MATHEMATICAL LITERACY PAPER 2
GRADE 12
NOVEMBER 2020
MEMORANDUM
NATIONAL SENIOR CERTIFICATE

NOTE:

• If a candidate answers a question TWICE, mark only 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 provided at least one of the values is correct; 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 penalise for every extra item presented.

QUESTION 1 [39 MARKS]
1.1.1 Slovakia (diff. 2015 -16):
163 740 – 161 906
= 1 834
1 RT correct values
1MA method of subtraction
1CA answer (3)
D
L2

1.1.2 Range = highest – lowest
2 947 664 = 2 970 436 – N
N = 22 772
1M Range concept
1RT highest value
1CA simplification
AO (3)
D
L2

1.1.3 Number of learners enrolled decreased from 2014/2015/2016
OR The number of learners decreased every year
1O decrease
1O time (2)
D
L4

1.1.4 % increase(Turkey)
= 1 221165 - 1 064 190 x 100%
          1 064 190
= 1 064 190 x 100%
      156 975
= 14,75 %
% increase(United Kingdom)
= 2 248 162 - 1 596 803 x 100%
             1 596 803
= 40,79 %
United Kingdom has the biggest percentage increase
1M using correct formula
1MA subtractingcorrect values
1CAsimplification
1MA subtractingcorrect values
1CAsimplificationas a percentage
1CA county
OR
Turkey: (1 221 165 ÷ 1 064 190) ×100%
= 114,75%
% increase (Turkey) = 114,75% –100%
= 14,75%
(United Kingdom): (2 248 162 ÷ 1 596 803) ×100%
= 140,79%
% increase United Kingdom = 140,79% - 100%
= 40,79%
United Kingdom has the biggest percentage increase
1MA subtracting correct values
1M using correct formula
1CA simplification
1MA subtracting correct values
1CAsimplification as a percentage
1CA county
NPR (6)
D
L3

1.1.5 Probability (decline 2015-2016)
= ³/₁₁
≈ 0,27
1A numerator
1Adenominator
1CA as decimal
NPR (3)
P
L3

1.1.6 Denmark cost
= €520,83 × 284 655
= €148 256 863,70
Slovenia cost
= €350 × 85 407 = €29 892 450
€148 256 863,70 : €29 892 450
4,959… : 1
The statement is NOT VALID
1RT correct values
1A cost
1RT correct values
1A cost
1CA simplified ratioin correct order
1O verification
OR
Accept per year or per month
2016 Denmark : 2016 Slovenia
284 655 × 520,83 × 12 : 85 407 × 350 × 12
1 779 082 364 : 358 709 400
4,959… : 1
The statement is NOT VALID
1RT Denmark values
1RT Slovenia values
1A cost
1A cost
1CA simplified ratio in correct order
1O verification
OR
Denmark: € 520,83 × 12= € 6249.96 per year
€ 6 249,96 × 284 655
= € 1 779 082 364
Slovenia : € 350 × 12 = € 4 200 per year
€ 4 200 × 85 407
= € 358 709 400
Denmark: Slovenia
€ 1 779 082 364 : € 358709 400
(€ 1 779 082 364 ÷ € 358 709 400) : (€ 358 709 400 ÷€ 358 709 400)
= 4,9596 : 1
The statement is NOT VALID
1RT correct values
1A cost
1RT correct values
1A cost
1CA simplified ratioin correct order
1O verification
NPR (6)
D
L4

1.2.1 Profit= R30 × 120% = R36
Profit per marble = R36 = R0,36
                               100
Cost price per marble = R30 = R0,30
                                      100
Selling price
= R0,36 + R0,30 = R0,66 per marble
1MA calculating profit
1CA profit per marble
1A price per marble
1MCA simplification
OR
R30 per 100 marbles 100%
Profit on 100 marbles to yield 120% per pack
= R30 x 120%
       100%
= R36 per pack
Price of selling 1 marble is/Verkoopprys per albaster is:
R30 + R36
    100
= R0,66
1MA calculating profit
1MCA cost plus profit
1M dividing by 100
1CA simplification
OR
Selling price = R30 × 220% = R66
Price per marble
= R66 = R0,66
   100
1MA calculating % increase
1MCA selling price
1M dividing by 100
1CA simplification
OR
Price per marble = ³⁰/₁₀₀ = R0,30
Selling price = 0,3 x 2,2 = R0,66
1MA dividing by 100
1M calculating % increase
1MCA selling price
1CA simplification
OR
Selling price = 30 2,2 = R66
Price per marble = ⁶⁶/₁₀₀
= R0,66
1MA calculating % increase
1MCA selling price
1M dividing by 100
1CA simplification
NPR (4)
F
L3

1.2.2 Radius container =6,4
                                        2
= 3,2 cm
Volume of a cylinder
= π × radius² height
= 3,142 × (3,2 cm)² × 30cm
= 965,2224 cm³
Volume of 2 bags of marbles
= 2 × 2 cm³ × 100
= 400 cm³
Vol. Water to fill container
= 965,2224 cm³ – 400 cm³
= 565,2224 cm³
½ℓ = 500 cm³
Statement is valid
1C conversion
1MCA finding the radius
1SF both radius and height
1CA simplification
1MA Vol. of total marbles
1CA simplification
1MCA subtraction
1CA simplification
1O conclusion
OR
Radius of container = ^6.4/₂ = 3,2 cm
Volume of a cylinder
= radius² height= 3,142 3,2 cm 3,2 cm 30 cm
= 965,2224 cm³ OR 0,9652224 litres
Volume of 2 bags of marbles
= 2 2 cm³ 100
= 400 cmv OR 0,4 litres
Vol. Water to fill container
= 965,2224 cm³ – 400 cm³
= 565,2224 cm³
OR
= 0,9652224 ℓ – 0,4 ℓ = 0,5652224 ℓ
More than 0,5 ℓ VALID / meer as 0,5ℓ GELDIG
1C conversion
1MCA finding the radius
1SF both radius and height
1CA simplification
1MA Vol. of total marbles
1CA simplification
1MCA subtraction of volumes
1CA simplification
1O conclusion
M
L4
OR
Radius of container = = 3,2 cm
Volume of a cylinder
= radius2 height
= 3,142 3,2 cm 3,2 cm 30 cm
= 965,2224 cm³ OR 0,9652224 litres
Volume of 2 bags of marbles = 2 2 cm³ 100
= 400 cm³ OR 0,4 litres
400 cm³ + 500 cm3 = 900 cm³
This is less than 965,2224 cm³ of the cylinder , VALID
1C conversion
1MCA finding the radius
1SF both radius and height
1CA simplification
1MA Vol. of total marbles
1CA simplification
1MCA addition
1CA simplification
1O conclusion
(9)

1.2.3 Outer diameter
= 64 mm + 2×0,5 mm = 65 mm
Circumference = π × diameter 
= 3,142 ×(6,5) cm
= 20,423 cm
OR
Radius = 32 mm + 0,5 mm = 32,5 mm
= 3,25 cm
Circumference = 2 × π × radius
= 2 × 3,142 × 3,25 cm
= 20,423 cm
1MA increased diameter
1SF substitution
1CA simplification
OR/OF
1MA increased radius
1SF substitution
1CA simplification
NPR (3)
M
L2
[39]

QUESTION 2 [38 MARKS]
2.1.1 Total = 2 × (79 × R244,35)
= R38 607,30
1A number of personnel
1A tariff
1CA simplification
OR
Amount claimed per person:
CM/HM = 79 × R244,35 = R19 303,65
IM = 79 × R244,35 = R19 303,65
Total = R19 303,65 + R19 303,65
= R38 607,30
OR
1A CM amount
1A IM amount
1CA simplification
(3)
F
L2

2.1.2 A (Hours worked by SM)
= R13 763,75
   R211,75/h
= 65 hours
1MA numerator and denominator
1CA simplification (2)
M
L2

2.1.3

1. Number of marking hours
   = 2 808 28
        23 60
   = 56,97391303 hours≈ 57 hours
   1st day (Monday): 14:00 to 20:00 = 5 hours
   Tuesday to Saturday: 50 hours
   Sunday = 2 hours
   Total/Totaal 5 + 50 + 2 = 57 hrs
   Finish at 10:00 on Sunday.
   1SF correct numerator
   1SF correct denominator
   1CA simplification/hours
   1A hours of 1st day
   1A hours of complete days to last day
   1CA day& time
   OR
   Number of marking hours
   =2808 x 28 = 56,97391303 hours ≈ 57 hours
       23 x 60
   Actual marking time per day
   = 12 hrs – 2 hrs= 10 hrs
   Start
   Mon + Tue + Wed + Thu + Fri + Sat + Sun
   = 5h + 10h + 10h + 10h + 10h + 10h + 2h
   = 57 hours /ure
   Sunday/Sondag = 08:00 + 2h
   = 10:00
   1SF correct numerator
   1SF correct denominator
   1CA simplification/hours
   1A hours of 1st day
   1A hours of complete days to last day
   1CA day and time
   M
   L3
   OR
   Number of marking hours = 56,97391303 hours ≈ 57 hours
   57hours: Monday = 5hrs
   Rest of the days = 57hrs – 5 hrs
   = 52 hrs
   Full marking days = 5,2 days
   Therefore 5 days + 0,2 days
   5 days Tuesday to Saturday
   0,2 days 10 = 2hrs for Sunday
   Ends  Sunday 10:00
   1SF correct numerator
   1SF correct denominator
   1CA simplification/hours
   1A hours of 1st day
   1A hours of complete days to last day
   1CA day& time
   OR
   Number of marking hours
   =2808 x 28 = 56,97391303 hours ≈ 57 hours
       23 x 60
   14:00 to 14:00 = 10 working hours
   Monday 14:00 to Saturday 14:00 = 50 hours
   Saturday 14:00 to Sunday 10:00 = 7 hours
   Finish at 10:00 on Sunday
   1SF correct numerator
   1SF correct denominator
   1CA simplification/hours
   1A full day's work
   1A hours of complete days
   to last day
   1CA day and time
   (6)
   [Accept Tues 10:00]
2. 57 – 52 hours/ure = 5 working hours earlier
   2 hrs of Sunday and last 3 hrs of Saturday not worked
   20:00 – 16:00 = 3 hrs excluding supper
   Finish at 16:15 on Saturday om 16:15
   (Including tea break)
   OR
   52 hours claimed = 5 (Monday) + 40 (Tue to Fri) + 7 (Sat)
   Finish Saturday
   8:00 + 7 hours + 15 min (tea 1) + 45 min (lunch) + 15 min (tea 2) = 16:15
   [also accept 16:00 since they are not paid for tea time]
   1MCA hrs less from marking
   [ CA from 2.1.3 (a)]
   1A separation of hrs
   1CA time
   1CA day
   OR/OF
   1MA breaking up the time
   1A the hours per day
   1CA day
   1CA time
   AO
   (4)
   M
   L3
3. Some candidates omitted some questions or sub-sections.
   OR
   Some candidates wrote short answers (skipping other steps or lines or sentences).
   OR
   Responses were very clear to follow.
   OR
   Some markers mark fast.
   OR
   Markers took shorter breaks
   2O reason (2)

2.1.4 Transport = 11 542 km × R3,26 / km
= R 37 626,92
Marking:
= 2 × 79 × R244,35 + 5 × 65 × R211,75 + 23 × 52 × R195,50
= 2 × R19 303,65 + 5 × R13 763,75 + 23 × R10 166
= R38 607,3 + R68 818,75 + R233 818
= R341 244,05
Total = R341 244,05 + R 37 626,92
= R378 870, 97.
R400 000 budget will be enough
1MA calculation
1CA amount
1MCA multiply correct number of persons by amount claimed
1CA simplification
1CA total
1O conclusion (6)
F
L4

2.2.1 Diameter = 1 m + 0,8 m + 0,8 m = 2,6 m
Area of big circle
= 3,142 ×(^2.6m/₂)²
= 5,30998 m2
Area of the small circle = 3,142 × (0,5 m)²
= 0,7855 m²
Area of the wood= 2,7 m × 2,7 m
= 7,29 m²
Cut-off = 7,29 m²– 5,30998 m2 + 0,7855 m²
= 1,98002 m2 + 0,7855 m²
≈ 2,77 m2
Statement is NOT valid
1A diameter
1SF circle formula
1CA area big circle
1MAarea small circle
1A area of the wood
1MCA subtracting total circles from square area wood
1CA area
1O conclusion
OR
Cut-off wood (in m²)
= Area(square) – [Area (big circle) – Area(small circle)]
= 2,7 × 2,7 – [ 3,142 (0,8 + 0,5)² – 3,142 (0,5)²]
= 7,29 – [ 5,30998– 0,7855]
= 7,29 – 4,52448
= 2,76552.
Which is more than 2,01. Hence, the statement is not valid
1A radius big circle
1SF circle formula
1CA area big circle
1MAarea small circle
1A area of the wood
1M subtracting total circles from square area wood
1CA area
1O conclusion
M
L4
OR
Area of semi-circle = ½ r²
Outer circle = ½ × 3,142 × (1,3 m)²
= 2, 65499m²
Inner circle = ½ × 3,142 × (0,5 m)²
= 0,39275 m²
Desk = 2,65488m²– 0,39275m²
= 2,26224m²
Total area= 2,26224 m² × 2
= 4,52448 m²
Cut-off Area = 7,29 m2 – 4,452448 m²
= 2,7552 m²
Statement not valid
1A diameter/ radius
1SF circle formula
1CA area big circle
1MA area small circle
1CA area of the wood
1MCA total circles area
1CA area
1O conclusion
OR
Area of big semi-circle 
= 3,142 × 1,32 ÷ 2 = 2,65499 m²
Area of small semi-circle
= 3,142 × 0,52 ÷ 2 = 0,3927 m²
One semi-circular top
= 2,65499 – 0,3927 = 2,26224 m²
Area of two semi-circular tops= 2,26224 × 2 = 4,52448 m²
Square Board = 2,7 × 2,7 = 7,29 m²
Cut-off /Afsny = 7,29 m2– 4,52448 m2≈ 2,77 m²
Statement not valid
1A diameter/ radius
1SF circle formula
1CA area big circle
1MA area small circle
1MCA total circles area
1A area of the wood
1CA area
1O conclusion
(8)

2.2.2
Volume wood = 2,7 m × 2,7 m × 0,038 m
= 0,27702 m³
Price of one piece of wood excl.VAT
= 0,27702 m³ × R1 215 = R336,58
Price including VAT= R336,58 × 1,15
= R387,07
12 semi-circles cut form 6 boards
Cost = R387,07 × 6
= R2 322,40
1SF volume of wood
1C conversion
1CA simplification
1MA calculating cost
1MCA adding 15%
1A 6 boards
1CA cost
OR
Volumewood  = 2,7 m × 2,7 m × 0,038 m
= 0,27702 m³
Volume of 6 woodenboards = 0,27702 m³ × 6
= 1,66212 m³
Cost of 6 boards = 1,66212 × R1 215
= R2 019,48
Cost with VAT/Koste met BTW
= R2 019,48 + (15% ×R2 019,48)
= R2 322,40
1SF volume of wood
1C conversion
1CA simplification
1A 6 boards
1MA calculating cost
1MCA adding 15%
1CA simplification
OR
Price of wood including VAT
= R1 215 × 1,15 = R1 397,25
Volume wood/hout = 2,7 m × 2,7 m × 0,038 m
= 0,27702 m³
Cost = R1 397,25 × 0,27702
= R387,07
Cost for 12 semicircles
= R387,07 × 6
= R2 322,40
1MCA adding 15%
1SF volume of wood
1C conversion
1CA simplification
1MA calculating cost
1A 6 boards
1CA simplification (7)
F
L3
[38]

QUESTION 3 [39 MARKS]
3.1.1   The data is discrete.
Percentages run from 0 to 100 and depends on the total of the test and the mark obtained. It is presented as whole numbers.
1Adiscrete
2O opinion (3)
D
L4

3.1.2  Median score test 2
= 66 + 67
        2
= 66,5
1RT correct value
1M median concept
1CA simplification (3)
D
L2

3.1.3 Mean = Y (% mark) + 1 443 = 84
                                 18
Y (% mark) = 18 × 84 – 1 443
= 69%
1MA adding all known% marks
1MA mean concept
1M changing the subject
1CA simplification
OR
18 × 84 = 1 512 MA
Y + 1443 = 1 512
Y = 1 512 – 1 443 M
= 69% 
OR
1MA mean concept
1MA adding all known %
marks
1M changing the subject
1CA simplification (4)
D
L3

Related Items

3.1.4  Helen : 87% – 57% = 30%
Kevin : 97% –67% = 30%
[Note: Afrikaans scripts the answers will be Paul &Oscar]
2RT candidate
1RT candidate (3)
D
L3

3.1.5 Q3 = 71%
Q1 = 61%
IQR = Q3 – Q1 = K3 – K1
= 71% – 61%
= 10%
1A quartile 3
1A quartile Q1
1MCA IQR concept
1CA simplification (4)
D
L3

3.1.6  P(non distinction) = ⁸/₁₈
= ⁴/₉
CA value of Y from 3.1.3
1A numerator
1A denominator
1CAsimplification
OR
P(non distinction) = ¹⁰/₁₈ = ⁵/₉
P(not distinction) = 1 – ⁵/₉ = ⁴/₉
OR
1A numerator
1MA subtracting from 1
1CA simplification (3)
P
L3

3.1.7 Mode = 73%
2A modal value (2)
D
L2

3.2.1 View Terrace OR View OR Terrace
2RT Reading from the map (2)
MP
L2

3.2.2 Facing oncoming traffic
OR
One way road
2O reason (2)
MP
L4

3.2.3 North west or NW
2A correctdirection (2)
MP
L2

3.2.4 21 mm = 110 yards
XY = 50 x 110
             21
XY = 261,904…yards
≈ 262 yards
[Bar scale accept measurements in the range 20 mm to 23 mm
For XY measurements in the range 47 mm to 53 mm]
1A measuring scale
1A measuring distance
1M working with scale
1CA answer
NPR (4)
MP
L3

3.2.5

Parking offence
Street parking is limited to 1 hour before 5 pm
Exceeded allowable duration of parking.
Free parking time was over 
Parked for more than 1 hour
2O Reason for charge (2)
MP
L4

From/Vanaf 12:00 - 15:25 = (3 – 1) +²⁵/₆₀
= 2 ,4166666667 hours
Rate per hour =      £79,75         
                           2,4166666667
= £33
1M subtracting free hour
1C conversion minutes into hours
1CA total charged hours
1M division by hours
1CA simplification rounded to the nearest pound
OR
From 12:00 - 15:25 = 3 h 25 min
Hours she was charged for
3 h 25 min – 1 h = 2 h 25 min
2h 25 min = 145 min
Rate per hour = 79.75 x 60 
                               145
=4785
   145
= £33
OR
1M subtracting free hour
1CA total charged hours
1C conversion hours into minutes
1M division by minutes
1CA simplification rounded to the nearest pound (5)
F
L3
[39]

QUESTION 4 [34 MARKS]
4.1.1 P(odd seat) =  288 × 100%
                                  2
= 0,69%
1A numerator
1Atotal seats
1CA simplification (3)
L2
P

4.1.2 D10
1RT row
1RT seat (2)
L2
MP

4.1.3 Person at D7:

• Turn left walk towards the corridor.
• Turn right walk towards the stage.
• At end of the corridor turn left.
• Walk towards the last seat in the front of section B

1A turn left and walk
1A turn right towards stage
1A turn left end of corridor
1A last seat; section B
(4)
L3
MP

4.1.4 Collection:
Adults: 150 × $28,60 = $4 290
Students: 57 × $26,40 = $1 504,80
Kids: 33 × $17,60 = $ 580,80
Total collection
= $4 290 + $1 504,80 + $580,80
= $6 375,60
Excluding VAT =$6 375,60
                              1,10
= $5 796
Claim is CORRECT
1MA multiply tariff by relevant total patrons.
1CA amount
1CA amount
1CA amount
1MCA total collection
1MCA dividing by 1,10
1CA amount excl. VAT
1O conclusion
F
L4
OR
Adults = 53 + 57 + 40 = 150
Cost = $28,60 × 150 = $4 290
Cost excl VAT = $4 290 ÷ 1,10
= $3 900
Students = 15 + 32 + 10 = 57
Cost = $26,40 × 57 = $1 504,80
Cost excl VAT = $1 504,80 ÷ 1,10
= $1 368
Children = 9+15+9 = 33
Cost/Koste = $17,60 × 33 = $580,80
Cost excl VAT = $580,80 ÷ 1,10
= $528
Total = $3 900 + $1 368 + $528
= $5 796
The claim is correct
1MA multiply tariff by relevant total patrons.
1MCA dividing by 1,10
1CA amount
1CA amount
1CA amount
1MCA total collection
1CA amount excl. VAT
1O conclusion
OR
Section A:
= 53 × 28,60 + 15 × 26,40 + 9 × 17,60
= 1 515,80 + 396,00 + 158,40= 2 070,20
Section B:
= 57 × 28,60 + 32 × 26,40 + 15 × 17,60
= 1 630,20 + 844,80 + 264,00= 2 739,00
Section C/ Afdeling C:
= 40 × 28,60 + 10 × 26,40 + 9 × 17,60
= 1 144,00 + 264,00 + 158,40= 1 566,40
Total amount of Sections = 2 070,20 + 2 739,00 + 1 566,40
= $6 375,60
Excluding VAT = $6 375,60
                              1,10
= $5 796
or/of
$5 796 × 1,1 = $6 375,60 which equals total collection
Claim is CORRECT
1MA multiply tariff by
relevant total patrons.
1CA amount
1CA amount
1CA amount
1MCA total collection
1MCA dividing by 1,10
1CA amount excl. VAT
1O conclusion
OR
Adult 
Price excl. VAT= $28,60 × 100
                                          110
= $26
Total amount = 26 × 150 = $3 900
Student /Studente
Price excl.VAT /Prys sonder BTW=$26,40 ×
110
100
= $24
Total amount/Totale bedrag = $24 × 57 = $1 368
Children/Kinders
Price excl. VAT/ Prys sonder BTW = $17,60 ×
110
100
= $16
Total amount/Totale bedrag = $16 × 33 = &528
Total collection/ Totale insameling = 3 900 + 1 368 + 528
= $5 796
Claim is CORRECT
OR
1MCA dividing by 1,10
1MA multiply tariff by relevant total patrons.
1CA amount
1CA amount
1CA amount
1MCA total collection
1CA amount excl. VAT
1O conclusion
(8)

4.1.5
Cost in USD
= $30,50 × 0,71
= 21,655 USD/VSD
Cost in rand
= $21,655 × R14,43/$
= R312,48
1A Conversion factor
1RT ticket price
1MCA answer in rand
OR
Conversion factor ZAR to AUD :
R14,43 × 0,71 = R10,2453
$30,50 × R10,2453
=R312,48
1RT ticket price
1MCA Conversion
1MCA answer in rand
OR
Conversion to ZAR
= $30,50 × 0,71 × R14,43
= R312,48
1RTticket price
1MCA answer in USD
1MCA answer in rand
(3)
L2
F

4.2.1

5 × A for each correct bar (5)
L2
D

4.2.2 June
Difference = 2,87% – 1,63% = 1,24%
1A correct month
1MCA subtracting values
1CA simplification (3)
L3
F

4.2.3 Inflation Nov = AUD 156 831,36 × 2,18 %
= AUD3418,92
Dec cost of car= AUD 156 831,36 + AUD3418,92
= AUD 160 250,28
Inflation Dec/Inflasie Des = AUD 160 250,28 × 1,91 %
= AUD 3 060,78
Jan. cost of car.
= AUD 160 250,28 + AUD 3 060,78
= AUD 163 311,06
Increase = AUD 163 311,06 – AUD 156 831,36
= AUD 6 479,70
He is incorrect
1RT correct rate
1MCA Increasing
1CA simplification
1CA simplification second month cost
1CA increase
1O opinion
F
L4
Inflation Nov = $156 831,36 × 2,18 %
= $3418,92
Dec. cost of car = $ 156 831,36 + $3418,92
= $ 160 250,28
Inflation Dec = $ 160 250,28 × 1,91 %
= $ 3 060,78
Price increase = Inflation Nov + Inflation Dec
= $3418,92 + $ 3 060,78
= $ 6 479,70
He is incorrect
1RT correct rate
1MCA Increasing
1CA simplification
1CA simplification second month inflation
1CA increase
1O opinion
OR
December:
Cost of car = $156 831,36 × 102,18 %
= $160 250,28
January
Cost of car= $ 160 250,28 × 101,91 %
= $ 163 311,06
Increase = $ 163 311,06 – $156 831,36
= $ 6 479,70
He is incorrect
1RT correct rate
1MCA Increasing by %
1CA simplification
1CA simplification
1CA increase
1O opinion
OR
Price in January
= AUD 156 831,36 1,0218 1,0191
= AUD 163 311,0641
Increase = AUD 163 311,06 – AUD 156 831,36
= AUD 6 479,70
Incorrect
1RT correct rate
1MCA Increasing
1CA Increasing
1CA simplification
1CA increase
1O opinion
F
L4
OR
December price  = AUD 156 831,36 1,0218
= AUD 160 250,28
Januaryprice = AUD 160 250,28 1,0191
= AUD 163 311,06
Adding the increase to the price in November
= AUD 156 831,36 + AUD 6 500
= AUD 163 331,36
Therefore AUD 163 331,36 ≠ AUD 163 311,06
Incorrect 
1RT correct rate
1MCA Increasing by %
1CA simplification
1CA simplification
1CA increase
1O opinion
OR
Price end October = AUD 156 831,36
January price
= AUD 156 831,36 1,0218 1,0191
= AUD 163 311,0641
Subtracting stated increase AUD 163 311,0641 – AUD 6 500
= AUD 156 811,06
Therefore AUD 156 831,36 ≠ AUD 156 811,06
Incorrect
1RT correct rate
1M Increasing by %
1M Increasing by %
1CA simplification
1CA comparing values
1O opinion
(6)
[34]
TOTAL:150