MATHEMATICAL LITERACY PAPER 2
GRADE 12
MEMORANDUM
NSC EXAMINATIONS
NOVEMBER 2021
QUESTION1 [29 MARKS]Answer Only AO - full marks
1.1.1
Total mass
= 6 × 110g
= 660 g
1MA multiply mass by 6
1A mass
(2)
1.1.2
Radius = 32 mm
2A radius (2)
1.1.3
A
2A correct letter
[accept: mm3]
(2)
1.1.4
Total No. of days
= 11 Jan to 31 Mar
= (31 – 10) + 28 + 31
= 21 + 28 + 31 = 80
1MA days in Jan
1MCA adding days in 3 months
1CA simplification
(3)
1.1.5
Price for 2 Pringles
= 2 (R100)
6
= 2 × R16,666
= R33,33
1MA dividing price by 6 and multiplying by 2
1CA simplification
NPR
(2)
1.2.1
A
2A correct letter
(2)
1.2.2
D
2A correct letter
Accept 60 km/h
(2)
1.3.1
7,3 m = 7,3 × 100 cm
= 730 cm
1MA multiplying correct value by 100
1A simplification
(2)
1.3.2
D = 7,3 m – 5,2 m
= 2,1 m
1MA difference of correct lengths
1CA simplification
(2)
1.3.3
0,5m
2A height
(2)
1.4.1
A layout plan is a top view that shows the arrangement of features.
OR
A layout plan is the structural arrangement of items within a certain space.
OR
Plan of the entire inside cabin, showing location of seats, exit doors etc.
OR
Drawing to scale showing physical arrangements of all resources that consume space within facilities.
2Aexplanation
(2)
1.4.2
28
2A number of seats
(2)
1.4.3
G1
1A correct seat
1A correct row
(2)
1.4.4
6
2A correct number
(2)
[29]
QUESTION 2 [24MARKS]
2.1
3
2A correct number
(2)
2.2
Living room
2A correct room
(2)
2.3
North East or NE
2A direction
(2)
2.4
Pnot interior= Pexterior
= 2/6
= 1/3
2RT numerator
1A denominator
1CA simplification
OR
Pnot interior = 1 – 4/6
= 2/6
= 1/3
1MA probability of NOT
1RT numerator
1A denominator
1CA simplification (4)
2.5
Jan is wrong, the kitchen is on the Southern side.In South Africa it does not get a lot of sun.
1A wrong
2O reasoning
(3)
2.6
It cannot be the view showing the kitchen and dining room, as it does not show the extra window for the bathroom.
It does not show the other rooms on both sides of the windows.
OR
It shows the veranda, door, bedroom and livingroom windows.
OR
2O reason
OR
Because there is no veranda on the side of the kitchen and the picture shows the veranda.
OR
The drawing shows the SE elevation and the kitchen is on the SW side.
OR
The windows shown does not look like kitchen windows, they are too big.
OR
The drawing represents the front view.
OR
Kitchen should be on the left-hand side with the window and door / The door knob is on the right- hand side and not on the left-hand side of the door adjacent to the kitchen window. (2)
2.7.1
10 mm : 1 000 mm
= 1 : 100
OR
1 cm : 100 cm
= 1 : 100
1A correct ratio and conversion
1CA simplification
OR
1A correct ratio and conversion
1CA simplification
AO (2)
2.7.2
Length on floor plan/Lengte op die vloerplan = 4,4 cm
1 cm = 100 cm
4,4 cm = 4,4 × 100 cm
= 440 cm
= 4,4 m
CA from 2.7.1
1A correct measurement
1MCA using the scale
1CA simplification
1C conversion
Accept 4,3 m to 4,5 m
OR
1 cm is 1 000 mm
4,4 cm is 4 400 mm
4 400 mm = 4,4 m
1A correct measurement
1MCA using the scale
1CA simplification
1C conversion
OR
1cm : 1 000 mm
1cm : 1 m
4,4 cm : 4,4 m
1MCA using the scale
1C conversion
1A correct measurement
1CA simplification
(4)
2.7.3
Jan is correct.
When a photocopy is made the size of the plan may change while the number scale remains the same.
1A opinion
2O verification (3)
[24]
QUESTION 3 [35 MARKS]
3.1.1
A = 162 cm + 1,5 cm + 1,5 cm
= 162 cm + (1,5 cm × 2)
= 165 cm
1RT inside length
1MA adding both sides
1CA simplification
(3)
3.1.2
B = 80 cm – (40 cm + 4,5 cm + 1,5 cm + 1,5 cm)
= 32,5 cm
1RT both heights
1MA subtracting
1CA simplification
(3)
3.2
31,496 inches = 80 cm
1 inch = 80 cm
31,496
= 2,54 cm
1RTheight 80 cm
1MA dividing by 31,496
1A simplification
(3)
3.3.1
Area of a rectangle = length × width
= 165 cm × 80 cm
= 13 200 cm2
CA from 3.1.1
1MCA substitution
1CA simplification
(2)
3.3.2
Area of a rectangle = 13 200 cm2
= 13200 m2
(100)2
= 1,32 m2
CA from 3.3.1
1MCA dividing by 1002 or 10 000
1CA simplification
AO
(2)
3.3.3
1ℓ covers 6,9 m2
n = 1,32
6,9
= 0,1913… ℓ
To paint three coats
0,1913... ℓ × 3 = 0,57 ℓ
1MA multiplying with 3
1CA simplification
1MA dividing by 6,9
1CA simplification
1R rounding
OR
Total area to cover
= 1,32 m2 × 3 = 3,96 m2
1ℓ covers 6,9 m2
x ℓ covers 3,96 m2
x = 3,96 = 0,57 ℓ
6,9
CA from 3.3.2
1MA dividing by 6,9
1CA simplification
1MA multiplying with 3
1CA simplification
1R rounding
OR
Paint needed
= 1,32 x 2ℓ + 1,32
6,9 6,9
= 0,38 ℓ + 0,19 ℓ
= 0,57 ℓ
1MA dividing by 6,9
1MA adding the 2 coats and 1
1CA simplification
1CA simplification
1R rounding
OR
Total area to cover
= 1,32 m2 × 3 = 3,96 m2
Spread rate = 1ℓ
6.9m2
= 0,144… ℓ/m2
Total amount of litres
= 0,144 × 3,96
= 0,57 ℓ
1MA multiplying with 3
1CA simplification
1MA dividing by 6,9
1CA simplification
1R rounding
OR
Spread rate = 1ℓ
6.9m2
= 0,144… ℓ/m2
Paint needed for 1 coat
= 0,144 × 1,32 = 0,19… ℓ
Paint needed for 3 coats
= 0,19.. × 3
= 0,57 ℓ
1MA dividing by 6,9
1CA simplification
1MA multiplying with 3
1CA simplification
1R rounding
(5)
3.3.4
0,57 ℓ × 1 000
= 570 mℓ
Not valid
1MCA (from Q3.3.3 multiply by 1 000)
1CA simplification
1O verification
OR
500 mℓ ÷ 1 000
= 0,5 ℓ less than 0,57ℓ
Tsidi’s statement is invalid
1MCA (from Q3.3.3 dividing by 1 000)
1CA simplification
1O verification
OR
1ℓ covers 6,9 m2
500 mℓ covers 6,9 = 3,45 m2
2
Area to paint = 1,32 m2 × 3 = 3,96 m2
The paint is not enough / invalid
1MCAarea
1CA simplification
1O verification
OR
Coverage per coat
= 500mℓ = 0.5ℓ = 0,166..
3 3
Coverage = 0,166 × 6,9
= 1,15 m2
1,32 m2 needs to be covered per coat
Not valid
1MCAdividing
1CA simplification
1O verification
(3)
3.4.1
Number of boxes
= 162cm
34,5cm
= 4,695…
∴ 4 boxes
1MA dividing
1C conversion
1CA simplification
1R rounding down
OR
Number of boxes
=1 620 mm
345mm
= 4,695…
∴ 4 boxes
1C conversion
1MA dividing
1CA simplification
1R rounding down
(4)
3.4.2
Number of single files
= 162cm
8,1cm
= 20
Number of files in boxes
= 4 × 4
= 16
Difference in the number of files
= 20 – 16
= 4
CA number of boxes from 3.4.1
1MA dividing
1A simplification
1RT number of files in a box
1CA simplification
1CA difference in files (5)
3.4.3
Neater storage
OR
Files stand up straight
OR
Prevents dust on documents in the files
OR
It is easier to separate the files accordingly
OR
To categorise /organise her files
OR
Prevent files from breaking/ damage/protect files
2O reason
(2)
3.4.4
P = 1/16 x 100%
= 6,25%
CA denominator from 3.4.2
1A numerator
1MCA denominator
1CA simplification
(3)
[35]
QUESTION 4 [33 MARKS]
4.1.1
Perennial garden bed.
OR
Compost
2A correct feature
(2)
4.1.2
Water is scarce
OR
Rain water is free compared to tap water
OR
Pay less water bills
OR
Water storage
OR
To save water for future use
OR
To harvest rain water
2A Reason
(2)
4.1.3
Greenhouse roof/ gutters
OR
Livestock Barnroof/ gutters
OR
Solar greenhouseroof / gutters
1A correct structure
1A 2nd correct structure
Accept roof and gutter /pipe full marks
(Any 2 structures)
(2)
4.1.4
Area = ½ x 17,024 m 19,5 m
= 165,984 m2
1RT correct height
1RTcorrect base
1CAarea of a triangle
NPR
(3)
4.1.5
Option A = R1 154 × 2
= R2 308
Option = R127,30 × 19
= R2 418,70
Option A.
1MA multiply by 2
1CA option A cost
1MA multiply by 19
1CA option B cost
1O best option
(5)
4.2
Volume = 3,142 × r2 × height
5000 ℓ = 3,142 × r2 × 220 cm
5000 000 = 691,24 × r2
5 000 000 = r2
691,24
7233,377698 = r2
√7233,377698 = r2
85,05 cm = r
1SF substituting 5000
1C converting ℓ to cm3
1M dividing by 691,24
1S simplification
1M finding square root
1CA radius value
NPR
(6)
4.3.1
18 : 42
= 3 : 7
1A correct order and values
1CA only if one value is correct or reversed order
(2)
4.3.2
Height = 42"/12" = 3,5 feet
3,28084 feet = 1 000 mm
3,5 feet = 3.5 x 1 000
3,28084
= 1 066,799…mm
1MA converting to feet
1C converting to mm
1CA simplification
OR
3, 28084feet = 1 000mm
1 foot = n
n = 304,79999mm
1 foot = 12 inches
Then 12 inches = 304,79999mm
1 inch = 304,79999mm
12
= 25,39999 mm
Therefore 42 inches = 42 × 25,39999mm
= 1066, 7999mm
= 1 066,8 mm
1MA converting to feet
1C converting to mm
1CA simplification
NPR
(3)
4.3.3
3A correct Roman numeral
(3)
4.3.4
Q
2A correct letter
(2)
4.3.5
The notch labelled S is placed against B and the notch labelled R is placed against C
2A mentioning the position of the 1st notche
1A second notch
(3)
[33]
QUESTION 5 [29MARKS]
5.1.1
5.1.2*
5.1.3
1A Newburyport
1A Lawrence
1A Boston
1A Worcester
(4)
5.1.4
Number of litres in 23 gallons
= 3,785 × 23
= 87,055 litre
Cost of 87,055 litre
= 87,055 × R15,97
= R1 390,27
Valid
1C gallons to litre
1S simplification
1CA cost of fuel
1O conclusion
OR
Number of litres
=R 1400
R 15,97
= 87,664.. litre
Number of gallons
= 87,664
3,785
=23,16 gallons
Can buy more with R1 400
Valid
1S simplification
1C gallons to litre
1CA cost of fuel
1O conclusion
NPR
(4)
5.1.5
1 full tank of fuel = 23 gallons
He can travel = 23 × 18 = 414 miles
Distance
Greenfield - Fitchburg = 49 miles
Number of trips on 1 full tank
=414 = 8,448..
49
8 trips on 1 full tank
So, then he will fill tank back to 23 gallons
Amount of fuel for 1 return trip
= 98 = 5,44 gallon
18
Left in a tank is 23 – 5,44 = 17,56 gallons.
1A travel distance
1RT trip distance
1MA dividing
1CA number of trips
1MA dividing
1CA simplification
1MA subtracting
1CA simplification
OR
Distance(Greenfield and Fitchburg) = 49 miles
Weekly must travel
= 5 × 2 = 10 trips
He can travel = 23 × 18 = 414 mileswith a full tank.
8 trips is 49 × 8 = 392 miles – now he needs to refill after Thursday’s trips
With the full tank he only needs to travel Friday return trip
Friday trip: 49 × 2 = 98 miles
Used = 98 = 5,44 gallon
18
Left in a tank is 23 – 5,44 = 17,56 gallons.
18 miles on 1 gallon
49 miles on x gallon
x = 48 = 2,722… gallon per trip
18
Number of trips on 1st full tank
= 23 = 8,44…
2,722...
∴ 8 trips before he fills up again
∴ 2 trips with second full tank
Fuel used
= 2,722…× 2 = 5,44… gallon
Left in the tank
= 23 – 5,44… = 17,56 gallon
1RT trip distance
1MA weekly miles
1MA multiply
1A travel distance
1MA dividing
1CA usage on last day
1MA subtracting
1CA diff. between capacity and used gallons
OR
Single Trip = 49 miles
Number of gallons for 1 trip
= 49 ÷ 18 = 2,72
Number of gallons for return trip
= 2,72 × 2 = 5,44
23 gallons ÷ 5,44 = 4,22 days
≈ 4 days
No of gallons left
= 23 – 5,44 = 17,56 gallons
1RT trip distance
1MA dividing
1A travel distance
1CA number of trips
1MAmultiplying
1CA simplification
1MA subtracting
1CA simplification
23 × 18 = 414 miles
Monday : 49 × 2 = 98 miles
Tuesday: 98 miles
Wednesday: 98 miles
Thursday: 98 miles
Total = 392 miles
Fill up the tank on Thursday
Used per day
= 98 ÷ 18 = 5,44gallons
Petrol left in tank
= 23 – 5,44
= 17,56 gallons
Miles that can be travelled after Friday
= 414 – 98
= 316 miles
Petrol left in tank = 316 ÷ 18
= 17,56 gallons
1A travel distance
1RT trip distance
1MA multiplying
1CA number of trips
1MA dividing
1CA simplification
1MA subtracting
1CA simplification
(8)
5.2
ºC = 5/9 (ºF – 32 )
–7 = 5/9 (ºF – 32 )
ºF = 9/5 x -7 + 32
= 19,4
≈ 20ºF
1SF substitution
1S simplification
1CA simplification
1R rounding
(4)
[29]
TOTAL: 150