national senior certificate grade 12 · this question paper consists of 16 pages and 3 annexures....

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This question paper consists of 16 pages and 3 annexures.

MATHEMATICAL LITERACY P1

NOVEMBER 2012

NATIONAL SENIOR CERTIFICATE

GRADE 12

Mathematical Literacy/P1 2 DBE/November 2012 NSC


INSTRUCTIONS AND INFORMATION 1. This question paper consists of SIX questions. Answer ALL the questions. 2. Answer QUESTION 4.1.7, QUESTION 6.3.3 and QUESTION 6.4.1 on the attached

ANNEXURES. Write your centre number and examination number in the spaces on the ANNEXURES and hand in the ANNEXURES 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 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. Maps and diagrams are NOT necessarily drawn to scale, unless stated otherwise. 10. Write neatly and legibly.



QUESTION 1

1.1

1.1.1 Simplify: 1 441,62 – 13,268,72 −

(2) 1.1.2 Write 0,0528 as a common fraction in simplified form. (2) 1.1.3 Convert 23,005 litres to millilitres. (2) 1.1.4 Determine the total price of 2,5 kilograms of meat costing R63,99 per

kilogram.

(2) 1.1.5 Shameeg had to attend a meeting that was scheduled to start at 13:15.

At what time did he arrive at the meeting if he arrived 1 hour 18 minutes early?

(2) 1.1.6 Convert R3 850 to euros (€) if the exchange rate is €1 = R10,2584. (2) 1.1.7 State whether the following event is CERTAIN, MOST LIKELY or

IMPOSSIBLE: Christmas Day is on 25 December in South Africa. (2)

1.1.8 The price per litre of diesel at nine different garages is:

R9,97 R9,97 R10,12 R10,17 R10,29 R10,79 R10,79 R10,79 R10,95 Determine the median price per litre of diesel.

(2)

1.2 Miss Lena asked all the learners in her class what their favourite fruit juice was. She

illustrated the data in the bar graph below.

How many learners does she have in her class? (3)

Mango

Orange

Pineapple

0 12 24 36

FAVOURITE FRUIT JUICE

Number of learners

Typ

es o

f jui

ce



1.3 Mrs Rose received a cash-sale slip after she bought some goods at CT-Haven at the

Cape Town International Airport. Below is a copy of the cash-sale slip with some of the details omitted. NOTE: VAT is value-added tax.

1.3.1 How much did Mrs Rose pay in total for the THREE slabs of chocolate? (2) 1.3.2 How many bangles did Mrs Rose buy? (2) 1.3.3 A Joy magazine costs R21,89 excluding VAT. Calculate the amount of

VAT paid on the Joy magazine.

(2) 1.3.4 Calculate the total (excluding VAT) for the goods bought. (3)

CT-HAVEN

Cape Town International Airport Domestic Departures, Opposite Gate 8

Tel: (+2721) 1234567 VAT Reg# 461010565

TAX INVOICE

1705359 Reg 1 ID 41 14:54 01/11/11 CHOCOLATE SLAB 3 @ 14,95 … BANGLES ... @ 13,95 97,65 JOY MAGAZINE 1 @ 24,95 24,95 SUBTOTAL 167,45 TOTAL (EXCLUDING VAT) ... TOTAL (INCLUDING VAT) 167,45 CASH PAYMENT 167,45 AMOUNT TENDERED 200,00 CHANGE 32,55

Receipt total includes 14% VAT

RETAIN AS PROOF OF PURCHASE



1.4 South Africa imports crude oil from different countries. TABLE 1 below shows

crude oil imports during 2010 and 2011. TABLE 1: Crude oil imports during 2010 and 2011

COUNTRY AMOUNT OF CRUDE OIL (IN MILLIONS OF TONS)

2010 2011 Angola 3,409 1,948 Iran 5,528 4,874 Nigeria 3,594 3,755 Saudi Arabia 4,584 4,793 Other countries 2,139 2,264

[Source: Business Times, 1 April 2012]

1.4.1 Calculate the total amount of crude oil imported during 2011. (2) 1.4.2 From which country did South Africa import most of its crude oil during

2010 and 2011?

(2) 1.4.3 Which country showed the largest increase in the amount of crude oil

exported to South Africa between 2010 and 2011?

(2) [34]



QUESTION 2 2.1 Didi is a contestant in a game show where they spin a wheel. She can win a prize if

the arrow points to a specific colour after she spins the wheel and it stops. The diagram below shows a spin wheel that is divided into 24 equal parts called sectors. When someone spins the wheel, it is equally likely for the arrow to point to any one of the sectors when the wheel stops.

One half of the sectors are grey, one third of the sectors are white, 81 of the sectors are

black and 241 of the sectors are spotted.

2.1.1 How many white sectors are there on the spin wheel? (2) 2.1.2 Didi spins the wheel. Which sector is the arrow LEAST likely to be

pointing at when the wheel stops?

(2) 2.1.3 The wheel has a radius of 60 cm. (a) Calculate the circumference of the wheel.

Use the formula: Circumference of a circle = 2 π× × radius, using π = 3,14

(2) (b) Calculate the area of ONE of the sectors of the wheel.

Use the formula:

Area of a sector of a circle =n

2)radius(×π

where π = 3,14 and n = number of sectors

(3)

Pointer (arrow)



2.2 South Africa's Road Traffic Management Corporation reported that sending an SMS

(short message service) from a cellphone while driving, increases the reaction time needed to stop a vehicle in an emergency from 1,2 seconds to 1,56 seconds.

2.2.1 Calculate the percentage increase in the reaction time it takes to stop a

vehicle when sending an SMS while driving.

Use the formula:

Percentage increase in reaction time = 100%time original

time in difference×

(3) 2.2.2 Calculate the distance (in metres) that a car will travel in 1,36 seconds if it

is travelling at an average speed of 27,95 m/s. Use the formula: Distance = average speed × time

(2)

2.3 Two businessmen, Mr Nobi and Mr Khoza, travel from their home towns to Pretoria.

The distance from Pretoria and the time is indicated in the graph below:

2.3.1 At what time did Mr Khoza leave his home town? (2) 2.3.2 Which ONE of the two businessmen lives closer to Pretoria? (1) 2.3.3 How long did Mr Nobi take to travel to Pretoria? (2) 2.3.4 Estimate Mr Khoza's arrival time in Pretoria. (2) 2.3.5 At what time were the two businessmen exactly 100 km apart? (2)

0

100

200

300

07:30 08:30 09:30 10:30 11:30

Dis

tanc

e fr

om P

reto

ria

(in k

m)

Time

TRAVELLING TO PRETORIA

Mr Khoza

Mr Nobi



2.4 Kedibone has a cheque account with Iziko Bank. The bank charges a service fee up to

a maximum of R31,50 (VAT included) on all transaction amounts. TABLE 2 below shows five different transactions on Kedibone's cheque account. TABLE 2: Transactions on Kedibone's cheque account

NO. DESCRIPTION OF TRANSACTION

TRANSACTION AMOUNT

(IN R)

SERVICE FEE (IN R)

1 Debit order for car repayment 4 250,00 31,50 2 Debit order for cellphone contract 344,50 A 3 Personal loan repayment 924,00 14,59 4 Vehicle and household insurance B 11,85 5 Cheque payment 403,46 8,34

2.4.1 Calculate the missing value A, using the following formula:

Service fee (in rand) = 3,50 + 1,20% of the transaction amount

(3)

2.4.2 Calculate the missing value B, using the following formula:

Amount (in rand) =1,20%

3,50 fee service −

(3)

[29]



QUESTION 3 3.1 Mr De Haan and his family live in Mossel Bay and he intends buying a new car.

He sees the advertisement below for one of the cars that he is interested in.

R199 000 cash

or R19 900 deposit + R3 599,85 × 60 months

on hire purchase

3.1.1 Calculate the total cost of the car in the advertisement if it is bought on

hire purchase.

(2) 3.1.2 Mr De Haan decides to buy a new car in two years' time instead. He will

then sell his current car and use that money as the deposit for the new car. Currently the value of his car is R51 600. The value of the car depreciates at a rate of 13,5% per annum. Calculate (rounded off to the nearest R100) the depreciated value of his car in TWO years' time. Use the formula: A = P(1 – i )n where A = depreciated value P = current value i = annual depreciation rate n = number of years

(3)

3.2 Petrol consumption can be calculated using the following formula:

Petrol consumption (in litres per 100 km) = 5,12100

covered distance×

3.2.1 How many litres of petrol will Mr De Haan's car use to travel 100 km? (1) 3.2.2 Calculate the petrol consumption (in litres per 100 km) if Mr De Haan

covered a distance of 325 km.

(2)



3.3 Below is a street map of a part of the area where Mr De Haan lives.

3.3.1 Give the grid reference of the Van Riebeeck Sport Stadium. (2) 3.3.2 Write down the names of the streets on either side of the City Hall

Complex.

(2) 3.3.3 Mr De Haan drives out of the parking area of the Van Riebeeck Sports

Stadium and then turns right into George Street. He then turns left into Montague Street and continues driving until he reaches Marsh Street. In which direction must he turn if he wants to go directly to the entrance of the police station?

(2) 3.3.4 The distance measured on the map from Mr De Haan's house to the

entrance of the Bayview Hospital is 8,9 cm. Calculate the actual distance (in km) if 1 cm on the map represents 0,3 km.

(2) [16]



QUESTION 4 4.1 Lunje's dog gave birth to 9 puppies (6 males and 3 females).

Lunje's dog with her puppies

Lunje collected data from 10 of his friends whose dogs had puppies and summarised the data (including his own) in the table below. TABLE 3: Number of puppies in a litter* NAME OF DOG A B C D E F G H I J K

Litter size 14 6 7 9 14 12 11 8 14 8 11

Number of males 13 5 6 6 10 8 9 1 6 0 2

Number of females 1 1 1 3 4 4 2 7 8 8 9

*A litter is the number of puppies born at one birth.

4.1.1 Arrange the litter sizes in ascending order. (2) 4.1.2 Which dog had seven more females than males? (2) 4.1.3 Give the modal litter size. (2) 4.1.4 Determine the range of the number of females. (2) 4.1.5 Calculate the mean (average) number of males. (3) 4.1.6 Determine the ratio (in simplified form) of males to females for dog E. (2) 4.1.7 Use the information in TABLE 3 to complete the compound bar graph on

ANNEXURE A.

(7)



4.2 Lunje made a rectangular box for his dog to sleep in. This helps to keep the puppies

safe and comfortable. The dimensions of the box are as follows: • The width is the same as the length of the dog. • The length is 125% of the length of the dog. • The height is 6 inches.

Lunje's dog is 105 centimetres long.

Give the following dimensions in centimetres: 4.2.1 The length of the box (2) 4.2.2 The height of the box if 1 inch = 2,5 cm (2) [24]



QUESTION 5 5.1 Maria has a house in Qwaqwa. The floor plan of Maria's house showing the actual

exterior measurements is given below:

5.1.1 How many windows does Maria's house have? (1) 5.1.2 On the floor plan the exterior length of the northern wall is 70 mm.

Determine the scale of the floor plan in the form 1 : ...

(2) 5.1.3 Calculate the exterior side length of the house excluding the step section. (2) 5.1.4 The area of the kitchen is 72% less than the area of the living room.

Calculate the area (in m2) of the kitchen if the area of the living room is 39,54 m2.

(3)

Window

Living Area

N

Bathroom Bedroom

7 000 mm

Kitchen

Living room

Step 1 200 mm

10 714 mm

Front door



5.2 The step at the front door of Maria's house is in the shape of a symmetrical trapezium

based prism as shown below. The step is made of concrete. The top (A) and sides (B and C) will be tiled.

The dimensions of the step are as follows: f = length of the front of the step = 1,3 m s = length of the slanting side = 1,6 m h = height of the step = 0,12 m

A = Area of the trapezium = 2,52 m2 B = Area of the slanting side of the step C = Area of the front of the step

5.2.2 Calculate the volume of concrete (in m3) required for the step.

Use the formula: Volume of the step = area of the trapezium × height of the step

(2)

5.2.3 Maria wants to tile the top and side surfaces of the step. Calculate,

rounded off to ONE decimal place, the total area that will be tiled. Use the formula: Total tiled area (in m2) of the step = A + (2s + f) × h

(4)

5.2.4 Maria decides to put a metal strip on the top edge of the step. Calculate

the length of the strip. Use the formula: Total length of the strip = f + 2s

(2)

[19]

5.2.1 Concrete is made by adding water to a mixture of cement, sand and stone in the ratio: cement : sand : stone = 1 : 2 : 4 How many wheelbarrows of stone will Maria need for 2

11 bags of cement if one bag of cement equals one wheelbarrow of cement?

(3)

A B B

C

f

h

s



QUESTION 6

6.1 Gracia is an athlete and is training for a 42,2 km standard marathon to be held in four weeks' time. She wants to finish the race in less than 3 hours. Gracia's training schedule involves endurance training and speed training. To build muscle strength she does strength exercises and long-distance running at a slow pace.

Gracia runs 450 metres in 4 minutes at a constant pace. Calculate the distance she will

cover in 9 minutes if she runs at the same constant pace.

(2)

6.2 Other preparation for the race involves 'carbo-loading'. Carbo-loading means following a special diet that will increase the amount of glycogen in your muscles so that the muscles can endure long periods of physical strain/activity. According to the Tips For Endurance Athletes (www.beginnertriathlete.com), an athlete requires between 1,4 and 2,27 grams of carbohydrates per kilogram of body mass per meal.

Calculate the MAXIMUM number of grams of carbohydrates Gracia requires per

meal if she weighs 65 kg.

(3)

6.3 Gracia is sure that her training will allow her to finish the race in less than 3 hours. She doesn't want to start the race too fast and fade (grow tired and run slowly) at the end or start too slowly and then finish later than her targeted time. In order to plan her race, Gracia constructed a table showing the time (in minutes) and the required distance (in km) she needs to cover during the race. TABLE 4: Gracia's plan for the race

Time after start of race (in minutes)

15 30 45 60 75 90 105 120 135 150 165

Distance she needs to cover (in km)

3 6 9 13 17 21 26 31 35 39 42,2

6.3.1 Gracia managed to complete the race in her planned time. How many minutes

did she take to finish the race?

(1) 6.3.2 Calculate the average pace (in kilometres per minute) she needs to maintain

from the 60th to the 90th minute of the race. Use the formula:

timestwothebetweendifferencedistancestwothebetweendifference

timeinchangedistanceinchangeminute)perkm(inpaceAverage

=

=

(4)

6.3.3 Use TABLE 4 to draw a line graph on ANNEXURE B representing Gracia's

plan for the race.

(8)


Copyright reserved

6.4 Titus, who was a marshal at the race, was stationed at the halfway point. 6.4.1 Titus kept the following record of the athletics clubs of the first 20

athletes who ran past him. Athletics Clubs: Liberty Striders Harmony Ramblers Striders Harmony Striders Ramblers Ramblers Harmony Liberty Harmony Liberty Liberty Striders Liberty Harmony Ramblers Striders Harmony

Complete, on ANNEXURE C, the frequency table representing the

athletic clubs of the first 20 athletes.

(4) 6.4.2 The data of the club membership of the top 300 athletes that finished the

race is represented in the pie chart below.

Club membership of the top 300 athletes

Key to the chart

A Other B Striders C Harmony D Ramblers E Liberty

(a) What percentage of the top 300 athletes belonged to the Striders

Club?

(2) (b) Which club had the second largest number of athletes in the top 300? (2) (c) Calculate the actual number of Ramblers athletes that finished in the

top 300.

(2) [28]

TOTAL: 150

A 8%

B

C 35%

D 12%

E 29%

Mathematical Literacy/P1 DBE/November 2012 NSC

Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE A QUESTION 4.1.7 TABLE 3: Number of puppies in a litter NAME OF DOG A B C D E F G H I J K

Litter size 14 6 7 9 14 12 11 8 14 8 11

Number of males 13 5 6 6 10 8 9 1 6 0 2

Number of females 1 1 1 3 4 4 2 7 8 8 9

0

2

4

6

8

10

12

14

16

A B C D E F G H I J K

Num

ber

of p

uppi

es

Name of dog

THE LITTER SIZE OF 11 DOGS

Females Males


Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE B QUESTION 6.3.3 TABLE 4: Gracia's plan for the race Time after start of race (in minutes)

15 30 45 60 75 90 105 120 135 150 165

Distance she needs to cover (in km)

3 6 9 13 17 21 26 31 35 39 42,2

0

5

10

15

20

25

30

35

40

45

50

0 15 30 45 60 75 90 105 120 135 150 165 180

Dis

tanc

e (in

km

)

Time (in minutes)

GRACIA'S PLAN FOR THE RACE


Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE C QUESTION 6.4.1 ATHLETICS CLUB FREQUENCY Liberty Striders Ramblers Harmony


MARKS: 150

Symbol Explanation M Method M/A 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

PLEASE NOTE: 1. If a candidate deletes a solution to a question without providing another solution, then the deleted solution must be marked. 2. If a candidate provides more than one solution to a question, then only the first solution must be marked and a line drawn through any other solutions to the question.

This memorandum consists of 15 pages.

_______________________ ______________________ ______________________ EXTERNAL MODERATOR EXTERNAL MODERATOR INTERNAL MODERATOR MR MA HENDRICKS MR RI SINGH MRS J SCHEIBER 15 NOVEMBER 2012 15 NOVEMBER 2012 15 NOVEMBER 2012


NOVEMBER 2012

FINAL MEMORANDUM


GRADE 12

Mathematical Literacy/P1 2 DBE/November 2012 NSC – Final Memorandum 13 November


Rounding off penalty once only in question 5 QUESTION 1 [34 MARKS] Correct answer only: Full marks Ques Solution Explanation AS/L 1.1.1

1 441,62 – 26,137,8 2 −

= 1441,62 – 43,62 = 1441,62 – 7,9012... = 1 433,718734 ≈1 433,72

1S simplifying 1CA simplification

(2)

12.1.1 L1

1.1.2

0,0528 =00010

528 = 62533

1A writing as a common fraction 1CA simplifying

(2)

12.1.1 L1

1.1.3

23,005 = 23,005× 1 000 m = 23 005 m

1M/A multiplying by 1 000 1CA simplification if multiplied by power of 10

(2)

12.3.2 L2

1.1.4

R63,99/kg× 2,5 kg = R159,975 ≈R159,98 (accept R159,97 - no rounding penalty)

1M/A multiplication 1CA simplification to nearest cent

(2)

12.1.1 L1

1.1.5

13h15 min – 1h18 min = 11h57 min Shameeg arrived at 11:57. OR 3 minutes to 12

1M/A subtracting 1h18 min 1CA arrival time

(2) (Accept 11H57)

12.3.2 L2

1.1.6

2584,108503€

= €375,30

1M/A dividing 1CA simplification

(2)

12.1.3 L2

1.1.7

CERTAIN

2A conclusion

(2)

12.4.5 L2

1.1.8

R10,29

2A median

(2)

12.4.3 L1

S

CA

CA M/A

CA

CA

M/A

CA

M/A

CA

M/A

A

A

A



Ques Solution Explanation AS/L 1.2

21 + 30 + 9 = 60

1A one correct reading from graph 1A correct reading of the other two values from graph 1CA total of the three (values within the range)

(3)

12.4.4 L1 (1) L2 (1)

1.3.1

3 × R14,95 = R44,85 OR R167,45 – 24,95 – 97,65 = R44,85

1M/A multiplying 1CA simplification (CA only when using R14,95 or multiplying 3 with a price on the slip) OR 1M/A subtracting the values from the total 1CA the amount

(2)

12.1.3 L1

1.3.2

95,1365,97

= 7 bangles

1M/A dividing 1CA simplification

(2)

12.1.3 L1

1.3.3

R24,95 – R21,89 OR 14% of R21,89 = R3,06 OR

R24,95 × 11414 = R3,06

1M/A subtracting/ calculating percentage 1CA simplification to the nearest cent OR 1 M/A multiplying 1 CA simplification to the nearest cent

(2)

12.1.3 L1

M/A

CA

CA

M/A

M/A

CA

A CA

M/A

CA

M/A

CA



Ques Solution Explanation AS/L 1.3.4

%114R167,45

= R 146,89 OR

×114100 R167,45

= R146,89 OR

VAT = R167,45 × 11414 = R20,56

Total without VAT = R167,45 – R20,56 = R146,89

1M dividing 1A correct values 1CA simplification OR 1M dividing 1A correct values 1CA simplification OR 1 M calculating VAT 1A correct values 1CA simplification (if 14% is calculated : 0 marks)

(3)

12.1.3 L2

1.4.1

(1,948 + 4,874 + 3,755 + 4,793 + 2,264) millions of tons = 17,634 millions of tons OR 17 634 000 tons

1 M/A adding 1CA total ( if using the wrong data set: max 1 mark)

(2)

12.1.2(1) 12.4.4(1) L1

1.4.2

Iran

2A correct country (extra country: 0 marks)

(2)

12.4.4 L 1

1.4.3

Saudi Arabia

2A correct country

(2)

12.4.4 L1

[34]

M/A

CA

A

A

CA

A M

CA

A M

M A

CA



QUESTION 2 [29 MARKS] Ques Solution Explanation AS/L 2.1.1

×31 24 = 8

1M multiplying 1A simplification Correct answer only: full marks

(2)

12.1.1 L1

2.1.2

Spotted sector

2A correct sector (accept dotted sector, black & white sector)

(2)

12.4.5 L2

2.1.3 (a)

Circumference = 2 × 3,14 × 60 cm = 376,8 cm (Using π: 376,99 cm)

1SF substitution 1CA simplification

(2)

12.3.1 L1

2.1.3 (b)

Area of a sector of a circle = cm²24

60 14,3 2×

= cm²24304 11

= 471 cm² (using π: 471,24 cm²)

1SF substitution [refer to radius used in 2.1.3 (a)] 1CA simplification 1A square unit shown anywhere in solution

(3)

12.3.1 L1

2.2.1

%30

%1002,1

2,156,1

in time increasePercentage

=

×−

=

×= 100%time original

time in Difference

OR 0,3

1SF difference in time 1SF substituting 1,2 1CA simplification ( no subtraction no CA)

(3)

12.1.1 L2

2.2.2

Distance = (27,95 × 1,36) m = 38,012 m ≈38,01 m

1SF substitution

1A simplification (2)

12.2.1 L1

M A

SF

A

SF

SF

SF

SF

CA

A (any one)

CA A

CA




09:00 or nine o' clock or 9 am

1RG reading from graph

(2)

12.2.3 L1

2.3.2

Mr Nobi


(1)

12.2.3 L2

2.3.3

2 hours or 3 hours


(2)

12.2.3 L2

2.3.4

10:47 (accept any time from 10:45 to 10:50)


(2)

12.2.3 L2

2.3.5

09:00 or nine o' clock or 9 am


(2)

12.2.3 L2

2.4.1

Service fee (in rand)

= 3,50 + 1,20% of the transaction amount

= 3,50 + 1,20% × 344,50

= 3,50 + 4,134

≈ 7,63

1SF substituting 344,50

1A simplification

1CA amount to the nearest cent Correct answer only if correctly rounded : full marks

(3)

12.2.1 L1 (2) L2 (1)

2.4.2

Amount (in rand) =%20,1

3,50 fee Service −

= %20,13,50 11,85 −

= 012,035,8

≈ 695,83

1SF substitution of 11,85 1A simplification 1CA amount to the nearest cent

(3)

12.2.3 L1

[29]

CA

A

SF

CA

SF

A

RG

RG

RG

RG

RG




R19 900 deposit + R3 599,85 × 60 months = R19 900 + R215 991 = R235 891

1S simplification 1CA simplification Correct answer only: full marks

(2)

12.1.3 L1

3.1.2

( )

600 R38608,41 R38

%5,311600 R51

)P(1A2

n

≈=

−=

−= i

1 SF correct substitution 1CA simplification

1 R rounding to the nearest R100 Correct answer only: full marks

(3)

12.1.3 L2

3.2.1

12,5

1A conclusion

(1)

12.2.1 L1

3.2.2

Petrol consumption (in litre per 100 km)

= 5,12100

covered distance×

= 5,12100325

×

= 40,625 ≈ 40,63 OR Petrol consumption (in litre per 100 km) = 12,5 × 3,25 = 40,625 ≈ 40,63

1SF substitution 1CA simplification 1SF substitution of factor 3,25 1CA simplification Correct answer only: full marks

(2)

12.2.1 L2

SF

R CA

S

CA

SF

CA (any one)

A

SF

CA (any one)




C 4 OR 4 C

1A C 1A 4

(2)

12.3.4 L2

3.3.2

Long Street and Marsh Street (or High Street)

2A any two correct (1 Penalty if other street names are given)

(2)

12.3.4 L1

3.3.3

Right (accept Easterly direction)

2A conclusion

(2)

12.3.4 L2

3.3.4

1 cm represents 0,3 km ∴8,9 cm represents 0,3 km × 8,9 = 2,67 km OR 1 : 0,3 ∴ 8,9: 0,3 × 8,9 ∴ 8,9 : 2,67

1M multiplying by 8,9 1 A simplification 1M multiplying by 8,9 1 A simplification (If unit is incorrect: 1 mark)

(2)

12.3.3 L2

[16]

M A

M A

A

A A A A

A A




6 7 8 8 9 11 11 12 14 14 14

1M ascending order 1 A all correct (descending order: 1 mark, one number omitted: 1 mark, Using names of the dogs: 1 mark)

(2)

12.4.3 L1

4.1.2

Dog K

2A conclusion (Dog G: give 1 mark)

(2)

12.1.1 (1) 12.4.4 (1) L1

4.1.3

14

2A mode OR CA from 4.1.1

(2)

12.4.3 L1

4.1.4

Range = 9 – 1 = 8

1M identifying 1 and 9 1CA range

(2)

12.4.3 L2

4.1.5

Mean = 11

2061981066513 ++++++++++

= 1166

= 6

1M sum of the values (no penalty for omitting 0) 1M dividing by 11 1CA mean Correct answer only: full marks

(3)

12.4.3 L2

4.1.6

10 : 4 = 5 : 2

1A correct ratio 1CA simplified ratio (unit ratio 1: 0,4 or 2,5 : 1 give 1 mark; written as a fraction 0 marks; Inverting the ratio 1 mark) Correct answer only: full marks

(2)

12.1.1 (1) 12.4.4 (1) L1

A

CA

A

CA

M

M

CA

M A

A

M



Ques Solution Explanation AS/L

4.1.7

1A for each bar drawn correctly (correct litter size only, max 3 marks)

(7)

12.4.2 L2

4.2.1

105 cm × 1,25 OR 105 cm ×100125

= 131,25 cm = 131,25 cm

1M multiplying 1A length Correct answer only: full marks

(2)

12.3.1 L1

4.2.2

6 × 2,5 cm = 15 cm

1M multiplying 1A height Correct answer only: full marks

(2)

12.3.2 L2

[24]

0

2

4

6

8

10

12

14

16

A B C D E F G H I J K

Num

ber

of p

uppi

es

Name of Dog

THE LITTER SIZE OF 11 DOGS

Female

Male

M

A

A

M M

A

A A

A

A

A

A

A



QUESTION 5 [19 MARKS] Once off penalty for rounding off Ques Solution Explanation AS/L

5.1.1 7

1A conclusion

(1)

12.3.1 L1

5.1.2 70 mm : 7 000 mm = 1: 100

1M/A correct ratio 1CA simplification

(2) Note: AFRIKAANS additional options

12.3.1 L1

5.1.3

10 714 mm – 1 200 mm = 9 514 mm OR Perimeter = 7 000 + 9 514 + 7 000 + 9 514 = 33 028 mm

1M/A subtraction 1CA simplification OR 1 M finding perimeter 1 CA simplification (no penalty for units)

(2)

12.3.1 L1

5.1.4

72% × 39,54 m2 ≈ 28,47 m2 ∴ area of the kitchen = 39,54 m2– 28,47 m2

= 11,07 m2

OR 100% – 72% = 28% ∴ area of the kitchen =28% × 39,54 m2

≈ 11,07 m2

1M % concept 1M concept of decrease of area 1CA simplification OR 1M concept of decrease of % 1M % concept 1CA simplification (no penalty for units)

(3)

12.3.1 L2

M

CA

M

CA

A

M/A CA

M/A CA

M

M

M CA




5.2.1 cement : stone = 1 : 4 1,5 bags of cement = 1,5 wheelbarrows of cement For 1 2

1 wheelbarrows of cement,

she will need 4 × 1 21 wheelbarrows of stone

= 6 wheelbarrows of stone

1M concept 1M multiply by 4 1CA simplification Correct answer only: full marks

(3)

12.3.1 L2

5.2.2 Volume of the step = Area of the trapezium × height of the step = 2,52 m2 × 0,12 m = 0,3024 m3

≈ 0,30 m3 or 0,3

1SF substitution 1A simplification (no penalty for units)

(2)

12.3.1 L2

5.2.3 Total tiled area (in m2) = A + (2s+f)× h = 2,52 + (2 ×1,6+1,3) × 0,12 = 3,06 ≈ 3,1

1 SF substitution two correct 1 SF substitution another two correct 1CA simplification 1R rounding

(4)

12.3.1 L2

5.2.4

Total length of the strip = 1,3 m + 2 × 1,6 m = 4,5 m

1SF substitution 1CA simplification

(2)

12.2.1 L1

[19]

SF CA R

SF CA

SF

A

M M

CA



QUESTION 6 [28 MARKS] Ques Solution Explanation AS/L 6.1

In 4 minutes she covers 450 m

∴1 minute she covers 4

450 m = 112,5 m

∴in 9 minutes she covers 112,5 ×9 m = 1 012,5 m OR 4 minutes: 450 m

9 minutes: 4

9450 × m = 1012,5 m

1M working with ratio 1CA simplification OR 1M working with ratio 1CA simplification

(2)

12.1.1 L1

6.2

Grams of carbohydrate = 2,27×65 = 147,55

1A using 2,27 1M multiplying 1CA simplification Correct answer only: full marks

(3)

12.1.1 L2

6.3.1

165 minutes RT

1RT reading from table

(1)

12.2.3 L1

6.3.2

Average pace (in km per minute) = 60901321

−−

= 308 =

154

≈ 0,27

1SF distances 1SF times 1S simplification 1CA average pace (if inverted, max 2 marks; if using other values from the table, max 2 marks)

(4)

12.2.3 L1

M

CA

M A

CA

SF SF

CA

S

M CA




6.3.3

No penalty for omitting (0;0) and joining 6A any 6 points plotted correctly 1A all correct points joined 1M correct shape (not a straight line) If only a Bar graph is correctly drawn - max 4 marks

(8)

12.2.2 L1

0

5

10

15

20

25

30

35

40

45

50

0 15 30 45 60 75 90 105 120 135 150 165 180

Dis

tanc

e (in

km

)

Time (in minutes)

GRACIA'S PLAN FOR THE RACE

A A

A

A

A

A

A

M




6.4.1

ATHLETIC CLUB FREQUENCY Liberty 5 Striders 5 Ramblers 4 Harmony 6

4A one mark for each correct frequency (just tallies or frequencies as fractions :MAX 2 marks)

(4)

12.4.2 L1

6.4.2 (a)

Striders Club = 100% – (8 + 35 + 12 + 29)% = 16%

1M/A subtracting from 100% 1CA simplification Correct answer only: full marks

(2)

12.4.2 L1

6.4.2 (b)

Liberty or club E or E

2A correct club

(2)

12.4.4 L1

6.4.2 (c)

Actual number of Ramblers athletes = 12% × 300 = 36

1M/A calculating actual number 1CA simplification

(2)

12.4.4 L1

[28]

TOTAL: 150

M/A CA

A A A

A

M/A CA

A


MARKS: 150 TIME: 3 hours

This question paper consists of 15 pages and 3 annexures.

GRAAD 12


NOVEMBER 2012


GRADE 12



INSTRUCTIONS AND INFORMATION

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

2. Answer QUESTION 3.1.2(c), QUESTION 3.2.3 and QUESTION 4.2.2 on the attached ANNEXURES. Write your examination number and centre number in the spaces provided on the ANNEXURES and hand in the ANNEXURES 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 to TWO decimal places, 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.



QUESTION 1

1.1 The Nel family lives in Klerksdorp in North West. They travelled by car to George in the Western Cape for a holiday. A map of South Africa is provided below.

MAP OF SOUTH AFRICA SHOWING THE NATIONAL ROADS

KEY: N1–N12, N17 represent national roads.

Use the map above to answer the following questions.

1.1.1 In which general direction is George from Klerksdorp? (2)

1.1.2 Identify the national road that passes through only ONE province. (2)

1.1.3 The family travelled along the N12 to Kimberley. When they reached Kimberley, they took a wrong turn and found themselves travelling on the N8 towards Bloemfontein. Describe TWO possible routes, without turning back to Kimberley, that the family could follow to travel from Bloemfontein to George. Name the national roads and any relevant towns in the description of the two routes.

(4)

N



1.2 The Nel family (two adults and two children) were on holiday for nearly one week.

• They left home after breakfast on Saturday morning and arrived at the guesthouse in time for supper.

• On Sunday and Wednesday they ate all their meals at the guesthouse. • On Monday they visited a game park. • On Tuesday they went on a nature walk. • On Thursday they went on a boat cruise. • They left George after breakfast on Friday and returned to Klerksdorp. TABLE 1: The Nel family's holiday costs

ITEM COST*

1 Accommodation only R1 050 per day per family 2 Meals at the guesthouse: Breakfast R60 per person per day

Lunch R90 per person per day Supper R120 per person per day

3 Travelling costs: Long distance driving (to and from

Klerksdorp) and meal costs en route R1 602,86 for the return trip

Local driving (in and around George) R513,60 for the duration of the holiday

4 Entertainment costs: Nature walk, including breakfast R120 per adult and

R100 per child Visit to the game park, including lunch

R200 per person

Boat cruise, including supper R200 per adult and R150 per child

Other entertainment R2 000 *All the costs above include value-added tax (VAT).

Use the information above to answer the following questions.

1.2.1 Determine the total amount that they paid for accommodation. (2)

1.2.2 (a) Write down an equation that could be used to calculate the total cost of meals eaten at the guesthouse in the form: Total cost (in rand) = ...

(3)

(b) Use TABLE 1 and the equation obtained in QUESTION 1.2.2(a) to

calculate the total cost of the meals that they ate at the guesthouse if they ate THREE meals daily.

(4)

1.2.3 Mr Nel stated that the total cost of the holiday was less than R20 000.

Verify whether or not Mr Nel's statement is correct. ALL calculations must be shown.

(9)

[26]



QUESTION 2

2.1 On 14 February 2012 there was a queue of customers waiting to eat at Danny's Diner, a popular eating place in Matatiele. The time (in minutes) that 16 of Danny's customers had to wait in the queue is given below:

30 15 45 36 A 40 34 B B 42 26 32 38 35 41 28

B is a value greater than 20.

2.1.1 The range of the waiting times was 37 minutes and the mean (average)

waiting time was 34 minutes.

(a) Calculate the missing value A, the longest waiting time. (2)

(b) Hence, calculate the value of B. (4)

(c) Hence, determine the median waiting time. (3)

2.1.2 The lower quartile and the upper quartile of the waiting times are 27 minutes and 41,5 minutes respectively. How many of the 16 customers had to wait in the queue for a shorter time than the lower quartile?

(2)

2.1.3 Danny's previous records, for 16 customers on 7 February 2012, showed

that the median, range and the mean (average) of the waiting times were 10 minutes, 5 minutes and 10 minutes respectively. Compare the statistical measures relating to the waiting times on 7 and 14 February and then identify TWO possible reasons to explain the difference in these waiting times.

(4)



2.2 The pie chart below shows the percentage of customers who ordered different meals at Danny's Diner on 14 February 2012.

Percentage of customers who ordered different meals

Lamb25%

Fish30%

Beef20%

ChickenSausage

10%

2.2.1 If 40 customers ordered beef meals, determine how many customers

ordered chicken meals.

(4)

2.2.2 A customer is randomly selected. What is the probability that the customer would NOT have ordered a lamb meal?

(2)



2.3 Danny bought a braai drum to cater for those customers who wanted 'shisanyama' or grilled meat. The braai drum is made by cutting a cylindrical drum in half and placing it on a stand, as shown in the picture below. The semi-cylindrical braai drum has a diameter of 572 mm and a volume of 108 ℓ. A rectangular metal grid with dimensions 1% greater than the dimensions of the braai drum is fitted on top.

H = Height of the drum D = Diameter of the drum The following formulae may be used:

Volume of a cylinder = π × (radius) 2 × (height) where π = 3,14 Area of a rectangle = length × breadth 1ℓ = 1 000 000 mm3 = 0,001 m3

2.3.1 Danny filled 3

1 of the base of the drum with sand. Give TWO practical reasons why sand was placed in the braai drum.

(4)

2.3.2 Calculate the length (in mm) of the rectangular metal grid. Show ALL

your calculations.

(9) [34]

H D



QUESTION 3

Longhorn Heights High School needs R7 000,00 to buy a new computer. The finance committee decides to sell raffle tickets to raise funds. A food hamper donated by one of the school's suppliers will be the prize in the raffle. A raffle is a way of raising funds by selling numbered tickets. A ticket is randomly drawn and the lucky ticket holder wins a prize.

3.1 The committee decides to sell the raffle tickets at R2,00 each. The tickets will be

divided evenly amongst a number of ticket sellers.

3.1.1 Write down a formula that can be used to calculate the number of tickets

to be given to each ticket seller in the form: Number of R2,00 tickets per seller = …

(2)

3.1.2 TABLE 2 below shows the relationship between the number of ticket

sellers and the number of tickets to be sold by each seller. TABLE 2: Sale of R2,00 raffle tickets

Number of ticket sellers P 20 25 35 50 100 125 140

Number of tickets per seller 250 175 140 100 70 35 Q 25

(a) Identify the type of proportion represented in TABLE 2 above. (1)

(b) Calculate the missing values P and Q. (4)

(c) Use the information in TABLE 2 or the formula obtained in QUESTION 3.1.1 to draw a curve on ANNEXURE A to represent the number of ticket sellers and the number of tickets sold by each seller.

(4)

3.2 The finance committee changed their plan and decided to sell the tickets at R5,00 each instead.

3.2.1 Give a possible reason why they made this decision. (2) 3.2.2 State ONE possible disadvantage of increasing the price of the tickets. (2) 3.2.3 On ANNEXURE A, draw another curve representing the number of ticket

sellers and the number of R5,00 tickets sold by each seller. Show ALL the necessary calculations.

(8)

3.2.4 Use your graph, or otherwise, to calculate the difference between the

number of R2,00 and R5,00 tickets that must be sold by 70 ticket sellers, assuming the ticket sellers sell all their tickets.

(3)

[26]



QUESTION 4

A local airline company uses three types of aircraft for its domestic and international flights, namely Jetstreams, Sukhois and Avros. Below is a picture of the Jetstream aircraft as well as a table showing information on the three types of aircraft.

TABLE 3: Information on the three types of aircraft

TYPE OF AIRCRAFT JETSTREAM SUKHOI AVRO Maximum number of passengers 29 37 83 Length 19,25 m 26,34 m 28,69 m Wing span* 18,29 m 20,04 m 21,21 m Height 5,74 m 6,75 m 8,61 m Fuel capacity (in kg)** 2 600 kg 5 000 kg 9 362 kg Maximum operating altitude*** 25 000 ft (feet) 37 000 ft (feet) 35 000 ft (feet) Maximum cruising speed**** 500 km/h 800 km/h 780 km/h

[Source: Skyway, November 2011] * The distance from the tip of the left wing to the tip of the right wing ** The mass of the fuel in the tank *** The recommended maximum height that the aircraft should fly at for best fuel efficiency ****The maximum average speed that the aircraft flies at its maximum height

4.1 Use TABLE 3, which is also given on ANNEXURE B, to answer the following.

4.1.1 Mr September flew from Johannesburg to Polokwane along with 37 other

passengers. In which aircraft was he travelling? Explain your answer.

(3)

4.1.2 The length of the Jetstream in the picture is 9,9 cm, while its actual length

is 19,25 m. Determine the scale (rounded off to the nearest 10) of the picture in the form 1: …

(4)

Length of the Jetstream



4.1.3 A nautical mile is a unit of measurement based on the circumference of the earth. 1 nautical mile = 1,1507 miles = 6 076 feet = 1,852 kilometres

Calculate the maximum operating altitude (to the nearest nautical mile) of the Jetstream.

(3)

4.1.4 Ms Bobe travelled in an aircraft that covered a distance of 510 km in

39 minutes. Determine, showing ALL calculations, in which ONE of the three aircraft she could have been travelling. The following formula may be used: Distance = average cruising speed × time

(4)

4.1.5 Determine the fuel capacity (to the nearest litre) of the Avro aircraft.

Use the formula:

Fuel capacity (in litres) = g820

kg)(incapacityfuel

(3)



4.2 The table below shows the schedule of flights between Johannesburg and Polokwane. TABLE 4: Schedule of South African Airways flights between Johannesburg and Polokwane

FLIGHT NUMBER

ROUTE DEPARTURE TIME

ARRIVAL TIME

OPERATING DAYS

SA 8801 JNB–POL 06:35 07:25 1 2 3 4 5 SA 8802 POL–JNB 07:55 08:50 1 2 3 4 5 SA 8809 JNB–POL 11:40 12:40 1 2 3 4 5 6 SA 8809 JNB–POL 11:40 12:30 7 SA 8810 POL–JNB 13:00 14:05 1 2 3 4 5 6 SA 8810 POL–JNB 13:00 13:55 7 SA 8817 JNB–POL 13:15 14:05 1 2 3 4 5 6 7 SA 8818 POL–JNB 14:25 15:20 1 2 3 4 5 6 7 SA 8815 JNB–POL 16:30 17:20 1 2 3 4 5 7 SA 8816 POL–JNB 17:45 18:40 1 2 3 4 5 7

[Source: Skyways, November 2011]

KEY: JNB = Johannesburg; POL = Polokwane 1 = Monday 2 = Tuesday 3 = Wednesday 4 = Thursday 5 = Friday 6 = Saturday 7 = Sunday

Use TABLE 4 above to answer the following questions.

4.2.1 Mr Likobe has to fly from Johannesburg to Polokwane on a Thursday to attend a business meeting that starts at exactly 13:00 and finishes at exactly 15:30. He needs to be present for the full duration of the meeting. He has to attend a 1-hour meeting at 08:30 with a client in his office in Johannesburg before his flight. His office is 30 minutes' drive from the OR Tambo International Airport in Johannesburg. The meeting venue in Polokwane is a 5-minute drive from the airport. Passengers need to check in at the airport at least 1 hour before the departure time of their flight. Which flight numbers should he book for his trip if he has to return to Johannesburg on the same day?

(3)

4.2.2 On ANNEXURE B a line graph representing the number of flights

available daily for the Johannesburg-to-Nelspruit route has been drawn.

(a) Use ANNEXURE B and the information in TABLE 4 above to draw

a line graph representing the number of flights available daily for the Johannesburg-to-Polokwane route.

(4)

(b) Use the line graphs on ANNEXURE B to determine on which day

each route has the lowest number of flights available. Give ONE reason why there are fewer flights on this particular day.

(3)

[27]



QUESTION 5 5.1 Mr Stanford owns a company that sells health care products. The company pays

R50 per item plus R3 500 for shipping and packaging. They sell the items at R120 each. The graph below shows the company's costs and income according to the number of items sold.

0

2000

4000

6000

8000

0 20 40 60

Am

ount

(in

rand

)

Number of items

COSTS AND INCOME OF HEALTH CARE PRODUCTS

5.1.1 Use the graph above to determine the exact number of items sold that will

give a loss of R1 400.

(3) 5.1.2 Mr Stanford stated that the company would break even if 40 items were

sold at R137,50 each. Verify whether Mr Stanford's statement is correct or not. Show ALL the necessary calculations.

(4)

Costs

Income



5.2 Mr Stanford employed eight salespersons in his company. He budgeted R300 000 for bonuses at the end of 2010 for his salespersons. He allocated the bonuses according to each salesperson's contribution to the total sales for the year. TABLE 5 below shows the total annual sales of health care products for each salesperson during 2010 and 2011 with some information omitted. TABLE 5: Total annual sales of health care products during 2010 and 2011

2010 2011 NAME OF

SALESPERSON SALES

(IN THOUSANDS OF RANDS)

SALES AS A PERCENTAGE

SALES (IN

THOUSANDS OF RANDS)


Carl 350 7 440 8 Themba 750 K 715 13 Mabel 1 050 21 1 320 24 Vanessa L 17 935 17 Henry 800 16 1 100 20 Vivesh 900 M 660 12 Peter 200 4 220 4 Cindy 100 2 110 2 TOTAL N 100 5 500 100

Use the information above to answer the following questions.

5.2.1 Calculate the missing values N, K and L. (7)

5.2.2 Vivesh received a bonus of R50 000 in 2010. The other salespeople objected and claimed that he should have received less than this amount. Verify, showing ALL the necessary calculations, whether this objection was valid or not.

(5)



5.2.3 For 2011 Mr Stanford decided to allocate 6,5% of the total sales to bonuses and that each salesperson would be paid a basic bonus as shown in TABLE 6 below. The remaining budgeted amount for bonuses would then be shared equally amongst all the salespersons. TABLE 6: Basic bonus structure for 2011

CATEGORY AMOUNT IN RAND

Sales up to and including 10% 10 000 Sales of more than 10% up to and including 20% 50 000 Sales of more than 20% 100 000

(a) Use TABLE 5 and TABLE 6 on ANNEXURE C to determine

Henry's basic bonus.

(2)

(b) Verify, showing ALL calculations, whether Mabel's total bonus is more than R104 000.

(8)


Copyright reserved

5.3 Mr Stanford was given the following graph by his sales director showing the percentage sales for each salesperson in 2011 and 2012.

PERCENTAGE SALES IN 2011 AND 2012

0 10 20 30 40 50

Carl

Themba

Mabel

Vanessa

Henry

Vivesh

Peter

Cindy

Nam

e of

sale

sper

son

Percentage sales

20112012

5.3.1 Interpret the change in the percentage sales for Vivesh from 2011 to 2012. (2)

5.3.2 After he looked at the graph, Mr Stanford identified Henry and Mabel as the two top salespeople for 2012 with sales of 45% each. What errors did Mr Stanford make in his interpretation of the graph? Explain your answer.

(4)

5.3.3 Name TWO other types of graphs that the sales director could have used

so that Mr Stanford would not misinterpret the graph so easily.

(2) [37]

TOTAL: 150


Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER: ANNEXURE A QUESTION 3.1.2(c) and QUESTION 3.2.3

SALE OF RAFFLE TICKETS

0

40

80

120

160

200

240

280

0 40 80 120 160

Number of ticket sellers

Num

ber

of ti

cket

s sol

d by

eac

h se

ller


Copyright reserved

CENTRE NUMBER: EXAMINATION NUMBER:

ANNEXURE B QUESTION 4.1

TABLE 3: Information on the three types of aircraft

TYPE OF AIRCRAFT JETSTREAM SUKHOI AVRO Maximum number of passengers 29 37 83 Length 19,25 m 26,34 m 28,69 m Wingspan* 18,29 m 20,04 m 21,21 m Height 5,74 m 6,75 m 8,61 m Fuel capacity (in kg)** 2 600 kg 5 000 kg 9 362 kg Maximum operating altitude*** 25 000 ft (feet) 37 000 ft (feet) 35 000 ft (feet) Maximum cruising speed**** 500 km/h 800 km/h 780 km/h

[Source: Skyway, November 2011] QUESTION 4.2.2

0

2

4

6

Mon

day

Tues

day

Wed

nesd

ay

Thur

sday

Frid

ay

Satu

rday

Sund

ay

Num

ber o

f flig

hts

Day

JNB - NEL

NUMBER OF FLIGHTS AVAILABLE PER DAY


Copyright reserved

NOTE: THIS IS AN INFORMATION SHEET ONLY. DO NOT ANSWER

QUESTION 5.2 ON THIS ANNEXURE AND DO NOT HAND IT IN. ANNEXURE C: INFORMATION SHEET QUESTION 5.2 TABLE 5: Total annual sales of health care products during 2010 and 2011

2010 2011 NAME OF

SALESPERSON SALES

(IN THOUSANDS OF RANDS)


SALES (IN THOUSANDS

OF RANDS)


Carl 350 7 440 8 Themba 750 K 715 13 Mabel 1 050 21 1 320 24 Vanessa L 17 935 17 Henry 800 16 1 100 20 Vivesh 900 M 660 12 Peter 200 4 220 4 Cindy 100 2 110 2 TOTAL N 100 5 500 100 QUESTION 5.2.3(a) TABLE 6: Basic bonus structure for 2011

CATEGORY AMOUNT IN RAND Sales up to and including 10% 10 000 Sales of more than 10% up to and including 20% 50 000 Sales of more than 20% 100 000

NSC – Final Memorandum


+

MARKS: 150

Symbol Explanation M Method M/A 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 J Justification

PLEASE NOTE: 1. If a candidate deletes a solution to a question without providing another solution, then the deleted solution must be marked. 2. If a candidate provides more than one solution to a question, then only the first solution must be marked and a line drawn through any other solutions to the question.

This memorandum consists of 19 pages.


NOVEMBER 2012

FINAL MEMORANDUM


GRADE 12

Mathematical Literacy/P2 2 DBE/November 2012 Final Memorandum


QUESTION 1 [26 MARKS] Ques Solution Explanation AS 1.1.1

South-westerly (accept abreviations for compass directions)

2A correct direction 1A Southerly 1A Westerly

(2)

12.3.4 L3

1.1.2

N5 OR N17

2A correct national road N17 accepted due to unclear provincial boundaries

(2)

12.3.4 L3

1.1.3

One possible route: From Bloemfontein turn onto the N1 and travel south until Beaufort West. Then turn onto the N12 until George. A second possible route: From Bloemfontein turn onto the N1 and travel south until the intersection with the N9. Then follow the N9 until George. A third possible route: From Bloemfontein turn onto the N1 and travel south until the intersection with N10. Then follow the N10 in a south easterly direction until the N2. Then follow the N2 in a westerly direction until George. A fourth possible route: From Bloemfontein turn onto the N1 and later turn onto the N6 to East London. Then follow the N2 in a westerly direction until George. A fifth possible route: From Bloemfontein turn north onto the N1, turn right unto N5, take a right unto N3 pass Pietermaritzburg to Durban. Then at Durban turn south unto the N2, pass East London, Port Elizabeth and continue until George. NOTE: Follow the learners route. But leaners cannot go back to Kimberley (No N8 route).

1A N1 1A N12 and Beaufort West

OR 1A N1 1A N9 OR

1A N1 1A N10, N2

OR 1A (N1) N6 and East London, 1A N2 OR

1A N1; N5 and 1A N3 Durban; N2

(4)

12.3.4 L2

A

A

A

A

A

A

A

A

A

A

A

A



Ques Solution Explanation AS

1.2.1

Total amount for accommodation = R1 050 × 6 = R6 300

OR (due to language interpretation)

Total amount for accommodation = R1 050 × 7 = R7 350

1A rate × 6 1CA simplification Correct answer only– full marks

(2)

12.1.3 L2

1.2.2 (a)

Total cost (in rand) = (60 × 4 × number of breakfasts) + (90 × 4 × number of lunches) + (120 × 4 × number of suppers) OR Total cost (in rand) = (60 × x + 90 × y + 120 × z) × 4 Where x = number of breakfasts y = number of lunches and z = number of suppers OR Total cost (in rand) = (number of days × n × 60) + (number of days × n × 90) + (number of days × n × 120) Where n = number of people OR Total cost (in rand) = (Sat + Sun + Mon + Tues + Wed + Thurs + Fri) cost = 120n + 270n + 180n + 210n + 270n + 150 n + 60n) = 1 260 n Where n = number of people

Note: Equation must have a variable 1M adding 1M multiplying cost 1M multiplying by 4 or number of people OR 1M adding 1M costs in terms of meals 1M variables explained OR 1M adding 1M costs in terms of meals 1M variable explained OR 1M adding 1M costs in terms of days 1M variable explained 270 × number of people/meals - (1 mark only)

(3)

12.2.3 L3

1.2.2 (b)

Total cost (in rand)

= (60 × 4 × 5) + (90 × 4 × 4) + (120 × 4 × 5)

= 1 200 + 1 440 + 2 400

= 5 040

OR

REFER TO CANDIDATE’S FORMULA Correct answer only– full marks 1S correct substitution of number of people 1S correct substitution of number of meals 1CA simplification 1CA total

12.2.3 L3

CA

A

S

CA

M

S

M M

CA

M

M

M

CA

A

M

M

M

M

M M




OR Total cost (in rand) = (60 × x + 90 × y + 120 × z) × 4 = (60 × 5 + 90 × 4 + 120 × 5) × 4 = 1 260 × 4 = 5 040 OR (using equation from 1.2.2 (a) working with daily cost) Total cost (in rand) = 1 260 × 4 = 5 040 OR (calculating total daily costs) Cost of meals: Saturday = R120 × 4 = R480 Sunday = (R60 + R90 + R120) × 4 = R1 080 Monday = (R60 + R120) × 4 = R720 Tuesday = (R90 + R120) × 4 = R840 Wednesday = (R60 + R90 + R120) × 4 = R1 080 Thursday = (R60 + R90) × 4 = R600 Friday = R60 × 4 = R240 Total cost (in rand) = 480 +1 080 +720 +840 + 1 080 + 600 + 240 = 5 040 OR (calculating total cost of types of meals) Total cost of breakfast = R60 × 5 × 4 = R1 200 Total cost of lunches = R90 × 4 × 4 = R1 440 Total cost of suppers = R120 × 5 × 4 = R2 400 Total cost (in rand) = 1 200 + 1 440 + 2 400 = 5 040

1S correct subst. no. of people 1S correct subst. no. of meals 1CA simplification 1CA total 2S substitution of no. of people 2CA total 2S correct subst. daily cost 1CA simplification 1CA total 2S correct subst. meal cost 1CA simplification 1CA total

(4)

S

CA

S S CA CA

S

S

CA CA

CA

S

S S

CA CA



Ques Solution Explanation AS 1.2.3

Cost for nature walk = (R120 × 2) +(R100 ×2) = R440 Cost for game park = R200 × 4 = R800 Cost for boat cruise = (R200 × 2) + (R150 × 2) = R700 Total entertainment cost = R440 + R800 + R700 + R2 000 = R3 940 Six day option: Total cost for the trip (accom. + meals + long dist. + local + ent) =R6 300 + R5 040 + R1 602,86 + R513,60 + R3 940 = R17 396,46 OR Seven day option: Total cost for the trip (accom. + meals + long dist. + local + ent) =R7 350 + R5 040 + R1 602,86 + R513,60 + R3 940 = R18 446,46 ∴ Mr Nel's estimate was CORRECT

1M/A expression for cost 1CA simplification 1A cost for game park 1M/A expression for cost 1CA simplification 1CA total cost 1M/A adding all costs 1CA total cost

1M/A adding all costs 1CA total cost

1J verification

(9)

12.1.3 L4

[26]

M/A CA

A

M/A CA

CA

M/A

CA

J

M/A

CA



QUESTION 2 [34 MARKS] Ques Solution Explanation AS 2.1.1(a)

A – 15 = 37 A = 52

A = 37 + 15 = 52

1M concept of range 1A simplification Correct answer only– full marks

(2)

12.4.3 L3

2.1.1(b)

The mean for 16 customers is 34 minutes ∴ total waiting time = 16 × 34 = 544 Total of known waiting times = 2841353832264234405236451530 +++++++++++++ = 494 Difference is 544 – 494 = 50 ∴ 2 customers have a total waiting time of 50 minutes

∴ B = 2

50 = 25

OR Mean =

1628413538322642BB34405236451530 +++++++++++++++

= 34

16B2494 + = 34

2B = (34 × 16) – 494 = 50 ∴B = 25

OR B =

2494)1634( −×

= 25

Refer to value of A in 2.1.1(a) 1M total waiting time 1M total of known times 1S difference of the totals 1CA value of B OR

1M adding all the values 1M dividing by 16

1S simplification 1CA value of B

Correct answer only - full marks

(4)

12.4.3 L3

2.1.1 (c)

Waiting times are: 15; 25; 25; 26; 28; 30; 32; 34; 35; 36; 38; 40; 41; 42; 45; 52

Median = 2

3534+

= 34,5

(Using A and B values calculated above) 1M/A arranging 16 terms in ascending order 1M median concept (even number of terms) 1CA simplification

(3)

12.4.3 L3

M

M/A

CA

M

S

CA

M

S

CA

M M A OR A

M

M

S

CA




2.1.2

4

2CA correct number Note if B is greater than 27 answer can be 2

(2)

2.1.3

The mean, median and range for 7 February are less than those for 14 February.

This means that his customers had to wait for a shorter time on 7 February than on 14 February. Any two of the reasons below: • It could be that more people came to eat at his eating

place on 14 February, because of Valentine's Day.

• He had less staff on the 14th, • He had the same number of staff but did not anticipate

the increased number of customers. • His equipment was faulty on the 14th – people had to

wait longer to be served • The electicity was off for a while

OR The mean, median and range for 14 February are more than those for 7 February.

This means that his customers had to wait for a longer time on 14 February than on 7 February. Any two of the reasons below: • It could be that less people came to eat at his eating

place on 7 February, because of Valentine's Day.

• He had more staff on the 7th, • He had the same number of staff but did not anticipate

the difference in number of customers. • His equipment was working well on the 7th – people did

not wait long to be served • No electicity problems on the 7th

OR Any other valid, well thought out reason will be accepted

2O comparing the measures Accept a comparison table of correct values 2J conclusion

(4)

12.4.4 L4 O

J

O

J

J J

J

CA

O

J

O

J

J J

J




2.2.1

Percentage ordering chicken = 15% If 20% of the total = 40

∴ 1% of the total = 2040 = 2

∴ 15% of the total = 15 ×2 = 30 OR 20% : 40 = 15% : x

x = 40%20%15

×

= 30 OR 20% of total = 40

Total = %20

40

= 200 ∴ 15% of 200 = 30

1A percentage ordering chicken 1M finding 1% 1A multiplying by 15 1CA simplification

OR

1M using proportion 1A percentage ordering chicken 1S expression for x 1CA simplification

OR

1M finding total no. of customers 1A total number of customers 1A percentage ordering chicken 1CA simplification Correct answer only– full marks

(4)

12.1.1 (2) 12.4.4 (2) L2 (2) L3 (2)

2.2.2

P(not lamb) = 1 – 25% = 75% OR 0,75 OR 43

OR Percentage not ordering lamb = 10 + 15 + 20 + 30 = 75

P(not lamb) = 75% OR 0,75 OR 43

OR Number of people not ordering lamb = 20 + 30 + 40 + 60 = 150

P(not lamb) = 200150 =

43 OR 0,75 OR 75%

1M subtracting from100 % 1A simplification

1M adding percentages 1A simplification 1M adding actual numbers 1A simplification Correct answer only - Full marks

(2)

CA

M

A A

CA

M

A

CA

S

M A

A

M A

M A

M

A




Two of the following possible reasons:

• To protect the base of the drum from burning. • To bring the fire closer to the grid. • To spread the coals evenly. (Perfect the braaing) • To use less coal. • To stabilise the drum. • To retain the heat of the burning coals. • The sand can be used to put out the fire.

Accept any two valid reasons.

2O reason 2O reason

(4)

2.3.2

Volume of the braai drum = 108 ℓ = 108 × 1 000 000 mm 3 = 108 000 000 mm 3

Radius of the braai drum = 2mm572

= 286 mm

Volume of the braai drum = 2

1 × π × (radius) 2 × (height) 108 000 000 mm 3 = 2

1 × 3,14 × (286 mm) 2 × (height)

Height = 2

3

)mm286(14,3mm0000001082

××

= 840,99 mm (840,56... mm using π ) ≈841 mm But length of grid = 1% more than height of drum 1% of 840,99 mm = 8,4099 ∴Length of grid = 840,99 mm + 8,4099 = 849,41 mm OR ∴Length of grid = 101% of 840,99 mm = 849,40 mm

1C volume in mm 3 1A value of radius 1M using 2

1 cylinder 1SF substitution into formula 1M Finding expression for height 1CA for height only 1M calculation percentage

1M increasing by 1% 1CA length of grid

OR

1M increasing by 1% 1M calculation percentage 1CA length of grid

(9)

12.3.1 L4

[34]

A

C

SF

M

CA

M

M

CA M

No penalty if answer is rounded to 850 mm

M CA

M

O O




Number of R2,00 tickets per seller = sellers ofnumber

5003

OR

Number of R2,00 ticket per seller = sellersofnumber2

0007×

OR

Number of R2,00 tickets per seller = =2n0007

n5003

where n = number of sellers

1A using 3 500 1A dividing by number of sellers

OR 1A using 7 000 ÷ 2 1A dividing by number of sellers

(2)

12.2.1 L3

3.1.2 (a)

Indirect/Inverse proportion

1A correct type of proportion two answers zero marks

(1)

12.1.1 L2

3.1.2 (b)

P = 2505003 OR P : 70 = 50 : 250

= 14 = 50 × 25070 = 14

Q = 1255003 = 28

1A finding the number of tickets 1M dividing by 250 1CA correct value of P 1CA correct value of Q Correct answer only - Full marks

(4)

12.2.1 L2

A A

A

M CA

CA

A

A

A

A A

CA



3.1.2 (c)

1A correct plotting of point (20;175) 1A correct plotting of point (140;25) 1A one other point plotted correctly 1CA joining the plotted points by a "smooth" curve (section from 20 ticket sellers to 100 ticket sellers) (4)

12.2.2 L2

3.2.1

Fewer tickets have to be sold. OR To reduce the number of sellers. OR To raise the money faster (in a shorter time) OR To raise more money/to buy more computers

2J reason for decision

(2)

12.1.2 (1) 12.2.3 (1) L4

3.2.2

Fewer people can afford (too expensive) to buy the R5,00 tickets. OR Some of the sellers might not be able to sell all their tickets

2J disadvantage

(2)

12.1.2 (1) 12.2.3 (1) L4

A

A

CA

A

R2 Tickets

J

J

J

J




3.2.3

Number of tickets to be sold = 5R

00,0007R

= 1 400

Number of tickets per person = sellersofnumber

4001

1M dividing by R5 1A number of tickets to be sold 1CA formula OR Showing values in a table/co-ordinates - 3 marks

12.2.1 (3) 12.2.2 (5) L3 (4) L4 (4)

The possible points learners can use: (other point values can be used)

10 20 35 50 100 140 140 70 40 28 14 10

4CA any 4 points plotted correctly 1CA joining the plotted points by a smooth curve

(8)

M

A

CA

R5 Tickets

R2 Tickets

A

A A

A

CA




At R2 per ticket 50 tickets must be sold At R5 per ticket 20 tickets must be sold Difference = 50 – 20 = 30 tickets

OR

Number of R2,00 tickets per person = 705003

= 50

Number of R5,00 tickets per person = 704001

= 20 Difference = 50 – 20 tickets = 30 tickets

1RG reading from graph 1RG reading from graph 1 CA difference in number of tickets

OR

1M calculating the number of R2,00 tickets 1M calculating the number of R5,00 tickets

1CA difference in number of tickets

Accept values from 29 to 32. (refer to candidate's graph)

(3)

12.1.1 (1) 12.2.3 (2) L3

[26]

RG

CA

RG

M

CA

Answer only – Full marks

M



QUESTION 4 [27 MARKS] Ques Solution Explanation AS

4.1.1 Avro It is the only one that can take MORE than 37 passengers (himself plus 37 others)

1A correct aircraft 2J justification

(3)

12.4.4 L4

4.1.2

Scale is 9,9 cm to 19,25 m or 9,9 cm to 1 925 cm OR 0,099 m : 19,25 m

Scale = 1 : 9,9

1925

= 1 : 194,44 = 1 : 190

OR 1 : 099,025,19

1M scale concept 1C converting to the same unit 1CA dividing to bring to a unit ratio 1CA rounding off Reversed ratio maximum 2 marks No conversion maximum 2 marks

Correct answer only- full marks

(4)

12.3.2 (1) 12.3.3 (3) L3

4.1.3

Maximum Operating Altitude = 25 000 feet

= 076600025

nautical miles

= 4,1145… nautical miles ≈4 nautical miles

1RT reading from the table 1M dividing by 6076 ft

1CA nearest nautical mile

(3)

12.3.2 L3

4.1.4

Distance = average cruising speed × time 510 km = average cruising speed × 39 minutes

Average cruising speed = minutes39

km510

= h65,0

km510

= 784,62 km/h Ms Bobe was travelling in the SUKHOI OR

Distance (Jetstream) = (500 ×6039 )km = 325 km

Distance (Sukhoi) = (800 ×6039 )km = 520 km

Distance (Avro) = (780 ×6039 )km = 507 km

Ms Bobe was travelling in the SUKHOI

1SF substitution 1C converting to hours 1CA average speed 1J identification of Aircraft OR

1SF substitution 1C converting to hours 1CA distance travel 1J identification of Aircraft

12.2.1 L3 (2) L4 (2)

A

M C

RT M

CA

SF

C

CA

J

J

CA

CA CA

SF C

J

CA



Ques Solution AS Ques

4.1.4 cont

OR Comparing time

Time = speed

distance

Time (Jetstream) = 500510 h = 1,02 hours = 61,2 minutes

Time (Sukhoi) =800510 h = 0,6375 hours = 38,25 minutes

Time (Avro) = 780510 h = 0,6538... hours = 39,23 minutes

Ms Bobe was travelling in the SUKHOI

1SF substitution 1CA time taken 1C converting to minutes 1J identification of Aircraft

(4)

4.1.5


kg)(in capacity fuel

= g820kg3629

= g820

g0003629

= 11 417,07317 ≈ 11 417

OR


kg)(in capacity fuel

= g820kg3629

= kg0,820kg3629

= 11 417,07317 ≈ 11 417

1SF substitution 1C converting to grams 1CA nearest litre

1SF substitution 1C converting to kilograms 1CA nearest litre No conversion - maximum 2 marks

(3)

12.3.2 L2 (2) L3 (1)

4.2.1

Johannesburg to Polokwane: SA 8809 Polokwane to Johannesburg: SA 8816

2A correct flight number 1A correct flight number

(3)

12.4.4 L3 A

A

SF

C

CA

SF

C

CA

J

SF CA C



Ques Solution AS 4.2.2(a)

1A drawing the horizontal line at 4 1A plotting (Saturday; 2 ) 1A plotting (Sunday; 3) 1CA joining the plotted points

(4)

12.4.2 L3

4.2.2 (b)

Saturday Not many people travel on Saturday, as most business meetings are scheduled during the week. OR If people go away for the weekend on holiday, they travel there on a Friday and travel back on Sunday. OR Possible religious reason OR Any other valid reason

1A correct day 2O own opinion based on candidates graph

(3)

12.4.4 L4

[27]

A

O

A

A

A CA

O

O

O




For 30 items: Cost = R5 000 Income = R3 600 Loss = R5 000 – R3 600 = R1 400 ∴ 30 items

1RG cost 1RG income 1A number of items Correct answer only - full marks

(3)

12.2.2 L3

5.1.2

Cost of 40 items = R5 500 OR 40 × R50,00 + R3 500 Income from 40 items = R137,50 × 40 = R5 500 At 40 items, Cost = Income ∴ Mr Stanford's statement is CORRECT.

1RG/A cost Or Cost = income 1M finding total income 1Asimplification 1CA verification

(4)

12.2.2 L4

5.2.1

N is the total sales. 16 % of N = 800

N = 800 × 16100

= 5 000 OR

16% of the sales = 800

1% of the sales = 16

800

∴100 % of the sales = 10016

800×

∴N = 5 000

OR

21 % of total sales = 1 050

Total sales = 1 050 × 21

100

∴ N = 5 000

K = ×0005

750 100

= 15

1M concept 1M finding an expression for N 1A total sales

OR 1M finding unit value 1M finding 100% 1A total sales OR 1M concept 1M finding an expression for N 1A total sales

1M concept 1CA simplification

12.1.1 L2 (4) L3 (3) M

A

M

A

M

M

M

M

A

M

CA

RG RG

A

RG

M A

CA




L = 17% of total sales

L = 000510017

×

= 850

OR

16% of the total is 800

1% of the total is 16

800

∴17% of the total is 1716

800×

∴L = 850 Please note If L is found first: N = 350 + 750 + 1 050 + 850 + 800 + 900 + 200 + 100 = 5 000

1M finding 17 %

1CA simplification OR 1M finding unit value 1CA simplification Correct answer only full marks The values need not be a calculated in the same order as on the memo (7)

5.2.2

Vivesh's % (value of M)

= ×0000005000900 100% OR

= 18%

%1000005900

×

= 18%

OR 100% – (7 + 15+ 21 + 17 + 4 + 2 + 16)%

= 18% Vivesh's bonus = 18% of R300 000

= R54 000 ∴ The objection is NOT VALID. CA

1M expression for % 1CA simplification 1M calculating percentage 1CA simplification

1CA conclusion (5)

12.1.1 L4

5.2.3 (a)

R50 000

2A correct basic bonus

(2)

12.1.1 L3

CA

CA

M

CA

CA

M

A

M

CA

CA

M

M

CA

M

M

CA

Mathematical Literacy/P2 19 DBE/November 2012 NSC – Memorandum

Copyright reserved

Ques Solution Explanation AS 5.2.3 (b)

Total bonus amount =6,5 % × R5 500 000 = R357 500 Sales up to and including 10% : 3 persons Sales of more than 10% up to and including 20% : 4 persons Sales of more than 20% : 1 person Bonus amount remaining = R357 500 – (3 × R10 000 + 4 × R50 000 + R100 000) = R357 500 – R330 000 = R27 500

Amount each will receive = 850027R

= R3 437,50 Mabel's total bonus = R100 000 + R3 437,50 = R103 437,50 ∴ Mabel's bonus is NOT MORE THAN than R104 000.

1A total bonus 1 M finding the total basic bonus 1M finding the difference 1CA simplification 1M dividing by 8 1CA simplification 1CA Mabel's bonus (must include R100 000)

1O verification

(8)

12.1.1 L4

5.3.1

Vivesh's sales in 2012 was more than double his sales in 2011. Vivesh was the top salesperson in 2012. OR There is an increase in percentage sales from 12% to 28% OR Any other numerical comparison

2O interpretation

(2)

12.4.6L4

5.3.2

He read Mabel's and Henry's combined sales of 2011 and 2012 as the sales for 2012. Henry's sales for 2012 were only 25%, Mabel's sales were 21% and the person with the highest sales was Vivesh with 28%

2O errors

1J Henry & Mabel 1J mention Vivesh as highest

(4)

12.4.6 L4

5.3.3

Any TWO of the following: • Different type of Bar graphs • Line graphs • Pie charts

1O bar graphs 1O line graphs OR 1O pie charts

(2)

12.4.6L2

[37]

TOTAL: 150

A

CA

CA

O

J

O

O

M

CA O

M

M

O O

J

national senior certificate grade 12 · this question paper consists of 16 pages and 3 annexures....

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