Grade 11 Mathematical literacy / Graad 11 Wiskundige...

Grade 11 Mathematical literacy / Graad 11 Wiskundige Geletterdheid

Work that must have been done by 11 June (for the year):

Werk wat klaar gedoen moes gewees het teen 11 Junie (vir die jaar):

General Revision / Algemene Herining: P. 8-11

Chapter 1 / Hoofstuk 1: Revision Exercise (Rev ex.) / Hersienings Oefening (Rev ex.) – P. 16+17 Exercise (Ex.) / Oefening (Ex.) 1 – P. 19 Ex. 2 – P. 21+22 Ex. 3 – P. 24+25 Ex. 4 – P. 30+31

Chapter 2 / Hoofstuk 2: Rev ex. – P38+39 Ex. 1 – P. 41 Ex. 2 – P. 41 Ex. 3 – P. 42 Ex. 4 – P. 43 Activity (Act.) / Aktiwiteit (Act.) 1 – P. 45 Act. 2 – P. 46

Chapter 5 / Hoofstuk 5: Rev ex. – P. 112+113 Act. 1 – P. 114+115 Ex. 1 – P. 115+116 Ex. 2 – P. 119 Ex. 3 – P. 121 Ex. 4 – P. 125

Chapter 3 / Hoofstuk 3: Rev ex. – P. 52+53 Act. 1 – P.56 Act. 3 – P. 59 Act. 4 – P. 62+63 Ex. 1 – P. 67 Ex. 2 – P. 72+73 Ex. 3 – P. 76+77

Chapter 4 / Hoofstuk 4: Rev ex. – P. 86+87 Ex. 1 – P. 91+92 Ex. 2 – P. 93 Ex. 3 – P. 97 Ex. 4 – P. 101 Ex. 5 – P. 108+109

Chapter 9 / Hoofstuk 9: Rev ex. – P. 212+213 Summary and questions / Opsomming en vrae – P. 218+219

Chapter 11 / Hoofstuk 11: Rev ex. – P. 240+241 Ex. 1 – P. 244+245 Ex. 2 – P. 247

Chapter 6 / Hoofstuk 6: Rev ex. – P. 128+129 Ex. 1 – P. 133 Ex. 2 – P. 139 Ex. 3 – P. 144 Ex. 4 – P. 148+149 Ex. 5 – P. 152

Chapter 7 / Hoofstuk 7: Rev ex. – P170+171 Ex. 1 – P. 175 Ex. 2 – P.179-181

Work for period from 12 June – 3 July:

Werk vir die tydperk van 12 Junie – 3 Julie:

Chapter 8 / Hoofstuk 8:

Rev ex. – P. 184+185

Ex. 1 – P. 194+195

Ex. 2 – P. 199

Act. 4 – P. 203

(Must be done by 17 June)

Chapter 10 / Hoofstuke 10:

Rev ex. – P. 220+221

Ex. 1 – P. 224+225

Ex. 2 – P. 228+229

Ex. 3 – P. 232+233

(Must be done by 24 June)

Chapter 12 / Hoofstuk 12:

Rev ex. – P. 250+251

Act. 1 – P. 253

Act. 2 – P. 255

Ex. 1 – P. 256

Ex. 2 – P. 257

Act. 3 – P. 261

(Must be done by 3 July)

This chapter will be continued in class if school reopens for grade 11s the week of 6 July.

1

GR 11 Handboekoefeninge antwoorde:

Gr11 Text book exercises answers:

INDEX:

Gr 10 general revision / Gr 10 algemene hersiening – P. 2 - 4

Chapter 1 / Hoofstuk 1 - P. 5 - 13







Chapter 8 / Hoofstuk 8 - P.





2

Gr10 Algemene hersiening (p.8-11)

Gr10 General revision (p.8-11)

Vraag 1 / Question 1

1.2) 1+2+6 = 9

R684 000 ÷ 9 = R76 000

So 1 x R76 000 = R76 000

2 x R76 000 = R152 000

6 x R76 000 = R456 000

R76 000 : R152 000 : R456 000

1.3) 500 ÷ 18 = 27,777

So 27 sakkies/bags

1.4.1) 370 ÷ 30 = 12,333

So 13 houer/trays

1.4.2) 370 ÷ 1,5 = 246 omelette(s)

1.5) 11 – (-7)

= 18o verskil/difference


2.1) 16 000 000 000

2.2) 13,5h = 13h 30m

13 x 60 = 780 + 30 = 810min

810 x 60 = 48 600s

2.3) 2 720 000 000 ÷ 9 000

= 302 222km/s

2.4) Drie honderd en twee duisend twee honderd twee en twintig kilometer per sekonde.

3

Three hundred and two thousand two hundred and twenty two kilometers per

second.

2.5) 300 000km/s

2.6) 300 000 ÷ 40 000

= 7,5 keer/times

2.7.1) 5 : 11

2.7.2) 1,7m ÷ 5

= 0,34m x 11

= 3,74m

2.8.1) 23

40

2.8.2) 23 ÷ 40

= 0,575 x 100

= 57,5%


3.1.1) R 1299,95 + R 200

= R 1499,95

3.1.2) R 200 ÷ R 1499,95

= 0,13 x 100

= 13 %

3.1.3) R 300 ÷ R 2599,95

= 0,115 x 100

= 12 %

3.1.4) Omdat R 300 ‘n kleiner persentasie van R 2599,95 opmaak as wat R 200 van

R 1499,95 opmaak.

4

Because R300 makes up a smaller percentage of R2599,95 than R200 makes up of

R1499,95.

3.2.1) Omdat die kosprys 100% van die prys opmaak en die wins die ander 25%

Because the cost price makes up 100% and the profit the other 25%.

3.2.2) R 6999,95 ÷ 1,25

= R 5599,96


4.1.1) 672 x 0,41953

= R 281,92

4.1.2) R 345,69 ÷ R 0,41953

= 824 kWh

4.2.1) R 26,10 ÷ 5,2

= R 5,01923/kl

4.2.2) 13 – 5,2

= 7,8 x 10,13

= R 79,014 + R 26,10

= R 105,11

5

Hoofstuk 1 / Chapter 1

Hersiening / Revision (p.16+17)

1.1) R 5000, want die heining kan enige lengte wees, maar die hek se prys sal dieselfde

bly.

R 5000, because the fence can be any length, but the gate’s price stays the same.

1.2) R 450/meter, want dit hang van die hoeveelheid meter palissande wat benodig word

af.

R450/metre, because it is dependent on the length of the palisade that will be

needed.

2) (R 450 x 20) + R 5000

= R 9000 + R 5000

= R 14 000

3) Omdat die vergelyking wys dat die hek koste R5000 is en dat die koste van die om

heining afhang van die hoeveelheid meter wat benodig word en saam gee dit vir

jou die totale koste.

Because the equation shows that the gate costs R5000 and that the cost of the

fence depends on the length of fencing that is needed and together it will give

you the total cost.

4)

Aantal meter omheining/ Amount of meters fencing

0 20 40 70 100 150 200

Totale koste/ Total cost (in Rand)

R5000 R14000 R23000 R36500 R50000 R72500 R95000

5.1) Koste, dit hang af van die lengte van die omheining

The cost because it is dependent on the length of the fence.

5.2) Omdat dit moontlik is om die Koste uit te werk vir enige hoeveelheid meter omheining.

Because it is possible to determine the cost for any length of fencing.

6

6)

7.1) R 81 500

7.2) 89m

8.1) Omdat die hek R 5000 kos, al word daar geen omheining op gesit nie.

Because the gate will still cost R5000 even if no fencing was required.

8.2) Omdat daar ‘n konstante tarief vir die omheining is.

Because there is a constant rate for the fencing.

8.3) Nee, omdat dit blyk om direk te wees, maar die grafiek begin nie by die punt 0;0 nie.

No, it seems to be a direct proportion but the graph does not start at the point 0;0.

9) stel die koste van die heining as R500/m en weet net dat jy eers na 100m die hek ingesluit

by hierdie koste kry.

Make the cost of the fence R500/m, but know that only after you have purchased

100m of fence, the gate will be included in the price.

10.1) R40

10.2) R49

10.3) 49 ÷ 75

= R 0,65/Mb

10.4) 49 ÷ 5

= R 9,80/Mb

0

20000

40000

60000

80000

100000

0 40 80 120 160 200

Ko

ste

in r

and

/ c

ost

in r

and

Lengte in meter / lenght in meter

Koste van omheining/cost of fencing

Cost

7

10.6) Nee, want die nie in bondel prys sal my goedkoper uitwerk.

No, because the out of bundle price will work out cheaper.

10.7) Omdat die relatiewe prys per Mb afneem as ek meer Mb van die bondel gebruik.

Because the relative price per Mb will decrease if I use more of the bundle.

Oefening 1 / exercise 1 (p.19)

1.1a)

0

1000

2000

3000

4000

5000

6000

7000

8000

0 1 2 3 4 5

Tota

al /

To

tal (

in R

and

)

Jare / Years

Belegging 1 / Investment 1

groei/growth

8

1.1b)

1.2.1) Belegging 2 / Investment 2

1.2.2) Belegging 1, omdat daar elke keer net R450 bygetel word.

Investment 1, because R450 is added every year.

1.3) Belegging 1: R 7 250.00 + R 450.00

Inventment 1: = R 7 700.00 (J/Y 6)

Belegging 2: R 7 693.12 x 0,08

Investment 2: = R 615,45 + R 7 693.12

= R 8 308,57 (J/Y 6)

Oefening 2 / Exercise 2 (p.21 + 22)

1.1) R4,35 + (0,0137 x R750)

= R 4,35 + R 10,28

= R 14,63

0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 1 2 3 4 5

Tota

al /

To

tal (

in R

and

)

Jare / Years

Belegging 2 / Investment 2

Groei / Growth

9

1.2) R 4,35 + (0,0137 x R 2 383,00)

= R 4,35 + R 32,65

= R 37,00

1.3) R 4,35 + (0,0137 x R 2 500)

= R 4,35 + R 34,25

= R 38,60 --> R37.00(max fooi/fee)

2)

3) ‘n Grafiek wys vir ons dat die tarief vir elke debiet order meer raak, hoe groter die debiet

order is, tot by ‘n maksimum tarief, waar dit van daar af dieselfde bly.

A graph shows us that the fee for each debit order increases as the amount of the debit

order increases, up to a maximum fee, from there it stays at a fixed fee.

oefening 3 / Exercise 3 (p.24+25)

1.1) Reguitlyn: Konstante verskil

Straight line: Constant difference

Afnemendebalans: Konstante verhouding

Diminishing balance: Constant ratio

1.2) Tabel 1: Jaar 4 = R 12 000; Jaar 5 = R0

Table 1: Year 4 = R 12 000; Year 5 = R0

Tabel 2: Jaar 4 = R 24 576; Jaar 5 = R19 660,80

Table 2: Year 4 = R24 576; Year 5 = R19 660,80

1.3) Reguitlyn, omdat die waarde vinniger verminder

Straight line, because the value decreases faster.

Bedrag van debietorder / Debit order amount

R 0,00

R 500, 00

R 1000, 00

R 2000, 00

R 2383, 21

R 2500, 00

R 3000, 00

R 3500, 00

Totale terieffooi / Total fee

R 4,35

R 11, 20

R 18, 05

R 31, 75

R 36, 99

R 37, 00

R 37, 00

R 37, 00

10

2.1) R 440 000

2.2) R 140 000

2.3) R 300 000

2.4) R 20 000 afr

R40 000 eng

2.5) R 120 000 afr

R100 000 eng

2.6) Ja want die voertuig verloor elke jaar naastenby 30% van sy waarde.

Yes, because the vehicle looses about 30% of the value every year.

2.7) Aan die begin / At the beginning

2.8) Wanneer die motor omtrent 5 jaar oud is, want dit sal redelik goedkoop wees en stadig

sy waarde verloor van daar af.

When the car is about 5years old, because it will be relatively cheap and loose its

value slowly from there.

3.1)

3.2) R 235.00

0

50

100

150

200

250

300

350

400

0 1 2 3 4 5 6 7 8 9 10 10+

Tota

l mo

nth

ly d

isco

un

t

No. of gym visits per month

Total gym discount

11

3.3) minder as 3 keer

Less then 3 times

3.4) Die tabel, omdat dit makliker is om vinnig te verstaan.

The table, because it is easier to understand quickly

Oefening 4 / Excerice 4 (p.30 + 31)

1.1) R 2.50/minute

1.2) R 2.90/minute

1.3) +/- 52 seconds

1.4) Die per sekonde tarief, Omdat hy omtrent 40sent per oproep gaan spaar.

The per second option, because he is going to save about 40 cents per call.

2.1) 60min x R 2,85

= R 171,00

2.2) R 115 + (60min x R 2,50)

= R 115 + R 150

= R 265

2.3)

0 60 120 180 240 300 360 420

R0 R171,00 R342 R513 R684 R855 R1026 R1197

R115 R265,00 R415 R565 R715 R865 R1015 R1165

12

2.5) Omtrent 330minute en dit is +/- R950.

About 330 minutes and +/- R950.

2.6) Voorafbetaald, omdat dit R42 goedkoper is.

Prepaid, because it will work out R42 cheaper.

3.1) R 1 500

3.2) Opsie 2, omdat nie in ‘n reguit lyn loop nie, maar eerder in ‘n toenemende kurwe

Option 2, because it is not a straight line, but rather a postive curve.

3.3) Opsie 1 / option 1

3.4) Jaar 8, net onder R 3 000

Year 8, just below R3 000

3.5) Opsie 2, op ‘n stadium raak sy helling baie steil positief en gaan hy opsie 1 verby.

Option 2, at one stage its gradient becomes very steep and it passes option 1.

4.1) R 3 000 ÷ 5

= R 600

115

265

415

565

715

865

1015

1165

0

200

400

600

800

1000

1200

1400

0 60 120 180 240 300 360 420

voorafbetaal

kontrak

13

4.2) R 250 + (R 1 500 ÷ 5)

= R 250 + R 300

= R 550

4.3)

4.4)

4.5) 6 = R500

4.6) Bart se bote, want dit is aansienlik goedkoper per duiker.

Bart’s boats, because it is a lot cheaper per diver.

0

200

400

600

800

1000

1200

1400

1600

2 4 6 8 10 12

Bart

Sally

2 R1500 R1000

4 R 750 R625

5 R600,00 R550.00

6 R500 R500

8 R375 R437.50

10 R300 R400

12 R250 R375

14


Gr10 hersiening / Gr10 revision (p.38+39)

1.1) 1200cm

1.2) 1350g

1.3) 0,0235liter

1.4) 0,135km

1.5) 0,843m

1.6) 0,356kg

1.7) 1650kg

1.8) 1500ml

2.1)1250ml

2.2) 562,5ml

2.3) 105ml

2.4) 775ml

2.5) 10 eetlepels/ tablespoon

2.6) 15teelepels/teaspoons

2.7) 10koppies/cups

2.8)1koppie en 30teelepels

1 cup and 10 tablespoons3)

In woorde In vm/nm –formaat In 24 uur- formaat

Tien uur in die oggend 10:00 vm 10:00

Sewe uur in die aand 7:00 nm 19:00

Drie uur in die middag 3:00nm 15:00

Halfsewe in die oggend 6:30 vm 06:30

Kwart voor twaalf in die oggend 11:45 vm 11:45

Vyf en twintig oor agt in die aand 8:25 nm 20:25

In words In a.m. / p.m. format In 24 hour format

10 o’clock in the morning 10:00 am 10:00

Seven o’clock at night 7:00 p.m. 19:00

Three o’clock in the afternoon 3:00 pm 15:00

Half past 7 in the morning 7:30 am 07:30

Quarter to twelve in the morning 11:45 a.m. 11:45

Twenty five past eight in the evening 8:25pm 20:25

15

4.1a) Twintig voor een in die nag/oggend – 00:40 vm

Twenty too one at night – 00:40 am

b) Sewe minute voor vyf in die oggend – 4:53 vm

Seven minutes too five in the morning – 4:53am

4.2a) Twintig voor een in die middag – 12:40 nm

Twenty to one in the afternoon – 12:40 pm

b) Sewe minute voor vyf in die middag – 4:53 nm

Seven minutes to five in the afternoon – 4:53 pm

5.1) Globe Trekker

5.2) 06:00 – Six o’clock in the morning

5.3) 06:45

5.4) 1 uur en 30 minute

1 hour and 30 minutes

5.5) ‘n uur / an hour

5.6) 45min

6.1) 10ml

6.2) 125ml

6.3) 437,5ml

Oefening 1 / Exercise 1 (p.41)

1) 23dium/inches x 25,4

= 584,2mm ÷ 10

= 58,42cm

2) 6voet/feet x 304,8

= 1828,8mm ÷ 1000

= 1,8288m

16

3) 150myl/miles x 1,609

= 241,35km

4) 6onse/ounces x 28,35

= 170,1g

5) 12pond/pounds x 453,59

= 5443,08g ÷ 1000

= 5,44308kg

6) 40VSA-gelling/US-gallons x 3,785

= 151,4liter

7) 30cm x 0,394

= 11,82duim/inches

8) 1,83m x 100

= 183cm x 0,394

= 72,102 duim/inches ÷ 12

= 6,0085voet/feet

9) 250km x 0,621

= 155,25myl/miles

10) 150g x 0,035

= 5,25onse/ounces

11) 25kg x 2,205

= 55,125pond/pounds

12) 15liter x 0,264

= 3,96 VSA-gellings/US-gallons

17

Oefening 2/Exercise 2 (p.41)

1.1) *F = (1,8 x 23) + 32

= 73,4*F

1.2) *C = (84 - 32) ÷ 1,8

= 28,9*C

1.3) *F = (1,8 x 0) + 32

= 32*F

1.4) *C = (0 – 32) ÷ 1,8

= - 17,8*C

2) *F = (1,8 x 100) + 32

= 212*F


1) Gis/yeast: 10g ÷ 4

= 2,5 x 5

= 12,5ml

Margarien/margarine: 60 ÷ 5

= 12 x 5

= 60ml

2) Sout/salt: 3 x 5

= 15ml ÷ 5

= 3 x 5.5

= 16,5g

Appelkooskonfyt/apricot jam: 90 ÷ 5

= 18 x 6

= 108g

18

3) Gis/yeast: 12,5ml ÷ 5

= 2,5teelepels/teaspoons

Margarien: 60 ÷ 15

= 4 eetlepels/tablespoons


1.1) 45cm3 ÷ 1

= 45 x 1

= 45ml

1.2) 3050cm3 ÷ 1000

= 3,05 x 1

= 3,05Liter

1.3) 45m3 ÷ 1

= 45 x 1000

= 45 000Liter

1.4) 375m3 ÷ 1

= 375 x 1000

= 375 000liter ÷1000

= 375kl

1.5) 350ml ÷ 1

= 350 x 1000

= 350 000mm3

1.6) 20 500ml ÷ 1

= 20 500 x 1

= 20 500cm3

19

1.7) 20 750liter ÷ 1000

= 20,75 x 1

= 20,75m3

1.8) 30kl x 1000

= 30 000liter ÷ 1000

= 30 x 1

= 30m3

2.1) 150liter x 1000

= 150 000ml ÷ 1

= 150 000 x 1

= 150 000cm3

2.2) 150liter ÷ 1000

= 0,15 x 1

= 0,15m3

3.1) 32m3 ÷ 1

= 32 x 1000

= 32 000liter

3.2) 32 000liter ÷ 1000

= 32kl

20


Hersienings oef./Revision exer. (P.52+53)

1.1.1) 29/11/2007 – 29 November 2007

1.1.2) R358,86

1.1.3) 926kWh

1.2.1) 40 x R2,80

= R112,00

1.2.2) R112,00 x 0,14

= R15,68

1.2.3) R112,00 + R15,68

= R127,68

1.2.4) 926 x R0,219

= R202,79

1.2.5) R127,68 + R231,18

= R358,86

1.3) R0,4034 x 926

= R373,55

Nee dit sou duurder uitwerk / No it would be more expensive

2.1.1) Inkomste: Geld wat deur ‘n persoon of organisasie ontvang word.

Income: Money that is received by a person or organisation.

2.1.2) Uitgawe: Geld wat deur ‘n persoon of organisasie spandeer word.

Expenditure: Money that is spent by a person or organisation.

2.1.3) Wins: Wanneer die inkomste meer as die uitgawes is.

Profit: When the income is more than the expenses.

2.1.4) Verlies: Wanneer uitgawes meer as inkomstes is.

Loss: When the expenses are more than the income.

21

2.2) ‘n Inkomste- en uitgawestaat werk met die werlike syfers van geld wat in kom en geld

wat uitgegee word, waar ‘n begroting met geskatte waardes werk.

An income-and-expenditure statement works with the actual amounts of money that

was received and money that was spent, where as a budget works with estimated

values.

2.3.1) Begroting, al die waardes is afgerond na getalle wat maklik is om mee te werk, ipv die

werklike syfers van wat alles gekos het.

Budget, Because all the values are rounded to values that are easy to work with,

instead the actual values of what everything costs.

2.3.2) Dit kan ‘n persoon ‘n idiee gee van waaraan al sy geld spandeer gaan word en of daar

moontlik gaan spaar geld oorbly.

It will give you an idea of what you spent your money on and if there might be any

money left over.

2.3.3) Omdat jy maklik kan begroot vir vaste uitgawes, omdat hulle waarde nie deur die

maand verander nie, waar veranderlike uitgawes moeilik is om voor te begroot,

omdat hulle waarde kostant sal verskil

Because it is easy to budget for fixed expences, because their values will not change

throughout the month, where as variable expences are difficult to budget for,

because their values keep changing.

2.3.4) Totale uitgawes – Ja

Geld wat oorbly – Nee, dit moet R526,85 wees.

Total Expenditure – Yes

Money left over – no, it should be R526,85

2.3.5) Ja,want sy begroot vir alle moontlike uitgawes en daar sal steeds spaar geld oorbly.

Yes, because she is budgeting for all posible expences and there will still be money

left over.

2.3.6) “Enige 2 geldige opsies is korrek”

“Any 2 good options is correct”

22

Oefening 1 / Exercise 1 (p. 67)

1) Oranje krale / Orange beads:

R26,00 ÷ 40

= R0,65 x 8

= R5,20

Pers krale / Purple beads:

R14,95 ÷ 40

= R0,37 x 30

= R11,10

Grys krale / Gray beads:

R16,95 ÷ 30

= R0,57 x 24

= R13,68

Wynrooi krale / red beads:

R18,95 ÷ 20

= R0,95 x 4

= R3,80

Metaalknippie / Metal clasp:

R4,95 ÷ 2

= R2,48

Metaalhakies / Metal hooks:

R4,95 ÷ 50

= R0,10 x 2

= R0,20

Rek / Elastic:

3m x 100 = 300cm

23

So R18,95 ÷ 300

= R0,06 x 45 = R2,70

R5,20 + R 11,10 + R13,68 + R3,80 + R2,48 + R0,20 + R2,70

= R39,16

2) Omdat al die pryse afgerond is tot die naaste sent en dit sluit ook nie enige ander

bedryfsuitgawes in nie.

Because all the prices are rounded to the closest cent and this does not include any

other bussiness expences.

3) R80,00, omdat dit ‘n ronde bedrag is en dit behoort nie te duur te wees vir meeste

moontlike koopers nie.

R80,00 , Because it is a round number and it should not be too expensive for most

possible buyers.

4) R80,00 – R39,16 = R40,84 wins/profit

So R40,84 ÷ R39,16

= 1,04 x 100

= 104% wins/profit

5.1) Nee, die hoeveelheid wins sal slegs R9,50 wees

No, the amount of profit will only be R9,50

5.2) R9,50 ÷ R25,50

= 0,37 x 100

= 37% wins/profit

Dus sal dit ‘n kleiner persentasie wins oplewer.

Thus it will generate a smaller percentage profit.

5.3) Dit beteken dat die produk teen ‘n laer koste gemaak en verkoop word, maar dat daar

‘n groter persentasie wins op elke produk is.

It means that the product is produced and sold at a lower cost, but that there is a

higher percentage profit on each item.

24

Oefening 2 / Exercise 2 (p.72+73)

1.1)

1.2) Nee, omdat haar uitgawes vir 15 halsnoere aansienlik meer sal wees as haar inkomste.

No, because her expences for 15 necklaces will be much more than her income.

1.3) 49 necklaces ; R2 450

1.4.1) Uitgawes = Huur + Vervoer + (halsnoere x R37,40)

= R500 + R120 + (H x R37,40)

Expences = Rent + transport + (necklaces x R37,40)

= R500 + R120 + (N x R37,40)

1.4.2) Inkomstes = Halsnoere x R50

Income = Necklaces x R50

1.5) H/N = 46

1.6) 50H = R670 + 37,4H

13,6H = R670

H = 49

1.7.1) 60H = R620 + 37,4H

23,6H = 620

H = 26

0

500

1000

1500

2000

2500

3000

3500

4000

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60

25

1.7.2) “sien eerste grafiek” / “see first graph”

1.8) Sodat hulle weet hoeveel van hulle produk hulle moet verkoop voordat daar ‘n wins

gemaak word.

So that they know how many of their products they need to sell before there is going

to be a profit made.

2.1) Twee verskillende selfoonkontrakte wat met mekaar vergelyk word.

Two different cell phone contracts that are being compaired

2.2) Dit is om die gratis minute te illustreer

It is to illustrate the free minutes

2.3) Dit wys hoe die maandlikse koste vir die minute wat gebruik is toeneem en dit is ‘n

reguitlyn omdat dit teen ‘n konstante koste toeneem.

It shows how the monthly cost for the minutes use increases and it is a straight

because it increases at a constant rate

2.4) Dit beteken dat die per minuut koste hoër is vir die blou grafiek as vir die rooi grafiek.

It means that the cost per minute is higher for the blue graph than it is for the red

graph.

2.5) (40minutes ; R100)

2.6) Dit gee mens ‘n idiee van hoeveel minute jy sal moet gebruik voordat dit goedkoper sal

wees om die rooi grafiek se kontrak te gebruik, as die blou grafiek sin.

It gives you an idea of how many minutes you need to use before it will be cheaper

to use the contract shown by the red graph than the one shown by the blue graph.

2.7) Indien jy weet hoeveel minute jy per maand gebruik, kan die grafiek jou help om te

weet watter kontrak vir jou beter waarde vir geld sal gee.

If you know how many minutes you use in a month, the graph help you decide which

option will give you more value for your money

Oefening 3/ Exercise 3 (p. 76 + 77)

1.1) (6 x R0) + (2 x R4,92)

= R9,84

26

1.2) (6 x R0) + (4 x R4,92) + (3 x R10,51)

= R51,21

1.3) (6xR0)+(4xR4,92)+(10xR10,51)+ (15xR15,57)+(4,2xR18,99)

= R438,09

1.4) (6xR0)+(4xR4,92)+(10xR10,51)+ (15xR15,57)+(15xR18,99)+ (18,5xR25,37)

= R1 112,53

2)

Verbruik/consumption (kl) Tarief/tariff (rand)

Van/from Tot/to Tarief/ tariff in 2010

Tarief/tariff in 2011 % Toename/ increase in tarief/ tariff

0 6 Gratis Gratis 0

6 10 R4,55 R4,92 8,13

10 20 R9,70 R10,51 8,35

20 35 R14,38 R15,57 8,28

35 50 R17,76 R18,99 6,93

50 R23,42 R25,37 8,32

Gemiddelde % toename:

Average % increase:

(8,13 + 8,35 + 8,28 + 6,93 + 8,32) ÷ 5

= 8,002 , so dus was die artikel korrek/ So the article was correct

3.1) Omdat die toename aanvanklik teen ‘n lae tarief plaasvind.

Because the initial increase is at a lower tariff

3.2) +/- R190

3.3) +/- R650

3.4) Nee, omdat die tarief vir hoë water verbruik aansienlik meer is as die vir relatiewe lae

water verbruik.

No, because the tariff for high water consumption is a lot higher than the tariff for

low water consumption.

3.5) Om vir hulle te wys dat dit die moeite werd is om te probeer water spaar of minder

water te gebruik.

To show them that it is worth the effort to try and save water or to use less water

27

4.1) Om hulle begroting en ook betaling daarvoor makliker te maak.

To make their budgeting and payment for the water easier.

4.2) 500kl x R9,25

= R4 625

4.3.1) Omdat die koste vir water by besighede teen ‘n konstante faktor vermeerder.

Because the cost of water for businesses increase at constant factor.

4.3.2) +/- 37kl

4.3.3) Om te kan sien waar dit meer voordelig is om as ‘n besigheid te regestreer teenoor ‘n

privaat huishouding.

To see when it becomes more advantages to register as a business, as opposed to

being a private residence.

28


Hersienings oefening/ Revision exercise

1.1.1) ob/oa: R500

Rk/ir: 5%

Rente/interest: R25

1.1.2) ob/oa: R3 785,50

Rk/ir: 1,8%

Rente/interest: R68,14

1.1.3) ob/oa: R17 125

Rk/ir: 5,3%

Rente/interest: R907,63

1.2a) R500 x 0,05

= R25 (korrek/corect)

b) R3 785,50 x 0,018

= R68,14 (korrek/corect)

c) R17 125 x 0,053

= R907,63 (korrek/corect)

1.3) rente is die fisiese geld waarde waarmee ‘n belegging groei, waar rentekoers is die

persentasie waarde waarteen ‘n belegging groei.

Interest is the actual money value at which the investment grows, whereas interest

rate is the percentage value at which the investment grows.

1.4.1) R1 800

1.4.2) 0,5% per maand/month

1.4.3) R1 800 x 0,005

= R9,00

1.4.4) R1800 + R9

= R1 809

29

1.4.5) R1 809 x 0,005

= R9,05

1.4.6) R1 827,14 + R9,14

= R1 836,28

1.4.7) R1 836,28 x 0,005

= R9,18 + R1 836,28

= R1 845,46

2.1.1) R800

2.1.2) R14,15

2.2) - R532,60 uit rekening vir Vodacom selfoon debiet order/ R532,60 out of account for

Vodacom cell phone debit order

- R1478,32 uit rekening vir aankope by Checkers/ R1478,32 out of the account for

purchase made at Checkers.

2.3) Dit is die fooie wat op elke transaksie gehef is / It is the fees charges on all the

transactions.

2.4) Debiet is geld wat uit die rekeing uit is

Krediet is geld wat in die rekening in betaal is.

Debit is money going out of the account

Credit is money going into the account.

2.5) R5 284,40 – (R14,15 + R800)

= R5 284,40 – R814,15

= R4 470,25

2.6) R 4 470,25 – (R11,07 + R532,60)

= R3 926,58 – (R4,15 + R1 478,32)

= R2 444,11

30

2.7) R195,71 – R98,59 + R8 585,55

= R8 682,67 – (R17,75 + R1500)

= R7 164,92

2.8) OTM-onttrekking/ ATM withdrawl:

(R800 x 0,0125) + R4,15

= R14,15

Debietorder aan ander maatskappye/ Debit order to another company:

(R532,60 x 0,013) + R4,15

= R11,07

Debietkaartaankope/ Debit card purchase:

R4,15

Onttrekking by tak met OTM-kaart/ Withdrawl at branch with ATM card:

(R2 200 x 0,012) + R22

= R48,40

Kontantdeposito/cash deposit:

(R8 585,55 x 0,011) + R4,15

= R98,59

Elektroniese betaling/Elecrtonic payment:

(R1 500 x 0,008) + R4,15

= R17,75

Oefening 1/ Exercise 1 (p91+92)

1.1) R500 x 0,02

= R10 x 3

= R30 + R500

= R530

31

1.2) R3 200 x 1,018

= R3 257,60 x 1,018

= R3 316,24

1.3) 4,5% ÷ 12

= 0,375 x 3

= 1,125%

R8 750 x 1,01125

= R8 848,44

2.1) R1 357,50 x 1,01758

= R1 381,36(month 1) x 1,01758

= R1 405,65(month 2) x 1,01758

= R1 430,36(month 3)

2.2) R24 998,55 x 1,000457

= R25 009,97(month 1) x 1,00129

= R25 042,24

2.3) Omdat hulle meer geld moet maak uit geld wat hulle uitleen as wat hulle rente moet

betaal aan mense wat hul geld by hulle belê / Because they need to make more

money out of the money that they lend out, then the interest that they need to pay

to the people investing with them.

3.1) R600 for 24months

3.2) R21

3.3) R21 ÷ R600

= 0,035 x 100

= 3,5%

3.4) Soos wat die tyd verloop ,groei die geld wat belê is al steiler, soos wat die kapitaal in die

rekening meer raak.

As time goes on, the growth of the money invested becomes bigger, as the capital in

the account becomes more.

32

3.5) R1 278,91 x 1,022

= R1 307,05(month 23) x 1,022

= R1 335,80(maand 24)

4) R340 000 x 0,8

= R272 000(after 1 year) x 0,88

= R239 360(after 2 years)


1.1) A=P(1+i)n

= 500(1+0,012)3

= R518,22

1.2) 4,5% ÷ 12

= 0,375% per month

A = P(1+i)n

= 12750(1+0,00375)4

= R12 942,33

2) R500 x 1,012

= R506(month 1) x 1,012

= R512,07(month 2) x 1,012

= R518,22(month 3)

3)a) R518,22 – R500

= R18,22

b) R12 942,33 – R12 750

= R192,33

33

4.1)

Maand / Month

Bedrag geld in die rekening aan die begin van die maand / Amount of money in the account at the beginning of the month

Rentekoers = 3% per jaar (maandeliks saamgestel) / Interest rate = 3% per year (compounded monthly

Bedrag geld in die rekening aan die einde van die maand / Amount of money in the account at the end of the month

Month 1

R15 000,00 0,25% = R37,50 R15 037,50

Month 2

R15 037,50 0,25% = R37,59 R15 075,09

Month 3

R15 075,09 0,25% =R37,69 R15 112,78

4.2) A = P(1 + r)n

= 15000(1 + 0,0025)6

= R15 226,41

Oefening 3 / Exercise 3 (bl. 97)

1.1) 12% ÷ 365

= 0,032877% per dag/day

1.2) 0,25% x 12

= 3% per year ÷ 365

= 0,008219% per dag/day

2.1) 1,2% ÷ 365

= 0,003288% per dag/day

34

3.1)

Maand / Month

Dae / Days

Openingsaldo / Opening balance

Daaglikse rente / Daily interest

Totale rente vir die maand / Total interest for the month

Eindsaldo / Closing balance

Jan. 31 R8 000,00 R1,424658 R44,16 R8 044,16

Feb. 28 R8 044,16 R1,532522 R40,11 R8 084,28

Mrt. 31 R8 084,28 R1,439665 R44,63 R8 128,90

Apr. 30 R8 128,90 R1,447612 R43,43 R8 172,33

Mei. 31 R8 172,33 R1,455346 R45,12 R8 217,45

3.2) R8 217,45 x 0,00017808

= R1,463382 X 30

= R43,90 + R8 217,45

= R8261,35

Rente / Interest:

R8 261,35 – R8 000

= R261,35

Oefening 4 / Exercise 4 (p. 101)

1.1) T.V.: R3 550 x 0,15

= R532,50

Motor/car: R105 000 x 0,08

= R8 400

Huis/house: R 750 000 x 0,1

= R75 000

35

1.2) T.V.: R385,20 x 24

= R9 244,80

Motor/car: R2 052,75 x 60

= R123 165

Huis/house: R5 663,25 x 300

= R1 698 975

1.3) T.V.: R532,50 + R9 244,80

= R9 777,30

Motor/car: R8 400 + R123 165

= R131 565

Huis/House: R75 000 + R1 698 975

= R1 773 975

1.4) T.V.: R9 777,30 – R3 550

= R6 227,30

Motor/car: R131 565 – R105 000

= R26 565

Huis/house: R1 773 975 – R750 000

= R1 023 975

2.1) Deposito/Deposit: R3 699 x 0,05

= R184,85

Werklike koste/ real cost: (R205,85 + R12,50) x 48

= R218,35 x 48

= R10 480,80

Totale koste / Total cost: R184,85 + R10 480,80

= R10 665,65

36

2.2) R10 665,65 – R3 699

= R6 966,65

2.3) Nee, omdat jy amper 3 keer meer betaal op huurkoop as wanneer jy dieselfde item

kontant gekoop het.

No, Because you pay almost 3 times as much for the item on hire purchase then

you would have when buying it cash.

Oefening 5 / Exercise 5 (p. 108-109)

1.1) R6,20 x 1,03

= R6,39

1.2) R9,49 x 1,05

= R9,96

1.3) R580 000 x 1,083

= R628 140

1.4) R1,2247 x 1,252

= R1,5333

2) R1,2247 x 950

= R1 163,47

R1,5333 x 950

= R1 456,64

So R1 456,64 – R1 163,47

= R293,17 meer elke maand/ more every month

3.1) Duurder, want alhoewel die persentasie inflasie per jaar afgeneem het, het die prys bly

styg.

More expensive, because even though the percentage inflation per year

decreased, the price still increased.

37

3.2) R7,40 x 1,034

= R7,65 x 1,013

= R7,75 x 1,006

= R7,80

4) Sodat hulle weet watter persentasie verhoging hulle moet kry, sodat hulle nie tegnies

minder verdien as gevolg van inflasie nie.

So that they know what percentage increase they need, so that they aren’t

technicaly loosing money because of inflation

5.1)

Maand / Month Beginsaldo / Opening balance

Rente / Interest

Eindsaldo / Closing balance

22 R4383,08 R20,09 R4403,17

23 R4403,17 R20,18 R4423,35

24 R4423,35 R20,27 R4443,62

5.2) R4 000 x 1,072

= R4 288 x 1,113

= R4 772,54

5.3) Afgeneem, omdat selfs met die rente wat die geld verdien het, kon dit nie die prys van

goedere met inflasie klop nie.

Decreased, Because even with the interest earned, the money in the investment

could not beat the price of goods with inflation.

38


Hersienings oefening / Revision exercise (p. 112 – 113)

1.1) Liniaal / ruler

1.2) Maatband / measuring tape

1.3) Odometer

1.4) Winkelskaal / scale

1.5) Maatbeker / measuring cup

1.6) Maatkoppie of koppie / measuring cup or cup

1.7) Eetlepel of teelepel / Table spoon or teaspoon

1.8) Termometer / thermometer

2.1) cm / mm

2.2) m

2.3) km

2.4) g / kg

2.5) liter / milliliter

2.6) liter / kiloliter

2.7) oC

2.8) oF 2.9) g / ml

3.1)a) ‘n kamer wat geteël moet word / A room that needs to be tiled.

b) Die afstand van Johannesburg tot Kaapstad / The distance from Johannesburg to

Cape Town.

3.2) a) Die gewig van ‘n pakkie wat internasionaal gepos moet word / The weight of a

package that is sent internationally.

b) Die gewig van ‘n vrag sand wat by ‘n bouperseel afgelewer moet word / the

weight of a load of sand delivered to a building site.

39

3.3)a) Die hoeveelheid medisyne wat vir ‘n persoon moet gegee word / the amount of

medicine given to a patient.

b) Die hoeveelheid water wat in ‘n emmer moet gegooi en met

vloerskoonmaakmiddel gemeng moet word / The amount of water that needs to

be mixed with tile cleaner in bucket.

3.4)a) Die temperatuur waarteen ‘n eksperiment gedoen moet word / the temperature at

which an experiment needs to be done.

b) Hoe warm die dag is / how warm the day is.

4) A – 2cm

B – 4,4cm / 4,3cm

C – 7,3cm / 7,2cm

D – 10,4cm / 8,3cm

5.1) 23,5kg

5.2) 700g

5.3) 700ml

5.4) 88oF en/and 31oC

6.1) R10,95 x 6,75

= R73,91

6.2) R13,45 ÷ R12,65

= 1,06kg

7) 250ml ÷ 150g

= 1,666 x 350

= 583,33ml

40

Aktiwiteit 1 / Activity 1 (p. 114 + 115)

1) (27,5cm x 43) + 17,5cm

= 1 185,5cm + 17,5 cm

= 1 200cm ÷ 100

= 12m

2.1) 40 x 30cm

= 1 200cm ÷ 100

= 12m

2.2) ja / yes

2.3) (33 x 30cm) + 10cm

= 990cm + 10cm

= 1 000cm ÷ 100

= 10m

2.4) 1 000cm ÷ 27,5cm

= 36voet/feet en/and 10cm

Oefening 1 / Exercise 1 (p.115+116)

1.1) Hulle kan die maatband vat om 3m van die lengte meet, ‘n merk te maak en weer van

daar af 3m te meet, indien daar minder as 3m oor is nadat hulle ‘n merk gemaak het,

kan hulle net die oorblywende stuk meet en al die korter lengtes by mekaar tel.

They can use the tape measure to measure 3m, mark it and then measure 3m

again and again, if there is less then 3m left after marking, they can measure that

and add all the shorter lenghts together.

1.2) 40 x 30cm

= 1200cm ÷ 100

= 12m ÷ 3

= 4 maatband lengtes wat wel ‘n heelgetal is / 4 tape measure lengths and this is a

whole number.

41

1.3) 33 x 30cm + 10cm

= 1000cm ÷ 100

= 10m ÷3

= 3,3 maatband lengtes, wat nie ‘n heelgetal is nie / 3,3 tape measure lenghts and

this is not a whole number.

2.1.1) 100 tree / steps

2.1.2) 70 tree / steps

2.1.3) 144 tree / steps

2.2) Doel tot doel / Try line to try line: 100 x 0,95 = 95m

Breedte / width: 70 x 0,95 = 66,5m

Totale lengte / total lenght: 144 x 0,95 = 136,8m

2.3) Doel tot doel / try line to try line: 100 ÷ 0,95 = 105 tree / steps

Breedte / Width: 70 ÷ 0,95 = 74tree / steps

Totale lengte / total lenght: 144 ÷ 0,95 = 152 tree / steps


1) S2 = 3,92 + 1,92

= 15,21 + 3,61

√𝑠2 = √18,82

S = 4,34m per koord / cord

42

2) S2 = 32 + 1,252 = 9 + 1,56

S = √10,56 = 3,25m per lang koord / longer cord

S2 = 1,52 + 0,62 = 2,25 + 0,36

S = √2,61 = 1,62m per kort koord / shorter cord

Dus die totale hoeveelheid hout benodig / The total amount of wood needed:

4(1,5m) + 2(0,6m) + 2(1,62m) + 2(3,25m) + 1,25m

= 6m + 1,2m + 3,24m + 6,5m + 1,25m

= 18,19m


1) Nee / No

2) Omdat 50oC wel warm vir die dag temperatuur is, maar dit sal selde nodig wees om die

dag se temperatuur bo 50oC te meet.

Because 50oC is warm for the day temperature, but it will seldomly be neccesary to

measure the day temperature above 50oC.

3) Omdat die dag se temperatuur gereeld negatief is in OC, veral in die winter in die

Noordelikke halfrond.

Because the day temperature regularly goes into negative oC, especially in the winter

in the Northen Hemisphere.

4) Nee, die dag se temperatuur gaan selde in die negatiewe waardes in met oF

No, the day temperature rarely goes into negative oF.

5) Sondag/Sunday & Maandag/Monday:

maksimum: +/- 83oF

Minimum: +/- 55oF

6) Maksimum: oF = (1,8 x 26) + 32 = 78

Minimum: oF = (1,8 x 9) + 32 = 48,2

43

7) Somersklere, maar dalk ‘n trui inpak vir die aand, maar ‘n sambreel sal nie nodig wees nie.

Summer clothes and perhaps a jersey for the evenings, but he does not need an

umbrella.

8) Ja, want dit gaan sonnig en lekker wees

Yes, because it will be sunny and pleasant.

44

Hoofstuk / chapter 6

Hersieningsoefening / revision exercise (p.128 + 129)

1.1) 3 of 1

1.2) 4

1.3) Nuwe veld of laer-damveld / New field or lower dam field

2) 15

3.1) Nee, dis is net oop tot 14:00 / No, that gate closes at 14:00

3.2) Nee, die hek is gesluit oor naweke

No, That gate is closed over weekends

3.3) - Ry verby City lodge aan jou linkerkant totdat jy by Peter place kom. / Drive past City

lodge on your left, untill you get to Peter place.

- Draai regs in Peter place in / Turn right into Peter place.

- Volg die pad totdat jy by ‘n verkeerslig kom, waar Coachman’s Herberg aan jou

linkerkant sit / Follow the road untill you get to a traffic light where Coachman’s Inn

will be on your left.

- Draai dan regs, ingang 1 sal dan voor jou wees / Then turn right, entrance 1 will be in

front of you.

4) Sodat mense wat nie die skoolgronde ken nie, hul pad kan vind binne die skool.

So that people who aren’t familiar with the school grounds can find their way within

the school grounds.

5) Sodat mense kan uitwerk hoever dit van een plek na ‘n ander op die kaart is.

So that people can determine how far it is from one place with in the school to

another.


1.1) Begin in Johannesburg en eindig in George / starts in Johannesburg and ends in George.

1.2) Johannesburg, Potchefstroom, Klerksdorp, Kimberley, Beaufort-wes en George.

2) 2

45

3) 3

4) N3

5) N1 Suid tot Colesberg, dan N9 tot in

Middelburg, van daar af die N10 tot waar dit by die N2 aansluit en dan die N2 wes tot

in Port Elizabeth / N1 South up to Colesberg, then N9 to Middleburg, from there N10

to where in merges with the N2 and from there N2 to Port Elizabeth

OF / OR

N6 suid tot in Oos-Londen dan vat jy die N2 wes tot in Port Elizabeth / N6 south up to

East Londen then take the N2 west to Port Elizabeth.

6.1) N2

6.2) 984km

7) 1402km

8.1) Durban - Bloemfontein - Kaapstad: 634km + 1004km = 1638km

Durban - Oos-Londen - Port Elizabeth - Kaapstad: 674km + 310km + 769km

= 1753km

Dus is die eerste roete korter/ the first route is shorter

8.2) Die Durban-Bloemfontein-Kaapstad roete / The Durban-Bloemfontein- Cape Town

route.

9.1) 578km

9.2) 578km ÷ 90km/h

= 6,42

So ongeveer 6 en ‘n half uur. / Around about 6 and a half hours.


1.1) 210km + 618km

= 828km

1.2) 1 keer voor hy/sy wegtrek en 1 keer oppad

1 time before he starts the journey and 1 time on route

46

1.3) Bethlehem of Harrismith want dit is omtrent hoe ver die motor kan ry op een tenk.

Bethlehem or Harrismith, because that is about the distance that can be driven on 1

tank of petrol.

2.1) Omdat die vervoer koste om die brandstof op die verskillende plekke te kry verskil.

Because the transport cost to get the fuel to the different places will differ.

2.2) 550km ÷ 10km/l

= 55liter tank

So (55 x R10,46) + (55 x R9,49)

= R575,30 + R521,95

= R1097,25


1) Capital

2)10 of/or 11

3.1) AS20

3.2) AR20

3.3) AS18

3.4) AS21

4) Ry met die straat wat die hospitaal verlaat, tot dat jy by Hyslop straat kom.

Draai links in Hyslop straat in en volg dit tot jy by ‘n sirkel kom.

Vat die derde uitgang uit die sirkel uit en draai in Chatterton straat in.

Volg Chatterton straat oor sewe verkeersligte.

By die agtste verkeerlig sal die hospitaal aan jou regterkant wees.

Follow the road that leaves the hospital until you get to Hyslop street.

Turn left into Hyslop street and follow it until you get to a circle.

Take the third exit into Chatterton street.

Follow Chatterton street past seven traffic lights.

When you get to the 8th traffic light, the hospital will be on your right hand side.

47

5) - Konstruksie werke op die pad / construction on the road

- Verkeer / traffic

- Die gerieflikheid van die roete / the convenience of the route

6) Omdat hulle weet watter paaie rondom hulle die gerieflikste is en watter die minste

verkeer sal hê, asook watter roetes die minste tyd sal neem.

Because they know which routes around them are the most convenient and which will

have the least traffic aswell as which routes will be the quickest.

Oefening 4 / Exercise 4 (p.148-149)

1.1.1) 300 000cm

1.1.2) 300 000cm ÷ 100

= 3 000m ÷ 1000

= 3km

1.2.1) 12,5km x 1000

= 12 500m x 100

= 1 250 000cm ÷ 20 000

= 62,5cm

1.2.2) Nee, omdat die kaart nie 62,5cm lank of wyd is nie, indien die roete baie kronkels het

sal jy moontlik kan.

No, because the map is neither 62,5cm long or wide, only if the route has a lot of

turns, will you be able to find both.

2.1) 15mm

2.2) 1000m ÷ 15

= 66,667m per 1mm

So 66,667 x 182,5mm

= 12 166,667m ÷ 1000

= 12,2km --> 12km

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2.3.1) 89km x 1,5cm

= 133,5cm

2.3.2) Nee, omdat dit nie op meeste papiere of kaarte sal pas nie.

No, because it won’t fit on most paper sizes or maps.

3) “Enige geldige roete”


1.1) Pietermaritzburg na/to Durban

Want die begin is in Pietermaritzburg / Because the starting point is in

Pietermaritzburg.

1.2) 9

1.3) - Hulle kan beplan hoeveel keer hulle gaan water drink sovel as waar.

- Hulle kan hul ondersteuners laat weet waar om te staan om hulle aan te moedig.

- They can plan how many times they want to drink water aswell as where.

- They can tell their supporters where to stand to cheer them on.

2.1) Afwedloop, want die algehele helling van die roete is aftraande.

It’s a down run, because the general slope of the route is down hill.

2.2)310m – Umlaas 20km in.

2.3) Botha’s hill

2.4) Optraande/up hill – between Polly shorts and Umlaas

Aftraande/down hill – between Botha’s hill and Pinetown

2.5) Hulle kan sien waar die groot optraandes is en hulle energie spaar daarvoor

Hulle kan sien waar is die groot aftraandes is en dus waar hulle weer tyd kan opmaak.

They can see where the biggest up hills are and save their energy for them.

They can see where the biggest down hills and so where they can make up time again.

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Hersienings-oefening / Revision exercise (p.170 – 171)

1.1) 2cm + 2cm + 3cm + 1,5cm + 5cm + 3,5cm

= 17cm

1.2) 2 x 3,142 x 1,5cm

= 9,429cm

1.3) 7cm + 2cm + 2,1cm + 4cm + 4cm + 4cm + 3,8cm + 1,9cm + 1,3cm + 5,8cm

= 35,9cm

1.4) (3,142 x 18mm) + 2(47mm) + 36mm

= 56,556mm + 94mm + 36mm

= 186,556mm

2.1) 0,5 x 4 x 3

= 6cm2

2.2) 50cm x 72cm

= 3 600cm2

2.3) 3,142 x 2,5cm x 2,5cm

= 19,64cm2

2.4) 0,5 x 7cm x 2cm

= 7cm2

2.5) (2cm x 4cm) + (3cm x 1,5cm)

= 8cm2 + 4,5cm2

= 12,5cm2

2.6) (4mm x 5,5mm) + (0,5 x 3,142 x 2mm x 2mm)

= 22mm2 + 6,28mm2

= 28,28mm2

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3.1.1) Per/Omt F = 45cm + 5cm + 20cm + 10cm + 5cm + 10cm + 15cm + 15cm + 5cm + 20cm

= 150cm

Per/Omt E = 45cm + 20cm + 5cm + 15cm + 15cm + 10cm + 5cm + 10cm + 15cm + 15cm

+ 5cm + 20cm

= 180cm

Per/Omt T = 5cm + 7,5cm + 40cm + 5cm + 40cm + 7,5cm + 5cm + 20cm

= 130cm

So Per/Omt F + E + T = 150cm + 180cm + 130cm

= 460cm

3.1.2) 5m

3.2) Area/Opv F=(45x5) + (10x5) + (15x5)

= 225 + 50 + 75

= 350cm2

Area/Opv E=(45x5) + (10 x 5) + 2(15 x 5)

= 225 + 50 + 150

= 425cm2

Area/Opv T = (40 x 5) + (20 x 5)

= 200 + 100

= 300cm2

So Area/Opv F + E + T = 350 + 425 + 300

= 1075cm2

1075 ÷ 100

= 10,75 ÷ 100

= 0,1075m2, dus gaan hulle net een 500ml blik verf benodig/ So they

will only need one 500ml can of paint

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4.1)Per/Omt of/van pool/swembad:

2(2,5) + 2(6) + 4

= 5 + 12 + 4

= 21m x R295

= R6 195 (exl. VAT) x 1,14

= R7 062,30 incl. VAT

4.2) Per/Omt of/van fence/heining:

(0,5 x 2 x 3,142 x 4) + 2(8) + 8 - 1

= 12,568 + 16 + 8 – 1

= 35,568m

4.3) Opv van swembad/Area of pool:

Area of triangle/Opv van driehoek + Area of rectangle/opv van reghoek.

So 2(0,5 x 2 x 2)= 2(2) = 4m2 + 6x4 = 24m2 = 24 + 4 = 28m2

Opv van area binne heining/ Area inside the fence:

Area of half circle/Opv van half sirkel

+ Area of rectangle/opv van reghoek

So 0,5 x 3,142 x 4 x 4 = 25,136m2

+ 8x8 = 64m2

= 64 + 25,136

= 89,136m2

Area of paving/Opv van plaveisel =

Area inside fence/opv binne heining – Area of pool/opv van swembad

= 89,136 – 28

= 61,136m2

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1.1) T.b.o/T.o.a: 2(4 x 5,5) + 2(4 x 7,1) + 2(5,5 x 7,1)

= 2(22) + 2(28,4) + 2(39,05)

= 44 + 56,8 + 78,1

= 178,9cm2 X 10 x 10

= 17 890mm2

1.2) T.b.o/T.o.a: 2(3,142 x 22,5 x 22,5) + (2 x 3,142 x 22,5 x 25)

= 2(1590,6375) + 3524,75

= 3181,275 + 3524,75

= 6706,025cm2 x 10 x 10

= 670 602,5mm2

2.1) T.b.o / T.o.a: 2(9 x 19) + 2(9 x 6) + 2(6x19)

= 2(171) + 2(54) + 2(114)

= 342 + 108 + 228

= 678cm2

2.2) T.b.o/ T.o.a: 2(35 x 35) + 4(35 x 47)

= 2(1225) + 4(1645)

= 2450 + 6580

= 9030cm2

2.3) T.b.o / T.o.a: 2(3,142 x 3,5 x 3,5) + ( 2 x 3,142 x 3,5 x 10,5)

= 2(38,4895) + 230,937

= 76,979 + 230,937

= 307,916 = 307,9cm2

2.4) T.b.o / T.o.a: (3,142 x 5 x 5) + ( 2 x 3,142 x 5 x 16)

= 78,55 + 502,72

= 581,27 = 581,3cm2

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3.1) 20m2 ÷ 10m2/liter

= 2liter

3.2) 24m2 ÷ 8m2/liter

= 3liter

3.3) 30m2 ÷ 9m2/liter

= 3,333 → 4liter

3.4) 120 000cm2 ÷ 60 000cm2/liter

= 2liter

3.5)380 000cm2 ÷ 40 000cm2/liter

= 9,5 → 10liter

3.6) 500 000cm2 ÷ 100 ÷ 100

= 5m2 ÷ 5m2/liter

= 1liter

4.1) Die totale buite oppervlakte van die laaikas

The total outside area of the chest of drawers

4.2) Lengte, breedte en hooghte van die laaikas

The length, width and height of the chest of drawers

4.3) T.b.o / T.o.a: 2(59 x 45) + 2(59 x 115) + 2(45 x 115)

= 2(2655) + 2(6785) + 2(5175)

= 5 310 + 13 570 + 10 350

= 29 230cm2

4.4) 29 230cm2 ÷ 100 ÷ 100

= 2,923m2 x 2

= 5,846m2 wat geverf moet word/ that needs to be painted

So 500ml verf gaan te min wees

500ml of paint will not be enough

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Oefening 2 / Exercise 2 (p.179 – 181)

1.1a) Volume: 4 x 5,5 x 7,1

= 156,2cm3

b) Volume: 40 x 55 x 71

= 156 200mm3

1.2a) Volume: 3,142 x 22,5 x 22,5 x 25

= 39 765,9cm3

b) Volume: 3,142 x 225 x 225 x 250

= 39 765 937,5mm3

2.1a) Volume: 9 x 6 x 19

= 1 026cm3

b) Volume: 90 x 60 x 190

= 1 026 000mm3

2.2a) Volume: 35 x 35 x 47

= 57 575cm3

b) Volume: 350 x 350 x 470

= 57 575 000mm3

2.3a) Volume: 3,142 x 3,5 x 3,5 x 10,5

= 404,1cm3

b) Volume: 3,142 x 35 x 35 x 105

= 404 139,8mm3

2.4a) Volume: 3,142 x 5 x 5 x 16

= 1 256,8cm3

b) Volume: 3,142 x 50 x 50 x 160

= 1 256 800cm3

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3) Volume: 1 726 x 595 x 651

= 668 557 470mm3 ÷ 1000

= 668 577,47ml ÷ 1000

= 668,6liter kapasiteit/capacity

4.1.1) Volume: 90 x 72 x 45

= 291 600cm3

= 291 600ml ÷ 1000

= 291,6liter

4.1.2) Volume: 3,142 x 45 x 45 x 72

= 458 103,6cm3

= 458 103,6ml ÷ 1000

= 458,1 liter

4.2) Die silindriese sak

The cylindrical bag

5.1) mm vir die afmetings en liter vir die volume

Mm for the dimentions and litre for the volume

5.2a) Volume: 3,142 x 550 x 550 x 1300

= 1 235 591 500mm3 ÷ 1000

= 1 235 591,5ml ÷ 1000

= 1 235,6liter

b) Volume: 3,142 x 575 x 575 x 1700

= 1 766 000 375mm3 ÷ 1000

= 1 766 000,4ml ÷ 1000

= 1 766liter

5.3) “enige geldige antwoord”/”any appropriate answer”

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5.4) Hou die houer onder water en kyk hoeveel water dit verplaas

Press the container under water and see how much water it displaces.

6.1) “teken boks”/ “draw a box”

6.2) L: 50 ÷ 10 = 5

W: 30 ÷ 10 = 3

H: 25 ÷ 12,5 = 2

So 5 x 3 x 2 = 30

6.3) L: 50 ÷ 15 = 3

W: 30 ÷ 10 = 3

H: 25 ÷ 12,5 = 2

So 3 x 3 x 2 = 18

6.4) L: 50 ÷ 10 = 5

W: 30 ÷ 10 = 3

H: 25 ÷ 12,5 = 2

So 5 x 3 x 2 = 30

6.5) L: 50 ÷ 12,5 = 4

W: 30 ÷ 12,5 = 2

H: 25 ÷ 12,5 = 2

So 4 x 2 x 2 = 16

Grade 11 Mathematical literacy / Graad 11 Wiskundige...

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