Symbol

Explanation

M

Method

MA

Method with accuracy

CA

Consistent accuracy

RCA

Rounding consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

SF

Correct substitution in a formula

J

Justification

O

Opinion/Example/Definition/Explanation/Justification/Verification

RT/RG/RM

Reading from a table/graph/map

P

Penalty, e.g. for no units, incorrect rounding off etc.

R

Rounding off

NPR

No penalty rounding or omitting units

AO

Answer only, full marks

KEY TO TOPIC SYMBOL:
F = Finance; M = Measurement; MP = Maps, plans and other representations; P= Probability

QUESTION 1 [26 MARKS]

Quest

Solution

Explanation

Level

1.1.1

25 mg : 1 000 mg √M   (divide by 25)
= 1 : 40 √MA

1M ratio concept
1MA Simplification  (2)

M
L1

1.1.2

Tablets per day:
25 × 2 √MA
= 50

Tablets per week:
50 × 7  √MA
= 350 mg √A

1MA multiplying by 2
1MA multiplying by 7
1A answer (3)

M
L1

1.1.3

Number of days to use tablets:
60 tablets ÷ 2 = 30 days √M
Last day: 30 March 2021 √M
Therefore, from the 31st March 2021 a refill is needed. √O

1M finding the number of days
1M Referring to last date
1O Conclusion   (3)

M
L1

1.1.4

25 mg × 120 √RT

= 3 000 mg ÷ 1 000 √C

= 3 g √A

1RT multiply correct values
1C divide by 1 000 
1A answer    (3)

M
L1

1.2.1

The “North Elevation” shows the side view of the building from the northern direction. √√A

2A correct explanation.  (2)

MP
L1

1.2.2

(a)

A floor plan shows a top view of the inside of a building. √√A

2A correct explanation (2)

MP

L1

(b)

An elevation plan shows the side view of the outside of the building. √√A

2A correct explanation (2)

MP

L1

1.2.3

Scale = 1 : 100
Measured length: 5 cm
Actual length: 5 × 100 cm √M
= 500 cm √MA

1M multiply by scale
1MA answer   (2)

MP
L1

1.3.1

90 km × 1000 × 100 × 10 √C
= 90 000 000 mm √A

OR

                              √C
90 km × 1 000 000 = 90 000 000 mm √A

1A correct answer    (2)

M
L1

1.3.2

No of tiles per box = 1,44 𝑚² √M
0,36 𝑚²
= 4 tiles per box √A

1M dividing by 0,36 𝑚²
1A number of tiles   (2)

M
L1

1.3.3

1,125 hours
1 hour
0,125 hr. × 60 = 7,5 min √C
0,5 min × 60 = 30 seconds √C
1,125 hours     = 1 hr. 7 min 30 seconds √A

1C finding seconds
1A correct answer   (3)

M
L1

[26]

QUESTION 2 [28 MARKS]

Quest

Solution

Explanation

Level

2.1.1

George; Knysna; Mossel Bay √√RT
(Any two of the three)

2RT names of towns   (2)

MP
L1

2.1.2

South East √ √A (accept East)
North East √√A

2A Direction
2A Direction  (4)

MP
L1

2.1.3

National roads connect major cities. √√O
Save on fuel consumption. √√O
Fewer traffic lights √√O
Travel faster √√O
[Any two or any acceptable explanation.]

2O Reason 1
2O Reason 2   (4)

MP
L4

2.1.4

2,3 cm = 300 km √A
Measured distance = 7 cm     √A
Actual distance = 7 𝑐𝑚   × 300 km   √M
                           2,3 𝑐𝑚
≈ 913 km √CA

1A measured bar scale
1A measured distance range for bar 2.2cm to 2,4cm measured map range distance from 6,9 to 7,1
1M concept for ratio
1CA rounded answer  (4)

MP L2

2.1.5

Distance = Speed × Time
913 = 100 × Time√SF
=       913 𝑘𝑚              √M
       100 𝑘𝑚/ℎ
= 9,13 hrs. √CA
= 0,13 × 60
= 7,8 minutes ≈ 8 minutes √C
Time of arrival = 9hrs 08 min +1hr +08:00  √M
= 18:08       √CA
They will arrive later than the planned time. √J

Accept CA from Q 2.1.4
1SF substitute correct values
1M changing the subject of the formula
1CA time in hours
1 C hours to minutes
1M adding time
1CA arrival time
1J justification   (7)

MP L4

Quest

Solution

Explanation

Level

2.2.1

7,6 litres = 100 km
Petrol Consumption = 913𝑘𝑚  × 7,6 𝑙i𝑡𝑟𝑒𝑠 √M
                                  100 𝑘𝑚
= 9,13 × 7,6 𝑙i𝑡𝑟𝑒𝑠
= 69,388 litres      √S
Return trip Consumption = 69,388 × 2 √M
= 138,776 litres  √CA

Accept CA from Q 2.1.4
1M concept of ratio
1S simplification
1M multiply by 2
1CA Answer (4)

MP
L2

2.2.2

1 642 cents ÷ 100 = R16,42     √M
Petrol cost = R16,42 × 138,776 litres √CA
= R 2 278,701
= R 2 278,70 √CA

Accept CA from Q 2.2.1
1M Conversion
1CA multiply by 138,78 litres
1CA Answer  (3)

MP
L2

[28]

QUESTION 3 [34 MARKS]

Quest

Solution

Explanation

Level

3.1.1

Radius: 18 cm ÷ 2 = 9 cm √M
Conversion to mm = 9 cm × 10 = 90 mm √C
Area = 𝜋r²
= 3,142 × 90 mm × 90 mm √SF
= 25 450,2 √S
= 25 450 mm² √R

1M radius
1C conversion
1SF correct substitution
1S simplification
1R rounding  (5)

M
L3

3.1.2

Radius: 250 mm ÷ 2 = 125 mm √MA
Circumference = 2 × π × r
= 2 × 3,142 × 125 mm √SF
= 785,5 + 50 mm √M
= 835,5 mm √CA

OR

Radius: 250 mm ÷ 2 = 125 mm √MA
Circumference = π × D
= 3,142 × 250 √SF
= 785,5 + 50 √M
= 835,5 mm √CA

1MA finding radius
1SF correct values
1M adding 50 mm
1CA answer

OR

1MA finding radius
1SF correct values
1M adding 50 mm
1CA answer

M
L2

3.1.3

Conversion: length = 400 mm ÷ 10 = 40 cm √C
Volume           = l × b × h
42 000 cm³      = 40 cm × 30 cm × h √SF
Height             = 42 000 cm³ ÷ 1200 cm² √S
= 35 cm √CA

1C conversion
1SF in formula
1S changing the subject of the formula
1CA answer  (4)

M
L3

3.1.4

25,4 mm = 1 inches
Diameter in mm          = 250 mm √RT
Diameter in inches      = 250 ÷ 25,4
= 9,84 inches √MA
Statement is not valid √O

1RT using correct values 250 mm and 25,4
1MA convert to inches
1 O   (3)

M
L4

3.1.5

Radius value  
= 250 mm ÷ 2
= 125 mm √MA
Height
= 0,64 × 125 mm √M
= 80 mm √CA

1MA finding the radius
1M percentage calculation
1A answer  (3)

M
L2

3.1.6

Flour : Sugar (3 cups)
7 cups : 3 cups (× 2)
14 cups : 6 cups
= 6 √MA
   3
= 2
= 2 × 7 √MA = 14 cups of flour √A

OR

7 √MA × 6 √MA = 14 cups of flour √A
3

1MA multiplying by 7
1A answer
1MA dividing by 3
1MA multiplying by 6
1A answer  (3)

M
L2

Quest

Solution

Explanation

Level

3.1.7

°C        = (°F – 32°) ÷ 1,8       SF
= (365 °F – 32°) ÷ 1,8
= 333 °F ÷ 1,8   √ S
= 185°C
≈ 190 °C √ R

1SF correct value
1S simplify
1R rounding to 10 degrees  (3)

M
L2

3.2.1

Tree diagram √√A

2A correct answer  (2)

P
L1

3.2.2

Missing value P: Boy     √RT
Missing value Z: GBB  √RT

1RT correct answer
1RT correct answer  (2)

P
L1

3.2.3

(a)

Probability (2 girls at least)     = ⁴/₈  √RT√RT
= ½    √CA

1RT Numerator
1RT Denominator
1CA simplification  (3)

P
L2

3.2.3

(b)

Probability (BGB) = 0 % √√A

2A correct percentage  (2)

P
L2

[34]

QUESTION 4 [32 MARKS]

Quest

Solution

Explanation

Level

4.1.1

Basin / Wash basin √√RT

2RT correct feature (2)

MP
L1

4.1.2

1 Window √√RT

2RT correct number (2)

MP
L1

4.1.3

Width (bedroom 1 and bedroom 2) = 4 680 + 5 130 √MA
= 9 810 √A

Length of bathroom    = 13 680 – 9810 √ M
= 3870 √CA

Wall (minus door opening)     = (3 870 – 860) mm √ M
= 3 010 mm √ CA

1MA adding correct values
1A length
1M finding length of bathroom
1CA answer
1M subtracting door width
1CA answer     (6)

M
L2

1A finding area
1C ÷ by 1 000 000
1CA for the area

1SF correct values
1A finding area
1C ÷ by 1 000 000
1CA for the kitchen area
1CA total area                  (9)
NPR

4.2.2

Area of 1 tile: 500 mm × 500 mm
= 250 000 mm² √MA (¸1000 000) √C
= 0,25 m² √A
No. of tiles needed     = 39,6414 m² ÷ 0,25 m² √MCA
=158,5656 × 1,05√M
=166,49388 tiles
= 167 tiles √CA
Statement not valid. √O

CA from 4.2.1
1M area of tile
1C divide by 1 000 000
1MCA ÷ total area by 0,25 m²
1M increasing by 5%
1CA rounded no. of tiles
1O conclusion     (7)

M
L4

4.2.3

No of boxes    = 167 ÷ 4 √MCA
= 41,75
= 42 boxes√CA
Amount for tiles          = 42 × 249,90 √M
= R10 495,80 √CA
Total amount               = 10 495,80 + 8 186,09
= R18 681,89 √M
Claim is valid. √O

CA tiles from 4.2.2
1MCA dividing tiles by 4
1 number of boxes
1M multiply by cost
1CA cost of tiles
1M adding labour cost
1O conclusion  (6)

M
L4

[32]

QUESTION 5 [30 MARKS]

Quest

Solution

Explanation

Leve l

5.1.1

Convert: 175 ÷ 100 = 1,75 m √C
BMI = Mass in kg
         Height × m²
25,1 =    Mass in kg       √SF
          1,75m ×1,75m
Mass   = BMI × (height)²
= 25,1 × (1,75)²     √S
Mass in (kg)    = 76,86875 kg
= 76,87 kg √CA

1C conversion
1SF correct values
1S changing subject of formula
1CA answer   (4)

M
L3

5.1.2

Overweight √√ J

1J conclusion  (2)

M
L1

5.1.3

The young promising rugby player can eat healthier. √√O
OR
He should exercise more regularly √√O

2O Explanation   (2)

M
L4

5.2.1

3 laps = 5 000 m ÷ 1000 √C
1 lap  = 5 km ÷ 3 √MA
= 1,67 km √CA
OR
1 lap  = 5000 m ÷ 3 √MA
= 1 666,666667 ÷ 1 000 √C
≈ 1,67 km√CA

1C divide by 1 000
1MA divide by 3
1CA distance per lap
1MA divide by 3
1C divide by 1000
1CA distance per lap  (3)

MP
L2

5.2.2

Incomplete runs: = 4 laps × 1,67 km √M
= 6,68 km √A
Complete runs  = 3 × 5 km
= 15 km + 6,68 km √MA
= 21,68 km √CA
His statement is not valid. √O
OR
          √M       √M
(4 × 1,67) + (9 × 1,67)
      √A       √A
= 6,68 + 15,03
= 21,71 km √CA
His statement is not valid. √O

1M multiply by lap distance
1A correct distance
1MA finding complete run distance
1CA finding total distance
1O correct conclusion
1M multiply 4 by lap distance
1M multiply 9 by lap distance
1A correct distance
1A correct distance
1CA finding total distance
1O correct conclusion  (5)

MP
L4

5.2.3

Difference in Time = 19 min 30 sec − 15 min 45 sec √M
= 3 min 45 sec √A
Difference in Time per lap = 3 min 45 sec ÷ 3 √M
= 1 min 15 sec √A
Statement is valid √

1M subtract time
1A difference in time
1M divide by 3
1A total time
1O reason       (5)

MP
L3

5.3.1

Number of candles
Length : 24 cm ÷ 8 cm = 3 √MA
Width : 16 cm ÷ 8 cm = 2 √MA
Height : 24 cm ÷ (1 +11 cm) = 2     √M
Total number of candles: 3 × 2 × 2 = 12 candles √CA
OR
Length: 8 cm × 3 = 24 therefore 3 will fit √MA
Width: 8 cm × 2 = 16 therefore 2 will fit √MA
Height: (11 cm + 1 cm + 11 cm) = 23 cm, therefore 2 will fit √MA
                                                       √S
Total number of candles: 3 × 2 × 2 = 12 candles √CA

1MA ÷ by correct values
1MA for 2 correct values 3 and 2
1M adding 1cm
1S simplify no of candles
1CA conclusion

OR

1MA × by correct values
1MA for 2 correct values 3 and 2
1M adding 1cm
1S simplify no of candles
1CA conclusion    (5)

MP
L3

5.3.2

Total area = 2 (H × L) + 2 (W × H) √C
= 2 (0,24 m × 0,24 m) + 2 (0,16 m × 0,24 m)  √SF
= 0,1152 m² + 0,0768 m² √S
= 0,192 m² √CA

1C converting to m
1SF correct values
1S simplification
1CA answer  (4)

M
L2

[30]

TOTAL: 150