Answers to Homework Book 9 - Hamilton Maths Blog · PDF fileGraduated Assessment for OCR GCSE Mathematics © Hodder Murray 2007 209 Answers to Homework Book 9 STAGE 9 1 Checking answers

Graduated Assessment for OCR GCSE Mathematics © Hodder Murray 2007 209

Answers to Homework Book 9

STAGE

9

1 Checking answers

Exercise 1.1H (page 1)1 a) 0·7 ÷ 0·1 = 7, 5·317

b) (7 – 4) � 50 = 150 (or, better, (7·5 – 3·5) � 50 = 200), 181·89

c) 502 = 2500, 2652·25d) 60 ÷ 6 = 10, 10·64e) 4 � 8 = 32, 29·97f) 20 � 30 = 600, 656·81g) 70 � 50 = 3500, 3435·12h) 90 � 62 ≈ 90 � 40 = 3600, 3042·394

2 a) 40 � 7 ÷ 5 = 56, 59·997b) (50 + 70) ÷ (10 – 6) = 30 (or, better,

(50 + 70) ÷ (15 – 5) = 12), 14·59c) 20 � 6 ÷ 40 = 3, 2·588d) 0·6 � 70 ÷ 6 = 7; 6·956e) 0·06 ÷ 1 � 1 = 0·06, 0·057f) 3 � 4 ÷ 1 = 12, 10·484g) 3 � 5 � 6 ÷ 32 = 90 ÷ 9 ≈

81 ÷ 9 = 1, 1·09273 a) 1·5 � 102, 1·433 � 102

b) 30 � 108, 27·542 � 108

c) 63 � 103, 61·61 � 103

d) 1·1 � 102, 1·2024 � 102

e) 3 � 105, 3·8441 � 105

f) 3·6 � 10–2, 4·93 � 10–2

g) 1·67 � 104, 1·6832 � 104

h) 2·5 � 103, 2·118 72 � 103

i) 1·8 � 103, 1·816 18 � 103

2 Algebraicmanipulation

Exercise 2.1H (page 3)1 x2 – 9x + 202 x2 – 4x – 213 4x2 – 33x + 84 6x2 + 16x + 85 3x2 – x – 106 8x2 – 10x – 127 14x2 – 53x + 148 6x2 – x – 159 9x2 + 42x + 49

10 15a2 + 26ab + 8b2

11 6m2 – 5mn – 6n2

12 10p2 – 11pq + 3q2

13 3a2 + 7ab – 6b2

14 6x2 + xy – 12y2

15 25a2 – 4b2

Exercise 2.2H (page 4)1 6a5

2 18x6

3 2a3b4

4 Cannot be simplified5 20a4b5

6 14c3

7 16y10

8 2p2q9 4p–3

10 2p–2

11

12

13 2p13

14 36a5b5

15 12b4

Exercise 2.3H (page 5)1 3(2x + y)2 2(2a – 5b)3 a(5 + 7a)4 2x(3x – 2)5 a(3a – b + c)6 y(4x + 2y – 1)7 2y(1 + 4x)8 3a(3a + b)9 4ab(2 – b)

10 a2(a + b)11 x2(2 – y)12 3x(2x2 – 5y2)13 5ab(a + 3b)14 3a(4bc + 5b2 – ac)15 4xyz(2x + y + 3)

Exercise 2.4H (page 5)1 (x – 6)(x + 6)2 (7 – y)(7 + y)3 (3x – 5)(3x + 5)4 (2y – 3)(2y + 3)5 (1 – 8t)(1 + 8t)6 (x – 11y)(x + 11y)7 (9a – 4b)(9a + 4b)8 (6a – 5b)(6a + 5b)9 (10 – 7y)(10 + 7y)

10 2(x – 5y)(x + 5y)

4pq2––––

3

3mn––––2

18_ANSWERS_BK9.QXD 12/6/07 16:16 Page 209

210

Exercise 2.5H (page 6)1 (x + 2)(x + 6)2 (2x + 1)(x + 3)3 (2x – 1)(x – 2)4 (2x – 1)(3x – 2)5 (2x – 3)(2x – 1)6 (x + 2)(4x + 5)7 (4x + 1)(x + 5)8 (3x + 1)(x + 5)9 (x – 3)(5x – 1)

10 (3x – 5)(2x – 3)

Exercise 2.6H (page 7)1 (x – 4)(x + 3)2 (2x + 1)(x – 3)3 (2x + 3)(x – 4)4 (3x – 1)(x + 3)5 (3x – 1)(2x + 1)6 (2x – 3)(2x + 1)7 (4x – 3)(2x + 1)8 (6x + 1)(x – 1)9 (6x + 5)(x – 2)

10 (5x – 1)(2x + 3)

Exercise 2.7H (page 7)1 2ab6

2

3

4

5

6

7

8

9

10

Exercise 2.8H (page 8)1 x = 3 or 1

2 x = –

or 2

3 x = –

or 7

4 x = –

or –5

5 x = or 4

6 x = –

or 3

7 x = or –3

8 x = –

or 3

9 x = or –4

10 x = –

or –4

3 Proportion and variation

Exercise 3.1H (page 9)1 a) y � x b) t �

c) d � t d) r �

e) p �

2 a) y � x b) y �

c) y � d) y � x

e) y �

Exercise 3.2H (page 10)1 a) y = or x = 4y

b) y = or xy = 40

y

40

y = 40x

40 x

10

1

40––x

y

5

y = x4

200 x

x–4

1–x

1–x

1–x

1–t

1–d

1–y

3–2

5–6

2–3

2–5

1–2

3–5

2–5

1–2

4–3

3x – 2––––––x – 1

3x – 1––––––x + 2

x–––––2x – 1

x – 4–––––x + 3

x + 4–––––x2

x – 1–––––x + 5

2x–––––x + 2

7––––x – 5

xy––2

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c) y = or xy = 30

d) y = x or 3y = 5x

e) y = or xy = 2400

2 a) PV = 1000b) P = 2

Exercise 3.3H (page 10)1 81·252 483 324 750

5

6 0·8

7 a) y � x3 b) y �

c) y � x2 d) y �

e) y � x3

Exercise 3.4H (page 11)1 y =

2 y = 12x2

3 y = 4x3

y

256

y = 4x3

40 x

y

192

y = 12x2

40 x

y

52

y = 13x2

4

40 x

13x2––––

4

1––x2

1––x2

1–8

y

2400

240

y = 2400x

1010 x

2400––––x

y = x

y

50

53

300 x

5–3

30x

y

30

y =

40 x

7.5

1

30––x

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4 y = 6x3

5 y =

6 y =

7 a) y = 0·1x3

b) y =

c) y = 12·5x2

d) y =

e) y =

y

8

y = x3

8

40 x

x3––8

y

500

20

y = 500x2

50 x1

500–––x2

y

200

y = 12.5x2

40 x

y

36

y = 36x2

40 x

2.25

1

36––x2

y

100

y = 0.1x3

100 x

y

20

y = 20x2

410 x

1.25

20–x2

y

8

y = 8x2

410 x

0.5

8–x2

y

384

y = 6x3

40 x

212

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Exercise 3.5H (page 12)1 a) (i) y � x3

(ii) y = 0·25x3

(iii) 2b) (i) y � x2

(ii) y = 3x2

(iii) 48

c) (i) y �

(ii) y =

(iii) 0·5

d) (i) y �

(ii) y =

(iii) 10e) (i) y � x

(ii) y =

(iii) 0·0625

2 a) It is of its value at the Earth’s surface.

b) 6400 km3 18 G

4 Indices

Exercise 4.1H (page 14)1 a) n b) n2 a) b) 1

c) 5 d) 25e) 16

3 a) 243 b) 100c) 16 d)e) 32

4 a) 36 b)c) 1 d) 8

e) 575 a) 21·8 b) 128

c) 0·0234 d) 7·17e) 1·89

6 a) 19·3 b) 5·94c) –22·3 d) 9e) 0·247

Exercise 4.2H (page 15)1 a) 26 b) 23

c) 2–3 d) 20

e) 2 f) 22n+1

2 a) 23 � 32 b) 32 � 2–4

c) 2 � 3 d) 2 � 3e) 2n � 3n f) 23n � 36n

3 a) 26 � 3 b) 23 � 52 � 7 c) 32 � 53 � 11

4 a) 25 � 32 b) 3c) 52 � 3–4 d) 211

e) 24 � 34 f) 2n � 52n

5 Rearranging formulae

Exercise 5.1H (page 16)1 r =

2 y = z2 – x2

3 x = (tv)2 or t2v 2

4 x = f 2y

5 r = 3

6 x =

7 x = p2 – q2 + y2

8 x = p2 + q2 + y2

9 x =

10 x =

11 x =

12 g = L( )2or

6 Arcs, sectors andvolumes

Exercise 6.1H (page 17)1 a) 6·03 cm b) 41·4 cm

c) 25·9 cm d) 13·7 cme) 47·6 cm

2 a) 14·5 cm2 b) 161 cm2

c) 123 cm2 d) 55·5 cm2

e) 295 cm2

3 a) 19·5 cm b) 19·3 cmc) 58·8 cm

4 a) 57° b) 244°c) 74° d) 65°e) 108° f) 213°

5 a) 10·7 cm b) 6·2 cmc) 8·3 cm d) 3·6 cme) 7·2 cm f) 5·9 cm

6 a) 147° b) 63·0 cm2

7 a) 38·2 cm b) 57°

4πL–––T 2

2π––T

yb––a

y––f 2

p––aq

y2

––m

3V–4π

V–4π

3–2

2–3

4–3

1–2

1–2

7–2

9––20

1––75

1––64

1–5

3–4

1–5

1–9

x––40

250–––x2

1––x2

1–x

1–x

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214

Exercise 6.2H (page 21)1 a) 75 cm3 b) 40 cm3

c) 46·8 cm3 d) 13·8 cm3

2 a) 39·3 cm3 b) 127 cm3

c) 61·6 cm3 d) 311·7 cm3

3 a) 408 cm3 b) 2140 mm3

4 19·9 cm5 8·95 cm6 405 cm3

7 6·45 cm

7 Upper and lower bounds

Exercise 7.1H (page 24)1 a) 125·0 cm b) 29·49 seconds2 a) 124·8 cm b) 29·47 seconds3 a) 38 m b) 0·299 kg4 a) 36 m b) 0·297 kg5 5 hours 1 minute, 4 hours 59 minutes6 1505·5 mm, 1494·5 mm7 0·2 m8 Yes. The upper bound of the combined weight

of the parcels is 500 g, so the weight will beless than this.

Exercise 7.2H (page 25)1 a) 27·1875 m2 b) 26·697 375 m2

2 a) 26·1375 m2 b) 26·592 275 m2

3 a) 64·29 km/h b) 7·733 cm/s4 a) 61·37 km/h b) 7·731 cm/s5 a) 25·8 cm, 27·6 cm b) 4·25 cm, 4·35 cm6 a) 9·37 cm, 6·92 cm b) 3·68 m, 3·47 m7 a) 28·5 cm b) 270·75 cm2

8 100 seconds

8 Similarity andenlargement

Exercise 8.1H (page 27)1 a) 64 b) 2252 a) 216 b) 80003 a) 6 b) 24 364·5 cm2

5 172·8 cm2

6 0·4 m3

7 11·7 cm8 a) 0·75 m b) 1·25 litres

c) 24 000 kg9 a) 4·75 cm b) 256 cm2

c) 2432 cm3

10 a) 3·2 cm b) 3·73 cm2

c) 14·18 cm3

Exercise 8.2H (page 29)1 a) (2, 6) b) –22 a) (–4, 7) b) –3

3

4

9 Probability

Exercise 9.1H (page 31)1 0·72 a) b)3 0·4245 a) 0·04 b) 0·326 a) 0·21 b) 0·097 a) 0·42

b) P(stop once) = P(stop at lights, don’t stopat level crossing) + P(don’t stop at lights,stop at level crossing)= (0·4 � 0·7) + (0·6 � 0·3)= 0·28 + 0·18= 0·46Rami has only calculated the probabilitythat she has to stop at the lights and not atthe level crossing. She has forgotten toadd the probability that she doesn’t haveto stop at the lights but does have to stopat the level crossing.

8 a) 0·225 b) 0·775 c) 0·275

1––12

3––40

9––20

y

x

6

4

2

–2–4–6–8

–10

642–2–4–6–8–10 0

B

C

A

A’B’

C’

10

8

6

4

2

y

181614121086

A

D C

B

420 x

C’

B’

D’

A’

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Exercise 9.2H (page 33)1 a)

b) (i) (ii)

2 a)

b) (i) (ii)

3 a)

b) (i) 0·48 (ii) 0·28

4 a)

b) 0·515 b) (i) (ii)6 0·14

10 Working in two andthree dimensions

Exercise 10.1H (page 35)1 a) 7·21 units b) 6·40 units

c) 7·28 units d) 8·06 unitse) 7·62 units

2 a) 5·39 units b) 8·60 unitsc) 9·06 units d) 5 unitse) 13 units f) 18·79 units

Exercise 10.2H (page 36)1 a) (i) 7·21 cm (ii) 33·7°

(iii) 7·81 cm (iv) 22·6°b) A(0, 0, 3), B(0, 6, 3), C(4, 6, 3), D(4, 0, 3),

E(0, 0, 0), F(0, 6, 0), G(4, 6, 0), H(4, 0, 0)2 a) 17·0 cm b) 55·6°

c) 13·7 cm d) 64·1°3 a) 4·47 cm b) 48·2°

c) 9·22 cm d) 29·0°4 63·8°5 5·29 cm6 15·3°

Exercise 10.3H (page 38)1 a) 5·20 units

b) 10·10 unitsc) 3·46 unitsd) 7·87 unitse) 11·05 units

2 2 or 4


b)

2 a) 32·0° b) 44·9°3 35·3°; all cubes are similar.4 a) 23·4° b) 49·3°5 62·1°6 a) 10·1 cm b) 8·2 cm7 a) 22·9 cm

b) (i) 12·6° (ii) 60·8°

E F

C

A C

G

10––21

31–––105

Rain

Rain0.3

0.7

0.7

0.3

0.5

0.5

No rain

Saturday Sunday

Rain

No rain

No rain

M

M0.6

0.4

0.6

0.4

0.2

0.8

L

Year 1 Year 2

M

L

L

19––26

7––26

512

712

R

R

B

First card Second card

612

612

613

713

R

B

B

48––95

33––95

719

1219

B

B

G

Student 1 Student 2

819

1119

820

1220

B

G

G

215

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11 Histograms

Exercise 11.1H (page 41)1

2

2

1

0 10 20 30Age (years)

Women

Freq

uenc

y de

nsity

40 50 60 70 80

2

3

1

0 10 20 30Age (years)

Men

Freq

uenc

y de

nsity

40 50 60 70 80

2.5

2.0

1.5

1.0

0.5

060 80 100 120

Height (cm)

Freq

uenc

y de

nsity

140 160 180

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3

4

40

60

80

100

120

20

0 200Amount earned (£)

Freq

uenc

y / £

100

400 600 800 1000

0.5

0.4

0.3

0.2

0.1

0 100 200 300Money raised (£)

Freq

uenc

y de

nsity

(p

eop

le p

er £

)

400 500 600 700 800 900 1000

217

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b) 18·8 hours2 a) 82

b) 55·5 years

3 a)

b) £34 200 c) More people took part in the second race:

133 in the second race compared with 116in the first.The distributions were similarly shaped.Both distributions were positively skewedwith just over half the people in each raceraising less than £200, about one thirdraising between £200 and £500 and around13% raising more than £500.

4 a)

b) 60·2 minutes

12 Circle propertiesOther reasons may be possible in some cases.

Exercise 12.1H (page 46)1 a = 150° (angle at centre = 2 � angle at

circumference)2 b = 65° (angle at centre = 2 � angle at

circumference)c = 25° (angle sum of a triangle = 180° and

base angles of an isosceles triangleare equal)

3 d = 35° (base angles of an isosceles triangleare equal)

e = 70° (angle at centre = 2 � angle atcircumference or exterior angle of atriangle = sum of interior oppositeangles)

4 f = 128° (angle at centre = 2 � angle atcircumference)

5 g = 40° (angle in a semicircle and exteriorangle of a triangle)

6 h = 151° (angle at centre = 2 � angle atcircumference and angles on astraight line)

7 k = 55° (angle sum of a triangle and baseangles of an isosceles triangles giveangle at centre as 110°; angle atcentre = 2 � angle at circumference)

8 w = 152° (angle at the centre = 2 � angle atcircumference)

x = 208 (angles at a point)y = 104° (angle at the centre = 2 � angle at

circumference)

Exercise 12.2H (page 48)1 a = 44° (angles in the same segment)2 b = c = 110° (angles at centre = 2 � angle at

circumference and angles in thesame segment)

d = 70° (opposite angles of a cyclicquadrilateral)

3 e = 90° (angle in a semicircle)f = 38° (angles in the same segment)g = 52° (angle sum of a triangle)

4 j = 106° (opposite angles of a cyclicquadrilateral)

k = 98° (opposite angles of a cyclicquadrilateral)

5 m = 26° (angle at centre = 2 � angle atcircumference and angles in thesame segment)

6 p = 31° (angles in the same segment)q = 59° (angle at centre = 2 � angle at

circumference and isoscelestriangle)

Exercise 12.3H (page 49)1 a = 62° (angles in the same segment)

b = 32° (angle BAC = angle BDC = 48°,angles in the same segment; anglesum of a triangle)

2 angle BAC = 33° (base angles of anisosceles triangle)

angle ABC = 114° (angle sum of atriangle)

reflex angle AOC = 228° (angle at centre = 2 � angle atcircumference)

c = 132° (angles at a point)

Waiting time Frequency

(w minutes)

0 � w � 20 50

20 � w � 40 82

40 � w � 60 130

60 � w � 90 153

90 � w � 120 42

120 � w � 180 24

Amount raised Frequency

(£x)

0 � x � 50 15

50 � x � 100 25

100 � x � 200 30

200 � x � 500 45

500 � x � 800 18

Time (t hours) Frequency

0 � t � 5 3

5 � t � 10 10

10 � t � 20 30

20 � t � 30 24

30 � t � 35 5

35 � t � 40 3

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3 d = 28° (exterior angle of a triangle andisosceles triangle)

e = 112° (angle at centre = 2 � angle atcircumference)

4 f = 43° (angle in a semicircle and angle sumof a triangle)

g = 137° (opposite angles of a cyclicquadrilateral)

5 h = 22 ° (3h + h = 90°, angle in a semicircleand angle sum of a triangle)

6 k = 43° (isosceles triangle and angles in thesame segment)

Exercise 12.4H (page 50)1 a = 55° (angle between radius and

tangent and angle sum of atriangle)

2 b = 40° (perpendicular from centre tochord and angle sum of atriangle)

c = 50° (angle between radius andtangent and angle sum of atriangle)

3 d = 53° (isosceles triangle and anglebetween radius and tangent)

e = 53° (angle in a semicircle and anglesum of a triangle)

4 f = 24° (angle at centre = 2 � angle atcircumference)

g = 66° (angle between radius andtangent)

5 n = 9 cm (perpendicular from chord tocentre and Pythagoras)

6 OM = 12 cm (perpendicular from chord tocentre and Pythagoras)

ON = 5 cm (perpendicular from chord tocentre and Pythagoras)

k = 17 cm

Exercise 12.5H (page 51)1 a = 50° (angle in the alternate segment)

b = 80° (angle in the alternate segment)2 c = 65° (angle in the alternate segment)

d = 55° (angle in the alternate segment)3 e = 25° (angle between radius and tangent)

f = 65° (angle in the alternate segment)4 g = 63° (angle in the alternate segment)

h = 42° (angle in the alternate segment)5 n = 77° (opposite angles of a cyclic

quadrilateral)p = 42° (angle in the alternate segment)q = 61° (angles on a straight line)

6 u = 51° (angles in the alternate segment)v = 66° (angles on a straight line)

13 Straight-line graphs

Exercise 13.1H (page 52)1 2y = 3x + 42 y = 3x + 53 8x + 3y = 244 x + 3y = 95 3y = 2x + 56 4y + 3x = 97 y = x + 38 y = 2x – 1 9 3x + y = 2

10 a)

b) Gradient = 12 and represents accelerationin m/s2; Intercept = 20 and represents the initialvelocity in m/s

c) v = 20 + 12t

Exercise 13.2H (page 53)1 a) –2

b)c) –1

2 y = 2x – 23 4x + 2y = 14 or 2x + y = 74 a)

–

b) 3y = 5x – 25 x + 4y = 86 3x + y = 57 2y = 7x – 48 a) None of them

b) y = 3x + 5 and x + 3y = 59 x + y = 7

10 a) 2x + y = 7b) 2y = x + 4c) (4, 4) and (0, 2)

3–5

3–4

200v

100

150

50

0 5Time (seconds)

Velo

city

(m

/s)

10 15 t

1–2

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14 Surveys and sampling

Exercise 14.1H (page 55)1 For example:

How old are you? or What school year are youin? together with an appropriate tick box list.How often do you visit the cinema? togetherwith an appropriate tick box list.

2 a) Give a list of programme types e.g. pop,rock, easy listening, classical and so on.

b) What time of day? e.g. breakfast, morning,afternoon, evening, night.

c) What type of competition? Would youtake part?

d) Irrelevant?Other possible questions:Age. How long do you listen for? Do you listen to other stations? Why?

Exercise 14.2H (page 56)1 a) Would be random but excludes those

without phones and those who are ex-directory. Response depends on whoanswers the phone.

b) May not be from your town. Excludes those who do not travel by train.

c) Will only include those who do go torestaurants.

2 a) Systematic random sampling or stratifiedrandom sampling

b) Simple random sampling or systematicrandom sampling

c) Systematic random sampling

Exercise 14.3H (page 56)1 a) Select three from each class at random.

b) Choose one boy and two girls from eachclass at random.

2 50 out of 600 is 1 in 12. Select the followingnumber of employees from each department.

Exercise 14.4H (page 57) 1 250 2 Check students’ work.

The actual total number of nests is 477. 3 a) and b) Start at 98 (1st value):

mean = 109·55 cmStart at 125 (2nd value): mean = 108·05 cmStart at 77 (3rd value): mean = 101·3 cm Start at 102 (4th value): mean = 104·65 cm Start at 105 (5th value): mean = 104·45 cm

c) Students’ own sample. d) Approximately the same. e) Average of all 100 is 105·6 cm

Exercise 14.5H (page 60)1 a) Likely to involve bias as only football

supporters will be chosen. Also a very small sample.

b) Satisfactory as the cars are likely to havebeen parked randomly.

c) Could be biased as quality could alterduring the shift.

2 a) Not random and likely to be biased asmainly car owners use car parks.

b) Satisfactory as register is arranged inhouse order, which is not a relevantattribute. Only bias is that under-18s are notincluded.

c) Likely to be biased as many use the busbecause they do not own a car.

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Department Number of employees

A 15

B 4

C 21

D 10


Documents

Answers to Homework Book 9 - Hamilton Maths Blog · PDF fileGraduated Assessment for OCR GCSE Mathematics © Hodder Murray 2007 209 Answers to Homework Book 9 STAGE 9 1 Checking answers