NATIONAL SENIOR CERTIFICATE GRADE 11 - Mindset Learn
嚜燒ATIONAL
SENIOR CERTIFICATE
GRADE 11
MATHEMATICAL LITERACY P1
EXEMPLAR 2013
MEMORANDUM
MARKS: 100
SYMBOL
M
MA
CA
A
C
S
RT /RG
F
SF
O
P
R
EXPLANATION
Method
Method with accuracy
Consistent accuracy
Accuracy
Conversion
Simplification
Reading from a table/Reading from a graph
Choosing the correct formula
Substitution in a formula
Opinion
Penalty, e.g. for no units, incorrect rounding off etc.
Rounding off/Reason
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Mathematical Literacy/P1
2
NSC 每 Grade 11 Exemplar 每 Memorandum
DBE/2013
QUESTION 1 [19 marks]
Ques
Solution
Explanation
Topic
1.1.1
2,35 ? = 2,35 ℅ 1 000 m ?
= 2 350 m ? ?A
1A answer
M
L1
1.1.2
1.2.1
R 19,99
3 kg ?M
= R6,66 ?A
Cost per kg =
?SF
R 53,75 ? R 43,00
Percentage mark-up =
℅ 100%
R 43,00
R10,75
=
℅ 100% ?S
R 43
= 25% ?CA
(1)
1M dividing by 3 kg
F
L1
1A answer
(2)
F
L2
1SF substitution
1S simplification
1CA answer
(3)
1.2.2
1.2.3
1.2.4
R2 000 ?M
R54
= 37,04 ?A
Number of kilograms =
A = R76,00 + 30% of R76,00 ?M
= R76,00 + R22,80 ?A
= R98,80 ?CA
OR
?A
A = 1,3 ℅ R76,00 ?M
= R98,80 ?CA
?M
?M
Cost = 1,2 kg ℅ R53,75 + 0,5 kg ℅ R85,00
= R64,50 + R42,50 ?S
= R107,00 ?CA
F
L1
1M dividing
1A answer
(2)
F
L2
1M adding 30%
1A mark up
1CA answer
(3)
F
L1(2)
L2(2)
2M using correct
selling prices
1S simplification
1CA answer
(4)
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Ques
Solution
1.2.5
?M
Cost price = 50 ℅ R43,00
= R2 150 ?S
Selling price = 50 ℅ R53,75
= R2 687,50 ?S
Profit
= R2 687,50 每 R2 150
= R537,50 ?CA
DBE/2013
Explanation
Topic
1M/A multiplying
1S simplifying
F
L1(2)
L2(1)
L3(2)
1S selling price
1CA answer
OR
Profit per kilogram = R53,75 每 R43,00 ?M
= R10,75 ?S
1M/A multiplying
1S simplifying
Profit on 50 kg = 50 ℅ R10,75 ?M
1M/A multiplying
= R537,50 ?CA
1CA answer
(4)
[19]
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QUESTION 2 [15 marks]
Ques
Solution
2.1
Deposit = 10% of R6 599,99
10
=
℅ R6 599,99 ?M
100
= R659,999
> R660 ?CA
2.2
2.4
Topic
F
L1(2)
1M finding 10%
1CA answer
(2)
Amount to be financed (P) = R6 599,99 每 R660,00
= R5 939,99 ?A
?A
?A
Interest = R5 939,99 ℅ 0,115 ℅ 2
= R1 366,20 for two years
Total due = R5 939,99 + R1 366,20
= R7 306,19
2.3
Explanation
?CA
R 7 306,19
Amount =
?M
24
= R304,4245.....
> R304,42 ?A
?A
?M
Height = 12 ℅ 2,54 cm = 30,48 cm
F
L1(2)
1A subtracting deposit L2(2)
1A value of i.
1A two years
1CA adding i to
outstanding amount
(4)
F
L2
1M dividing by 24
1A answer
(2)
M
L1
1M multiplying by
2,54
1A height
(2)
2.5.1
Amount = 12 ℅ R300?M
= R3 600 ?A
2.5.2
?M
?M
Amount = R300 每 (R120 + R150)
= R30 ?A
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F
1M multiplying by 12 L1
1A answer
(2)
F
1M subtracting from
L1(2)
R300
L2(1)
1M adding costs
1A answer
(3)
[15]
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NSC 每 Grade 11 Exemplar 每 Memorandum
DBE/2013
QUESTION 3 [20 marks]
Ques
Solution
Explanation
Topic
3.1.1
Height of tin = 430 mm = 43 cm
1A length
1C conversion
M
L1
(2)
3.1.2
Distance = 4 570 mm 每 1 780 mm
= 2 790 mm
?A
3.1.3
Length = 4 260 每 2 610
= 1 650 mm ?A
Area = 1 600 ℅ 1 650 ?SF
= 2 640 000 mm2 ?CA
3.1.4
3.2.1
3.2.2
?M
?SF
?C
?SF
?SF
A = 4,830 m (2 ℅ 9,75mm + 6,4 m) 每 6,4 m ℅ 0,43 m
= 4,83 m (25,9 m) 每 2,752 m2
?S
= 122,345 m2 ?CA
122,345
? ?M
8
= 15,293.. ? ?CA
> 16 ? ?R
Amount of paint =
16 ?
?M
5?
= 3,2 = 4 ?CA
Number of 5 ? =
Cost = 4 ℅ R215,85 ?M
= R863,40 ?CA
M
1M subtracting correct
L1
values
1A correct answer
(2)
M
1 A value of length
L1(1)
L2(2)
1SF substitution
1CA answer
(3)
1C conversion
M
1SF value of k
L1(2)
1SF value of b
L2(2)
1SF value of t and p.
L3(2)
1S simplification
1CA area
(6)
M
1M dividing by 8
L1(2)
L2(1)
1CA amount of paint
1R rounding up
(3)
M/F
L1(2)
1M dividing by 5 ?
L2(2)
1 CA number of
containers
1M multiplying by cost
1CA cost of paint
(4)
[20]
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