NATIONAL
SENIOR CERTIFICATE
GRADE 12
JUNE 2022
MATHEMATICAL LITERACY P1
MARKING GUIDELINE
MARKS: 100

 Symbol   Explanation 
 M  Method
 MA  Method with accuracy 
 CA  Consistent accuarcy
  A  Accuracy
 C  Conversion
 S Simplification
 RT/RG/RM   Reading from a table/ graph/ map 
 F  Choosing the correct formula
 SF  Correct substitution in a formula
 J  Justification
 P  Penalty, e.g for no units, incorrect rounding off etc. 
 R  Rounding off/ Reason
 AO  Answers only
 NPR  No penalty for correct rounding off to minimum of two decimal places

 
MARKING GUIDELINES

NOTE:

  • If a candidate answers a question TWICE, only mark 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; 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 incorrect item presented.

LET WEL:

  • As ʼn kandidaat ʼn vraag TWEE keer beantwoord merk slegs die EERSTE poging.
  • As ʼn kandidaat ʼn antwoord van ʼn vraag doodtrek (kanselleer) en nie oordoen nie, merk die doodgetrekte (gekanselleerde) poging.
  • Volgehoue akkuraatheid (CA) word in ALLE aspekte van die nasienriglyn toegepas, maar dit hou by die tweede berekeningsfout op.
  • Wanneer ʼn kandidaat aflees van ʼn grafiek, tabel, uitlegplan en kaart en ekstra antwoorde gee, penaliseer vir elke ekstra item.

 

 QUESTION 1 (20 Marks)      
 Que  Solution   Explanation/Marks 
 AO: FULL MARKS 
 T/L 
 1.1.1 18,25  =     1825      ✓M 
  100        10000 
= 73         ✓A
   400
1M fraction
1A answer in a reduced form
(2)
 F
 L1
 *
 1.1.2

% of price = 100 – 18,25%
                 = 81,75%
Price =  81,75  × 380  ✓M
               100
 = R310,65 ✓ CA

                OR
Reduction = 18,25 × 380
                      100
= R69,35  ✓M
Price = R380 – 69,35  ✓M
= R310,65  ✓CA

1M subtraction
1M % calculation
1CA answer

    OR

1M % calculation
1M subtraction
1CA answer

                                       (3)

 F
 L1
 *
 1.2.1 Difference = R469 – (– R447)  ✓CA 
= R916 million  ✓RT
1 RT for the two correct values 
1 CA answer (2)
  F
 L1
 1.2.2  Total = 265+277+326+390+447+458+486 – (469+300)
= 1880 million  ✓CA
1M addition (+) and
subtraction (–) of the
values
1CA             (2)
 F
 L1
 1.3.1  Weekend wage rate = 3/2 × 25   ✓MA
= R37,50   ✓A
 1MA multiplication
 1A answer
 F
 L1
 *
 1.3.2                     ✓M
Earnings = 6 × 25 + 37,50 × 4   ✓MA

        = R300  ✓CA
 1M multiplications
1MA addition
1CA answer (3)
 F
 L1
 *
 1.4.1  Discrete  ✓✓A    D
L1
 1.4.2  Game ✓✓RT    D
L1
 1.4.3  Total games = 4 + 6 + 5 + 4 + 1 + 2 + 2 = 24 games ✓M ✓CA   1M adding the games
1CA answer (2)
 D
L1
                                                       [20]   

Related Items

 

 QUESTION 2 [18 MARKS]       
 Que Solution    Explanation/Marks
 AO: FULL MARKS  
 T/L 
 2.1.1  Time 4 hours   ✓✓RT   2RT    (2)  F
 L2
 2.1.2  From graph:
2 welders complete 1 frame in 4 hours
2 : 1
20 : ? frame in 4 hours
Frames =  20×1 ✓✓M
                   2
              = 10 frames ✓
                        OR
n × t = 8
20 × t = 8 ✓
t = 8/20
= 0,4 hours to make 1 frame by 20 welders ✓S
In four hours= 4/0,4 ✓M = 10 frames ✓A
 1M value from graph
1M numerator
1M denominator
1A answer
             OR
1SF substitution
1S simplification for
2,5 frames done in 1
hour by 20 welders
1M multiplication
1A answer (4)
 F
 L3
 2.2.1        ✓M
A = 28−25,81  × 100%  ✓MA
           25,81

= 8,485%
= 8,5% ✓CA
 1M correct values for
numerator and
denominator
M % calculation
1CA                    (3)
(NPR)
 F
 L2
 2.2.2  Cost: Up to 6 kℓ = R0 = R0 ✓M

6 – 25 kℓ = 19 k × R23,60 = R448,40  ✓M

25 – 30 kℓ = 5 kℓ × R32,20 = R161,00   ✓M
                                 ✓M
TOTAL COST = R448,40+R161,00 = R606,40   ✓CA 
 1M cost in block 1
1M cost in block 2
1M cost in block 3
1M addition all costs
1CA answer           (5)
F
 L3 
 2.3.1  Salary B = R3 192,05+15 761,80   ✓M
= R18 953,85   ✓CA
 1M adding the two
balances
1 CA answer (2)
 F
 L2
 2.3.2   Bank fees for March = 42,37+17,47+100,88  ✓M
= R160,72   ✓CA
 1M adding fees of
March
1CA answer     (2)
 F
 L1
                                   [18]  

 

 QUESTION 3 [21 MARKS]       
 Que Solution    Explanation/Marks
 AO: FULL MARKS  
 T/L 
 3.1  2020  ✓ A Reason: Covid-19 pandemic    ✓J   1A year
1J reason (2)
 D
L1
 3.2               ✓M
C = 25 285,1 – (2093,5+2092,8+2249,4+ 1988,8+1750,5
+1964,7+2067,1+2204,4+2308,0+2267,8+2493,4)
= 1804,7  ✓M
     ✓CA
 1M subtracting from
25 285,1
1M addition of all other
values
1CA answer (3)
 D
L2
*
 3.3  descending order:            ✓RT
2493,4; 2308,0; 2267,8; 2249,4; 2204,4; 2093,5; 2092,8
2067,1; 1988,8; 1964,7: 1804,7;1750,5
 1RT all values including
value from 3.2
1CA order with value
from 3.2 (2)
 D
L2
*
 3.4                    ✓RT
 Range = 2 262,3 – 33,8   ✓M
= 2 228,5 million    ✓CA
 1RT highest and lowest
values
1M concept of range
1CA answer (3)
 D
L2

 3.5                                          ✓M
Mean income for 2018 =  24846,4  = 2070,53 million  ✓A
Mean income for 2020 = 9818,5 = 818,21 million  ✓A
Double mean income for 2020 = 818,21×2 = 1636,42
Million  
Mean income for 2018 (2 070,53) is greater than double
mean income for 2020 (1636,42)
Statement Valid  ✓J
 1M concept of mean
1A mean for 2018
1A mean for 2020
1M comparing values of
mean 2018 and double
mean income for 2020
1J valid statement. NPR
 D
L4
*
 3.6  From 2018 December income dropped right through up to
July 2019; then increased from August 2019 to December
2019. It remained high up to March 2020.   ✓J
Then it dropped drastically in from April 2020 and remained
low in 2020.  ✓J
 1J justification for the
period Dec 2018 to July
2019
1J justification for the
period August 2019 to
2020 (2)
 D
L4
3.7 May
and June
1A first months
1A second months.
CA from 3.2 (2)
D
L2
                          [20]           

 

 

 QUESTION 4 [20 MARKS]       
 Que Solution    Explanation/Marks
 AO: FULL MARKS  
 T/L 
 4.1.1   Values of dependent variable at break-even point
Income = R300   ✓RT
Expenses = R300   ✓RT
 1RT value for income
1RT value for expenses 
 F
L2
 4.1.2  Total sales in a week = 37 packets  ✓RT
From Graph: Income = R555  ✓RT
Expenses = R385   ✓RT
Profit = R555 – R385 = R170   ✓CA
                      OR
Total sales = 37  ✓RT
Income=37 × 15 = R555  ✓SF   
Expenses = 200+37× 5 = R385   ✓SF
Profit = R555 – R385 = R170   ✓CA
 1RT adding sales from
table
1RT reading income
from graph
1RT expenses from
graph
1CA answer for profit
                OR
1RT total sales
1SF for income
1SF for expenses
1CA answer for profit (4)
 F
L2
 4.2.1   Year 2009 ✓✓RT  2RT for the year    (2)  F
L2
 4.2.2  Fees in 2015 = 1,093 × R12 500 = R13 662,50  ✓M
Cost of fridge in 2015 = 1,04×R12 500 = R13 000   ✓M

Difference = R13 662,50 – R12 500 = R662,50   ✓CA
 1M value from
multiplication with
education inflation rate.
1M value from
multiplication with
general inflation rate
1CA answer (3)
 F
L4
 4.2.3  The graph shows education has constantly outstripped
general inflation.  ✓✓J
   F
L4
 4.3.1 Arrangement of currencies: £; €; $; P; R; ¥ ✓RT ✓A

1RT all currencies
1A order according to
strength (2)

F
L3
 4.3.2 1¥ = R0,1383
3974,85 = R?
Cost of 1 in Rands = 3974,85× 0,1383  ✓M
= R549,72  ✓A
Cost of 500 DVD players = 500 × 549,72
= R274 860,88  ✓CA
1M converting the
Japanese yens to Rands
1A cost of one DVD
1CA answer for cost of
500 DVDs (3)
F
L2
                                   [18]  

 

 QUESTION 5 [25 MARKS]       
 Que Solution    Explanation/Marks
 AO: FULL MARKS  
 T/L 
 5.1.1  Tax bracket = 4 ✓✓ RT   2RT bracket   (2)  F
L1
 5.1.2  R128 650 ✓✓ RT  2RT value of threshold  (2)   F
L2
 5.1.3

 Monthly income = R35 455
Annual income = R35 455× 12 = R425 460,00 ✓MA
Pension: 7,5% of R425 460 =  7,5  × R425 460,00
                                                100
                                          = R31 909,50  ✓A
Taxable Income = R425 460,00 – R31 909,50
                           = R393 550,50  ✓CA
Tax = R67 144 +  31  × (393 550,50 – 321 600)  ✓M
                           100
= R67 144 +  31  × 71 950,50
                     100
= R67 144 + 22 304,655
= R89 448,655    ✓CA
                                                                        ✓RT
Tax less the rebates = R89 448,655 – (R14 958+R8199)
Annual tax payable = R66 291,655   ✓M

 1MA multiplication by
12 and annual income
1A the annual pension
1CA taxable income
1M use of correct tax
bracket
1CA tax payable before
rebates
1RT Total value of
rebates
1M subtracting rebates
and tax after rebates     (7)
 F
L4
 5.2.1                                         ✓RT
 2,27%; 5,04%; 5,05%; 5,90%; 6,68%; 7,24%; 13,38%;
16,15%; 38,28%.   ✓M
Median value = 6,68% giving EC  ✓CA
 1RT all values from
graph
1M arranging in order
descending or ascending
1CA median value: EC
 D
L2
 5.2.2   Q1 = 5,04+5,05  ✓M
                   2
        = 5,045%   ✓A
Q3 = 13,38 + 16,15
                  2 
       = 14,765%   ✓A
IQR = Q3–Q1
       = 14,765% – 5,045%    ✓M
         = 9,72%   ✓CA
 1M concept of getting
Quartile 1
1A for Q1
1A for Q3
1M method of
subtracting Q3-Q1
1CA answer (5)
 D
L3
 5.2.3 Probability is the chance that an event is likely to
happen.  ✓✓A
 2A explanation   (2)  P
L1
 5.2.4 Probability for GP = 0,3828  ✓CA
Probability for EC = 0,0668   ✓CA
Probability for a car to be in GP OR EC = 0,3828 +
0,0668 = 0,4496   ✓A
 1CA converting 5 to
decimal for QP
1CA converting to
decimal for EC
1A answer                         (3)
 
                                                        [24]  
                                                                          TOTAL:                                     100  
Last modified on Friday, 16 December 2022 12:56