Tuesday, 14 September 2021 08:28

MATHEMATICAL LITERACY PAPER 1 GRADE 12 MEMORANDUM - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

Download this page as PDF Share via Whatsapp Join our WhatsApp Group Join our Telegram Group

MATHEMATICAL LITERACY PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM

SEPTEMBER 2018

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


QUESTION 1 [30 MARKS] INTEGRATED QUESTION

Question Solution Explanation/Marks
AO: FULL MARKS 
Lev.
1.1
1.1.1 
Inflation is the increase in the price of goods
and services over time.✓✓A
OR
The decrease in the purchasing value of the
money over time. ✓✓A
2A Explanation
(2)
F
L1
1.1.2 Difference = R125 – R120 = R5 ✓M
= 5 × 100 = 500 cents ✓A
1M Difference
1CA In cents (2)
F
L1
1.1.3 Price = 112.5 x 125
             100
✓M
= R140,60 ✓CA
OR
Profit = 12,5 × 125 = R15,625✓M OR
             100
Price = 125 × 1,12
Price = 125 + 15,625 = R140,625 = 140, 625
= R140,60 ✓CA
= R140,60
1M Multiply112,5 by 125 and
dividing by 100
1CA Price
OR
1M For profit
1CA Price
NPR
(Concept of money)
(2)
F
L1
1.2
1.2.1
Cost of a dozen = 90 x 12 ✓ MA
                             60
= R18,00 ✓ CA
OR
Dozens = 60
                12
= 5 ✓
Cost of a dozen =90
                             5
= R18 ✓
1MA Divide by 60 and
multiply by 12
1CA Cost
1MA Number of dozens
1CA Cost (2)
F
L1
1.2.2 % Profit = (( 60 + 75 ) 90 ) x 100
                           90
= 45 x 100% ✓M
   90
= 50% ✓CA
1M Subtraction for profit
1M Fraction 45/90 multiply
by 100
1CA Percentage
(3)
FL1
1.3 Time = 24:00 – 16:25 = 7h 35 min ✓M
From midnight to 3:30 am = 3h30 min ✓M
∴ Total time = 7 hours 35 min + 3h30min
= 11h 05 min ✓CA
1M For 7h 35min
1M For 3h 30min
1CA Time
(3)
M
L1
1.4 Plastic cups = 2 x 1000 ✓M
                          275
= 7,2727
= 7 ✓A
1M Correct numerator and
denominator
1A Rounding down
(2)
ML1
1.5
1.5.1
Length = 2𝑐𝑚 × 75 𝑚 ✓✓2M
                 100 𝑐𝑚
= 1,5 m ✓
1M for Numerator
1M for Denominator
1CA With correct unit
(3)
ML1
1.5.2 Scale = 100 : 7 500 ✓✓2M
= 1 : 75 ✓ 1CA
1M Ratio form
1C Conversion
1CA Answer
(3)
MP
L1
1.6
1.6.1
 72,67% = Nelson Mandela District ✓✓ A 2A District
(2)
DL1
1.6.2 Districts are: Chris Hani East, OR Tambo
Coastal, ✓A
Amathole East and Amathole West ✓A
1A First 2 districts
1A Second 2 districts
(2)
DL 1
1.6.3 Nelson Mandela, Sarah Baartman OR Tambo
Inland, Chris Hani West, Alfred Nzo West, Joe
Gqabi, Buffalo City ✓M
∴ 𝑀𝑖𝑑𝑑𝑙𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑐𝑡 = CHRIS HANI WEST ✓CA
1M Identifying 7 best
performing districts in
order
1CA for middle district
(2)
DL1
1.6.4 Candidates failed = 26.1 × 313 030 ✓M
                                100
= 81 701 ✓CA
OR
Passed = 73,9 × 313 030
                 100
= 231 329 ✓M
Number failed = 313 030 – 231 329
= 81 701✓CA
1M Correct values
1CA Candidates
1MA Candidates passed
1CA Candidates failed
Accept 81 700 (2)
DL1
      [30]


QUESTION 2 [46 MARKS] FINANCE

Question Solution Explanation/Marks
AO: FULL MARKS 
Lev.
2.1
2.1.1 
Value of A = R3 250 + R4 500 + R1 200 ✓ M
= R8 950 ✓ A
OR
Value of A = R7 950 + R1 000 ✓ M
= R8 950✓ A
1M Addition
1A Value of A
1M Addition
1A Value of A (2) 
L2
2.1.2 Value of B = R1 440 – R1 500 ✓M
= -R60 ✓ A
Value of B = R1 440 – R1 500 ✓M
= (R60) ✓ A
1M Subtracting correct
values
1A Negative answer
1M Subtracting correct
values
1A Answer in brackets
(2)
 
2.1.3  Value of C = R2 600 – R200 ✓ M
= R2 400 ✓ CA
OR
Value of C = R5 780 – (R880 + R1 000 + R1 500) ✓M
= R2 400 ✓CA
1M Subtraction
1CA Value of C
1M Subtraction
1CA Value of C (2)
L1
2.1.4  Delivery ✓✓A 2A Correct answer (2) L1
2.2
2.2.1
Annual gross salary = R65 000 ×12 ✓M
= R780 000 ✓CA
1M Multiply by 12
1CA Annual salary
(2)
L1
2.2.2 Mrs John’s annual pension fund contribution:
=    7.5
    (100 × 65 000) ×12 ✓M
= R58 500 ✓CA
1M Multiply by 12
CA Pension contr.
(2)
L1
2.2.3  Medical Aid contribution = R1 050 ×12 ✓M
= R12 600 ✓A
1M Multiply by 12
1A Contribution (2)
L1
2.2.4 Mrs John’s performance bonus = 75 
                                                    100 ×65 000 ✓M

= R48 750 ✓A
1M Correct values
1M Multiply by 75%
(2)
L1
2.2.5 Annual taxable income
= R780 000 + R48 750 – (58 500 + R12 600) ✓✓M
= R828 750 – R71 100 ✓M
= R757 650 ✓CA
CA from 2.2.1, 2.2.2 and 2.2.3
1M Total income
1M Total contributions
1M Subtraction
1CA Taxable income (4)
L2
2.3
2.3.1
R701 301 and above ✓✓RT 2RT Correct group
(2)
L1
2.3.2 Rebate Mrs John will receive = R13 500 + R7 407 ✓M
= R20 907 ✓A
1M Adding correct rebates
1A Total rebate (2) L1
2.3.3 Actual tax = Income tax calculated on taxable
income Rebate
= 206 964 + 41 % of the amount above 701 300 –
R20 907
= 206 964 + 0,41 × (757 650 – 701 300) –20 907 ✓SF
= 206 964 + 0,41 × 56 350 – 20 907 ✓S
= 206 964 + 23103,50 – 20 907 ✓S
= R209 160,50 ✓CA
CA from 2.2.5, 2.3.1 and 2.3.2
1SF Substitution
1S Simplification
1S Simplification
1CA Nearest rand
(4)
L3
2.3.4 Net annual salary = Annual taxable income –
Actual annual tax
= R757 650 – R209 160,50 ✓SF✓M
= R548 489,50 ✓CA
CA from 2.2.5 and 2.3.3
1SF Substitution
1M Subtraction
1CA Net salary (3)
L1
2.4
2.4.1
Interest 1st year = 15.5
                              100 × 400 000
= R62 000 ✓A
2nd year amount = R400 000 + R62 000
= R462 000 S✓
Interest 2nd year = 15.5
                                100 × R462 000
= R71 610✓S
Total interest = R62 000 + R71 610
= R133 610 ✓A
1A Interest
1S New amount 2nd year
1S Interest in 2nd year
1A Total interest
(4)
L2
2.4.2 Phone D:
Cost excluding VAT = 100
                                   115 × 1 750
= R1 521,74 ✓M
VAT = R1 750 – R1 521, 74 ✓M
= R228,26 ✓A
OR
VAT = 15  ✓M × 1 750 ✓M
         115
= R228,26 ✓A
1M Cost without VAT
1M Subtraction
1A VAT value
1M Fraction
1M Multiplication
1A VAT value (3)
L2
2.4.3 Phone D : Phone E
3 : 2
60 : E ✓M
E = 60 x 2
          3
= 40 ✓CA
1M Ratio form
1CA Number of phones
(2)
 
2.4.4  Total cost = 60 × 1 750 + 40 × 2 000
= 105 000 ✓M + 80 000 ✓M
= R185 000 ✓CA
CA from 2.4.3
1M Cost for phones D
1M Cost for phones E
1CA Total cost (3)
L1
2.4.5 1: 0,52709
185 000 : Total cost in CYN
Total cost = 185 000 × 0,52709 ✓M
= CYN 97511,65 ✓A✓A
1M Conversion
1A Total cost
1A Answer in Yuan
(3)
L2
    [46]  


QUESTION 3 [25 MARKS] MEASUREMENT

Question Solution Explanation/Marks
AO: FULL MARKS 
Lev.
3.1.1  Length = 4 880 mm ÷ 1 000 ✓C
= 4,88 m ✓A
1C Divide by 1000
1A Metres
(2)
L1
3.1.2 Distance A = 3 (150 mm) ✓M
= 450 mm ✓A
1M Multiplication by 3
1A Distance
(2)
L1
3.1.3 Height of the wall = 2,1 m + 450 mm ÷ 1 000 ✓C
= 2,1 m + 0,45 m ✓S
= 2,55 m ✓CA
CA from 3.1.2
1C Conversion
1S Simplification
1CA Height (3)
L1
3.1.4 Area = Length × Width
= 4,88 ✓ × 2,1 m ✓M
      2 
= 5,124 m2 ✓CA ✓Unit
CA from 3.1.1
1M Dividing by 2
1M Multiplication
1CA Area
1A Unit (4)
L2
       
3.1.5 Area covered by bricks
= Area of garage – Area of double door
= (2,55 m × 5,18 m) ✓ – (4,88 m × 2,1 m) ✓M✓M
= 13,209 m2 – 10,248 m2 ✓S
= 2,961 m2 ✓CA
CA from 3.1.1 and 3.1.3
1M Area of garage
1M Area of double door
1M Subtraction
1S Simplification
1CA Area covered by bricks (5)
L2
3.2.1 Height of the bricks and cement = (12 x 2) + 76 mm ✓M
= 100 mm✓CA
1M Multiplication by 2 and addition
1CA Height
(2)
L1
3.2.2 Number of rows of bricks = 2 500 ✓M ✓M
                                             100 
= 25 ✓CA
CA from 3.2.1
1M Conversion
1M Division
1CA Number of rows of bricks (3)
L1
3.2.3 Volume = 23 cm ×11 cm × 7,6 cm ✓SF ✓C
= 1 922,8 ✓ A cm3 ✓A
1SF Substitution
1C Conversion
1A Volume
1A Units
(4)
L2
      [25]


QUESTION 4 [31 MARKS] DATA HANDLING

Question Solution Explanation/Marks
AO: FULL MARKS 
Lev.
4.1.  Range = 3,316 kg – 0,182 kg ✓RT ✓M
= 3,134 kg ✓CA
1RT Correct values
1M Subtraction
1CA Range
(3) 
L2
4.2 1,668 kg ✓✓A 2A Median
(2)
 L2
4.3 Average =
1,26 ×2 +1,371 ×9 +1,668 ×8 + 1,746 ×4 + 1,849 ×8 + 2,163 +2,333 + 3,128 ×2 ✓ M
= 60,731 ✓M
      35
= 1,735
= 2 kg ✓R
1M Adding
1M Concept of mean
1R Average to nearest kg
(3)
L2
4.4 22 ✓✓ A 2A Explanation
(2)
L1
4.5 Probability = 8  ✓M × 100%✓M
                   35
= 22,9 % ✓CA
1M Fraction value
1M Multiply by 100
1CA Percentage
(3)
L1
4.6 1
4M For the first 4 bars plotted correctly
1M For the next 2 bars plotted correctly
1M For the last 2 bars plotted correctly (6) 
L2
4.7 0,182; 0,182; 0,182; 0,309; 0,729; 0,729; 0,729; 0,856; 0,856; 0,856; 0,856
0,936; 2,448; 2,448; 2,449; 3,038; 3,316; 3,316; 3,316; 3,316; 3,316; 3,316
Q2
Q2=Median= 0,856+0,936 = 0,896 ✓✓ MM
                               2
Q3=3,316 ✓ CA
1M Correct values
1M Value or position of median
1CA Q3
(3)
L1
4.8 3,316 kg ✓✓

2A RT Modal value
(2)
L1
4.9 % of total fishes =  ✓× 100 ✓M
                              57
= 14,04 % ✓
1M Fraction
1M Multiply by 100
1CA Percentage
(3)
L1
4.10 3rd hour ✓✓ 2RT Correct hour
(2)
L1
4.11 1,1,1,1,2,3,3,4,6 ✓✓ 2M Arrangement
(2)
L1
      [31]

 
QUESTION 5 [18 MARKS] MAPS, PLANS and OTHER REPRESENTATIONS

Question Solution Explanation/Marks
AO: FULL MARKS 
Lev.
5.1 There is no seat for Lundi here ✓✓A  2A No seat
(2)
MP
L1
5.2 South ✓✓A 2A Direction
(2)
L1
5.3 A8 ✓ RP
A11 ✓ RP
A15 ✓ RP
1RP First seat
1RP Second seat
1RP Third seat (3)
L1
5.4 35 ✓✓RP 2RP Available seats
(2)
L1
5.5 B14 ✓✓ RP 2RP Asi’s seat no.
(2)
L1
5.6 Row J ✓✓ RP 2RP Furthest row
(2)
L1
5.7  Side BB ✓✓RP 2RP Correct side
(2)
L1
5.8 P(seat from side AA) = 20  ✓A ✓A
                                   19 4
= 0,103 ✓CA
1A Numerator
1A Denominator
1CA Answer
(3)
P
L2
      [18]
  TOTAL: 150  
Last modified on Tuesday, 14 September 2021 13:06