Examination Preparation for Grade 12 Mathematical Literacy...

for the learner

Examination Preparation for Grade 12

Mathematical Literacy

Foundational Knowledge for Paper 1 & 2

Learner Booklet

Grade 12 Mathematical Literacy Foundational Knowledge Examination Preparation Guidelines

Winning Teams

CONTENTS

Learning Outcome 1: Number and Operations in Context (Code A1 – A37) 4

1. Number Operations 2. Decimal fractions, common fractions and percentage 3. Rate, ratio and proportion 4. Working with finances 5. Foreign exchange rates 6. Examination questions Learning Outcome 2: Functions and Algebra (Code B1 – B25)

1. Reading information from a table 2. Reading and drawing graphs 3. Working with formulae 4. Finding the break-even point Learning Outcome 3: Space, shape and measurement (Code C1 –C26) 19 1. Length and distance 2. Perimeter 3. Area 4. Surface area 5. Volume 6. Conversions 7. Maps, grids and compass direction Learning Outcome 4: Data handling (Code D1 – D36) 2 1. Representing data 2. Reading data from tables and graphs 3. Measures of central tendency and spread 4. Probability

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Mathematical Literacy Grade 12 Foundational Knowledge Examination Preparation Guidelines

Winning Teams Grade 12 Mathematical Literacy

Learning Outcome 1: Number and Operations in Context 1. Number operations When you multiply a positive number by a positive number, you get a positive answer. For example, 3 5 = 15 When you multiply a positive number by a negative number, you get a negative answer. For example, 3 –5 = –15 When you multiply a negative number by a positive number, you get a negative answer. For example, –3 5 = –15 When you multiply a negative number by a negative number, you get a positive answer. For example, –3 –5 = 15 When you complete a calculation, remember the order of the operations: Work out brackets first, then multiply or divide and only add or subtract after this.

Example: Simplify the following: [(3)3 – 4] + √81 ÷ 3

[(3)3 – 3] + √81 ÷ 3

= [27 – 3] + √81 ÷ 3 simplify the inside brackets

= 24 + √81 ÷ 3 simplify the outer brackets

= 18 + 9 ÷ 3 multiply and work out the square root = 18 + 3 divide = 21 add 2. Decimal fractions, common fractions and percentage

1. Percent means ‘out of 100’. For example, is 63%. The symbol used for percent is %.

2. To change a percentage to a decimal fraction or a common fraction, divide by 100: 9% =

To change a fraction to a decimal, divide the numerator by the denominator. You can use a calculator: = 9 ÷ 100 = 0,09

3. We can convert between decimal numbers, fractions and percentages:

0,09 = = 9% 63,3% = , = 0,633 = 0,733 = 73,3% 68% = =

4. To find one quantity as a percentage of another, they both need the same units.

70c as a percentage of R8,00: 100 = 8,75% or ,, 100 = 8,75%

5. To find the percentage of a quantity, convert to a decimal or common fraction and multiply:

33% of R240 = R240 = R79,20 or 33% of R240 = 0,33 R240 = R79,20

6. Percentage decrease on a price means the normal price less the decrease. Example: A pair of jeans is on sale with a mark down of 20%. If the original price of the jeans was R199,00, what is the sale price? Sale price is normal price less 20%, so sale price is 80% of normal price. 80% of R199 = 0,8 R199 = R159,20

7. Percentage increase on a price means the normal price plus increase. Example 1: The price of petrol increases by 12%. The original price was R7,70 per litre. What is the new petrol price? Method 1: New price = old price + 12% of old price New price = R7,70 + 12% R7,70 = R7,70 + 0,12 R7,70 = R7,70 + R0,92 = R8,62

Method 2:New price = old price + 12% of old price = 112% of old price = 1,12 old price = 1,12 R7,70 = R8,62

+ + = +

+ – = –

– + = –

– – = +

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Example 2: Jane is offered a salary increase of 7%. If her new salary is R10 600, what was her salary before the increase? New salary = original salary + 7% of original salary So 107% of original salary = R10 600

1% of original salary is = R99,07

So the original salary is R99,07 100 = R9 907.

More examples of working with percentage: 1. I pay 14% VAT on my telephone bill. If the total before VAT is R378,50, what is my telephone bill after VAT?

I will have to pay R378,50 114% = R378,50 1,14 = R431,49.

2. A clothes shop advertises their sale of 33% off all marked prices. If a jersey is marked R205, what price will I pay on the sale? I will pay 100% – 33% = 67% R205 = R137,35.

8. Percentage increase or decrease So far the examples we have looked at have asked you to find values after a percentage increase or decrease. We have not looked at finding what actual percentage increase or decrease takes place. We can find this using:

Percentage increase = – 100%

Percentage decrease = – 100% Example: 1. The number of learners in the school increased from 500 to 700. By what percentage have they

increased?

Percentage increase = – 100% = 100% = 100% = 100% = 40%

2. The number of elephants in the area have decreased from 800 to 600. By what percentage have they

decreased?

Percentage decrease = – 100% = 100% = 100% = 100% = 25%

3. Rate, ratio and proportion A ratio is used to compare quantities with each other. For example, if I share 21 sweets between Thoko and

Thandi in the ratio 2 : 5, then for every 2 sweets I give Thoko, I must give Thandi 5 sweets. Thoko will get 6 sweets and Thandi will get 15 sweets.

The quantities compared need to have the same units (the ratio of litres to litres, or rands to rands etc). You may have to convert the units to make them the same (the ratio of 5 cm to 20 mm = 50 mm : 20 mm = 5 : 2)

We can work with ratios by multiplying or dividing each part of the ratio by the same number. The ratio 1 : 4 is the same as the ratio 5 : 20 and they are called equivalent ratios.

A proportion is an equation stating that two ratios are equal (or ‘in proportion’). For example, = .

Rate compares quantities with different measuring units. We usually reduce the rate to a quantity per one unit. For example, a car’s speed might be 60 (km/h) kilometres per hour – for every hour of driving, a distance of 60 km is covered.

Scale is a ratio which shows the relationship between the actual distance on the ground and the distance on a map. We deal with this in Learning Outcome 3.

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4. Working with finances Use the Winning Teams financial literacy cards to make sure that you understand the following vocabulary: Budget: A plan for what income you expect to receive and what expenses you expect to have over a period of time. Interest: The price for using someone else’s money or the price for investing your money with someone else. For example, you can borrow money from the bank and pay it back over time. You have to pay back the whole amount borrowed and a percentage interest (interest rate) on top of that. Or if you invest money in an account at the bank, the bank will pay you a percentage interest (interest rate) on your money. VAT (Value Added Tax): VAT is money charged by the government on top of the price of any goods or services made

in South Africa. Profit: The money you make on selling something after you have subtracted your expenses. Loss: The money you lose on selling something after you have subtracted your expenses. Mark up: The amount you add onto the cost price of goods before you sell them. Discount: The amount by which you reduce the selling price of goods. Break even: The point at which expenses and income are equal, so there is no profit or loss. Example (March 2009 Paper 2, Question 2): Annabel started working for a construction company on 3 July 2007. She earned a net income of R144 000 per annum without any bonus. She set aside R8 400 per month for her monthly expenses, and each month set aside 90% of the balance towards a deposit for a car. Table 1: Annabel’s monthly expenditure before buying a car

Items Monthly expenditure

Rent and electricity R2 850Groceries R1 500Student loan repayment R900Public transport to work R700Clothing R350Household insurance R420Entertainment R350Life insurance R300Other R1 030

TOTAL R8 400 2.1 2.1.1 Calculate Annabel’s net monthly salary. (2) 2.1.2 How much did Annabel save towards the deposit for a car each month? (3) 2.2 Annabel was advised to invest some of her monthly savings in a special monthly savings account that pays more interest. She thus invested R3 000 of her monthly savings each month in this account. This special savings account paid an interest rate of 10,8% per annum, compounded monthly. Use the formula F = ( ) to calculate the total amount she will have for her deposit if she saves monthly for 11 months, where:

F = total amount received x = monthly amount invested i = monthly interest rate n = number of months for which the money was invested (5)

2.3 On 1 July 2008, Annabel receives a 10% salary increase. Calculate Annabel’s new monthly net salary. (4) 2.4 On 1 July 2008 Annabel buys a car. She finds that she has to budget R3 900 per month for the car to cover the cost of petrol, repayments, insurance and maintenance. However, she no longer has to pay for public transport. Determine her new total monthly expenditure. (3)

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Mathematica

Winning Team

Solutions:

2.1.1 Net2.1.2 Amo

90% 2.2 F =

F =

F =

F =

F =

2.3 New 2.4 New

5. Foreign• The exch

Countrie• The rand• A weak r• A strong On 25 Novem

httpExamples: Rands to USDIf I have R69 USD to RandIf I have 2 00Rands to INRIf I have R69 INR to RandsIf I have 45 0

l Literacy Grade

ms Grade 12 Ma

monthly salaount remainin

% of R3 600 = 0

? x = R3 00( ) ( , ), ( , ), ( , ),w monthly sal

w monthly exp

n exchangehange rate gives often compd dollar excharand means th

g rand means

mber 2009, thOne u

1 US

1 Chi

One

1 Ran

1 Ran

p://www.sa-ven

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00 USD, this isR:

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e 12 Foundatio

athematical Lite

ry = =ng each mont0,9 R3 600 =

00 i =

) = R34 525

ary = 112% of

penditure = R8

e rates ves us a way tpare their currange rate is thhat it will costthat it will cos

hese were the unit of foreign dollar (USD) = 7

nese yuan renm

rand equals x cnd = 0,0808727

nd = 6,23588 In

nues.com/excha

69 900 ÷ 7,46

2 000 7,46

69 900 ÷ 1,09

his is 45 000

onal Knowledge

eracy

= R12 000 per h = R12 000 –= R3 240

, % = 0,9% =

5,83

f R12 000 = 1,

8 400 + R3 90

o convert cosrency with thee number of rt you more rast you less ran

currency exccurrency equa7,46 ZA Rands

minbi (CNY)= 1,

currency 7 Pound sterling

ndia Rupees (IN

ange.htm

= 9 369,97 US

= R14 920. M

= 64 128,44 I

1,09 = R49 05

e

month. – R8 400 = R3

= 0,009 n =

12 R12 000

00 – R700 = R1

st in one curree US dollar (Urands we neend to buy 1 Und to buy 1 US

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09 ZA Rands

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.

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6. Examination questions Fractions, decimals and percentages; rate, ratio and proportion; finances and exchange rates are integrated in the questions from examination papers below. Example (March 2009 Paper 1, Question 1.1): Do the following calculations. Show all calculations in full.

1.1.1 Write 47% as a common fraction.

1.1.2 Write as a decimal fraction.

1.1.3 Simplify: 1,2 m + (23,5 m 5) – 4,7 m

1.1.4 Simplify: (3)3 + √64

1.1.5 Calculate 14% VAT on R24 650,00. 1.1.6 Convert R1 500 into Euros (�). Use the conversion R1 = � 0,11. 1.1.7 Increase R1 250,00 by 24%. 1.1.8 Calculate the number of 30 g portions of jam that can be obtained from a 450 g tin. 1.1.9 Determine the cost of 6 bus tickets using the formula:

Cost of bus tickets = number of bus tickets R12,15. Solutions:

1.1.1 47% =

1.1.2 = = = 0,65

1.1.3 1,2 m + (23,5 m 5) – 4,7 m = 1,2 m + 117,5 m – 4,7 m = 114 m

1.1.4 (3)3 + √64 = 27 + 8 = 9 + 8 = 17

1.1.5 R24 650,00 14% = R24 650,00 0,14 = R3 451,00 1.1.6 R1 = � 0,11. R1 500 0,11 = �165 1.1.7 R1 250 124% = R1 250 1,24 = R1 550 1.1.8 450 g ÷ 30 g = 15 portions 1.1.9 Cost of bus tickets = 6 R12,15 = R72,90

Example (November 2008 Paper 1, Question 1):1.1 1.1.1 Write 20% as a common fraction in its simplest form. 1.1.2 Write as a percentage. 1.1.3 Simplify 120 : 150 1.1.4 Decrease 500 kg by 12%.

1.2 Calculate the following: 1.2.1 R450 – R32,40 10 1.2.2 52 – √36 1.2.3 34% of 450 km 1.3 Marie wants to bake coconut tarts for the school fete using her

grandmother’s old recipe. They have to be baked at 350F for 20 minutes. 1.3.1 Convert ½ lb to grams (1 lb = 450 g) 1.3.2 Convert 9 oz to grams (1 oz = 30 g) 1.3.3 One cup of sugar is equal to 250 ml. How many ml of sugar are needed for this recipe? 1.3.4 Convert 350F to C using the formula:

Temp in C = (temp in F – 32) Round your answer off to the nearest 10 degrees.

1.3.5 How many eggs does Marie need to bake 72 coconut tarts?

Coconut tarts(makes 3 dozen)

Ingredients: 1 lb self-raising flour 9 oz margarine ¾ cup sugar 4 eggs 10 oz coconut ½ lb apricot jam 1 teaspoon vanilla essence

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Solutions: 1.1.1 20% = =

1.1.2 = = 68% 1.1.3 120 : 150 = 12 : 15 = 4 : 5 1.1.4 500 kg – 12% of 500 kg

= 500 – ( 500) = 500 – 60 = 440 kg

OR 100% - 12% = 88% and 88% of 500 kg = 0,88 500 kg = 440 kg. 1.2.1 R450 – R32,40 10 = R450 – R324 = R126 1.2.2 52 – √36 = 25 – 6 = 19 1.2.3 34% 0f 450 km = 0,34 450 = 153 km 1.3.1 ½ lb = ½ 450 g = 225 g 1.3.2 9 oz = 9 30 g = 270 g 1.3.3 ¾ cup = ¾ 250 ml = 187,50 ml 1.3.4 Temp in C = (temp in F – 32)

Temp in C = (350F – 32)

Temp in C = 318F = 176,67C 180C to the nearest 10C. 1.3.5 4 eggs makes 3 dozen (36) tarts, so 8 eggs make 72 tarts. Example (March 2009 Paper 1, Question 3.3): Jabulani starts a small business making and selling money boxes. It costs him R25,50 to make ONE money box, including labour and overheads. He intends selling each money box for R30,00. 3.3.1 Calculate the percentage profit Jabulani will make on each money box. 3.3.2 How many money boxes will Jabulani need to sell in order to make a profit of at least R400,00? Solutions: 3.3.1 He makes a profit of R30,00 – R25,50 = R4,50 on each money box. , , × 100% = 17,65%

3.3.2 ,, = 88,89 He will need to sell 89 boxes to make a profit of R400,00.

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Learning Outcome 2: Functions and Algebra When we want to find out about how one changing value affects another changing value, we can do this using a table of values, a formula or a graph. 1. Reading information from a table:

In the example below, use the columns and the rows of the table to read information about times and destinations.

Example (March 2009 Paper 1, Question 4): A train timetable is shown for travelling from Durban to Bloemfontein and then to Kimberley. Table 2: The train journey from Durban to Kimberley via Bloemfontein Frequency: From Durban on Wednesdays

Town Arrival Departure Time in minutes stopped at

station Durban 18:30Pietermaritzburg 20:53 21:10 17Ladysmith 00:33 27Harrismith 03:23 03:53 30Bethlehem 05:20 05:40 20Kroonstad 07:49 08:19 30Hennenman 08:57 08:59 2Virginia 09:17 09:19 2Theunissen 09:50 09:52 2Brandfort 10:25 10:27 2Bloemfontein 11:15 11:45 30Kimberley 14:50

4.1 On which day of the week does this train from Durban arrive in Kimberley? (2) 4.2 How long did the total journey take? Write your answer in hours. (3) 4.3 At what time did the train leave Ladysmith? (2) 4.4 Calculate the total time taken for stops between Durban and Kimberley. Give the answer in hours and minutes. (3) 4.5 Not counting stops, the actual travel time for the train journey is 17,6 hours and the distance between Durban and Kimberley is 842 km. Calculate the average speed at which the train is travelling. Use the formula: Speed = (3) 4.6 James travels from Durban to Brandfort on the same train. He needs to board a bus in Brandfort that is leaving the bus station at 11:00. It takes 5 minutes to walk from the train station to the bus station. Determine whether or not James will be in time to board the bus. Show ALL the necessary calculations. (3) Solutions: 4.1 The train leaves Durban on Wednesday evening and travels through the night to the next afternoon, so it arrives in Kimberley on a Thursday. 4.2 From 18:30 to 14:50. From 18:30 to midnight is 5½ hours. From midnight to 14:50 is 14 hours 50 minutes. So the total time taken is: 5 hours 30 minutes + 14 hours 50 minutes = 19 hours 80 minutes = 20 hours 20 minutes = 20 = 20,33 hours. 4.3 It arrived at 00:33 and waited 27 minutes, so it left at 01:00. 4.4 Total time taken for stops = 17 + 27 + 30 + 20 + 30 + 2 + 2 + 2 + 2 + 30 = 162 minutes = 2 hours 42 minutes 4.5 speed = speed = , = 47,84 km/h. 4.6 The train arrives in Brandfort at 10:25. James only needs 5 minutes to walk to the bus station and there is 35 minutes before the bus leaves at 11:00. So he has plenty of time to board the bus.

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2. Reading a graph: In the example below, Jane completes a 20 kilometre walk. We want to see how the distance she covers changes over time. The graph gives us this information. We can read information about time and distance from the graph. Example (March 2009 Paper 1, Question 1.2): Jane participated in a sponsored 20 km walk to raise funds for Aids orphans. The organiser encouraged the walkers to have a fifteen-minute rest during the walk. The graph showing the distance covered by Jane against the time take by her, is given below. Time taken for 20 km walk

20

19

18

17

16

15

14

13

12

11

10

9

8

7

6

5

4

3

2

1

0

07:00 07:30 08:00 8:30 09:00 09:30 10:00 10:30

Time of day 1.2.1 At what time did the walk start? (1) 1.2.2 How many kilometres did Jane walk during the first hour? (2) 1.2.3 How far had Jane walked by 10:00? (2) 1.2.4 How long did Jane take to walk the first 9 km? (2) 1.2.5 After how many hours of walking did Jane rest? (2) 1.2.6 Give an estimate of the time at which Jane finished the walk. (2) Solutions: 1.2.1 Read this from the graph: At 07:00, Jane had not started walking yet. 1.2.2 She had walked 6 km. Read this at A on the graph. 1.2.3 She had walked 16,5 km. Read this at B on the graph. 1.2.4 She took 1½ hours to walk the first 9 km. Read this at C on the graph. 1.2.5 Jane rested after two hours. Read this from the graph at D. 1.2.6 She finished the walk just before 10:30, probably at about 10:25.

Dist

ance

cov

ered

in k

m

A

B

C

D

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3. Working with formulae A formula shows a relationship between two sets of values. If you are given a formula, you can use it to determine and calculate particular values. You can substitute numbers into a formula to find a particular value. You can also substitute numbers into a formula to be able to complete a table of values or plot points on a graph. Example 1: If y = 2x + 3 and x = 5, determine the value of y. y = 2x + 3 y = 2(5) + 3 = 13 Example 2: If y = 2x + 3, complete the table of values below and use this to draw the graph.

x –2 –1 0 1 2 y = 2x + 3 2(–2) + 3 = –1 2(–1) + 3 = 1 2(0) + 3 = 3 2(1) + 3 = 5 2(2) + 3 = 7

We can plot the points from the table of values onto a graph. The points are (–2; –1), (–1; 1), (0; 3), (1; 5) and (2; 7). These points show the pattern for the rest of the points represented by this formula, and so we can join them to form a straight line graph.

7 6

5

4

3

2

1

–2 –1 0 1 2

–1

Example 3: Tim rides his motorbike for 4 hours and travels 280 km in that time. Work out his average speed.

Use the formula: speed = Speed = = 70 kilometres per hour (km/h)

y = 2x + 3

y

x

Drawing a line graph: - Use a scale for each axis that

works for your data. - Label the axes. - Plot the coordinates. - Join the plotted points if this

works for your data.

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Example (March 2009 Paper 1, Question 6): Lourens and his five friends are planning a five-day holiday. As none of them own a car, they have decided to hire a Toyota Avanza from Oom Piet’s car hire for 5 days. They will have to pay a rental cost plus an additional cost for petrol used. As part of their rental costs, Oom Piet’s car hire offers 200 km free per day. Total rental cost for 5 days = R2 000 + (number of kilometres over 1 000 km) R3,00 Lourens worked out the total rental cost for 5 days for various distances travelled. He then drew a graph to illustrate his calculations.

Total rental cost 4 000

3 500

3 000

2 500

2 000

1 500

1 000

500

0 200 400 600 800 1 000 1 200 1 400 1 600

6.1 Use the graph to: 6.1.1 Write down the total rental cost over the five days if Lourens and his friends travel: (a) 950 km (1) (b) 1 300 km (2) 6.1.2 Determine the maximum number of kilometres they could travel in five days if they set aside R3 500 to pay for car hire. (2) 6.2 A Toyota Avanza covers 10 km on 1 litre of petrol. Suppose the friends travelled 1 400 km and petrol costs R10,40 a litre, calculate the petrol bill for the journey, using the following formula: Petrol bill = cost of 1 l of petrol (3) 6.3 On a particular day Lourens and his friends covered a distance of 360 km. Table 3 illustrates the time taken to travel the 360 km at different speeds. Table 3: Time taken to travel 360 km for different speeds

Speed in km per hours

10 20 30 45 B 90 100 120

Time taken in hours

36 18 A 8 6 4 3,6 3

6.3.1 Determine the values of A and B. Use the formulae: time = or speed = (4)

Cost

in ra

nd

Kilometres travelled

A

B

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6.3.2 Use the grid below to draw a line graph representing the information in Table 3.

Time taken to travel 360 km 40

35

30

25

20

15

10

5

0 20 40 60 80 100 120 140

6.4 The bill for the car hire came to R4 236,00. The six friends decided to share the bill equally. How much did each one have to pay? (2) Solutions: 6.1.1 (a) If they travel 950 km, this is below the 1 000 km free, so it is R2 000.

You can read this from the graph at A. (b) If they travel 1 300 km, they pay R2 000 + (300 km R3,00) = R2 000 + R900 = R2 900

You can read this from the graph at B. 6.1.2 R3 500 – R2 000 = R1 500 R1 500 ÷ R3,00 = 500 km They can travel 1 000 km + 500 km = 1 500 km. 6.2 Petrol bill = cost of 1 l of petrol

Petrol bill = R10,40 per litre = R1 456,00 6.3.1 For A, time = = / = 12 hours

For B, speed = = = 60 km per hour

Speed in kilometres per hour

Tim

e in

hou

rs

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6.3.2 Time taken to travel 360 km

40

35

30

25

20

15

10

5

0 20 40 60 80 100 120 140

6.4 R4 236 ÷ 6 friends = R706 each

Speed in kilometres per hour

Tim

e in

hou

rs

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Example (March 2009 Paper 2, Question 1): The Hospitality Studies department of Ses’fikile High School bakes brown bread in order to raise funds for the shortfall incurred in their day-to-day expenses. The school charges the Hospitaliy Studies department a fixed weekly cost of R400,00 for water and electricity. The cost of producing one loaf of brown bread, including labour and ingredients, is R3,50. The brown bread is sold at R6,00 a loaf. 1.1 If one loaf of brown bread requires 450 g of flour, determine the maximum number of loaves of brown bread that can be baked from a 12,5 kg bag of flour. (4) 1.2 The table below shows the weekly cost of making the bread. Table 4: Weekly cost of making brown bread

Number of loaves 0 40 80 120 160 B 300 Total cost (in rand) 400 540 680 A 960 1 240 1 450

The formula used to calculate the total cost per week is: Total cost per week = fixed weekly cost + (number of loaves of bread cost per loaf) Use the formula to determine the values of A and B in Table 4. (4) 1.3 The table shows the weekly income from selling the bread.

Table 5: Weekly income received from selling bread Number of loaves 0 40 120 150 D 250 300 Total income (in rand) 0 240 C 900 960 1 500 1 800

Determine the values of C and D in Table 5. (4) 1.4 Use the values from Table 4 and Table 5 to draw TWO straight line graphs on the same grid (see next page),

showing the total COST per week of making bread and the INCOME per week from selling bread. Label the graphs ‘COSTS’ and ‘INCOME’. (8)

1.5 Use the tables or the graphs to answer the following questions:

1.5.1 How many loaves of bread must they sell to break even and describe what is happening at the break-even point? (3) 1.5.2 What income would they receive if 230 loaves were sold? (2) 1.5.3 Estimate the number of loaves baked if the total cost is R840. (2) 1.5.4 Determine, by calculation, whether Ses’fikile High School will make a profit or a loss if they bake 300 loaves of bread during the week, but only sell 250 of these loaves of bread. (3)

Solutions: 1.1 12,5 kg = 12 500 g

12 500 g ÷ 450 g = 27,77 So you can make 27 loaves of bread from 12,5 kg of flour. 1.2 Total cost per week = fixed weekly cost + (number of loaves of bread cost per loaf)

At A, Total cost per week = R400 + (120 loaves R3,50) = R400 + R420 = R820 At B, R1 240 = R400 + (number of loaves R3,50) R1 240 – R400 = (number of loaves R3,50) R840 = (number of loaves R3,50) R840 ÷ R3,50 = 240 loaves

1.3 A loaf of bread is sold at R6,00 per loaf.

At C, 120 loaves R6,00 = R720 At D, R960 = number of loaves R6,00 R960 ÷ R6,00 = 160 loaves

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1.4

Income and costs 1800

1700

1600

1500

1400

1300

1200

1100

1000

900

800

700

600

500

400

300

200

100

0 50 100 150 200 250 300

Number of loaves of bread

Amou

nt in

rand

s

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1.4

Income and costs 1800

1700

1600

1500

1400

1300

1200

1100

1000

900

800

700

600

500

400

300

200

100

0 50 100 150 200 250 300

1.5.1 160 loaves must be sold. At this point, both the cost and the income are the same and are equal to R960. 1.5.2 230 loaves will sell for R1 380 (at B) 1.5.3 Read this from the graph at A. It is about 125 loaves. 1.5.4 Cost of making 300 loaves = 400 + 300 R3,50 = R1 450

Income from selling 250 loaves = 250 R6,00 = R1 500. So they will make a profit of R50.

Number of loaves of bread

A

Amou

nt in

rand

s

Income Costs

B

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Learning Outcome 3: Space, shape and measurement 1. Perimeter The perimeter of a flat shape is the length around its outer edge. Perimeter is measured in units of length e.g. mm, cm, m and km. perimeter of a rectangle = 2l + 2b perimeter of a square = 4l

circumference of a circle = 2πr where π 3,14 perimeter of a triangle = a + b + c and r is the radius. 2. Area Area is measured in square units: mm2; cm2; m2; km2 This square has an area of 1 cm by 1 cm. It is one square cm or 1 cm2 in area.

area of a rectangle = length x breadth area of a square = length x length area of a rhombus = base height area of a parallelogram = base height area of a trapezium = ½h (a + b) area of a triangle = ½b x h

area of a circle = πr2

where π 3,14 and r is the radius.

ab

c

r

r

b

h

b

h

h

a

b

h

b b

h

length (l)

breadth (b) length (l)

length (l)

length (l)

breadth (b) length (l)

length (l)

Any formulae for perimeter, area, surface area or volume that are needed to calculate answers will be provided in the examination paper.

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3. Surface area

Surface area is the total amount of area on the surface, the exposed part of a 3-D object. If you were asked to paint the object, the paint would cover the surface area. To calculate the surface area, find the area of each face of the object and add the areas together. Rectangular prism: A right rectangular prism is an object with equal parallel bases; all faces are rectangles;

side faces which are perpendicular to the bases.

You can unfold the prism to find its net. This is one possible net, depending on how you unfold the prism. Surface area = 2(length breadth) + 2(length height) + 2(breadth x height) = 2lb + 2lh + 2bh Triangular prism: Surface area = 2(½ b h) + 2(H S) + b H

Cylinder: The length of the rectangular part of the net fits around the circumference of the circular part, so Length = 2πr Surface area = 2πr h + 2(πr)2 = 2πrh + 2πr2 where r is the radius, h is the height and π 3,14.

length (l)breadth (b)

h

r

2πr

r

h

height (h)

lb

length (l) breadth (b)

vertex

edge

face

Hh

b

S

H

H

H

H

b

h

S

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4. Volume: The amount of space a solid shape takes up is called its volume or its capacity. The standard unit of measurement for volume is a unit cube, with each edge one unit long. Volume is measured in cubic units e.g. mm3; cm3; m3; km3 Example: Volume of a unit cm cube = 1 cm 1 cm 1 cm = 1 cm3 Volume of a rectangular prism = area of base height = length breadth height

Volume of triangular prism = area of base height = (½ b h) H

Volume of a cylinder = area of circle base height = r2 h where π 3,14 and r is the radius.

5. Conversions We use the following metric units of length and distance. We usually use kilometres (km) for very long distances, metres (m) and centimetres (cm) for shorter lengths and millimetres (mm) for very small measurements.

thousands hundreds tens ones tenths hundredths thousandthskilometre

(km) hectometre decametre metre

(m) decimetre centimetre

(cm) millimetre

(mm) 1. To convert to a smaller unit of length, multiply by 10, 100, 1 000 etc.

Examples: From kilometres to metres: 1,34 km = 1,34 1 000 m = 1 340 m From metres to centimetres: 7,4 m = 7,4 100 cm = 740 cm From centimetres to millimetres: 33 cm = 33 10 mm = 330 mm.

2. To convert to a bigger unit of length, divide by 10, 100, 1 000 etc. Examples: From metres to kilometres: 1 340 m = 1 340 ÷ 1 000 km = 1,34 km From centimetres to metres: 740 cm = 740 ÷ 100 m = 7,4 m From millimetres to centimetres: 330 mm = 330 ÷ 10 cm = 33 cm.

3. To convert units of area, multiply or divide by 102, 1002, (1 000)2 etc. Examples: From km2 to m2: 1,34 km2 = 1,34 (1 000 1 000)m = 1 340 000 m2 From m2 to cm2: 7,4 m2 = 7,4 (100 100)cm = 74 000 cm2 From cm2 to m2: 7,4 cm2 = 7,4 ÷ (100 100)m2 = 0,00074 m2

4. To convert units of volume, multiply or divide by 103, 1003, (1 000)3 etc. Examples: From m3 to cm3: 7,4 m3 = 7,4 (100 cm)3 = 7 400 000 cm3

5. From cm3 to m3: 7,4 cm3 = 7,4 ÷ (100 m)3 = 0,0000074 m3 The same conversions apply to units of capacity: kilolitres (kl), litres (l), milliliters (ml) The same conversions also apply to units of mass: kilograms (kg), grams (g) and milligrams (mg). Sometimes we need to convert between different measurement systems. For example, measurements in recipes are sometimes written in pounds (lb) and ounces (oz). You will be given the conversion factor needed.

r

h

1 cm

1 cm1 cm

length (l) breadth (b)

height (h)

Hh

b

S

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Volume can also be written in litres. It is useful to know 1 000 cm3 = 1 litre. 1 m3 = 1 m 1 m 1 m = 100 cm 100 cm 100 cm = 1 000 000 cm3 = 1 000 litres = 1 kilolitre. Example: If a cylindrical bucket has a radius of 10 cm and a height of 20 cm, how many litres will it contain Volume = r2 h = 3,14 (10)2 20 = 6 280 cm3 = 6,28 litres. Examples: 1. A recipe uses 2,2 lbs of flour. Convert this to grams if 1 lb = 450 g.

2,2 lbs = 2,2 450 g = 990 g From pounds to grams: multiply

2. How many lbs is 2 kg of flour? 2 kg = 2 000 g 2 000 g ÷ 450 g = 4,44 lb From grams to pounds: divide

3. Grandpa says that he is 5 foot 7 inches tall. How tall is this in metres if 1 foot = 12 inches and 1 inch = 2,54 cm? 5 foot 7 inches = (5 12) + 7 = 67 inches 67 2,54 cm = 170,18 cm From inches to cm: multiply 170,18 ÷ 100 = 1,7018 m 1,7 m

4. How many inches long is a 30 cm ruler?

30 cm ÷ 2,54 cm = 11,8 inches. From cm to inches: divide

Example (March 2009 Paper 2, Question 3): As a result of load shedding, Wayne, a chicken farmer, goes back to using a generator to provide dependable power for his chicken shed and his farmhouse. He buys a second-hand diesel tank with a radius of 1 m and a length of 2 m to store the fuel for the generator. 3.1 He decides to paint both the outside surface area of the tank and the stand on which it rests. The surface area of the stand is 1 m2. It takes 1 litre of paint to paint 3 m2 of the surface area. 3.1.1 Calculate the surface area (SA) of the tank in m2. Use the formula: SA = 2πrh + 2πr2 where r is the radius, h is the height and π 3,14. (3) 3.1.2 Calculate the quantity of paint (in litres) needed to paint both the outside of the tank and the stand. Round off your answer to the nearest litre. (5) 3.1.3 If a one litre tin of paint costs R23,63 and a 5 litre tin of paint costs R113,15, calculate the most economical way to purchase the amount of paint needed in Question 3.1.2. (3) 3.2 3.2.1 Calculate the capacity (volume) of the diesel tank in litres where 1 m3 = 1 000 litres. Use the formula: V = r2 h where π 3,14, h is the height and r is the radius. (4) 3.2.2 Farmer Wayne fills the diesel tank to 80% of its capacity. The generator used 72 litres of diesel in 36 hours. Calculate the amount of diesel in litres remaining in the tank after 7 days of the generator running continuously. (8) Solutions: 3.1.1 SA = 2πrh + 2πr2 r = 1 m and h = 2 m SA = 2(3,14)(1)(2) + 2(3,14)(1)2 SA = 12,56 m2 + 6,28 m2 = 18,84 m2

10 cm

20cm

?

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3.1.2 Area to be painted = surface area of stand + surface area of tank = 1 m2 + 18,84 m2 = 19,84 m2 1 litre paints 3 m2 19,84 m2 ÷ 3 m2 = 6,613 litres of paint 7 litres of paint. 3.1.3 If you buy 7 one litre tins, it will cost 7 R23,63 = R165,41. If you buy a 5 litre tin and 2 one litre tins, it will cost R113,15 + 2 R23,63 = R160,41. This is more economical than buying 7 one litre tins. 3.2.1 V = r2 h V = (3,14)(1 m)2(2 m) = 6,28 m3 = 6,28 1 000 litres = 6 280 litres. 3.2.2 In one hour the generator uses = 2 litres. 7 days = 7 24 hours = 168 hours So in 7 days, the generator will use 168 2 litres = 336 litres. Amount of diesel in the tank to start: 80% of 6 280 litres = 0,8 6 280 = 5 024 litres. So after 7 days, the amount of diesel remaining in the tank is 5 024 – 336 = 4 688 litres.

6. Maps, grids and compass direction A map represents an area of land, a city or even a street on paper. The map uses a scale factor to reduce the real size of the area to a representation on paper. A scale on a map shows how the lengths on the ground have been reduced to be represented on the map. For example, if a map has a scale of 1 : 50 000, this means that every unit of measurement on the map represents 50 000 units on the ground. So 1 cm on the map represents 50 000 cm (which is 5 000 metres or 5 km) on the ground. Example (November 2008 Paper 2, Question 5): Naledi High School decides to tile the floor of their school hall using black and grey tiles. The hall is L-shaped. A rectangular stage is located against one wall of the hall as shown in the sketch. The width of the stage is 5 m and the length is 10 m. The stage is not going to be tiled. The size of a square tile is 50 cm by 50 cm. The school needs to calculate how many tiles to buy. 5.1 A scale drawing is to be made of the hall. Determine the scale (in simplified form) to be used if the length of the north wall of the hall is 60 mm. (2) 5.2 Calculate the area of the floor (excluding the stage) to be tiled. (8) Area of a rectangle = length breadth 5.3 The tiler requires that 5% more tiles must be purchased to allow for cutting and breakages. Calculate how many tiles must be bought. (7) 5.4 The ratio of black tiles to grey tiles needed is 4 : 1. Tiles come in boxes of 12. Calculate the number of full (complete) boxes of black tiles that must be bought. (4)

18 m

18 m

10 m

6 m

9 m

NFloor plan of school hall

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Solutions: 5.1 60 mm represents 18 m. Convert to the same units of measurement:

18 m = 18 100 cm = 1 800 cm = 1 800 10 mm = 18 000 mm. So the scale is 60 : 18 000. This can be simplified to 1 : 300.

5.2 Total area of floor (including stage area) Total area of floor (excluding stage area) = (18 m 18 m) – (9 m 6 m) = 270 m2 – (10 m 5 m) = 324 m2 – 54 m2 = 270 m2 – 50 m2 = 270 m2 = 220 m2

5.3 The tiles must cover 220 m2. One tile is 50 cm 50 cm = 0,5 m 0,5 m = 0,25 m2 , = 880 tiles

5% more tiles is 0,05 880 = 44 tiles. So 880 + 44 = 924 tiles must be bought.

5.4 924 tiles divided in a ratio of 4 : 1 Number of black tiles needed is of 924.

924 = 739,2 tiles 740 black tiles will be needed. Number of boxes needed: 740 ÷ 12 = 61,67 So 62 boxes of black tiles will be needed. When we give directions or describe a route, it helps to use direction words such as ‘left’ and ‘right’. It also helps to use the compass directions e.g ‘The church is North-east of the school’. It is useful to place a grid of lines that cross each other at right angles over the map. This allows us to find a specific point on a map and label it with its grid reference. We name a point using its horizontal coordinate first and then its vertical coordinate.

Learn these compass directions!

North (N)

South (S)

West (W) East (E)

North-East (NE)

South-East (SE)

North-West (NW)

South-West (SW)

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Mathematica

Winning Team

Example (Ma5.5 Ger

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25 of 35



Solutions: 5.5.1 C3 5.5.2 South-East 5.5.3 Here are some possible answers:

Turn left in 4th Street. Turn left into Buiten Street. After passing Gerrie Visser Street, turn right into the next street. You will see the petrol station ahead of you. OR Turn left into 4th Street. Turn left into Wishart Street. Turn right into Gerrie Visser Street. Turn left into Buiten Street. At the next street turn right. You will see the petrol station ahead of you. OR Turn in a northerly direction along 4th Street. Turn in a westerly direction along Buiten Street. After passing Gerrie Visser Street, turn in a northerly direction into the next street you come to. You will see the petrol station ahead of you.

5.5.4 (a) Total walking distance on the map = (1,8 cm + 2,7 cm + 1,8 cm + 1,5 cm) = 7,8 cm

(b) Scale is 1 : 11 000. So 7,8 cm on the map is 7,8 11 000 cm on the ground. 7,8 11 000 cm = 85 800 cm = 85 800 ÷ 100 m = 858 m = 858 ÷ 1 000 km = 0,858 km.

Learning Outcome 4: Data handling Data handling is the study of statistics, or data. We collect, organise, analyse and interpret data and we can interpret the data in different ways. 1. Representing data

1.1 Frequency tables Example: In a mathematics class, 30 learners completed a test that was out of 20 marks. Here is a list of their results: 14; 10; 11; 19; 15; 11; 13; 11; 9; 11; 12; 17; 10; 14; 13; 17; 7; 14; 17; 13; 13; 9; 12; 16; 6; 9; 11; 11; 13; 20.

We can show these results in a frequency table (horizontal or vertical). A frequency table shows how many learners there are in each category. We can ‘tally’ the number of learners getting each possible mark out of 20 and then write this as a frequency.

Mark out of 20

Tally Frequency (number of

learners) 6 / 1 7 / 1 8 0 9 /// 3

10 // 2 11 //// / 6 12 // 2 13 //// 5 14 /// 3 15 / 1 16 / 1 17 /// 3 18 0 19 / 1 20 / 1

With this data set, it is more useful to group the data. We can use class intervals of 5 and make this frequency table (including a column for the tallies):

Class interval

Frequency

1 – 5 0 6 – 10 1+1+0+3+2 = 7 11 – 15 6+2+5+3+1 = 17 16 – 20 1+3+0+1+1 = 6

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Data can be represented with a graph. This helps us to read all the data easily and to compare parts of the data with each other.

1.2 Bar graphs

Example (March 2009 Paper 1, Question 2.2): Learners were invited to enter a national essay-writing competition. The 70 winners attended a Youth Forum in Johannesburg. A survey was done to find out how many winners came from each province. The results are given in Table 6. Table 6: Number of winners from each province

Province Number of winners

Eastern Cape 8 Free State 6 Gauteng 10 Kwazulu-Natal 11 Limpopo 8 Mpumalanga 7 Northern Cape 5 North West 6 Western Cape 9 70

2.2.1 What percentage of the winners came from Mpumalanga? (2) 2.2.2 Calculate the ratio of the number of winners from Gauteng to the number of winners from Northern Cape. Give your answer in simplified form. (2) 2.2.3 Suppose one of the winners is chosen randomly. What is the probability that the learner is from: (a) The Eastern Cape (b) South Africa (4) 2.2.4 Draw a vertical bar graph to represent the data in Table 6 on the grid provided. (5) Number of winners attending the Youth Forum

12

10

8

6

4

2

0 EC FS GP KZN LP MP NC NW WC

0

5

10

15

20

1 - 5 6 - 10 11 - 15 16 - 20

Leave equal spaces between the bars. Use a scale with equal intervals to mark

the vertical axis. The graph needs a heading. Both axes need labels. The height or length of the bar represents

the frequency of that category of information.

The bars can be drawn vertically or horizontally.

Class intervals of test marks

Test marks

Province

Num

ber o

f lea

rner

s

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Solutions: 2.2.1 Mpumulanga: 100% = 10% 2.2.2 Gauteng : Northern Cape = 10 : 5 = 2 : 1 2.2.3 (a) P(learners from Eastern Cape) = = = 0,114 or 11,43%

(b) P(learners from South Africa) = = = 1 or 100% 2.2.4 bar graph

Number of winners attending the Youth Forum 12

10

8

6

4

2

0 EC FS GP KZN LP MP NC NW WC

1.3 Pie Charts A pie chart uses sectors of the circle to represent the categories of data. It is easy to compare the categories of data to each other and to the total data in a pie chart. A pie chart shows a value as a fraction of the total. Here is the pie chart for the test results of the mathematics class (refer to frequency table and bar graph). Each frequency has been converted to a percentage in this pie chart.

2. Reading data from tables and graphs When working with data, we can read and interpret the data represented in tables and graphs. We can also compare two sets of data with each other using two tables or two graphs.

1 - 50%

6 - 10 23%

11 - 1557%

16 - 20 20%

Province

Num

ber o

f lea

rner

s

Class intervals for test results

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2.1 Comparing two sets of data represented in tables Example (March 2009 Paper 2, Question 4): One of the aims of the Arrive Alive Campaign is to increase safety on South African roads. The Arrive Alive team decide that one of the ways of alerting the public to the dangers of road travel is to publish the data on fatalities on South African roads. Table 7: Number of fatalities per province

YEAR PROVINCE GAU KZN WC EC FS MPU NW LIM NC RSA

2005 2 959 2 906 1 589 1 366 1 012 1 473 1 156 1 320 354 14 135 2006 3 412 2 967 1 637 1 779 1 175 1 530 1 241 1 262 389 15 392

[Source: www.statssa.gov.za ]The Arrive Alive team also decide to compare the estimated million vehicle kilometres (mvk) travelled in a province to the number of fatalities in that province. Table 8: Million vehicle kilometres (mvk) travelled per province

YEAR PROVINCE GAU KZN WC EC FS MPU NW LIM NC RSA

2005 43 408 20 227 19 514 9 192 7 226 9 767 7 225 6 186 2 731 125 476 2006 44 042 20 750 19 884 9 226 7 517 10 397 7 320 6 056 2 894 128 086

[Source: www.statssa.gov.za ]Note: This means that a total of 43 408 000 000 km were travelled by all the vehicles on Gauteng roads in 2005. Use the data in TABLE 7 and TABLE 8 to answer the following questions: 4.1 Which province had a decrease in both the number of fatalities from 2005 to 2006 and the million vehicle kilometres (mvk) travelled? (1) 4.2 4.2.1 Which TWO provinces had the highest number of fatalities in 2005 and 2006? (2) 4.2.2 In which TWO provinces were the highest million vehicle kilometres (mvk) travelled? (2) 4.2.3 Describe the possible relationship between the number of fatalities and the number of million vehicle kilometres (mvk) travelled per province in the two provinces indicated in Question 4.2.2. (2) 4.3 4.3.1 What percentage of the total number of fatalities in South Africa in 2006 occurred in Gauteng (GP)? (3) 4.3.2 Calculate the number of fatalities per million vehicle kilometres travelled in 2006 in: (a) Gauteng (rounded off to THREE decimal places) (4) (b) The province with the lowest number of fatalities (rounded off to THREE decimal places) (4) 4.3.3 Which of these two provinces do you think is the safest in terms of kilometres travelled and fatalities suffered? Give ONE valid reason for your answer. (3) Solutions: 4.1 Limpopo 4.2.1 Gauteng and KwaZulu-Natal 4.2.2 Gauteng and KwaZulu-Natal 4.2.3 The higher the number of million vehicle kilometres (mvk) travelled, the higher the number of fatalities.

4.3.1 Fatalities in Gauteng = 100% = 22,17%

4.3.2 (a) Gauteng: = = 0,077

(b) Northern Cape: = = 0,134

4.3.3 0,077 < 0,134 so fewer fatalities occur in Gauteng than in Northern Cape per million vehicle km. So Gauteng is safer.

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2.2 Comparing two sets of data represented in two pie charts, bar graphs or broken line graphs Example (November 2008 Paper 2, Question 4.1):The graphs below represent the provincial data on average annual incomes (in thousands of rand) and unemployment rates (as a percentage).

Key: WC = Western Cape KZN = KwaZulu-Natal MP = Mpumalanga EC = Eastern Cape NW = North West LIM = Limpopo NC = Northern Cape GP = Gauteng FS = Free State RSA = whole of South Africa Use the graphs to answer the questions below: 4.1.1 Calculate the difference in value between the average annual income for Gauteng and the Eastern Cape. (4) 4.1.2 What relationship, if any, exists between the unemployment rate and the average annual household income for the provinces? (2) 4.1.3 Give a valid reason why a person would choose to work in Gauteng rather than in the Eastern Cape. (2) Solutions: 4.1.1 Difference between average annual income between Gauteng and Eastern Cape = R80 000 – R28 000 = R52 000 4.1.2 The higher the average income, the lower the unemployment rate. 4.1.3 Gauteng has a higher average annual income and lower unemployment rate than the Eastern Cape. It is likely that a person will earn a better salary in Gauteng and the chances of being employed in Gauteng are better.

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2.3 Interpreting a compound bar graph Compound and multiple bar graphs are used to compare two or more sets of data. Example (March 2009 Paper 1, Question 2.3): The compound bar graph below shows the percentage of South African children from age seven to thirteen, enrolled in primary schools during 1996 and 2007. Percentage enrolment of South African children in primary school

1996 73,1 81,8 87,9 91,3 93,6 94,4 94,8 2007 94,8 95,6 95,9 96,3 96,3 96 95,8

Ages in years [Source: www.busrep.co.za 8 November 2007] Use the graph to answer the following questions: 2.3.1 What percentage of the 10-year-olds was enrolled during 1996? (1) 2.3.2 Calculate the percentage increase in enrolment of 11-year-olds from 1996 to 2007. (3) 2.3.3 Which age group had: (a) The largest percentage enrolment during 1996. (1) (b) The smallest percentage enrolment during 2007 (1) (c) The greatest increase in percentage enrolment between 1996 and 2007. (2) 2.3.4 If there were 240 000 10-year-old children in South Africa in 1996, calculate the number of 10-year-old children enrolled in primary school. (2) Solutions: 2.3.1 91,3% (read this from the table) 2.3.2 Percentage increase = 96,3% – 93,6% = 2,7% 2.3.3 (a) 13-year-olds (b) 7-year-olds (c) 7-year-olds 2.3.4 91,3% of 240 000 = , 240 000 = 219 120 learners

Perc

enta

ge

31 of 35



2.4 Using graphs to represent a point of view Sometimes graph misrepresent or appear to change the facts This may be done purposely to mislead or persuade the audience to a point of view. Some of the reasons why graphs can be misleading:

The scales used are too big or too small for the data Scales that use changing intervals between markings on the axes Scales that do not start at zero Unlabelled axes Leaving out important information

Example (March 2009 Paper 2, Question 6.1): Bathwizz is a company that installs and renovates bathrooms. The general manager had to present the company’s earnings for the first three quarters of the year to the company directors. He drew the graphs below. GRAPH 1 GRAPH 2

Use the graphs to answer the following questions: 6.1.1 What possible trend do you notice with regard to Bathwizz’s quarterly income? (2) 6.1.2 Calculate the average (mean) monthly income for Bathwizz for the first nine months of the financial year. (4) 6.1.3 The general manager wanted to prove to the company directors that Bathwizz’s income was increasing and that the company was doing well. Which graph would be the better one to show to the company directors? Give a reason for your answer. (3) Solutions: 6.1.1 There is a steady increase in income. 6.1.2 The three quarters shown are nine months.

Mean monthly income (in R100 000s) = , , = = R1,22 hundred thousand = R122 000

6.1.3 It will be better to show the company directors Graph 2 because the increase between quarters looks bigger. This is because the scale of the graph starts at 2,5 and not at 1.

s

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5. Measures of central tendency and spread: These are different measures of finding the ‘middle’ of a set of data and how much the set of data is ‘spread out’. In a mathematics class, 13 learners completed a test out of 25 marks. Here is a list of their results:

14; 10; 23; 21; 11; 19; 13; 11; 20; 21; 9; 11; 17.

There are three ways to find the ‘average’ of their results. We can use the mean, the median or the mode.

5.1 To find the mean, we add up all the values (test marks in this example) and divide by the number of values (number of learners in this example)

mean = ( ) ( ) =

=

= 15,3846… = 15,38 rounded off to two decimal places

5.2 The median is the middle number in an ordered data set. If you order the results in ascending order, you get:

9; 10; 11; 11; 11; 13; 14; 17; 19; 20; 21; 21; 23. The middle number is the 7th number out of 13 numbers which is 14. So the median is 14. When there is an even number of values in the data set, the median lies halfway between the middle two values. We can add these two values and divide by 2. Suppose another learner wrote the test and her result was 9. Then we can add this to the ordered data set:

9; 9; 10; 11; 11; 11; 13; 14; 17; 19; 20; 21; 21; 23. Now the middle two numbers are 13 and 14. = = 13,5 So the median is 13,5.

5.3 The mode is the number or value that appears most frequently in the data set.

The mode of the results of the test is 11.

The measures of dispersion or spread give us information about how spread out the data is. The range is the difference between the highest value (or maximum) and the lowest value (or minimum) in a data set. It shows how spread out the whole data set is. Range = largest value in the data set – smallest value in the data set For example, the range of the test results: 23 – 9 = 14.

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Example (November 2008 Paper 2, Question 1.3): Dina and Mpho wrote some English Olympiad practice tests at school. Their marks, in percentages, are given in the table. Table 9: Percentage scored in practice tests

Dina 48 48 48 53 58 62 70 72 80 86Mpho 36 42 48 58 60 61 62 76 86

1.3.1 Mpho’s median mark is 60%. Calculate Dina’s median mark. (2) 1.3.2 Dina’s mean mark is 62,5%. Calculate Mpho’s mean mark. (3) 1.3.3 Calculate the range of Dina’s marks. (2) 1.3.4 Dina stated that she did better in her practice tests than Mpho. Give TWO reasons to support Dina’s claim. (4) Solutions:

1.3.1 Marks are already in ascending order. Dina’s median mark = = = 60%

1.3.2 Mpho’s mean mark = = = 58,78%

1.3.3 Range of Dina’s marks = 86% – 48% = 38% 1.3.4 Dina and Mpho have the same median mark, but Dina’s mean mark of 62,5% is higher than Mpho’s mean mark of 58,78%. The range of Dina’s marks is 38%, while the range of Mpho’s marks is 86% – 36% = 50%, which shows that Dina’s marks are more consistent than Mpho’s marks. 3 Probability Probability is a part of data handling that has to do with the chances of something happening. Probability is the likelihood of an event happening and we can write down this likelihood using numbers. For example, there is a 50% chance that if you flip a coin, it will land on ‘heads’ and not on ‘tails’.

We can write this as P(heads if you flip a coin) = or 50%.

When an event will definitely happen, it has a probability of 1 (or 100%). For example, the probability of the sun rising tomorrow in South Africa is 1. When an event will never happen, it has a probability of 0. For example, the probability of the day before Tuesday being Friday is 0. Any other probabilities lie between 0 and 1, or between 0% and 100%.

We can calculate probability:

For example, what is the probability of throwing a 3 on a dice? There are 6 possible numbers you can get if you throw a dice and only one of them will be a 3.

P(throwing a 3) =

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Example (November 2008 Paper 2, Question 6.4): Shaya FC plays two matches in March. There are three possible outcomes for each match: win (W), lose (L) or draw (D). A tree diagram is drawn to work out the possible outcomes for the two matches. Match 1 Match 2 Possible Outcomes for the two matches 6.4.1 Complete the tree diagram to show all the possible outcomes of the two matches. (4) 6.4.2 Use the completed tree diagram to predict the probability that Shaya FC will: (a) win both matches (2) (b) win only one of the matches (2) (c) draw at least one of the matches (3) Solutions: 6.4.1

6.4.2 (a) There are 9 possible outcomes. P(win both matches) = = 0,1111 = 11,11%

(b) There are 4 possible outcomes or events when they win only one match.

P(win only one of the matches) = = 0,4444 = 44,44%

(c) There are 5 events when they draw at least one (or more) of the matches.

P(draw at least one of the matches) = = 0,5555 = 55,55% 56%

W

L

D

W

L

D

WW

WL

WD

W

L

D

WW

WL

WD

W

L

D

LW

LL

LD

W

L

D

DW

DL

DD

W

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D

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This material has been prepared for schools who participated in the Winning Teams’ World of Work Education Programme in 2009. Only these schools may reproduce this material without prior permission.

For further information, contact: Winning Teams on 011 883 6412 PO Box 786676, Sandton 2146.

Written, adapted and compiled by Susan Jobson, edited by Lynn Bowie.

Examination Preparation for Grade 12 Mathematical Literacy...

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