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F

Topic 2001 2002 2003 2004 2005

I II I II I II I II I II

Calculations 1 1 1 1 1

Fractions 2 2 2 2 2

Scientific Notation 1 1 1 1 1

% Calculations 3 2 6 4 3

Circle Geometry 6 10 8 10 10

Similarity 12 9 6

Area/Volume 5 8 11 5 13 4 12 9 12 3 8

Speed/Distance/Time 11

Triangle Calculations 6 10 7 4 3 6 7 5 6 7 5 7

Trig Equations & Graphs 7 8 9 10 11

Patterns 9 11 11 8

Brackets/ Factorising 5a 3 5

Quadratics 8 9 3 8 11 4

Surds & Indices 10 10 11 12 11 11

Algebraic Fractions 5b 4

Formulae 11 5 3 9

Ratio/Proportion/Variation 9 7 10 10

The Straight Line 6 4 12 6 10 2 5 9

Equations/Inequations 4 3 6

Simultaneous Equations 13 9 7 8

Change The Subject 6

Functions 3 4 4

Statistics 5 7 2 8 8 9 2 5 7 3 4 7 2

20

06

- 2

01

0

Topic 2006 2007 2008 2009 2010

I II I II I II I II I II

Calculations 1 1 1

Fractions 2 2

Scientific Notation 1

% Calculations 3 1 5 1 3

Circle Geometry 8 12 7 5 9

Similarity 11 8

Area/Volume 7 12

Speed/Distance/Time

Triangle Calculations 10 5 6 6 8 7 8

Trig Equations & Graphs 10 13 10 12

Patterns 7

Brackets/ Factorising 5a 4a 5 2

Quadratics 8 2 11 13 8 12 11

Surds & Indices 4bc 7 9 9 11 10

Algebraic Fractions 5b 5

Formulae 9 10 14 10

Ratio/Proportion/Variation 7 9 6

The Straight Line 4 6 4

Equations/Inequations 6 11 4 6 13

Simultaneous Equations 9 11 4

Change The Subject 4 3

Functions 3

Statistics 2 3 3 2

Ma

in G

rid

a

acbbxcbxax

2

4 are 0 of roots The

22

Sine rule:

Cosine rule:

Area of a triangle:

Standard deviation:

bc

acbbccba

2A cosor A cos 2

222222

C

c

B

b

A

a

sinsinsin

Cabsin2

1 Area

size. sample theis where

,1

/

1

222

n

n

nxx

n

xxs

Main Grid

Main Grid

Solution

F

Q1 BODMAS

3.1 + 2.6 × 4 2.6 × 4 = 10.4

3.1 + 10.4 = 13.5

Main Grid

Main Grid

Solution

F

Q2

24

78

24

199

24

11287

38

)148()293(

3

14

8

29

3

24

8

53

Main Grid

Main Grid

Solution

F

Q3

401525)5(

)15(25)5(

)53()5()5(

3)(

2

2

f

f

f

mmmf

Main Grid

Main Grid

Solution

F

Q4

3

155

16138

16)13(8

)4( 44

)13(2

x

x

xx

xx

xx

Main Grid

Main Grid

Solution

F

Q5

10 15 20 25 30 35 40 45 50 55

Time in Days

60

Timberplan

Allwoods

Furniture Delivery Time

For consistency of delivery the furniture maker should use

Timberplan, because a smaller interquartile range suggests a

smaller range of delivery times, therefore, more consistent.

Main Grid

Main Grid

Solution

F

Q6

atM

atat

atM

at

atM

AT

AT

AT

1

))((

22

Main Grid

Main Grid

Solution

F

Q7

60

31

600

310 years) 3 than P(less

600

201608050 years) 3 than P(less

60

1

600

10 y Probabilit

cars 70604200 :4200 sample aIn

Main Grid

Main Grid

Solution

F

Q8

4

321

35.045.04

5.0at valuemin.symmetry of Because

)0,5.0( )0,5.1(

5.1 and 5.0

0)32)(12(

0344

:0 Cuts

)3,0(

330404

:0 Cuts

2

2

2

y

y

y

x

CB

xx

xx

xx

y x-axis at

A

y

x y-axis at

Main Grid

Main Grid

Solution

F

Q9

)124)(12(18

28 )12)2)((12(18

)1)(1(1

)177)(17(17

23

3 323

23

23

pppp

pppppp

nnnn

Main Grid

Main Grid

Solution

F

Q10

4

2

24

26

24

236

24

72

24

24

24

3

Main Grid

Main Grid

Solution

F

Q11

20

12

minimum. a is 2 when valueMaximum

1

22

20210

2

2010

8

20

2

20

0

c

3

I

c

I

c

c

c

Main Grid

Main Grid

Solution

F

Q1

910256.5

5256000000

100006024365

Main Grid

Main Grid

Solution

F

Q2

28.1647.1

9

489.7111531.71130

110

10

3.84331.71130

1

22

2

s

s

n

n

xx

s

3.8410

3.843mean

In rural areas petrol prices are higher on average

and there is a greater variation on prices.

Main Grid

x x2

81.0 6561

83.9 7039.21

84.2 7089.64

84.2 7089.64

84.4 7123.36

84.8 7191.04

84.9 7208.01

85.1 7242.01

85.2 7259.04

85.6 7327.36

843.3 71130.31

Main Grid

Solution

F

Q3

907.53 £150721.28 46186.25 104 :Value

28.721 46£)92.0(000 60 :Contents

25.186 104£)05.1(000 90 :House

3

3

Main Grid

Main Grid

Solution

F

Q4

63

63

)3( 23

1

2 3

1

12

4

012

26

xy

xy

xy

cMcMxy

6

11419

1381512

24412

4654

63

x

x

xy

xy

xy

xy

4,6

4

123

663

y

y

y

Q4b

Main Grid

Main Grid

Solution

F

Q5

cmd

hr

hr

cmhrV

3.726.763.32

63.32.131214.3

49.49749.497

49.497

49.4971525.325.314.3

2

32

Main Grid

Main Grid

Solution

F

Q6 We need to calculate angle W. Use

the sine rule to calculate angle P.

o

o

o

W

P

k

KpP

K

k

P

p

8.1572.22180 P WBearing

2.228.27130180

8.27)467.0(sin

467.0410

130sin250sinsin

sinsin

1

Main Grid

Main Grid

Solution

F

Q7

oooo

o

o

o

o

o

o

oo

xx

x

x

x

x

x

x

4.3556.4360 and 6.1846.4180

6.4)081.0(sin Angle Related

081.0sin

2161.0sin

161.0sin2

1839.0sin2

1sin2839.0

1sin240tan

1

Main Grid

Main Grid

Solution

F

Q8 Volume = end area × depth

3

2

75.275

515.55

15.55

100sin1485.0

sin2

1

cmV

V

cmA

A

cabA

o

o

Main Grid

Main Grid

Solution

F

Q9

4

27

427

94

3

32

3

22

22

L

Lkk

Lkk

Lkk

RR

d

kLR

d

LR

Main Grid

Main Grid

Solution

F

Q10

oA

A

bc

acbA

5.107301.0cos

301.0336

101cos

12142

211214

2cos

1

222222

h 70cm

(180 – 107.5 = 72.5º)

cmh

h

h

o

o

8.66

5.72sin70

705.72sin

Main Grid

Main Grid

Solution 11c

Solution 11ab

F

Q11ab

60050

2030600

)20)(30( )

30 )

2

2

xxA

xxxA

xxAb

xla

Main Grid

Solution 11c

Q11c

cmcm

cbaa

acbb

xx

xx

cm

25by 35 dimensions Minimum

24.41by 41.34 dimensions Minimum

41.54 41.4

2

346050

2

346050

2

)960(250050

12

)24014(5050

240 50 1 2

4

024050

24050

2406000.4 600 of 40%

6002030 Area Original

2

2

2

2

2

Main Grid

Main Grid

Solution

F

Q1 BODMAS

7.18 – 2.1 × 3

2.1 × 3 = 6.3

7.18 – 6.3 = 0.88

Main Grid

Main Grid

Solution

F

Q2

2

11

2

3

24

36

3

4

8

9

4

3

8

9

4

3

8

11

Main Grid

Main Grid

Solution

F

Q3

1

33

225

)1(25

x

x

xx

xx

Main Grid

Main Grid

Solution

F

Q4

6)3(

)15(9)3(

)35()3()3(

5)(

2

2

f

f

f

xxxf

Main Grid

Main Grid

Solution

F

Q5

3

2

23

22

63

4

22

4

22

22

qp

qp

qpqp

qp

qp

qpqp

qp

Main Grid

Main Grid

Solution

F

Q6

tLh

htL

thL

thL

2

2

2

2)( 2

1

Main Grid

Main Grid

Solution

F

Q7

8

1

40

5cos

40

361625cos

452

645cos

2cos

222

222

A

A

A

bc

acbA

Main Grid

Main Grid

Solution

F

Q8

10 15 20 25 30 35 40 45 50 55

Number of replies

60

Posted

Handed Out

Medical Questionnaires

You could also draw a back-to-back stem and

leaf diagram. Main Grid

Main Grid

Solution

F

Q9

4 and 1

0)4)(1(

043

3512

)()(

2

2

xx

xx

xx

xxx

xgxf

Main Grid

Main Grid

Solution

F

Q10

35

3233

3239

3227

Main Grid

Main Grid

Solution

F

Q11

2

68

238

y

yy

yy

Main Grid

Main Grid

Solution

F

Q12

129

7

12 9

7

90

70

090

1282

hg

cM

cMhg

Main Grid

Main Grid

Solution

F

Q13

22.0

20.210

80.4168

60.268

20.142

30.134

g

g

gp

gp

gp

gp

16.0

64.04

30.166.04

30.1)22.03(4

p

p

p

p

92.0£44.048.0

)22.02()16.03(23

gp

Main Grid

Main Grid

Solution

F

Q1

3

5

104308.3

0034308.0

1006.1918

Main Grid

Main Grid

Solution

F

Q2

66.127£175.1150

Main Grid

Main Grid

Solution

F

Q3

8.2 .31

766.2 .2661

4

653

4

653

4

653

4

)56(93

22

)724(33

7 3 2

2

4 0732

2

22

cba

a

acbbxx

Main Grid

Main Grid

Solution

F

Q4

kmt

S

Tst

S

s

T

t

S

o

o

o

9.296105sin

35sin500

sin

sin

sinsin

1054035180 Angle

kmh

h

h

o

o

8.190

40sin9.296

9.29640sin

V

S

296.9km

40°

h

Main Grid

Main Grid

Solution

F

Q5

3

2

2

2

2

2.11652.142913.0

2913.01413.015.0

1413.0

3.03.014.35.05.0

15.025.06.0

length area endVolume

mVolume

mA

m

rA

mA

TOTAL

CIRCLESEMI

RECTANGLE

Main Grid

Main Grid

Solution

F

Q6

3.4 - 2.1 = 1.3m

a

x = 2a

mx

ma

3.365.12

65.172.23.11.2

Pythagoras Using

22

m8.03.11.2 Main Grid

Main Grid

Solution

F

Q7

produced. be

can tinskilogram-one 41 Therefore,

67.4160025000

5040020000

Colombian. of 600g and

Brazilian of 400g contains tin One

Main Grid

Main Grid

Solution

F

)4.0,4.156B( )4.0,6.23A(

4.1566.23180 and 6.23

)4.0(sin

4.0sin

1

oo

ooo

o

o

x

x

x

Q8

Main Grid

Main Grid Solution F

Q9

minutes 6 Therefore,

33.5

163

276560

276 560

4280 15575 )

)2(280 )

)3(575 )

110)75()325( )

m

m

mm

mcmc

mcmcd

mcc

mcb

pa

Main Grid

Main Grid

Solution

F

Q10

22

r

kvT

r

vT

18.by multiplied isTension

180.59

0.5by divide and 3by Multiply 2

Main Grid

Main Grid

Solution

F

Q11

12

132168421

5n

32)2222(2 322

n

5

)(

Main Grid

Main Grid

Solution

F

Q12 Similar triangles. A

B

P

B 1m

1.5m 6m

mAPmPBPB

246 45.1

16 so

5.1

1

6

mB

B

height

height

32

16

2

6

1

again. ianglessimilar tr Using

Main Grid

Main Grid

Solution

F

Q1 BODMAS

2.1

4.87 1 5.04 + 1.2

6.24

Main Grid

Main Grid

Solution

F

Q2

28

17

56

34

56

6

56

28

56

6

28

14

8

3

4

7

7

2

8

3

4

31

7

2

Main Grid

Main Grid

Solution

F

Q3

166

412126

)13(4)42(3

x

xx

xx

Main Grid

Main Grid

Solution

F

Q4

2

1

24

974

947)(

15)2(

)8(7)2(

)24(7)2(

47)(

t

t

t

ttf

f

f

f

xxf

Main Grid

Main Grid

Solution

F

Q5

)5)(32(

1572 2

xx

xx

Main Grid

Main Grid

Solution

F

Q6

1

55

56

56

532

52

25

10

14

73

k

k

kk

kk

kk

xy

M AB

Main Grid

Main Grid

Solution

F

Q7

£5 isbreakfast one ofCost

5

720915

7251015

24035

14523

b

bn

bn

bn

bn

Main Grid

Main Grid

Solution

F

Q8

40

1

10

1

40

4

Main Grid

Main Grid

Solution

F

Q9

25% of matchboxes contain fewer than 50

matches.

50 is the lower quartile and every quartile

contains 25% of the sample.

Main Grid

Main Grid

Solution

F

Q10

pupils 75 75:15:5

teachers9 45:9:3

Main Grid

Main Grid

Solution

F

Q11

12

12

)1(

9

22

22

2

n

nnn

nn

n

Main Grid

Main Grid

Solution

F

Q12

323412

122

122

2

24

46488 33 23

2

Main Grid

Main Grid

Solution

F

Q13

xx

xTD

xxxDB

DB

xTD

xDBTD

xx

xxx

32

6

23

6

32

1

3124

1A

1243A

2

2

2

22

POCKET

2

CLIPBAOARD

Main Grid

Main Grid

Solution

F

Q1

509054.5090006.150003

Main Grid

Main Grid

Solution

F

Q2

x x2

41 1681

43 1849

44 1936

47 2209

49 2401

52 2704

276 12780

1.48.16

5

1269612780

16

6

27612780

1

22

2

s

s

n

n

xx

s

466

276mean

There is more variation in the price of milk than

there is in the price of sugar.

Main Grid

Main Grid

Solution

F

Q3

kmh

h

h

h

h

Habbah

o

o

7.47

2.2276

2.2276

8.11239002500

68cos305023050

cos2

6872140 H Angle

2

2

222

222

Main Grid

Main Grid

Solution

F

Q4

cmr

h

hrV

cmml

cmhrV

6.75.78

600

5514.3

600600

600

600600

1099145514.3

2

2

3

32

Main Grid

Main Grid

Solution

F

Q5

sides. 8 hasPolygon The

5or 8

058

0403

340

340

2

320

2

2

nn

nn

nn

nn

nn

nn

Main Grid

Main Grid

Solution

F

Q6 SOH CAH TOA

cmSW

SW

SW

cmSV

SV

SV

o

o

o

o

65.6

25cos33.7

33.725cos

33.7

34sin1.13

1.1334sin

Main Grid

Main Grid

Solution

F

Q7

oB

B

B

B

BacArea

1.37603.0sin

603.063

38sin

sin6338

sin9145.038

sin2

1

1

Main Grid

Main Grid

Solution

F

Q8

8,1at Point Turning

8222

31112

312

1at T.P.symmetry By

2

36

6,0at 30106

31

3 and 1

y

y

xxy

x

k

k

k

xxky

ba

Main Grid

Main Grid

Solution

F

Q9

ml25.101375.330 Volume Large

375.38

27 Factor Scale

27:8

3:2 VolumeFor

3:2 Height For

33

Main Grid

Main Grid

Solution

F

Q10

2

x

3

mx

x

x

xxx

xx

4

13

4

522 OB So

4

5

54

944

32

Pythagoras Using

22

222

Main Grid

Main Grid

Solution

F

Q11

mphx

x

xx

xxxxxx

x

x

S

DT

6030

2 Speed Average

302 Speed Average

Time Total

Distance Total Speed Average

30150

5

150

32

5075 Time Total

2 Distance Total

75

Main Grid

Main Grid

Solution

F

Q1 BODMAS

77.243.32.6

43.31.153.4

Main Grid

Main Grid

Solution

F

Q2 BODMAS

5

12

10

22

10

22

10

814

5

4

10

14

10

14

2

7

5

2

2

13 of

5

2

Main Grid

Main Grid

Solution

F

Q3

2

1618

432 22

A

A

A

Main Grid

Main Grid

Solution

F

Q4

1

37

1

433

1

413

1

43

mm

m

mm

mm

mm

mm

mm

Main Grid

Main Grid

Solution

F

0 3 6 9 12 15 18 21 24 27 30

Average Monthly Temperature in Holiday Resort

Q5

Temp (ºC)

Main Grid

Main Grid

Solution

F

Q6

g4009

8450

8

9450 :So

8

9

8

11 isjar so

8

1 is %5.12

Main Grid

Main Grid

Solution

F

Q7

.90

1 than chancebetter a is

80

1 because

raffle School in the chancebetter a has He

90

1

1800

20 P(win)

80

1

1200

15 P(win)

Main Grid

Main Grid

Solution

F

Q8

1735

17 :So

17 , , , , ,

532

5 :So

5 , , , ,

yx

xyxxyxyx

xyxxyxyxxyxy

yx

yxyyx

yxyyxyx

4

123

1735

532

x

x

yx

yx

1 and 4 So

1

33

538

5342

yx

y

y

y

y

Main Grid

Main Grid

Solution

F

Q9

.360in curves complete 4 be will therebecause 4

2.- to2 from goesgraph because 2

o

b

a

Main Grid

Main Grid

Solution

F

Q10

The line must have a negative gradient (going down).

The line must cut the y-axis below zero.

y

x

y = –ax – b

Main Grid

Main Grid

Solution

F

Q11

3

11

3

1132

3103523252752

10

Main Grid

Main Grid

Solution

F

Q12

252555

52

10 so

10

10

2

2

rA

rd

d

dC

Main Grid

Main Grid

Solution

F

Q1

m12

8

1064.8

)60608(103STD

Main Grid

Main Grid

Solution

F

Q2

s453

230t

30t3

2

100t3

2-70

70v therefore litres30 loses container The

100t3

2v

100c 3

2

150

100M cMtv

min

Main Grid

Main Grid

Solution

F

Q3

x x2

49 2401

50 2500

50 2500

51 2601

52 2704

52 2704

53 2809

357 18219

41.12

6

1820718219

17

7

35718219

1

22

2

s

s

n

n

xx

s

517

357mean

Main Grid

Main Grid

Solution

F

Q4

mg1288.02503

Main Grid

Main Grid

Solution

F

Q5 Find side a because it is opposite the smaller angle and will

therefore be the shortest distance.

ma

a

H

Aha

H

h

A

a

o

o

6.1189

75sin

50sin1500

sin

sin

sinsin

Main Grid

Main Grid

Solution

F

Q6 SOH CAH TOA

mPQ

PQ

o

o

1045tan10

1045tan

mPQPB 202 :So

ooo

oo

o

.

S

PS

PBS

4.1845463 degrees More

4.632tan

210

20tan

1

Main Grid

Main Grid

Solution

F

Q7

cm

Abccba

o

800.3228 length Rod

7.6451 length Rod

7.6451

3.154816006400

76cos408024080length Rod

cos2

222

222

Main Grid

Main Grid

Solution

F

Q8 2000 – 1800 = 200mm

500 – 200 = 300mm

C

A B

mm

mmAB

AB

AB

BCACAB

8004002 door ofWidth

400160000

160000

300500

Pythagoras Using

2

222

222

Main Grid

Main Grid

Solution

F

Q9

3

2

2

1902875.237 Volume

depth area Volume

75.23755.475 area Total

55.47

72sin10105.0

sin2

1 triangleof Area

725360 centreat Angle

cm

cm

cm

Cab

o

o

Main Grid

Main Grid

Solution

F

Q10

ooo

oo

oo

o

o

o

o

x

x

x

x

x

x

4.311,6.228

4.3116.48360

6.2286.48180

quadrants.4th and 3rdin

be willsolutions so negative isSin

6.4875.0sin angle Related

75.0sin

3sin4

21sin4

1

A S

T C

48.6º 48.6º

Main Grid

Main Grid

Solution

F

Q11

m

xx

xx

xx

xx

xx

xxx

xxxx

623 lawn ofLength

2or 3

1

0213

0253

0253

253

33 :Lawn

251123111 :Path

2

2

2

2

Main Grid

Main Grid

Solution

F

Q1

92.0

36.781

3.8 – 0.92

2.88

BODMAS

Main Grid

Main Grid

Solution

F

Q2 BODMAS

30

35

5

7

6

5

5

21 of

6

5

30

105

30

3570

30

35

3

7

30

35

3

12

2

13

6

33

6

21

Main Grid

Main Grid

Solution

F

Q3

10% = 14 so 2.5% = 3.5

12.5% = 14 + 3.5 = £17.50

Main Grid

Main Grid

Solution

F

Q4

Possible outcomes = 36

Possible ways of getting 8 or 9 = 9

4

1

36

9 y Probabilit

Main Grid

Main Grid

Solution

F

Q5

24

8

0-4

6-2- M

C = 6

y = Mx + C

y = -2x + 6

Main Grid

Main Grid

Solution

F

Q6

5

2

52

52

612

x

x

x

x

Main Grid

Main Grid

Solution

F

Q7 Speed Frequency Cuml. Frq.

30 1 1

40 4 5

50 9 14

60 14 28

70 38 66

80 47 113

90 51 164

100 32 196

110 4 200

200 cars therefore median between 100th and 101st

So median = 80km/hr Main Grid

Main Grid

Solution

F

Q8

4th term = 52 – 32

nth term = (n + 1)2 – (n – 1)2

= (n2 + 2n +1) – (n2 – 2n + 1)

= n2 + 2n +1 – n2 + 2n – 1

= 4n

Main Grid

Main Grid

Solution

F

Q9

Litres put in car = 3000 ÷ 75 = 40 litres

Litres used = 5 × 3 = 15 litres

Litres remaining = 40 – 15 = 25 litres

Litres used = kt

Litres put in car = c

2000

ktc

R 2000

Main Grid

Main Grid

Solution

F

Q10

5cm

4cm

a 345

Pythagoras Using

22 a

5cm 7 – 3 = 4cm

b

345

Pythagoras Using

22 b

Therefore base = 2 × 3 = 6cm Main Grid

Main Grid

Solution

F

Q11

225

2224

2264

22364

2724)72(

f

2

1

4

2

2

2

2

2

4

22

224

2234

2324)(

2

t

t

t

t

t

ttf

Main Grid

Main Grid

Solution

F

Q12

2

13

1or 2

7

0)1(or 0)72(

0)1)(72(

0752

752

7)52(

7)52)(2(2

1

72

1

2

2

x

xx

xx

xx

xx

xx

xx

xx

bhA

Main Grid

Main Grid

Solution

F

Q1

15

882

1024.3

103103106.3

E

E

Main Grid

Main Grid

Solution

F

Q2

5.796

477

6

757971849177

Mean

x x – x (x – x)2

71 -8.5 72.25

75 -4.5 20.25

77 -2.5 6.25

79 -0.5 0.25

84 4.5 20.25

91 11.5 132.25

251.5

09.73.505

5.251

16

5.251

s

Main Grid

Main Grid

Solution

F

Q3

25.187

110sin21195.0

sin2

1

cmArea

Area

PqrArea

o

Main Grid

Main Grid

Solution

F

Q4

2.4 2.2

162.4 162.2

2

402

2

402

2

402-

2

)36(42-

12

)914(22-

9 2 1

2

4

092

92

2

2

2

2

cba

a

acbb

xx

xx

Main Grid

Main Grid

Solution

F

Q5

Using the converse of Pythagoras if 902 + 602 = 1102

then the triangle is right angled.

902 + 602 = 8100 + 3600 = 11700

1102 = 12100

11700 12100

So the slab is not a right angled triangle.

Main Grid

Main Grid

Solution

F

Q6

12m

5m

20m

h

Using similar triangles:

mh

h

485

1220

5

20

12

Main Grid

Main Grid

Solution

F

Q7 Angle A = 40°

Angle B = 294 – 270 = 24°

Angle V = 180 – 24 – 40 = 116°

kmv

B

Vbv

B

b

V

v

05.1124sin

116sin5

sin

sin

sinsin

0

0

Main Grid

Main Grid

Solution

F

Q8

cmx

xx

xx

xx

xx

xx

xxxxx

62length side

3or 0

0)3(or 08

0)3(8

0248

248

)22(6222

2

2

23

23

Main Grid

Main Grid

Solution

F

Q9

Fixed rental = £10

Call charge per minute = gradient

05.060

3

060

1013

gradient

Call charge per minute = £0.05 = 5p

Main Grid

Main Grid

Solution

F

Q10

12.5cm 11.5cm

xº

oo

oo

o

x

x

x

23

074.23)92.0(cos

92.05.12

5.11cos

1

mlengtharc

dlengtharc

d

lengtharc

0.10

036.10360

2546

360

46

360

46

Main Grid

Main Grid

Solution

F

Q11

o144.7or 3.35

3

1sin

01sin3

oo

o

o

x

x

x

ooo

ooo

x

x

72.35or 65.17

7.144or 3.352

Main Grid

Main Grid

Solution

F

Q1

1.25 × 40 = 50

56.4 – 50 = 6.4

BODMAS

Main Grid

Main Grid

Solution

F

Q2

35

64

35

146

35

9056

7

18

5

8

7

42

5

31

Main Grid

Main Grid

Solution

F

Q3

4 – (-3)2

= 4 – 9

= -5

Main Grid

Main Grid

Solution

F

Q4

8x3

2y

8c

3

2

6

4

06

812m

Main Grid

Main Grid

Solution

F

Q5

Difference of 2 squares

4x2 – y2 = (2x + y)(2x – y)

3

yx2

yx23

yx2yx2

y3x6

yx4

22

Main Grid

Main Grid

Solution

F

Q6

x – 2(x + 1) = 8

x – 2x – 2 = 8

–x = 10

x = –10

Main Grid

Main Grid

Solution

F

Q7

ml540

2720

8

27160Cup Large of Volume

8

27

2

3Factor Scale Volume

2

3

14

21Factor Scale Length

3

3

Main Grid

Main Grid

Solution

F

Q8

y = (x – 1)2 – 4

Main Grid

Main Grid

Solution

F

Q9

a) x + y = 20

b) 5x + 2y = 79

c) 2x + 2y = 40

5x + 2y = 79

3x = 39

x = 13

Euan won 13 games. Main Grid

Main Grid

Solution

F

Q10

m12h

25300h

300h25

150h252

1

m15015202

1Area 2

Main Grid

Main Grid

Solution

F

Q11 a) C = 3x

b) 20 + 2 × 9 = £38 2(x – 6) + 20

c) 2(x – 6) + 20 < 3x

2x – 12 + 20 < 3x

2x + 8 < 3x

2x – 3x < –8

–x < –8

x > 8 so 9 sessions Main Grid

F Main Grid

Solution

Q1

c = d

c = 3.14 × 2 × 4.96 × 107

c = 3.11488 × 108 km

Main Grid

F Main Grid

Solution

Q2

74.65.45

5

5.3511335341

16

6

45935341

35341788272867368

5.766

459

6

788272867368

2

2222222

s

s

x

x

On average children's pulse rates are faster but

there is less variation. Main Grid

F Main Grid

Solution

Q3

324 ÷ 1.05 = £300

Main Grid

F Main Grid

Solution

Q4

5

5354

53542 )c

m2m )b

4x11x3

4x12x3x )a

2

5

2

1

2

2

Main Grid

F Main Grid

Solution

Q5

o1o

o

22

31120M

2010

2M

1068MS

..tan

.tan

Main Grid

F

Main Grid

Solution

Q6 B° = 74° + 50° = 124°

m30544305b

1693295b

1693295b

162829565000b

1242301102230110b

Bac2cab

2

2

o222

222

.

.

.

.

cos

cos

Main Grid

F Main Grid

Solution

Q7

28 × 18 = 504cm3

cm632770143

504

r

504L

504LrV

22

2

...

Main Grid

F Main Grid

Solution

Q8

2228g10022.28

22.28244.59

cm5944360

18143284Length Arc

..

Main Grid

F Main Grid

Solution

Q9

13n

10n 13n

10n13n )c

0130n3n

130n3n

1303nn

2 653nn2

1 )b

1443.5 )a

2

2

Main Grid

F

Main Grid

Solution

Q10 a) -31 × cos20 + 33 = 3.87m

b) -31cost + 33 = 60

-31cost = 27

cost = 27 ÷ -31 = -0.871

R.A. = cos-1(0.871) = 29.4

so: 180 – 29.4 = 150.6 seconds

c) 180 + 29.4 = 209.4 seconds

Main Grid

F

Main Grid

Solution

Q11

A

A

M

Q P B C 8

6 3

x

x3

44

6

x824PQ

6

x38

AC

AQBCPQ so

AC

AQ

BC

PQ b)

x3 AQ )a

Main Grid

Main Grid F Solution

Q1

3.72 × 20 = 74.4

6.04 + 74.4 = 80.44

BODMAS

Main Grid

Q2

10

91

30

57

5

3

6

19

3

5

6

19

3

21

6

13

Main Grid

Q3

2508

2000

8

4005

Main Grid

Q4

2

83

832

823

423

3)( 3

42

Pm

Pm

mP

mP

mP

Main Grid

Q5

(2x + 3)2 – 3(x2 – 6)

4x2 + 12x + 9 – 3x2 + 18

x2 + 12x + 27

Main Grid

Q6

25

4

2

5

4

05

26

df

c

m

Main Grid

Q7

aa

aa

aa

2

2

2

2

1

2

1

2

1

Main Grid

Main Grid

F

Solution

Q8

48cm

40cm

Length

75cm

9080 because Yes

9040

3600

40

4875 so

40

75

48

L b)

ianglessimilar tr Using

L

Main Grid

Q9

23

29

18

18

362

36

2

2

22

x

x

x

x

x

xx

Main Grid

Q10

value.its of 8

1 becomes T

doubled is L when Therefore,

8233

L

k

L

kT

Main Grid

Q11 a) x + y = 300

b) 4x + 6y = 1380

c) 4x + 6y = 1380

4x + 4y = 1200

2y = 180

y = 90

x = 300 – 90 = 210

210 standard seats were sold

90 deluxe seats were sold. Main Grid

Main Grid

F

Solution

Q12

cm

cmx

x

x

x

235 d

39

9

1625

45

Pythagoras Using

2

2

222

d

5

4

5

4

x

Main Grid

Q13

b = 2 because there are 2 complete cos curves

within 360°.

c = 3 because the graph has been moved up 3

units from its normal position.

Main Grid

Q14

4n

813

813

8013 (b)

813 (a)

4

n

n

2

2

S

Main Grid

Main Grid

Solution

F

Q1

70.684£045.16003

Main Grid

Main Grid

Solution

F

Q2

5.1 2.2

6

1242

6

1242

6

1242

6

)120(42

32

)1034(2-2

10 2 3

2

4

2

2

cba

a

acbb

Main Grid

Main Grid

Solution

F

Q3

749

6

40324326

17

7

1684326

432631261819143228

247

168

7

31261819143228

2

22222222

s

s

x

x

On average Erin's recordings are higher on

average but there is less variation. Main Grid

Main Grid

Solution

F

Q4

22

5 2

4 5 2

1

4

x

x

x

Main Grid

Main Grid

Solution

F

Q5

135£1.15.148

Main Grid

Main Grid

Solution

F

Q6

kmBC

BC

BC

26

88sin

65sin30

88sin

30

65sin

Main Grid

Main Grid

Solution

F

Q7

2

2

82.55496.13

96.13360

514.364

cm

Main Grid

Main Grid

Solution

F

Q8

cmPR

PR

PR

Qpr

105.1

15

155.1

1530sin62

1

15sin2

1

Main Grid

Main Grid

Solution

F

Q9

384g224160

224 : 160

7 : 5

G : C

357245

325160

Main Grid

Main Grid

Solution

F

Q10

ooo

ooo

o

o

o

x

x

x

x

9.216,1.143

9.2169.35180 and 1.1439.36180

9.365

4cosAngle Related

5

4cos

04cos5

1

Main Grid

Main Grid

Solution

F

Q11

a) (10 × 6) – x(6 – x) – 10x

= 60 – 6x + x2 – 10x

= x2 – 16x + 60

b) x2 – 16x + 60 = 12

x2 – 16x + 48 = 0

(x – 12)(x – 4) = 0

x = 12 x = 4 Main Grid

Main Grid

Solution

F

Q12

cmr

r

r

r

r

cmhrV

8.354

54

09.204.113

04.11309.2

04.1133

2

04.1134314.3

3

3

3

3

3

322

Main Grid

Main Grid Solution F

Q13

600,19£

70280

)70140)(704( :70at

70 ismidpoint therefore

140 and 0

y

y

yx

x

xx

Main Grid

Q1

6.3 × 3 = 18.9

24.7 – 18.9 = 5.8

BODMAS

Main Grid

Q2

)3)(3(5

)9(5

455

2

2

xx

x

x

Main Grid

Q3

B

WH

HB

W

BBHW

root) (square

)(

2

2

Main Grid

Q4

182

18

29

18

xy

c

m

Main Grid

Q5

)5(

53

)5(

25

)5(

21

pp

p

pp

pp

pp

Main Grid

Q6a

kphx

x

x

x

xx

xdstd

xxdstd

12

605

302

5

46162

5

462

1162

2

1

162)8(2

Q6b

Q6c

Main Grid

Q7a

Q7b

Q7c

775126

1265

)42(3)5(2

)42(3

)5(2

63

)6(3

3

183

3

)11()7(4 Term

53

15

3

8614 Term

3

3

3

3

xxxT

xTx

xTxx

xTxx

xxx

xxx

Main Grid

Q8a

Q8b

25169

9

33

)52)(85( :5at

5at is ofpoint -Mid

)0,8( )0,2(

h

y

y

yx

xQR

RQ

Main Grid

Q9

27

21

21

3

33

m

m

mmmm

Main Grid

Q10a

Q10b

4

16

16 :)16,2(at

)1,0(

2

a

a

a

C

Main Grid

Q11

23

29

18

18)(

3250)(

3250)(

2

2

222

AC

AC

AC

AC

AC

AC

Main Grid

Q12

7

1825 5

185 102

18 102

21810

)(8110

2

2

222

22

b

ba

ba

baax

baaxxxx

baxxx

Main Grid

Q13

3

514823

248351

)24(2)17(3

3

2

24

17

x

xx

xx

xx

x

x

Main Grid

Q1

52900

9.52907)08.1(42000 3

Main Grid

Q2a

Q2b

30

1140)least P(at

34

often)(most 29 Mode 3533 :Median

Main Grid

Q3

25.56£8.045

458.0

C

C

Main Grid

Q4

a) x + y = 60

b) 0.5x + 0.2y = 17.40 (×5)

c) x + y = 60

2.5x + y = 87

1.5x = 27

x = 18

Aaron has 18 50p coins in his bank. Main Grid

Q5a

Q5b

units 32.6

40

40)(

2565)(

565)(

tangent.meets radius

because triangleangled-Right

6518

2

2

22

2

22

PT

PT

PT

PT

PT

OP

Main Grid

Q6

5.3

1.2220 because no 414

1.22 1614

405.3:40

5.3

k

k

dk

dhhkd

hdhd

Main Grid

Q7

mb

b

b

b

b

Baccab

o

62.5

6.31

6.31

3.1289.1041.8

130cos3.39.223.39.2

cos2

2

2

222

222

c

b

a

B

A C

130º

Main Grid

Q8a

Q8b 190sin because 90

86.126

70sin18155.0

sin2

1

2

oo

o

o

mA

A

cabA

Main Grid

Q9a

Q9b

cmr

r

r

r

ncecircumfere

arcangle

o

6.4514.32

288

2882

150

3601202

2

120

360

150

360

circle. theof radius the

is handclock theoflength The

150512

360 AOB

Main Grid

Q10a

Q10b

20012.04

80.152£12.0440425

mdC

Main Grid

Q11a

Q11b

Q11c

10 11

01011

0110

110

1101

2 5512

1

21675.0

1775.0

2

2

nn

nn

nn

nn

nn

nn

r

r

Main Grid

Q12a

Q12b

ooo

ooo

o

o

R

Q

P

7.4381807.258

7.2581807.78

7.78

7.785tan 1

Main Grid

Main Grid F

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