Symbol/

Explanation

M

Method

MA

Method with accuracy

CA

Consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

RT

Reading from a table/graph/document/diagram

SF

Correct substitution in a formula

O

Opinion/Explanation

P

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

R/RCA

Rounding off /Rounding with CA

NPR

No penalty for rounding

AO

Answer only

MCA

Method with constant accuracy

QUESTION 1 [38 MARKS]

Q/V

Solution

Explanation

T&L

1.1.1

Discount percentage
                     ✔RT
=      R6140     ×   100% ✔ MA
  R160 087,72
= 3,835397...  ≈ 3,8% ✔ A

1RT numerator and denominator
1MA  multiply correct values with  100 %
1A  simplification rounded to one decimal place
AO  (3)

F

L2

1.1.2

Sub Total
✔ M                        ✔RT                ✔ MA
= R160 087,72 - R6 140 + (2 × R3 500 + R4 298,25 + R1 315,79)
= R166 561,76

1M subtracting discount
1RT all values
1MA adding accessories, on roads & transaction fee (3)

F

L2

1.1.3

                                ✔✔ O
Safety reason/as a safety feature - protect against thieves / hijackers /sunlight / door against damages
OR
                                ✔✔ O
Beautification of the car / reduce sunlight
OR
                                ✔✔ O
Longer lasting /
OR
                                ✔✔ O
Convenience
OR
                                ✔✔ O
For insurance purposes

2O reason (2)

F

L4

Q/V

Solution

Explanation

T&L

1.1.4

Interest Year 1
= 6% × R 1 250 000 = R 75 000 ✔ MA
Interest Year 2
✔ CA
= 6% × (R 1250 000 + R 75 000)  = R 79 500 ✔ CA
Interest rate 3 Months      ✔C
= 6% ÷ 4 = 1,5%    or 6% × 3/12 = 15%   ✔ M
Interest 3 Months
= 1,5% × (R 1 325 000 + R 79 500) = R 21 067,50 ✔CA
Interest earned
✔ M                         ✔CA
= R 75 000+ R 79 500 + R 21 067,50 = R 175 567,50
Interest earned is not enough / not sufficient to cover the price of the bakkie.     ✔O
OR
27 months = 2 years and 3 months or  2¹/₄ years          ✔C
1st year value
✔ MA                              ✔CA
= R1 250 000 × 6% + R1 250 000 = R1 325 000
2^nd year value     ✔ CA
= R1 325 000 × 6% + R1 325 000 = R1 404 500
Last 3 months
✔ M                                  ✔ CA
= R1 404 500 × ^6%/₄ + R1 404 500 = R1 425 567,50
Difference
✔ MA                      ✔CA
= R1 425 567,50 - R1 250 000 = R175 567,50 
                                                 ✔  O

It is not enough / not sufficient 
OR
Value the interest after 27 months
✔✔M  ✔ M  ✔✔CA                            ✔ C
= R1 250 000 × 1,06 × 1,06 × 1,015 - R1 250 000
= R1 425 567,50 -  R1 250 000    ✔MA
✔CA
= R175 567,50
Not enough / not sufficient ✔O

1MA calculating interest
1CA 1^st year value
1CA 2^nd year interest
1C conversion to years (allocated since there are 3 periods)
1M dividing % value by 4 (or the interest by 4)
1CA last 3 months interest
1M adding the interest values
1CA available amount
1O conclusion

OR 

1C conversion to years
1MA calculating interest
1CA 1^st year value
1CA 2^nd year value
1M dividing % value by 4
1CA last 3 months value
1MA subtracting
1CA available amount
1O conclusion

OR

2M multiply the principal with 106 %
1M 2^nd year value
2CA 3months rate and value
1C conversion to years
1MA subtracting
1CA available amount
1O conclusion  (9)

F

L3

Q

Solution

Explanation

T&L

1.1.5

                                                                         ✔O 
Mistake: calc. 14% on original price AND an extra 1% on accumulated price
Correct calculation should be 15% on original price
New selling price 
= R160 087,72 + 15% of R160 087,72 ✔ MA
= R160 087,72 + R24 013,16          ✔MA
= R184 100,88  ✔CA

OR

The dealer added 1% on the VAT inclusive price of ✔ O R182 500 / Calculating VAT on VAT
He should have calculated the 15% directly on the original selling price excluding VAT.
New selling price inl. VAT   ✔ A  ✔ MA
= 115% × R160 087,72
= R184 100,88    ✔ CA

OR

Mistake is calculating the increased 1% on the VAT inculsive amount.    ✔ O
The 1% must be added to the original price
Increased price incl. VAT / Verhoogde prys met BTW ✔ MA
= R182 500 + R160 087,72 × 1%
= R184 100,88 ✔CA

1O reason
1MA calculating 15%
1MA adding
1CA simplification

OR

1O stating the error or the solution
1A 115%
1MA multiplying
1CA simplification

OR

1O describing the error
1MA calculating 1% on original amount
1MA adding to VAT incl. amount
1CA simplification (4)

F

L4

Q

Solution

Explanation

T&L

1.2.1

(a)

Surface area of an open box    ✔SF
= Width × length  + 2(length × height + width × height) ✔A
= 1,374 m × 1,807 m + 2(1,807 m × 0,535 m +1,374 m × 0,535 m)
= 2,482818 m² + 2 (1,701835 m²)        ✔S
= 5,886488m²      ✔CA
Surface area of bin (bakkie) ✔MCA
= 5,886488 m²   + 2% × 5,886488 m²
= 5,886488 m²   + 0,11772976 m²
= 6, 00421776  m²    ✔CA

Or

= 1,02 × 5,886488 m²
= 6, 00421776  m²
Number of litres required
= 6,00421776 m²                ✔MA
       0,25 m²/L
=  24,01687104  ≈ 25 ℓ ✔R

1SF Substitution
1A correct values used
1S simplification
1CA total area
1MCA increasing by 2%
1CA simplification
1MA dividing with spread rate
1R rounding up litres (8)

M

L3

1.2.1

(b)

Cost = Number of 5 litre × 2 coats × Price per 5 litre         ✔ CA
= ²⁵/₅ × 2 × R549,00
= R5 490,00 ✔CA

OR
                          MCA✔
For two coats of paint = 25 × 2 = 50 litres
                                          ✔ CA
Number of 5 litre tins  = ⁵⁰/₅ = 10
Cost = 10 × R5 49 = R5 490      ✔ CA

CA from 1.2.1(a)
1CA number of 5 litres
1MCA multiply 2 by price
1CA cost for 10 litres

OR

1MCA multiply by 2
1CA number of 5 litres
1CA cost
AO (3)

F

L2

Q

Solution

Explanation

T&L

1.2.2

To protect the cargo bin's surface from scratching/rusting/ being damaged.    ✔✔O

OR

Extend the life span of a bakkie's loading box   ✔✔O

OR

To stop goods from slipping/protection of goods ✔✔O

2O reason (2)

M

L4

1.3

Time:

• Apply = 20 min × 2 coats = 40 min
  Re-coat = 4 hours = 240 min      ✔ C
  Drying time = 2 hours = 120 min

Total time needed
               ✔ M                                  ✔CA
= 40 min + 240 min + 120 min = 400 min = 6 hours 40 min
Completion
= 8 h 15 + 6 h 40 = 14 h 55
✔Time 14:55            ✔CA

OR

Apply 1^st coat (20 min)  8:15 - 8:35      ✔M
Waiting time  (4 hours)  8:35 - 12:35          ✔ MCA

Apply 2^nd coat (20 min) 12:35 - 12:55  ✔ MCA
Drying time (2 hours)      12:55 - 14:55   ✔ CA
∴ Time 14:55   or 2:55 p.m.  or   five to three in the afternoon

1C converting
1M adding times
1CA time needed
1CA time

OR

1M adding times
1MCA adding correct hours
1MCA adding correct times
1 CA time
AO (4)

M

L2

[38]

QUESTION2 [38MARKS]

Q

Solution

Explanation

T&L

2.1.1

(a)

         ✔ MA
A =216 329 - 227 665 × 100%
               227 665   ✔A
= - 4,979%  ✔A
≈ - 5%       ✔ RCA

1MA subtracting correct values
1A denominator
1A negative simplification
1RCA value of A (4)

D

L2

2.1.1

(b)

                  ✔MCA
-12  ;-5 ; -2 ; -1 ; 0 ; 2 ;  5 ;  10 ;  13 ;  13 ;  16  ; 18 ;  19  ;  40   
                    ✔ M
Median/Mediaan = 5% + 10%
                                     2
= 7,5%                       ✔ CA

CA from 2.1.1(a)
1MCA arranging
1M median concept
1CA median (3)

D

L3

2.1.2

                      ✔ A
As the year increased the value of the imports of make-up and skincare increased.     

1A year increased
1A  value increased (2)

D

L4

2.1.3

 ✔A                               ✔O
Fr:  import  share  increased  from 2013  to 2014 ,  but decreased in 2015.
Pf: import share decreased from 2013 to 2014, but increased in 2015

1A product
1O reasoning
1A product
1O reasoning (4)

D

L4

2.1.4

 ✔O                                                                             ✔O
No. Too many sectors and one pie chart cannot be used as different years need to be shown.
OR
 ✔ O                                                   ✔O
No. Too many sectors/columns; some are too small /negligible.
OR 
✔ O                                                     ✔ O
No. Negative values will be difficult to indicate.   ✔O                                   
OR/OF 
 ✔ O
No. Percentages do not add up to 100%.

1O No
1O reason (2)

D

L4

Q/V

Solution/Oplossing

Explanation/Verduideliking

T&L

2.1.5

Percentage imports and average growth of Personal Care and Cosmetics to Australia

1A first point
1A last point
3 × 1A Every other two points correctly plotted
1A Joining (6)

D

L3

Q

Solution

Explanation

T&L

2.2.1

Total cost = Basefare+ 10 × cost per mile
✔ RT                ✔ RT
= $20,00 + 10× $5,00 per mile
= $70,00      ✔CA

2RT using correct values
1CA value of B if only 1 value is incorrect (3)

F

L2

2.2.2

Maximum distance (in miles)/Maksimum afstand(in myl)
= $4,65 ✔ RT
   $0,90 ✔ M
= 5,166… ✔ CA
≈ 5      ✔R

1RT reading correct values from table
1M dividing
1CA simplification
1R rounding (4)

F

L3

2.2.3

1 hour 9 minutes = 69 minutes ✔ C
Post trip cost    ✔SF
= 69 min × $0,45 / min + 29,73 mi × $3,55 /mi
=$31,05+ $105,5415 ✔ S
= $136,59 ✔ CA
Upfront cost     ✔SF
= $8 + 29,73mi × $3,55/mi
= $113,54      ✔CA
Difference = $136,59 - $113,54 = $23,05 ✔ S
The statement is correct ✔O

OR

Difference = Post trip cost - Upfront cost
✔ C                              ✔SF                            ✔SF
= 69 min × $0,45 / min + 29,73 mi × $3,55 /mi - ($8 + 29,73mi  × $3,55/mi)
= 69 min × $0,45 / min  - $8 ✔✔✔S
           ✔ CA
= $23,05
The statement is correct/Die stelling is korrek.      ✔O

1 C converting to minutes
1SF substituting correct values
1S simplification
1CA post trip cost
1SF substituting correct values
1CA upfront trip cost
1S difference
1O conclusion

OR

1C time to minutes
1SF values into 1^st formula
1SF values into 2^nd formula
3S simplification
1CA difference
1O conclusion (8)

F

L4

Q

Solution

Explanation

T&L

2.2.4

To cover cost for idle/wasted time when a vehicle could have been used to assist someone when you cancel the booking.  ✔✔O
OR
Penalty for booking made if one does not finally use the vehicle (time wasting).      ✔✔ O
OR
Prevent hoax calls         ✔✔ O
OR
To cover petrol costs and wear and tear of the vehicle ✔✔ O
OR
For the company to make a profit / avoid losses      ✔✔ O

2O reasoning  (2)

F

L4

[38]

QUESTION 3 [39MARKS]

Q

Solution

Explanation

T&L

3.1.1

P(Coke & water) = [4/9] ✔A
= 0,44✔CA

1A numerator
1A denominator
1CA decimal number
NPR (3)

P

L2

3.1.2

South East OR East of South OR SE.    ✔✔A

2A direction (2

MP

L2

3.1.3

(a)

                 ✔A
The start is at 1 400 m running to 1 565 m at the 5 km mark and then 1 708 m at the 10 km mark.   ✔A

1A for height 1 400 m
1 A for height 1 708 m
[Accept increase in height above sea level/altitude] (2)

MP

L4

3.1.3

(b)

Lowest point : highest point
✔ RT          ✔RT
= 1 166 m : 1 708 m
= 1 : 1,464837... ✔CA
≈  1 : 1,46  or 1 : 1,5

2RT correct values
1CA ratio
NPR (3)

MP

L2

3.1.4

To take struggling runners out of the race because they are not coping.                    ✔✔O
OR
Security reasons (guards and health personnel deployed in strategic sections along the race course during specific times).   ✔✔O
OR
For runners to know whether they have a realistic chance of finishing race within the time allowed for the race. ✔✔O
OR
Also helps organisers to plan appropriately for other scheduled events. ✔✔O
OR
If the road was closed it needs to be opened.      ✔✔O

2O understanding/reason (2)

MP

L4

Q

Solution

Explanation

T&L

3.1.5

The average speed required to beat the cut-off 2:
Speed(marathon) = 31,5km ✔RT
                               5h15min ✔M
= 6 km/h ✔CA
Speed(½ marathon) =16,5km      ✔MA
                                      5h
= 3,3 km/h ✔CA
The claim is correct (6 - 3,3 = 2,7 km/h).   ✔O
OR
                                    ✔M
Speed/Spoed(½ marathon)  = 16,5 km ÷ 5h = 3,3 km/h ✔CA

Increased  speed for full marathon  = (3,3 + 2,7) km/h = 6km/h ✔MA
               ✔MA
Distance  = 6 km/h × 5,25h = 31,5 km ✔CA
Correct   ✔O
OR

         ✔M                          ✔CA
Speed (½ marathon)  = 16,5 km ÷ 5h = 3,3 km/h
                     ✔MA
Increased  speed for full marathon  = (3,3 + 2,7) km/h = 6km/h
                                 ✔MA
Time to cut-off = 31,5km  = 5,25 h ✔CA
                             6km/h
Correct/Korrek      ✔O

1RT  correct values (dist. & time)
1M   calculating speed / change the subject
1CA simplification
1MA calculating speed
1CA 2nd speed
1O conclusion

OR

1M   calculating speed / change the subject
1CA simplification
1MA calculating incr. speed
1MA calculating distance
1CA distance
1O conclusion

OR

1M   calculating speed / change the subject 
1CA simplification
1MA calculating incr. speed
1MA calculating time
1CA time
1O conclusion

(6)

MP

L4

3.2.1

20 ℓ  =  20 × 1 000 cm³ ✔ C
Inner diameter   = 31,2 cm - 2 × 0,2 cm
= 30,8 cm    ✔A
                        ✔MCA
V = 3,142 × (30,8cm ÷ 2)² × height
                             ✔SF
20 000 cm³ = 3,142 × (30,8 cm)²  × H
                                       2
H  =                20000 cm³          ✔ M
      3,142 ×  237,16cm cm²      ✔ S
=  20000    
 745,15672
= 26,84 cm        ✔CA

1C conversion
1A calculating inner diameter
1MCA radius
1SF correct values
1M changing the subject
1S simplification
1CA height (7)

M

L3

Q

Solution

Explanation

T&L

3.2.2

(a)

Area of base of 1 bucket
                       ✔ A

= 3,142 × (15,6 cm)²
= 764,63712 cm²          ✔CA
Area of base of 11 buckets
= 11 × 764,63712 cm² = 8 411,00832 cm²    ✔ CA
Area of base of pallet
                           ✔ SF
= 100 cm × 120 cm  = 12 000 cm² ✔ A
Difference  = 12 000 cm² - 8 411,00832 cm²
= 3 588,99168 cm²        ✔CA

1A radius
1CA simplification
1CA multiply by 11
1SF correct values
1A rectangular area
1CA area unused
NPR (6)

M

L3

3.2.2

(b)

                 ✔A
120 cm = 31,2 × 3 + C
C = 120 cm - 31,2 cm × 3 ✔ M
= 26,4cm              ✔ CA

1A 120 cm
1M multiplying and subtracting
1CA finding C (3)

M

L4

3.2.3

Length occupied by 4 buckets
                  ✔ MA                ✔ A
= 4 × 31,2 cm  = 124,8 cm
Length should be increased by
        ✔CA
= 124,8 - 120   ×  100%      ✔ M
          120
= 4%  ✔CA

OR

Length occupied by 4 buckets
                  ✔ MA                     ✔A
= 4 × 31,2 cm  = 124,8 cm
120 cm is 100% 
                     ✔ M
124,8 cm is 124,8 × 100% = 104%      ✔CA
                     120
∴ 4% increase    ✔CA

1MA multiplying
1A correct length
1CA substituting
1M % change
1CA simplification

OR

1MA multiplying
1A correct length
1M multiply with 100%
1CA simplification
1CA simplification  (5)

MP

L3

[39]

QUESTION 4 [35 MARKS]

Q

Solution

Explanation

T&L

4.1.1

Total for these capsules
  ✔ MA            ✔✔MA              ✔MA          ✔ MA
= 23 ×  £27 + 5 × £27 × 90% + 8 × £22 + 7 × £25,50
= £621 + £121,50 + £176 + £178,50
= £1 097      ✔ CA
Rand value/waarde = £1 097 × R16,58/ £
= R18 188,26        ✔ C
                       ✔ O
∴ the statement is not correct

OR

Without discount for 5
                     ✔MA        ✔ MA
= 28 × £27 + 8 × £22 + 7 × £25,50 ✔MA
=  £756 + £176  + £178,50
= £1 110,50 ✔CA
Discount for 5/ = 5 × £27 × 10%
= £13,50      ✔A
Total ticket price
= £1 110,50 - £13,50= £1 097          ✔CA
Rand value
= £1 097 × R16,58 /£  = R18 188,26    ✔ C
                        ✔ O
NOT correct

OR

Cost of Capsule 24 + Cost of Capsule 30 - Discount for 5 Adults
✔ MA                ✔ M               ✔ MA            ✔ A
(18 × £27 + 7 × £22 + 2 × £25,50) + (10 × £27 + 1 × £22 + 5 × £25,50) - 5 × £27 ×10% 
           ✔ CA
=  £691 + £419,5 - £13,5 = £1097 ✔CA
Rand value
= £1 097 × R16,58 /£ = R18 188,26      ✔ C
NOT correct  ✔O

3MA multiply tickets by price
2MA discount for 5
1CA total for 2 capsules
1C pounds to rand
1O conclusion

OR

3MA multiply tickets by price
1CA simplification
1A discount
1CA total
1C pounds to rand
1O conclusion

OR

2MA multiply tickets by price
1M adding costs
1A discount
1CA simplification
1CA total
1C pounds to rand
1O conclusion

F

L4

Q/V

Solution/Oplossing

Explanation/Verduideliking

T&L

OR
Ticket price in rand:
Adult: 27 × 16,58 = R447,66  ✔ C
Children: 22 × 16,58 = R364,76
Senior citizens: 25,5 × 16,58 = R422,79
Discount adult = R44,77 ✔ A
Online ticket price = R402,89
 ✔ MA                  ✔MA                  ✔ MA                ✔ MA
Total price = (23 × R447,66) + (5 × R402,89) + (8 × R364,76) + (7 × R422,79)
= R18 188,24    ✔ CA
NOT correct     ✔ O

OR

1C conversion
1A discount
4×1MA multiply tickets by price
1CA total
1O conclusion (8)

4.1.2

(a)

Circumference of the wheel
= 2× π × radius
= 2 × 3,142 × 197        ✔ SF
=  1 237,948 feet    ✔ CA

1SF correct values
1CA circumference
NPR (2)

M

L2

4.1.2

(b)

Distance = 1237,948  feet    ✔ MA
                        32
= 38,685875 feet
=  38,685875 m      ✔C
         3,28
= 11,794…m ≈  11 m    ✔ R

OR

Circumference in metre
= 1237,948 = 377,4231707m    ✔ C
          3,28
Distance apart
=  377,4231707  ✔ MA
            32
= 11,794 m
≈11 m      ✔ R

CA from 4.1.2(a)
1MA dividing by 32
1C conversion
1R rounded distance [also accept 12m]

OR

1C conversion
1MA dividing by 32
1R rounded distance (3)

M

L2

4.2.1

✔ M            ✔ RT
Difference= 624 000 - 312 600
                ✔ CA
= 311 400  or/of  311,4 thousand

1RT correct values
1M subtraction
1CA difference in thousands (3)

D

L2

Q

Solution

Explanation

T&L

4.2.2

       ✔RT
P(Midlands West & East) = 609600 + 295000  × 100%
                                                 7146600
             ✔S
= 904 600  × 100% ✔ M
7 146 600
= 12,65776….%
≈  12,66%      ✔CA

1RT numerator & denominator
1S simplification
1M multiply by 100%
1CA probability
NPR
AO  (4)

P

L3

4.2.3

Ratio =1 157,0  ✔RT
             378,3
= 3,0584  ✔CA
∴ The statement is valid/Die bewering is geldig.      ✔O

OR

Number of business visitors  = 378,3 thousand 
And holiday visitors= 1 157 thousand
  ✔ RT                      ✔CA
378,3 thousand × 3 = 1 134,9 thousand
∴The statement is valid/Die bewering is geldig. ✔O

OR
    ✔ RT                ✔CA
1 157 000 ÷ 3 ≈ 385 667
∴The statement is valid   ✔✔O

1RT values
1CA simplification
1O conclusion

OR

1RT values
1CA simplification
1O conclusion

OR

1RT values
1CA simplification
1O conclusion
[No penalty for omitting thousand]      (3)

D

L4

4.2.4

175,1     324,8        405,7 480,5           562,7        600,8       ✔MA
762,6     806,8      856,2      1594,0      3 556,0     
       ✔A
Q1/K1 = 405,7          Q3/K3 = 856,2              ✔A
                          ✔M
IQR/IKO = (856,2 - 405,7) × 1 000
=  450,5 × 1000
= 450 500            ✔CA

1MA order, ascending or descending 
2A Q1 and Q3
1M subtracting quartiles
1CA  IQR value
[No penalty for omitting thousand] (5)

D

L3

Q

Solution

Explanation

T&L

4.2.5

Tourism boosts the economy (selling and buying) of the country.  ✔✔O
OR
Tourism assists people to know the places they want to visit and  be  prepared/  exposes  the  goods  and  services  of  a country      ✔✔ O
OR
Brings income to the country and more tourist stimulate the economy. / GDP grows. ✔✔ O
OR
Help to promote  Social and Cultural interaction.        ✔✔ O

2O reason financial
2O environmental reason
2O economic reason
2O humanitarian  reason (2)

D

L4

4.2.6

                           ✔M
Total = 162 666,5455 × 11 ≈  1 789 332  ✔ R
Known data total = 471 928 + 170 113 + 119 639 + 107 230 + 76 496 + 120 343 + 179 450 + 226 003 + 172 282
= 1 643 484    ✔ A
Wales = NE + 30 440
NE + NE + 30 440 + 1 643 484 = 1 789 332    ✔  MA
2NE = 115 408
NE = 57 704    ✔ CA

OR
Mean value  ✔  M
=   NorthEast + Wales + Other
                      11
                   ✔  MA
NE +  NE+ 30440 + 1 643484 =  162 666,5455
                   11
                ✔  S
2NE + 1 673924    =  1 789 332,001
2NE =  115408,001    ✔M
   2             2   
NE=  57 704,00025
Direct employment of North East = 57 704      ✔ R

1M multiplying with 11

1R rounding

1A known total

1MA two unknowns

1CA simplification

OR/OF

1M concept of mean

1MA two unknowns

1S simplification

1M dividing by 2

1R rounding

(5)

D

L4

[35]

TOTAL:150

1.1.1

If the values are swopped, give only 1 mark

1.1.2

If the candidate starts with R153 947,72  and not show how it was calculated, Max 2 marks
If they start with R189 880,76 and do a reverse VAT calculation, 0 marks.

1.1.5

Only calculation done and no explanation, Max 3 marks

1.2.1 (a)

Early rounding leading to a surface area of 6 and the litres required 24, Max 6 marks

1.2.1 (a)

Changing the formula by replacing a + with a ×, max 6

1.3

14:35 is worth 3 marks showing calculations; 18:55 is worth 3 marks with calculations. No calculations shown for these answers, 0 marks.

2.1.1 (b)

Omitting the value of A, max 2 marks provided it is arranged.
Using the % share columns' data is a break-down, 0 marks, since not all data is shown.

2.1.2

"constantly increasing" is worth 1 mark.

2.1.3

"Both Pf and Fr fluctuate", max 3 marks.

2.1.5

One or two points plotted wrong, max 5 marks. Three or four plotted wrong, max 4 marks etc.

2.2.1

Adding the costs on the table is a break-down,0 marks.
Wrong formula, 0 marks

2.2.2

Wrong formula,0 marks. Two wrong values,0 marks.
Incorrect order,max 1 mark.

2.2.3

Converting mark must be given if it is substituted without showing the time conversion.

2.2.3

After calculating both Post and Upfront costs the difference need not be shown, then the conclusion carries 2 marks.

3.1.1

Written as 4:9 or 4 out of 9,  give 2 marks,

3.1.3 (b)

If ratio values are swopped, max 2 marks.

3.1.5

If they use 42km and 7 hours or 25,5 km and 4h15min, max 4 marks.

3.2.1

If both thicknesses not subtracted, H = 26,156 cm, max 6 marks

3.2.2(a)

Max of 4 marks if only one bucket's area is subtracted from pallet's area.

3.2.2(b)

No unit was specified, answer can be in mm or cm, thus 264mm is accepted.
C= 120 cm - 100 cm = 20 cm, 3 marks.

4.1.1

Calculating discount on senior citizen, max 7 marks.
Calculation:

• Adults 1 mark
• Discounted adults 2 marks
• Children 1 mark
• Senior citizens 1 mark
• Adding 1mark
• Currency conversion 1 mark
• Conclusion  1 mark

4.2.1

If the values are swopped and the answer is negative, max 2 marks

4.2.2

With only 1 value in the numerator, max 2 marks.

4.2.4

• VFR -not ordered:  1594,0 - 762,6 = 831,4 thousand, 2 marks
• Business ordered:  609,6 - 273,0 = 336,6 thousand, 3 marks
• Holiday ordered: 1 157,0 - 273,0 = 884 thousand, 3 marks
• Wrong column used and not ordered, 0 marks