Adele

Adele

PHYSICAL SCIENCES
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

INSTRUCTIONS AND INFORMATION

  1. Write your examination number and centre number in the appropriate spaces on the ANSWER BOOK.
  2. This question paper consists of TEN questions. Answer ALL the questions in the ANSWER BOOK.
  3. Start EACH question on a NEW page in the ANSWER BOOK.
  4. Number the answers correctly according to the numbering system used in this question paper.
  5. Leave ONE line between two subquestions, e.g. between QUESTION 2.1 and QUESTION 2.2.
  6. You may use a non-programmable calculator.
  7. You may use appropriate mathematical instruments.
  8. Show ALL formulae and substitutions in ALL calculations.
  9. Round off your FINAL numerical answers to a minimum of TWO decimal places.
  10. Give brief motivations, discussions, etc. where required.
  11. You are advised to use the attached DATA SHEETS.
  12. Write neatly and legibly.

QUESTIONS 
QUESTION 1: MULTIPLE-CHOICE QUESTIONS
Various options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question numbers (1.1 to 1.10) in the ANSWER BOOK, e.g. 1.11 D.
1.1Which ONE of the following is the structural formula of the functional group of the KETONES?
1 uyhsuyhf(2)
1.2 Which ONE of the formulae below represents an ALKANE?

  1.  C2H4
  2.  C5H10
  3.   C14H30
  4.  C8H14     (2)

1.3 Consider the organic compound below.
2 uygyaghdu
The IUPAC name of this compound is …

  1. 2,3-dimethyl butane.
  2. 3,3-dimethyl butane.
  3. 2,2-dimethyl butane.
  4. 1,1,1-trimethyl propane.  (2)

1.4   Activation energy can best be described as the minimum energy required to …

  1.  cause effective collisions.
  2. make reactant molecules collide.
  3. change the orientation of reactant molecules.
  4.   increase the kinetic energy of reactant molecules.      (2)

1.5  Which statement is CORRECT for a system in DYNAMIC EQUILIBRIUM?

  1.   All reactants are used up.
  2. The forward reaction is equal to the reverse reaction.
  3. All substances in the reaction are of equal concentration.
  4. The concentration of the reactants and products remain constant.         (2)

1.6 Initially,  a  certain  amount  of  P(g)  was  placed  in  an  empty  container. The hypothetical reaction reaches equilibrium in a closed container according to the following balanced equation:

P(g)  ⇌  2Q(g)                       ΔH  <  0

At time t, the temperature is increased. Which graph below best illustrates the resulting changes in the rates of the forward and reverse reactions after the temperature is increased?
3 jkhuyadh
1.7 Reactions   I   and   II   below   have   equilibrium   constants       (Kc)   greater than 1.

  1. :    H3X +  HCO-3  ⇌  H2X  +  H2CO3              Kc > 1
  2. :   H3O+  +  H2X  ⇌  H2O  +  H3X                     Kc > 1

Based  on  the  reactions  above,  the  ACIDS  in  order  of  INCREASING STRENGTH (weakest to strongest) are …

  1.  H3X,  H2X,  H3O+
  2. H2CO3, H3X,  H3O+
  3.  H3X,  H2CO3,  H3O+
  4. H3X,  H3O+,  H2CO3     (2)

1.8  Consider the cell notation for a galvanic cell below.

Ni(s) | Ni2+(aq) || H+(aq) | H2(g) | Pt(s)

Which ONE of the following half-reactions takes place  at  the  ANODE of this cell?

  1.  2H+(aq) + 2e  →  H2(g)
  2. H2(g)  →  2H+(aq) + 2e
  3.   Ni2+(aq) + 2e  →  Ni(s)
  4. Ni(s)  →  Ni2+(aq) + 2e     (2)

1.9  Which ONE of the following is applicable to an ELECTROLYTIC CELL?

  1.  Reduction takes place at the anode.
  2. Oxidation takes place at the cathode.
  3. It uses alternating current.
  4. A battery is used for the cell to function.    (2)

1.10 The flow diagram below shows four stages (A, B, C and D) in the conversion of sulphur to sulphuric acid.
4 uygua77ygdu
At which stage is a catalyst used?

  1.  A
  2. B
  3. C
  4. D   (2) [20]

QUESTION 2 (Start on a new page.)
A test tube containing a straight chain organic acid X, ethanol and a catalyst is heated in a water bath, as illustrated below.
5 uygayuguhd
Organic compound Y is produced according to the following equation:
                                                    X  +  C2H5OH  ⟶  Y  +  H2O

2.1 Give a reason why the test tube is heated in a water bath instead of directly over the flame.   (1)
2.2 Write down the:
2.2.1  Type of reaction that takes place here     (1)
2.2.2 FORMULA of the catalyst needed    (1)
2.2.3 Homologous series to which compound Y belongs       (1)
The molecular mass of compound Y is 144 g∙mol-1 and its empirical formula is C4H8O.
2.3 Determine the molecular formula of compound Y.   (2)
2.4  Write down the IUPAC name of compound Y.      (2)
2.5  Write down the structural formula of the organic acid X.  (2) [10]

QUESTION 3 (Start on a new page.)
The boiling points of different organic compounds are given below.

COMPOUND

BOILING POINT (°C)

A

HCOOH

101

B

CH3COOH

118

C

CH3CH2COOH

141

D

CH3CH2CH2COOH

164

3.1 Define boiling point.      (2)
3.2 Write down the:
3.2.1 Name of the FUNCTIONAL GROUP of these compounds    (1)
3.2.2 IUPAC name of compound C    (1)
3.2.3  Structural formula of the FUNCTIONAL isomer of compound B   (2)
3.3  Which ONE of the compounds, A or B or C, has the highest vapour pressure? Refer to the data in the table to give a reason for the answer.   (2)
3.4 The boiling point of compound B is now compared with of compound X.

COMPOUND

BOILING POINT (°C)

B

CH3COOH

118

X

CH3CH2CH2OH

98

3.4.1 Besides the conditions used to determine boiling points, give a reason why this is a fair comparison.   (1)
3.4.2 Is compound X a PRIMARY, SECONDARY or TERTIARY alcohol? Give a reason for the answer.     (2)
3.4.3 Fully explain the difference between the boiling points by referring to the types of intermolecular forces present in each of these compounds.    (4)  [15]

QUESTION 4 (Start on a new page.)
4.1 Three reactions of organic compounds from the same homologous series are shown below.
6 jkgujayd
 4.1.1 Define a homologous series.   (2)
4.1.2 Name the type of reaction represented by I.    (1)
4.1.3 Write down the formula of the inorganic compound P.  (1)
4.1.4 Give   the   structural   formula   of   a   POSITIONAL   isomer  of 2-bromobutane.      (2)
4.1.5 Using molecular formulae, write down the balanced equation for reaction II.    (3)
Reaction III is an example of a cracking reaction.
4.1.6 Define a cracking reaction.   (2)
4.1.7 Give the structural formula of organic compound Q.  (2)
4.2 Study the flow diagram below.
7 khayuhd
4.2.1 Write down the IUPAC name of compound R.    (2)
4.2.2  Compound R reacts in the presence of concentrated phosphoric acid to form an alkene. Write down the structural formula of the MAJOR PRODUCT in this reaction.   (2)  [17]

QUESTION 5 (Start on a new page.)
The reaction of zinc and EXCESS dilute hydrochloric acid is used to investigate factors that affect reaction rate. The balanced equation for the reaction is:

Zn(s)  +  2HCℓ(aq)  ⟶  ZnCℓ2(aq)  +  H2(g)

The  reaction  conditions  used  and  the  results  obtained for  each  experiment are summarised in the table below. The same mass of zinc is used in all the experiments. The zinc is completely covered in all reactions. The reaction time is the time it takes the reaction to be completed.

EXPERIMENT

CONCENTRATION

OF HCℓ (mol∙dm-3)

VOLUME OF

HCℓ (cm3)

STATE OF DIVISION

OF Zn

TEMPERATURE

OF HCℓ (°C)

REACTION

TIME (min.)

1

2,0

200

powder

25

7

2

1,5

200

granules

25

14

3

5,0

200

powder

25

5

4

1,5

400

granules

25

x

5

2,0

200

powder

35

4

 5.1 Experiment 1 and experiment 5 are compared. Write down the independent  variable.    (1)
5.2 Define reaction rate.        (2)
5.3 Write down the value of x in experiment 4.             (2)
5.4 The Maxwell-Boltzmann energy distribution curves for particles in each of experiments 1, 3 and 5 are shown below.
8 kuhaiyhd
Identify the graph (A or B or C) that represents the following:
5.4.1 Experiment 3 Give a reason for the answer.  (2)
5.4.2 Experiment 5 Give a reason for the answer.  (2)
5.5  Experiment 6 is now conducted using a catalyst and the SAME reaction conditions as for Experiment 1.
5.5.1 What is the function of the catalyst in this experiment?  (1)
5.5.2 How will the heat of reaction in experiment 6 compare to that in experiment 1? Choose from: GREATER THAN, EQUAL TO or LESS THAN.  (1)
5.6 Calculate the average rate of the reaction (in mol·min-1) with respect to zinc for experiment 2 if 1,5 g of zinc is used.  (4) [15]

QUESTION 6 (Start on a new page.)
Dinitrogen tetraoxide, N2O4(g), decomposes to nitrogen dioxide, NO2(g), in a sealed syringe of volume 2 dm3.
9 kuhuyadu
The  mixture  reaches  equilibrium  at    325 °C  according  to  the  following  balanced equation:

  • N2O4(g)  ⇌  2NO2 (g)
    colourless       brown

When equilibrium is reached, it is observed that the colour of the gas in the syringe is brown.
6.1 State Le Chatelier's principle.    (2)
6.2 The syringe is now dipped into a beaker of ice water. After a while the brown colour disappears. Is  the  forward  reaction  EXOTHERMIC  or  ENDOTHERMIC?  Explain  the answer using Le Chatelier's principle. (3)
6.3 The volume of the syringe is now decreased while the temperature is kept constant. How will EACH of the following be affected? Choose from: INCREASES, DECREASES or REMAINS THE SAME.
6.3.1 The number of moles of N2O4(g)  (1)
6.3.2 The value of the equilibrium constant    (1)
6.3.3 The rate of the forward and reverse reactions   (1)
6.4 Initially X moles of N2O4(g) were placed in the syringe of volume 2 dm3.  When equilibrium was reached, it was found that 20% of the N2O4(g) had decomposed.  If the equilibrium constant, Kc, for the reaction is 0,16 at 325 °C, calculate the value of X.  (8)  [16]

QUESTION 7 (Start on a new page.)
7.1  Sulphuric acid is a strong acid present in acid rain. It ionises in two steps as follows:

  1. :         H2SO4(aq)  +  H2O(ℓ)  ⇌  H3O+(aq)  + HSO-4 (aq)
  2. :         HSO-4 (aq) +  H2O(ℓ)  ⇌  H3O+(aq)  + SO2-4          (aq)

7.1.1Define an acid in terms of the Lowry-Brønsted theory.       (2)
7.1.2 Write down the FORMULA of the conjugate base of H3O+(aq).  (1)
7.1.3  Write  down  the  FORMULA  of  the  substance  that  acts  as  an ampholyte in the ionisation of sulphuric acid.    (2)
7.2 Acid  rain  does  not  cause  damage  to  lakes  that  have  rocks  containing limestone (CaCO3). Hydrolysis of CaCO3 results in the formation of ions, which neutralise the acid.
7.2.1  Define hydrolysis of a salt.    (2)
7.2.2 Explain, with the aid of the relevant HYDROLYSIS reaction, how limestone can neutralise the acid.   (3)
7.3 The water in a certain lake has a pH of 5.
7.3.1 Calculate the concentration of the hydronium ions in the water.  (3)
The volume of water in the lake is 4 x 109 dm3. Lime, CaO, is added to the water to neutralise the acid according to the following reaction:

CaO  +  2H3O+  ⇌  Ca2+  +  3H2O

7.3.2  If the final amount of hydronium ions is 1,26 x 103 moles, calculate the mass of lime that was added to the lake.    (7)  [20]

QUESTION 8 (Start on a new page.)
8.1Corrosion is a redox reaction that takes place in the presence of oxygen and water. Rusting is the corrosion of iron leading to the formation of iron(III) ions.
8.1.1 Define oxidation in terms of electron transfer.    (2)
A cleaned copper rod and a cleaned iron nail  are placed in a beaker containing water at 25°C, as shown below.
10 iuhiuyahd
After a while it was observed that the iron nail was coated with rust. The copper rod showed no visible signs of corrosion.
8.1.2Write down the half-reaction for the iron nail.           (2)
8.1.3 Does iron act as REDUCING AGENT or OXIDISING AGENT in the beaker?  (1)
8.1.4 Explain the above observation by referring to the Table of Standard Reduction Potentials.  (3)
To prevent rusting of an underground iron pipe, the pipe is connected to a metal (Q) that corrodes easily.
11 nkjhghuyadgh
8.1.5 You  are  given  two  metals,  Zn  and  Cu,  to  use  as  metal  Q. Which metal would more suitable? Give a reason.   (2)
8.2 A galvanic cell is constructed using a Fe | Fe3+ half-cell and a Cu | Cu2+ half-cell.
8.2.1 Write down the overall (net) cell reaction that takes place when the cell is functioning.  (3)
8.2.2  Calculate the cell potential of this cell under standard conditions. (4) [17]

QUESTION 9 (Start on a new page.)
The electrolytic cell below is set up to obtain pure copper from a piece of impure copper.
12 uygauydg
The impure copper contains other metals, such as platinum, iron, cobalt, silver and nickel.
The cell potential of the power source is adjusted so that only copper is deposited on electrode B.
9.1 Define an electrolytic cell.         (2)
9.2 Write down the FORMULA of a suitable electrolyte for this cell.         (1)
9.3 Which electrode (A or B) is the cathode? Write down the relevant half-reaction taking place at this electrode.  (3)
9.4 Sludge forms below one of the electrodes while the cell above is in operation. Which of the metals, PLATINUM, IRON, COBALT, SILVER or NICKEL, will be present in the sludge?  (2)  [8]

QUESTION 10 (Start on a new page.)
In the flow diagram below, I and II represent industrial processes used in the fertiliser industry. P and Q are chemical reactions that take place to produce ammonium sulphate and fertiliser Y respectively.
13 uygyagdu
10.1 Write down the name of the industrial process:
10.1.1   I     (1)
10.1.2  II   (1)
10.2  Write down the NAME or FORMULA of:
10.2.1 Fertiliser Y   (1)
10.2.2The catalyst used in process I   (1)
10.3  In reaction P, NH3(g) reacts with another substance. Write down a balanced equation for this reaction.    (3)
10.4 The following substances are present in a bag of fertiliser:

  • 20 kg ammonium nitrate (NH4NO3)
  • 12 kg sodium phosphate (Na3PO4)
  • 18 kg potassium chloride (KCℓ)

Calculate the NPK ratio of the fertiliser.  (5)  [12]

TOTAL: 150

DATA FOR PHYSICAL SCIENCES GRADE 12  PAPER 2 (CHEMISTRY)
TABLE 1:  PHYSICAL CONSTANTS

NAME

SYMBOL

VALUE

Standard pressure

pθ

1,013 x 105 Pa

Molar gas volume at STP

Vm

22,4 dm3∙mol-1

Standard temperature

Tθ

273 K

Charge on electron

e

-1,6 x 10-19 C

Avogadro's constant

NA

6,02 x 1023 mol-1

 14 ujygauygdsu
15 nhvugavduya
16 jkygauydgh
17 uygaydg

PHYSICAL SCIENCES
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

MEMORANDUM 

QUESTION 1
1.1 C ✔✔ (2)
1.2 C ✔✔ (2)
1.3 C ✔✔ (2)
1.4 A ✔✔ (2)
1.5 D ✔✔ (2)
1.6 B ✔✔ (2)
1.7 B ✔✔ (2)
1.8 D ✔✔ (2)
1.9 D ✔✔ (2)
1.10 B ✔✔ (2) [20]

QUESTION 2
2.1 ANY ONE

  • (Alcohol/ethanol) is flammable/catches fire easily.
  • To heat it evenly.
  • Water bath is used for low heat/low temperature.
  • Alcohol/ethanol will evaporate too quickly.
    Accept/
    (Alcohol/ethanol) is volatile.  (1)

2.2
2.2.1  Esterification/condensation  (1)
2.2.2  H2SO4  (1)
2.2.3  Esters   (1)
2.3

  • M(ester)       = 144   =  2
    M(C4H8O)        72
    ∴  2 x C4H8O = C8H16O2 

Marking guidelines:

  • If only answer given, award 2 marks on final
  • If 72 g·mol-1 calculated without substituting, no   mark   is   awarded (2)

2.4  Ethyl hexanoate   (2)
Note

  • Accept any other ethyl ESTER from QUESTION 2.3.

2.5 POSITIVE MARKING FROM QUESTION 2.4.
1 kgujygayud
Marking criteria

  •  Whole structure correct 2/2
  • Only functional group correct : Max: 1/2
  • Accept  -OH as condensed  (2)[10]

QUESTION 3
3.1 Marking guidelines/

  • If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark
  • The temperature at which the vapour pressure of a substance equals atmospheric/external pressure.  (2)

3.2
3.2.1Carboxyl (group)
Accept

  • Carboxylic  (1)

3.2.2 Propanoic acid/propanoësuur  (1)
3.2.3
2 khuiahdik
Marking criteria

  • Whole structure correct: 2/2
  • Only functional group correct: 1/2

IF

  • More than one functional group/wrong functional group 0/2
  • If condensed structural formulae used: Max: 1/2

3 khaghkudghuia
3.3 A  -  Lowest boiling point./Shortest chain length. 
3.4
3.4.1 The same molecular mass/molecular size. 
3.4.2 Primary

  • OH group is bonded to a C atom bonded to one other C atom. 
    OR
  • OH group is bonded to a C atom that has two H atoms.

3.4.3 Marking guidelines

  • BOTH have hydrogen bonding.
  • Compare number of sites for hydrogen bonding.
  • Compare strength of IMFs.
  • Compare energy required.
  • Both compounds/X and B have (in addition to London forces and dipole-dipole forces) hydrogen bonding.
  • Compound X/CH3CH2CH2OH/propan-1-ol/alcohol has one site for hydrogen bonding and compound B/ethanoic acid/carboxylic acid has two/more sites for hydrogen bonding OR B/ethanoic acid/carboxylic acid has two/more sites for hydrogen bonding. 
  • Intermolecular forces in compound B/ethanoic acid/carboxylic acid are stronger than intermolecular forces in compound X/CH3CH2CH2OH/ propan-1-ol/alcohol.
    OR
    Intermolecular forces in compound X/CH3CH2CH2OH/ propan-1-ol/alcohol are weaker than intermolecular forces in compound B/ethanoic acid/carboxylic acid.
  • More energy is needed to overcome/break intermolecular forces in compound B/ethanoic acid/carboxylic acid than in compound X/CH3CH2CH2OH/ propan-1-ol/alcohol.
    OR
    Less energy is needed to overcome/break intermolecular forces in compound X/CH3CH2CH2OH/propan-1-ol/alcohol than in compound B/ethanoic acid/carboxylic acid.   (4)  [15]

QUESTION 4
4.1
4.1.

  • (A series of organic) compounds that can be described by the same general formula/functional group.   (2 or 0)
    OR
    (A series of organic) compounds in which one member differs from the next by a CH2  group   (2)

4.1.2 Substitution/halogenation/bromination  (1)
4.1.3  HBr  (1)
4.1.4  
4 jhgujygauygd
Marking criteria

  • Br on first C atom: Max: 1/2
  • Whole structure correct   2/2

IF

  • Br2 but rest of structure correct  (2)

4.1.5 C5H12  +  8O2 → 5CO2  +  6H2O     Bal 
Marking guidelines

  • Reactants             Products               Balancing
  • Ignore double arrows and phases.
  • Marking rule 6.3.10/Nasienreël 6.3.10.
  • If condensed structural formulae used : Max: 2/3   (3)

4.1.6 Marking guidelines/Nasienriglyne

  • If any one of the underlined key phrases in the correct context is omitted, deduct 1 mark

The (chemical) process in which longer chain hydrocarbons/longer chain alkanes are broken down to shorter/more useful hydrocarbons/molecules/ chains/alkanes and alkenes.  (2)
4.1.7
5 jhgvyagyud
Marking guidelines

  • One or more H atoms omitted : Max: 1/2
  • Condensed or semi-structural formula:Max: 1/2  (2)

4.2
4.2.1 Butan-2-ol OR  2-butanol
IF:

  • Butanol  or butan-1-ol 1/2  (2)

4.2.2
6 jhgygauygd
Marking criteria

  • Only functional group correct: Max/Maks: 1/2
  • Whole structure correct: 1/2  (2)  [17]

QUESTION 5
5.1Temperature (1)
5. 2 NOTE
Give the mark for per unit time only if in context of reaction rate.
 ANY ONE

  • Change in concentration  of products/reactants per (unit) time. 
  • Change in amount/number of moles/volume/mass of products or reactants  per (unit) time.
  • Amount/number of moles/volume/mass of products formed/reactants used per (unit) time.
  • Rate of change in concentration/amount/number of moles/volume/mass. (2 or/of 0)   (2)

5.3 14 (min)  (2)
5.4
5.4.1 Graph  B 

  •  (Experiment 3) has the highest (acid) concentration/more particles/higher number of moles.   (2)

5.4.2 (Graph/grafiek) C 

  • (Experiment 5) is at highest temperature/more particles with sufficient kinetic energy/HCℓ is at 35oC    (2)

5.5
5.5.1 Speeds  up  the  reaction./Increases  the  reaction  rate./Provides  alternate pathway./Lowers the (net) activation energy. (1)
5.5.2 Equal to  (1)
5.6 
7 jhkuahukda
Marking guidelines

  • Substitute 65 g∙mol-1 in n = m
                                                M
  • Substitute change in mol to calculate  rate.
  • Substitute change in time to calculate  rate.
  • Final answer:
         1,65 x 10-3 mol∙min-1 
    Range/Gebied:
    1,43 x 10-3 to/tot 1,65 x 10-3 (mol∙min-1)

Notes

  • Ignore if zeros omitted in calculation of reaction rate.
  • Accept negative answer i.e. -1,65 x 10-3 mol·min-1  (4)  [15]

QUESTION 6
6.1

  • When the equilibrium in a closed system is disturbed, the system will re- instate a (new) equilibrium  by favouring the reaction that will cancel/oppose the disturbance.     (2)

6.2 Endothermic

  • Decrease in temperature favours the exothermic reaction.
  • The   reverse   reaction   is   favoured.
    OR
    Number of moles/amount/concentration of N2O4/colourless gas increases.
    OR
    Number of moles/amount of NO2/brown gas decreases. (3)

6.3
6.3.1 Increases (1)
6.3.2 Remains the same (1)
6.3.3 Increases  (1)
6.4  CALCULATIONS USING NUMBER OF MOLES
Marking guidelines/Nasienriglyne

  •  ∆n(N2O4) = 20% of x/0,2x. 
  • USE ratio: N2O4 : NO2 : = 1 : 2. 
  • n(N2O4)eq/ewe = n(N2O4)initial/begin - ∆n(N2O4).
  • n(NO2)eq/ewe = n(NO2)initial/begin + ∆n(NO2).
  • Divide equilibrium moles by 2 dm3
  • Correct Kc expression (formulae in square brackets).
  • Substitution of Kc value
  • Substitution of concentrations into correct Kc expression. 
  • Final answer/Finale antwoord: 1,6 (mol) 

8 khgyjuagdua
CALCULATIONS USING CONCENTRATION
Marking guidelines

  • Initial n(N2O4)/x divide by 2 dm3.
  • ∆c(N2O4) = 20% of initial concentration/0,1x.
  • USE ratio/GEBRUIK verhouding: c(N2O4) : c(NO2)  = 1 : 2.
  • c(N2O4)eq/ewe = c(N2O4)initial/begin - ∆c(N2O4).
  • c(NO2)eq/ewe = c(NO2)initial/begin + ∆c(NO2).
  • Correct Kc expression (formulae in square brackets).
  • Substitution of Kc value/Vervanging van Kc-waarde.
  • Substitution of concentrations into Kc expression.
  • Final answer/Finale antwoord: 1,6 (mol) 

OPTION 3
9 jkhgjyagyudh(8) [16]

QUESTION 7
7.1
7.1.1 An acid is a proton donor. (2)
7.1.2 H2O  (1)
7.1.3 HSO-4    (2)
7.2
7.2.1 Reaction of a salt with water/H2O. 
Accept

  • Reaction of cations or anions with water  (2)

7.2.2 
ans kjgkygad
Marking guidelines

  • Reactants    Products
  • The formation of OH  (aq) neutralises the excess acid.
  • Ignore single arrows and phases
  • Marking rule 6.3.10
  • Ignore balancing.

7.3
7.3.1

  • pH = -log[H3O+]
    5  = -log[H3O+]
    [H3O+] = 1 x 10-5 mol·dm-3 (3)

7.3.2 POSITIVE MARKING FROM QUESTION 7.3.1.
Marking guidelines

  • Any formula : c = n / n = m / Ca x Va =na /c = m
                                V        M    Cb x Vb   nb       MV
  • Substitute V = 4 x 109 dm3 
  • Calculate na(reacted) = na(initial) - na(final) 
  • Use  n(CaO) : n(H3O+) = 1:2
  • Substitution of 56 g∙mol-1 
  • Final answer : m = 1,08 x 106 g to/tot 1,09 x 106 g  

IF final answer is negative: Max: 6/7
10 ighaiygduyag(7) [20]

QUESTION 8
8.1
8.1.1 Loss of electrons./Verlies aan elektrone.  (2 or/of 0)            (2)
8.1.2 Fe → Fe3+ + 3e
Marking guidelines

  •  ans kljhkuhad
  • Ignore if charge omitted on electron.
  • If charge (+) omitted on Fe3+: Example:  Fe → Fe3 + 3e- 
    Max1/2  (2)

8.1.3 Reducing agent   (1)
8.1.4

  • Fe is a stronger reducing agent  than Cu  and (Fe) will be oxidised  (to Fe3+).
    OR
  • Cu is a weaker reducing agent  than Fe and (Cu) will not be oxidised  (to Cu2+). (3)

8.1.5 Zinc/Zn 

  • Stronger reducing agent (than Fe).
    OR
  • Zn will undergo oxidation (before Fe).
    OR
  • Cu is a weaker reducing agent (than Fe). (2)

8.2
8.2.1 3Cu2+ + 2Fe  → 3Cu + 2Fe3+                Bal.
Marking guidelines/Nasienriglyne

  • Reactants              Products               Balancing 
  • Ignore double arrows.
  • Marking rule 6.3.10   (3)

8.2.2
jgjagdug
(4) [17]

QUESTION 9
9.1

  • A cell in which electrical energy is converted to chemical energy.  (2 or 0)
    OR
  • A cell in which electrical energy/electricity is used to obtain a chemical change/reaction. (2 or 0) (2)

9.2 Any soluble copper(II) salt e.g

  • CuSO4/Cu(NO3)2/CuCℓ2  (1)

9.3  Marking guidelines

  • alisjiouad
  • Ignore if charge on electron is omitted.
  • If a charge of an ion is omitted e.g. Cu2 + 2e- → Cu  Max.:  1/2 (3)

9.4 Platinum/Pt  AND  silver/Ag/  (2) [8]

QUESTION 10
10.1
10.1.1 Haber (process)
10.1.2 Ostwald (process)
10.2
10.2.1Ammonium nitrate/Ammoniumnitraat/NH4NO3 
10.2.2  Iron/iron oxide/Fe/FeO 
10.3       

  • 2NH3    +  H2SO4   →  (NH4)2SO4     Bal 

Marking guidelines

  • Reactants              Products             Balancing
  • Ignore double arrows.
  • Marking rule 6.3.10

10.4 khuygaud
10.4 b kbkahbda  (5)

[12]
TOTAL: 150 

PHYSICAL SCIENCES
PAPER 1
GRADE 12
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

MEMORANDUM

QUESTION 1
1.1 C /D✔✔ (2)
1.2 C ✔✔ (2)
1.3 C ✔✔ (2)
1.4 B ✔✔ (2)
1.5 B ✔✔ (2)
1.6 A ✔✔ (2)
1.7 A ✔✔(2)
1.8 D ✔✔ (2)
1.9 D ✔✔ (2) 
1.10 C ✔✔ (2) [20]

QUESTION 2
2.1  (2) 

When a (non-zero) resultant/net force acts on an object, the object will  accelerate in the direction of the force with an acceleration that is directly  proportional to the force and inversely proportional to the mass of the object.  ✔✔
OR
The (non-zero) resultant/net force acting on an object is equal to the rate of  change of momentum of the object in the direction of the resultant/net force.  ✔✔ (2 or 0) 
ACCEPT
Acceleration is directly proportional to the net force and inversely proportional  to the mass of the object. 

NOTE

  • If any of the underlined key words in the correct context is omitted deduct 1  mark. 

2.2 
1 JYGUAGUD(4) 

Notes

  • Mark is awarded for label and arrow
  • Do not penalise for length of arrows 
  • If T is not shown but T|| and T⊥ are shown, give 1 mark for both
  • If force(s) do not make contact with body : Max/Maks: 3/4
  • Deduct 1 mark for any additional force  
 

Accept the following symbols 

FN; Normal;Normal force ✔

f

Ff / fk / frictional force ✔

w

Fg; mg; Weight;FEarth on block;Fw  ;Gravitational force ✔

Tension ; FT /FA, F /16,96 N ✔

2.3.1  (1) 

The 2/8 kg block /system is accelerating ✔
OR
The acceleration is not zero / a ≠ 0 (m·s-2) / a = 1,32 m.s-2✔
OR
Velocity is /increasing/changing/not constant ✔
OR
Fnet is not equal to zero  ✔
OR
The acceleration is changing  ✔
Accept

  • An unbalanced force is acting on it  ✔ 

2.3.2 (3) 

For 2 kg 
Fnet = ma  {1 mark for any }    Fnet = ma  
mg -T= ma                              mg + T = ma  ✔ 
(2)(-9,8) + T = 2(-1,32) ✔   (2)(9,8) – T = 2(1,32) ✔ 
T = 16,96 N ✔                       T = 16,96 N ✔ 

2.3.3  (4) 

POSITIVE MARKING FROM 2.3.2
2 JYGUYAgduj
OR
3 HFGATYD

2.4  (1) 

  • ANY ONE
    Normal force changes/decreases ✔
  • The angle (between string and horizontal) changes/increases. 
  • The vertical component of the tension changes/increases 

2.5  (2)   [17] 

Yes✔
The frictional force (coefficient of friction) depends on the nature of the surfaces  in contact. ✔ 
ACCEPT
The nature of the surface changes / µk changes  

QUESTION 3
3.1 (2) 

Downwards ✔ 
The only force acting on the object is the gravitational force/weight which acts  downwards.✔
ACCEPT

  • The only force acting is gravitational/weight.✔
    OR
  • Gravitational force/weight acts downwards.✔
    OR
  • The ball is in free-fall ✔✔
    OR
  • (Gravitational) acceleration is downwards✔✔

OPTION 1
Upward positive
vf = vi + aΔt ✔ 
0 = 7,5 + (-9,8)Δt ✔ 
Δt = 0,77 s ✔ 

Downward positive
vf = vi + aΔt ✔ 
0 = -7,5 + (9,8)Δt ✔
Δt = 0,77 s ✔

3.2  (3)

OPTION 2
Upward positive 
At highest point vf is zero 
vf2 = vi2 + 2aΔy
0 = (7,5)2 + (2)(-9,8)Δy
Δy = 2,87 (2,869)m
4 jhgayaduguy
Δt = 0,77 s✔ 

OPTION 2
Downward positive 
At highest point vf is zero 
vf2 = vi2 + 2aΔy
0 = (-7,5)2 + (2)(9,8)Δy
Δy = 2,87 (-2,869)m
5 jyguatygduja
Δt = 0,77 s✔ 

OPTION 3
Upward positive 
FnetΔt = m(vf – vi) ✔
mgΔt = m(vf – vi) 
(-9,8)Δt = 0 – 7,5 ✔
∴ Δt = 0,76531 s (0,77 s) 

OPTION 3
Downward positive 
FnetΔt = m(vf – vi) ✔
mgΔt = m(vf – vi) 
(9,8)Δt = 0 –(- 7,5) ✔
∴ Δt = 0,76531 s (0,77 s)✔

OPTION 4
Upward positive 
(Top to Bottom ) 
vf = vi + aΔt ✔
-7,5 = 0 + (-9,8)Δt ✔
∴ Δt = 0,76531 s (0,77 s) ✔

OPTION 4
Downward positive 
(Top to Bottom ) 
vf = vi + Δt ✔
7,5 = 0 + (9,8)Δt✔
∴ Δt = 0,76531 s (0,77 s) ✔

OPTION 5
Upward positive 
(Top to Bottom) 
vf2 = vi2 + 2aΔy 
(7,5)2 = (0)2 + 2(-9,8)Δy 
Δy = -2,87 m 
Δy = viΔt + ½ aΔt2
-2,87 = (0)Δt + ½ (-9,8)(Δt)2 
Δt = 0,765 s ✔

OPTION 5
Downward positive 
(Top to Bottom ) 
vf2 = vi2 + 2aΔy 
(7,5)2 = (0)2 + 2(9,8)Δy 
Δy = 2,87 m Δy = viΔt + ½ aΔt2 ✔
2,87 = (0)Δt + ½ (9,8)Δt2
Δt = 0,765 s ✔

NOTES for marking QUESTION 3.3 

 

Formula mark

Substitution mark

✔✔

Mark for height/distance  

Mark for comparison

Mark for conclusion

3.3  (6) 

OPTION 1
Upward positive
At highest point vf is zero
vf2 = vi2 + 2aΔy
0✔ = (7,5)2 + (2)(-9,8)Δy ✔ 
Δy = 2,87 (2,869) m✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass point T.✔ 

Downward positive
At highest point vf is zero
vf2 = vi2 + 2aΔy
0✔ = (-7,5)2 + (2)(9,8)Δy✔ 
Δy = - 2,87 (-2,869) m✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass the target. ✔ 

OPTION 2
(POSITIVE MARKING FROM 3.2) 
Upward positive
Δy = viΔt + ½ aΔt2✔ 
Δy = (7,5)(0,77) ✔ + ½ (-9,8)(0,77)2✔  
Δy = 2,87 m (2,86 m)✔ 
This is higher than height needed to reach point T (2,1 m)✔therefore the ball will pass point T. ✔ 
Downward positive
Δy = viΔt + ½ aΔt2✔ 
Δy = (-7,5)(0,77) ✔ + ½ (9,8)(0,77)2✔  
Δy = -2,87 m (2,869 m)✔ 
This is higher than the height needed to reach point T (2,1 m)✔therefore the  ball will pass point T.

OPTION 3
(Emech)Top = (Emech)Ground
1 mark for any 
(EP +EK)Top = (EP +EK)Bottom
(mgh + ½ mv2)Top = (mgh + ½ mv2)Bottom 
(9,8)(h) + 0 ✔= 0 + (½ )(7,5)2 ✔ 
h = 2,87 m (2,869 m)✔ 
This is higher than height needed to pass the target (2,1 m)✔therefore the ball will pass the target.✔ 

OPTION 4
Wnet = ∆EK 
mgΔxcosθ = ½ mvf2- ½ mvi2 ✔ 
(9,8)Δxcos180º✔ = 0 - ½(7,5)2✔ 
Δx = 2,87 m (2,869 m)✔ 
This is higher than point height needed to pass point T (2,1 m)✔therefore the  ball will pass point T. ✔ 

OPTION 5
Upward positive
If the highest point is yf then Δy = (yf – y1,6) At highest point vf is zero
vf2 = vi2 + 2aΔy
0✔ = [(7,5)2 + (2)(-9,8)(yf – 1,6)]✔ 
yf = 4,47 (4,469) m✔ 

Yes ✔ ✔ 
OR
This point (4,47m) is higher than point T ✔ ✔  (or even the required height of  2,1 m) therefore the ball will pass point T. 

Downward positive
If the highest point is yf then Δy = (yf – y1,6)
At highest point vf is zero
vf2 = vi2 + 2aΔy
0✔ = [(-7,5)2 + (2)(9,8){yf – (-1,6)}] ✔ 
yf = -4,47 (-4,469) m✔ 
height is  is 4,47 m. 
This point (4,47 m) is higher than point T✔✔ (or even the required height of  2,1 m) therefore the ball will pass point T. 

OPTION 6 (POSITIVE MARKING FROM 3.2) 
Upward positive
If the highest point is yf then Δy = (yf – y1,6) At highest point vf is zero
Δy = viΔt + ½ aΔt2✔ 
(yf – 1,6) = (7,5)(0,77) ✔ + ½ (-9,8)(0,77)2✔  
yf = 4,47 m (4,469 m)✔ 
This point (4,47m) is higher than point T ✔✔ (or even the required height  of 2,1 m) therefore the ball will pass point T. 

Downward positive
If the highest point is yf then Δy = (yf – y1,6) At highest point vf is zero 
Δy = viΔt + ½ aΔt2✔ 
{yf – (-1,6)} = (-7,5)(0,765) ✔ + ½ (9,8)(0,765)2✔  
yf = -4,47 m (-4,469 m)✔ 
This point (4,47m) is higher than point T✔✔ (or even the required height of  (2,1m) therefore the ball will pass point T. 

OPTION 7 (POSITIVE  MARKING FROM 3.2) 
Upward positive
6 ujy gayd
This is higher than height needed to  pass the target (2,1 m)✔therefore the  ball will pass the target.✔ 

OPTION 7(POSITIVE  MARKING FROM 3.2 
Downward positive
7 jkguaygdgi
Height  is 2,89m 
This is higher than height needed to  pass the target (2,1 m)✔therefore the  ball will pass the target.✔ 

OPTION 8 
Upward positive
At highest point vf is zero
vf2 = vi2 + 2aΔy
0✔ = vi2 – (2)(9,8)(2,1) ✔ 
vi = 6,42 m∙s-1 ✔ 

  • This is the actual velocity needed to reach the target.
  • The given velocity is greater than the actual velocity needed. ✔
  • The ball will pass the target. ✔ 

Downward positive
At highest point vf is zero 
vf2 = vi2 + 2aΔy
0✔ = vi2 + (2)(9,8)(-2,1) ✔ 
vi = 6,42 m∙s-1 ✔ 

  • This is the actual velocity needed to pass the target.
  • The given velocity is greater than the actual velocity needed. ✔
  • The ball will reach the target. ✔ 

OPTION 9
Wnc = ΔEp + ΔE✔ 
0 = mghf – mghi + ½ mvf2 – ½ mvi2 
0 ✔ = (9,8)hf – (9,8)(1,6) + ½ (0) 2 – ½ (7,5)2✔ 
 0 = (9,8)hf – 43,805 
∴ hf = 4,47 m ✔ 
∴ The ball will pass point T ✔ ✔ 

OPTION 10
POSITIVE MARKING FROM 3.2
Upward positive
Δt(max. height/maks. hoogte) = 0,77 s 
Δy = viΔt + ½ aΔt2 ✔ 
2,1✔  = (7,5)Δt + ½ (-9,8)Δt2 ✔ 
∴ Δt = 0,36 s ✔ 
∴ Δt (max height, 0,77 s) > Δt (to pass point T 0,36 s) ✔ 
∴ The ball passed point T ✔ 

Downward positive
Δt (max height) = 0,77 s 
Δy = viΔt + ½ a Δt2 ✔ 

2,1✔ = (7,5)Δt + ½ (-9,8)Δt2 ✔ 
∴ Δt = 0,36 s✔ 
∴ Δt (max height, 0,77 s) > Δt (to reach point T, 0,36 s) ✔ 
∴ The ball passed point T ✔ 

OPTION 11
Upward positive
∆y = vi∆t + ½ a∆t2✔ 
(3,7 – 1,6)✔  = 7,5 ∆t + ½ (- 9,8) ∆t2 ✔ 
∆t = 0,375 s ✔ 
The time to pass point T is less than time to reach maximum height ✔ . Ball  will pass point T.✔ 

Downward positive
∆y = vi∆t + ½ a∆t2✔ 
(3,7 – 1,6)✔  = -7,5 ∆t + ½ (9,8) ∆t2 ✔ 
∆t = 0,375 s ✔ 
The time to reach point T is less than time to reach maximum height ✔ . Ball  will pass point T.✔ 

OPTION 12
Upward positive
vf2 = vi2 + 2aΔy ✔ 
vf2 = (7,5)2 ✔ + 2(-9,8)(2,1) ✔ 
vf = 3,88 m·s-1 ✔ 
Velocity at T is 3,88 m·s-1therefore the ball still moving towards its maximum height ✔ ✔ 

Downward positive
vf2 = vi2 + 2aΔy ✔ 
vf2 = (-7,5)2 ✔ + 2(9,8)(-2,1) ✔ 
vf = -3,88 m·s-1 ✔ 

Velocity at T is -3,88 m·s-1therefore the ball is still moving towards its  maximum height ✔✔

3.4 POSITIVE MARKING FROM 3.2 
8 kuhkahgduyha (2) [13] 

Notes

 

Initial velocity and time for final velocity shown  

Correct straight line (including orientation) drawn  

   

QUESTION 4
4.1 Momentum is the product of the mass of an object and its velocity ✔✔
[NOTE: 2 or/of 0] (2)
4.2 To the left ✔
Newton's third law✔
ACCEPT:

  • Principle of conservation of linear momentum / law of action-reaction✔
  • Newton's third law and Newton’s second law ✔ (2) 

NOTE: For QUESTION 4.3 and 4.4 motion to the right has been taken as positive. Candidates may use the opposite direction. 
9 gujyagdyug
4.5  (2)  [14]

POSITIVE MARKING FROM 4.4 /POSITIEWE NASIEN VANAF 4.4 32 kg⋅m⋅s-1/ N·s  to the right/opposite direction /na regs /teenoorgestelde  rigting 

QUESTION 5 
5.1

  • A force is non-conservative if the work it does on an object which is moving  between two points depends on the path taken✔✔
    OR
  • A force is non-conservative if the work it does on an object depends on the  path taken.✔✔
    OR
  • A force is non-conservative if the work it does in moving an object around a  closed path is non-zero ✔✔. (2) 

NOTE

  • If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. If the word work is omitted 0 marks  .

5.2 No ✔ (1) 
5.3 (3)

OPTION 1
P =  W✔ 
       Δt
= 4,8×106✔ 
      (90) 
 = 53 333,33 W  
 = 5,33 x 104 W (53,33 kW) ✔ 

OPTION 2
10 ikjgajyd
 WF = FΔxcosθ 
4,80 x 106 = F(1 125)cos0º
 F = 4 266,667 N 
Pave = Fvave 
 = (4 266,667)(12,5) 
 = 53 333, 33 W 

5.4

  • The net/total work done on an object is equal to the change in the object's  kinetic energy✔✔
    OR
    The work done on an object by a net force ✔is equal to the change in the  object's kinetic energy. ✔  (2) 

NOTE

If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. .

5.5 (5) (5) 

OPTION 1
Wnet = ΔK✔ 
Ww + Wf + WF = ½ mvf2- ½ mvi2 
mgΔxcosθ + Wf + WF = ½ mvf2- ½ mvi2 
(1 500)(9,8)200cos180º✔ + Wf + 4,8 x 106 ✔= ½ (1 500)(252 – 0) ✔
-2 940 000 + Wf +4,8 x 106 = 468 750 
Wf = -1 391 250 J 
 = -1,39 x 106 J✔ 

OR

Wnet = ΔK✔ 
Ww + Wf + WF = ½ mvf2- ½ mvi

-ΔEp + Wf + WF = ½ mvf2- ½ mvi2 
-(1 500)(9,8)(200 – 0)✔ + Wf + 4,8 x 106 ✔= ½ (1 500)(252 – 0) ✔
-2 940 000 + Wf +4,8 x 106 = 468 750 
Wf = -1 391 250 J 
 = -1,39 x 106 J ✔ 

NOTE

  • 0 can be omitted in above substitutions. 

OPTION 2
1 mark for any of  these
Wnc = ΔK + ΔU 
Wnc = ½ mvf2- ½ mvi2 + mghf - mgh
 = ½ m (vf2- vi2) + mg(hf - hi) 
Wnc = ½ mvf2 + mghf - ½ mvi2- mgh
 Wf + WF = ½ mvf2- ½ mvi2 + mghf - mgh
Wf + 4,8 x 106 ✔= [½ (1 500)(25)2 + -0] ✔+[(1 500)(9,8)(200) - 0]✔ 
Wf = - 1,39 x 106 J (-1,40 x 106J )✔ 

OR

1 mark for any of  these
Wnc = ΔK + ΔU 
Wnc = ½ mvf2- ½ mvi2 + mghf - mghi 
 = ½ m (vf2- vi2) + mg(hf - hi) 
Wnc = ½ mvf2 + mghf - ½ mvi2- mghi 
Wf + 4,8 x 106 ✔= [½ (1500)(25)2 + (1500)(9,8)(200)] - [0 + 0] 
Wf = - 4,8 x 106 + 3,4 x 106 
 = - 1,39 x 106 J (-1,40 x 106J )✔ 

ACCEPT THE FOLLOWING FOR: 3 /5
POSITIVE MARKING FROM 5.3
vf = vi + aΔt 
25 = 0 + a(90) 
a = 0,277…m\s-2 
Fnet = ma  
 =(1 500)(0,2777…) = 416,66… N 
F + (w||) + (-fk) = 416,666… 
4 266,6667 – 1 500(9,8)sinθ - fk = 416,666… 
fk = 1 236,6667 N 
Wf = fkΔxcosθ 
 = (1 236,6667)(1 125)(cos180°)
 = - 1 391 250 J 

(5) [13]

QUESTION 6
6.1

  • The change in frequency (or pitch), of the sound detected by a listener because the sound source and the listener have different velocities relative to the medium of sound propagation.
    OR
  • An (apparent) change in observed/detected frequency (pitch), as a result of the  relative motion between a source and an observer  (listener).  (2) 

NOTE

If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

6.2 Away from✔ 
Observed frequency lower  (2) 
6.3 (3) 

v = fλ ✔ 
340 = f (0,34) ✔ 
 f = 1 000 Hz ✔ 

6.4 

POSITIVE MARKING FROM 6.3
11 jhgvagtdhd
Accept:
12 hafysdd

(6) [13] 

QUESTION 7
7.1  (2)   

Qnet  = Q1 + Q2 + Q3 
                     3 
-3 x 10-9 = 15 × 10-9 +  Q  + 2x10-9 
                                  3 
Q = + 4 x 10-9 C✔ 
NOTE

  • for addition of the three correct charges 
  • correct answer 

13 jgajydgud
NOTES

  • Correct shape ✔
  • Correct direction ✔
  • Lines must not cross and must touch spheres ✔ (3) 

7.3 The magnitude of the electrostatic force exerted by one point charge (Q1) on  another point charge (Q2) is directly proportional to the product of the  (magnitudes) of the charges and inversely proportional to the square of the  distance (r) between them ✔✔ (2)

NOTE

  • If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. If masses used (0/2) 

7.4 (5) 
OPTION 1
14 jhgaydguagd
OPTION 2
15 ytfajydtafd
7.5  (3)

POSITIVE MARKING FROM 7.4 
OPTION 1
E = F
       q
= 8,15 × 10-6
    3 ×  10-9
 = 2,72 x 103 N.C-1 ✔ 

OPTION 2
ES = KQ
         r2
= (9×109)(3×10-9
         (0,1)2
 = 2 700 N.C-1 

ET = KQ
         r2
= (9×109)(3×10-9
         (0,3)2
 = 300 N.C-1 

Enet = √ (ES2 + ET2)
= √  (2700)2 +(30)2
 = 2 716,62 N.C-1  

NOTE
Mark Allocation

  • Correct formula
  • both subsitutions
  • correct answer

If calculation done in 7.4 award full marks for answer written here.

7.6.1 Sphere P or/of T✔ (1) 
7.6.2 
16 ujyaudyd(3)[19]

QUESTION 8 
8.1

  • The battery supplies 12 J per coulomb/12 J per unit charge. ✔ ✔OR
  • The potential difference of the battery in an open circuit is 12 V. ✔ ✔OR
  • The battery does 12 J of work per coulomb of charge.✔ ✔ OR
  • Maximum work done by the battery per unit charge is 12 J  ✔ ✔ OR
  • Maximum energy supplied by the battery per unit charge is 12 J  ✔ ✔ OR
  • The battery supplies 12 J of energy per coulomb/ 12 J of energy per unit  charge ✔ ✔ OR
  • The greatest potential difference that can be generated by a battery is 12V ✔ ✔ OR
  • The total energy transferred by a battery to a unit electric charge is 12 J  ✔ ✔ OR
  • The total amount of electric energy supplied by the battery per coulomb/per  unit charge is 12 J ✔ ✔  (2) 

NOTE
If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

8.2.1  (3) 

OPTION 1
Vlost = I r 
 = (2) (0,5) 
 = 1 V 
Vext  = Emf – Vlost 
 = (12 - 1) 
 = 11 V

OPTION 2
ε = I(R + r) 
12 = Vext +(2)(0,5) 
Vext = 11 V

OPTION 3
ε = I(R + r)
12 = 2(R + 0,5)
R = 5,5 Ω 
V = IR  
 = 2(5,5)
 = 11 V 

8.2.2  (2)

POSITIVE MARKING FROM 8.2.1
OPTION 1
R =
       I
 =  11 
      2
 = 5,5 Ω

OPTION 2
0,5:R 
 1:11 
R = 5,5 Ω 

OPTION 3
 1  =  11 
0,5    R 
R = 5,5 Ω 

OPTION 4
Vtotal = IRtotal 
12 = (2)Rtotal 
Rtotal = 6 Ω
R = 6 – 0,5 
 = 5,5 Ω 

OPTION 5 
ε = I(R + r) 
12 = 2(R + 0,5) 
R = 5,5 Ω 

8.3 Decreases 

  • Total resistance decreases ✔
  • Current increases ✔
  • "Lost volts" increases, (emf the same) ✔
  • External potential difference decreases ✔
    OR
  • Decreases ✔
  • Total resistance decreases  ✔
  • Current increases ✔
  • ε = Vext + Ir ✔
  • Ir increases ✔
  • ε is constant ✔
    ∴Vext/eks decreases (4) [11] 

QUESTION 9
9.1 Temperature/Temperatuur ✔ (1)
9.2.1 r = 3 Ω or/of 1,5 Ω ✔✔

  • Accept for one mark only:
    r = -3 Ω ✔ or -1,5 Ω (2) 

9.2.2 (3) [6] 

ε = slope (gradient) of the graph✔

  • Accept any correct values from the graph 

ε = 7,5 -(-3)
       1,5 -0 
 = 7 V  ✔

OR

POSITIVE MARKING FROM 9.2.1 
R = ε- r 
       I

  • Accept any correct values on the line from the graph 

7,5 = 1,5ε -3
ε= 7 V  ✔

OR

ε = I(R + r) 
 = 0,5(11 + 3) 
ε= 7 V  ✔

QUESTION 10
10.1.1 Y to/na X ✔ (1) 
10.1.2

  • Faraday’s Law Electromagnetic Induction ✔
    OR
  • Electromagnetic induction/Faraday's Law (1) 

10.1.3 Mechanical (kinetic) energy ✔to electrical energy ✔(2)
10.2.1 340 V ✔(1) 
Accept  -340 V 
10.2.2  POSITIVE MARKING FROM 10.2.1
17 ujygajydgyad(3) 

10.2.3 POSITIVE MARKING FROM  10.2.2 (3) [11] 
OPTION 1                                                                         OPTION 2
18 igajkgddc

OPTION 3
Pave =  VrmsIrms 
1600 = (240,416) Irms 
Irms = 6,66 A 
R = Vrms
       Irms 
= 240,416 
       6,66 
 = 36,1 Ω (36,09 Ω )

(Do not penalise if rms is omitted V) in R = Vrms
                                                                     Irms  

OPTION 4
Pave  =  VmaxImax
                    2 
1600 = 340Imax 
              2 
Imax = 9,412 A 
R = Vrms
       Irms 

 340    
    9,412 
 = 36, 12 Ω 

(Do not penalise if rms is omitted V) in R = Vrms
                                                                     Irms 

QUESTION 11
11.1  (2) 

Work function of a metal is the minimum energy needed to eject an electron  from the metal surface ✔✔

NOTE
If any of the underlined key words/phrases in the correct context is omitted  deduct 1 mark. 

11.2 (2) 

Potassium / Kalium / K ✔
fo for potassium is greater than fo for caesium 
OR
Work function is directly proportional to threshold frequency ✔
ACCEPT
Wo = hfo

Wo α fo

11.3  (3) 

OPTION 1
c = fλ 
3 x 108 = f(5,5 x 10-7
f = 5,45 x 1014 Hz 
fuv < fo of K(potassium) 
∴Ammeter in circuit B will not show a reading 

OPTION 2
E = hc (6,63 × 10-34)(3 × 108
       λ                5,5 × 10-7

 = 3,6164 x 10-19 J 
Wo = hfo = (6,63 x 10-34)(5,55 x 1014) = 3,68 x 10-19
Wo > E or hfo > hf 
∴The ammeter will not register a current 
Mark allocation

  • both correct formulae: E = hc and Wo = hfo 
                                                λ
  • both substitutions:  (6,63 × 10-34)(3 × 108
                                              5,5 × 10-7
     (6,63 x 10-34)(5,55 x 1014)
  • correct conclusion

OPTION 3
c = foλo
3 x 108 = (5,55 x 1014)λ 
λo = 5,41 x 10-7
λo (threshold wavelength) < λ (incident wavelength) 
∴ the ammeter will not register a current  

11.4 (5) 

OPTION 1
19 iaguydguyag

NOTE: If EK of the incorrect   photocell is calculated, candidate   forfeit the mark for the final  answer. 
(6,63 × 10-34 )(3 x 108= (6,63 x 10-34)(5,07 x 1014) + Ek(max) 
      5,5  ×107
 EK = 2,55 x 10-20 J  (Range: 2,52 x 10-20 – 2,6 x 10-20 J) 

OPTION 2
POSITIVE MARKING FROM 11.3
20 iaghkjdgya

NOTE: If EK of the incorrect  photocell is calculated, candidate  forfeit the mark for the final  answer. 
(6,63 x 10-34)(5,45 x 1014) = (6,63 x 10-34)(5,07 x 1014) + Ek(max) 
EK = 2,52 x 10-20 J  (Range: 2,52 x 10-20 – 2,6 x 10-20 J) 

11.5 Remains the same (1) [13] 

TOTAL: 150

ADDENDUM 
QUESTION 7.2 
Accept the following electric field diagram which would be formed if the effect of the third  charge is considered. 
21 jhgajdyguagd

MATHEMATICAL LITERACY
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

INSTRUCTIONS AND INFORMATION

  1. This question paper consists of FOUR questions. Answer ALL the questions.
  2. 2.1 Use the ANNEXURES in the ADDENDUM to answer the following questions:
    • ANNEXURE A for QUESTION 1.2
    • ANNEXURE B for QUESTION 2.1
    • ANNEXURE C for QUESTION 3.1
    • ANNEXURE D for QUESTION 4.1
      2.2 Answer QUESTION 2.1.5 on the ANSWER SHEET attached. Write your centre number and examination number in the spaces on the ANSWER SHEET. Hand in the ANSWER SHEET with your ANSWER BOOK.
  3. Number the answers correctly according to the numbering system used in thisquestion paper.
  4. Start EACH question on a NEW page.
  5. You may use an approved calculator (non-programmable and non-graphical), unless stated otherwise.
  6. Show ALL calculations clearly.
  7. Round off ALL final answers appropriately according to the given context, unless stated otherwise.
  8. Indicate units of measurement, where applicable.
  9. Maps and diagrams are NOT drawn to scale, unless stated otherwise.
  10. Write neatly and legibly.

QUESTIONS 

QUESTION 1
1.1 Mapotjo plans to purchase a bakkie (motor vehicle). She received the followingquotation from a car dealer on 28 February 2018:
Extract of a quotation for a bakkie from a dealer:

  Rand (R) 
Selling price (excluding VAT) without accessories (extras)   160 087,72
Discount  6 140,00
ACCESSORIES (EXTRAS)  
Smash-and-grab film  3 500,00
Door protector  3 500,00
OTHER CHARGES  
On-road charges  4 298,25
Transaction fee  1 315,79
SUBTOTAL  166 561,76
VALUE-ADDED TAX  14%
TOTAL DUE  189 880,41

[Source: Group 1 Nissan and The Glen]
Use the information above to answer the questions that follow.
1.1.1 Calculate (rounded off to ONE decimal place) the percentage discountgiven on the bakkie's selling price, excluding VAT. (3)
1.1.2 Show how the amount of R166 561,76 was calculated. (3)
1.1.3 Give ONE reason why customers would prefer to install the accessories(extras), as shown in the quotation. (2)
1.1.4 Mapotjo has an investment of R1,25 million. The money was invested as follows: 

  • Twenty-seven (27) months investment period
  •  6% interest per annum, compounded annually

Show whether the interest earned on this investment is sufficient to cover the total purchase price of R189 880,41. (9)
1.1.5 VAT in South Africa increased to 15% with effect from 1 April 2018.
The following shows how the dealer calculated the new increased VAT incorrectly:

  • Selling price, including 14% VAT
    = R160 087,72 + 14% of R160 087,72
    = R160 087,72 + R22 412,28
    = R182 500
  • VAT increased with 1%.
    New selling price, including 15% VAT = R182 500 + R182 500 x 1%
    = R182 500 + R1 825
    = R184 325

Identify the mistake the dealer has made in calculating the new selling price. Hence, calculate the new selling price, including 15% VAT (excluding accessories and other charges). (4)
1.2

After Mapotjo had purchased the bakkie, she decided to paint the cargo bin (loading box) of her bakkie with rubberising paint. The spread rate of the paint is 0,25 m2/f. The rubberising paint is sold in 5 f tins.
A photograph of the cargo bin of a bakkie and a simplified model of the cargo bin with dimensions are given in ANNEXURE A.
NOTE: Rubberising paint — a special paint used to coat the surface of a cargo bin 

 Use the information above and ANNEXURE A to answer the questions that follow.
1.2.1 The cargo bin does not have a flat surface area and therefore the surface area must be increased by 2% to accommodate the uneven surface area.

  1. Calculate how many litres of rubberising paint Mapotjo needs to purchase in order to rubberise the cargo bin of her bakkie.
    You may use the formula:
    • Surface area of an open box
      = Width x length + 2(length x height + width x height)
  2. Calculate the cost of applying TWO coats of rubberising paint (excluding labour) if a 5 f tin costs R549, including VAT.

1.2.2 Explain the significance of applying rubberising paint to the cargo bin of a bakkie.
1.3 It takes 20 minutes to apply a layer of rubberising paint. There is a 4-hour waiting period before the second coat of rubberising paint can be applied. In addition, a further drying time of 2 hours is required after the second coat has been applied.
Determine, with calculations, at what time the bakkie would be ready if the workshop started rubberising the cargo bin at 08:15. (4) [38]

QUESTION 2
2.1 TABLE 1 in ANNEXURE B shows data relating to the import of personal care and cosmetic products in Australia for the period 2013 to 2015.
Use the information in ANNEXURE B to answer the questions that follow.
2.1.1 Determine:

  1. Missing value A (rounded off to a whole percentage) using the following formula:
    • % change (A) = 2015 imports -  2013 imports  x 100%  (4)
                                            2013 imports
  2. The median of the percentage change for the period 2013 to 2015. (3)

2.1.2 Describe the trend in the imports of make-up and skincare products. (2)
2.1.3 The negative value of the percentage change does not necessarily imply that the import value of the products decreased continuously over the three-year period. Name TWO different products in TABLE 1 and explain how they support the statement above. (4)
2.1.4 State, with a reason, whether the data in TABLE 1 can be represented using a single pie chart. (2)
2.1.5 A line graph showing the % share of 2015 imports for the first eight products in TABLE 1 has been drawn on the ANSWER SHEET. On the same grid, draw another line graph representing the percentage change for the period 2013 to 2015 for the same eight products. (6)
2.2 Nomsa plans to visit Los Angeles for an educational conference. She will be travelling from her hotel to the conference venue with an Uber taxi.
An Uber taxi is operated using two options, as shown below.

  • Option 1: UPFRONT fare
    = base fare (call-out fee) + (number of miles x per mile fare)
  • Option 2: POST-TRIP fare
    = (number of minutes x per minute fare) + (number of miles x per mile fare)

TABLE 2 below shows the different Uber taxis and their respective rates in Los Angeles for both UPFRONT and POST-TRIP fare options, including an example of a 10-mile trip using the UPFRONT fare option.

COST

UberX  UberBLACK UberLUX 
Base fare (call-out fee)  $0,00 $ 8,00 $20,00
Per minute fare $0,15 $ 0,45 $ 0,60

Per mile fare

$0,90 $ 3,55 $ 5,00
*Minimum fare $4,65 $15,00 $30,00
Cancellation fee $5,00 $10,00 $10,00
Total fare (for a 10-mile trip using the UPFRONT option) $9,00 $43,50 B

[Adapted from www.uber.com, March 2016]
*Minimum fare: the lowest fare one would be charged per trip
Use the information above to answer the questions that follow.
2.2.1 Calculate the missing value B. (3)
2.2.2 Calculate (rounded off to the nearest mile) the maximum distance for which a person can use the UberX taxi if you pay the minimum upfront fare. (4)
2.2.3 Nomsa travelled a distance of 29,73 miles with UberBLACK. The post-trip fare option Was used and the trip took 1 hour and 9 minutes to complete. Nomsa stated that she would have saved more than $20,00 if she had used the upfront fare option. Show, with calculations, whether her statement is correct. (8)
2.2.4 Explain the importance of a cancellation fee for the Uber service provider. (2) [38]

QUESTION 3
3.1

The Big Five Marathon is an annual event in South Africa. It can be run as a full 42 km marathon or as a half-marathon of 21 km.
The race has specific cut-off times (certain compulsory distances to be covered within specific times). Runners who do not meet the cut-off times are forced to withdraw from the race.
Below are the cut-off times for the full marathon.

FULL MARATHON   
  CUT-OFF 1  CUT-OFF 2 CUT-OFF 3
Distance from start  25,5 km  31,5 km  Finish line
Time from start  4 hours 15 min  5 hours 15 min  7 hours

[Source: http://aublog.southafrica]

ANNEXURE C contains the Big Five 42 km full marathon map.

Use the information above and ANNEXURE C to answer the questions that follow.
3.1.1 Determine (as a decimal fraction) the probability of a runner of the Big Five marathon route accessing a refreshment station that offers ONLY Coke and water. (3)
3.1.2 Give the general direction in which a marathon runner is heading when passing the 20 km mark. (2)
3.1.3 Consider the heights above sea level for this race.

  1. Explain why a runner was CORRECT when he stated that he was running uphill from the start to the 10 km mark. (2)
  2. Express, in the form 1 : ..., the lowest possible height above sea level to the highest height above sea level. (3)

3.1.4 Explain why there are cut-off times for a marathon. (2)
3.1.5 For the half-marathon a runner must cover a distance of 16,5 km in a time of 5 hours from the start of the race to beat the cut-off 2 time for the half-marathon. A runner of the full marathon compared his speed with the speed of a half-marathon runner and stated that he had to run 2,7 km/h faster in order to beat the cut-off 2 time of the full marathon.
Verify, showing ALL calculations, whether he is CORRECT. You may use the formula:

  • Distance = speed x time (6)

3.2 

 More water must be taken to the refreshment stations. The water will be transported in cylindrical buckets (with lids) with a maximum capacity of 20 litres of water.
The cylindrical buckets, containing water, with lids are shown below.
MATHS 1

[Source: www.meyaplast.co.uk] 

Use theinformation and picture above to answer the questions that follow.
3.2.1 Determine the maximum height (in cm) of the water in the bucket if the outside diameter of the bucket is 31,2 cm.
You may use the formula:

  • Volume of a cylinder = n x (radius)2 x height
    where π = 3,142 and 1 f = 1 000 cm3 (7)

3.2.2 Buckets are placed on the pallet, as shown in the diagram above.

  1. Calculate the unused area (in cm2) of the rectangular floor of the solid pallet.
    You may use the formula:
    • Area of a circle = 7 x (radius)2, where π = 3,142 (6)
  2. Determine length C, as shown in the diagram above. (3)

3.2.3 The organiser would have preferred each pallet to have 12 buckets arranged in three rows of four each, as shown in the diagram alongside.
MATHS 2 IUHYUAHD
Calculate the percentage by which the length of the pallet should be increased to accommodate this new arrangement. (5) [39]

QUESTION 4
4.1 ANNEXURE D shows photographs of the London Eye, which is a *Ferris wheel.
The following information represents features of the Ferris wheel:

  • The structure is 443 feet tall.
  • The radius of the wheel is 197 feet.
  • It has 32 capsules that are spaced apart evenly.
  • Each capsule can carry a maximum of 28 passengers.
  • Ticket prices for all capsules are identical.

The following are the ticket prices per person:

TICKET CATEGORY PRICE
Adults (16+)  £27,00
Children aged 3-15 £22,00
Senior citizens £25,50
10% discount for tickets bought online

 [Source: www.londoneye.com]

  • Conversions: 1 pound (£) = R16,58
    1 metre = 3,28 feet

*Ferris wheel: a very large upright wheel with capsules on its circumference for people to ride in
Use the information above to answer the questions that follow.
4.1.1 Capsule 24 and capsule 30 have the following number of occupants.

  ADULTS (16+)  CHILDREN AGED 3-15    SENIOR CITIZENS
Capsule 24   18  7  2
Capsule 30  10  1  5

Only 5 adults of the total number of occupants in the table above bought tickets online. A South African tourist at the London Eye remarked that more than R18 400 was spent on these tickets. Verify, with calculations, whether this remark is CORRECT. (8)
4.1.2 The capsules are mounted on the circumference of the wheel.

  1. Calculate (in feet) the circumference of the wheel. 
    You may use the formula:
    • Circumference = 2 x π x radius, where π = 3,142 (2)
  2. Hence, calculate the distance (to the nearest metre) on the circumference between any TWO capsules right next to each other. (3)

4.2 Tourism creates many employment opportunities in the United Kingdom. Tourists are most likely business visitors, holiday visitors or visitors to friends and relatives (VFR).
TABLE 3 below shows information regarding the number of visitors (in thousands) and the reasons for their visits. It also shows the number of employment opportunities for the different tourist regions.
TABLE 3: NUMBER OF VISITORS (IN THOUSANDS) AND THE REASONS FOR VISITS TO DIFFERENT TOURISM REGIONS IN THE UNITED KINGDOM AND OPPORTUNITIES

REGIONS 

NUMBER OF VISITORS (IN THOUSANDS) ACCORDING TO REASON FOR VISIT

DIRECT EMPLOYMENT

 
HOLIDAY VFR BUSINESS
London 7 575,9  3 556,0  3 092,2  471 928
North East  115,3  175,1  115,3  ...
North West  624,0  762,6  531,5  170 113
Yorkshire  273,0  480,5  273,0  119 639
West Midlands  312,6  562,7  609,6  107 230
East Midlands  166,0  405,7  295,0  76 496
East of England 467,0 856,2 447,6

120 343

South West 766,5 806,8 302,6

179 450

South East 1 335,5 1 594,0 1 033,9

226 003

Scotland 1 157,0 600,8 378,3

172 282

Wales 324,5 324,5 87,6 ...
TOTAL 13 117,3 10 124,8 7 146,6 ...

[Adapted from visitbritain.org/research]
Use the information above to answer the questions that follow.
4.2.1 Calculate the difference between the number of holiday visitors to the North West and the number to the West Midlands. (3)
4.2.2 Determine (as a percentage) the probability of randomly selecting a business visitor to the Midlands from the total business visitors. (4)
4.2.3 A visitor stated that there are more than 3 times more holiday visitors than business visitors to Scotland. Verify, with a calculation, whether this statement is valid. (3)
4.2.4 Calculate the interquartile range for VFR visitors. (5)
4.2.5 Give ONE other reason, besides employment opportunities, why tourism is important to a country. (2)
4.2.6 The mean direct employment in the United Kingdom is 162 666,5455. The North East employs 30 440 fewer people than Wales. Calculate the direct employment for the North East. (5) [35]

TOTAL: 150

ANSWER SHEET QUESTION 2.1.5
CENTRE NUMBER
EXAMINATION NUMBER
Percentage imports and average growth of personal care and cosmetic products in Australia
MATHS 3 IUHYAUHD

MATHEMATICAL LITERACY
PAPER 1
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

INSTRUCTIONS AND INFORMATION

  1. This question paper consists of FIVE questions. Answer ALL the questions.
  2. 2.1 Use the ANNEXURES in the ADDENDUM to answer the following questions:
    • ANNEXURE A for QUESTION 2.1
    • ANNEXURE B for QUESTION 3.1
    • ANNEXURE C for QUESTION 4
      2.2 Answer QUESTION 5.2.7 on the attached ANSWER SHEET.
      2.3 Write your centre number and examination number in the spaces on the ANSWER SHEET. Hand in the ANSWER SHEET with your ANSWER BOOK.
  3. Number the answers correctly according to the numbering system used in this question paper.
  4. Start EACH question on a NEW page.
  5. You may use an approved calculator (non-programmable and non-graphical), unless stated otherwise.
  6. Show ALL calculations clearly.
  7. Round off ALL final answers appropriately according to the given context, unless stated otherwise.
  8. Indicate units of measurement, where applicable.
  9. Maps and diagrams are NOT necessarily drawn to scale, unless stated otherwise.
  10. Write neatly and legibly.

QUESTIONS 

QUESTION 1
1.1 Happy Life Superstore advertised the specials below for the annual Black Friday in 2017.
1.1 yghugad
NOTE:

  • 1 L =1000ML
  •  ALL amounts given INCLUDE the discount

Study the advertisement above to answer the questions that follow.
1.1.1 Write down the number of day(s) on which these prices are valid. (2)
1.1.2 Calculate the original price of hth before the saving. (2)
1.1.3 Write down the name of the product which is now half price. (2)
1.1.4 Convert 750 mt to litres. (2)
1.1.5 Calculate the total price of ONE 2 L-bottle of Coca Cola and TWO 2 L-bottles of Fanta. (2)
1.1.6 Arrange ALL the sale prices in ascending order. (2)
1.2 The picture below is a scaled drawing of a T-shirt for Grade 12 learners.
1.2 kughiuyahd
Scale = 1 : 25
1.2.1 Calculate the number of letters needed to print the logo on the front of the T-shirt. (2)
1.2.2 Write down the temperature displayed on the thermometer in 'C. (2)
1.2.3 Explain the meaning of the scale in the drawing above. (2)
1.2.4 Measure the length of the back of the T-shirt in mm, as indicated in the drawing. (2)
1.3 The Two Oceans Marathon and the Comrades Marathon are two of the most popular ultramarathons in the world.
TABLE 1 below shows the dates, distances and entry fees of these marathons.
TABLE 1: TWO OCEANS MARATHON VS COMRADES MARATHON 

  TWO OCEANS   COMRADES 
Date (2017)   15 April 2017  4 June 2017
Distance  56 km  89 km
Entry fee  R520,00  R460,00

[Adapted from www.capetownmagazine.com and www.news.comrades.com]
Use TABLE 1 above to answer the questions that follow.
1.3.1 Which race took place first? (2)
1.3.2 Which one of the two races had the longest distance? (2)
1.3.3 Determine the difference between the entrance fee of the Two Oceans Marathon and the entrance fee of the Comrades Marathon. (2)
1.4

The Comrades Marathon Association (CMA) has issued its medical statistics for the
race held on Sunday 4 June 2017.
Start of the race: 05:30
End of the race: 17:30
TABLE 2 shows the medical statistics on race day.
TABLE 2: MEDICAL STATISTICS

Athletes starting the race 17 031 
Athletes finishing the race 13 852
Athletes treated in the medical tent  400
Hospital-treated athletes 90
Hospital-admitted athletes  40

[Adapted from http://www.runnersworld.co.za]
Use TABLE 2 above to answer the questions that follow.
1.4.1 Write down the maximum time given to the athletes to complete the Comrades Marathon. (2)
1.4.2 State if the medical statistics data is discrete or continuous. (2)
1.4.3 Write down the ratio of athletes starting the race to the athletes finishing the race. (2) [32]

QUESTION 2
2.1 ANNEXURE A shows the student fees statement for Tamryn Abrahams, a second-year Architecture student at the University of Cape Town (UCT).
Use ANNEXURE A to answer the questions that follow.
2.1.1 Explain the meaning of the term interest with reference to the student fees statement.  (2)
2.1.2 Write down the balance (excluding interest) that was brought forward on the last day of the previous year.  (2)
2.1.3 Calculate the monthly interest rate that was used on the overdue fees for the previous year.  (3)
2.1.4 Write down the code and the name of the module/course that is the most expensive.  (2)
2.1.5 Show how the amount of R6 317,70 was calculated.  (2)
2.1.6 Calculate the total amount debited to this account for the courses studied in the 2017 academic year including interest on overdue fees in 2017.  (3)
2.1.7 State the payment method used to transfer money into this account.  (2)
2.1.8 A family friend paid the balance of R40 386,60 on condition that the amount could be paid back in equal monthly instalments, interest free. Show how the monthly instalment of R8 077,32 was calculated if the first payment was due on 1 November 2017 and the last payment was due on 1 March 2018. (2)
2.2
2.2 jugyjagd
Use the information above to answer the questions that follow.
2.2.1 Explain the term inflation within the given context. (2)
2.2.2 Write down the price of a Spur burger in 1970. (2)
2.2.3 Calculate by how much the cost, in rand, of a trolley had increased from 2000 to 2005. (3)
2.2.4 Calculate the percentage increase of Ricoffy from 1970 to 2015. You may use the following formula:

  • Percentage increase = new amount - original amount x 100 %  (3)
                                                  original amount

2.2.5 A cheddamelt steak was sold for R104,90 at a percentage profit of 17,5%. Determine the cost price.   (2)
2.3 TABLE 3 below shows the national budget and education budget of South Africa for 2017/18.
TABLE 3: NATIONAL BUDGET AND EDUCATION BUDGET OF SOUTH AFRICA FOR 2017/2018

NATIONAL BUDGET OF SOUTH AFRICA (IN RAND)  EDUCATION BUDGET OF SOUTH AFRICA (IN RAND

Economic affairs and agriculture

241,6 billion   Basic education 216,7 billion 

Defence and public safety

198,7 billion  University subsidies 31,6 billion

Health

187,5 billion  Education administration 15,8 billion

General admin

70,7 billion Skills development levy institutions 21,1 billion

Local development and infrastructure

195,8 billion  National student financial aid scheme (NSFAS) 15,3 billion

Debt service costs

162,4 billion  Technical and vocational education and training 7,5 billion

Socialprotection

180,0 billion  Other 12,5 billion
Education 320,5 billion    

[Adapted from www.graphics24.com]
Use TABLE 3 above to answer the questions that follow.
2.3.1 Which of the amounts below represents the economic affairs and agriculture budgets?

  1. 24 160 000
  2. 241 600 000 000
  3. 241 600 000
  4. 24 160 000 000 000 (2)

2.3.2 Explain the term budget within the context above. (2)
2.3.3 Write down the item which receives the third most money from the education budget. (2)
2.3.4 Calculate the percentage of the total education budget that is allocated to the NSFAS. (3)
2.3.5 University subsidies comprise about 9,86% of the total education budget. Estimate the combined budget, as a percentage, for education administration and the NSFAS. (2) [41]

QUESTION 3
3.1 Liam and Amy are planning their wedding. Amy wants a four-layer red velvet wedding cake. She must still decide between a cylindrical or rectangular cake as shown on ANNEXURE B.
Use ANNEXURE B to answer the questions that follow.
3.1.1 Determine the total height of the cylindrical cake in millimetres.
3.1.2 The base (bottom) layer of the cylindrical cake has a radius of 14 cm.

  1. Determine the diameter of the base layer in cm.
  2. Calculate the volume (in cm3) of the base layer.

You may use the following formula:

  • Volume of a cylinder = π x (radius)2 × height, and using π = 3,142 (3)

3.1.3 Define the term perimeter. (2)
3.1.4 Calculate the area (in cm2) of the base of the pan needed to bake the top layer of the rectangular cake.
You may use the following formula:

  • Area = length x width (2)

3.2 Aunt Abby will bake the wedding cake. She will be using a recipe from a recipe book published in England.
NOTE:

  •  1 kg = 2,25 pounds
  •  1 ml flour = 0,7 g flour

3.2.1 Aunt Abby needs 3 and a half pounds of butter. Determine the mass of butter, in kilogram. (2)
3.2.2 Aunt Abby only has a kitchen scale available. If aunt Abby needs 625 ml, of flour, determine the mass of the flour in grams. (2)
3.2.3 The cake must be baked at 356 °F. Determine to what degree Celsius the oven should be turned. You may use the following formula: 

  • ºC = (ºF - 32º) ÷ 1,8   (2)  [18]

QUESTION 4
4.1 A parkrun is a weekly 5 km run. A group of runners drove from Upington to Springbok to take part in the weekly parkrun in Springbok.
ANNEXURE C shows a route map from Upington to Springbok.
Use ANNEXURE C to answer the questions that follow.
4.1.1 Give the general direction from Upington to Springbok. (2)
4.1.2 Write down the name of the national park close to Kamieskroon. (2)
4.1.3 Name TWO towns the runners will pass through on their way to Springbok, following the N14. (3)
4.1.4 Identify the type of scale used on the map. (2)
4.1.5 Use the given scale to determine the actual distance (to the nearest km) between Upington and Springbok. (4)
4.2 

On arrival in Springbok the runners must first pick up Joe, a fellow runner, before heading to the parkrun (B).
ANNEXURE C shows a street map indicating the route from entering Springbok (A) to the parkrun (B). 

Use ANNEXURE C to answer the questions that follow.
4.2.1 Name the road by which they will enter Springbok. (2)
4.2.2 Joe gives them the following directions to his home:

  • Enter Springbok from Upington.
  • Turn right into Uitspan Street.
  • Turn left into Lukhof Street.
  • Turn left into the first street.

Use the directions above to determine in which street Joe lives. (2)
4.2.3 Name of the lodge near the parkrun. (2)
4.2.4 The distance from Joe's house to the parkrun is 2,34 km. They travel at an average speed of 40 km/h. Determine how long it will take them (in minutes) to get from Joe's house to the parkrun.
You may use the following formula:

  • Time = distance  (3)
                 speed

4.2.5 29 of the 42 athletes who participated in the parkrun were female. Determine the probability of randomly selecting a male athlete from this group. (2) [24]

QUESTION 5
5.1
5.1 UIHIUAYKHD
[Adapted from: http://www.fpasa.co.za]
Study TABLE 4 above to answer the questions that follow.
5.1.1 Write down the total loss, in rand, caused by fire during 2011. (2)
5.1.2 Calculate the mean total loss, in rand, caused by fires for the period 2010 to 2015. (3)
5.1.3 Identify the maximum number of fires for the period 2010 to 2015. (2)
5.1.4 Calculate the value of A, the fire loss as a percentage of the GNI for 2013. Round your answer to TWO decimal places. (4)
5.2 
5.2 ikhuakhed
[Adapted from: www.statssa.co.za]
Use TABLE 5 above to answer the questions that follow.
5.2.1 Explain the meaning of the term unemployed within the context of the table above. (2)
5.2.2 Determine the value of X, the number of people employed in Mpumalanga. (2)
5.2.3 Name ONE data collection instrument that could be used to collect the data above. (2)
5.2.4 Calculate the percentage of people in the Western Cape who are NOT economically active (NEA). (3)
5.2.5 Write down the ratio of employed people to unemployed people in South Africa in the form ... : 1. (2)
5.2.6 Determine the probability (as a decimal) of randomly selecting a person in the Free State who is NOT economically active (NEA). (3)
5.2.7 The graph on the ANSWER SHEET represents the number of economically active people, as well as those who are not economically active (NEA) in South Africa. The bars for ALL economically active persons and only the bar for the people in the Eastern Cape who are NOT economically active (NEA) are drawn.
Use the ANSWER SHEET to draw the graphs for the rest of the provinces. (6)
5.2.8 Determine the probability, as a simplified fraction, of selecting a province where fewer than 350 000 people are unemployed. (4) [35]

TOTAL: 150

ANSWER SHEET
QUESTION 5.2.7 
CENTRE NUMBER:
EXAMINATION NUMBER
graphs iuhiuhad

MATHEMATICAL LITERACY
PAPER 1
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

MARKS: 150  

Symbol 

Explanation

Method

MA 

Method with accuracy

CA 

Consistent accuracy

Accuracy

Conversion

Simplification

RT 

Reading from a table/graph/document/diagram

SF 

Correct substitution in a formula

Opinion/Explanation

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

Rounding off

NPR 

No penalty for rounding

AO 

Answer only

MCA 

Method with constant accuracy

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. 

MEMORANDUM 

QUESTION 1 [32 MARKS] ANSWER ONLY FULL MARKS

Q/V 

Solution

Explanation

T&L

1.1.1 

✔✔A    1 / one
OR
A day / 'n dag   ✔✔A 
OR  ✔✔A
One day 

2A for correct day  (2)

L1

1.1.2 

Price before saving  ✔M 
R70 + R250  ✔A
= R320 

1M adding correct values 
1A simplification  (2)

L1

1.1.3

Ariel ✔✔A

2A product   (2)

L1

1.1.4 

                  ✔MA 
750 m ℓ ÷ 1 000  
= 0,75 ℓ  ✔A 
OR
                 ✔MA 
750 m ℓ × 0,001  
= 0,75 ℓ   ✔A

1MA for dividing by 1 000 
1A simplification only if division

OR

1MA for multiplying by 0,001 
1A simplification only if multiplied (2)

L1

1.1.5 

Price  
= R11 × 3  ✔MA
= R33,00✔CA 

1MA multiplying correct values 1CA simplification  (only if R7,70× 3)  (2)

L1

Solution

Explanation

T&L

1.1.6 

R11; R15; R18; R22; R30; R43; R44; R45; R65;  R250✔✔A

2A arranging in correct order
If names used max 1 mark (2)

L1

1.2.1 

English = 35 letters OR 15 letters  ✔✔A

2A correct number  WC, FS, NC Provinces  accept both  (2)

L1

1.2.2 

44 0C ✔✔A

2A correct reading Accept 44 - 45 0C  (2)

L1

1.2.3 

One unit on the drawing represents twenty five units  in reality ✔✔A
OR
Scale in this context means that the drawing of the  T-shirt is 25 times smaller than in reality . ✔✔A 
OR
On the picture the shirt is 25 times smaller  ✔✔A 

2A correct definition 
Accept no units (2)

MP 

L1

1.2.4 

± 61 mm ✔✔A

2A correct measurement (Accept 59 mm – 64 mm) \
Correct answer in cm =  max 1 mark (2)

L1

Solution

Explanation

T&L

1.3.1

Two Oceans Marathon ✔✔RT

2RT reading from table 
Accept: 

  • Race on 15 April 2017
  • Race of 56 km
  • Race with an entry fee of  R520,00  (2)

L1

1.3.2

Comrades Marathon  ✔✔RT

2RT reading from table 
Accept: 

  • Race on 4 June 2017
  • Race of 89 km
  • Race with an entry fee of  R460,00   (2)

L1

1.3.3

R520,00 – R460,00  ✔RT 
= R60,00 ✔A

1RT correct values from the   table 
1A answer  (2)

L1

1.4.1 

12 Hours  ✔✔A

OR

Half a day ✔✔A 

2A correct time 
Accept: 

  • 12:00 OR/OF
  • 12 o’clock 

Max 1 mark (2)

L1

1.4.2 

Discrete ✔✔A

2A discrete  (2)

L1

1.4.3

   ✔RT      ✔A
17 031 : 13 852 

1RT correct values from  table 
1A correct values in correct   order 
Accept answer as unit   ratios: 

  • 1 : 0,813
  • 1,229 : 1 

Accept answer in fraction  form NPR  (2)

L1

     

[32]

QUESTION 2 [41 MARKS] 

Q

Solution 

Explanation

T&L

2.1.1

Interest refers to the amount that will be added to an  ✔A  account that is not settled yet
OR
Extra amount is charged on the late payments  ✔A 
OR
Extra money to be charged on overdue fees ✔A 
OR
Money charged for not paying fees on time ✔A 
OR
Interest in this context is the charge levied because of  unpaid fees or late payment of fees ✔A 

1A amount charged 
1A reason  (2)

L1

2.1.2 

R14 819,50 ✔✔RT

2RT balance  (2)

L1

2.1.3

                   ✔RT 
   148,20   ×   100 %  ✔M
14 819,50         1 
= 1,000033739329937 
≈ 1%  ✔CA 

1RT correct values 
1M multiply by 100 
1CA answer  (3)

L2

Solution

Explanation

T&L

2.1.4 

                          ✔RT
APG 2039W Design & Theory Studio II✔RT 

1RT code 
1RT name 
If APG omitted = full marks (2)

L1

2.1.5

                   ✔RT                              ✔M 
R14 967,70 – R8 650,00  = R6 317,70 

1RT correct values 
1M subtracting deposit  (2)

L1

2.1.6 

                                 ✔RT                ✔M 
Total amount  = R3 030 + R3 030 + R2 280 + R2 280 + R9 580 +   R4 530 + R29 460 + R2 087 + R395,95 +   R395,95   ✔CA 
= R57 068,90  
OR
Total amount 
               ✔M                   ✔RT 
= R62 594 – R6 317,70 + 2 × R395,95   = R57 068,90  ✔CA 
OR
Total amount 
               ✔RT               ✔M 
R40 386 + R23000 + R8650 – R14819,50 – R148,50 = R57 068,90   ✔CA 
OR
Total amount 
            ✔RT                                 ✔M 
R3 030 + R3 030 + R2 280 + R2 280 + R9 580 +  R4 530 + R29 460 + R395,95 + R395,95 = R54 981,90    ✔CA 
OR
Total amount 
                      ✔M                                     ✔RT 
= R62 594 – R6 317,70 + 2 × R395,95 − R2 087 = R54 981,90  ✔CA 
OR
Total amount 
                ✔RT                     ✔M  
R40 386 + R23000 + R8650 – R14819,50 – ✔CA 
R148,50 − R2 087 = R54 981,90 

AO 
1RT reading all correct values
1M adding values 
1CA simplification 

OR

1RT reading all correct values
1M subtracting values 
1CA simplification 

OR

1RT reading all correct values
1M subtracting values 
1CA simplification 

OR

1RT reading all correct values
1M adding values 
1CA simplification 

OR

1RT reading all correct values
1M subtracting values 
1CA simplification 

OR

1RT reading all correct values
1M subtracting values 
1CA simplification 

L1

Solution

Explanation

T&L

2.1.7 

Direct deposit \ ✔✔RT

2RT reading correctly 
Accept deposit only  (2)

L1

2.1.8 

Monthly instalment 
R40 386,60 ÷ 5  ✔A 
= R8 077,32  ✔M
OR
Monthly instalment ✔A  
R8 077,32 × 5  ✔M 
= R40 386,60 
OR
Monthly instalment 
R40386,60  ✔M 
R8077,32  
= 5  ✔A

1A calculating 5 
1M dividing by 5 
OR
1A calculating 5 
1M multiply by 5 
OR
1M dividing correct values in  correct order 
1A calculating 5  (2)

L1

2.2.1 

Inflation is a measure of rate at which the cost of goods  is changing over a period of time and is usually  ✔A  expressed as a percentage ✔A
OR
The percentage increase of the food prices over the  ✔A  period 1970 – 2015 ✔A
OR
Percentage increase of price over a period of time  ✔A
OR
Inflation is the rising price of goods/items over time ✔A

1A percentage increase 
1A time  (2)

L1

Q

Solution

Explanation

T&L

2.2.2 

R0,30 OR/OF 30c ✔✔RT

2RT correct value 
Accept 0,30 
If the candidates only   wrote 30 = max 1 mark  (2)

L1

2.2.3 

✔M   R557,00 – R418,00   ✔RG 
= R139,00✔CA 

AO 
1RG correct amount 
1M subtracting 
1CA simplification (one of the 2  values must be correct) (3)

L1

2.2.4 

Percentage change
        ✔RT 
R75,00 R0,25  ×    100%    ✔SF 
      R0,25                 1 
= 29 900 %   ✔CA 
OR
Percentage change 
          ✔RT
  75   × 100% = 30 000%  ✔M 
0,25 
Therefore % increase = 30 000% - 100%  ✔CA
= 29 900% 

AO 
1RT all correct values 
1SF substitute correct values 
1CA correct percentage 
OR
1RT all correct values 
1M subtracting 
1CA correct percentage  (3)

L2

Q

Solution

Explanation

T&L

2.2.5 

Cost price
  100   ×  104,90  ✔MA 
117,5 
= R89,28   ✔A 
OR
Cost price 
104,90  ✔MA 
117,5% 
= R89,28   ✔A 
OR
Cost price  
104,90  ✔MA 
1,175 
= R89,28  ✔A 
OR
Cost price 
 17,5 × R104,90 = R15,62  ✔MA 
117,5 
R104,90 – R15,62 
=R89,28  ✔A

AO 
1MA multiplying correct values
1A answer 
OR
1MA dividing correct values    in the correct order
1A answer 
OR
1MA dividing correct values  in the correct order 
1A answer 
OR
1MA multiplying and subtracting  correct values 
1A answer  (2)

L2

2.3.1 

B OR/OF R241 600 000 000 ✔✔A 

2A correct value  (2)

L1

Q

Solution

Explanation

T&L

2.3.2 

Budget is the proposed way in which money will be  spent on different items ✔✔A 
OR
A plan on how money is going to be spent on  estimated income ✔✔A 
OR
A plan in how money is going to be spent ✔✔A 
OR
Financial plan how to spend money/finance  ✔✔A 
OR
Estimated income and expenditure of money ✔✔A

2A definition  (2)

L1

2.3.3 

Skills development levy institutions s✔✔RT

2RT correct sector  (2)

L1

2.3.4 

Percentage of the total education budget
✔RG/RT 
 15,3  × 100%   ✔M 
320,5 
= 4,77%   ✔CA 
OR
Percentage of the total education budget
✔RG/RT 
 R15 300 000 000    × 100 %  ✔M
R 320 500 000 000       1 
= 4,77%   ✔CA 

1RG/RT correct values 
1M multiply by 100 
1CA answer 
OR
1RG/RT correct values
1M multiply by 100 
1CA answer 
NPR  (3)

L2

Solution

Explanation/ 

T&L

2.3.5 

Education Administration plus NSFAS amount  to 31,1 billion rand 
9,7% ✔✔A 
Accept any estimation from 9,5% but less than 9,86% Aanvaar enige skatting vanaf 9,5% maar minder as  9,86% 
OR
15,8 + 15,3 = 31,1 billion ✔M
= 9,7%   ✔A 
Accept any estimation from 9,5% but less than 9,86% 

AO 
2A correct estimation
OR
1M adding values 
1A estimated value  (2)

L2

     

[41]

QUESTION 3 [18 MARKS] 

Q/V 

Solution

Explanation

T&L

3.1.1 

15 cm + 17 cm + 19 cm + 21 cm  ✔A 
= 72 cm × 10  ✔CA
= 720 mm  ✔M 

1A adding of correct values
1CA conversion 
1CA answer in mm  (3)

L1

3.1.2a 

Diameter  = 2 × radius 
 = 2 × 14 cm  ✔M
 = 28 cm✔A 

AO 
1M multiplying by 2 
1A diameter  (2)

L1

3.1.2b 

Volume of a cylinder = π × r2 × height 
Volume of a cylinder = 3,142 × (14)2 × 15 cm   ✔SF 
= 3,142 × 196 cm2 × 15 cm  ✔S
= 9 237,48 cm3  ✔CA 

AO 
1SF substitution 
1S squaring 14 
1CA simplification  (3)

L2

3.1.3

The perimeter of a shape is the total distance around  the edges defining the outline of that shape  ✔✔A 
OR
Total distance around the shape  ✔✔A

2A explanation  (2)

L1

3.1.4

Area of a rectangle = length × width  ✔SF
 = 15 cm × 12 cm 
 = 180 cm2✔CA 

1SF correct substitution  
1CA simplification  

  • 35 cm × 33 cm = 1 155 cm2  Max 1 mark   (2)

L2

3.2.1

Amount  in kg = 3,5 ÷ 2,25  ✔C 
 = 1,556  ✔A

1C conversion 
1A simplification 
Accept

  • 1,56 kg ; 
  •  1,6 kg
  • 1,5 only = 0 marks  (2)

L2

Solution/

Explanation

T&L

3.2.2

1 mℓ flour = 0,7 g flour 
625 × 0,7 g ✔C
  1 
= 437,5 g✔A 

1C conversion 
1A simplification (2)

L2

3.2.3

°C = (°F – 32°) ÷ 1,8 
°C = (356° – 32°) ÷ 1,8  ✔SF
°C = (324°) ÷ 1,8  ✔A
 = 180 °C  

1SF correct substitution
1A simplification  (2)

L2

   

[18]

 

QUESTION [24 MARKS]

Q

Solution

Explanation

T&L

4.1.1 

South West OR SW ✔✔A

2A direction (2)

MPL 

2

4.1.2 

Namaqua National Park✔✔RM

2RM national Park (2)

MPL 

1

4.1.3 

✔✔RM ✔RM
Keimoes, Kakamas, Pofadder 
(Any 2 of the 3) 

2RM first correct town 
1RM second correct town (3)

MPL 

1

4.1.4 

Ratio scale OR number scale OR numerical  scale ✔✔A

2A ratio / number / numerical 
Accept unit ratio  (2)

MP 

L1

4.1.5 

Measured distance /Gemete afstand = 135 mm ✔A 
 1 : 3 007 874 ✔M 
135 mm × 3 007 874 
= 406 062 990 mm 
= 406 062 990 ✔C
    1 000 000 
= 406 km  ✔R 
OR
         ✔A            ✔M 
13,5 cm × 3 007 874 
= 40606299cm   ✔C 
     100 000 
= 406,06299 km  ✔R
≈ 406 km 

1A measures distance 
1M using scale 
1C conversion 
1R to the nearest km 
(Range: 130 mm to 140 mm) 
OR
1A measures distance 
1M using scale 
1C conversion 
1R to the nearest km 
(Range: 13 cm to 14 cm) (4)

MPL 

3

4.2.1 

Voortrekker Road ✔✔RM 
OR
N14  ✔✔RM 

2RM correct road  (2)

MPL 

1

Q

Solution

Explanation

T&L

4.2.2 

Rivier Street ✔✔RM

2RM correct road   (2)

MP 

L2

4.2.3 

Debs-Lodge  ✔✔RM

2RM correct road  (2)

MP 

L2

4.2.4

Time  = 2,34 km   ✔SF 
             40 km/h  
= 0,0585 h × 60  ✔C
= 3,51 minutes✔CA 

1SF calculating time 
1C multiply by 60 
1CA simplification 
NPR  (3)

MP 

L2

4.2.5

P = 13    ✔A
      42
OR
0,310
OR
31% 
OR
1 − 29= 13  ✔A
      42   42

1A numerator (independent) 
1A denominator  
OR
1MA subtracting from 1 
1A simplification  (2)

L2

   

[24]

 

QUESTION 5 [35 MARKS] 

Q

Solution

Explanation 

T&L

5.1.1

R2 085 600 000 ✔✔RT 
OR
R2 085,6 million✔✔RT 
OR
R2,0856 billion✔✔RT 

2RT correct amount  
Table value = max 1   mark  (2)

L1

5.1.2

R1 323+ R2 085,6 +R3 162+R2 158 +R1 847 + R2 732  million    ✔RT  ✔CA 
                                            6 
= R2 217 933 333 OR R2 217,933333 million

AO 
1RT correct values 
1M concept of mean 
1CA simplification 
NPR  (3)

L2

5.1.3

                ✔A                ✔A
Maximum = 46,1 thousand
OR
Maximum = 46 100✔✔RT 

1A correct value 
1A unit 
OR
2RT correct maximum  (2)

L1

5.1.4

                ✔RT 
A =      2 158 000 000     × 100 %   ✔M
       3 441 000 000 000 
 = 0,062714327%  ✔CA 
 = 0,06%  ✔R 

AO 
1RT correct values 
1M multiply by 100 
1CA simplification 
1R rounding 
If omitted zeros =  max 3 marks  (4)

L2

5.2.1

A person who is able and willing to work, but cannot find  work  ✔✔A
OR

2A explanation

L1

Q

Solution

Explanation

T&L

 

People who are without work ✔✔A 
OR
People who are jobless ✔✔A 
OR
Not earning a salary / wage / income  ✔✔A 
OR
Retrenched ✔✔A

2A explanation (2)

 

5.2.2

X = 1 748 – 506  ✔M
 =1 242  ✔A 
OR
X = 16 172 – (1 391+ 806 + 4 991+ 2 513 + 1 417 + 321 +   999 + 2 492)  ✔M
= 1 242 ✔A 

1M subtracting correct values
1A simplification 
1M subtracting correct values
1A simplification 
No penalty for  including zeros  (2)

L1

5.2.3

Questionnaire ✔✔A 
OR
Survey  ✔✔A 
OR
Population census ✔✔A 
OR 
Document analysis✔✔A 
OR
Interview  ✔✔A

2A correct answer  (2)

L1

Solution

Explanation

T&L

5.2.4

Percentage of people
          ✔RT
1412 000  ×   100 %    ✔M 
4507 000          1 
= 31,329 %✔CA 

1RT using both correct values
1M percentage calculation
1CA simplification 
If omitted zeros   = full marks
NPR  (3) 

L2

5.2.5

16 172 000 : 5 882 000   ✔RT 
2,7494 : 1  ✔A

1RT both correct values 
1A ratio in unit form 
Accept: 

  • 2,749 / 2,75 / 2,7  (2)

L2

5.2.6 

                                        ✔RT 
Probability (NEA) =    697 000    ✔RT
                                 1 893 000 
 = 0,368  ✔CA 
OR
Probability (NEA) =    697 000     ✔RT 
                               15 475 000 
 = 0,045  ✔CA 

AO 
2RT correct values 
1CA simplification 
OR
2RT correct values 
1CA simplification 
If omitted zeros = full marks 
NPR   (3)

L2

Solution 

Explanation 

T&L

5.2.7

Do not mark this question.   

5.2.8

✔✔A
9 ✔A 
= 31 ✔CA 

AO 
2A numerator 
1A denominator 
1CA simplification (4)

L3

 

[35]

 
 

TOTAL: 150

 

Upscaling of Question 5 mark 

QUESTION   5 

Mark out of 29 

Mark out of 35 

29 

35 

28 

34 

27 

33 

26 

31 

25 

30 

24 

29 

23 

28 

22 

27 

21 

25 

20 

24 

19 

23

18 

22

17 

21

16 

19

15 

18

14 

17

13 

16

12 

14

11 

13

QUESTION  5

Mark out of 29 

Mark out of 35

10 

12

11

10

8

7

6

5

4

2

1

MATHEMATICAL LITERACY
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

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

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.

MEMORANDUM 

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
2nd 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 1st year value
1CA 2nd 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 1st year value
1CA 2nd 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 2nd 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 1st coat (20 min)  8:15 - 8:35      ✔M
Waiting time  (4 hours)  8:35 - 12:35          ✔ MCA

Apply 2nd 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
MEMO 1 IKUGHIYUHGD
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 1st formula
1SF values into 2nd 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

 

Notes to the Marking Guideline Mathematical Literacy P2 November 2018

Note: In any verification/opinion question, some form of calculation must be shown in order to give a mark for conclusion.

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

The following tolerance range was agreed upon during marking guideline discussions: Questions  1.1.4 , 1.2.1 , 3.2.1 , 4.1.1 (1 mark each)

RELIGION STUDIES
PAPER 1
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

INSTRUCTIONS AND INFORMATION 

  1. This question paper consists of SECTION A and SECTION B.
  2. SECTION A: COMPULSORY
    SECTION B: Answer any TWO questions in this section. 
  3. Read ALL the questions carefully.
  4. Start EACH question on a NEW page. 
  5. The length of your answers must be in accordance with the marks allocated to  each question. 
  6. Number the answers correctly according to the numbering system used in this  question paper. 
  7. Write neatly and legibly.

QUESTIONS 

SECTION A (COMPULSORY) 
QUESTION 1  
1.1 Various options are provided as possible answers to the following questions.  Choose the answer and write only the letter (A–D) next to the question  numbers (1.1.1 to 1.1.10) in the ANSWER BOOK, e.g. 1.1.11 D. 
1.1.1 The Tao Te Ching was written by … 

  1. Chuang Tzu. 
  2. Yang. 
  3. Lao Tzu. 
  4. Confucius. (1) 

1.1.2 In the African Traditional Religion this person may throw bones in  the performance of certain rituals: 

  1. Induna 
  2. Chief 
  3. Sangoma/Isangoma 
  4. Intlabi (1) 

1.1.3 The oldest branch of Buddhism is … 

  1. Theravada Buddhism. 
  2. Zen Buddhism.
  3. Mahayana Buddhism. 
  4. Tibetan Buddhism. (1) 

1.1.4 The great Hindu sage, Krishna Dwipayana, compiled these sacred  books: 

  1. The Kitáb-i-Aqdas and the Mahabharata
  2. Ecclesiastes and the Apocalypse
  3. The Pali Canon and the Tripitaka 
  4. The Rig-Veda and the Samaveda (1) 

1.1.5 Holding on to traditional, conventional values: 

  1. Conservatism
  2. Secularism
  3. Confucianism
  4. Heresy (1)

1.1.6 The Tenach in Judaism corresponds with this part of the Christian  Bible: 

  1. The New Testament 
  2. The Pentateuch 
  3. The Acts of the Apostles 
  4. The Old Testament (1) 

1.1.7 A spiritual exercise to attain moksha, which includes breath control  and specific body postures: 

  1. Mysticism 
  2. Meditation 
  3. Mantra 
  4. Yoga (1) 

1.1.8 Every person in the universe controls his/her own destiny and is  not controlled by any other person or any superior being, like a  god. This is a teaching of … 

  1. the African Traditional Religion. 
  2. Hinduism. 
  3. Buddhism. 
  4. the Bahá'i faith. (1) 

1.1.9 A declaration of the Parliament of the World's Religions adopted  in 1993: 

  1. The Charter for Compassion 
  2. Towards a Global Ethic 
  3. Millennium Development Goals 
  4. The Freedom Charter (1) 

1.1.10 Two leaders of the Bahá'i faith were … 

  1. Abdu'l-Baha and Bahá'u'lláh. 
  2. Abdu'l-Baha and Abu Bakr. 
  3. Bahá'u'lláh and Hermann Zimmer. 
  4. Bahá'u'lláh and Maimonides. (1)

1.2 Choose the word in each list below that does NOT match the rest. Write down  the word next to the question numbers (1.2.1 to 1.2.5) in the ANSWER BOOK  and give a reason why it does NOT fit. 
EXAMPLE: Banana; Apple; Potato; Grape 
ANSWER: 1.2.6 Potato. The others are all fruit. 
1.2.1 Roman Catholic; Baptist; Presbyterian; Methodist (2)
1.2.2 Fatwa; Shahada; Ilima; Wudu (2)
1.2.3 Psalms; Myths; Parables; Legends (2)
1.2.4 Copernicus; Darwin; Galileo; Kepler (2)
1.2.5 Imam; Dominee; Rabbi; Martyr (2) 
1.3 Choose an item from COLUMN B that matches the description in COLUMN A.  Write only the letter (A–H) next to the question numbers (1.3.1 to 1.3.6) in the  ANSWER BOOK. Do NOT use any letter more than ONCE. 

COLUMN A 

COLUMN B

1.3.1 On this mountain the Ten  Commandments were given to Moses 
1.3.2 This form of Christianity has its  origin in Africa 
1.3.3 Family members who have died and joined the spirit world 
1.3.4 The Dalai Lama, leader of Tibetan  Buddhism 
1.3.5 The Roman Catholic Church  established this law court in Spain to guard against heresy in the 15th century 
1.3.6 All religions are paths towards the  truth and are equally valid

  1. ancestors 
  2. Tenzin Gyatso 
  3. Inquisition 
  4. pluralism 
  5. atheism 
  6. Sinai 
  7. African Traditional
  8. Religion 
  9. African Initiated Churches

(6 x 1) (6)

1.4 Indicate whether the following statements are TRUE or FALSE. Write 'true' or  'false' next to the question numbers (1.4.1 to 1.4.6) in the ANSWER BOOK.  Correct the statement if it is FALSE. 
1.4.1 Martin Luther King started the Reformation. (2) 
1.4.2 The Archbishop of Canterbury is the hereditary head of the  Anglican Church. (2) 
1.4.3 Mixing elements of one religion with elements of another to form a  new religion, is called fundamentalism. (2) 
1.4.4 The Bahá'i faith is a clan-based religion. (2)
1.4.5 Conservative Judaism is the oldest subdivision of Judaism. (2) 
1.4.6 Compassion is a religious observance intended to remind of  a sacred occasion. (2) 
1.5 Complete the following sentences by filling in the missing word. Write only  the word next to the question numbers (1.5.1 to 1.5.6) in the ANSWER  BOOK. 
1.5.1 The Crusades of the Middle Ages were religious wars fought  between the Muslims and the … (2) 
1.5.2 The followers of Hinduism call their divinity … (2) 
1.5.3 In the 4th century, Christianity became the official religion of the  Roman Empire under Emperor … (2) 
1.5.4 A statement about how something is observed to be in a religion, is  called a descriptive statement, but a statement about how  something ought to be, is called a … statement. (2) 
1.5.5 According to Taoism, the 'ten thousand things' are produced by the  Three and the Three are produced by the Two. The Two are the …  and … (2)
1.5.6 The primary source of the Sharia is the holy scripture called the … (2)

TOTAL SECTION A: 50

SECTION B 
Answer any TWO questions in this section. 
QUESTION 2 
2.1 Read the extract below and answer the questions that follow. 

PARLIAMENT OF THE WORLD'S RELIGIONS 

The first session Parliament of the World's Religions was held in Chicago in  1893 with the aim of creating a global dialogue of faiths. Since that event,  there have been several further Parliament session. 

[Source: Shuters Top Class Religion Studies Grade 12]

2.1.1 What are THREE positive outcomes of the first Parliament of the  World's Religions? (6) 
2.1.2 In TWO sentences, explain the phrase 'global dialogue of faiths'. (4) 
2.1.3 How is the parliament of a country different from the Parliament of  the World's Religions? (6) 
2.1.4 The Parliament of the World's Religions is but one interreligious  organisation. Name ONE other interreligious organisation and  discuss its successes and failures. (12) 
2.1.5 The Parliament of the World's Religions was held in Cape Town in  1999. Give THREE reasons why Cape Town was chosen. (6) 
2.2 Explain the relationship between religious teaching and religious belief. (4)
2.3 State TWO differences between doctrine and dogma. (4)
2.4 Explain the following in the context of religion: 
2.4.1 Uniqueness (2)
2.4.2 Unity (2)
2.5 Give ONE example of the following: 
2.5.1 Unity within a religion (2) 
2.5.2 Unity between different religions (2) [50]

QUESTION 3 
Read the extract below and answer the questions that follow. 

RELIGION AND MEDIA 

In several studies of religious coverage in the media the main conclusions were the  following: 

  1. Coverage had increased significantly over the last decade but was very  negative.
  2. Towards orthodox faiths, the tone is mostly hostile, but towards minority  religions, it is favourable.
  3. The people interviewed most often are those who question orthodox religions.
  4. The political elements of religious ideas and institutions receive the most  coverage.
  5. The struggles of women and minorities also receive a lot of attention. 
  6. Issues involving sexual morality also receive a lot of attention. 
  7. Theological and spiritual questions are rarely presented as newsworthy.

[Adapted from Religion Studies Grade 12, Steyn et al.]

3.1 What do you understand by the word media? Include TWO different kinds of  media in your answer. (6) 
3.2 Discuss conclusion (a) in the extract above, and indicate whether it is  beneficial to religion. (6) 
3.3 Refer to conclusion (b) in the extract above and answer the following  questions: 
3.3.1 Explain, in your own words, what is meant by orthodox faiths and  minority religions in this context. (4) 
3.3.2 Suggest TWO reasons why 'the tone is mostly hostile' towards  orthodox faiths, but 'favourable' towards minority religions. (4) 
3.4 Refer to conclusion (f) in the extract above. Discuss ONE example of this you  have noticed in the media. (10) 
3.5 With regard to religious coverage in the media, explain what is meant by  EACH of the following and give a suitable example in EACH case: 
3.5.1 Stereotyping (4)
3.5.2 Bias (4)
3.5.3 Impartiality (4)
3.5.4 Sensationalism (4)
3.5.5 Factual account (4) [50]

QUESTION 4 
Read the extract below and answer the questions that follow. 

HUMAN RIGHTS 

In 1948 the United Nations' Universal Declaration of Human Rights was produced and  signed by governments all over the world. In this document human rights are based on  the inherent dignity of human beings. 

 [Source: Religion Studies Grade 12, Steyn et al.]

4.1 Religious freedom is a human right. Name any TWO other human rights and  discuss the responsibilities associated with EACH of them. (8) 
4.2 What does the South African Constitution say about religious freedom? (8)
4.3 How does South Africa demonstrate religious freedom? (10) 
4.4 Name ONE country in the world where religious freedom has been violated. Explain the nature of the violation. (8) 
4.5 With reference to ANY of the religions you have studied, state FOUR  teachings that show support for EACH of the following: 
4.5.1 Human rights (8) 
4.5.2 Religious freedom (8) [50] 

QUESTION 5 
5.1 Explain EACH of the following and name ONE religion with which it is  associated: 
5.1.1 Monotheism (4)
5.1.2 Sannyasin (4)
5.1.3 Ubuntu (4)
5.1.4 Dharma (4) 
5.2 Identify ONE common feature of EACH of the following groups:
5.2.1 Christianity, Islam and Judaism (2)
5.2.2 Hinduism and Buddhism (2)
5.2.3 African Traditional Religion and the African Initiated Churches (2) 
5.3 All Muslims share a common set of beliefs (imaan). List SIX of these beliefs. (12)
5.4 Compare the concepts of religion and ideology. (4) 
5.5 Choose ONE religion and discuss THREE of its unique features. (12) [50] 

TOTAL SECTION B: 100 
GRAND TOTAL: 150

RELIGION STUDIES
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

INSTRUCTIONS AND INFORMATION 

  1. This question paper consists of FIVE questions. 
  2. Answer any THREE questions.
  3. Start EACH question on a NEW page. 
  4. Read ALL the questions carefully. 
  5. Number the answers correctly according to the numbering system used in this  question paper. 
  6. The length of your answers must be in accordance with the marks allocated to  each question. 
  7. Write neatly and legibly.

QUESTIONS 

QUESTION 1 
1.1 Read the extract below and answer the questions that follow. 

EUTHANASIA 

Euthanasia is normally thought of as providing a gentle and easy death to  someone suffering from a painful, terminal disease, which results in a poor  quality of life. Euthanasia can be done by assisted suicide, voluntary  euthanasia or non-voluntary euthanasia. 

[Source: Religion and Life by V Watton]

1.1.1 Briefly explain TWO forms of euthanasia. (4)
1.1.2 List THREE advantages and THREE disadvantages of euthanasia. (12) 
1.1.3 With reference to the relevant teachings of any ONE religion you  have studied, discuss its response to euthanasia. (10) 
1.2 Read the extract below and answer the questions that follow. 

Not only are religions different from each other, but there are also internal  differences within religions themselves. 

[Source: Focus on Religion Studies, Grade 12]

Name any TWO branches that exist within ONE religion and discuss their  differences in terms of: 
1.2.1 Teachings (12) 
1.2.2 Practices (12) [50] 

QUESTION 2 
2.1 Discuss the central teachings of any ONE religion under the following  headings: 
2.1.1 The nature of the world (10)
2.1.2 Reward and punishment (10)
2.2 Compare the role of sacred texts in Abrahamic and non-Abrahamic religions. (14) 
2.3 Name and discuss any FOUR hermeneutical principles. (16) [50]

QUESTION 3  
3.1 What is the role of inspiration in the African Traditional Religion? (10)
3.2 Name and discuss the teachings of any ONE secular worldview. (14) 
3.3 What influence did secularism have on the current South African  Constitution? (10) 
3.4 Do you think secularism has a negative impact on the moral values of  society? Give reasons for your answer. (16) [50] 

QUESTION 4 
Read the statement below and answer the questions that follow. 

I don't see why my religion and science can't work together positively.

4.1 Compare the scientific and the religious accounts of how the universe came  into being. (10) 
4.2 Discuss the scientific explanation of how human beings came into existence. (14) 
4.3 Discuss the different responses of the following religions to Darwin's theory  evolution: 
4.3.1 Hinduism (12) 
4.3.2 Christianity (14) [50] 

QUESTION 5 
Read the extract below and answer the questions that follow. 

In many countries around the world today we find people living together who have  different beliefs and practices. This can sometimes cause conflict between religious  communities, especially if one of them has more power or authority in a society. 

[Source: Focus on Religion Studies, Grade 12]

5.1 Identify ONE area of possible religious conflict anywhere in the world. (2) 
5.2 Briefly discuss the causes of conflict in the area you identified in  QUESTION 5.1. (12) 
5.3 Does religion play a major role in the conflict? Give reasons for your answer. (10)
5.4 Describe the current situation in this conflict. (10) 
5.5 What practical steps can religious organisations follow to resolve this conflict? (16) [50] 

TOTAL: 150

RELIGION STUDIES
PAPER 2
GRADE 12 
NSC EXAMS
PAST PAPERS AND MEMOS NOVEMBER 2018

MEMORANDUM 

QUESTION 1 
1.1.1

  • Non-voluntary euthanasia means ending someone's life painlessly when  they are unable to ask for death, but there is sound medical reason for thinking they would want death. 
  • Voluntary euthanasia refers to the situation where a person who is  terminally ill and dying in pain, asks another person to end his/her life  painlessly. 
  • Assisted suicide refers to providing a terminally ill person with the means  to commit suicide. (4) 

1.1.2 ADVANTAGES 
It may help the following patients: 

  •  People who are terminally ill and for whom there is no relief from  excruciating pain. Such a person's suffering will be ended. 
  • It can also help those who are too weak to ask for voluntary euthanasia. 
  • It can help those who cannot respond to treatment as a result of organ  failure. 

In addition: 

  • It will ease the stress of people who have been bedridden for a very long  time. 
  • The economic benefit is that this may help save the family and the state  money, as some people will never recover, based on medical evidence. 
  • Medicine and medical treatment are very expensive, and the resources  can be used for those who have a better chance of recovery. 
  • The family is relieved from helplessly watching the suffering of their loved  one.  

DISADVANTAGES 

  • Conflict often arises due to differing opinions and beliefs of various family  members. 
  • The family of the terminally ill person may choose euthanasia and assisted  suicide for economic and inheritance reasons, and benefit from the  inheritance. 
  • Some people may regret their decision at a later date, leading to  unresolved grieving and guilt. 
  • Euthanasia could be used by unscrupulous doctors as an excuse for organ  harvesting. 
  • Euthanasia and assisted suicide may send signals to other family  members that choosing death is an acceptable action, and lead to  suicides. 
  • Doctors may refuse to perform euthanasia on the basis of their religious  beliefs. 

NOTE: Other relevant responses must be credited. (12)
1.1.3 EXAMPLE 1: CHRISTIANITY 

  • Christianity is against euthanasia because it values the sanctity of life.
  • It teaches that life is created by God. 
  • The sacredness of life is embedded in the Word, so no one can tamper  with life in any way. 
  • It is up to God, not humans, to determine when people would die. 
  • Euthanasia is to put oneself on par with God. This is condemned in the  Bible. 
  • In the Christian faith the option of euthanasia is regarded as murder. 
  • 'Why should you die before your time?' (Ecclesiastes 7:17). 
  • This means euthanasia is against God's will because it causes somebody to die before his/her time. 
  • 'Your eyes saw my unformed substance; in your book were written, even  one of them, the days that were formed for me, when as yet there was  none of them'. (Psalm 139:16). 
  • This means that God alone determines the time of death for every one of  us. Therefore euthanasia is against God's plan for men. 
  • Thou shall not commit murder' (Exodus 20:13). This commandment  forbids euthanasia. 

EXAMPLE 2: HINDUISM 

  •  In Hinduism, there are differences of opinion, due to different traditions.
  • Hindus embrace karma (a belief that every action has a consequence,  which may show up only in a later reincarnation). If one practises euthanasia there will be negative consequences for the soul.
  • Hindus believe that whatever suffering a person experiences in his/her  current life is the result of something one did in a past life.
  • If one circumvents karma by taking action to stop suffering, one will pay for  it later. 
  • Some Hindus believe that there are circumstances that justify a hastening  of death.
  • Some believe that if a person has reached a stage in life in which s/he can  no longer worship God, s/he may ask a doctor to hasten the end of his/her  life.
  • However, most Hindus do not subscribe to the concept of euthanasia.
  • In general, Hinduism is against euthanasia because it breaches the  teaching of ahimsa (doing no harm). 

NOTE: Other relevant responses must be credited. 
Credit is given for teachings from only the first religion identified. (10)
1.2
1.2.1 NOTE: A total of FOUR marks may be awarded for naming the  two branches in either 1.2.1 or 1.2.2, but not both. The same  two branches must be discussed in 1.2.1 and 1.2.2. 
EXAMPLE 1: BUDDHISM 
The learner may choose any TWO of the following branches: 

  • Theravada Buddhism
  • Theravada Buddhism believes in the doctrine of anatman. 
  •  'Anatman' means that the ego/self prevents us from becoming enlightened and reaching nirvana.
  • Theravada Buddhism teaches that enlightenment comes  through an individual's efforts, and not by the intervention of  others or any gods. 
  • Mahayana Buddhism 
  • This branch believes that there is no such thing as an individual,  autonomous self. All beings are connected.
  • Followers of Mahayana believe in collective enlightenment.
  • Mahayana Buddhism teaches that the motivation for  enlightenment is compassion for all living things. 
  • The followers strive to become bodhisattva, (one who is at the  service of the enlightenment of others). 
  • Tibetan Buddhism 
  • The followers of Tibetan Buddhism believe in the reincarnation  of lineage of certain lamas (teachers), such as the Dalai Lama. 
  • They believe that Buddha can be shown in human form. 
  • They believe in a pantheon of Buddhas, bodhisattvas and  dharma protectors. 
  • Zen Buddhism 
  • They teach that direct communication, and not scriptural study, is the way of experiencing enlightenment. 
  • They believe that Zen is the way of life and not solely a state of  conscience. 
  • They believe that Buddha can be shown in a human form, such  as in the person of Padmasambhava (one who brought Zen  Buddhism in Tibet). 

NOTE: Other relevant responses must be credited. (12)
1.2.2 EXAMPLE 1: BUDDHISM 
Theravada Buddhism 

  • Monks live in monasteries, they beg for their food and meditate most of the day. 
  • Theravada monks also assist and co-operate with laity in important  life events such as births, weddings and funerals. 
  • Building stupas (tower- like structures) is a widespread practice. 
  • Most practices are performed in the home although there is also  individual and communal meditation and chanting in temples. 
  •  They leave altar offerings of food, incense, candles and flowers at  the temple, in front of the statue of Buddha. 

Mahayana Buddhism 

  • They practise more ritual elements than Theravada Buddhism, such as prostrating themselves. 
  • Only people who live in monasteries are allowed to practise  meditation. 
  • Lay people are allowed to practise chanting and prayer. 

Tibetan Buddhism 

  • Practices used include meditation, rituals and chanting. 
  • They use methods like trances to recover hidden ancient  scriptures. 
  • Tibetan Buddhist monks wear orange robes in their temples. 

Zen Buddhism 

  • They practise meditation in order to attain enlightenment. 
  • They choose Zen teachers to guide students in meditation and  performing rituals. 
  • They practise dharma transmission (passing of Buddha's teachings  from the master to the students). 

NOTE: Other relevant responses must be credited (12) 
1.2.1 EXAMPLE 2: CHRISTIANITY 

  • Roman Catholic Church 
  • This branch sees itself as the original church of Christ. 
  • The Holy Spirit comes from God the Father and the Son. 
  • Eastern Orthodox Church 
  • According to the Eastern Orthodox Church, the Holy Spirit comes  from God the Father only, and not the Son. 
  • They believe that they are the rightly guided church. 
  • They teach that no one has the power to change Christian  teachings and traditions. 
  • They see the Bible as canonical.
  • Protestantism 
  • In Protestantism, the Bible has more authority than the Pope. 
  • Rituals are less important than belief. 
  • They believe that faith is the key to salvation. 
  • Salvation is a gift given freely through the work of Jesus Christ,  who died for sinners.  

NOTE: Other relevant responses must be credited. (12)
1.2.2 EXAMPLE 2: CHRISTIANITY 
The learner may choose any TWO of the following branches: Roman Catholic Church 

  • They ordain their priests, who practise celibacy. 
  • They confess their sins to a priest. 
  • Nuns and monks do not marry. 
  • There is no dissolution of marriage. 
  • They practise last rites before the death of a person. 

Eastern Orthodox Church 

  • They express piety by kissing images of Saints, called icons. 
  • They see the entire Bible as canonical. 
  • Ordained priests can get married. 

Protestantism 

  •  Attendance of Sunday services is important in Protestantism. 
  • Holy Communion service is also practised in many forms of  Protestantism. 
  • Speaking in tongues in the Pentecostal/Charismatic churches is  a characteristic. 
  • They have revival tent crusades with the gospel,to reach the  lost. 
  • Healing services are also common. 
  • Charity is seen as an outward sign of inner grace, and not as  'buying one's way to heaven'. 

NOTE: Other relevant responses must be credited. (12) [50]

QUESTION 2
2.1
2.1.1 EXAMPLE 1: AFRICAN TRADITIONAL RELIGION 

  • For the indigenous traditional African, the nature of the world  includes all that exists. 
  • The world is understood more as the cosmos than simply as the  planet Earth. 
  • All that exists is looked upon as a living wholeness manifesting  itself in a variety of forms and phases. 
  • Life is understood as a perpetual exchange of interconnected  forces among and between all beings. 
  • According to this understanding, the cosmos or the universe  does not have a centre. 

2.1.2 NOTE: If ONLY reward or ONLY punishment is discussed,  award a maximum of SIX marks.  
EXAMPLE 1: AFRICAN TRADITIONAL RELIGION 

  • Reward and punishment in ATR occurs in the present life, not  after death. 
  • The living dead/ancestors are believed to be able to punish  evildoers and reward those who do good. 
  • Punishment from the living dead, comes in various forms of  affliction. 
  • The person experiencing these afflictions, is said to have bad  blood. 
  • The remedy for bad blood is an appropriate form of cleansing  which is intended to appease the living dead. 
  • Whenever these rituals are needed, the traditional healer holds  the remedy for the cleansing of bad blood. 
  • Those who lived a good life are rewarded by becoming  ancestors. 
  • Those who venerate the ancestors are rewarded by prosperity  and a good life. 

2.1.1 EXAMPLE 2: ISLAM 

  • According to the Qur'an, Allah created the universe in six  phases. 
  • Allah also provides the energy that everything in the universe  needs for survival and growth. 
  • The existence of the universe is not accidental and life is not  purposeless, but planned and purposeful.
  • The universe, having been created in time, is not eternal but has  a fixed time span. 
  • Since Allah alone is eternal, to believe in the eternity of the  universe would mean equating creation to the Creator. This is a  major sin (shirk).
  • NOTE: Other relevant responses must be credited. (10)

2.1.2 EXAMPLE 2: ISLAM 

  • Islam teaches that reward and punishment exist in the present  world and after death. 
  • On the Day of Judgment all humanity will be resurrected to be  judged by Allah.
  • Those who believe in Allah and have fulfilled their obligations to  the Creator, as well as to humanity, will be rewarded and  admitted to paradise. 
  • Those who deny the existence of Allah will be punished and  consigned to hell. 
  • Those who believe but have failed to fulfill their obligations, will  be either forgiven or punished in accordance with the nature of  their wrongdoing. 
  • The giving of alms to the poor is said to bring a person closer to  Allah and will be rewarded with Allah's mercy and kindness.
  • NOTE: Other relevant responses must be credited. (10) 

2.2 NOTE: If there is no comparison, award a maximum of EIGHT marks.

  • In Abrahamic religions sacred texts are central to their practices and  beliefs. That is why they are called 'religions of the Book'. 
  • Abrahamic religions' teachings are fixed in their written sacred scriptures.
  • In non-Abrahamic religions, like Hinduism, absolute authority is awarded to  the original sound of the spoken word, not to the text. 
  • Therefore non-Abrahamic religions see their beliefs and teachings as  being objects of fluidity/seeking. 
  • In Abrahamic religions sacred texts evoke a deeper connection with the Divine. 
  • In non-Abrahamic religions, such as Buddhism, scriptures are not seen as  divine dictates.
  • The Abrahamic religions are characteristically exclusivist because their  religion is fundamentally rooted in their holy scriptures. 
  • The non-Abrahamic religions are characteristically pluralist, e.g. Hinduism.
  • In Abrahamic religions, sacred texts foster communal identity, while in  religions like Hinduism, there is a variety of texts and one can follow any  text.
  • NOTE: Other relevant responses must be credited. (14)

2.3

  • Historical context 
  • This means that the writings must be understood within the context of the  time and circumstances in which the text was written. 
  • Clearest meaning 
  • The meaning that is clearest to the reader should be considered. 
  • Plan, purpose and context 
  • The writing plan or structure of the whole document must be taken into  account. 
  • An extract must be seen as part of the whole. 
  • E.g. Is the writing in the form of poetry or prose? 
  • Meaning of words
  • The meaning of words often changes over time and context. 
  • For the correct interpretation the original meaning must be considered.
  • Figurative language 
  • Figurative language is used widely in sacred texts.
  • This must be identified as such, so that it is not interpreted literally. 
  • Figurative language requires the application of all the other hermeneutical  principles for correct interpretation. 
  • Other sacred texts 
  • One sacred text may be used to interpret other sacred texts from the same  religion. 
  • This is because there is consistency between the teachings of the religion  and its sacred texts. (16) [50]

QUESTION 3 
3.1

  • In ATR, people communicate with their ancestors through inspiration.
  • Mediums and diviners specialise in communicating with the ancestors and  spirits. 
  • Mediums often go through a process of death and rebirth during their  training. This happens because of inspiration. 
  • In this process, they believe that personality is terminated and the medium  receives a new personality, dedicated to the supernatural forces. 
  • Such mediums have an intermediary function. African Traditional Religion includes contemporary and divine inspiration. NOTE: Other relevant responses must be credited.  (10) 

3.2 EXAMPLE 1:  
AGNOSTICISM 

  •  Agnosticism comes from the Greek, 'a', which means 'without' and 'gnosis' which means 'knowledge'. 
  • This term was first used by a philosopher called TH Huxley in 1869. 
  • Agnostics believe that it is not possible to either prove or disprove the  existence of God or a supernatural being. 
  • This refers to uncertainty about God-knowledge. 
  • Agnostics are sceptical of religious teachings. 
  • They reject religious doctrine, especially religions that claim to have  spiritual knowledge. 

EXAMPLE 2: 
ATHEISM 

  •  Atheists reject the belief that divine or supernatural powers exist. 
  • There are different degrees of atheism. 
  • Soft or neutral atheists do not actively reject the existence of a  supernatural being. 
  • Strong or positive atheists believe there is evidence to support their  atheistic views. 
  • In some cases soft atheists reject both theism and strong atheism.
  • This is because they feel both world views depend on proof to support their  claims. 
  • Atheists often turn to science to explain the nature of the universe, rather than relying on faith.   (14)
  • NOTE: Other relevant responses must be credited. 

3.3

  • The Constitution of South Africa is not based on any religious teachings.
  • The Constitution of South Africa promotes equal recognition of all religions.
  • The Constitution also promotes religious freedom. 
  • Secularism helped South Africans to design a constitution that promotes  religious tolerance. 
  • Freedom of religions, beliefs and opinion are enshrined in the Constitution.
  • In the South African Constitution, morals and ethical values are based on  reasoning and not on religious teachings. 
  • There is no state religion.
  • NOTE: Other relevant responses must be credited. (10)

3.4 YES  

  • Today society relies on the faculty of reason in order to make decisions.
  • They ignore the authority of the Supreme Being. 
  • They rely on reason, evidence and scientific approach in dealing with life.
  • They undermine the dictates of sacred texts. 
  • Religious places of worship are no longer respected. 
  • The majority of youth do not attend religious celebrations and rituals,  because these are not based on human reasoning. 
  • Secular humanists put more emphasis on liberal views of human rights than traditional religious beliefs. 
  • As a result there is an increase in social challenges, such as crime and  teenage pregnancy, which result from a degeneration of moral values. 

NO 

  • Secularism promoted the development of the Bill of Human Rights that  promote moral values. 
  • Moral values are promoted through the Constitution in secular states.
  • There is no need for sacred texts to develop a moral code. The rule of law  will enforce moral values. 
  • Families have the responsibility of developing moral values in their  children. This can be done independently of religion. 
  • Secularism has contributed through secular states to end religious wars, thus restoring human dignity. 
  • Today people of different faiths are able to coexist and help each other during difficult times because one religion does not dominate another. 
  • This does not mean that people who hold a secular worldview do not have  morals and values. 
  • Secularism instils a sense of self awakening. Some people do not belong to any religion, but still have a deep concern for their soul or spirit, thus developing high moral values.  (16)
  • NOTE: Other relevant responses must be credited. A combination of both perspectives may be accepted. [50]

QUESTION 4 
4.1 NOTE: If there is no comparison, award a maximum of FOUR marks.

  • The Big Bang Theory teaches that there was an enormous explosion from  which the different planets were formed. 
  • This happened about 13,7 billion years ago. 
  • Small temperature differences in the initial explosion led to varying  densities through the universe. 
  • These eventually formed into clusters of matter and energy. 
  • The clusters continued to condense in a lumpy way and formed the vast  collection of stars called galaxy. 
  • Some of the lumps in galaxies condensed into a combination of stars and  planets. 
  • Earth was formed about 4,5 billion years ago. 
  • Traditional religions maintain that the universe was created perfect. 
  • In Abrahamic religions the Creator is male. 
  • At first, only the Creator exists. The Creator then makes the universe from nothing. 
  • Abrahamic religions believe that God/Allah/Yahweh created the universe in  six days. 
  • For Hindus the Creator exists in either an active or passive state. 
  • The passive state is a state of rest when nothing happens. 
  • At rest, the universe has no form, is undifferentiated, and 'flat' all over.
  • After a long time, the Creator stirs and become active. This is when the  creation begins. 
  • Hinduism upholds both physical and spiritual evolution but science is  limited to physical evolution.
  • NB: Any relevant response must be awarded (10) 

4.2

  • The theory proposes that all life evolved from primitive forms and  continues to adapt and evolve. 
  • It asserts that humans have evolved from an ancestor shared with apes.
  • The theory of evolution, according to Charles Darwin, consists of the  following ideas: 
    • Species contain a great variety of minor differences. 
    • In the fight for survival, better adapted variations will be favoured while  those that are not fit will struggle to survive. 
    •  This applies to change in humans as well. 
    • In the fight for survival, humans adapted and gradually change from  apes to human beings. 
    • The process of change and adaptation happened over a very long period of time. 
    • Science offers no explanation of creation as part of a divine plan. (14)
    • NB: Other relevant response must be awarded. 

4.3 4.3.1 HINDUISM 

  •  Hindus have no problem with evolution. 
  • They believe that the universe is based on evolution. 
  • They believe that they have a more advanced theory of  evolution than the scientific one. 
  • The scientific theory is limited. 
  • It focuses only on the physical perspective of creation.
  • Hindus believe that humans have control over their spiritual and  physical evolution. 
  • Hindus believe that if one lives a good life one will evolve  through many rebirths until one is physically and spiritually  advanced.
  • Ultimately, one will achieve liberation from the physical and be  one with God. 
  • Hindus believe that there is an intelligent designer behind  everything like evolution of humanity and of the universe. (12) 

4.3.2 CHRISTIANITY 

  •  In contrast, there is no connection between the traditional  Christians and the theory of evolution.
  • Traditional Christians only accept that God created Adam and  Eve as it is narrated in the Book of Genesis.
  • They argue that there is no evidence in history of human beings  evolving from apes. 
  • God created perfect human beings and gave them authority to  rule and manage the world. 
  • However there is a connection between the theory of evolution  and liberal/progressive Christians. 
  • They argue that everything in the universe was created by God,  and the process of evolution is also part of the Divine Plan. (It is  referred as 'intelligent design')
  • They say God was responsible for the Big Bang. 
  • Progressive Christians accept the idea of guided/theistic  evolution, and add that the creation myth must be seen as  symbolic. 
  • They admit that there were some stages of creation which  involved evolution, as explained by scientists. (14) [50]

QUESTION 5 
5.1 EXAMPLE 1: WORLD CONFLICT (2) 

  •  Israel-Palestine Conflict 

5.2

  • The on-going conflict started with the establishment of Israel in 1948.
  • Hard-line Israelis and Zionists claim that, according to their scripture,  Palestine rightfully belongs to the Jews. 
  • This claim is rejected by Palestinians, as well as many Jewish  organisations, including some Orthodox Jews. 
  • The land of Israel is sacred to Jews, Christians and Muslims. 
  • The Al Aqsa Mosque in Jerusalem is Islam's third holiest site. 
  • Jerusalem is also sacred to Christians, as it is where Christ was crucified.
  • Jews regard Jerusalem as the location of the Temple Mount/Temple of  Solomon. 
  • All these sites existed long before the creation of the state of Israel, and  were occupied by various tribes throughout history. 
  • Jews and Palestinians are fighting for the control over land. (12) 

5.3

  • Religion does not play a major role in this conflict.
  • Religion is used as an excuse for the conflict. 
  • There is no attempt by any side to convert people to their faith.
  • The continuous occupation of Palestinian land by Jewish settlers is a major  factor in the conflict. 
  • The building of illegal settlements on this land is also a major factor.(10)

5.4

  • Numerous UN resolutions have been passed condemning this occupation.
  • However, Israel continues to defy UN resolutions and is building more  settlements on Palestinian land. 
  • There are sporadic attacks on Palestinians, as well as on Jews. 
  • The American president, Donald Trump, recently declared that America  recognises Jerusalem as the capital city of Israel. 
  • This has stirred up the conflict between the Israelis and Palestinians. 
  • Most of the world does not accept Trump's declaration. (10)

5.5 NOTE: A maximum of EIGHT marks may be awarded for social  intervention. 

  •  Religious leaders can call an inter-religious summit to discuss the conflict.
  • The dialogue should lead to practical steps that can be taken on both  sides, regardless of religion, towards alleviating the situation. 
  • Religious leaders are still respected and can act as mediators between  conflicting parties. 
  • Religious organisations can use their religious conviction about peace in  the world and the sacredness of life, to lead and model peaceful  behaviour. 
  • Religious leaders can mediate and hold a dialogue with all parties with  influence over the situation in Israel/Palestine. E.g. America, which is not  one of the parties in the conflict, but has huge influence over the situation. 
  • Religious organisations can co-operate to alleviate human suffering such  as shortage of food, medical supplies or other necessities. 
  • Religious organisations have a network of contacts with other parts of the  world, and can publicise the situation. (16) 
  • They can mobilise people to pray for and support the affected people.

5.1 EXAMPLE 2: AFRICA 

  •  Darfur in Sudan (2) 

5.2

  • Fault lines in Darfur society can be traced back well over 100 years.
  • The main divisions in this country are ethnic and cultural. 
  • Religion does not seem to be the main source of conflict although it is  always perceived. 
  • Two broad groupings can be found in Darfur, namely Arab tribes called  Baggara and many African tribes. 
  • The African groups include the Fur, the Masalit and the Zaghawa. 
  • Another set of divisions relates to issues of culture. 
  • Most people in the area are Sunni Muslims. 
  • There are also Christians and animists. 
  • For many centuries, nomads and farmers made an effort to coexist, but  when food became scarce, competition for scarce resources led to conflict.
  • The oil resources of the Heglig fields are also a major factor in the conflict.
  • Darfur was also the centre of the slave trade in North East Africa. (12) 

5.3

  • Religion plays only a minor role in the conflict, and therefore this is not a  religious war. No attempt is made by one religion to convert another. 
  • Muslims also enlisted in the SPLA. 
  • There was also evidence of infighting in the SPLA soon after it was  formed. 
  • The conflict in Sudan is known as the longest running civil war in history. (Focus p. 94). The imposition of Sharia law (1983) was therefore not the  cause of the conflict, but it worsened the fragile situation. 
  • Civil war had started as far back as 1955. 
  • There are numerous other divisions (economic, cultural and tribal), which  continue even after establishment of South Sudan. 
  • Even after South Sudan gained independence from Khartoum in 2011,  armed conflict still continues in the region. (10)

5.4

  • There is still conflict in South Sudan, even after independence which was  gained on 9 July 2011. 
  • The 2013–2015 civil war displaced 2,2 million people to many parts of the  world. 
  • Around 6 million people are currently at risk of going hungry. 
  • About 70% of schools have been closed due to fighting. 
  • There is a new element of ethnic hatred and conflict adding to what had  been going on before.
  • There are disturbing indicators of potential impending genocide inside the  country. 
  • Human rights groups say the evidence of war crimes grows by the day.
  • The people in South Sudan are now totally dependent on aid groups. (10) 

5.5

  • Religious leaders can mediate and have dialogue with all parties. 
  • Religious leaders are respected and through their religion usually have  contacts and connections in many parts of the world. 
  • They must liaise with political parties to find a solution, as the conflict is not  religious. 
  • They have access to a network of people from other parts of the world that can assist in the resolution of the conflict.
  • Most religions teach peace and compassion. Therefore members are  willing to contribute materially and financially to the victims of conflict. 
  • Religious leaders can call an inter-religious summit for dialogue on the  conflict. 
  • Dialogue means to talk and listen with respect so as to try and understand  the problem and seek a solution. 
  • The dialogue should lead to practical steps that can be taken on both  sides, regardless of religion, towards alleviating the situation. 
  • Religious organisations can use their religious conviction about peace in  the world and the sacredness of life, to lead and model peaceful  behaviour. 
  • Religious organisations can co-operate to alleviate human sufferings such  as shortage of food, medical supplies or other necessities. 
  • The African Union, the United Nations and many international aid agencies  must try to protect and support civilians. 
  • Organisations like the Islamic Relief Agency and the Catholic Agency for  Overseas Development should focus on practical support, such as medical  supply, shelter for victims and the supply of food. (16) [50] 

 TOTAL: 150