1,5 cm A (Accept 1,4 cm – 1,6 cm)
A 1A Measure line
Scale is 4,2 cm = 1 500 km (Accept 4,1 cm – 4,4 cm)
1,5 𝑐𝑚 1A Measure scale
Distance = × 1 500 km
4,2 𝑐𝑚
= 535,714 km CA 1CA Kilometres
(7)
4.2.2 Traveller 1 and 2 = $670,36 × 2 1MA Amount for 2 F
= $1 340,72 MA travellers L3
Traveller 3 and 4 = 0,239 × 670,36 M 1M 23,9% of 670,36
= $160,22 × 2 1CA Amount
= $320,44 CA 1MA Tax amount
Tax Amount = $188,64 × 4 1MCA Adding
= $754,56 MA amount
Total amount = $1 340,72 + $320,44 + $754,56MCA 1CA Total amount
= $2 415,72 CA (6)
4.3.1 Range = Highest value – Lowest value 1M Concept of range D
34°F = A - 54°F M 1CA Value of A L2
A = 54°F + 34°F (2)
= 88°F CA
5
4.3.2 Probability = A 1A Numerator D
12 A
1A Denominator L2
= 0,416… × 100
= 41,6…
1CA Nearest %
≈ 42% CA
(3)
4.3.3 °Celsius = (°F – 32) ÷ 1,8 M
= (34°F – 32) ÷ 1,8SF 1SF Substitution L2&4
= 2 ÷ 1,8 S 1S Simplification
= 1,1…°C CA 1CA °C
Statement not valid O 1O Not valid
(4)
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(EC/SEPTEMBER 2020) MATHEMATICAL LITERACY P2 11
4.3.4 1A Jan – Apr D
Average Minimum and Maximum 1CA May – L2
temperatures in North Carolina in 2018 Aug from 4.3.1
100 1A Sept – Dec
A
Temperature in °Fahrentheit
80 A
60
A
40
20
0
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Months
(3)
TOTAL: 150
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