MATHEMATICAL LITERACY
PAPER 2
GRADE 12 
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2018

SYMBOL

EXPLANATION

M

Method

MA

Method with accuracy

CA

Consistent accuracy

RCA

Rounding consistent accuracy

A

Accuracy

C

Conversion

S

Simplification

RT/RG

Reading from a table/graph/diagram

SF

Correct substitution in a formula

O

Opinion/Example/Definition/Explanation/Justification/Verification

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

 

MEMORANDUM 

QUESTION 1 [37 MARKS]

Ques

Solution

Explanation

T/L

1.1.1

Number of days = 10              ✔A
Number of hours per day = 10          ✔A
Total hours = 10 x 10 =100 ✔CA

1A 10 days
1A 10 hours
1CA 100 hours
AO  (3)

M
L2

1.1.2

VAT on teens ticket
                    ✔RT
=     R50    x     14         ✔MA
                        114
= R6,14035
≈ R6,14          ✔'RCA
OR
                                ✔
RT

Price without VAT = R50   or     R50
                                114%          1,14   ✔MA
≈ R43,86
VAT = R50 – R43,86
= R6,14                              ✔'RCA

1RT using correct value
1MA for  by multiplying
1RCA VAT rounded to nearest cent

OR

1RT using correct value
1MA for dividing by 114% (1,14)
1RCA VAT rounded to nearest cent  (3)

F
L2

1.1.3

                              ✔(A
P(Friday) =          ✔ CA
                  10
1/5 or 20% or 0,2   ✔CA

1A numerator
1CA denominator (Q 1.1.1)
1CA simplification
AO (3)

P
L2

Ques

Solution

Explanation

T/L

1.1.4

For 23 April: 
                                               ✔RT                    ✔M
Total ticket cost = 2 x R150 + R50 + R50 + R20
= R420 ✔CA

For 20 April:
                                              ✔A
Total ticket cost = 2 x R75 + R25 + R50 + R20
= R245 ✔CA                                             ✔CA
Amount saved in rand = R420 – R245 = R175
Percentage savings  =  175/420 x  100%   ✔M
= 41,66...% ✔CA

Mrs Abrahams statement is VALID       ✔O

OR

For 23 April:
                      ✔RT                                        ✔M
Total ticket cost = 2 x R150 + R50 + R50 + R20
= R420 ✔CA

For 20 April:
                                             ✔A
Total ticket cost = 2 x R75 + R25 + R50 + R20
= R245  ✔CA
Percentage of original = 245/420 x 100%   ✔M
= 58,333...% ✔CA
Percentage savings = 100% – 58,333...%
= 41,66...%            ✔CA
Mrs Abrahams statement is VALID ✔O

1RT all correct values
1M adding values
1CA total cost
1A calculating adult and pensioner ticket price
1CA total cost
1CA amount saved
1M multiplying by 100%
1CA percentage
1O verification

OR

1RT all correct values
1M adding values
1CA total cost
1A calculating adult and pensioner ticket price
1CA total cost
1M multiplying by 100%
1CA simplification
1CA percentage
1O verification
NPR (9)

F
L4

1.2.1

Eastern Cape or EC ✔✔RT

2RT correct province (2)

Data
L2

Ques

Solution

Explanation

T/L

1.2.2

Supporting the needy /poor / sick / elderly / orphaned ✔✔O
OR
Supporting the physically / mentally challenged ✔✔O
OR
Any other suitable reason to explain why grants are given. ✔✔O

2O reason (2)          

Data
L4

1.2.3

                                                          ✔✔O
No or The data cannot be represented by a single pie chart
Two categories / types / topics of data          ✔✔O
OR
There are too many sectors (18) to be accurately/ easily represented using a single pie chart.    ✔✔O

OR
                      ✔O
                                 ✔✔O
Not easy to compare if it is a single pie chart.

1O opinion
2O reason (3)          

Data
L4

1.2.4

Total number of citizens receiving social grants =                                             ✔M
2 756 621 + 2 405 846 + 3 921 846 + 463 599 + 1 205 069 + 987 337 + 1 429 411 + 1 506 147 + 2 474 055      ✔'RT
= 17 149 931    ✔CA
Limpopo percentage   ✔CA
=  2 405 846  ×  100%   ✔M
   17 149 931

≈ 14 ,028313 % ✔ CA
OR also accept
                                       ✔ 'M              ✔RT
Total number in Limpopo = 2 405 846 + 1 324 000
= 3 729 846  ✔CA

Limpopo percentage
= 2405846    ×  100%      ✔ CA
    3729846 ✔ M
= 64,50%     ✔CA

1M adding
1RT for all correct values
1CA for number of people
1CA for dividing in correct order
1M calculating %
1CA simplification
OR
1M adding
1RT for all correct values
1CA for number of people
1CA for dividing in correct order
1M calculating %
1CA simplification
NPR (6)

Data
L3

Ques

Solution

Explanation

Topic/L

1.2.5

Gauteng
Employed citizens : social grants recipients
4 942 000 : 2 474 055 ✔RT         ✔M
1 : 0,500 6  ✔CA

Western Cape
2 266 000 : 1 506 147      ✔RT
1 : 0,664672                   ✔CA
Gauteng✔O
OR
Gauteng                                   ✔M
Employed citizens : social grants recipients
4 942 000 : 2 474 055 ✔RT
1,99753 : 1✔CA

Western Cape  ✔RT
2 266 000 : 1 506 147  ✔CA
1,5045  : 1
Gauteng     ✔O

1M writing as a ratio
1RI ratio with correct values
1CA Unit ratio
1RI ratio with correct values
1CA Simplification
1O conclusion
OR
1M writing as ratio
1RI ratio with correct values
1CA Unit ratio
1RI ratio with correct values
1CA simplification
1O conclusion (6)

Data
L4

   

[37]

 

QUESTION 2 [40 MARKS]

Ques

Solution

Explanation

T/L

2.1.1

32 OR 31               v'v'A

2A correct number of days (2)

M

L2

2.1.2

Total credit                                                        ✔MA
= – R37,81 + (–R200,00) + (–R0,01) = – R237,82    ✔CA
Total debit                                            ✔MA
= R200,00 + R4,00 + R31 716,69 + R10 770,00 = R42 690,69    ✔CA
                                    ✔MA
Closing balance = R42 690,69 + (– R237,82) = R42 452,87
OR 

                                   ✔MA
R37,81 + R200 + R0,01 = R237,82 credit        ✔CA
Total debit                                                 ✔MA
= R200,00 + R4,00 + R31 716,69 + R10 770,00 = R42 690,69 ✔CA
                                    ✔MA
Closing balance = R42 690,69 – R237,82 = R42 452,87

1MA adding credits
1CA simplification
1MA adding debits
1CA simplification
1MA adding credits to debits
OR
1MA adding credits
1CA simplification
1MA adding debits
1CA simplification
1MA adding credits to debits

 [Using the Account Summary: Closing Balance
= 42 690,69 – 200,01 – 37,81 = 42 452,87

max 4 marks] (5)

F

L3

2.1.3

                                                                     ✔✔O
Safety reasons OR prevent Fraud / Confidentially/ Account number private to Mr Son only

2O Explanation (2)

F

L4

2.1.4

Insurance premium
= R42 452,87 ÷ R1 000          ✔M
= 42,45287       ✔CA
≈ 43 ✔R

Insurance cost
= 43 × R3,50  ✔MA
= R150,50       ✔CA

1M dividing by 1 000
1CA simplification
1R rounding up
1MA multiplying correct values
1CA correct premium
[not rounding up max 3 marks] (5)

F
L3

Related Items

Ques

Solution

Explanation

 Marks

T/L

2.1.5

The bank owes Mr Son R 37,81            ✔✔O
OR
The account has a credit balance                  ✔✔O
OR
Over-payment from previous months.                ✔✔O

2Oreason

(2)

F
L4

2.1.6

Does not have large amounts of cash to purchase expensive goods    ✔✔O
OR
Easier / convenient to settle expensive items with smaller monthly payments ✔✔O
OR
Loyalty points  ✔✔O
OR
Safety          ✔✔O
OR
Did not have money when he saw something he likes. ✔✔O
OR
To be able to see on what he spent his money. ✔✔O
OR
Credit card could be used in times of crisis.    ✔✔O
OR
Some people use credit merely because it is easily accessible (available) ✔✔O
OR
To build a good credit record.                  ✔✔O
OR
He is using the interest free period.                  ✔✔O

2O reason

(2)

F L4

Ques

Solution

Explanation

T/L

2.2

Distance = average speed × time    ✔SF
34 km         = 85 km per hour × time
Time = 0,4 hours    ✔A
= 24 minutes    ✔C
Mr Son left home at 24 minutes before 12:10
= 11:46          ✔CA
He did NOT leave at 11:40    ✔O
OR
Time diff. = 12:10 – 11:40 = 30 min = 0,5 hours    ✔A
✔SF                                ✔CA            ✔O
Distance = 85 km/h × 0,5 h = 42,5 km more than 34 km
Mr Son did NOT leave at 11:40 but a bit later    ✔o

1SF substitution of both values
1A time in hours
1C time in minutes
1CA simplification
1O conclusion
OR
1A time in hours
1SF substitution
1CA distance
1O comparing
1O conclusion (5)  

M
L4

2.3.1

No data was available for Japan      ✔✔O
OR
Japan did not provide data  ✔✔O
OR
The books were not published in time                         ✔✔O

2O no data available (2)  

Data
L4

2.3.2

Range = maximum – minimum I              ✔M
463 223                 = maximum – 4 612    ✔A
Maximum = 463 223 + 4 612
= 467 835    ✔A

1M range concept
1A identifying minimum
1A calculating the maximum  (3)  

Data
L2

2.3.3

                                        ✔MA
4 612; 6 373; 8 870; 24 177; 43 146; 47 352; 64 117; 76 434; 77 910; 93 600 ; 95 483; 184 000; 304 912; 444 000; 467 835   ✔CA
Median = 76 434            ✔✔CA

CA from2.3.2
1MA all values in correct order
1CA maximum value
2CA median
AO (4)  

Data
L2

2.3.4

no mode ✔✔A

2A no mode (2)

Data
L2

Ques

Solution

Explanation

T/L

2.3.5

7 countries ✔✔✔A

3A correct number of countries
[Listing ALL 7 without counting
max 2 marks] (3)

Data
L2

2.3.6

            ✔A
P = 12/15  × 100%   ✔A
= 80%   ✔CA

1A numerator
1A denominator 
1CA probability as a percentage (3)

Prob.
L2

   

[40]

 

QUESTION 3 [36 MARKS]

Ques

Solution

Explanation

T/L

3.1

Area of display = length x width
                  ✔SF
= 48 inches x 36 inches
     ✔C                ✔C
= 48 x 25 mm ÷ 10 x 36 x 25 mm ÷ 10
= 120 cm x 90 cm = 10 800 cm2         ✔CA
Total area of 25 displays
= 10 800 cm2 x 25     ✔M
= 270 000 cm2         ✔CA
Amount of whiteboard paint needed   
                                      ✔M
= 270 000 cm2 ÷ 50 cm
                    ✔
CA     ✔C
= 5 400 mℓ ÷ 1 000 = 5,4 litres
5ℓ is not enough.      ✔O
OR
5 litres of paint can cover
5 ℓ × 1 000                 ✔C
= 5 000 m ℓ × 50  ✔M
= 250 000 cm2             ✔CA
Area of display = length x width 
                      ✔SF
= 48 inches x 36 inches         
                                    ✔C                ✔C
= 48 x 25 mm ÷ 10 x 36 x 25 mm ÷ 10
= 120 cm x 90 cm = 10 800 cm2    ✔CA
Total area of 25 displays
= 10 800 cm2 x 25          ✔M
= 270 000 cm2        ✔CA
5 ℓ is NOT enough.   ✔O
OR
5 litres of paint can cover
5 ℓ × 1 000  ✔C
= 5 000 m ℓ × 50        ✔M
= 250 000 cm2          ✔CA
Display
✔C                                  ✔C
48 x 25 mm = 1200 mm = 120 cm
36 x 25 mm = 900 mm = 90 cm
Area = 120 cm x 90 cm              ✔SF
= 10 800 cm2                      ✔CA
Spray paint is enough for = 250 000  ✔M
                                            10 800
= 23,148 boards
5 ℓ is not enough              ✔O
OR
Area of display

= 48 inches x 36 inches  ✔SF
= 1 728 inches2    ✔CA
= 1 728 x 625 mm2    ✔C
= 1 080 000 mm2
=1 080 000 ÷ 100 cm2
= 10 800 cm2            ✔CA
Total area of 25 displays   
                         ✔M                    ✔CA
= 10 800 cm2 x 25 = 270 000 cm2
Amount of whiteboard paint needed   
                         ✔M
= 270 000 cm2 ÷ 50 cm2
= 5 400 mℓ ~ 1 000
= 5,4 litres          ✔C
5ℓ is not enough.        ✔O

 

1SF substituting correct values
1C converting inches to mm
1C converting mm to cm
1CA area of one display
1M multiplying by 25
1CA calculating total area
1M working with ratio
1CA calculating paint used
1C converting to litres
1O conclusion
OR
1C converting to mℓ
1M multiplying by 50
1CA area
1SF substitution
1C conversion to mm
1C conversion to cm
1CA one display board area
1M multiplying by 25
1CA total area
1O conclusion
OR
1C converting to mℓ
1M working with ratio
1CA calculating area that paint can cover
1C converting inches to mm
1C converting mm to cm
1SF substituting correct values
1CA area of one display board
1M dividing
1CA number of boards
1O conclusion
OR
1SF substitution
1CA area in inches
1C converting to mm2
1CA area of one display board
1M multiplying by 25
1CA total area
1M dividing by rate
1CA ml needs
1C converting to litre
1O conclusion (10)

 

M
L4

Ques

Solution

Explanation

T/L

3.2

Total Surface Area of cylinder A
= π × diameter × height
= 3,142 × 30 × 30        ✔SF
= 2 827,80 cm2✔CA
Total Surface Area of decorative sticker for cylinder A   = 2 827,80 cm2 + (1 × 30) cm2 ✔M
= 2 857,80 cm2 ✔CA
Total Surface Area of cylinder B
= π × diameter × height
= 3,142 × 40 × 20
= 2 513,60 cm2 ✔CA
Total Surface Area of decorative sticker for cylinder B
= 2 513,60 cm2  + (1 × 20) cm2
= 2 533,60 cm2 ✔CA
Correct, B will require less ✔O
OR
Total Surface Area of sticker for cylinder A
= [(π × diameter) + 1] × height ✔M
                                  ✔M
= [(3,142 × 30) + 1] × 30 ✔SF
= 2 857,8 cm2 ✔CA
Total Surface Area of sticker for cylinder B
= [(π × diameter) + 1] × height 
                                   ✔SF
= [(3,142 × 40) + 1] × 20
= 2 533,6 cm2 ✔CA
Correct, B will require less ✔O

1SF correct values
1CA calculating area
1M adding area of overlap
1CA calculating area of sticker
1CA area of cylinder B
1CA area of sticker B
1O conclusion
OR
1M formula
1M for adding 1 to circumference
1SF substitution
1CA calculating area
1SF correct values
1CA calculating area
1O conclusion

[Max 5 marks if the overlap is left out] (7)

M

L4

Ques

Solution

Explanation

T/L

3.3.1

Easily accessible to all stands ✔✔R
OR
Would not waste any time looking for the stand.✔✔R
OR
Any other suitable reason

2R reason (2)

Maps

L4

3.3.2

Maximum number of HEI from the USA
= 6 × 6 ✔M ✔A
= 36 ✔CA

1M multiplying by 6
1A correct USA’s stands
1CA simplification
AO   (3)

Maps

L2

3.3.3

                  ✔A
P(Not China) =288/324 ✔  A
                        ✔ CA
=    8    or 0,89 or 88,89%
     9
OR
P(Not China)48 ✔A
                      54  ✔ A
=  8  or 0,89 or 88,9% ✔CA
    9
OR
      ✔M                                 ✔A                        ✔M         ✔A
P(not China) =   54 - 6  × 100%  OR  P(Not China) =  1 -  6      
                         54                                                       54
            ✔CA                                    ✔ CA
 = 88,89%                                       = 8/9

1A numerator
1A denominator
1CA simplification
OR
1A numerator
1A denominator
1CA simplification
OR
1M subtracting from whole
1A numerator
1CA simplification (3)

P

L2

3.3.4

Delivery entrance 3 ✔✔A
L01 ✔✔A

2A Delivery entrance
2A stand (4)

Maps

L3

3.3.5

L 42 ✔✔A

2A stand number (2)

Maps

L2

Ques

Solution

Explanation

T/L

3.3.6

                                                                ✔A
Length of Information centre on plan = 70 mm
                                          ✔C
Scale  = 70 mm : 24 500 mm ✔M
= 70mm : 24 500 mm    ✔M
  70mm     70 mm
=  1  :  350      ✔CA
OR
                          ✔A
Scale :  70 mm = 24,5 m        ✔M
1mm = 0,35 m    ✔M
                                     ✔C
Scale         = 1 : 350              ✔CA

1A measuring with ruler  (Accept a range of 66 74; dependant on provincial printing)
1M concept of ratio
1C converting to sameunit of measurement
1M dividing by 70 mm
1CA simplified scale
OR
1A measurement
1M ratio concept
1M unit ratio
1C converting to like units
1CA simplified scale
NPR (5)

Maps

L3

   

[36]

 

QUESTION 4 [37 MARKS]

Ques

Solution

Explanation

T/L

4.1.1

(a)

R105  = cost of T-shirt + cost of Shorts + printing
        ✔A          ✔A                ✔A
= R50,00 + R35,00 + 2 × R10
OR
= (R50 + R10) + (R35 + R10)
= R60 + R45        ✔✔✔A

1A cost of T-shirt
1A cost of short
1A printing  (3)

F

L2

4.1.1

(b)

                                   ✔SF
Total cost = R10 000 + R105 × 500
= R62 500    ✔A

1SF substitution
1A simplification
AO

[Using the selling price 0 marks] (2)

F

L2

4.1.2

87,5 thousand rand = R87 500 ✔A
              ✔M
A  = R87500,00
         R125,00
= 700 ✔CA

B = 800  × 125 ✔M
           1 000 ✔A
=  100✔CA
OR
                          ✔RT ✔M
A = 500 × 87,5  = 700        ✔CA
          62,5
         ✔RT         ✔M
B = 800 × 62,5   = 100        ✔CA
          500
OR
                             ✔✔✔A
A = 500 + 200 = 700
(because 25 + 62,5 = 87,5) 
                            ✔✔✔A
B = 50 × 2 = 100
(because 400 × 2 = 800)

1A writing value in full
1M dividing
1CA value of A
1M multiplying by 125
1A dividing by 1 000
1CA value of B
OR
1RT values from table
1M using ratio
1CA value of A
1RT values from table
1M using ratio
1CA value of B
OR
3A value of A
3A value of B
AO (6)

F

L2

Ques

Solution

Explanation

T/L

4.1.3

INCOME GRAPH
3 A for each graph (start point; any other point; correct straight line)
1A Graphs clearly labelled.

F

L3

(7)

Ques

Solution

Explanation

Topic/L

4.1.4

(a)

Number of Sets = 500 ✔CA
Income at break-even point
= R62 500  or  R62,5 thousand ✔CA

1CA number of sets
1CA income

[Accept values between R62 000 to R63 000] (2)

F

L3

4.1.4

(b)

Number of sets = 800      ✔✔✔RT
OR
x = number of sets
Profit = Income -  Expenses
R6 000 = 125 × x - (10 000 + 105 × x)      ✔M
R6 000 = 20 x - R10 000
x = 800 ✔✔CA
OR
x = number of sets
Income = 125 × X
Expenses = 10 000 + 105 × x
Profit = 20 x - R10 000 ✔M
20 x - R10 000 = R6 000
x = 800 ✔✔CA

3RT number of sets from graph (CA from graph)
OR
1M using thousand rand
2CA number of sets
OR
1M using thousand rand
2CA number of sets  (3)

F

L3

Ques

Solution

Explanation

T/L

4.2

For  Scale:  1 : 3
                   ✔M                  ✔A
Total length of the set = 71 cm + 34 cm = 105 cm
Scaled length of the set = 105 cm ÷ 3    ✔M
 ✔CA
= 35 cm
Length of page is 29,6 cm (does not fit) ✔O
The width of the T-shirt = 57 cm
Scaled width = 19 cm  ✔O
Hence the scale 1 : 3 should NOT be used 
OR
For Scale: 1 : 4 
                ✔                        ✔A
Total length of the set = 71 cm + 34 cm = 105 cm
Scaled length of the set = 105 cm ÷ 4 ✔M
= 26,25 cm    ✔CA
Length of page is 29,6 cm (does fit) ✔O
The width of the T-shirt = 57 cm
Scaled width = 14,25 cm
The scale 1 : 4 SHOULD be used    ✔O

1M adding lengths
1A total length
1M concept of ratio
1CA scaled length
1O does not fit
1O conclusion
OR
1M adding lengths
1A total length
1M concept of ratio
1CA scaled length
1O does fit
1O conclusion (6)

Maps

L4

4.3.1

Convenient ✔✔O
OR

Cheaper ✔✔O
OR                                                  ✔✔O
Save time to go to the shop / transport cost
OR
No need to drive and look and pay for parking ✔✔O
OR
Your purchases gets delivered to you ✔✔O
OR
Availability of stock in stores - if it is sold out. ✔✔O
OR
Greater choice    ✔✔O

2O reason (2)

F

L4

Ques

Solution

Explanation

Topic/L

4.3.2

                                  ✔A
Electronics 51% - 43% = 8%      ✔M
                  ✔A
Sports equipment 44% - 36% = 8%

1A Electronics
1A Sports equipment
1M difference of 8% (3)

F

L2

4.3.3

Groceries      ✔A
Fresh produce like bread and milk is immediately available. ✔✔O
OR
                                                   ✔✔O
Wrong items will not be delivered to your home
OR
You can pay cash for your groceries ✔✔O
OR
You can taste or test some products before you buy them. ✔✔O
OR
                                                         ✔✔O
Frozen goods may melt before they reach you.
OR
Better   comparison   can   be   made   if   you   buy groceries in store.    ✔✔O
OR
                     ✔A
Clothing and footwear - it has to be tried to see whether it fits correctly. ✔✔O
OR
                                 ✔A 
Jewellery - to fit the size of a ring. ✔✔O
OR
                                      ✔A
Electronic goods - it can be tested in the shop before buying. ✔✔O
OR
Or any other items where instore graph is higher than the internet graph with a valid reason.

1A Item
2O opinion  (3)

F

L4

   

[37]

 
   

TOTAL: 150

 
Last modified on Thursday, 12 August 2021 11:45