national senior certificate grade 12€¦ · marks: 150 symbol explanation m method ma method with...

MARKS: 150

Symbol Explanation

M Method

MA Method with accuracy

CA Consistent accuracy

A Accuracy

C Conversion

S Simplification

RT/RG Reading from a table/Reading from a graph

SF Correct substitution in a formula

O Opinion/Example

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

R Rounding off

This memorandum consists of 18 pages.

MATHEMATICAL LITERACY P1

NOVEMBER 2010

MEMORANDUM

NATIONAL

SENIOR CERTIFICATE

GRADE 12

Mathematical Literacy/P1 2 DBE/08 November 2010

NSC – FINAL Memorandum

Copyright reserved

QUESTION 1 [33 MARKS]

ANSWER ONLY: If totally correct – Full marks; Otherwise 0

No penalties if units of measurement are omitted

Ques Solution Explanation AS

1.1.1

(a)

15,43 + 46,08 × 15,6875

= 15,43 + 722,88

= 738,31

1A multiplying

1CA simplifying

(2)

NO MARKS – If order of

operation is incorrect

12.1.1

1.1.1

(b)

3

517 − × (29,35 – 10,63) =

3

12 × 18,72

= 74,88

1A simplifying both the

bracket and fraction

1CA simplifying

NO penalty for rounding

(2)

12.1.1

1.1.2

2,875 =0001

8752

= 28

7 OR

8

23

OR

2,875 = 21000

875

= 28

7

1M Changing from decimal

to fraction form

1A simplified fraction

No marks if 1 000

2 875 used.

(2)

12.1.1

1.1.3

ZAR 110,35

= 110,35 × 9,48 DZD

= 1 046,118 DZD OR 1 046,12 DZD

1M multiplication

1A amount in dinar

No rounding off penalties

Max 1 mark if given in

rand

(2)

12.1.1

�A

�M

�A

�M

�A

�CA

�CA

�A

�M

�A



Copyright reserved

ANSWER ONLY: If totally correct – Full marks ; Otherwise 0



1.1.4

3 024 cm = 3 024 ÷ 100 m

= 30,24 m

1M division by 100

1A correct simplification

No penalty if incorrect

units are given

(2)

12.3.2

1.1.5

64

1% of 420 000

= 100

25,6 × 420 000

= 0,0625 × 420 000

= 26 250

OR

64

1% of 420 000 =

4

25% of 420 000

= ×400

25 420 000

= 26 250

1M multiplication with

correct percentage

1A correct simplification

Do not accept 630 000

(2)

12.1.1

1.1.6

Percentage Profit = 1501R

1501R0841R − ×100%

= 60% OR 0,6 OR 100

60

1M correct substitution

1A percentage profit

No marks for – 37,5%

Max 1 mark for – 60%

(2)

12.1.3

1.2.1

21

1A number of classes

(1)

12.1.1

1.2.2

(a)

3 learners

2A mode

(2)

12.4.3

1.2.2

(b)

3 learners

2A median

(2)

12.4.3

�M

�A

�M

�A

�A

��A

��A

�M

�A

�M

�A



Copyright reserved


ANSWER ONLY: If totally correct – Full marks ; Otherwise 0


1.3.1

Volume = 50 cm × 40 cm × 45 cm �M

= 90 000 cm3

1M substituting correct values

1CA volume

(2)

12.3.1

1.3.2

Height of liquid = cm40cm50

cm0003 3

×

= 12

1 cm OR 1,5 cm OR

3

2 cm

1M/A accurate substitution

1A simplification

(2)

12.3.1

1.4.1

Daily payment = R12,50 × 821

= R 106,25

OR

Daily payment = R12,50 × 8 + 2

50,12R

= R 106,25

OR

Daily payment = (R12,50 × 8) + (R12,50 ÷ 2)

= R 100 + R6,25

= R106,25

1S substitution

1CA simplification

Max 1 mark if rounded off

to 9 hours

(2)

12.2.1

1.4.2

Number of hours worked = 50,12R

75,218R

= 17,5 or 172

1

1M dividing by correct values

1A simplification

Full marks if set out as:

R12,50 × 17,5 = R218,75

(2)

12.1.1

1.5.1

10 hours

2A reading from graph

(2)

12.2.3

1.5.2

5

2A reading from graph

(2)

12.2.3

1.5.3

7 hours and 30 minutes

OR

7,5 hrs

2A correct number of hours

Accept any time between 7

and 8 hours

(2)

12.2.3

[33]

�M/A

�A

��A

��A

�A

�M

�S

�CA

�S

�CA

�S

�CA

��A

��A

�CA



Copyright reserved




2.1.1

(a)

Lateral surface area of the cylindrical holder

= 2 × 3,14 × 5 cm × 15 cm

= 471 cm²

1SF substitution of correct radius

and height

1A total surface area

Accept 471,24 cm² or 471,43 cm²

(2)

12.3.1

2.1.1

(b)

Lateral surface area of the rectangular holder

= 2 × (8 + 10) cm × 15 cm

= 2 × 18 cm × 15 cm

= 540 cm²

1SF substitution

1S correct addition

1CA total surface area in cm2

Max 1 mark if incorrect formula

is used

Penalty if units omitted

(3)

12.3.1

2.2.1

33 minutes

1RG correct reading

(1)

12.2.3

2.2.2

6 minutes

2RG correct reading

(2)

12.2.3

2.2.3

12 minutes – 6 minutes

= 6 minutes

1RT correct values from the table

1A correct minutes

(2)

12.2.3

2.2.4

2 500 m

2RG correct reading

Accept any value greater than 0

up to and including 3 000 m

(2)

12.2.3

2.2.5

27 minutes

2RG correct reading

(2)

12.2.3

�RG

��RG

�RT

�A

��RG

��RG

�S

�CA

�SF

�A

�SF



Copyright reserved



2.2.6

10:55 + 12 minutes

= 11:07

1M adding

1A solution

Max 1 mark if given as 10:67

(2)

12.3.1

2.2.7

Average speed = min6

m0003

= 500 m/min

1A correct distance

1A correct time

1CA simplifying

2 marks if using 1 000 m

No penalty if units omitted.

Max 2 marks if answer in km/h

(3)

12.2.1

2.3.1

47,1 % – 42,7%

= 4,4 %

1RT correct values selected

1CA percentage decrease

Accept – 4,4%

No penalty if % is omitted

(2)

12.1.1

12.4.4

�M

�A

�A

�CA

�A

�RT

�CA



Copyright reserved



2.3.2

(a)

A = %0,10

0007204 OR

10,0

0007204

= 47 200 000

OR

10,0% of the population is 4 720 000

∴1% of the population is %0,10

0007204

∴100% of the population is %0,10

0007204 × 100%

= 47 200 000

OR

10% of the population is 4 720 000

100% of the population = 10 × 4 720 000

= 47 200 000

1M method


1CA correct population

(3)

12.1.1

12.4.4

2.3.2

(b)

B = 45,0% × 621 600

= 0,450 × 621 600

= 279 720

� 279 700

1M method


1CA rounded to nearest hundred

(3)

12.1.1

12.4.4

2.3.2

(c)

C = 80065348

0000605 × 100

= 10,40000987

� 10,4


1CA rounded to 1 decimal place

(3)

No penalty if given as 10,4%

12.1.1

12.4.4

�RT �M

�CA

�RT

�CA

�M

�RT

�M

�CA

�CA

�RT

�M

�RT

�RT

�CA



Copyright reserved



2.3.3

49 320 500 : 5 210 000

= 1 : 50032049

0002105

= 1 : 0,105 635 5

� 1 : 0,1

1RT reading correct values

1M correct ratio

1CA simplifying ratio

rounded to one decimal

place

Max 2 marks if order is

changed and the answer is

1 : 9,5

Max 1 mark if written as a

fraction

(3)

12.1.11

2.4.4

[33]

�CA

�RT

�M



Copyright reserved




3.1.1

R25 460 000 000 + R22 670 000 000 + R22 074 000 000

+ R25 458 000 000 + R26 978 000 000

= R122 640 000 000 or R122 640 million

1M adding correct

values

1A simplifying to the

correct value

Max 1 for

R 1 599 565 000 000

or

R1 599 565 million

Penalty of 1 mark if

million left out in

either 3.1.1 or 3.1.2

(2)

12.4.4

12.1.1

3.1.2

R 273 127 million R 292 079 million R 314 927 million

R 326 385 million R 393 047 million

1M arrangement in

ascending order

1A correct values

Penalty of 1 mark if

million left out in

either 3.1.1 or 3.1.2

NO marks for

descending order

Max 1 mark if

incorrect column

values are arranged

(2)

12.4.4

�M

�A

�A

�M



Copyright reserved


3.1.3

2002 2003 2004 2005 2006

8,1 % 8,3 % 7,6 % 7,8 % 6,9 %

AGRICULTURAL EXPO RTS AS A PERCENTAGE

OF THE TOTAL EXPO RTS

0

1

2

3

4

5

6

7

8

9

10

2002 2003 2004 2005 2006

Year

Percen

tag

e

OR

5A one for each correct

bar

Do NOT penalise if

no gaps between bars

Do NOT penalise if

the spaces between

bars are uneven and

the bar widths are

unequal

Maximum of 3 marks

if a line graph is

drawn

Bars can also be

represented as vertical

lines

12.4.2

AGRICULTURAL EXPO RTS AS A

PERCENTAGE O F THE TO TAL EXPORTS

0

1

2

3

4

5

6

7

8

9

10

2002 2003 2004 2005 2006

Year

Percen

tag

e

(5)

�A �A

�A �A

�A



Copyright reserved



3.2.1

450 000 m² = 00010

000450 ha

= 45 ha

OR

10 000 m² = 1ha

Therefore 450 000 m² = 45 × 10 000 m²

= 45 ha

1M division by 10 000

1A number of hectares

OR

1M concept


No penalty if units omitted

(2)

12.3.2

3.2.2

Number of hectares = 65,0

0005 ha

= 7 692,3 ha

� 7 692 ha

1M dividing


1CA rounding off

(3)

12.1.1

12.2.1

3.2.3

Fertiliser needed = 4,32 × 2 000 kg �M�A

= 8 640 kg �CA

2 M/A multiplication with correct

values

1CA simplifying

Max 2 marks if divided by 4,32

(3)

12.1.1

12.2.1

3.2.4

32,4

65,0 ×

1

%100

= 15,046 % OR ≈ 15,05 %

1M concept

1A solution

No penalty if % omitted

No penalty for rounding

(2)

12.1.1

[19]

�M

�M

�M

�A

�CA

�M

�A

�A

�A



Copyright reserved




4.1.1

Increasing

No

penalty

for

omitting

units

2A type of function

(2)

12.2.1

4.1.2

32 oF ��RG

2RG correct reading

Accept 31 F° to

33 F°

(2)

12.2.3

4.1.3

40 oC ��RG

2RG correct reading

(2)

12.2.3

4.1.4

21 °F ��RG �R

2RG correct reading

1R rounding

Accept 22 oF

(3)

12.2.3

4.1.5

Range = 17 oC – (–2

oC)

= 17 oC + 2

oC

= 19 oC

1M calculating the range

1A correct values

1CA range

max of 2 marks if :

–19 o

C or

15 o

C or

from –2 o

C to 17 o

C

or [–2 o

C ; 17 o

C]

(3)

12.2.3

12.1.2

12.4.3

4.2.1

Total Entrance fee

= (4 + 5) × R3,50 + 10 × R6,50

= R31,50 + R65,00

= R96,50

2A substitution of correct values

1CA solution

(3)

12.2.1

12.1.1

4.2.2

Perimeter = 3,14 × 5 m

= 15,7 m

1SF substitution

1A simplifying

Accept 15,71 m or 15,714 m

No marks if diameter is not 5 m

No penalty for omitting units

(2)

12.3.1

�A �A

�M

�CA

�A

�A

�CA

�A

�SF

�A



Copyright reserved



4.2.3

6 000 � = 546,4

0006 gallons

= 1 319,8416.. gallons

≈ 1 319,84 gallons

OR

1 � = 546,4

1 gallon

∴ 6000 � = 546,4

1 × 6000 gallons

= 546,4

0006gallons

= 1 319,8416.. gallons

≈ 1 319,84 gallons

1 M dividing

1A number of gallons

Accept up to 1 320 gallons

(2)

12.3.2

[19]

�M

�A

�M

�A



Copyright reserved




5.1.1

South Westerly

OR SW

OR SSW

OR WS

OR S45oW

OR West of South

OR South of West

1A direction

(1)

12.3.4

5.1.2

Perimeter of Mr Khoso’s plot:

= 224 m + 200 m + 150 m + 200 m + 250 m

= 1 024 m

1M Concept of perimeter

1A using correct values

1CA sum of the lengths

(3)

12.3.1

5.1.3

Volume = 3,14 × (10 m)² × 2 m

= 628 m³

1SF substitution

1A simplifying

1A correct units

Accept 628,32 m3 OR

628,57 m3

Max 1mark if radius not

squared

(3)

12.3.1

5.1.4

Area of a cattle kraal = 2

1 × 200 m × 200 m

= 20 000 m²

1A height

1SF substitution

1CA simplifying

Max 2 marks if area is

equal to 22 400 m2

(3)

12.3.1

� A

� SF

� A � A

� SF

� CA

� A

� M

� CA

� A



Copyright reserved



5.1.5

Area of Mr Khoso’s plot

= 2

1 × (200 m + 150 m + 250 m) × 200 m

= 2

1 × 600 m × 200 m

= 60 000 m²

OR

Area = Area of triangle + Area of trapezium

= 2

1× 200m × 200m +

2

1(150m + 250m) × 200m

= 20 000 m² + 2

1(400 m) × 200 m

= 20 000 m² + 40 000 m²

= 60 000 m²

1A adding correct parallel

sides

1SF substitution

1A correct values

1CA simplifying

Max 2 marks if area of

vegetable garden is

calculated as 5 625 m2

(4)

12.3.1

5.2.1

Total mass = 2 × 2 kg + 12 × 0,12 kg

= 4 kg + 1,44 kg

= 5,44 kg

1M multiplying and adding

1A simplifying

(2)

12.3.1

12.2.1

5.2.2

(a)

A = 2 × 12

= 24

OR

A = 4 × 6

= 24

1M multiplying

1A number of carrots

.

(2)

12.2.1

� CA

� A

� SF � A

� M

� A

� M

� A

�A � SF

� A

� CA

� M

� A



Copyright reserved



5.2.2

(b)

2 cabbages in 1 box

24 cabbages in 2

24 boxes

= 12 boxes

OR

B = 12

144

B = 12 boxes

OR

1 box = 14 vegetables

B = 14

168

= 12

OR

B = 10

245×

= 12

1M dividing correct values

1A number of boxes

(2)

12.2.1

5.2.3

12 cabbages in 6 boxes

Number of carrots = 6 × 12

= 72

OR

��

carrots12andcabbages2hasbox1

carrots60andcabbages10haveboxes5

∴ (60 + 12) carrots = 72 carrots

1M number of boxes

1CA number of carrots

1M both statements

1CA number of carrots

(2)

12.2.1

[22]

� M

� CA

� M

� CA

� A

� M

� M

� M

� M

� A

� A

� A



Copyright reserved




6.1.1

Mean

= 14

4037373340373746373734375525 +++++++++++++

= 14

532

= 38

1M sum

1M dividing the sum of

scores

1A simplifying

max 1 mark if

coffee mugs used

(3)

12.4.3

6.1.2

P(37 key rings) = 14

7

= 2

1

1A correct numerator

1A correct denominator

1CA simplified fraction

Max 2 marks for

50% or 0,5

(3)

12.4.5

12.1.1

6.1.3

(a)

Range = 38 – 25

= 13 coffee mugs

1A minimum & maximum

values

1A range

Accept –13

Max 1 mark if key

rings used

(2)

12.4.3

6.1.3

(b)

Mode = 35 and 37

2A mode

(2)

12.4.3

6.1.3

(c)

Median = 2

3535 +

= 35

1M finding median

1A median

(one value only)

(2)

12.4.3

6.2.1

Income = 128 × R7,00

= R896,00

1M calculating income

1CA income

1 mark for

128 × R4,80 = R614,40

(2)

12.1.3

� A

� M

�A

� A

� A

� A

� M

� M

� CA

�CA

�A

� M

� A � A



Copyright reserved

6.2.2

(a)

NOTE: To assist with marking, the graph that the learner has to draw is represented

as a dotted line. The learners DO NOT have to draw a dotted line.

A plotting (1;250) 1A plotting (3;370) 1A plotting (4;380)

1A plotting (2;350) 1A plotting (5;270) 1A plotting (6;350)

1A plotting (7;370)

1CA joining the points with a line

Max 6 marks if incorrect type of graph is drawn

(8)

12.2.2

6.2.2

(b)

Day 2

2RG/RT correct

days

(2)

12.4.4

[24]

TOTAL: 150

�� RG/RT

INCOME FROM ITEMS SOLD DURING THE THIRD WEEK

0

50

100

150

200

250

300

350

400

450

500

0 1 2 3 4 5 6 7

Day of week

Inco

me

( in

ran

d)

Coffee mugs

Key rings

�A

�A

�A

�A

�A

�A

�A

Copyright reserved Please turn over

MARKS: 150

TIME: 3 hours

This question paper consists of 15 pages and 2 annexures.


NOVEMBER 2010

NATIONAL SENIOR CERTIFICATE

GRADE 12

Mathematical Literacy/P1 2 DBE/November 2010 NSC


INSTRUCTIONS AND INFORMATION 1. This question paper consists of SIX questions. Answer ALL the questions. 2. QUESTION 3.1.3 and QUESTION 6.2.2(a) must be answered on the attached

ANNEXURES. Write your centre number and examination number in the spaces on the annexures and hand in the annexures with the 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. An approved calculator (non-programmable and non-graphical) may be used, unless

stated otherwise.

6. ALL the calculations must be clearly shown. 7. ALL the final answers must be rounded off to TWO decimal places, unless stated

otherwise.

8. Units of measurement must be indicated where applicable. 9. Write neatly and legibly.



QUESTION 1

1.1 1.1.1 Simplify: (a) 15,43 + 46,08 × 15,6875 (2)

(b) 3

517− × (29,35 – 10,63)

(2)

1.1.2 Write 2,875 as a common fraction in its simplest form. (2) 1.1.3 Convert R110,35 (South African rand/ZAR) to Algerian dinar (DZD) if

1 ZAR = 9,48 DZD.

(2) 1.1.4 Convert 3 024 cm to metres. (2) 1.1.5 Calculate 64

1 % of 420 000. (2) 1.1.6 It cost Ridge R1 150,00 to make a matric dance dress. He sold it for

R1 840,00. Use the following formula to calculate the percentage profit made on the dress:

Percentage profit = pricecost

pricecostpriceselling − × 100%

(2)

1.2 The principal of Hills Primary School compiled data of the number of learners who

receive social grants in each class. He arranged these numbers in ascending order, as follows: 0 0 1 1 1 2 2 2 3 3 3 3 4 4 5 5 6 6 6 7 7

1.2.1 How many different classes are there at Hills Primary School? (1) 1.2.2 Determine: (a) The mode (2) (b) The median (2)



1.3 During an experiment, an amount of liquid was poured into a calibrated rectangular

container, as shown in the diagram below. A calibrated container has accurate measurements marked on it. It is used to measure volume.

The dimensions of the container are: length = 50 cm, breadth = 40 cm and height = 45 cm A calibrated rectangular container

1.3.1 Calculate the volume, in cm3, of the container.

Use the following formula: Volume = length × breadth × height

(2) 1.3.2 3 000 cm3 of the liquid was poured into the calibrated container.

Calculate the height of the liquid in the container by using the following formula:

Height of liquid = breadth lengthliquidofvolume

×

(2) 1.4 Casual workers employed during the Soccer World Cup were paid an hourly rate of

R12,50. The following formula may be used:

Daily payment = hourly rate × number of hours worked

1.4.1 One casual worker worked 2

18 hours daily. How much did he/she earn

daily?

(2) 1.4.2 A casual worker was paid a total of R218,75. For how many hours did

he/she work?

(2)

1 000

2 000

3 000

4 000 cm3



1.5 Mrs White, the owner of a cleaning company, uses the graph below to determine

how long it would take different teams of workers to clean a block of offices.

CLEANING COMPANY PLANNING GRAPH

0

5

10

15

20

25

30

0 2 4 6 8 10

Number of workers in a team

Num

ber

of w

orki

ng h

ours

1.5.1 How long will it take THREE workers to clean the same block of offices? (2) 1.5.2 How many workers will she need to employ to complete cleaning the

same block of offices in exactly SIX hours?

(2) 1.5.3 Estimate the number of hours it will take FOUR workers to clean the

same block of offices.

(2) [33]



QUESTION 2 2.1 Thandiwe wants to make a new pencil holder. She has a choice of an open cylindrical

holder or an open rectangular holder. She wants to cover the outside of the holder to match the table cloth on her desk. A cylindrical holder with: radius = 5 cm and height = 15 cm

A rectangular holder with: length = 10 cm, breadth = 8 cm and height = 15 cm The following formulae may be used: Lateral surface area of a cylinder = 2π × radius × height, and using π = 3,14 Lateral surface area of a rectangular prism = 2 × ( length + breadth) × height

2.1.1 Determine the lateral surface area of:

(a) The cylindrical holder (b) The rectangular holder.

(2) (3)

Pencils



2.2 The graph below illustrates John's trip to the bicycle shop, 3 000 m away from his

home, to collect his bicycle that was sent for repairs. He first walked to the post office to buy stamps and then went to collect his bicycle. He rode the bicycle back home.

JOHN'S TRIP TO COLLECT HIS BICYCLE

0

1000

2000

3000

0 10 20 30 40

Time in minutes

Dis

tanc

e aw

ay fr

om h

ome

in m

etre

s

2.2.1 How many minutes was John away from home? (1) 2.2.2 How long did John take to reach the post office, 1 000 m away from his

home?

(2) 2.2.3 How many minutes did John spend at the post office? (2) 2.2.4 How far was John away from home after 21 minutes? (2) 2.2.5 After how many minutes did John begin his journey back home? (2) 2.2.6 John took 12 minutes to walk from the post office to the bicycle shop.

If he left the post office at 10:55, at what time did he arrive at the bicycle shop?

(2)

2.2.7 If the trip from the bicycle shop to his home took 6 minutes, calculate the

average speed, in metres per minute, at which John cycled. Use the formula:

Average speed = time

travelleddistance

(3)



2.3 TABLE 1 below shows the approximate number of South Africans living with HIV and Aids, and the number of Aids-related deaths from 2005 to 2009. TABLE 1: Data related to population, deaths, HIV and Aids from 2005 to 2009

SOUTH AFRICANS LIVING WITH HIV AND AIDS

AIDS-RELATED DEATHS

YEAR

SOUTH

AFRICAN POPULATION

NUMBER % OF POPULATION

TOTAL DEATHS

IN COUNTRY NUMBER

% OF TOTAL

DEATHS 2005 A 4 720 000 10,0 634 100 298 600 47,1 2006 47 821 700 4 830 000 10,1 628 600 289 800 46,1 2007 48 431 400 4 940 000 10,2 621 600 B 45,0 2008 48 653 800 5 060 000 C 602 800 257 500 42,7 2009 49 320 500 5 210 000 10,6 613 900 263 900 43,0

[Source: Statistics South Africa]

Use TABLE 1 to answer the following questions.

2.3.2 Calculate the following missing values: (a) A (3) (b) B, rounded off to the nearest 100 (3) (c) C, rounded off to ONE decimal place (3) 2.3.3 Determine the ratio between the total South African population during

2009 and the number of South Africans living with HIV and Aids during 2009. Write, rounded off to ONE decimal place, the ratio in the form 1 : …

(3)

[33]

2.3.1 Calculate the difference in percentage of Aids-related deaths between 2005 and 2008.

(2)



QUESTION 3 3.1 Each year South Africa generates income from exports (products sold to other

countries). The income generated from these exports varies from year to year. Part of the income generated by exports comes from agricultural products. The table below shows the total income from exports, as well as the percentages of the total earned from agricultural products. TABLE 2: Relationship between South African exports of agricultural and

other products

YEAR

Total income generated by South

African exports (in millions of rand)

Income generated by agricultural

exports (in millions of rand)

Percentage of the total income earned by

agricultural products

2002 314 927 25 460 8,1 2003 273 127 22 670 8,3 2004 292 079 22 074 7,6 2005 326 385 25 458 7,8 2006 393 047 26 978 6,9

[Source: South African Year Book, 2007]

3.1.1 Calculate the total income generated by agricultural exports from 2002 to

the end of 2006.

(2) 3.1.2 Arrange the total incomes for the different years in ascending order. (2) 3.1.3 Draw a bar graph on the grid on ANNEXURE A to represent the

percentage of the total income earned by agricultural products from 2002 to 2006.

(5) 3.2 The use of fertilisers for crops such as mealies, sorghum, fruit and vegetables, can

result in an increased harvest of these crops. In South Africa farmers use an average of 0,65 kg of fertiliser per hectare (ha), while farmers in Egypt use an average of kg32,4 of fertiliser per hectare, where 10 000 m2 = 1 ha.

3.2.1 Convert 450 000 m2 to hectares. (2) 3.2.2 Calculate the number of hectares that could be fertilised with 5 000 kg of

fertiliser by a farmer in South Africa. Give the answer rounded off to the nearest hectare.

(3) 3.2.3 Calculate the number of kilograms of fertiliser that would be needed in

Egypt to fertilise 2 000 ha.

(3) 3.2.4 Write the average amount of fertiliser used per hectare in South Africa as

a percentage of the average amount of fertiliser used per hectare in Egypt.

(2) [19]



QUESTION 4 4.1 Mrs Smith is visiting South Africa for the Soccer World Cup. In her country

temperature is measured in degrees Fahrenheit. She used the graph below to easily convert temperatures between degrees Fahrenheit (oF) and degrees Celsius (oC).

TEMPERATURE CONVERSION GRAPH

0

10

20

30

40

50

60

70

80

90

100

110

120

130

-10 0 10 20 30 40 50

Temperature in degrees Celsius

Tem

pera

ture

in d

egre

es F

ahre

nhei

t

Use the graph to answer the following questions: 4.1.1 Is this graph representing an increasing, a decreasing or a constant

function?

(2) 4.1.2 The melting point of ice is 0 oC. Write down the melting point of ice in

degrees Fahrenheit.

(2) 4.1.3 What is a temperature of 104 oF in oC? (2) 4.1.4 What is a temperature of –6 oC in oF? Round the answer off to the

nearest degree.

(3) 4.1.5 On a particular day, the minimum temperature was –2 oC and the

maximum temperature was 17 oC. Determine the temperature range for that day.

(3)



4.2 Mrs Smith and her touring party decide to visit an indoor swimming pool. 4.2.1 The entrance fee for the swimming pool is:

• R3,50 for children under 12 years and pensioners • R6,50 for adults and children 12 years and older There are four children under 12 years, five pensioners and ten adults in the group who visit the swimming pool. Calculate the total entrance fee paid by the group. Use the formula:

Total entrance fee = (number of children + pensioners) × R3,50 + (number of adults) × R6,50

(3) 4.2.2 The circular kiddies pool at the indoor pool has a diameter of 5 m. There

is a protective fence around the perimeter of the pool.

Determine the perimeter of the fence.

Use the formula: Perimeter = π × diameter, and using π = 3,14

(2) 4.2.3 The kiddies pool is filled with 6 000l of water. Mrs Smith wanted to

know what this volume of water would be in gallons. Convert the volume of water in the pool into gallons if 1 gallon = 4,546.l

(2) [19]



QUESTION 5

5.1 Mr J Khoso owns a plot, as shown in the diagram below (not drawn to scale). His house (D) is on the eastern side of the plot. Also on the plot is a cattle kraal (A), a circular water tank (B), and a vegetable garden (C).

Diagram of Mr J Khoso's plot

KEY DIMENSIONS

A Cattle kraal Height = 200 m Base (south ) = 200 m Slanting side = 224 m

B Water tank Radius = 10 m

C Vegetable garden Parallel sides = 100 m and 125 m Distance between parallel sides (height) = 50 m

D House Length = 25 m Breadth = 8 m

5.1.1 Give the general direction of the water tank from the house. (1)

5.1.2 Determine the perimeter of Mr Khoso's plot. (3)

5.1.3 Calculate the volume of water in the circular water tank, if the height of the water in the tank is 2 m.

Use the formula:

Volume = π × (radius)2 × height, and using π = 3,14

(3)

5.1.4 Determine the area of the cattle kraal.

Use the formula:

Area of a triangle = 21 × base × height

(3)

5.1.5 Calculate the total area of Mr Khoso's plot.

Use the formula:

Area of a trapezium = 21 × (sum of the parallel sides) × height

(4)

A

D

B

C

200 m

N

250 m

200

m

224 m

150 m



5.2

Mr Khoso grows cabbages and carrots in his vegetable garden. He sells them in boxes. TABLE 3 shows the relationship between the number of cabbages and carrots in each box. TABLE 3: Relationship between the number of cabbages and carrots in each box Number of boxes 1 2 5 B 15 Number of cabbages 2 4 10 24 30 Number of carrots 12 A 60 144 180

5.2.1 The average mass of a cabbage is 2 kg and the average mass of a carrot is

0,12 kg. Calculate the total average mass of the cabbages and carrots in one box.

(2)

5.2.2 Determine the following missing values: (a) A (2) (b) B (2) 5.2.3 A customer placed an order for a number of boxes of vegetables which

contained a total of 12 cabbages. How many carrots in total were included in these boxes?

(2)

[22]



QUESTION 6 Mr Francis sold two different types of souvenirs during the Soccer World Cup: key rings and coffee mugs. He kept a daily record of the number of items sold for each type. TABLE 4 represents the number of items sold daily during the first two weeks of the tournament. TABLE 4: Number of items sold during the first two weeks of the tournament ITEM SOLD

Day 1

Day 2

Day 3

Day 4

Day 5

Day 6

Day 7

Day 8

Day 9

Day 10

Day 11

Day 12

Day 13

Day 14

Key rings

25 55 37 34 37 37 46 37 37 40 33 37 37 40

Coffee mugs

25 26 27 29 35 35 35 35 37 37 37 37 38 38

6.1 Use TABLE 4 above to answer the following questions. 6.1.1 Calculate the mean (average) number of key rings sold daily. (3) 6.1.2 Suppose one of the days is chosen at random, what is the probability that

37 key rings were sold on that day? (Write the answer as a common fraction in the simplest form.)

(3)

6.1.3 For the number of coffee mugs sold, determine the following: (a) The range (2) (b) The mode (2) (c) The median (2)


Copyright reserved

6.2 Mr Francis buys the key rings for R4,80 each and the coffee mugs for R7,00 each.

He sells each key ring for R7,00 and each coffee mug for R10,00.

6.2.1 Calculate Mr Francis' income if he sold 128 key rings. (2) 6.2.2 Mr Francis' income for the items sold in the third week of the tournament is

given in the table below. TABLE 5: Income from items sold during the third week

DAY OF WEEK ITEM SOLD Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Key ring

R175 R385 R259 R231 R259 R259 R322

Coffee mug

R250 R350 R370 R380 R270 R350 R370

(a) ANNEXURE B already shows the graph representing the daily

income from the sale of key rings. Draw a second line graph on ANNEXURE B to represent the daily income from the sale of the coffee mugs. Clearly label your graph.

(8)

(b) Use the table or the graphs to determine on which day his income

from the sale of key rings was greater than his income from the sale of coffee mugs.

(2)

[24] TOTAL: 150

Mathematical Literacy/P1 DBE/November 2010 NSC

Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE A QUESTION 3.1.3

PERCENTAGE OF THE TO TAL INCOME EARNED BY AGRICULTURAL PRODUCTS

0

1

2

3

4

5

6

7

8

9

10

2002 2003 2004 2005 2006

Year

Per

cent

age

Mathematical Literacy/P1 DBE/November 2010 NSC

Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE B QUESTION 6.2.2(a)

INCOME FROM ITEMS SOLD DURING THE THIRD WEEK

0

50

100

150

200

250

300

350

400

450

500

0 1 2 3 4 5 6 7

Day of week

Inco

me

(in r

and)

Key rings


MARKS: 150

SYMBOL EXPLANATION

A Accuracy

CA Consistent accuracy

C Conversion

J Justification (Reason/Opinion)

M Method

MA Method with accuracy

P Penalty for no units, incorrect rounding off, etc.

R Rounding off

RT/RG Reading from a table/Reading from a graph

S Simplification

SF Correct substitution in a formula

O Own opinion

This memorandum consists of 21 pages.


NOVEMBER 2010

MEMORANDUM

NATIONAL

SENIOR CERTIFICATE

GRADE 12

Mathematical Literacy/P2 2 DBE/ 08 November 2010


Copyright reserved



1.1.1

(a)

Diameter of tablecloth

= 4 × 30 cm

= 120 cm

Radius of tablecloth

= 120 ÷ 2

= 60 cm

OR

Radius of placemat

= 30 ÷ 2 cm

= 15 cm

Radius of tablecloth

= 4 × 15 cm

= 60 cm

1M finding diameter or

radius

1CA radius of tablecloth

Circumference of table cloth = 2 ×π radius

= 2 × 3,14 × 60 cm

= 376,8 cm

1SF substitution into correct

formula

1CA circumference with

correct unit

Using π (376,99 cm)

Using22

7 (377,14 cm)

Max 2 marks if incorrect

radius

Max 1 mark if radius of

placemat is used

Answer only full marks

(4)

12.3.

1

1.1.1

(b)

Number of segments = 71,4

8,376

= 80

1M dividing by 4,71

1CA number of segments

80,04 OR 80,07



(2)

12.3.

1

12.1.

1

�M

�SF

�M

�CA

�CA

�CA �CA

�M



Copyright reserved


1.2.1

(a)

Total cost

= R300 + R0,50 × (number of minutes more than 500)

OR

Total cost = R300 + R0,50 × x,

Where x = number of minutes more than 500

1A constant value R300

1A second term

1A constant value R300

1A second term

No penalty if R omitted

(2)

12.2.1

1.2.1

(b)

Total cost = R300 + R0,50 × (510 – 500)

= R300 + R5

= R305

OR

Cost of calls = R0,50 × 10

= R5,00

Total cost = R300,00 + R5,00

= R305,00

1M use of formula from 1.2.1(a)

1SF substitution of minutes

1S simplifying

1CA solution

1M calculating extra cost

1CA simplifying

1M calculating total cost

1CA solution

No penalty for units


(4)

12.2.1

�M

�S

�CA

�SF

�M

�M �CA

�CA

�A �A

�A �A



Copyright reserved


1.2.2 NOTE: To assist with marking, the graph that the learners have to draw has been

given as a dotted line. The learners DO NOT have to draw a dotted line.

12.2.2

LANDLINE CALL PACKAGES

0

100

200

300

400

500

600

700

0 100 200 300 400 500 600 700 800 900 1000

Number of minutes used per month

To

tal

Co

st i

n R

an

d

PACKAGE 1: 1A point (0;150)

1A horizontal line from (0;150) to the point (100;150)

1A another correct point

1G for having a break-even point between (100;150) and (500;350)

1A for totally correct straight line that must be up to the point (1000; 600))

No penalty if label is omitted

(5)

CALL PACKAGE 1

CALL PACKAGE 2

�A �A

�A

�A

�A



Copyright reserved


1.2.3(a)

The break-even point is the point where:

• the two graphs intersect.

OR

• both packages cost the same

OR

• there is no profit/gain or loss

OR

• both situations are the same

2M description of break-even

point

(other correct definitions)

2 Marks or zero

(2)

12.2.3

1.2.3(b)

Number of minutes used = 400

Total cost = R 300

CA from graph

1RG number of minutes

1RG cost

Accept (400 ; 300)

Point may be calculated

algebraically

(2)

12.2.3

1.2.4

Package 2

Reading 900 minutes and 1 000 minutes

Showing difference

Package 2 gives 100 minutes more call time for

R550 than Package 1

OR

She must accept Package 2

Package 1: 550 = 150 + 0,50 × x,

550 – 150 = 0,50 x

x = 5,0

400 = 800

Total minutes = 100 + 800 = 900

Package 2: 550 = 300 + 0,50 × x

550 – 300 = 0,50 x

x = 5,0

250 = 500

Total minutes = 500 + 500 = 1 000

CA from graph

2CA selecting correct package

1RG reading from the graph

1M difference

1J motivation

2CA selecting correct package

1M using formula

1CA simplification

1CA simplification

(5)

12.2.3

��M

��M

�RG

�RG

��M

��M

�RG

�1J

��CA

�CA

�M

�CA

�CA

�M

�CA

�CA



Copyright reserved



2.1.1

C3 OR 3C

1A for C

1A for 3

(2)

12.3.4

2.1.2

SE OR South East OR East of South

OR South of East

2A correct direction

2 Marks or zero

(2)

12.3.4

2.1.3(a)

• Carry on along Selby Msimang Road in a

(North-Easterly) direction.:

• At the traffic lights turn right into Sutherland

Road

• then turn right into F.J. Sithole Road

• then turn left into Nkugwini Road

• entrance to the stadium is on the left.

1A recognising direction

1A turn into Sutherland Rd

1A turn into F.J. Sithole Rd

1A turn into Nkugwini Rd

Follow learner’s route on map.

If direction very long Max 2

marks

Max 3 marks if names of roads

listed only in correct order

(4)

12.3.4

2.1.3(b)

Distance on map = 145 mm

Actual distance = 145 mm × 20 000

= 2 900 000 mm

= 2, 9 km

1A distance on map

(Accept 130 mm – 150 mm)

1M multiplying by the scale

1CA distance in mm

1CA distance in km

Accept measurement in cm

Accept 2,6 km – 3,0 km


(4)

12.3.1

12.3.3

�A �A

�A

�A

�A

�A

�M

�CA �CA

�A �A

�A



Copyright reserved

2.1.4


distance

40 km/h = time

km9,2

Time = h/km40

km9,2

= 0,0725 hours

= 0,0725 × 60 minutes

= 4,35 minutes

Arrival = 09:15 + 4,35minutes

= 09H 19,35minutes OR 09:19:21

∴the bus driver’s estimated time of arrival is

correct.

OR

Speed = time

distance

40 km/h = minutes 5

distance

Distance = 40 × 60

5km

= 3,33 km

∴ it is possible for him to be at the stadium at

09:20

He can cover a longer distance than he need to

cover in 5 minutes

OR

Speed = time

distance

= minutes 5

km2,9

= 2,9 km × 5

60hour

= 34,8 km/h

∴ He has 5 minutes to get to the stadium and

can travel at 34,8 km/h and still arrrive on time

1SF/CA substitution

1M rearranging the formula

1S simplification

1C converting to minutes

Range from 3,9 to 4,5 minutes

1CA time of arrival

1CA conclusion

1SF/CA substitution

1M rearranging the formula


1CA simplification

1CA conclusion

OR

1CA conclusion

1CA substituting distance

1A substituting time


1CA simplification

1CA comparison of speed

1CA conclusion

(6)

12.3.1

12.3.2

12.2.1

�SF

�C

�CA

�S

�M

�CA

�SF �M

�C

�CA

�CA

�CA

�CA �A

�C

�CA

�CA

�CA



Copyright reserved


2.2.1

PANTS SHIRT TIE POSSIBLE

OUTCOMES

T � LP; LS; T

LS

NT � LP; LS; NT LP

T � LP; SS; T SS

NT � L�; SS; NT

T � SP; LS; T LS

NT � SP; LS; NT SP

T � SP; SS; T SS

NT � SP; SS; NT

1A LS and SS

2A T and NT

4A POSSIBLE OUTCOMES

Max 1 mark if only 1 or 2 possible outcomes are correct

Max 2 marks if 3 or 4 possible outcomes are correct

Max 3 marks if 5 or 6 possible outcomes are correct

Max 4 marks if all 7 possible outcomes are correct

Order of outcomes not important in this solution

(7)

12.4.5

2.2.2

P(correct uniform) = 8

2 OR

4

1

= 0,25

1A number of actual outcomes

(numerator)

1A number of possible

outcomes (denominator)

1CA decimal form

Max 2 marks if 4

1 or 25%


(3)

12.4.5

�CA

�A �A

�A

�CA �A

�CA

�CA

�CA

�A



Copyright reserved


No penalty for rounding off

Ques Solution Explanation ASs

3.1.1

Monthly medical aid costs

��M ��RT

= R1 152 + R816 + 2 × R424

= R2 816 ��CA

Member’s contribution = ×

3

1 R2 816 ��CA

= R938,67 ��CA

OR

Member’s contribution

��CA ��M ��RT

= 3

1(R1 152 + R816 + 2 × R424)

= ×

3

1 R 2 816 ��A

= R 938,67 ��CA

OR

Members subscription = ×

3

1 R1 152 = R 384

Wife’s subscription = ×

3

1 R816 = R272

Children subscriptions = 2 × ×

3

1 R424 = R282,67

Member’s contribution = R 384 + R272 + R282,67

= R 938,67

1M correct main member

from table

1RT cost for wife and

children

1CA total cost of the three

categories

1A multiplying correct

value by 3

1

1CA simplifying


from table


children

1CA total cost


value by 3

1

1CA simplifying


from table


value by 3

1


children

1CA children cost

1CA simplifying

Max 4 marks if

incorrect row used


(5)

12.1.3

�A �M

�RT

�CA

�CA



Copyright reserved


3.1.2

(a)

ANNEXURE D

MONTHLY

DEDUCTIONS 3.1.2 (a)

A

Union membership

R35,00

B

Pension

= 7,5% of gross salary

7,5% × R7 986,50 = R598,99

C

PAYE

= (gross salary

– R4 750) × 18%

(R7 986,50 – R4 750,00) × 18%

= R3 236,50 × 0,18 = R582,57

D

Medical Aid contribution

R938,67

E

Total

= A + B + C + D

Total deductions

= R35 + R598,99 + R582,57 +R938,67

= R2 155,23

12.1.3

12.2.3

1M multiplying

1A simplifying

1SF substitution

into formula

1CA simplifying

1CA total

No penalty for

rounding off

(5)

3.1.2

(b)

Net salary = Gross salary – total deductions

= R7 986,50 – R2 155,23 �M

= R5 831,27 �CA

Net annual salary = R5 831,27 × 12

= R69 975,24

1M difference of

correct values

1CA simplifying

1CA annual net

salary

(3)

�A

�SF

�CA

�CA

�CA

�M



Copyright reserved


3.1.3

(a)

ANNEXURE E

MONTHLY

DEDUCTIONS 3.1.3(a)

A

Union membership

R35,00

B

Pension


New salary

= 1,045 × R7 986,50 = R8 345,89

Pension

= 7,5% × R8 345,89 = R625,94

C

PAYE

= (gross salary

– R4 750) × 18%

(R8 345,89 – R4 750,00) × 18%

= R3 595,89 × 0,18 = R647,26

D

Medical Aid contribution

Medical Aid cost

= R1 256 + R900 + 2 × R468

= R3 092

Member contribution

= ×

3

1 R3 092 = R1 030,67

E

Total

= A + B + C + D

Total deductions

= R35 + R625,94 + R647,26

+ R1 030,67

= R2 338,87

Net salary = R8 345,89 – R2 338,87

= R6 007,02

Difference in net salaries = R6 007,02 – R5 831,27 = R175,75

∴Mr Riet’s argument is NOT valid.

12.1.3

12.2.3

1A increase in %

1CA new salary

1CA simplifying

1CA simplifying

1RT values

1A medical aid

costs

1CA simplifying

1CA total

deductions

1CA simplifying

1CA conclusion

No penalty for

rounding off

(10)

�CA

�CA

�A

�CA

�RT

�A

�CA

�CA

�CA

�CA



Copyright reserved


3.1.3(b)

% change = %10024,97569R

24,97569R24,08472R×

−

= 3,013%

≈ 3,01%

OR

% change = %10027,8315R

27,8315R02,0076R×

−

= 3,013%

≈ 3,01%

1M calculating % change

1CA using new and old salary

1CA simplifying

1M calculating % change

1CA using new and old salary

1CA simplifying

No penalty for leaving out

% symbol

Accept 0,0301


(3)

12.1.3

3.2.1

2009/2010 = 17% of R834,3 billion

= 0,17 × R834,3 billion

= R141,831 billion

2010/2011 = 18% of R900,9 billion

= 0,18 × R900,9 billion

= R162,162 billion

Difference = R162,162 billion – R141,831 billion

= R20,331 billion

= R20 331 000 000

R20 331 000 000 > R20 000 000 000

1M calculating 17%

1A simplifying

1A percentage expenditure in

2010/2011

1M calculating 18%

1CA simplifying

1M calculating the difference

1CA difference in rand

1C conversion

Numbers may be written

with zeros instead of the

word billion

(8)

12.1.1

12.4.4

�M

�A

�M

�CA

�M

�CA

�A

�M

�CA

�C

�CA

�CA

�M

�CA



Copyright reserved


3.2.2

* Increases in number of employees

* Increase in salaries

* Building new schools/libraries

* Increase in the number of “no fee” schools

* Teacher development initiatives

* Increase in expenditure per learner

* Demands of the new curriculum

* Cater for inflation

* Free stationery and textbooks

* Feeding scheme for all learners

* Free transport for all learners

* More money for bursaries

* Improvement of matric results

* Demand for Higher Education

2O any correct reason

2O any correct reason

(4)

12.4.4

�� O

�� O



Copyright reserved



4.1

Height of bottle = %102

mm143

= 02,1

mm143 OR

143mm100%

102%×

= 140,196… mm

� 140 mm

1M dividing


1CA/R simplifying to

nearest mm

ax 1 for rounding off if

method is incorrect


(3)

12.1.1

12.3.1

� A

� CA/R

� M



Copyright reserved


4.2

Area of base of bottle = 3,14 × (29 mm)2

= 2 640,74 mm2

Length of base of box

= 105% × 58 mm

= 1,05 × 58 mm

= 60,9 mm

OR

×

100

105 58 mm

Area of base of box = (side length)2

= (60,9 mm)2

= 3 708,81 mm2

Difference in area = 3 708,81 mm2 – 2 640,74 mm

2

= 1 068,07 mm2

≈ 10,68 cm2

The dimensions satisfy the guideline

1SF substitution into correct

formula

1A value of radius

1CA simplifying

2642,08 using pi

2643,14 using 22

7

1M increasing percentage

1A simplifying

1SF substitution into formula

1CA simplifying

1M subtracting

1CA simplifying

2CA conclusion

Length of base rounded off

to 61 mm, or use of pi/22

7

the difference in area

= 10,80 cm2

Answer can be calculated

using cm.

(11)

12.3.1

12.1.1

� SF

� A

� CA

� CA

� SF

� CA

� CA

� M

� A

� M

�



Copyright reserved


4.3.1

Area A = 143 mm × 60,9 mm

= 8 708,7 mm2

Area B = (60,9 mm)2

= 3 708,81 mm2

Area C = ×

2

13,14 ×

2

2

mm9,60��

��

�

= ×

2

1 2 911,41585 mm

2

= 1 455,71 mm2

Area of open box

= 4(A + D) + 2(B + C) + E

= 4 (8 708,7 + 1 832) mm2 + 2 (3 708,81 + 1 455,71) mm

2

+ 2 855 mm

2

= 55 346,84 mm2

= 55 346,84

1 000 000 m

2

= 0,055346....m2

Mass of box = 240 g/m 2 × 55 346,84

1 000 000m 2

= 13,2832.. g

= 14 g

OR

1M calculating area

1CA simplifying

1CA area B

1SF substitution into

correct formula

1CA simplifying

1SF(CA) substitution

1CA simplifying

1C converting to m 2

1M multiplication

1S simplifying

1R rounding

accept 13 g

If area rounded off

to 0,06 m2 then

mass = 15 g

12.3.1

� CA

� CA

� SF

� C

� M

� CA

� SF

� CA

� M

� R

� S



Copyright reserved


4.3.1

(cont)

Area A = 143 mm × 61 mm

= 8 723 mm2

Area B = 61 mm × 61 mm

= 3 721 mm2

Area C = ×

2

13,14 ×

2

2

mm61��

��

�

= ×

2

1 2 920,985 mm

2

= 1 460,49 mm2

Surface area

= 4(A + D) + 2(B + C) + E

= 4 (8 723 + 1 832) mm2 + 2 (3 721 + 1 460,49) mm

2 +

2 855 mm2

= 55 437,98 mm2

= 0000001

98,43755 m

2

= 0,055..m2

Mass of box = 240 g/m 2 × 0,055..

= 13,31 g

= 14 g

1SF substitution

1CA area A

1CA area B

1SF substitution

1CA area C

1SF substitution

1CA surface area

1C converting to m 2

1M multiplication

1S simplification

1R rounding

(11)

12.3.1

4.3.2

1 kg = 1 000 g

∴ 14 g = 0,014 kg

Cost = R 16,00 + 0,014 kg × R 20 per kg

= R16,00 + R0,28

= R16,28

1C converting to kg

1SF substitution of answer

from 4.3.1 into the correct

formula

1CA simplifying

Accept R16,26 to

R16,30

(3)

12.2.3

12.3.2

� CA

� CA

� SF

� M

� R

� C

� SF

� CA

� SF

� S

� SF

� CA

� C



Copyright reserved



5.1.1(a)

July and August

2A July and August

June and July 1 mark

August and Sept 1 mark

(2)

12.4.4

5.1.1(b)

February; May; September; December

1 A two months

1 A two months

Penalty of 1 mark if more

than four months

(2)

12.4.4

5.1.1(c)

October and November

2A October and November

Sept and Oct 1 mark

Nov and Dec 1 mark

(2)

12.4.4

5.1.2(a)

Interpretation as % difference:

Percentage change = – 4,1% – 3,9%

= – 8%

OR

Percentage change = 3,9% – (– 4,1%)

= 8%

Interpretation as % change:

Percentage change = 4,1 3,9

100%3,9

− −×

= – 205,13%

1RG reading from graph

1M subtracting

1CA simplifying

OR


1M subtracting

1CA simplifying

OR


1M calculating %

1CA simplifying


(3)

12.1.1

�A �A

�M

�CA

�A �A

�M

�CA

�RG

�RG

�A �A

�CA

�RG

�M



Copyright reserved


5.1.2

(b)

Cost in May = 92 % × R150

= 0,92 × R 150

= R 138

OR

Cost in May = R 150 – 8% of R 150

= R 150 – 0,08 × R 150

= R 138

1CA percentage

1M calculating cost

1CA simplifying

1CA percentage

1M calculating cost

1CA simplifying


(3)

12.1.3

5.2.1

Price of bicycle × 105,8% = R1 586,95

Price of bicycle

= %8,105

95,5861R OR

8,105

100

1

95,5861R×

= 058,1

95,5861R

= R1 499,95

OR

Let x be the price of the bicycle in November 2008

Price of bicycle: x + 5,8% of x = R1 586,95

1,058 x = R1 586,95

x = R1 499,95

1M dividing


1CA simplifying

1M use of equation


1CA simplifying

(3)

12.1.3

5.2.2

A = P( 1 + i )n

= R 5,45(1 + 0,058)6

= R 7,64

1SF substitution of P

1A value of i

1A value of n

1CA simplifying



(4)

12.1.3

�A

�CA

�A �A

�CA

�SF

�M

�M �A

�CA

�M

�CA

�M

�CA

�CA

�CA



Copyright reserved


5.3.1

BASKET OF FRUIT: MONTH-ON-MONTH CHANGES (2008)

-3

-4,5

1,9

-0,5

0,7

5

4,1

1,2

-0,5 -0,5

1

4,6

-6

-4

-2

0

2

4

6

8

Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sept. Oct. Nov. Dec.

Months

Per

cen

tag

e ch

an

ge

in p

rice

4A all points plotted correctly (1A for every three points plotted correctly)

1CA horizontal line between July and August

2CA joining point with straight lines

Maximum of 4 marks if bar graph drawn correctly

Maximum of 5 marks for correct shape but incorrect scale

(7)

12.2.2

�A

�A

�A

�A

�A

CURRENT FRUIT BASKET

OLD FRUIT BASKET

�A

�A



Copyright reserved


5.3.2(a)

The graphs show a similar trend of month-on-

month changes in prices as follows:

An increase from May to November

OR

A decrease from January to February;

OR

A decrease from April to May;

OR

An increase from May to July;

OR

An increase from May to August;

OR

Zero change from July to August

OR

An increase from September to November

OR

A decrease from November to December.

OR

NO trend from January to December

2 CA for the trend

(2)

12.4.4

5.3.2(b)

Prices are generally high in December and January

due to festive season, and tend to drop in February.

OR

Prices tend to increase in the winter months (May,

June, July) as fruit becomes scarce.

OR

Valid reasons like:

Political reason; economic; climatic; religious; no

trend-flactuations

2O Own opinion that is valid

for the trend chosen in 5.3.2(a)

(2)

12.4.4

TOTAL: 150

�CA

��O

�CA


MARKS: 150 TIME: 3 hours

This question paper consists of 12 pages and 6 annexures.

GRAAD 12


NOVEMBER 2010

NATIONAL SENIOR CERTIFICATE

GRADE 12

Mathematical Literacy/P2 2 DBE/November 2010

NSC


INSTRUCTIONS AND INFORMATION

1. This question paper consists of FIVE questions. Answer ALL the questions.

2. Answer QUESTIONS 1.2.2, 2.2.1, 3.1.2(a), 3.1.3(a) and 5.3.1 on the attached

ANNEXURES. Write your centre number and examination number in the spaces

provided on the ANNEXURES and hand in the ANNEXURES with your ANSWER

BOOK.

3. Number the answers exactly as they are numbered in the 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 the calculations clearly.

7. Round off ALL the final answers to TWO decimal places, unless stated otherwise.

8. Indicate units of measurement where applicable.

9. Write neatly and legibly.


NSC


QUESTION 1

1.1 Ma Ndlovu makes circular place mats and circular tablecloths out of material (fabric)

edged with beads and sells them. The place mats have a diameter of 30 cm.

The radius of the tablecloth is FOUR times the radius of a place mat.

The following formula may be used:

Circumference = 2π × radius, and using π = 3,14

1.1.1 (a) Calculate the circumference of the tablecloth. (4)

(b) She uses a beaded edging consisting of triangular segments to

decorate the edge of each tablecloth, as shown in the diagrams

below. Each segment of the beaded edging is 4,71 cm long.

Circular tablecloth Enlargement of beaded triangular segments

Calculate the number of beaded segments that she will need for

each tablecloth.

(2)

4,71 cm


NSC


1.2 Ma Ndlovu has a landline telephone. A service provider has offered her a choice of

two different call packages.

CALL PACKAGE 1:

• Monthly rental of R150 • First 100 minutes are free • Calls cost R0,50 per minute

CALL PACKAGE 2:

• Monthly rental of R300 • First 500 minutes are free • Calls cost R0,50 per minute

1.2.1 The total cost for CALL PACKAGE 1 is given by the following formula:

Total cost (in rand) = R150 + R0,50 × (number of minutes more than 100)

(a) Write down a formula which can be used to calculate the total cost

(in rand) for CALL PACKAGE 2.

(2)

(b) If CALL PACKAGE 2 is used, determine the total cost, in rand, if

Ma Ndlovu and her family have made calls with a total duration of

510 minutes.

(4)

1.2.2 The line graph illustrating the total cost for CALL PACKAGE 2 has

already been drawn on ANNEXURE A.

On the same system of axes, draw a line graph to illustrate the total cost

for CALL PACKAGE 1.

(5)

1.2.3 (a) Define what is meant by the concept break-even point. (2)

(b) Write down the number of minutes used per month as well as the

total cost at the break-even point.

(2)

1.2.4 Ma Ndlovu wants to spend a maximum of R550 per month on one of the

call packages.

Which CALL PACKAGE would you advise Ma Ndlovu to accept?

Motivate your answer by showing ALL the calculations.

(5)

[26]


NSC


QUESTION 2

Freedom High School's soccer team is taking part in a football tournament at iMbali in the

iMbali Soccer Stadium.

2.1 On his way to iMbali, while travelling in a north-easterly direction, the driver of the

school bus stopped in Selby Msimang Road (refer to the map on ANNEXURE B) to

consult his map for directions to the iMbali Soccer Stadium.

Use the map on ANNEXURE B to answer the following:

2.1.1 Give the grid reference for the iMbali Soccer Stadium. (2)

2.1.2 In which general direction is the iMbali Soccer Stadium from where the

bus stopped?

(2)

2.1.3 (a) Describe the shortest possible route that the bus driver should take

from the point where the bus stopped to the entrance of the iMbali

Soccer Stadium, which is in Nkugwini Road.

(4)

(b) Hence, use a ruler to measure (in millimetres) the approximate

distance of this shortest route on the map, and then calculate the

actual distance, in kilometres, using the given scale.

(4)

2.1.4 At 09:15, after looking at the map, the bus driver was ready to start

driving again. He contacted the tournament coordinator to inform her

that they would be at the stadium at 09:20. If the bus travelled at an

average speed of 40 km/h, verify by means of relevant calculations

whether the bus driver's estimated time of arrival was correct.

The following formula may be used:


distance

(6)


NSC


2.2 At Freedom High School the basic boys' uniform consists of a pair of pants and a shirt

with the option of wearing a tie. The pants may be either long or short, and the shirt

may be either long-sleeved or short-sleeved. They are allowed to wear any

combination of these three items of clothing when they are on a trip.

2.2.1 Complete the tree diagram on ANNEXURE C to illustrate ALL the possible

combinations of these three items of clothing that the boys may wear on a

trip.

(7)

2.2.2 When the boys are at school, they are only allowed to wear ONE of the

following combinations of the uniform:

• Long pants with a long-sleeved shirt and a tie • Short pants with a short-sleeved shirt and no tie

If ONE of the boys in the bus were randomly selected, use the completed tree

diagram on ANNEXURE C to determine the probability (in decimal form)

that he would be wearing one of these two combinations.

(3)

[28]


NSC


QUESTION 3

3.1 Mr Riet is a secretary at a school and currently earns a gross monthly salary of

R7 986,50.

The following amounts are deducted from his gross

monthly salary:

Union Membership Fee: R35

Pension Fund: 7,5% of gross salary

PAYE (Tax): (Gross salary – R4 750) × 18%

Medical Aid:

31 of the total medical aid subscription due, as shown in

TABLE 1 below.

Gross Salary The salary

before pension,

tax, medical aid,

etc. have been

deducted.

Net Salary The 'take-home'

salary after

pension, tax,

medical aid, etc.

have been

deducted.

TABLE 1: Medical aid membership subscription costs SUBSCRIPTION COSTS GROSS

MONTHLY SALARY

MAIN MEMBER

WIFE/PARTNER/ PARENT EACH CHILD

R0 – R8 000 R1 152 R816 R424

R8 001 – R10 500 R1 256 R900 R468

More than R10 500 R1 396 R992 R516

3.1.1 Mr Riet, his wife and two children belong to a medical aid fund.

Use TABLE 1 to calculate his monthly contribution to the medical aid

fund.

(5)

3.1.2 (a) Calculate the total deductions from Mr Riet's monthly salary. Show

ALL calculations on ANNEXURE D.

(5)

(b) Hence, calculate Mr Riet's net annual salary. (3)

3.1.3 Mr Riet receives a 4,5% salary increase. His union membership fee

remains the same. Mr Riet states that his salary increase makes no

difference to his net salary.

(a) Determine whether Mr Riet's statement is valid by showing ALL

relevant calculations on ANNEXURE E.

(10)

(b) Hence, calculate the percentage change in his net annual salary. (3)


NSC


3.2 Mr Riet wanted to show his colleagues that the South African government was spending

more on education than on most other departments.

The two graphs below show the budgeted government expenditure for the financial years

2009/2010 and 2010/2011.

The total expenditure budgeted for 2009/2010 was R834,3 billion and for 2010/2011 was

R900,9 billion.

Budgeted Government Expenditure 2009/2010

E17%

F14%

G4%

H15%

D10%

C9%

B22%

A9%

1 billion = 1 000 million

KEY A: Public order and safety E: Education

B: Economic affairs F: Social protection

C: Housing and community amenities G: Defence

D: Health H: Other [Source: www.sars.gov.za]

NOTE: The names of the national government departments have since changed as per

President's Minute 690.

3.2.1 Show that the difference between the amounts budgeted for education

for the financial years 2009/2010 and 2010/2011 is more than

R20 000 000 000.

(8)

3.2.2 Give TWO possible reasons why you think the South African government

should increase its budgeted expenditure for education.

(4)

[38]

Budgeted Government Expenditure 2010/2011

E

F15%

G4%

H15%

D11%

C10%

B17%

A10%


NSC


QUESTION 4

Triggers Enterprises was awarded the tender for making rectangular cardboard boxes to package

bottles of cough syrup. Each bottle is packed in a cardboard box with a square base, as shown

below.

• The diameter of the base of the bottle is 58 mm and the height of the box is 143 mm. • The length of the side of the base of the box must be approximately 105% of the diameter of

the base of the bottle. • The height of the box is approximately 102% of the height of the bottle.

The following formulae may be used:

Area of circle = π × (radius)² , and using π = 3,14 Area of square = (side length)² Area of rectangle = length × breadth Area of opened cardboard box = 4(A + D) + 2(B + C) + E (See design of open cardboard box in QUESTION 4.3)

The following conversions may be useful:

1 cm² = 100 mm²

1 m² = 10 000 cm²

4.1 Calculate the height of the bottle to the nearest millimetre. (3)

4.2 In order to minimise the cost of cardboard required for the box, the following

guideline is used:

The difference between the areas of the base of the cardboard box and the base of the bottle should not be more than 11 cm2.

Determine whether the dimensions of this cardboard box satisfy the above guideline.

Show ALL appropriate calculations.

(11)

Side length

Cough

Syrup 143 mm


NSC


4.3 To ensure that the box is strong enough, the cardboard used for the box has a mass of

240 grams per m2 (g/m

2).

The layout of the opened cardboard box is shown below.

Picture of opened box

Diagram of layout of opened box

• Section C is semicircular. • The area of each section D = 1 832 mm2. • The area of section E = 2 855 mm

2.

4.3.1 Calculate the total mass of the cardboard needed for one box, to the

nearest gram.

(11)

4.3.2 The total cost of the cough syrup includes the cost of the cardboard box.

Use the following formula to calculate the cost of a boxed bottle of cough

syrup:

Total cost = R16,00 + (mass of cardboard box) × R20,00 per kg

(3)

[28]

A D

D

A A A A E

B D

B D

C

C

A

D


NSC


QUESTION 5

The Consumer Price Index (CPI) is the price of a collection (basket) of goods and services. The

prices of these goods and services are collected every month, and the total cost of a basket is

compared to that of the previous month's total cost.

The CPI is the official measure of inflation . Inflation is generally given as a percentage, and is

the measure of how much the price of goods and services have increased over a period of time.

In 2008, Statistics South Africa decided that a fruit basket should consist of bananas, apples,

oranges and lemons since these types of fruit are generally available throughout the year.

The graph drawn on ANNEXURE F shows the month-on-month changes in the CPI from

January 2008 to December 2008 expressed as a percentage.

5.1 Use the graph drawn on ANNEXURE F to answer the following questions:

5.1.1 (a) Name the consecutive months between which there was no change

in the CPI.

(2)

(b) During which months was the CPI less than that of the previous

month?

(2)

(c) Name the consecutive months between which the increase in the

CPI was the greatest.

(2)

5.1.2 (a) Determine the percentage change in the CPI from April 2008 to

May 2008.

(3)

(b) Hence, calculate the price of the fruit basket in May 2008 if it cost

R150,00 in April 2008.

(3)

5.2 In November 2009 Statistics South Africa announced that the annual inflation rate

was 5,8%.

5.2.1 Determine the price of a bicycle in November 2008 if it cost R1 586,95 in

November 2009.

(3)

5.2.2 Calculate the projected cost of a loaf of brown bread in November 2014 if

it cost R5,45 in November 2008. Assume the annual inflation rate

remained at 5,8% over the given period.

The formula A = P(1 + i)n may be used, where:

A

P

n i

= projected cost

= current cost

= number of years

= annual inflation rate

(4)


NSC

Copyright reserved

5.3 Prior to 2008 a fruit basket consisted of fruit which was not available throughout the year.

Statistics South Africa refers to this fruit basket as the 'old' fruit basket.

In 2008, Statistics South Africa kept a record of the monthly costs of both the 'old' fruit basket

and the 'current' fruit basket.

The graph on ANNEXURE F shows the changes in the CPI for the 'current' fruit basket.

TABLE 2 below shows the month-on-month changes in the CPI from January 2008 to December

2008 for the 'old' fruit basket.

TABLE 2: Month-on-month changes in the CPI for the 'old' fruit basket from January 2008 to December 2008

Jan. Feb. Mar. Apr. May Jun. Jul. Aug. Sep. Oct. Nov. Dec.

Percentage

change 4,6 1,9 1,0 – 0,5 – 4,5 – 3 – 0,5 – 0,5 0,7 1,2 5,0 4,1

5.3.1 Use the table to draw a labelled graph, on the grid provided on

ANNEXURE F, to represent the month-on-month changes in the CPI of the

'old' fruit basket for the given period.

(7)

5.3.2 (a) Describe clearly any possible trend shown by the graphs of the two

fruit baskets.

(2)

(b) Give ONE possible reason for the trend identified in

QUESTION 5.3.2(a).

(2)

[30] TOTAL: 150

Mathematical Literacy/P2 DBE/November 2010

NSC

Copyright reserved

ANNEXURE A CENTRE NUMBER:

EXAMINATION NUMBER: QUESTION 1.2.2

LANDLINE CALL PACKAGES

0

100

200

300

400

500

600

700

0 100 200 300 400 500 600 700 800 900 1000

Number of minutes used per month

Tot

al c

ost i

n ra

nd

CALL PACKAGE 2


NSC

Copyright reserved

ANNEXURE B: MAP OF A SECTION OF IMBALI

QUESTION 2.1

Traffic lights

iMbali Soccer Stadium

C

A

B

1

2

3

4

N

School bus

Scale 1 : 20 000


NSC

Copyright reserved

ANNEXURE C

CENTRE NUMBER:


CODES FOR THE TREE DIAGRAM CODE EXPLANATION

LP Long pants

SP Short pants

LS Long sleeves

SS Short sleeves

T Tie

NT No tie

PANTS SHIRT TIE ALL POSSIBLE OUTCOMES

T → LP; LS; T LS NT →

LP → SS → → →

SP → →


NSC

Copyright reserved

ANNEXURE D

CENTRE NUMBER:

EXAMINATION NUMBER: QUESTION 3.1.2(a)

MONTHLY DEDUCTIONS 3.1.2(a)

A

Union membership fee

R35,00

B

Pension fund


C

PAYE

= (Gross salary – R4 750) × 18%

D

Medical aid contribution

E

TOTAL = A + B + C + D


NSC

Copyright reserved

ANNEXURE E

CENTRE NUMBER:

EXAMINATION NUMBER: QUESTION 3.1.3(a)

MONTHLY DEDUCTIONS 3.1.3(a)

A

Union membership fee

R35,00

B

Pension fund


C

PAYE

= (Gross salary – R4 750) × 18%

D

Medical aid contribution

E

TOTAL = A + B + C + D


NSC

Copyright reserved

ANNEXURE F

CENTRE NUMBER:


MONTH-ON-MONTH CHANGES IN THE CPI (2008) OF A FRUIT BASKET

-6

-4

-2

0

2

4

6

8

MONTHS

Per

cent

age

chan

ge in

pric

e

[Source: Statistics South Africa, 2009]

Jan. Feb. Mar. Apr. Dec. May Jun. Jul. Aug. Sep. Oct. Nov.

CURRENT FRUIT BASKET

– 2,5 – 1,8

– 4,1

– 2,3

– 1

– 0,5

3,9

5,8

4,9

national senior certificate grade 12€¦ · marks: 150 symbol explanation m method ma method with...

Documents