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MARKING SCHEME BANK

2002 – 2011

Compiled & Edited By

Dr. Eltayeb Abdul Rhman

www.drtayeb.tk

First Edition

2011

Mark Scheme Syllabus Paper

IGCSE EXAMINATIONS – JUNE 2003 0580/0581 2

QuestionNumber

Mark SchemePart

MarksNotes

QuestionTotal

1 0.049 < 5% < 5/98 o.e. 2 M1 for figs 51… seenafter 0, SC1 for 2 correct entries 2

2 (a)

(b)

7.85 to 8(.00…)

56.25 to 57.5(0)

1

1 2

3 194(.4) 2 M1 for 54 � 3600/1000or SC1 for figs 194….seen 2

4 4

7

� ��

�

�

c.a.o.1

1 2

5 38 2 M1 for 665/(17 � 18) s.o.i. byequivalent complete method 2

6 201.25 2 allow 201 or 201.3 in ans. space if201.25 seenM1 for 17.5 � 11.5 s.o.i. 2

7 4 < x <6 2 SC1 for either oneafter 0, M1 for 8<2x<12 s.o.i. 2

8 �11 – �133114 196 –-7 49 –

3 2 for 4 or 5 correct1 for 2 or 3 correct

3

17

9 (a)

(b)

1

6 or 0.16(…..) or 0.17

art 9.5(�)

1

2 M1 for correct use of tan o.e. 3

10 11

( 3)( 4)

x

x x

�

� �

o.e.3 M1 for denom. (x � 3)(x � 4) o.e.

M1 for 2(x � 4) � (x � 3) o.e. 3

11 integer )7/112(

rational nos. 2.6 4/17

irrational no. 12

1

111

accept 16 or 4

accept 0.235accept 3.46

412 (a)

(b)

18

30

2

2

M1 for 2p � 3p � 90 = 180 o.e.or SC1 for 36 or 54 seen www.M1 for q � 5q = 180 o.e.or SC1 for 150 seen 4

14

www.xtremepapers.net



13 (a)

(b)

(c)

100

1200 �

10 < x < 30 ht 30 mm60 < x < 100 ht 22 mm

1

1

11

� for (12 � their a)

4

14 (a)

(b)

10 17 4�6 �9 0

2

1

2 4

3 5

� ��

oe

2

2

SC1 if 4 or 5 correct

1 for 2

1 s.o.i., 1 for k

2 4

3 5

� ��

s.o.i. 4

15 (a)

(b) (i)

(ii)

50.3

4710000 or 4.71 � 106

7.087 � 106

2

1

1

M1 for (7087000 4714900)

4714900

�

o.e.

must be recognisable completecorrect method

accept 7.09 � 106 , ignoresuperfluous zeros 4

16 (a)

(b)

24.7

46.2

2

2

M1 for 80 � sin 18� seen

M1 for 3(4 � 11.4) o.e. (no MRs)3 � 3.8 does not imply 11.4 4

16

17 (a)

(b) (i)

(ii)

Correct shear �1mm

Correct stretch �1mm

1 0

0 3

� ��

cao

2

2

1

M1 for shear with either axisinvariant

M1 for stretch with either axisinvariant

5

18 (a)

(b) (i)

(ii)

1:1000

accurate perp bisectorof AD, with two pairs ofarcs

accurate bisector of<BCD, with two pairs ofarcsT marked in correctposition

1

2

2

1

SC1 if accurate but no arcsSC1 if accurate arcs but no line


Indep. 6

11




19 (a)

(b)

(c)

correct demonstration

x � 2y = 120 o.e. fullysimplified

straight line thr. (120,0)and (0,60)60 cars, 30 trucks

2

2

1�

1

M1 for 20x � 80y seen

M1 for 25x � 50y = 3000 seencondone inequality signs formethod mark. Ignore $

� from their b). Line must becomplete , and be on given gridalso allow 80,20; 100,10; 120,0or points on the correct section ofthe line (60� x� 120) 6

6

20 (a)

(b)

(c)

art 0.1, 0.3, 0.6, 1, 1.7and 3

correct curve drawn

1.6 � x �1.65

3

2

1

SC2 for 4 or 5 correctSC1 for 2 or 3 correct

P1 for correct or � 6 or 7 pointscorrectly plotted 1mm

6

6

TOTAL MARKS 70




Marks in brackets are totals for questions or part questions.

1 (a) ($) 3490 B1 (1)

(b) 16n + 1570 = 4018 o.e.n = 153 c.a.o.

M1A1 (2) ww2

(c) x + y = 319 o.e.10x + 16y = 3784 o.e.Correct method s.o.i.

x = 220y = 99

B1B1M1

A1A1 (5)

e.g. 1st � 10 and subtraction.Condone arith. error (availableon wrong eqtns providedcoefficients not equal.)or 220 $10 ticketsor 99 $16 tickets (ww Correctanswer�M1)

(d) 0.85 � $16 o.e.($)13.6(0) c.a.o.

M1A1 (2)

[$16 – 0,15 � $16]ww2

(e) 100 � $10 o.e.125

($)8

M1

A1 (2) ww2

TOTAL 12

2 (a) 1202 = 772 + 552 – 2.55.77cos xcos x = 772 + 552 - 1202

2.55.77

or - 5446 = cos x = -0.64(29752) 8470 s.o.i. (-0.643)

x = 130(.0)

M1M1

A1

A1 (4)

Implied by next line

Implied by correct answer whichrounds to 130o

Scale drawing�M0. ww�SC2

(b) sin y = 55 sin 45o

60

sin y = 0.648 (1812) s.o.i.

y = 40.4

M2

A1

A1 (4)

If not scored, allow M1 forcorrect implicit eqtn

Implied by answer 40o aftersome workingAccept more accuracy but notless. www4 (40.39o – 40.41o;40oww�SC2)

(c) (i) 225o

(ii)* 275o

B2

B2 �(4)

Correct method seen ORanswer 222-224o, allow Sc1� 405o – their x (provided <360o). Answer 291-293o, allowSC1

TOTAL 12




3 (a)

0.35

0.6

0.55

B1

B1

B1 (3)

Accept percentages or fractionsbut not ratios

(b) (i) 0.4 � 0.65 ONLY0.26 c.a.o.

(ii)* Either0.4 � 0.35� or 0.6� � 0.45

0.4 � 0.35� + 0.6� � 0.45 ONLY0.41 c.a.o.

(iii)* Either 1 – (.6� � .55�) or .26+ .14� + .27�

0.67 c.a.o.

M1A1

M1

M1A1

M1A1 (7)

www2

Accepting their � values for Mmarks

www3

www2

(c) (i) 18 c.a.o.(ii) 12 � (his 18 + 6) o.e.

30 c.a.o.

B1M1

A1 (3) SC1 for 34.3 after 18 in (c) (i)

(d) (i) 22.5(ii)* Realises probability “STOP.STOP”

0.33

B1M1

dep.

A1�(3)

Accept 22min 30secImplied by correct answer aftercorrect work. Dep. On 18 and22.5 (approx.)�1 – their (b) (iii) or (their 0.6) �(their 0.55)

TOTAL 16

4 (a) Scales correct9 points correctly plotted (1mm)

Reasonable curve through 9 points

S1P3

C1�(5)

-4 ��x ��4 and -8 ��y ��8Allow P2 for 7 or 8 correct, P1for 5 or 6 correct� provided shape maintained,curvature OK and not ruled

(b) -3.6 ��x ��–3.3, x = 0, 3.3 ��x �3.6

B2 (2) Allow B1 for 1 correct non-zerosolution; condone (-3.5, 0)(answers must be in range andcorrect for their graph)

(c) Line from (-4, -3) to (4, 5), andruled

B2 (2) If B0, allow B1 for gradient 1 orintercept 1 on single line

(d) g(1) = 2fg(1) = -8g-1(4) = 33.75 ��x ��3.9

B1B1B1

B1 (4)

Not (1, 2)

Lost if y-coordinate given.Answer must be OK for theirgraph




(e) Tangent drawn at x = 3 on curveVert./Horiz. using scale

Answer in range 5-10 and OK for theirs

B1M1

A1 (3)

Not chord or daylightDep. on reasonable approx totangent used at x = 3(N.B. Gradient = 4.5 + y-value oftangent at x = 4)

TOTAL 16

5 (a) ½ 10.10.sin60o o.e.

43.3 cm2 or 25 3

M1

A1 (2)

Any complete method including

5.5.5.15

ww2

(b) 2r = 10 s.o.i.r = 1.59 (15494cm)

M1A1 (2)

Accept D = 10ww2

(c) (i) Tetrahedron or TriangularPyramid

4 (his (a))

* 173(.2cm2) or 100 3

(ii) CylinderUses (any r)2

10� ONLY

Uses (his (b))210�

Correct or � in range 79.35-79.65cm3

(iii) Cone

10 h

rAppreciates hypotenuse = 10

h = 22 ))((10 bhis�

9.87(25362cm)

B1

M1

�A1(3)B1M1

M1dep.

A1 (4)

B1

M1

M1

A1 (4)

If not his (a) then correct ∆ areamethod needed�4 (a) to 3s.f.

Accept circular (based) prismNot 2r210 or any othermodificationsImplies M2

Accept circular/round (based)pyramid

e.g. right-angled ∆ drawn or cos

x = 10

...

TOTAL 15

6 (a) 2x(x + 4)(x + 1) (cm3)2x

3 + 10x2 + 8x (cm3)

B1B1 (2) Must see this. Ignore further

correct work.




(b) 2x – 2, x + 2, x

Internal volume = 2x3 + 2x2 – 4xWood = his (a) – his(Int. Vol.)Correctly simplifies to 8x2 + 12x

B3

B1M1

A1 (6)

B1 each correct answer, anyorder but in this form

(Both could be wrong)No errors

(c) (i) 8x2 + 12x = 19802x2 + 3x – 495 = 0

r

qp �form�p = -3 and r = 4 or

2�2 �

� q = 32 – 4.2 – 495

� x = 15 www

� x = -16.5 or -2

33 www

B1 (1)

B1

B1

B1

B1 (4)

No error seen. Needs = 0

Alt. method B2 (x –15)(2x + 33)or SC1 for sign error(s) inbrackets

Or q = 3969 or q = 63. Allow

for p �r

q

If factorising method used,answers only score if correctand from correct bracket

(ii) Uses +ve answer

* 30 by 19 by 16

B1

�B1(2)

Rejects –ve solution explicitly orimplicitly�2(his), (his) + 4, (his) +1

TOTAL 15

7 (a)(i) OS = 3a www

(ii) AB = b – a www

(iii) CD = a www

(iv) OR = 2a + 2b www

(v) CF = 2a – 2b www

B1

B1

B1

B2

B2 (7)

If B0, allow SC1 for correct butunsimplified seen


(b) (i) |b| = 5(ii) |a – b| = 5 www

B1B1 (2)




(c) (i) Enlargement, S.F. 3, Centre 0

(ii) ReflectionIn line CF o.e.

B2

M1A1 (4)

Allow SC1 for Enlargement or(S.F. 3 and Centre 0)

SC1 for ‘Mirrored in CF’ o.e.

(d) (i) 6 c.a.o.

(ii) 60o

B1

B1 (2)

TOTAL 15

8 (a) (i) $60-80(ii) Midpoints 10, 30, 50, 70, 90

+ 120�fx attempted (12880)

�fx �200

Final answer $64.40 c.a.o.

B1M1

M1*

M1

A1 (5)

Needs at least 4 correct s.o.i.

Dep. on previous M1 or theirmidpoints � 0.5Dep. on M1*

Needs 2 d.p., www4 (64.4�M3AO)

(b) (i) (�)20, (�)40, (�)60, (�)80, (�)100, (�)140

10, 42, 90, 144, 180, 200(ii) Scales correct and labelled or

used to 140 and 200 6 plots correct (20, 10)� (140,

200) Graph from (0, 0), line or curve

B1

B1S1

P2

C1 (6)

Not for 42

4020 � type

Vert. 20cm ≡ 200 and Horiz. ≡14cm 140. Reversed axes SOP1 for 4 or 5 correct. 1mmaccuracyThrough all 6 points. Dep. on P1

(c) (i) Median ($)63-64

(ii) U.Q. ($)82-84(iii) IQR ($)38-41(iv) Using $75 reading on Cum.

Freq. Graph –67 or 68 or 69 or 70

or 71 or 72

B1

B1B1M1

A1 (5)

All answers in (c) must also becorrect for their graph (1mm)

e.g. answer 130 implies this

Must be integer answer and OKfor their graph

TOTAL 16

9 (a) Diagram 1�25% c.a.o.

Diagram 2�12½% o.e.


Diagram 4�60% o.e.

B1

B2

B2

B2 (7)

For whole section reversed (a)or (b), treat as MR-1 per sectionFor Diagrams 2-4 accept non%equivalentsAlso in each case if 2 notscored, allow SC1 if correct ideaseen (e.g. ½h �4h forDiagram 2)




(b) Diagram 5�1/9 o.e. fraction Diagram 6�1/25 o.e.

Diagram 7�5/9 o.e.

B1B2

B3 (6)

In Diagrams 6 and 7, acceptnon-fraction equivalents. If B0,allow SC1 for ()52 seenIf B0, allow SC1 for (k)22 andSC1 for (k)32 seen (k =1 orx/360) N.B. 4 must be from 22

and not 22

TOTAL 13

FINAL TOTAL 130




QuestionNumber

Mark SchemePart

MarksNotes

QuestionTotal

1 0.049 < 5% < 5/98 o.e. 2 M1 for figs 51… seenafter 0, SC1 for 2 correct entries 2

2 (a)

(b)

7.85 to 8(.00…)

56.25 to 57.5(0)

1

1 2

3 194(.4) 2 M1 for 54 � 3600/1000or SC1 for figs 194….seen 2

4 4

7

� ��

�

�

c.a.o.1

1 2

5 38 2 M1 for 665/(17 � 18) s.o.i. byequivalent complete method 2

6 201.25 2 allow 201 or 201.3 in ans. space if201.25 seenM1 for 17.5 � 11.5 s.o.i. 2

7 4 < x <6 2 SC1 for either oneafter 0, M1 for 8<2x<12 s.o.i. 2

8 �11 – �133114 196 –-7 49 –

3 2 for 4 or 5 correct1 for 2 or 3 correct

3

17

9 (a)

(b)

1

6 or 0.16(…..) or 0.17

art 9.5(�)

1

2 M1 for correct use of tan o.e. 3

10 11

( 3)( 4)

x

x x

�

� �

o.e.3 M1 for denom. (x � 3)(x � 4) o.e.

M1 for 2(x � 4) � (x � 3) o.e. 3

11 integer )7/112(

rational nos. 2.6 4/17

irrational no. 12

1

111

accept 16 or 4

accept 0.235accept 3.46

412 (a)

(b)

18

30

2

2

M1 for 2p � 3p � 90 = 180 o.e.or SC1 for 36 or 54 seen www.M1 for q � 5q = 180 o.e.or SC1 for 150 seen 4

14



13 (a)

(b)

(c)

100

1200 �

10 < x < 30 ht 30 mm60 < x < 100 ht 22 mm

1

1

11

� for (12 � their a)

4

14 (a)

(b)

10 17 4�6 �9 0

2

1

2 4

3 5

� ��

oe

2

2

SC1 if 4 or 5 correct

1 for 2

1 s.o.i., 1 for k

2 4

3 5

� ��

s.o.i. 4

15 (a)

(b) (i)

(ii)

50.3

4710000 or 4.71 � 106

7.087 � 106

2

1

1

M1 for (7087000 4714900)

4714900

�

o.e.

must be recognisable completecorrect method

accept 7.09 � 106 , ignoresuperfluous zeros 4

16 (a)

(b)

24.7

46.2

2

2

M1 for 80 � sin 18� seen

M1 for 3(4 � 11.4) o.e. (no MRs)3 � 3.8 does not imply 11.4 4

16

17 (a)

(b) (i)

(ii)

Correct shear �1mm

Correct stretch �1mm

1 0

0 3

� ��

cao

2

2

1

M1 for shear with either axisinvariant

M1 for stretch with either axisinvariant

5

18 (a)

(b) (i)

(ii)

1:1000

accurate perp bisectorof AD, with two pairs ofarcs

accurate bisector of<BCD, with two pairs ofarcsT marked in correctposition

1

2

2

1



Indep. 6

11



19 (a)

(b)

(c)

correct demonstration

x � 2y = 120 o.e. fullysimplified

straight line thr. (120,0)and (0,60)60 cars, 30 trucks

2

2

1�

1

M1 for 20x � 80y seen

M1 for 25x � 50y = 3000 seencondone inequality signs formethod mark. Ignore $

� from their b). Line must becomplete , and be on given gridalso allow 80,20; 100,10; 120,0or points on the correct section ofthe line (60� x� 120) 6

6

20 (a)

(b)

(c)

art 0.1, 0.3, 0.6, 1, 1.7and 3

correct curve drawn

1.6 � x �1.65

3

2

1

SC2 for 4 or 5 correctSC1 for 2 or 3 correct

P1 for correct or � 6 or 7 pointscorrectly plotted 1mm

6

6

TOTAL MARKS 70




1 (a) ($) 3490 B1 (1)

(b) 16n + 1570 = 4018 o.e.n = 153 c.a.o.

M1A1 (2) ww2

(c) x + y = 319 o.e.10x + 16y = 3784 o.e.Correct method s.o.i.

x = 220y = 99

B1B1M1

A1A1 (5)

e.g. 1st � 10 and subtraction.Condone arith. error (availableon wrong eqtns providedcoefficients not equal.)or 220 $10 ticketsor 99 $16 tickets (ww Correctanswer�M1)

(d) 0.85 � $16 o.e.($)13.6(0) c.a.o.

M1A1 (2)

[$16 – 0,15 � $16]ww2

(e) 100 � $10 o.e.125

($)8

M1

A1 (2) ww2

TOTAL 12

2 (a) 1202 = 772 + 552 – 2.55.77cos xcos x = 772 + 552 - 1202

2.55.77

or - 5446 = cos x = -0.64(29752) 8470 s.o.i. (-0.643)

x = 130(.0)

M1M1

A1

A1 (4)

Implied by next line

Implied by correct answer whichrounds to 130o

Scale drawing�M0. ww�SC2

(b) sin y = 55 sin 45o

60

sin y = 0.648 (1812) s.o.i.

y = 40.4

M2

A1

A1 (4)

If not scored, allow M1 forcorrect implicit eqtn

Implied by answer 40o aftersome workingAccept more accuracy but notless. www4 (40.39o – 40.41o;40oww�SC2)

(c) (i) 225o

(ii)* 275o

B2

B2 �(4)

Correct method seen ORanswer 222-224o, allow Sc1� 405o – their x (provided <360o). Answer 291-293o, allowSC1

TOTAL 12



3 (a)

0.35

0.6

0.55

B1

B1

B1 (3)

Accept percentages or fractionsbut not ratios

(b) (i) 0.4 � 0.65 ONLY0.26 c.a.o.

(ii)* Either0.4 � 0.35� or 0.6� � 0.45

0.4 � 0.35� + 0.6� � 0.45 ONLY0.41 c.a.o.

(iii)* Either 1 – (.6� � .55�) or .26+ .14� + .27�

0.67 c.a.o.

M1A1

M1

M1A1

M1A1 (7)

www2

Accepting their � values for Mmarks

www3

www2

(c) (i) 18 c.a.o.(ii) 12 � (his 18 + 6) o.e.

30 c.a.o.

B1M1

A1 (3) SC1 for 34.3 after 18 in (c) (i)

(d) (i) 22.5(ii)* Realises probability “STOP.STOP”

0.33

B1M1

dep.

A1�(3)

Accept 22min 30secImplied by correct answer aftercorrect work. Dep. On 18 and22.5 (approx.)�1 – their (b) (iii) or (their 0.6) �(their 0.55)

TOTAL 16

4 (a) Scales correct9 points correctly plotted (1mm)


S1P3

C1�(5)

-4 ��x ��4 and -8 ��y ��8Allow P2 for 7 or 8 correct, P1for 5 or 6 correct� provided shape maintained,curvature OK and not ruled

(b) -3.6 ��x ��–3.3, x = 0, 3.3 ��x �3.6

B2 (2) Allow B1 for 1 correct non-zerosolution; condone (-3.5, 0)(answers must be in range andcorrect for their graph)

(c) Line from (-4, -3) to (4, 5), andruled

B2 (2) If B0, allow B1 for gradient 1 orintercept 1 on single line

(d) g(1) = 2fg(1) = -8g-1(4) = 33.75 ��x ��3.9

B1B1B1

B1 (4)

Not (1, 2)

Lost if y-coordinate given.Answer must be OK for theirgraph



(e) Tangent drawn at x = 3 on curveVert./Horiz. using scale

Answer in range 5-10 and OK for theirs

B1M1

A1 (3)

Not chord or daylightDep. on reasonable approx totangent used at x = 3(N.B. Gradient = 4.5 + y-value oftangent at x = 4)

TOTAL 16

5 (a) ½ 10.10.sin60o o.e.

43.3 cm2 or 25 3

M1

A1 (2)

Any complete method including

5.5.5.15

ww2

(b) 2r = 10 s.o.i.r = 1.59 (15494cm)

M1A1 (2)

Accept D = 10ww2

(c) (i) Tetrahedron or TriangularPyramid

4 (his (a))

* 173(.2cm2) or 100 3

(ii) CylinderUses (any r)2

10� ONLY

Uses (his (b))210�

Correct or � in range 79.35-79.65cm3

(iii) Cone

10 h

rAppreciates hypotenuse = 10

h = 22 ))((10 bhis�

9.87(25362cm)

B1

M1

�A1(3)B1M1

M1dep.

A1 (4)

B1

M1

M1

A1 (4)

If not his (a) then correct ∆ areamethod needed�4 (a) to 3s.f.

Accept circular (based) prismNot 2r210 or any othermodificationsImplies M2

Accept circular/round (based)pyramid

e.g. right-angled ∆ drawn or cos

x = 10

...

TOTAL 15

6 (a) 2x(x + 4)(x + 1) (cm3)2x

3 + 10x2 + 8x (cm3)

B1B1 (2) Must see this. Ignore further

correct work.



(b) 2x – 2, x + 2, x

Internal volume = 2x3 + 2x2 – 4xWood = his (a) – his(Int. Vol.)Correctly simplifies to 8x2 + 12x

B3

B1M1

A1 (6)

B1 each correct answer, anyorder but in this form

(Both could be wrong)No errors

(c) (i) 8x2 + 12x = 19802x2 + 3x – 495 = 0

r

qp �form�p = -3 and r = 4 or

2�2 �

� q = 32 – 4.2 – 495

� x = 15 www

� x = -16.5 or -2

33 www

B1 (1)

B1

B1

B1

B1 (4)

No error seen. Needs = 0

Alt. method B2 (x –15)(2x + 33)or SC1 for sign error(s) inbrackets

Or q = 3969 or q = 63. Allow

for p �r

q

If factorising method used,answers only score if correctand from correct bracket

(ii) Uses +ve answer

* 30 by 19 by 16

B1

�B1(2)

Rejects –ve solution explicitly orimplicitly�2(his), (his) + 4, (his) +1

TOTAL 15

7 (a)(i) OS = 3a www

(ii) AB = b – a www

(iii) CD = a www

(iv) OR = 2a + 2b www

(v) CF = 2a – 2b www

B1

B1

B1

B2

B2 (7)



(b) (i) |b| = 5(ii) |a – b| = 5 www

B1B1 (2)



(c) (i) Enlargement, S.F. 3, Centre 0

(ii) ReflectionIn line CF o.e.

B2

M1A1 (4)

Allow SC1 for Enlargement or(S.F. 3 and Centre 0)

SC1 for ‘Mirrored in CF’ o.e.

(d) (i) 6 c.a.o.

(ii) 60o

B1

B1 (2)

TOTAL 15

8 (a) (i) $60-80(ii) Midpoints 10, 30, 50, 70, 90

+ 120�fx attempted (12880)

�fx �200

Final answer $64.40 c.a.o.

B1M1

M1*

M1

A1 (5)

Needs at least 4 correct s.o.i.

Dep. on previous M1 or theirmidpoints � 0.5Dep. on M1*

Needs 2 d.p., www4 (64.4�M3AO)

(b) (i) (�)20, (�)40, (�)60, (�)80, (�)100, (�)140

10, 42, 90, 144, 180, 200(ii) Scales correct and labelled or

used to 140 and 200 6 plots correct (20, 10)� (140,

200) Graph from (0, 0), line or curve

B1

B1S1

P2

C1 (6)

Not for 42

4020 � type

Vert. 20cm ≡ 200 and Horiz. ≡14cm 140. Reversed axes SOP1 for 4 or 5 correct. 1mmaccuracyThrough all 6 points. Dep. on P1

(c) (i) Median ($)63-64

(ii) U.Q. ($)82-84(iii) IQR ($)38-41(iv) Using $75 reading on Cum.

Freq. Graph –67 or 68 or 69 or 70

or 71 or 72

B1

B1B1M1

A1 (5)

All answers in (c) must also becorrect for their graph (1mm)

e.g. answer 130 implies this

Must be integer answer and OKfor their graph

TOTAL 16

9 (a) Diagram 1�25% c.a.o.



Diagram 4�60% o.e.

B1

B2

B2

B2 (7)

For whole section reversed (a)or (b), treat as MR-1 per sectionFor Diagrams 2-4 accept non%equivalentsAlso in each case if 2 notscored, allow SC1 if correct ideaseen (e.g. ½h �4h forDiagram 2)



(b) Diagram 5�1/9 o.e. fraction Diagram 6�1/25 o.e.

Diagram 7�5/9 o.e.

B1B2

B3 (6)

In Diagrams 6 and 7, acceptnon-fraction equivalents. If B0,allow SC1 for ()52 seenIf B0, allow SC1 for (k)22 andSC1 for (k)32 seen (k =1 orx/360) N.B. 4 must be from 22

and not 22

TOTAL 13

FINAL TOTAL 130


MATHEMATICS – JUNE 2004 0580/0581 2

Question Number

Mark Scheme Notes

1 3h 20m 1 2 10.9 1 3 0.53 < 0.52 < √ 0.5 2* M1 for 0.25, 0.7.... and 0.125 seen matched

4

2

1p20

2 B1

2

1 or p20

5 24 2* M1 x/4 = 6 or x – 32 = -8 seen 6 6375 6385 1, 1 B1 correct but reversed 7 7 2* B1 for one of -7/8, -1/8, -14/16, -2/16, -0.875,

-0.125 8 (a) 4 1

(b) 4 1 Not 90 or

4

1 turn

9 450 2* M1 for 3000 x 7.5 x 2/100 10 (a)

(b) 80000 8 x 104

1

1 √

8 x 104

11 x = 8 y = 1 3* M1 double and add/subtract consistently A1 A1 or M1 rearrange and substitute correctly

12 50, 5, 3 1, 1, 1 13

√

−

k

ec

3* R1, R1 for any 2 correct steps moving e, k or √ Allow d2 = (c – e)/k to score R2 as a single step

14 (a)

1 Arc must not continue outside rectangle.

Radius of arc 4 cm ± 1 mm. Ignore shading

(b) 12.6 2* M1 for 4

1 x π x 42

15 4 3* M1 Area factor or ratio 9 M1 LSF 3 16 (a)

(b) a + c a – c or –c + a

1 1

(c)

2

1− a -

2

1c or

2

1− (a + c) 2* M1 A0 for answers simplifying to these seen

17

2* 2*

1

M1 2 arcs centre B and D, line drawn A1 M1 construction arcs on AD and CD and centre

these for the bisector, line drawn A1 Dependent on at least 1 + 1 in part (a) SC1, SC1 If accurate and no construction arcs

18 (a) (b)

114 (0)47 cao

2* 3*

M1 782 + 832

M1 for finding one angle by trigonometry correctly M1 for clearly identifying bearing angle Scale drawing and answers with no working score zero

19 (a) 11 1

(b) x + 2 2* M1 12

)1(2+

+x

(c) 3 2* M1 for explicit g(1) or g-1(x) =

2

1−x

20 (a) 3(2x – y)(2x + y) 2 B1 (6x – 3y)(2x + y) o.e.

(b) (i) x2 – 6x + 9 (ii) p = 3 q = 1

2* 2

M1 correct method B1, B1




21 (a)

1.8

2

M1 convincing gradient calculation or use of a = (v – u)/t

(b) 450 2* M1 for 20 x 18 +

2

1 x 10 x 18

(c) 13

3* M1 for finding total area under graph

((b) + 135) dep M1 for ÷ 45

If the vertical scale is consistently misread then M4 A0 is available

22 (a) BA or (iii) 2* M1 checking order of all 4 matrices correctly

(b)

380

038 2 M1 either column or row correct

(c)

−

− 38/2

38/6

25

64

38

1

5/38

4/38or

1

−

− 0526.0132.0

158.0105.0

19/138/5

19/319/2or

TOTAL 70




Q1(a)(i) (ii) (b) (c) (d)(i) (ii)

60 x 120 o.e. 100 ($) 132 c.a.o. their(a)(i) x 100 o.e. 120 110(%) Final answer, but may be explained using 10. 159.10 (x100) o.e. their 86 ($) 185 c.a.o. 156 x 52 o.e. 169 48(cm) c.a.o. 11 x 36 o.e. 20 19.8(km) c.a.o. 36 x 23 o.e. 2 414(km) c.a.o.

M1 A1 M1 A1 √ M1 A1 M1 A1 M1 A1 M1 A1

Implied by 72 seen and not spoilt. ww2 √ ft their (a)(i) x 100 120 Sc1 for 10 or their extra % or their(a)(i) − 120 x100 120 Allow any statement that equates 159.10 with 86% provided it is not contradicted later. ww2 Alt. Method 156 = x o.e. 156+169 x+52 ww2 Method not spoilt by also doing 9 x 36 20 ww2 Condone 19.8:16.2 16.2:19.8 is M1A0 ww2 12

Q2(a)(i) (ii) (iii) (b)(i) (ii) (c)(i) (ii)

p = 9 q = −3 r = 9 Scales correct Their 8 points plotted correctly (1mm) Reasonable curve through all 8 of their points ( 1mm tolerance) Tangent drawn at x = −1 on curve −3.5 to −2.5 Condone fractions u = 6.33 or better v = 6 Their 6 points plotted correctly (1mm) Reasonable curve through all 6 of their points (1mm tolerance) x2 − x − 3 = 6 − x3/3 o.e. to x3 + 3x

2 −3x −27 = 0 2.3 to 2.7 c.a.o.

1+1+1 S1 √ P2 √ C1 √ T1 B2 1+1 P3 √ C1 √ E1 B1

Must be seen. No feedback from graph. x from −3 to 4. y to accommodate their values. P1 √ for 6 or 7 of their points correct. Condone ruled line for x = 3 to 4 or –3 to –2. ft provided correct shape maintained. Or a parallel line drawn. If B2 not scored, give B1 for 2.5 to 3.5 after M1. Allow u = 19/3 P2 for 5 correct ( √ ). P1 for 4 correct ( √ ). Condone ruled line for x = 2 to 3. ft provided correct shape maintained At least 1 intermediate step and no errors seen. Not coordinates 18

Q3(a)(i) (ii) (iii) (iv) (b)(i) (ii) (c)

Median 36 to 37 (cm) IQR 19 to 21 (cm) Evidence of using 146 (approx) 32 to 33 (cm) 275 to 281 350 − 303 365 −350 Midpoints 5,15,25,35,45,55,65 ∑fx attempted (13065) ∑fx / 365 35.8 or 36 or 35.79 www 2.9 (cm) c.a.o Evidence of dividing by 30 o.e 4.9 (cm) c.a.o..

B1 B2 M1 A1 B2 B1 B1 M1 M1* M1 A1 B1 M1 A1

Sc1 for 45.5 to 46.5 or 25.5 to 26.5 seen. ww2 Sc1 for 84 to 90 seen At least 6 correct s.o.i. Dep. on first M1 or using midpoints ±0.5 Dep. on second M1* www4 [35.79452055] ISW subsequent rounding to 3 or 5 once seen. eg a factor of 1.5 used constructively. 16




Q4(a) (b) (c) (d)(i) (ii)

(AC2 =) 9.52 +11.12 −2x9.5x11.1cos70 square root of correct combination (141.3279…) or 11.888… 11.9 (cm) (Opp. angles of) cyclic quadrilateral (add to 180) 70 – 37 attempted s.o.i. AD = their(a) o.e. sin33 sin110 (AD=) their (a) x sin33 sin110 art 6.89 or 6.90 (cm) 70 (h =)their(a)x tan55or their(a) (8.497..)

2 2xtan35

(area =) 0.5 x their(a) x their(h) o.e. 50.4 to 50.8 (cm2)

M2 M1 A1 B1 M1 M1 M1 A1 B1 M1 M1 A1

Allow M1 for 9.52 + 11.12 − AC2 = cos70 2 x 9.5 x 11.1 Dep. on previous M2. Must be convinced that errors are due to slips not incorrect combination. www4 Scale drawing gets M0A0. Condone 180 − 70 = 110 o.e. (not spoilt) e.g. 32 or 34 or 43, but be convinced. Dep. on first M1 Dep. on M2 Would imply M3 if nothing incorrect seen earlier. Condone 6.9 www4 Scale drawing gets M0A0 If not 70, ft for method in (ii), but not from 90 or60 (EC or EA=) their(a) or their(a) (10.37…) 2 sin35 2 cos55 Dep. on first M1 (area =) 0.5 x EC x EA x sin70 or Hero’s Method www3 13

Q5(a) (b) (c) (d)

10/x or 10 ÷ x o.e. 10 10 = 1 o.e. x x +1 2 20( x +1) − 20x = x( x + 1) o.e. x2 + x − 20 = 0

(x + 5)(x − 4) (= 0) −5 and 4 c.a.o. Rejects negative solution 2.5 (hours) c.a.o.

B1 M2 MA1 E1 M1 A1 R1 B1

Ignore all units in answers to Question 5. Not x = 10/x Condone 30 for ½ If M0 give Sc1 for 10 s.o.i. x + 1 Dep on M2. No longer condoning 30 o.e. Sc1 for 20x – 20(x + 1) = x( x + 1) o.e. after B1Sc1 No error of any kind at any stage and sufficient working to convince you (at least 1 extra step) −1 ± √ [12 − 4.1.(−20)] No errors or ambiguities 2 www2 May be explicit or implicit and could be in (c) Condone 2 hrs 30 (mins) or 150 mins 9




Q6(a)(i) (ii) (b) (c)

2 x π x 73 + π x 72 x 13 3 3

1384.7 to 1386 or 1380 or 1390 (cm3) their(a)(i) x 0.94 1.3 (kg) (L =) √(132 + 72) π x 7 x theirL 324 to 326 (cm2) CSA of hemisphere=2 x π x 72 s.o.i. their(b) + their CSA 631.7 to 634 411.58 s.o.i. their total ($)0.649 to 0.652 or 64.9 to 65.2 cents

M1 A1 M1 A2√ M1 M1 A1 M1 M1 A1 M1 A1

www2 √ ft their(a)(i) x 0.94 1000 www3 If A2 not scored, allow A1 √ for 1.30… Implied by √ 218 or 14.7….. or 14.8 Dep. on first M1. www3 307.7 to 308 if no working Dep. on first M1 Seen or implied by subsequent working. Dep. on a total www5 13NB M1M1A0M1A1 is not possible.

Q7(a)(i) (ii) (iii) (iv) (b)(i) (ii) (iii) (iv)

Venn Diagram with 12, 8, 7, 3 or with 20 – x, x, 15 – x, 3 8

12 o.e 30 12 o.e. 20

3/9 x 4/10 12 o.e. c.a.o. 90 1 − their(b)(i) 78 o.e. c.a.o. 90 5/8 or 5/9 seen 6/9 x 5/8 x 6/10 x 5/9 seen 900 6480 o.e. c.a.o. p(4 blacks) 3/9 x 2/8 x 4/10 x 3/9 (=1/90) 1 − their(b)(iii) − their p(4 blacks) 5508 6480 o.e. c.a.o.

B2 B1√B2√ B2√ M1 A1 M1 A1√ M1 M1 A1 M1 M1 A1

-1 each error/omission. Condone lack of labels. √ ft their 8 on diagram, but not x √ ft (their 12)/30 from (i) or (ii) Sc1 for k/30 where k < 30 √ ft (their 12)/20 from (i) or (ii) if their12<20 Sc1 for m/20 where m < 20 In all of Q7, accept fractions, decimals or %. Mark as ISW for wrong cancelling. Dec. or % need to be exact or accurate to 3 sf. No ratios. Other inappropriate notation is −1 once. or 6/9 x 6/10 + 6/9 x 4/10 + 3/9 x 6/10 √ ft 1 – their (b)(i) Allow a slip in 1 digit, but must use 4 fractions multiplied. Simplest 5/36 Alt. method. Must see all 14 combinations. Dep. on first M1. Must add them Simplest 17/20 17




Q8(a)(i) (ii) (iii) (iv) (v) (vi) (b) (c)(i) (ii)

Rotation (only) 90 (anticlockwise)(about O) or ¼ turn Translation (only) −2 −5 o.e. Reflection (only) y = −x o.e . 180 (or ½ turn) Rotation (only) Centre (1, −1) Enlargement (only) Scale Factor 2 (centre O) Shear (only) y axis invariant or parallel to y axis B 1 0 0 −1 1 0 1 1

B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B1 B2 B2 B2

“only” --- no other transformation mentioned. Ignore all matrices, except in (v). Do not allow “turn” for rotation Accept 270 clockwise or −270 Not translocation,transformation,transportation. eg 2 to left and 5 down. Condone (–2 -5) and lack of brackets. Enlargement sf= −1 earns B2 Sc1 for “Point Symmetry” Accept 2 0 for scale factor 2 0 2 Ignore any mention of scale factor. Sc1 for a correct column Sc1 for a correct column 18

Q9 (a) (b) (c) (d)(i) (ii) (e) (f)

15x + 25y ≤ 2000 seen y ≤ x o.e. c.a.o. y ≥ 35 o.e. c.a.o. Scales correct and full length. 3x + 5y = 400 correct (1mm) at (0,80) and (100,20) and long enough. y = x correct y = 35 correct Shading correct (in or out) 38 c.a.o. Identifying any point(s) in their area (enclosed by 3 lines or 3 lines and 1 axis). (75 , 35) s.o.i. c.a.o. ($) 6.2(0) or 620 (cents)

B1 B2 B1 S1 B2 L1 L1 B1 √ B1 M1 A1 B1 √

Allow 0.15x + 0.25y ≤ 20 but no others. Sc1 for any other sign between x and y Reversed scales S0 Sc1 for either point correct. √ ft from slips in lines that do not compromise the idea of the triangle. Implies M1 √ ft their (75, 35) evaluated for whole numbers only. Condone lack of units but not wrong units. www3 14




* indicates that it is necessary to look in the working following a wrong answer

1 (a) (b)

25/32 0.781 (25)

1 1√

2 0.276 2* M1 sin 5° = h/3.17 0.28 may score M0

3 (a) (b)

0.016 1.6 × 10 -2

1 1√

Allow 2/125 x 10 essential

4 1(.00) or 0.9r 2 M1 A0 other answers in range 0.99 to 1.053

5 (a) (b)

3 3 lines

1 1

by eye

6 (a) (b)

5.66 32(.0)

2 1√

M1 42 + 42 or 4/sin45 or 4√2 or √32 (a)2 from the answer space

7

(a) (b)

21.5 22.5 172

1,1 1√

SC1 correct but reversed (a) least value x 8

8 x = 8 y = 6 3* M1 for multiplication and subtraction

9 (a) (b)

wf = 300000 oe 500

2 1√

M1 wf = k A1 k = 300000

10 (a) (b)

8/19 or 0.421 7/18 or 0.389

2 1√

M1 their prime number count/19

11

3 B1 for 8 in correct place B1 for 2 in correct place B1 for 4 and 7 in correct place SC2 2 4 8 7 or 2 6 6 7

12 (a) (b)

xx

xx

24

42

22

22

54

45

xx

xx

1 2*

M1

++

++

2222

2222

422

224

xxxx

xxxx

13 (a) (b) (c) (d)

8, 11, 14 3n + 2 182 29

1 1 1√ 1√

integers only

14 (a) (b)

20% 400%

2* 2*

M1 for310000

62000x 100

M1 for 62000

248000x 100

15 (a) (b)

3/2 oe y = 3/2x – 7

1 2*√

M1 correct method

16 )2(

)1(4

+

+

xx

x oe

3* M1 (x + 2)(x +2) - x2 B1 x2 + 4x + 4




17 (a) (b) (c)

0 –1.5 below the height at midday

1 2* 1

M1 for t = 7

18 (a) (b)

0.4,0.3, 0.3 0.46

1 3*

M1 0.6 × “0.3” or “0.4” × 0.7 dep M1 add

19 (a) –12 x > –1 cao

2* 3*

M1 0.2x = –2.4 B1 for every two moves completed correctly

20 (a) (b)

232o

175(.4)° 2* 4*

M1 for 360 – (63 + “65”)

M1 for 63sin

410=

xsin

400 A1 GAW = 60.4

M1 115 + GAW and no further working A1 √

21 (a) (b)

(i) (ii) (i) (ii)

20 70 3.49 8.73

1 1 2* 2*

M1 360

40x 2 x π x 5

M1 360

40 x π x 52

TOTAL 70



IGCSE – JUNE 2005 0580/0581 4

1 (a)

1.33 × 7

5 o.e

M1

Implied by figures 95 in answer

950 (kg) c.a.o. A2

A1 for figs 95

(b) 765 ×

)89(

9

+

o.e. M1

($) 405 c.a.o.

A1

(c) their (b) their (a)

M1

($) 0.43 or ($) 0.426 A1√

f.t. their (b) must be in dollars for A mark their (a)

(d)(i) 0.35 x 60 o.e 100

M1

($) 0.21 c.a.o. A1

(ii) 0.35 x 100 o.e 125

M1

($) 0.28 c.a.o A1 11

0.26(25) is M0

2 All measurements ±2 mm or ±2 ° (a) AB = 12cm B1

(b) Perp. Bisector with arcs-2 sets for AB B2√

SC1 if accurate without arcs

(c) Accurate trapezium c.a.o. B2

dep. on B1 in (a) and at least SC1 in (b). SC1 for DC = 9cm and parallel to AB

(d)

Strict ft of their angle ABC (±2 º)

B1√

(e) (tan B =)

5.1

7

B1

or (sinB=) 7 or (cosB=) 1.5__ √(7²+1.5²) √(7²+1.5²)

77.9 final answer B1

Indep

(f)(i)

Arc, centre D, radius 5 cm B1√

No gaps in the trapezium, but condone extra

(ii) Bisector of their angle D with arcs B2√ SC1 if accurate without arcs

(iii) Correct shading c.a.o. B1 12

dep. on B1 in (i) and at least SC1 in (ii) and a correct trapezium



IGCSE – JUNE 2005 0580/0581 4

3 If choice of transformations in (i), (ii), (iii), (iv) then lose the 1st two B marks in each part e.g. 6 left and 1 up. Condone -6 1

(a)(i) Translation (only) (T) B1

-6 o.e. 1

B1

(ii) Reflection (only) (M) in y = -x o.e.

B1 B1

must be equation

(iii) Enlargement (only) (E) Centre (0,6) Scale factor 3 o.e. seen

B1 B1 B1

(iv) Shear (H) x-axis (y = 0) invariant (Shear) factor 0.5 o.e. seen

B1 B1 B1

(b)(i) (ii)

0 -1 o.e.

-1 0

1 0.5 o.e.

0 1

B2

B2 14

SC1 for a correct column

SC1 for a correct column Allow embedded matrices in both answers

4 (a) p = 0.25 q = 1 r = 8

B1 B1 B1

Must be seen. No feedback from graph. If not labelled, must be in order

(b) Scales correct Their 7 points plotted correctly (within 1mm and in the correct square) Smooth curve through all 7 points (1mm)

S1 P3√ C1√

x from –2 to 4. y to accommodate their values. ft P2 for 6 points correct. P1 for 5 points correct. ft provided correct shape maintained

(c) 2.75 to 2.85 B1

(d) 0 B1

(e) Tangent drawn at x = 1.5 Uses increase in y (using scale) increase in x 1.7 to 2.2

T1 M1 A1

Not a chord and no daylight Dep on T1 or a near miss (not chord or clearly drawn at x = 1 or x = 2) If correct method seen, condone any answer in range, even with a slight slip

(f) Correct ruled straight line (complete for range 0 to 4)

B2

SC1 for freehand complete line or any ruled line of gradient 2 or y-intercept of 1 (not y=1)

(g) Correct for theirs(±0.05) dep. on at least SC1 in (f)

B2√ 17

SC1 if y-coordinate also given or x=0 also given (or both)



IGCSE – JUNE 2005 0580/0581 4

5 (a)(i) c – d final answer o.e.

B1

(ii) OD + DE or OC + their CD + DE o.e. d – 0.5c final answer o.e.

M1 A1

Must be seen if answer incorrect

(iii) OA + AB or OC + CB or OC + EO o.e. 1.5c – d final answer o.e.

M1 A1

Must be seen if answer incorrect

(b)(i) 120

B1 If 90 then only method marks in (iv) available If 60 only method marks in (ii) and (iv) available

(ii) 0.5 × 8 × 8 sin120 o.e. M1 e.g. perp. onto AC, then 8sin60 × 8cos60 art 27.7 (cm²) www A1

(16√3)

(iii) 82 + 82 − 2 × 8 × 8 cos120 Square root of correct combination

M1 M1

** Dep on first M1. Errors must be due to slips, not incorrect combination

( √192 or 13.8

6

5)

art 13.9 (cm) (13.856406)

A1

(8√3) ** Alternative methods e.g. perp onto AC, then 8sin60 M1 ×2 M1 Sine Rule Implicit M1 Explicit M1

(iv) ABC (×2) + OACD o.e. M1 Alt meth. 6 × ABX (X is centre) or 6 × ABC etc. their (ii) × 2 + their (iii) × 8 M1 6 × [0.5 × 8 × 8 sin60] or their (ii) × 6 etc. 166 to 167 (cm2) c.a.o. A1

(96√3)

14



IGCSE – JUNE 2005 0580/0581 4

USE OF RADIUS = 0.7 6 (a) Vol of cyl.= π × 0.352 × 16.5 (6.3…) M1 Use of radius = 0.7 loses all marks in (a)

Vol of cone = π x 3

35.02

x 1.5 (0.19...) M1

After that they can revert to 0.35 without penalty

a.r.t. 6.54 (cm3) A1

Any later use of 0.7 after 0.35 penalty 2 from the marks gained using 0.7

(b)(i) 4.2 1.4

B1 B1

8.4 2.8

B1 B1

(ii) 18 × their 4.2 × their 1.4 M1 18 × their 8.4 × their 2.8 M1 106 (cm3) (105.84) A1

423 (cm3) (423.36) A1

(iii) 12 × their (a) ×100 their (b)(ii)

M1

12 × their (a) ×100 their (b)(ii)

M1

74.(0) to 74.2 (%) c.a.o. A1 74.1 to 74.3 (%) A1

(c)(i) (l =) √( 1.52 + 0.352) M1 (l =) √( 1.52 + 0.72) M1 1.54 (cm) A1

1.66 (cm) A1

(ii) Circle = π × 0.352 M1 Circle = π × 0.72 M1 Cylinder = 2 × π × 0.35 × 16.5 M1 Cylinder = 2 × π × 0.7 × 16.5 M1 Cone = π × 0.35 × their (c)(i) M1 Cone = π × 0.7 × their (c)(i) M1 Any 2 correct areas B2 Any 2 correct areas B2 ( a.r.t. 0.385 a.r.t. 36.3 a.r.t. 1.69 ) (a.r.t. 1.54 72.5 to 72.6 a.r.t. 3.65) 0.1225π 11.55π 0.539π 0.49π 23.1π 1.162π 38.3 to 38.4 (cm2) c.a.o. A1 77.7 to 77.8 (cm2) A1

17



IGCSE – JUNE 2005 0580/0581 4

7 (a)(i) Median 46.5 B1

(ii) IQR 9.5 www B2 SC1 for 42 or 51.5 seen

(iii) 48

B2 SC1 for 102 seen

(b)(i) n = 32

B1

(ii) Midpts 32.5, 37.5, 42.5, 47.5, 52.5, 57.5 10x32.5 + 17x37.5 + 33x42.5 + 42x47.5 + their 32x52.5 + 16x57.5 [6960] ∑fx / 150 46.4

M1 M1* M1 A1

At least 5 correct s.o.i. Dep on first M1 or midpoints ±0.5 Allow 1 more slip Dep on 2nd M1*

(c) Horizontal Scale correct S1

Implied by correct use. Ignore vertical scale

3 correct widths on their scale (f.t ) W1√ no gaps For each block of correct width

2.7 cm 7.1(3) or 7.2 cm 3.2 cm

H1 H1 H1 15

For scale error double or half, award H1, H1, H1 for correct f.t heights After H0, SC1 for 3 correct frequency densities written or for heights 2.7cm, 7.1cm and 3.2cm drawn on doubled/ halved horizontal scale.

8 (a) (x – 3)(x – 1) [= 0]

M1

4 ± √ [(-4)2 − 4.1.3] or (x - 22) = 1 or better 2

1 and 3 A1

(b) Correct first step of rearrangement M1 e.g. y + 1 = 2x or x + 1 = 2y or better

2

1+x o.e.

A1 not for x = ( )

(c) x2 – 6x + 4 = 0 MA1 Can be implied by later work (method marks) p ± √ q with p = 6 and r = 2

r

M1√ f.t. if in the form ax² + bx + c (= 0) with a ≠ 0 [ (x-3)² - 5 = 0 M1 then x = (±)√5 + 3 M1 is the equivalent for completing the square.]

and q = (-6)² – 4.1.4 o.e. or 20

M1√

Indep.

5.24 c.a.o. www

A1

SC1 for both answers ‘correct’ but not to 2 dp

0.76 c.a.o. www ( 5.236067977 , 0.763932022 ). Can be truncated or correctly rounded

(d) 29 B2 SC1 for [ f(-2) =] 15 seen or 2x² -8x +5 o.e seen

(e) ( 2x − 1)² – 4( 2x − 1) + 3

M1

4x² – 12x + 8 or correctly factorised final answer

A2 14

After A0, SC1 for 4x² - 12x + 8 seen



IGCSE – JUNE 2005 0580/0581 4

9 (a) x + y ≤ 12 o.e B1

x + y < 13

(b) y ≥ 4 o.e. B1

y > 3

(c) Scales correct – full length S1 (d) x + y = 12 ruled and long enough L1 or broken line x + y = 13 y = 4 ruled and long enough L1 or broken line y = 3. F.t from x ≥ 4 only in (b) 5x + 3y = 45 ruled and long enough B2 SC1 for either point correct (1 mm at (9, 0) and (0, 15) if extended)

Unwanted regions shaded B2√

SC1 for wanted regions shaded f.t. from minor slips in the lines that do not compromise the shape and position of the triangle or from x ≥ 4 in (b) and x = 4 drawn

(e) 6 super, 5 mini and 5 super, 7 mini (no extras)

B3

SC2 for 1 correct and no more than 1 wrong SC1 for any point(s) in their region selected

Can write as (6, 5) and (5, 7) (enclosed by 3 lines or 2 lines + 1 axis)

(f)(i) (7, 4) or (6, 5) s.o.i. M1 ($) 274 A1 ($) 260 A1 If 0 scored, SC1 for evidence of 30x + 16y

written or used

(f)(ii) ($) 94 c.a.o. B1 16



IGCSE – May/June 2006 0580 and 0581 02

1 4.496 x 109 1

2 97 cao 1

3 (a) (–)590 1

(b) Neptune 1

4 1.73 2* Allow √3 M1 for 1.15.... or 0.666....

5 21.3 2* M1 ½ x 8 x 12 x sin26.4 oe

6 )1(

5

+

+

xx

x 2* M1 5(x + 1) – 4x or better

7 20 2* M1 2.5 Q 0.125 oe

8 1/√2, sin 47, ¾, π /4 2* M1 for correct conversion to decimals 0.78(53..) 0.70(71...) 0.75 0.73(13..)

9 75000 76200 2* B1 B1 or M1 6250, 6350 seen

10 (a) 4 + 1½n 2* B1 for 4 B1 for 1½n o.e

(b) 154 1 f.t

11 (a) 13 cao 1

(b) –4 2* M1 3x/4 + 3 = 0 or x = x/4 – 3 or 4(x + 3) = x or 1 + 3/x =1/4 or better WWW

12 x=10 y=3 3* M1 Multiplying and subtracting consistently or M1 rearrange and substitute

13 x = –5.2 3* M1 any two steps completed correctly M1 any other two steps completed correctly

14 (a) 55, 40 2 B1 B1

(b) 25

16 1

15 (a) 500 + 170x 1

(b) 11 2* M1 their part (a) = 2370

16 (a) 6000 2* M1 7200 Q=1.2 oe

(b) 12.5 2* M1 (8100 – 7200) Q=7200 oe




17 (a) 2 B1 numbers B1 labels

(b)

2 B1 numbers B1 labels

Allow 0 in an intersection of A and B

18 w = 30 x = 22 y = 30 z = 52

1,1 1 f.t. 1 f.t.

y = w w + 22

19 (a) (2x – 3)(2x + 3) 1

(b) x(4x – 9) 1

(c) (4x – 1)(x – 2) 2

20 (a)

(b)

m = –1 c = 8 1,1

2*

21 (a)

1

1

or

(b) plane of symmetry 1

(c) 3 1




22 (a) p = 7.2 q = 6.4 2,2*

(b) 2304π 2* M1 for x by vsf 64 allow 7240 for 2 marks

23 (a) a + b, a – b, 3a + b

1½a + ½b

1,1,2*

1 f.t.

M1 in (iii) for (i) + a + (ii) + b

½ TP

(b) 4 1

TOTAL 70




1 (a) (i) 850 ÷ 80

10.625 (hrs) Must be exact

M1

A1

(ii) 10 hours 37 mins 30 secs B1

(b) (i) (0)6 08 (a.m.) B1

(ii) 850 ÷ 10 hrs 48 mins M1

78.7 (km/hr) (78.7037037) A1

(c) (i) Increasing (more slowly) B1 Accept speed going from 15 to 20.

(ii) Decreasing B1 Accept accel. going from 12.5 to 0

(iii)

18.1

515

−

−

M1

12.5 (m/s2) A1

(iv) 20 x 7 or 203

2

1××

M1 Alt Meth. 20 x 10 or

2

1 x 3 x 20

Second area and addition s.o.i. dep M1 Sec. area and correct subtraction

170 (m) A1

(v) Areas above and below broken line are approx. equal o.e.

B1

(vi) (their 1 7 0 ÷ 10) x 3.6 o.e. M1

61.2 (km/hr) A1 16

2 (a) Arc length

4

24×

=

π

(18.8…) M1

Perimeter = 6 + 22 + 18 + 10 + their arc M1

74.8 to 74.9 (cm) A1

(b) Sector area

4

122

×

=

π

(113. …) M1

Area = (6 x 22) + (12 x 10) + their sector

o.e.

M1

365 to 365.2 (cm2) A1

(c) 14600 to 14605 (cm3) B1

(d) their (b) x 2

their (a) x 40

Addition

M1

M1

M1

indep.

indep.

dep.

3720 to 3730 (cm2) A1 11




3 (a) (i) 1 B1

(ii) –1 B1

(iii)

2

3 or

2

11 or 1.5

B1

(b) (i) (r =) 0.25

(s =) 1

(t =) 8

B1

B1

B1

These must be seen. No feedback from the graph.

(ii) Scales correct S1 x from –2 to 3

y to accommodate their values.

(iii) Their 9 points plotted correctly. They must be in correct square and within 1 mm.

P3

ft P2 for 7 or 8 points correct.

Ft P1 for 5 or 6 points correct.

Smooth curve through all 9 points (1 mm) C1 ft provided correct shape maintained.

(c) (i) Correct ruled straight line of full length. B2 SC1 for complete freehand line or for short correct ruled line crossing the curve and y-axis.

(ii) 1.52 to 1.57 (correct for their graph) B1 Spoilt if y coordinate also given.

(iii) 1 B1 15

4 (a) Circle radius 5 cm (± 2 mm)

Circle radius 2 cm (± 2 mm)

AB is perpendicular to CD (± 1°)

B1

B1

B1

Lines parallel to roads at 0.5 cm from them (all 4 pairs) (Within 1 mm)

B1

(b) (i) Accurate (± 1°) angle bisector with arcs B2

(ii) T correct (± 1 mm) and labelled T1

(c) Accurate (± 1° and ± 1 mm)

perpendicular bisector of TB (using their T)

P correct (2.9 to 3.1 cm from 0) and labelled

B2

B1

Ft SC1 if ± 2° and ± 2 mm

(d) Their TP measured with km (±0.1 km) B1 11




5 (a) 2

1

xy ∝ or

2x

ky = o.e

M1

k = 4.8 x 52 A1

(b) 30 B1

(c) 10x2 = 120 o.e. M1

3.46 (3.464101…..) A1

(d) x2 x x = 120 o.e. M1

4.93 (4.932424…..) A1

(e) Divided by 4 o.e. B2 SC1 for (2x)2y = 120 o.e. seen or a correct calculation using a value of x. e.g. x = 4, y = 7.5

x = 8, y = 1.875

(f) Increases by 25% o.e. B2 SC1 for 1.5625 seen

(g) Division by y

Square root

M1

M1

13

6 (a) (AC =) √(82 + 62) M1

(PE =) √(132 – [

2

1 their AC]2)

M1 dep.

12 A1

(b) ××× 86

3

1their PE

M1

192 (cm3) A1

(c)

13sin

PEPCA

their= (0.92307...)

M1

13

CEtheirCos =

CE

PE

their

theirTan =

a.r.t. 67.4° (67.380....) A1

(d)

4tan

PEPME

their= (71.6°)

M1

PEMPE

their

4tan = (18.4°)

180 – 2 x anglePME dep

36.8° to 36.9°

M1

A1

2 x angleMPE

(e) (i)

13

3cos =PBC

M1

76.7° (76.6576…) A1

(ii) (KC2 = ) 42 + 62 – 2 x 4 x 6 cos(theirPBC) M1

Square root of correct combination M1 dep on first M1.

√40.957... or 6.3998

6.40 (cm) A1 15




7 (a) (i) (5 , 3) B1

(ii) (3, 5) 1+1 ft from (a)(i)

(b)

01

10

B2 SC1 for a correct column

(c) M(Q) = (k – 3 , k – 2) seen M1 SC2 if a numerical value of k is

TM(Q) = (k – 3 + 3 , k – 2 + 2) seen M1 chosen and full working leads to (k , k)

= (k , k) so y = x E1 (k , k)

(d)

01

10

B2 SC1 for determinant = –1 or for “self-inverse”

(e) (i)

− 01

10

B2 SC1 for 3 correct numbers.

(ii) Rotation B1

Centre (0 , 0) B1

270° or clockwise 90° B1 15

8 (a) (i) (x2 – 40) + (x + 2) + (2x + 4) + x = 62 o.e. M1

x2 + 4x–96 = 0 o.e. A1

(ii) (x + 12)(x – 8) (=0) M1 ( )2

96.1.4442 −−√±−

or better

x = –12 and 8 c.a.o. A1

(iii) 8 B1

(iv) 0.5 [(2 x their 8 + 4) + (their 82 –40)] x their 8

M1 Accept 0.5[2x + 4 + x2 – 40] x x

176 c.a.o. A1

(b) (i) (2y – 1)2 = y2 + (y + 2)2 o.e. M1

4y2 – 4y + 1 = y2 + y2 + 4y + 4 o.e. M1 dep

2y2 – 8y – 3 = 0 E1 No error at any stage. =0 essential

(ii)

r

qp √± where p = –(–8) and r = 2 x 2 o.e

M1

and q = (–8)2 – 4.2. – 3 o.e M1

4.35 c.a.o. A1

–0.35 c.a.o. A1

(iii) 13.8 c.a.o. (13.81125) B2 SC1 for

( )2

2+yy seen 16




9 (a) (i) 1 B1

(ii) 3 B1

(iii) 6.3

10

29=

++ mktheir o.e.

M1

(m =) 4 A1

(iv) 9 B1

(b) (i) mid-values 10, 25, 32.5, 37.5, 45, 55, 70 seen

M1 At least 6 correct s.o.i.

(10 x 10) + (10 x 25) + (15 x 32.5) + (28 x 37.5) + (22 x 45) + (7 x 55) + (8 x 70)

[3822.5]

M1* Dep on first M1 or mid-values ±0.5 Allow 1 more slip.

Total ÷ 100 M1 Dep on second M1*

38.2 (38.225) A1

(ii)

99

14

100

15×

M1

9900

210 o.e.

A1

330

7 Final Answer

A1

(c) (i) p = 20 B1

q = 72 B1

(ii) Horizontal scale correct S1 Implied by correct use. Ignore the vertical scale.

For each block of correct width For scale error (halved), award

Height 3.3 cm H1 H1, H1, H1 for correct ft heights.

Height 12 cm H1

Height 2 cm H1 After H0, H0, H0, give SC1 for correct frequency densities written. (0.67, 2.4, 0.4) 18




1 (a) (i)

(ii)

2 400

520 000

B2

B2

SC1 for figures 24

SC1 for figures 52

(b) (i)

(ii)

1 : 5 000 000 or n = 5 000 000

Time = 2hrs 8 mins or 128 (mins)

= 2.13(33..) (hours) oe soi

1580 ÷ their time

738 – 742 cso

B2

B1

B1

M1

A1

SC1 for 5 000 000 seen in final answer

or n = figs 5 oe in final answer

Implies previous B1

Accept 60

128

soi is by correct answer

www 4 (12.3 seen earns B1M1)

[10]

2 (a)

Axes to correct scale

S1

Accept 2mm accuracy throughout

(b)

Correct triangle A(2,1)B(3,3)C(5,1)

B1

Condone absence of labels

(c)

A1(1,2), C1(1,5), B1(3,3)

ft their ABC

B2

B1 for 2 correct points

Condone absence of labels and sides but

not incorrect suffices

(d)

A2(–2,1), C2(–5,1), B2(–3,3)

ft their A1B1C1

B2

B1 for 2 correct points

Condone absence of labels and sides but

not incorrect suffices

SC1 for rotation of their A1B1C1 90°

clockwise about the origin

If triangle ABC is rotated correctly treat

as mis-read

(e)

Reflection

y-axis oe cso

B1

B1

Indep (Only possible answer)

(f)

(i)

(ii)

(iii)

A3(2, –1), C3(5, –4), B3(3,0)

Shear, y-axis invariant oe

1

0

1

1

B3

B1,B1

B2

B2 for 2 correct points plotted

Condone absence of labels and sides

If B0, M1 for any set up of matrix

multiplication seen for at least one point

and A1 for correct result

(If correct triangle A2B2C2 used treat as

MR, and the co-ords are (–2, 3), (–5, 6),

(–3, 6))

Allow factor of either +1 or –1 if

invariant line omitted, but dependent on

shear or stretch

B1 for the left hand column

[15]




3 (a) (i)

(ii)

(iii)

(iv)

(v)

(vi)

0.5×40.3×26.8sin92 oe

539.6 – 540

oe 55sin

3.40

92sin=

AB

( )55sin

92sin3.40 ×

=AB

49.2 or 49.16 – 49.18

55

Angles in the same segment oe

33 correct or ft

Similar or enlarged

8.26

1.20

3.40=

XD oe

30.2(25)

M1

A1

M1

M1

A1

B1

B1dep

B1

B1

M1

A1

Any other method must be complete

(s = 58.13 – 58.15)

ww scores zero

( )92cos8.263.402

8.263.40222

××−

+=AB M1

(AB = ) square root of above and a

correct combination M1 (dep)

Accept if found in (i)

ww scores zero

ft 88 – their 55, if answer is positive

)(sin

1.20

)(sin ivtheiriiitheir

XD=

30.2(309…) cao


ww scores zero

(b) (i)

(ii)

(iii)

12

1

2 −

+=

+ y

y

y

y oe

y(2y – 1) = (y + 1)(y + 2)

2y2 – y = y2 + y + 2y + 2

y2 – 4y – 2 = 0

2

8164 +±

-0.45, 4.45 cao

7.9(0) or better 7.8989.. ft

M1

M1

E1

B1,B1

B1,B1

B1ft

May be implied by next line

Accept correct ratio statement

May be implied by next line

Implies previous M2

Dep (no errors in any line)

If M0, SC1 for

y(2y – 1) – (y + 1)(y + 2) =

2y2 – y – y2 – y – 2y – 2 =

y2 – 4y – 2

If of form r

qorp )( −+

B1for 4 and 2, B1 for 42-4(1)(-2)

If of form r

qorp )( −+

B1 for 42-4(1)(-2) but may recover the

other B1 from answers

SC1 for rounding or truncating to 1 dp

or more – 0.44948…, 4.44948…

ww scores max of 2

ft 2 × a positive root -1

[19]




4 (a) (i)

(ii)

3

–4.25 to –4

B1

B1

(b) (i)

(ii)

–1.6, 2.0, 8.6 to 8.63

9.2

B2

B1

B1 for any one correct

(c)

–9, 3

B1,B1

–1 each extra incorrect value

(d)

0<x<6, ( i.e.0 to 6 only) oe

B2

Accept (0,6), [0,6], (0, 3) to (6, –9).

SC1 for other inequality errors or

answers using 0 and 6 as boundaries

(e)

(i)

(ii)

1 – x oe

3

B1

B1

If re-arranged it must be correct

equation with y or f(x) in it but exclude

f(x) + x – 1 = 0

[11]




5

(a)

Using a right-angled triangle with

25 and 7

252 - 72 oe (or 502 – 142)

(BD) = 48 (or 24 × 2)

M1

M1

E1

25 and 7 seen is sufficient (or 50, 14)

Must be a correct numerical calculation

oe includes trig methods, which can

round to 24, then 48 for the E mark

Dep on M2, correctly established

(b)

(i)

(ii)

225

7cos

1 ×

− oe

147° cao

air 32 -34 or ft

M1

A1

B1

If scale drawing seen then M0

www 2

147.47…. score M1 only

ft 180 – their 147

(c)

(i)

(ii)

q + p oe

q – p oe

B1

B1

(d)

CEOC + oe

e.g. their (q – p) + 2 × their (q + p)

p + 3q cao

M1

A1

any correct unsimplified expression

2q + their (c) (i)

www 2

(e)

OBOC2

1+ oe

0.5p + 2.5q cao

M1

A1

any correct unsimplified expression

2q + ½ their (c) (i)

www 2

(f)

(i)

(ii)

24

0

− 24

7

B1

B1

B1

Accept any reasonable notation in both

parts

(g)

50

B1

[16]




6

(a)

1.5 < x ≤ 2

B1

(b)

(8×0.25 + 27×0.75 + 45×1.25 +……..

……….. 3×3.75)

their 345.5 ÷ 200

1.7275, 1.727, 1.728 or 1.73 cso

M1

M1

M1

A1

For mid-values (allow two slips)

For Σfx (allow two slips) dep on first

M1, or mid-values ± 0.05

for ÷ 200 dep on second M1

www 4

(c)

8, 35, 80, 130, 169, 190, 197, 200

B2

If B0, allow M1 for clear attempt to add

accumulatively

(d)

axes correct scale

8 points plotted ft part (c)

(0.5, 8), (1, 35), (1.5, 80), (2, 130), (2.5,

169), (3, 190), (3.5, 197),

(4, 200)

curve (or polygon) either correct or

through 8 points and correct shape

S1

P3dep

C1

Not reversed and must reach 200

vertically, even if not labelled

dep on at least M1 in (c)

8 points from their values

For x-values (upper boundary values),

points must touch grid line For

y-values, even, must touch grid line,

odd must be inside square.

P2 for 6 or 7 points ft


Allow 1 mm tolerance

Ignore any bars drawn if they do not

compromise the points and graph

(e) (i)

(ii)

(iii)

1.65-1.75

1.5

23 – 29 integers only

B1

B1

B2

If B0 allow SC1 for non-integer in

correct range, or 172 – 177 seen (may

be written on graph)

(f)

54 – 56.5

B2

SC1 for figures 108 – 113 or 87 – 92

Accept if written on graph

www 2

[18]




7

(a)

1.2 × 0.3 × 3 oe

× 60 oe

64.8 cao

M1

M1dep

A1

(1.08) or 3 × 60 (180)

× 1.2 × 0.3 (0.36)

www 3

(b)

1.2 × 0.8 × 15 × 60 oe (= 864 seen)

Their 864 – their (a)

÷ their (a) × 100

1230 (%) or better (1233.3…) cao

M1

M1ind

M1dep

A1

Their (a) 3

8 × 5 oe seen

or their 864 ÷ their (a) × 100 (1333.3..)

subtract 100 (Dep on second M1)

www 4

(1330 or 1333.3…www M1M1M0)

(c)

πr2× figs13 = figs 2 oe

2 ÷ 0.0013

0013.0

2)( 2

×

=

π

r oe

22.1 or 22.12 – 22.14 cao

M1

M1ind

M1dep

A1

(implied by 1538.46…)

Dep on M2 (489.7..)

www 4 figs 221… imply first M1

(d)

0.8 + 1.2 + 0.8 = (2.8)

50.40 = area × 0.12 oe

Length × their perimeter = their area oe

150 cao

M1

M1ind

M1

A1

Accept 2.8 seen

Accept 420 seen

www 4

[15]

8

(a)

x

105

B1

Do not allow x = , but allow other letter

and condone presence of units

(b)

4

105

+x

B1

Do not allow x = , but allow other letter

and condone presence of units

(c)

8.0

4

105105=

+

−

xx

oe

105(x + 4) – 105x = 0.8x(x + 4) oe

0.8x2 + 3.2x – 420 = 0 oe

x2 + 4x – 525 = 0

M2

M1

E1

SC1 if ± signs between terms incorrect

or SC1 for their (a) – their (b) = 0.8 oe

if (a) and (b) are fractions with linear

denominators

Dep on M2 or SC1 and allow all over

x(x + 4) at this stage

Condone any sign error in any

expanding done first (this is taken into

account in the E mark)

Completed without any errors

dep on M3

(d) (i)

(ii)

(x + 25)(x – 21)

-25, 21

B2

B1

B1 for (x – 25)(x + 21)

ft - allow 25 and -21 from above only

(e)

46

B1 ft

ft 2 × a positive root + 4

(f)

210 ÷ ( their (e))

4.57 or better (4.565…) ft

M1

A1 ft

www 2, but 4.6 ww scores zero

[12]




9

(a)

Sketch of 4 by 4 diagram

B1

(b)

(i)

(ii)

25, 40

n2

(n + 1)2 oe

(n + 1)2 + n2 – 1 or 2n2 + 2n) or

2n(n + 1) oe

B1,B1

B1

B1

B2

Any one of these oe isw and if B0

allow

SC1 for their (n + 1)2 + their (n2) – 1 or

an expression containing 2n2, as the

highest order term, soi

(c) (i)

(ii)

(iii)

(iv)

43

2=++ gf

22223

3

2×+×+× gf oe

3

3224 =+ gf

3

2022 =+ gf

3

3224 =+ gf

3

4)(,2)( == gf oe cao

880 cao

B1

M1

E1

M1

A1A1

B1

ie for substituting 2

No errors Allow 10, 3

2 10., 10.7, …

for correctly setting up for elimination

of one variable

www 3 accept 3

6 for 2

[14]



IGCSE – May/June 2008 0580/0581 21

1 53 and 59 1, 1 independent of each other

2 18

11x 2

M1 18

5

18

10

18

6 xxx

−+ oe fractions with common denom. not

decimals

3 150 2 M1 10012

18×

4 (a) 2870

(b) (n + 3)2 + 1

1 1

cao

Allow n2 + 6n + 10, (n + 2 + 1)2 + 1, (n – 1 + 4)2 + 1 oe

5 $231.13 cao 2 M1 245 / 1.06 or 245 × 0.94(3...) Allow 231, 231.1, 231.13… for M1

6 601

598 401

399 701

698 2 M1 correct decimals seen

0.99501.... 0.9957(2…) 0.99500… First and third must be to at least 5sf Accept these decimals in answer space

7 (a) 1045.28 cao (b) 1000

1 1

Allow 1.0 × 103

8 9x2 2 B1 9 B1 x2 terms must be multiplied

9 y = 2

1 x+ 5 3 M1 (m=) 06

58

−

−

oe B1 (c=) 5

or

M1 A1 y – 8 = 2

1 ( x – 6 ) or y – 5 = 2

1 ( x – 0 )

Allow 3/6 for the 2

1

A1 y = x2

1+ 5 or 2y – x = 10 oe

10 r = 18 h = 42 cao www 3 M1 Length scale factor of 6 used or stated Al Al

11 (±) 7.94 3 M1 212 = (2x)2 + x2 – 2.2x.x.cos120 oe

M1 441 = 7x2

(a)

2

B1 P and S not intersecting. Two sets must be labelled Three intersecting circles will have P ∩ S empty.

12

(b) 4 1√ from the number of elements in the shaded area

13 2

123−<x or –23.5 3 M1 2 moves completed correctly

M1 2 more moves completed correctly

maigna

First variant



IGCSE – May/June 2008 0580/0581 21

14 5.5 cm 5.5 cm

2.5 cm

1 1 1

Line in correct place; bisects rectangle Line 2cm long in correct place

4

1 circles in correct place

Not freehand.

15

−

−

−

14

11

11

1 1 1

16 (1, 3) www 3 M1 consistent multiplication and subtraction/addition A1 A1 Allow x = 1 and y = 3 (1, k) or (k, 3) scores 2 marks ONLY if M1 is scored

17 20 4 B1 4

3

500

370=

+

+

x

x

oe fraction, decimal, percentage

M1 two moves completed correctly M1 two more correct moves completed

18 (a) –14 1

(b) 2x3 – 6x2

+ 12x – 9 2 M1 attempting to double f(x) and –1

(c)

2

1+x 2 M1 valid method

19 (a) (i) Triangle (–1, –2)(–1, –3)(–3, –2) 2 M1 for one correct vertex of the triangle drawn on the diagram

(ii) Reflection in y = –x 2

(b)

−

01

10

2

M1 for the word reflection A1 y = –x oe

Combined transformation must be fully correct to the final answer but –1 once for the detail (e.g. centre, angle, etc)

B1 each column or

M1 solving two pairs of sim. equations A1 all correct in answer space

20 (a) 12900 3 M1 (1602 or 1002 ) × π × 95/360

M1 subtracting the two areas above

(b) 23300 1√ (a) multiplied by 1.8

(c) (i) 2.33 × 1013 1√ (b) × 109

(ii) 1.55 × 1013 2√ M1 (c)(i) / 1.5

maigna

First variant



IGCSE – May/June 2008 0580/0581 21

21 (a) 11.3 5 B1 identifying angle FAC

M1 6002 + 8002 Al 1000 (for AC) M1 tanx = 200/their 1000

(or cosx = "1000"/"1020")

Alternative method via DF and AF

M1 "(2002 + 6002)" + 8002 Al 1020

M1 sinx/(sin90) = 200/"1020" oe cosine rule also possible

(b) 233 3 M1 tany = 800/600 oe siny, cosy

M1 an angle found in (b) + 180 written in working

maigna

First variant



IGCSE – May/June 2008 0580/0581 22

1 59 and 61 1, 1 independent of each other

2 18

13x 2

M1 18

7

18

14

18

6 xxx

−+ oe fractions with common denom. not

decimals

3 140 2 M1 10015

21×

4 (a) 1240

(b) (n + 4)2 + 1

1 1

cao

Allow n2 + 16n + 17, (n + 3 + 1)2 + 1, (n – 1 + 5)2 + 1 oe

5 $308.41 cao 2 M1 330 / 1.07 or 330 × 0.93(4579...) Allow M1 308, 308.4(1...)

6 601

598 401

399 701

698 2 M1 correct decimals seen

0.99501.... 0.9957(2…) 0.99500… First and third must be to at least 5sf Accept these decimals in answer space

7 (a) 2045.49 cao (b) 2000

1 1

Allow 2.0 × 103

8 8x3 2 B1 8 B1 x3 terms must be multiplied

9 y = 2

1 x+ 7 3 M1 (m=) 06

710

−

−

oe B1 (c=) 7

or

M1 A1 y – 10 = 2

1 ( x – 6 ) or y – 7 = 2

1 ( x – 0 )

Allow 3/6 for the 2

1

A1 y = x2

1+ 7 or 2y – x = 14 oe

10 r = 24 h = 36 cao www 3 M1 Length scale factor of 6 used or stated Al Al

11 (±) 7.21 3 M1 262 = (3x)2 + x2 – 2.3x.x.cos120 oe

M1 676 = 13x2

(a)

2

B1 P and S not intersecting. Two sets must be labelled Three intersecting circles will have P ∩ S empty.

12

(b) 4 1√ from the number of elements in the shaded area

13 2

123−<x or –23.5 3 M1 2 moves completed correctly


maigna

Second variant



IGCSE – May/June 2008 0580/0581 22

14 A

5.5 cm 5.5 cm

2.5 cm

1 1 1

Line A in correct place; bisects rectangle Line 2cm long in correct place

4

1 circles in correct place

Not freehand.

15

−

−

−

14

11

11

1 1 1

16 (1, 3) www 3 M1 consistent multiplication and subtraction/addition A1 A1 Allow x = 1 and y = 3 (1, k) or (k, 3) scores 2 marks ONLY if M1 is scored

17 20 4 B1 4

3

500

370=

+

+

x

x

oe fraction, decimal, percentage

M1 two moves completed correctly M1 two more correct moves completed

18 (a) –17 1

(b) 2x3 – 6x2

+ 12x – 17 2 M1 attempting to double f(x) and –3

(c)

2

3+x 2 M1 valid method

19 (a) Triangle (–1, –2)(–1, –3)(–3, –2) 2 M1 for one correct vertex of the triangle drawn on the diagram

Reflection in y = –x 2

(b)

−

01

10

2

M1 for the word reflection A1 y = –x oe

Combined transformation must be fully correct to the specified answer but –1 once for the details (e.g. centre, angle, etc)

B1 each column or

M1 solving two pairs of sim. equations A1 all correct in matrix

20 (a) 12900 3 M1 (1602 or 1002 ) × π × 95/360

M1 subtracting the two areas above

(b) 23300 1√ (a) multiplied by 1.8

(c) (i) 2.33 × 1013 1√ (b) × 109

(ii) 1.55 × 1013 2√ M1 (c)(i) / 1.5

maigna

Second variant



IGCSE – May/June 2008 0580/0581 22

21 (a) 11.3 5 B1 identifying angle FAC

M1 6002 + 8002 Al 1000 (for AC) M1 tanx = 200/their 1000

(or cosx = "1000"/"1020")

Alternative method via DF and AF

M1 "(2002 + 6002)" + 8002 Al 1020

M1 sinx/(sin90) = 200/"1020" oe cosine rule also possible

(b) 233 3 M1 tany = 800/600 oe siny, cosy

M1 an angle found in (b) + 180 written in working

maigna

Second variant



IGCSE – May/June 2008 0580, 0581 04

1 (a) (i) 250 B1

(ii) their (a)(i) ÷ 5× 52 o.e.

2600 ft www2

M1

A1 ft

SC1 for 12.5 ÷ 5× 52, implied by 130

(iii) 100

2450

2450×

−(a)(ii) their o.e.

6.1(22……..) ft www2

M1

A1ft

100100

2450

−×

(a)(ii) their,

x

150

100

2450=

ft M & A only if their (a)(ii) > 2450

(b) (i) 20 ÷ 5× 3

12 www2

M1

A1

Accept 12, 8 or 8, 12

(ii) their (b)(i) ÷ 3 and (20 – their (b)(i)) ÷ 2.5

7 hours 12 mins cao www2

M1

A1

4 and 3.2 or 7.2 or 7h 20 mins seen imply

M1

Condone poor notation e.g. 7-12

(iii) 2.78 (2.777–2.778) o.e. cao

o.e. in other units

B1 o.e. must have units stated e.g.

0.7716..m/s, 46.29 – 46.30 m/min

(iv) 16 07 o.e. ft B1 ft ft their (b)(ii) + 08 55 iff finishes on same

day and (b)(ii) has hours and mins

(c) 20× 100000 ÷ 80 o.e.

25 000 or 2.5× 104 www2

M1

A1

25 000 seen in final ans. After M0, SC1

for figs 25 or 0.00004 final answer [13]

2 (a) (i)

(ii)

(x + 4)(x – 5)

–4, 5 ft

B2

B1 ft

If B0, SC1 if of form (x ± 4)(x ± 5),

Only ft the SC

–4, and 5 not from (x – 4)(x + 5).

(b)

2.3

24.32)(2)( 2−−−±−−

B1,B1

B1 for (–2)2–4(3)(–2) (or better) seen

inside a square root.

The expression must be in the form

r

qp )(or−+

then B1 for p = –(–2) and

r = 2.3 or better

Allow recoveries from incomplete lines

–0.55, 1.22 cao B1,B1 If B0, SC1 for –0.5 and 1.2 or both

answers correct to 2 or more decimal

places (rounded or truncated).

–0.54858, 1.21525…

(c) (i) (m – 2n)(m + 2n) B1

(ii) –12 B1

(iii)

20x + 5 o.e. cao final ans

B2

B1 for (4x2 + 6x + 6x + 9) or

(x2 – x – x + 1) or

(2x + 3 – 2(x – 1))(2x + 3 + 2(x – 1))

(iv) 4n2 = m2 – y o.e.

4

2

2 ymn

−

= o.e.

4)(

2ym

n−

= o.e. www3

Mark final answer

M1

M1

M1

M1 for correct re-arrangement for n2 term

(may be –n2)

M1 for correct division by 4 or – 4

M1 for correctly taking square root of n²

term

SC2 for4

2my ±

or 4

2ym −

o.e. ww

(d) (i)

(ii)

4 or –4 or ±4

n(m4 – 16n4) or

(m2n – 4n3)(m2 + 4n2) or

(m2n + 4n3)(m2 – 4n2) or

n(m – 2n)(m + 2n)(m2 + 4n2)

B1

M1

A1

Correctly taking out n or a correct factor

with n still in one bracket

Must be final answer [17]



IGCSE – May/June 2008 0580, 0581 04

3 Accept all probability answers as

fractions (non-reduced or reduced),

decimals or percentages.

–1 once for 2 sf answers or correct

words.

Condone numerical errors in

simplifying or converting after correct

answers seen.

Ratio answers score zero throughout.

(a) (i)

3

1, 8

3, 8

6, 8

2 o.e.

B3 –1 each error bod if no letters given

(ii)

8

5

3

2×

12

5 o.e. www2

M1

A1

24

10, etc., 0.416(6…)

(iii) their

8

6

3

1

12

5×+

3

2 o.e. cao www2

M1

A1

24

16, 12

8, etc., 0.666(6….)

(b) (i)

8

1

9

2

10

3××

120

1 o.e. www2

M1

A1

720

6, etc., 0.00833(3…)

(ii)

120

119 o.e.

B1ft

720

714, etc., 0.991(6…) ft 1 – their (i) not

for 7/10

Could start again and have a correct

answer independently [10]

4 (a) (i) 36 (36.0–36.4) B1

(ii) 50 (50.0–50.4) B1

(iii) 29 (28.6–29.4) B1

(iv) 20 B2 If B0, SC1 for 19 or 21 or 180 seen

(b) (i) p = 16, q = 4 B1,B1 If B0, SC1 if p and q add up to 20

(ii)

200

7220 = 36.1 cso www4

B4

Answer 36 scores 4 marks after some

correct working shown with no incorrect

working seen

M1 for using mid-values at least four

correct from 5, 15, 25, 35, 45, 55, 65, 75

M1 (dep on correct mid values or mid-

values ±0.5) for ∑ fx (at least four

correct products)

M1 (dependent on 2nd M1) for dividing

sum by 200 or 180 + their p + their q

(c) 8.2 (8.19–8.20), 11.4, 5 (5.00–5.01) B4 B3 for 2 correct

or B2 for 1 correct

After B0, SC2 for fd’s 2.7(3…) o.e.,

3.8 o.e, 1.6(6...) o.e.

or SC1 for 2 of fd’s correct (15)

5 (a) (i) 360 ÷ 8 or (8 – 2)× 180 M1 allow 6× 180



IGCSE – May/June 2008 0580, 0581 04

180 – their (360 ÷ 8) o.e. ÷ 8 M1 dependent

(ii) 45° used or use implied o.e. E1 Accept sketch with values

(b) (i)

12

l = cos45 o.e.

(PH = ) 8.49 (8.485….) www2

M1

A1

For o.e. allow implicit expression

Accept 72 , 182 , 83 , 26

(ii) (PQ =) 2× their PH + 12 o.e.

(PQ =) 29.(0) (28.96–29.00) ft www2

M1

A1 ft

ft their PH accept surd form

(iii) their PH× their PH÷2 o.e.

(Area APH =) 36 (35.95–36.1) ft www2

M1

A1 ft

ft their PH

(iv) (their PQ)2 – 4 × their area of triangle o.e.

(Area octagon = ) 695 (694.0–697.1) cao

www3

M2

A1

If M0, M1 for a clear collection of areas

leading to the octagon possibly without

any calculation shown

(c) (i) 0.5 of their PQ o.e.

14.5 (14.47–14.53) cao www2

M1

A1

e.g. 6 + PH, 6tan67.5°

accept surd form

(ii) 2( r) their×π

100×

area octagon their

area circle their

94.8 (94.35 to 95.60) cao www3

M1

M1

A1

(660.5…)

Dependent on first M1 and circle smaller

than the octagon

[17]

6 (a) (i)

1

2

B1

Allow (2 1), condone omission of

brackets

(ii)

1

2 ft

B1ft

Allow (2 1), condone omission of

brackets

ft their (i) if a vector

(b) Translation

− 4

0 o.e.

B1, B1

Allow (0 –4), condone omission of

brackets, allow in words

Any extra transformation spoils both

marks

(c) y > 0 o.e.

x < 2 o.e.

y > 2

1x o.e.

y < 2x + 4 o.e.

B1

B1

B1

B2

For all four, condone strict inequalities

and only penalise first incorrect sign,

which may be = or an inequality sign

If B0, B1 for 2x or for 4 if other

co-efficient is not zero

y < 2

1x + 4 gets zero [9]

7 (a) (i) cyclic B1 Condone concyclic

(ii) Any one of 40, 45, 50

Any one of 20, 25, 30

Any one of 105, 110, 115

B1

B1

B1

Angle BCT = 40° is inconsistent with ST

parallel to OB. So different values of

angles x, y, z, OCT and AOC can be

arrived at, depending on route taken.

(iii) Any one of 80, 85, 90 B1

(iv) Any one of 210, 215, 220, 225, 230 B1

(b) (i) Similar (or enlargement) B1

(ii) 2

10

7or

2

7

10 o.e. seen

9.8 (9.79 to 9.81) www2

M1

A1

(0.49), (2.04)

It is possible to do (iii) then (ii) and full

marks can still be scored

(iii)

2

1×10×height = 20

4 www2

M1

A1

[11]



IGCSE – May/June 2008 0580, 0581 04

8 (a)

108(.16) (allow 108.2(0)) www2

M1

A1

M1 for method of compound interest used

(b) 148(.02…) 324(.3…) B1 B1

(c) Correct axes full domains

5 correct pts 100, 148 ft, 219, 324ft, 480

Smooth exponential curve, correct shape

through 5 points

S1

P3ft

C1

Condone absence of labels

P2ft for 4 correct, P1ft for 3 correct

Points must be in correct square vertically,

including on line

Scale error – remove that part and try to

mark the rest

(d) (i) 265 – 270 B1ft If out of range, then ft their graph at 25

years

(ii) 17 or 18 cao B1

(e) (i)

(100)

207(100) ××

o.e.

100 + 7 × 20 or better

M1

E1

No errors

(ii) 380 B1

(iii) Correct straight ruled line for x – range 0 to

35

L2 P1ft for 2 of (0,100), (20,240) (40,380)ft

correctly plotted

(f) 27 – 29 cao B1 [17]

9 (a) (i) p + r B1 Answers in bracketed column form

penalise only once throughout

(ii) –p + r B1

(iii) –p +

3

2r

B1

(iv) p +

2

1r

B1

(b) (i)

2

3× (–p +

3

2r) or –

2

3p + r isw after

correct answer seen

B1 ft ft only

2

3× their (a)(iii)

(ii) PSQP + o.e.

–2

3p www 2

M1

A1 ft

o.e. is any correct route of at least 2

vectors

ft their (b)(i) – r

(c) lie on a straight line B1 dependent on their (b)(ii) being a multiple

of p [8]

10(a) (i) 4 B1

(ii) 24 B1

(b) (i) x + 12, x + 14 o.e. B1,B1 Any order ignore ref to g and i

(ii) (x + 14 – x) and (x + 12 – (x + 2))

14 – 10 or 14 – 12 + 2 or 4

E1

x + 12 and x + 14 must be seen to be used

No errors seen

(iii) (x + 2)(x + 12) – x(x + 14)

24

B1

E1

Subtraction can be implied later

Dep on B1 and no errors anywhere for the

E mark

(c) (i) 4 B1

(ii) 20 B1

(d) (i) 4 B1

(ii) x + 2n o.e., x + 2+ 2n o.e. B1,B1

(iii) 4n B1 Allow 4×n, n×4, n4 [13]


Mark Scheme: Teachers’ version Syllabus Paper

IGCSE – May/June 2009 0580, 0581 21

Abbreviations

cao correct answer only ft follow through after an error oe or equivalent SC Special Case www without wrong working

1 (a)

(b)

2

1 1

Any length, can be freehand lines solid or dotted Mark lost if additional lines drawn or axes extended

2

7

5 72%

17

9

1

3

4−

2 M1 correct decimals 0.727(6… ) 0.71(4… ) 0.72 0.75

3 (a)

(b)

06 41 $204

1 1

Allow 6.41(am). 6:41 and 06:41 Not 6h41m or 641h or 6.41pm

4

1, 1

5

−

−

24

35

2

1 or

−

−

12

5.15.2

2 M1 det A or |A| or 5×–2 – 4×–3 = 2 or

−

−

24

35 or

dc

ba

2

1 seen

Allow 5/2, –3/2, 4/2, –2/2 in matrix

6 62225000 or 6.2225 × 107 or 62.225 million cao

2 M1 9.5(million) and 6.55 seen 3sf not appropriate for UB and not allowed for 2 marks

7 (4, 2) 2 M1

2

62 + and

2

95 +−

oe

or a drawing used correctly

maigna

First Variant Mark Scheme



IGCSE – May/June 2009 0580, 0581 21

8 (a) 2a – g cao 1 –g + 2a

(b) ga2

1

2

12 + oe cao 1 Allow 2.5 or 2

5 and 0.5

9 (9(1 – x))2 oe

3 M1 1 move completed correctly M1 1 more move completed correctly Mark 3rd move in answer space

10

c

2

3 M1 d + c – c + d or better M1 common denominator cd used

11 £3000 3 M1 1.96 × 25000 M1 “49000” / 1.75

12 x = 4 y = –3 3

M1 consistent multiplication and subtraction of their rearranged eqns. Any other answers must first score M1 to gain an A mark Substitution, matrix and equating methods also permitted

13 0.128 3 M1 t = k/d 2 k is any letter except t, d or α A1 k = 12.8

or M1 0.2 × 82 = 12.8

14 (a)

(b)

3 × 1011

5 000 000 or 5 × 106 or 5 million

2 2

M1 60 × 5 × 109 or better

M1 0.8 × 107 – 3 × 106 oe

or M1 5x = 4 × 107 – 15 × 106 oe If m is used for a million it must be used consistently

15 (a)

(b)

24.7 11.5

2 2

M1 sin18 = AB/80 or cos72 = AB/80 Allow AB/sin18 = 80/sin90 M1 tan25 = h/(a) or h/sin25 = (a)/sin65

16 Angle bisector of angle in the middle Second angle bisector drawn

4 W1 correct bisector drawn W1 at least two arcs drawn on the arms and one pair of correct crossing arcs W1 as above

W1 as above Accuracy ±1° but line must go from edge to edge.

maigna




IGCSE – May/June 2009 0580, 0581 21

17 (a)

(b)

Reflection in y = x Triangle at (4,6), (4, 7), (7, 7)

2 2

M1 Reflection A1 correct description of the line M1 Rotation 90° clockwise A1 position

18 (a)

(b)

320 567

2 2

M1 1080 × 8/27 or (2/3)3 or 1080 ÷ 27/8 or (3/2)3

M1 252 × 9/4 or (3/2)2 or 252 ÷ 4/9 or (2/3)2

19 314 4 M1 π. 182 . 40/360 or OAD = 113 identified

M1 π. 62 (or π. 62 . 40/360) or OBC …”…….

M1 2 × (OAD – OBC) + circle oe OR

M1 π. 182 . 40/360 M1 π. 62 . 140/360

M1 2 × OAD + 2 × BOE oe

20 draw 2x – y = 4 draw x + y = 6 draw y = 4 correct region identified by R

2 1 1 1

W1 Line through (2,0) or (0,–4)

21 (a)

++

15

63

14

122 xx

2 M1 for any correct row or column

Allow 2(x + 6), 3(x + 2)

(b) 5 3 M1

+

+

15

21

42

122

x

x

one row (or column) correct

M1 2x + 4 = 14 or 3x + 6 = 21

22 (a)

(b)

(c)

(d)

58 32 58 24

1 1 1 ft 2

= (a)

R

6 0

maigna




IGCSE – May/June 2009 0580, 0581 22

Abbreviations


1 (a)

(b)

2

1 1

Any length, can be freehand lines solid or dotted Mark lost if additional lines drawn or axes extended

2

25

18

15

8 74%

1

20

27−

2 M1 correct decimals 0.74 0.730(2… ) 0.72 0.740(7…)

3 (a)

(b)

06 43 $247

1 1

Allow 6.43(am) Not 6h43m or 643h or 6.43pm

4

1, 1

5

−

−

6

7

10

1

4

3 oe

2 M1 det A or |A| or –6×3 – 7×–4 = 10 or

−

−

6

7

4

3 or

dc

ba

10

1 seen

6 62225000 or 6.2225 × 107 or 62.225 million cao

2 M1 9.5(million) and 6.55 seen 3sf not appropriate for UB and not allowed for 2 marks

7 (6, 3) 2 M1

2

84 + and

2

137 +−

oe

or a drawing used correctly

maigna

Second Variant Mark Scheme



IGCSE – May/June 2009 0580, 0581 22

8 (a) 2a – g cao 1 –g + 2a

(b) ga2

1

2

12 + oe cao 1 Allow 2.5 or 2

5 and 0.5

9 (8(1 – x))2 oe

3 M1 1 move completed correctly M1 1 more move completed correctly Mark 3rd move in answer space

10

c

2

3 M1 d + c – c + d or better M1 common denominator cd used

11 £2400 3 M1 3.92 × 20000 M1 “78400” / 3.50

12 x = 5 y = –2 3

M1 consistent multiplication and subtraction of their rearranged eqns. Any other answers must first score M1 to gain an A mark Substitution, matrix and equating methods also permitted

13 0.625 or

8

5

3 M1 t = k/d2 or td2 = k or M1 0.4 × 52 = 10 A1 k = 10 k is any letter except t, d or α

14 (a)

(b)

4.8 × 1011

5 000 000 or 5 × 106 or 5 million

2 2

M1 60 × 8 × 109 or better

M1 0.8 × 107 – 3 × 106 oe

or M1 5x = 4 × 107 – 15 × 106 oe If m is used for a million it must be used consistently

15 (a)

(b)

24.7 11.5

2 2

M1 sin18 = AB/80 or cos72 = AB/80 Allow AB/sin18 = 80/sin90 M1 tan25 = h/(a) or h/sin25 = (a)/sin65

16 Angle bisector of angle in the middle Second angle bisector drawn

2 2

W1 correct bisector drawn W1 at least two arcs drawn on the arms and one pair of correct crossing arcs W1 as above

W1 as above Accuracy ±1° but line must go from edge to edge.

maigna




IGCSE – May/June 2009 0580, 0581 22

17 (a)

(b)

Reflection in y = x Triangle at (4,6), (4, 7), (7, 7)

2 2

M1 Reflection A1 correct description of the line M1 Rotation 90° clockwise A1 position

18 (a)

(b)

320 567

2 2

M1 1080 × 8/27 or (2/3)3 or 1080 ÷ 27/8 or (3/2)3

M1 252 × 9/4 or (3/2)2 or 252 ÷ 4/9 or (2/3)2

19 314 4 M1 π. 182 . 40/360 or OAD = 113 identified

M1 π. 62 (or π. 62 . 40/360) or OBC …”…….

M1 2 × (OAD – OBC) + circle oe OR

M1 π. 182 . 40/360 (=113.10) M1 π. 62 . 140/360 (=43.98)

M1 2 × OAD + 2 × BOE oe

20 (a)

(b)

draw 2x – y = 4 draw x + y = 6 draw y = 4 correct region identified by R

2 1 1 1

W1 Line through (2,0) or (0,-4)

21 (a)

++

15

63

14

122 xx

2 M1 for any correct row or column

Allow 2(x + 6), 3(x + 2)

(b) 5 3 M1

+

+

15

21

42

122

x

x

one row (or column) correct

M1 2x + 4 = 14 or 3x + 6 = 21

22 (a)

(b)

(c)

(d)

58 32 58 24

1 1 1 ft 2

= (a)

R

6 0

maigna




IGCSE – May/June 2009 0580, 0581 04

Abbreviations

cao correct answer only cso correct solution only dep dependent ft follow through after error isw ignore subsequent working oe or equivalent SC Special Case www without wrong working

1 (a) ($) 450 B2 M1 for 650 ÷ (9 + 4) × 9

(÷ 14 does not imply 9 + 4)

(b) (i) ($) 120 B2 M1 for 0.8 × 150 o.e.

(ii) ($) 80 ft B2 ft M1 for (150 – their(b)(i)) ÷ 0.375 o.e. only if +ve. After M0, SC1 for answer 320

(c) (i) ($) 441 B2 M1 for 400 × 1.052 o.e. or for answer 41

(ii)

100400)400(their 2

1×÷−(i) o.e.

5.125 or 5.13 or 5.12 c.a.o. www3

M2

A1

If use Simple Int in (i), M0, M0 in this

part

i.e. a full explicit method for r

If M0,

M1 for 400their100

2400−=

××

(i)r

or their (i) ÷ 400 × 100 then – 100

or 100400

400- their ×

(i) (s.o.i. by 10.25)

If still M0, SC1 for answers 55.125 or 55.12 or 55.13 or 55.1 or 0.05125 or 0.0512 or 0.0513 [11]

2 (a) 1 B1

(b) 2.5 o.e. B1

(c) 2.96 c.a.o. B2 If B0, M1 for 15 × 1 + 10 × 2 + 7 × 3 + 5 × 4 + 6 × 5 + 7 × 6 (allow one slip) implied by 148 seen Ignore subsequent rounding

(d)

60 × 2.95 (= 177) their 177 – their 148 (or 50 × their 2.96) (Mean of new rolls =) 2.9 c.a.o. www3

M1

M1

A1

Dependent on first M and only if positive or M1 for

95.260

)10or ()96.2their 50(148their=

+× xx

then M1 for 148their 95.260)10or ( −×=xx

(or 50 × their 2.96) and only if positive [7]



IGCSE – May/June 2009 0580, 0581 04

3 (a) (sin P) = 14105.0

48

××

o.e. fraction

P = 43.29…. cao

M2

A1

M1 for 0.5 × 10 × 14 sin P = 48 o.e. Allow 0.5 × 10 × 14 sin 43.3 = 48 for M1

but no further credit

(b) 102 + 142 – 2 × 10 × 14cos 43.3 (= 92.2) Evaluating square root (QR =) 9.6(0) (9.60 to 9.603…) c.a.o. ww2

M2

M1

A1

If M0, M1 for correct implicit statement M1 (dependent on M2) for square root of correct combination (not negative) i.e 16cos43.3 (11.64..) implies M2M0 [7]

4 (a) 23sin126sin

250)( ×=AB (s.o.i by 120…)

121 (120.7 to 121) (m) c.a.o. www3

M2

A1

M1 for 126sin

250

23sin=

AB o.e. (implicit)

(b) (i) 280 B1

(ii) (0)69 c.a.o. B2 SC1 for answer 249 [6]

5 (a) (i) 1.5, 3.75, –1.5 B1,B1,B1

(ii) 12 points plotted ft Curve through at least 10 points and correct shape over full domain Two separate branches, one on each side of y-axis, neither in contact with y-axis

P3 ft

C1

B1

P2 ft for 10 or 11 points, P1 ft for 8 or 9 points i.s.w. if two branches joined Independent

(b) –1.4 ≤ x ≤ –1.1 and 3.1 ≤ x ≤ 3.4 B1,B1 i.s.w. 3rd answer if curve cuts y = 1 again

(c) (i) Correct ruled tangent at x = 2 or x = –2 Evidence of rise/run 0.8 to 1.2

M1

M1

A1

Long enough to be able to find gradient Dependent – check their graph against gradient of 1 – must be correct side of 1 No tangent drawn M0M0

(ii) 0.8 to 1.2 inc. or same answer as (i) ft B1 ft

(d) (i) Correct ruled line to cut curve for all possible intersections (at least 2)

B1 Within ½ square of (–1, 1) and (1, –1)

(ii) –1.3 to –1.05, 1.05 to 1.3 inclusive B1, B1 i.s.w. any extra answers

(e) y = kx with k 2

1≥ o.e. or x = 0

B2

If B0, allow SC1 for y = kx with k <

2

1 or

for y-axis stated [19]



IGCSE – May/June 2009 0580, 0581 04

6 (a) (i) 0.5 [(x + 6) + (x + 2)] × (x +1) (= 40) or better 0.5(2x + 8)(x + 1) (= 40) o.e. 0.5(2x2 +10x + 8) (= 40) o.e. x2 + 5x + 4 = 40 o.e. x2 + 5x – 36 = 0

M1A1

E1

M1 for any algebraic use of half base × height (Brackets may be implied later) May be first line If this first line, then M0

Dependent on M1A1. Fully established – no errors throughout and at least 2 steps, one with 40 or 80, after first line

(ii) –9, 4 B1,B1 If B0, SC1 for +9 and –4

(iii) (BC2 = ) (their x + 1)2 + (their x + 2)2 (BC = ) 7.81(0…)) c.a.o. www2

M1

A1

Their x must be positive Ignore any extra solutions

(b) (i) 12

59 or

12

5108 + or

12

5129 +×

or 60

565

or 60

25609 +×

seen

E1 Must be fractional form Condone 113/12 × 60 = 565; 9 × 60 + 25 = 565 Not for decimals

(ii) 3

23 +y or

2

4+y o.e.

6

)4(3

6

)23(2 +

+

+ yy o.e.

B1

B1

or 6

123

6

46 +

+

+ yy o.e.

(iii) 12

113

12

)169(2=

+y o.e.

y = 4.5 c.a.o. www2

M1

A1

o.e. means with common denominator or better (Trial and error scores 2 or 0.)

(iv) (Total dist =) (3 × their y) + 2 + (their y) + 4

o.e.

(Average speed = ) 1259

24their o.e.

2.55 (km/h) (2.548 – 2.549) c.a.o. www 3

M1

M1

A1

(= 24)

(dependent) Must be km divided by hours o.e. for full method Accept fractions in range [15]



IGCSE – May/June 2009 0580, 0581 04

7 (a) 250x2 = 4840 o.e.

x² = 19.36 or (x =) 2504840 ÷ (= 4.4)

M1

E1

Allow M1 for 250 × 4.42 = 4840

Then E1 for 250 × 19.36 = 4840

(b) 42.6 (kg) cao (42.592 or 42.59) B2 SC1 for figures 426 or 4259…

(c) 26.4 (cm) c.a.o. B2 If B0, M1 for any of following 88 ÷ 4.4 = 20 and 120 ÷ 20 = 6 (accept 6 bars high o.e.) or 88h = 4.42 × 120 or 250 × 88 × h = 120 × 4840

(d) (i) 4840 ÷ 4200 (implied by 1.15(2))

÷ π34 (implied by 0.274 to 0.276)

3 (seen or implied by correct answer to

more than 2 dp) 0.649 – 0.651

M1

M1

M1

dep

A1

4200 × π34 r3 = 4840

(r3 =) 4840 ÷ (4200 × π34 )

3 Third M dependent on M1M1

Must be 3dp or better

(ii) 5.31 (5.306 – 5.31) (cm2) B1

(iii) 100

2504.444.42

their 4200

2×

××+×

× (ii)

501.9 – 503 (%) c.a.o. www4

M3

A1

If M0, M1 for 4200 × their (ii) (22299) and M1 (independent) for correct method for surface area of solid cuboid (4438.72)

[15]

8 Throughout the question ratios score zero. If using decimals, 2 s.f. correct answers to parts (c) and (d) – penalty of 1 once Use of words e.g. 1 in 400 or 1 out of 400, Correct answers – penalty of one For method marks only accept probabilities p and q between 0 and 1

(a) p = 20

1 , q = 20

19 o.e. B1 Could be on diagram

(b) (i) 400

1 o.e. c.a.o. B2 0.0025 allow M1 for (their p)2 o.e.

(ii) 400

38 o.e. c.a.o. B2 0.095 allow M1 for 2 (their p)( their q) o.e.

(c) 8000

38 o.e. c.a.o. B2 0.00475 allow M1 for 2(their p)² (their q) o.e. including their (ii) × their p

(d) their (b)(i) + their (c)

8000

58 o.e. c.a.o.

M1

A1

0.00725

(e) their (d) × 1000 = 7.25 o.e. ft B1 ft Accept 7 or 8 or an equivalent integer ft [10]



IGCSE – May/June 2009 0580, 0581 04

9 (a) (i) 174 to 174.25 (cm) c.a.o. B1

(ii) 167 (cm) c.a.o. B1

(iii) 12 (cm) c.a.o. B1

(iv) 37 c.a.o. B2 If B0, B1 for 63 seen in working space

(b) (i)

(ii)

10, 25

155, 165, 175, 185 (their 10 × 155 + their 25 × 165 + 47 × 175 + 18 × 185) ÷ 100 172 or 172.3 (cm) c.a.o. www 4

B1

M1

M1

M1

A1

s.o.i. allow 1 slip

Use of fxΣ where the x’s are in/on their

intervals (allow one more slip) (17 230)

(dependent on second M) ÷ 100

[10]

10 (a) (i) –2, B1

(ii) 26, B1

(iii) 8

1 o.e. B1

(b) )(2

1x

y=

+

(f—1(x) = ) 2

1+x o.e. www2

M1

A1

If switch x and y first then M1 for x = 2y – 1 or If use a diagram/chart then M1 for any evidence of +1 then result ÷ 2

(c)

2

2

1

1

xz

xz

=−

+=

(x = ) 1−z www2

M1

M1

Correct rearrangement at any stage for x or x2. Correct sq root at any stage

Ignore +, – or ± in front of

(d) 1)12(2

+−x

2442

+−= xx or )122(22

+− xx

www 2

M1

A1

Final answer but condone one minor factorising slip if first answer seen

(e) 9 B1

(f) 2(2x – 1) + x2 + 1 (= 0) or better (x2 + 4x – 1 = 0 )

12

)1)(1(444)(

2

×

−−±−=x ft

(x = ) – 4.24, 0.24 c.a.o. www 4

(final answers)

B1

M1

M1

A1,A1

)1)(1(442

−− or better seen

If in form r

qorp −+

for – 4 and 2 × 1

or better Ft their 1, 4 and –1 from quadratic equation seen After A0A0, SC1 for – 4.2 or – 4.235 or – 4.236… and 0.2 or 0.235 or 0.236….. The SC1’s www imply the M marks

(g) (i)

(ii)

Straight line with positive gradient and negative y-intercept U-shape Parabola vertex on positive y-axis

L1

C1

V1

Dependent [18]



IGCSE – May/June 2009 0580, 0581 04

11 (a) 15, 21, 28, 36 B2 B1 for 3 correct

(b) (i) 10 + 15 = 25, 15 + 21 = 36 etc B1 Any two complete and correct statements

(ii) Square B1

(c) (i) 2 B1

(ii) 102

54=

× o.e. E1

(iii) 16 290 c.a.o. B1

(d) (i)

2

)2)(1( ++ nn or

2

232

++ nn seen

2

)2)(1(

2

)1( +++

+ nnnn or

2

23

2

22++

++ nnnn

2)1(2

)1)(1(2

2

)22)(1(

)2(2

)1(

+=++

++

++

+

n

nn

nn

nn

n

2

2

2

)1(

12

2

242

+

++

++

n

nn

nn

M1

M1

E1

Denominator could be their k May be implied by next line

This line must be seen and at least one more step, without any error, to gain the E mark Dependent on M1M1. Fully established – no errors

(ii) 1711 and 1770 final answers c.a.o. B2 SC1 for 59 or 58 or 1711 or 1770 seen [12]

Graph for Question 5

876543210-1-2-3-4-5-6-7-8

4

3

2

1

0

-1

-2

-3

4

876543210-1-2-3-4-5-6-7-8

4

3

2

1

0

-1

-2

-3

4

876543210-1-2-3-4-5-6-7-8

4

3

2

1

0

-1

-2

-3

4

876543210-1-2-3-4-5-6-7-8

4

3

2

1

0

-1

-2

-3

4



IGCSE – May/June 2010 0580 21

Qu. Answers Mark Part Marks

1 3.14 π 7

22 √10 2 M1 3.1428(…) and 3.16(2…) seen

2 650 2 M1 4.2

600 (× 2.6)

3 44 2 M1 97 or 53 seen

4 30 2 M1 108 × 1000 / (60 × 60)

5 3.2(0) × 104 2 B1 32000 or 32 × 103 etc

6 (a) 0.461939(…) (b) 0.4619 or ft

1

1ft

7 1.62 3 M1 4

1 π 0.82

M1 adding (0.8 × 1.4) to their k π

8 (a) (i)

1

or

(ii)

1

(b) 2 1

9 Sunday (May) 25 1045 1, 1, 1 Independent

10 24.3(0788…) 3 M1 5 × 3.5 + 2 × 1.5 M1 (√) 1.52 + 3.52

11 5

42 wcw −

oe 3 M1 one correct move to clear fractions M1 second correct move to subtract term M1 third correct move dividing by 5 May be in any order

12

4 6 10 P 15

2

8

5

20

RQ

3 M1 15 only in small circle M1 10 only in the intersection A1 all correct including labels

13 x = 12 y = –10

3

M1 consistent addition (& mult) for x or consistent subtraction (& mult) for y A1 only earned if method correct

14 3.84 or 325

21 3 M1 y =

2

x

k oe A1 k = 96

http://www.xtremepapers.net



15 (a) 4 1

(b) y = –2x + 9 oe 3 M1

32

35

−

−

oe

M1 substitution of a point into their equation If M1 only then A1ft for y = “m”x + “c” used correctly with their numeric values

16 (a) 8

3p

or 0.125p3 1, 1 Independent marks for letter and no.

(b)

8

9q–1 1, 1 Independent marks for letter and no.

Allow 18

1q–1 or

q8

9

17 (a) 52 (b) 64 (c) 71

1

1

2

M1 angle CED = 19

18 (a) E, G (b) A, B

1, 1

1, 1

19 (a) 2p 3p + q ……….. 5p + 3q cao (b) (i) all 4 plotted correctly ft (ii) a (straight) line

1, 1, 1

2

1

B1 2 or 3 correct Allow linear, collinear

20 (a) 27 (b) 9x2 cao (c) 3√x + 1

2

2

2

M1 g(–1) = 4 seen or ((x – 1)2 – 1)3

M1 (3x + 1 – 1)2 or better

M1 interchange x, y & rearrange formula

21 (a) CB and BA cao 1, 1 Independent

(b)

−

−164

248 cao

3

M1 8

1

4

3

4

1

2

1×−× (=

32

1)=

M1

−

−

2

1

8

1

4

3

4

1

seen

(c) determinant is zero 1 Allow cannot divide by zero





1 (a) 1 (b) 1

1

1

Allow none

2 0 2 M1 4sin3120 evaluated and rounding to 2.6 or

better (2.598…) or 2

33

3 2 – 3 , 2 –2

3,

3

2, 3

2 M1 correct decimals seen

4 40

3215 +a oe 2 B1 15a + 32 seen

or SC1 40

32

40

15+

a

on answer line

5 210 2 M1 26 or 2–4 seen

6 6.4 × 107 2 M1 64 × 1002 × 102 or 64 000 000 oe

7 (A ∪ B ∪ C)' (A ∪ C)' ∩ B

1 1

or A' ∩ B' ∩ C' or A' ∩ (B ∪ C)' or A' ∩ C' ∩ B

8 (a) 43 to 47 (b) 64 to 68

1

2

SC1 23 to 27

9 63.84 cao 3 M1 figs 1995 M1 32 × their lower bound

10 1

3

−

=

Px 4 M1 for each of the four moves completed

correctly

11 (a) 10(.0..) (b) 9.80

1

3

M2 √((a)2 – 22) or M1 PT2 + 22 = (a)

2

12 (a) 440

(b) 3 min 20 sec

2

2

M1 sin 37.1 or cos 52.9 = 730

h oe

M1 65.3

730

13 (a)

+−54

36

x

x but not

+−5)(436

x

x 2 B1 6x – 3 or B1 4x + 5 in a (2 × 1) matrix on answer line

(b) (6x2 + x + 5) cao 2 M1 any 1 × 1 matrix in answer space

14

4

Mark the position of the letter R (or the worst unshaded region if R is missing) as follows




15 (a) (2, 4) (b) (6, 0) (c) (i) (4, 2) ft (ii) y = – 3x + 14 oe

1

1

1ft

2

From (a) and (b)

M1 sub their (c)(i) into y = –3x + c oe

16 16 4

1or 16.3 5 M1 finding the area under graph A1 130

M1 2

1 × 16 × v

M1 equating and solving

17 (a) 201 2 M1 π × 82

(b) 87.9 or 88.0 4

M1

360

45 × 2 × π × 12 ….. d

M1 2 × π × 8 ……………..e M1 ft for their (4d + e) which must come from multiples of π SC2 43.9 or 44.0

18 (a) (i) 11 (ii) 1 – 6x

1

2

M1 3(1 – 2x) – 2

(b) –1.65, 6.65 4 M1

2

5 k± M1 √[(–5)2 – 4 × 1 × (–11)]

or better A1 A1

19 (a) 6, 30, 70 (b) graph (c) 82.5 or ft ±1 (d) 108 or ft ±1

2

3

1ft

1ft

B1 for 2 correct

P2 7 plots correct from table P1 5 or 6 plots correct from table

C1 smooth curve through the points in the given range within one small square of the plots or the correct position





1 (a) –5

(b) 11

1

1

2 %482380.411

53>>> 2 M1 for decimals seen

4.7958… 0.48 (4.80) 4.81(…)

3 500 2 M1 for 600 × 0.6 ÷ 0.72 seen

4 70 2 M1 for 252 × 1000 ÷ 60 ÷ 60 oe

5 18 2 M1 for 21.6 ÷ 1.2 oe

6 x + 8 2 M1 38 seen

7

2 B1 for one correct Venn diagram

8 6

35 −x

2 B1 for 5x – 3 seen

SC1 6

3

6

5−x on answer line

9 5(.00) × 105 2 SC1 for 5 × 10k or 500 000 on answer line

10 220.5 cao 2 M1 for 73.5 seen

11 16.8 3 M2 tan17 = 55

h or tan73 =

h

55

or M1 tan17 = h

55 or tan73 =

55

h if angle seen in

wrong place at P

12 9 – 2x2 3 B1 for x2 – 3x – 3x + 9 or 2x

2 – 6x – 6x + 18

B1 for 4x2 – 6x – 6x + 9 or –4x

2 + 6x + 6x – 9

13 (a) 0

(b) 2

(c) plane across centre of shape

1

1

1

Three possibilities

14 6 3 M1 for one correct first step which leads towards

simplifying

92

123 =+−y

y

or 6(y – 4) + y = 18

or y – 4 + 6

y = 3

M1 correctly collecting their terms to py = q




15 (a) g – h 1

(b)

4

1g +

4

3h 2 M1 for OH + HN or h +

4

1 (a)

OG + GN or g – 4

3 (a)

16 25

−

r

A or

r

rA 25 −

3 M1 for correctly multiplying by 5

M1 for correctly dividing by r

M1 for correct subtraction

in any order

17 (a) 10.9 2 M1 for 2

6.5π360

40××

(b) 15.1 2 M1 for 6.52π360

40××× (= 3.91..)

18 (a) 64 2 B1 for evidence of f(–2) = 6

(b) 9 2 M1 for 3x – 5 = 22 or

3

5+x seen

19 (a) 4

3or 0.75 1

(b) 2.6 3 M1 for finding the area under the graph or

M1 for their 39 ÷ 15

20 x [ 0 1 L1 x R 0

y [

2

1x oe 2 L1 y R

2

1x

x + y Y 4 oe 2 L1 x + y R 4 where R is any one of = < > Y [

B2 all inequalities correct or B1 2 correct

21 (a) 18.7 3 M2 for 100

140sin50sin ×=R (= 0.3219…)

or M1 for 100

140sin

50

sin=

R oe

(b) 261(.3) 2ft M1 360 – 80 – their (a)

22 Perpendicular bisector of AC

Bisector of angle A

Shaded region inside triangle and

to left of perp bisector of AC and

above bisector of angle A

2

2

1

B1 accurate line

B1 two pairs of correct construction arcs

B1 accurate line

B1 two pairs of correct construction arcs

B1 dep on first B1 being scored for both lines

23 (a) ( )75− 2 B1 either correct in a (1 × 2) matrix

(b)

32

12

4

1 oe 2 M1 for

32

12 seen or 2 × 3 – –1 × –2 ( = 4)

(c)

10

01 or I cao 1




Abbreviations

cao correct answer only

cso correct solution only

dep dependent

ft follow through after error

isw ignore subsequent working

oe or equivalent

SC Special Case

www without wrong working


1 (a) 11:14 1

(b) 50 2 M1 for (220 + 280) ÷ 10 o.e.

(c) 12 2 M1 for 21 ÷ (4 + 3) × 4 (or 3) o.e.

(d) 280 3 M1 for 0.35 × their 500 (175)

M1 dependent × 1.60

(e) 240 2 M1 for dividing 264 by 1.1 oe

2 (a) (i) 4 1

(ii) 5 1

(iii) 4.75 3 M1 for 1 × 2 + 1 × 3 + 17 × 4 + 12 × 5 + 6 × 6 + 3 × 7

condone one slip then M1 dependent result

(190) ÷ 40

(b) n

n

+

+

40

3190 2 SC1 for their 190 + 3n

3 (a) Triangle drawn with co-ords at (1, 4),

(4, 2), (4, 4)

2 SC1 for 2 correct vertices or an enlargement sf

2

1 with wrong centre

(b) (i)

−−−

884

288 2 B1 each row

(ii) Triangle drawn at (–8, 4), (–8, 8), (–2, 8)

ft (i)

2ft SC1 for 2 correct ft vertices. Can also be

correct regardless of (i)

(iii) Reflection cao

y – axis or x = 0 cao

2

B1 Independent of (i) or (ii)

Extra transformations lose all marks

B1 Independent of (i) or (ii)

(c) (i) Translation B1 Extra transformations lose all marks

−−10

10 o.e. 2 B1

(ii) Rotation

(0, 0)

90° clockwise oe

3 B1 Extra transformations lose all marks

B1 Allow word origin for (0, 0)

B1 Allow – 90° or 270° (anti-clockwise)

(d)

− 01

10 2 B1 each column




4 In (b) and (c) isw any cancelling or changing

to other forms, after correct answer seen.

Penalty of – 1 for 2 sf decimals or

percentages. Do not accept ratio or worded

forms.

(a) B and 5

2, 4

1 oe 1 Allow any reasonable explanation, e.g. 2 out of

5 greater than 1 out of 4.

(b) (i) 5

3,

5

2,

4

3,

3

1 4 B1 B1 B1 B1

(ii) 12

6 oe cao www 2 2

2

1, 0.5 etc M1 for

4

3their

3

2× i.e. product of

correct branches on their tree

(iii) 60

42 oe cao www2 2

10

7, 0.7 etc

M1 for their (ii) + 5

3their

3

1their × from their

tree

(c) 60

2 oe cao www2 2

30

1, 0.0333(3……..) etc

M1 for 4

1

5

2

3

10

4

1

3

2××+

××

5 (a) 200.5… to 201 www 2 2 M1 for 0.5 × 24 × 26 sin 40 oe

A1

(b) 17.2 (0….) www 4 4 M2 for 262 + 242 – 2 × 26 × 24 cos 40

or M1 for 26242

242640cos

222

××

−+

=

BD

A2 or A1 for 295.976..

(c) 12.8 (12.77…) www 4 4 B1 for Angle C = 110 soi accept on diagram

M2 for 110sin

30sin24)( =BC oe or

M1 BC

30sin

24

110sin= oe i.e. a correct implicit

statement soi

A1

(d) 8.208 to 8.230 www 2 2 M1 for their (c) × sin40 oe




6 (a) 32.5 cao www4 4 M1 for mid-values seen

M1 for use of Σfx with x’s anywhere in each

interval

(10 × 15 + 30 × 30 + 20 × 45)

M1 ÷ 60 dependent on second M1

(b) Histogram drawn 3 B1 Bars correct positions and widths – no gaps

B2 Heights of bars 1, 1.5 and 2 (B1 for any

two correct or for heights in the ratio 2:3:4)

7 (a) 4.53 or 4.526 – 4.530…. 3 SC2 for figs 453 or 4526 – 4530

If SC0, M1 for π × (figs 31)2 × 15

(b) 3.62 to 3.624 ft 2ft M1 for their (a) × figs 8 oe

(c) (i) 360 – 2 × 90 – 60 oe 2 E2 The 90’s and the 60 must be clearly

justified. Accept in diagram.

SC1 for 60 or two 90’s soi in correct positions oe

e.g 360 ÷ 3 scores 0

(ii) 0.649 (0.6492 to 0.6493) 2 M1 for π × figs 62 ÷ 3

(iii) 7.53 (7.527 or 7.528….) 3 M1 for their (ii) × 3

M1 (indep) for 18 × figs 31

This M is spoiled by extra lengths.

(iv) 112.9 to 113 ft 1ft ft their (iii) × 15

8 (a) 0.25, 8, 16 3 B1 B1 B1

(b) – 5, 4 2 B1 B1

(c) (i) 7 points plotted ft

Curve through all 7 points exponential

shape

P2ft

C1ft


ft only if exponential shape

(ii) 6 points plotted ft

Curve through all 6 points parabola

shape

P2ft

C1ft

P1 for 5 points ft

ft only if parabola shape

(d) (i) 3.2 to 3.4 1

(ii) 0.3 to 0.4 and 2 2 B1 B1

(iii) 3.1 to 3.4 1

9 (a) (i) –2.5 oe 2 M1 for 5(w + 1) = 3w

(ii) –3 or 1 2 B1 B1 (If 0, SC1 for y + 1 = ± 2)

(iii) 9.5 oe B3 M2 for 1526355 ×=+−+ xx Condone one

slip (sign or numerical) on left hand side

or M1 for 15

)2(3

15

)1(5 −

−

+ xx

or better,

condoning one sign or numerical slip.




(b) (i) )1)(10( +− uu 2 SC1 for ))(( buau ++ where ab = –10 or

a + b = –9

(ii) –1, 10 1ft Only ft B2 or SC1 in (i) but can recover to

correct answer only if new working or if (i) not

attempted

(c) (i) 2

2

)2)(1(x

xx

=++

oe M1

( 22))2)(1(2

+++=++ xxxxx B1 Allow 3x for x + 2x

22

222 xxxx =+++

0232

=−− xx E1 Established without any omissions or errors

(ii) )1(2

)2)(1(4)3()3(2

−−−±−−

2

B1 for )2)(1(4)3(2

−−− or better seen

anywhere.

If in form r

qp + or

r

qp −then B1 for

3(−− ) and 2(1) or better

Brackets and full line may be implied later

–0.56, 3.56 2 B1 B1

SC1 for –0.6 or –0.562 to –0.561 and 3.6

or 3.561 to 3.562

(iii) 12.7 or 12.67 to 12.69 ft 1ft ft their positive x squared

10 (a) 20x + 100y Y 1200 1

(b)(i) x + y [ 40 1

(ii) y [ 2 1

(c) x + y = 40 cao

y = 2 cao

Required region only region left not

shaded or otherwise clearly indicated

cao

L1

L1

R2

Each line ruled and long enough to enclose

required region.

If L0, SC1 if freehand but otherwise accurate

and enclose region

SC1 if one boundary error – see diagrams

(d) 5 cao 1

(e) 50 cao, 2 cao

270 ft

2

1ft

B1 B1

ft 5 × their x + 10 × their y

11 (a) Reasonable diagram, 25, 13, 62 4 B1 B1 B1 B1 diagram may be freehand

(b) 64, 19, 146 3 B1 B1 B1

(c) n2 oe

2n + 3 oe

2

B1

B1

(d)(i) 2 1

(ii) 20202 ft 1ft ft 10101 × their k




Abbreviations



dep dependent



oe or equivalent

SC Special Case



1 (a) 240 ÷ 8 × 3 or 240 ÷ 8 × 5 or 8

3 of 240

or 8

5 of 240 oe

1 Accept reverse e.g. 90 : 150 = 3 : 5 and

90 + 150 = 240

(b) (i) 5 www 2 2 M1 for 290

9100

×

×

oe

(ii) 165 www 2 2 M1 for 99 ÷ 0.6 oe

(c) 162.24 final answer cao 2 M1 for 150 × 1.04 × 1.04 oe implied by

answer 162.2

(d) (i) 58.67 final answer cao 3 SC2 for 58.7 or

M1 for 100

204150 ××

oe (120)

then M1 (dependent on the first M1)

328.67 – 150 – their 120 oe

Answers of 208.67 or 208.7 imply first M1

(ii) 219 (.1….) www 2 2 M1 for 100150

67.328× oe

2 (a) (i)

8

15 2 B1 each component

(ii) 17 www 2 2ft ft their 15 and their 8.

M1 for (their 15)2 + (their 8)2

(b) (i) 2

1v –

2

1c or

2

1(v – c) cao 2 M1 for CV

2

1 soi

(ii) 2

1c +

2

1v again allowing brackets cao 2 M1 for OM e.g. CMOC + or better seen

or v – their (i)

or c + their (i)

(iii) 6

1v –

2

1c again allowing brackets cao 2 M1 for any correct route e.g. VLMV +

or their (i) – 3

1 v

or 3

2v – their (b)(ii)




3 Throughout this question isw any cancelling

or changing to other forms, after correct

answer seen. Penalty of –1 for 2 sf decimals

or percentages. Do not accept ratio or

worded forms.

(a) (i) 6

4 oe (0.667) 1 Allow 0.6666 – 0.6667

(ii) 6

3 oe 1

(iii) 6

2 oe (0.333) 1 Allow 0.3333…

(iv) 6

5 oe (0.833) 1 Allow 0.8333…

(b) (i) 36

1 oe (0.0278) 2 Allow 0.02777 – 0.02778, M1 for

6

1

6

1×

(ii) 36

6 oe (0.167) www 2 2 Allow 0.1666 – 0.1667, M1 for 2

6

1

6

3×× oe

(c) (i) 4

1 oe 1

(ii) 2

1 oe 1

(d) 5 (but not from rounding) 2 M1 for repeating × 6

4 oe e.g.

n

3

2

4 (a) (i) Triangle with vertices (–4, 4), (–1, 4),

(–1, 6)

2 SC1 for translation

−k

7 or

3

k

(ii) Triangle with vertices (1, –3), (1, –6),

(3, –6)

2 SC1 two correct vertices or 90° anticlockwise

about (0, 0)

(b) (i) Reflection only

y = –x oe

1

1

Marks independent but must be single

transformation to score any marks

(ii) Stretch only

x-axis oe invariant

(factor) 3

1

1

1

Marks independent but must be single

transformation to score any marks




(c) (i)

−

−01

10 2 B1 each column

(ii)

30

01 ft 2ft ft factor in (b)(ii) only if stretch and can

recover to correct matrix

SC1ft for right-hand column

(iii)

3

10

01 ft 1ft ft

n0

01 to

n

10

01 or

10

0n

to

10

01

n

n ≠ 0, ± 1

for3

1, allow 0.33 or better

5 (a) (cos) 1151802

90115180222

××

−+ M2 M1 for correct implicit expression 902 = ……

24.98 – 24.99 A2 A1 for (cos) = 0.9064…

(b) (i) 125(.0….) ft 1ft ft 150 – their (a)

(ii) 305(.0….) ft 1ft ft 180 + their (b)(i)

(c) 180sin (54.98 to 55)

or 180cos (35 to 35.02) oe

or 180sin (360 – their (b)(ii))

or 180cos(their (b)(i) – 90) oe

147(.4….) cao www 3

M2

A1

B1 for 54.98 to 55 or 35 to 35.02 soi in correct

position.

Provided either angle is acute

(d) 70sin

30sin90

M2

M1 for 70sin

90

30sin=

TR or other correct

implicit equation

47.9 (47.88 – 47.89) cao www 3 A1

(e) 2 000 000 oe 2 Allow 1 : 2 000 000 as answer.

SC1 figs 2 in answer which could be a ratio.

6 (a) 3

4.23

4×π M1 Must see method

57.87 – 57.92 to at least 4 figures

A1

(b) (i) 14.4, 9.6, 4.8 1, 1, 1 Any order

(ii) 664 (663.5 – 663.6) ft 1ft

(iii) 315 or 316 or 317 (315.2 – 316.8) ft 1ft ft their (b)(ii) – 6 × ‘57.9’ (only if positive)

(iv) 507 (506.8 – 506.9) ft 2ft M1 for (14.4 × 9.6 + 14.4 × 4.8 + 9.6 × 4.8) × 2

or their 3 lengths.




(c) (i) Height seen or implied as 6 × 4.8

or better

π × 2.42 × their height

521 (520.8 – 521.3) www 3

M1

M1

A1

Indep

(ii) 174 or 173 (173.2 – 174.1) ft 1ft ft their (c)(i) – 6 × ‘57.9’ only if positive

(iii) 470 – 471 cao www 3 3 M1 for 2 × π × 2.42 (36.17 to 36.2), and

M1 indep for π × 4.8 × their height from (c)(i)

7 (a) 12 × 2.5 + 15 × 7.5 + 23 × 12.5 + 30 ×

17.5 + 40 × 22.5 + 35 × 27.5 + 25 ×

32.5 + 20 × 37.5

M1

M1

mid-values any three soi

Use of Σfx dep on x anywhere in each interval

(including lower bound) – allow 2 slips or

omissions

÷ 200

21.9 www 4

M1

A1

Depend on second M

(b) 155, 180 1

(c) 8 points plotted ft, ignoring (0, 0)

Reasonable increasing curve or

polygon through their 8 points

P3ft

C1ft

P2ft for 6 or 7 plotted , P1ft for 4 or 5 plotted

Condone starting at (5, 12) and ft only if shape

correct.

(d) Either horizontal or vertical line at

least 1 cm long at y = 50 on the curve

1

(e) (i) 22 – 23 1

(ii) 13.5 – 14.5 1

(iii) 25.5 – 26.5 1

(iv) 136 – 140 must be integer 2 SC1 for 60 – 64 seen and must be integer

8 (a) (p + q)2 – 5 oe final answer 2 SC1 for (p + q)2 oe seen

(b) 6x + 9(x – 3) = 51 or better

5.2(0) final answer

B3

B1

B2 for 6x + 9(x – 3)

or B1 for 6x or 9(x – 3)

5.2(0) ww is B1 only

(c) a + c = 52 oe

3a + 2c = 139 oe

Correctly eliminating a or c.

35

17

B1

B1

M1

A1

A1

Condone consistent use of other variables

or M3 for 3a + 2(52 – a) = 139

or 3(52 – c) + 2c = 139 o.e.

Allow one numerical slip.

If A0, SC1 for 17, 35




9 (a) (i) Similar 1 Allow enlargement

(ii) 4.5 2 M1 for 6

9

3=

AX oe

(iii) 13.5 cao 2 M1 for

2

2

3

or

2

3

2

oe e.g. using base and

height but other methods must be complete

(iv) 180 – x – y oe

180 – x oe

B1

B1

(b) (i) 96 1

(ii) 48 ft 1ft ft 0.5 their (b)(i)

(iii) 97 ft 1ft ft 145 – their (b)(ii)

(iv) 35 1

(c) 20n = 360 oe or 160)2(180=

−

n

n

oe

or 180(n – 2) = 8 × 360 oe

or nn

360180

3608 −=

M2

M1 for 9e = 180 oe allow diagram to show this

if reasonably clear

or M1 for 8 × 360 or n

3608×

18 www 3 A1

10 (a) Pentagon

Octagon 20

1

1, 1

(b)(i) 35 1

(ii) 54 1

(c)(i) p = 2, q = 3 3 M1 for substituting a value of n e.g.

2)4(41

=− qp

3≥n

or M1 for number of diagonals from one

vertex is n – 3 (allow in words)

and B1 for one correct value. If 0, SC1 for

)3(2

−n

n

seen.

(ii) 4850 ft 1ft ft their (c)(i) allow only if ft calculates to a

positive integer.

(iii) 20 cao 2 SC1 for answer of 17

or M1 for their formula = 170

(d) 31 cao 1




Abbreviations



dep dependent



oe or equivalent

SC Special Case



1 (a) (i) 2 : 3 1

(ii) 30 ÷ 2 × 3 o.e. E1 Allow 2 : 3 (oe) = 30 : 45

(iii) 60 2 M1 for 3 ÷ 5 × 100 oe

(b) 31.83 3 SC2 for 31.827 as final answer or not spoiled.

or M1 for × 1.03 twice oe

(c) 1.5 2 M1 for 25.2100

530=

×× r

oe

or for 2.25 ÷ 5 then ÷ 30 × 100

2 (a) 5.83 (5.830 to 5.831) 2 M1 for 32 + 52


(b) 113. 6 (114 or 113.5 to 113.6) www 4 4 M2 for (cosC) = 852

1185222

××

−+

or M1 for correct implicit expression

A2 (A1 for –0.4 or –5

2)

(c) 25.8 (25.77 to 25.85) cao www 3 3 M1 for 0.5 × 5 × 8 × sin (their angle C) o.e

must be full method e.g. Hero’s formula.

M1 for 0.5 × 3 × 5 oe




3 Throughout this question isw any cancelling

or changing to other forms, after correct

answer seen. Do not accept ratio or worded

forms.

(a) 0.4, 0.1 oe 1

(b) (i) 1 1

(ii) 0.7 oe ft 1ft ft their first three probabilities

(c) (i) 0.04 oe 1

(ii) 0.03 oe ft 2ft M1 for their 0.1 × 0.3

(iii) 0.12 oe ft 3ft ft their 0.1, their 0.4 and their (c)(i)

M2 for their 0.4 × their 0.1 + their 0.1 × their

0.4 + 0.2 × 0.2 (or their (c)(i))

or M1 for any two of these products added or

two of each

(d) 0.147 oe ft 2ft ft their (b)(ii).

M1 for their 0.7 × their 0.7 × (1 – their 0.7)

4 (a) Triangle drawn , vertices (6, 10),

(10, 10), (10, 8)

2 SC1 reflects correctly in x = 6

(b) Triangle drawn , vertices (2, 8), (6, 8),

(6, 10)

2 SC1 for translation

−

k

4 or

6

k

(c) Translation 2 B1 All part marks spoiled if extra

transformation

− 64

o.e.

B1 Indep. Allow other clear forms or words

(d) (i) Enlargement

(centre) (4, 6)

(factor) 0.5

3 B1 All part marks spoiled if extra

transformation

B1 Indep.

B1 Indep.

(ii) 4

1 or 0.25 oe 1

(e) (i) Stretch

y-axis o.e invariant

(factor) 0.5

3 B1 All part marks spoiled if extra

transformation

B1 Indep

B1 Indep

(ii)

10

05.0 ft 2ft ft their factor in (e)(i) only if stretch

SC1 (also ft) for left-hand column




5 (a) (i) Similar 1 Accept enlargement

(ii) 2.7 2 M1 for 4

3

6.3=

PQ oe

(iii) 3.15 2 M1 for

2

4

3

or

2

3

4

o.e seen

If Cabsin2

1 used or base and height used then

must be full method for M1

(b) (i) 29 1

(ii) 61 ft 1ft ft 90 – their (i) if (i) is acute

(iii) 61 ft 1ft ft their (ii) if their (ii) is acute, but can recover

(iv) 119 ft 1ft ft 180 – their (iii)

(c) (i) 20 1

(ii) 110 3 M1 for adding 6 angles going up 4 each time

and

M1 (indep) for 720 seen and not spoiled

(6A + 60 = 720 o.e. scores M2)

6 (a) –2.5, –2, 2, 2.5 2 B1 for 3 correct

(b) 4 points correct ft

Correct shape curve through at least 9

points over full domain

Two branches either side of y-axis and

not touching it

P1ft

C1ft

B1

ft only if correct shape and isw any curve

outside domain (including crossing y-axis)

Independent

(c) –1, 0, 1 2 B1 for two correct, each extra –1

(d) (x) < –1 and (x) > 1 as final answer 2 B1 B1 Condone inclusive inequality, allow in

words, condone inclusion of – 4 and + 4 as

limits. 11 −<< x or 1− > x > 1 SC1

11 <<− x scores 0. Each extra –1 if more

than two answers.

(e) (i) Correct ruled line though (–2, –3) to

(1, 3)

2 SC1 for ruled line gradient 2 or y-intercept 1

from x = –2 to 1 or correct line but short or

good freehand full line.

(ii) Some reasonable indication on graph

for both points

1 e.g. points of intersection marked, or, lines

drawn from point of intersection to x-axis etc

(iii) x2 + 1 = 2x2 + x oe then x2 + x – 1 = 0 3 E2 Must be intermediate step before answer –

no errors or omissions

or 1

1+= x

x

then 1 = x2 + x

then x2 + x – 1 = 0

or E1 Either no intermediate step or one error

or omission.

1, –1 B1




7 (a) (Mode) = 11

(Median) = 12.5

(Mean) = 12.8 (0 ….)

1

2

3

B1

M1 for evidence of finding mid-value

e.g. (126 + 1) ÷ 2 oe, (condone 126 ÷ 2)

M1 for correct use of Σfx (allow one slip)

M1 (dependent) for ÷ 126

(b) (i) 15, 27, 30, ……. 3 B1 B1 B1

(ii) 9.67 (9.674 to 9.675) cao www 4 4 M1 for mid-values, condone one error or slip

M1 for use of Σfx, with x’s anywhere in

intervals and their frequencies (allow one slip)

M1 (dependent on second M) for ÷ 126 (or

their Σf)

isw any conversion into hours and minutes

8 (a) 40 ÷ 10 and 12 ÷ 6 (or 12 ÷ 3) and

6 ÷ 3 (or 6 ÷ 6) oe

4 × 2 × 2 = 16 reducing (seen) to 16

E2 M1 Allow drawing for M1 but must see

reaching 16 for E2

Reaching 16 without any errors or omissions

SC1 for (b)their

61240 ××

even if = 16

or 4 × 2 × 2 = 16 or 4 × 4 × 1 = 16 without

other working

(b) 180 1

(c) (i) 23 640 (allow 23 600) 2 M1 for their 180 × 8 × 16 + 600

(ii) 23.64 (or 23.6) ft 1ft ft their (i) ÷ 1000

(d) (i) 216 2 M1 for (10 × 6 + 10 × 3 + 6 × 3) × 2 oe

(ii) 8.64 3 M1 for their (i) × 16 × 25

M1(indep) for ÷ 1002

Figs 864 imply M1 only

(e) 75.3 (75.26 to 75.33….) 3 M1 for 3

5.0π3

4× (0.5235..) Implied also by

104.7….

then M1 (dep) for their (b) – 200 × their

35.0π

3

4× must be giving positive answer

(f) 0.842 (0.8419 – 0.8421) 3 M1 for 2050)π3

4(

3÷=r

then M1 for

π3

4

2050 ÷ (0.5966 to 0.5972)

After 0 scored SC1 for 3

π3

4

50 (implied by 2.29)




9 (a) 8w + 2j = 12

12w + 18j = 45

Correctly eliminating one variable

Water 1.05, Juice 1.8(0)

5 B1 condone consistent use of other variables

B1

M1 allow one numerical slip

A1 A1 If A0, SC1 for 1.80, 1.05

(b) (i) 60

40

4

42=

−

+

yy oe M2 M2 If M0, SC1 for

y

2 or

4

4

−y

)4(3

)4(2

)4(3

43

)4(3

)4(32

−

−=

−

×+

−

−×

yy

yy

yy

y

yy

y

oe or better

6(y – 4) + 12y = 2y(y – 4) oe

6y – 24 + 12y = 2y2 – 8y oe

0 = 2y2 – 26y + 24

y2 – 13y + 12 = 0

E2 E2 Correct conclusion reached without any

errors or omissions including at least 3

intermediate steps.

or E1 if any one slip, error or omission that is

recovered or correct with only two steps.

(ii) (y – 1)(y – 12) 2 SC1 for (y + a)(y + b) where ab = 12 or

a + b = –13

(iii) 1, 12 ft 1ft Only ft SC1 but can recover to correct answer

with new working or if (ii) not attempted

(iv) 8 ft 1ft ft a positive root –4 if positive answer

(c) )1(2

)4)(1(4)1()1(2

−−−±−−

2

B1 for )4)(1(4)1(2

−−− or better

If in form r

qp + or

r

qp −

then B1 for –(–1) and 2(1) or better

Brackets and full line may be implied later

–1.56, 2.56 2 B1 B1 If B0, SC1 for –1.6 or –1.562 to

–1.561 and 2.6 or 2.561 to 2.562

10 (a) Dots all correctly placed in Diagram 4 1

(b) Column 4 16, 25, 16, 41

Column 5 25, 41, 20, 61

Column n: n2, 4n, n2 + (n + 1)2 oe

7 B2 or B1 for three correct

B2 or B1 for three correct

B1 B1 B1 oe likely to be (n –1)2 + n2 + 4n or

2n2 + 2n + 1

After any correct answer for column n, apply isw

(c)(i) 79 601 cao 1

(ii) 800 ft 1ft ft their 4n linear expression only

(d) 12 cao 1




Abbreviations

cao correct answer only cso correct solution only dep dependent ft follow through after error isw ignore subsequent working oe or equivalent SC Special Case www without wrong working Qu. Answers Mark Part Mark

1 847 1

2 correct regions shaded 1, 1

3 48 2

B1 for 3 and 16 seen

4 (a) 10 (b) 5.5 oe

1

1

5 (a) 86400 (b) 8.64 × 104

1

1ft

6 108 2 M1 for 33 or 27 or 3

3

1

or 27

1 seen

7 13 3 B1 for 12, 5 seen M1 for (their 12)2 + (their 5)2

or M2 √[(–8 – 4)2 + (1 – 6)2] oe or M1 if √ missing

8 6.70 3 M1 for ( r3 = ) 1260 × π4

3 oe seen

M1 for 3 of their r3 seen or implied

9 22.5 oe 3 B2 180 = 5x +2x + x oe or better B1 for 2x or 6x marked in the correct place on the diagram.

10 x = 13 y = –9

3 M1 for consistent multiplication and addition/subtraction A1 for x = 13 or A1 for y = -9

11 (a) 85.8 (b) 456.8625 cao

2

1

M1 for 23.25 and 19.65 seen

12 (a) (0)8(.)01 (am) (b) 78.4 or 78.38 to 78.39

1

3

Not 8.01pm M2 for 827 ÷ 10.55 or M1 for figs 827 ÷ their time

13 (a) 0.54

(b) 1.61

2

2

M1 for 000010

000207.2 ×

oe

or SC1 for figs 54 in answer SC1 for figs 161 or M1 2002 or 20 0002 seen

http://www.xtremepapers.com



14 –2.64, 1.14 cao with working 4 B1 for ( )( )62432

−− or better seen anywhere

B1 for p = –3 and r = 2 × 2 or better as long as in

the form r

qp + or

r

qp −

After B0B0, SC1 for –2.6 or –2.637(45…) and 1.1 or 1.137(45…)

15 (a) 4

(b) (i) 36

12 oe 0.333

(ii) 36

11, 0.306 or 0.3055 to

0.3056

(c) 15

8 oe 0.533(3…)

1

1

1

1

16 (a) Answer given

(b) k = (±)( )π4

4

−

A or

( )π42

−

A

2

3

M1 (A =)k2 – π2

2

k

E1 A = k2 –4

2kπ

correctly completed to 4A = 4k2 – πk2

M1 factorising (must contain a π) M1 division (by coefficient of k2) M1 square root

17 (a) 66° (b) 33° (c) 123°

2

1

2

M1 for 90° clearly identified as A B1 for OBA or OAB = 57°

18 (a) (i) -r + q or q – r

(ii) ½(3q – r) oe (b) correct working

1

1

3

Must be simplified M1 for MX = ½ r + ¾ their (–r + q)

M1 using a different route for XS or ½ MS E1 dep correct simplification and conclusion

19 (a) 480 (b) 9900

(c) 0.125 or 8

1

1

3

2

M1 for attempt at area under graph M1 for 0.5 × 15 × (their (a) + 14 × 60) oe or 0.5 × 15 × (8 + 14) oe

M1 for numerical vertical/horizontal or numerical use of v = u + at but t ≤ 120 or t ≤ 2

20 (a) (i) 9 (ii) 8x3 cao

(b) 4 www

(c) 2

3+x

1

1

3

2

M1 for (2x – 3)3 = 125 M1 2x – 3 = 5

M1 for x ± 3 = 2y or x = 2

3±y




Abbreviations



dep dependent



oe or equivalent

SC Special Case


Qu. Answers Mark Part Mark

1 53.1 2 B1 C = 36.9 seen, must have C stated or marked

on the diagram

or M1 sinA = 5

4 or tanA = 3

4 but must have

A stated

2 63 + , π 2

–1 for each error or omission

3 Working must be shown 2 M1

9

14and

9

16 M1

8

7

16

14= oe

or visible cancelling

4 0.82 2 M1 conversion of

27

16 (= 0.5(9...)) and

0.82 (= 0.64) to decimals seen

5 (6)€ or euros (with correct working) 2 M1 one of 6 × 1.9037 or 11.5 ÷ 1.9037

or 11.5 ÷ 6 seen

6 3.322 cao 2

B1 3.3219(…) or 3.32(20) seen

7 1.85 × 104 3 B2 18500 oe seen or M1 4x = 74000

or x = 2 × 104 – 1.5 × 103

8 16 3 M1 p = qk

A1 k = 1.6 or 8/5

9 1275, 1425 3 B1 85 or 95 or 0.85 or 0.95

M1 their LB or UB × 1500

where 85 Y LB < 90 90 < UB Y 95

10 (a) (0)700 or 7 am

(b) 1700 or 5 pm

2

1

M1 100 – (5 × their(22 – 6) + their(13 – 8))

or better soi

11

c

bc+4 or b

c+

4 cao

3 M1 correct move completed

M1 second correct move completed

M1 third correct move completed

12 x = 1

y = 0.2 or 5

1 only

3 M1 consistent mult and add/subtraction

A1 one value correct after M awarded

13 (a) 72

(b) 36

(c) 54

1

1

2ft

ft 90 – (b) M1 POQ = 108




14 (a) 84

(b) 15

(c) 6.28

1

1

2

M1 3π2360

120××× oe

15

)5)(1(

31

++

−

xx

x

www 4 M1 (x + 1)2 – x(x + 5) oe B1 x2 + x + x + 1

B1 denominator(s) (x + 1)(x + 5)

or x2 + 6x + 5

16 (a) 2

1 a – 2

1 c oe

(b) 4

3 a + 4

3 c oe

2

2

M1 correct but unsimplified e.g. 2

1 a + – 2

1 c

M1 correct but unsimplified

17 (a) 4x–24 or

24

4

x

(b) 16

2x

2

2

B1 4xn B1 24

x

k or kx–24 for any numerical k, n

B1 k

x2

or B1 16

n

x SC1 (

4

x

)2

18 (a) (6, 1½)

(b) y = – 5

1 x + 4 oe

1

3

B1 correct numerical format B1 correct m

B1 correct c

19 (a) 8

(b) 4x – 9

(c) 22(x + 1) or 22x + 2 or 4x + 1

1

2

2

M1 2(2x – 3) – 3 seen

M1 (2x + 1)2 seen

20 (a) (i)

(ii)

R

(b)

2

2

1

B1 correct line

B1 2 sets of correct arcs

B1 correct line

B1 two sets of correct arcs

correct region, shaded or shown by the letter R

21 (a) (i) ( )0 brackets essential

(ii)

−− 128

1812

(b)

−

−

31

11

2

1

2

2

2

M1 6 × 2 + 3 × –4 or 12 + –12

M1 any 2 × 2 matrix with 2 elements correct

B1

db

ca

2

1 seen

or

B1

−

−

31

11k seen




Abbreviations



dep dependent



oe or equivalent

SC Special Case


Qu. Answers Mark Part Mark

1 2y(x – 2z) 2 B1 for y(2x – 4z) or 2(xy – 2yz)

2 (x =) 3(y – 5) oe final answer 2 M1 for correct first move

y – 5 = 3

x

or 3y = x + 15

M1 for their correct second move

3 (a)

(b)

14

1

1

4 816 cao 2 M1 197.5 and 210.5 seen

5 a any negative integer

n any even (positive) integer

2 B1 for one correct

6 (a)

(b)

1.646 × 107

3.32 × 10–2

1

2

B1 for 0.0332 seen or 3.3 × 10–2 as answer

or B1 for 3.32 × 10k

7 (a)

(b)

36

correct working

1

2

M1 for 6

7 oe improper fraction

M1 for 21

12 = 7

4 oe or visible cancelling

8 (x =) 5 (y =) –1 3 M1 for consistent multiplication and add/subtract

as appropriate

A1 for 1 correct answer

9 127.31 cao 3 M1 for 120 × 1.032

A1 for 127.308

If M0 award SC2 for 7.31 or 247.31

10 120 3 M1 7t + 11(t + 5) = 2215

A1 18t + 55 = 2215

11 500 3 M1 V = kL3 any letters may be used for V, k and L

A1 k = 4




12

d

x−840 or

d

x

d−

840

3 M1 400 × 2.1

M1 “400 × 2.1” – x

13

R

3 Give the mark for R shown in region below

2

3 1

2

2 1 0

14 y = 4x + 1 3 B1 correct numerical y = mx + c

B1 c = 1

B1 m = 4

15 4.94 3 M1 π r2 × 12 = 920

M1 (r2) = )12π(their

920

×

16

)2)(2(

25

+−

−

xx

x

3 M1 2(x + 2) + 3(x – 2) seen

B1 (x – 2)(x + 2) common denom. seen

17 (a)

(b)

4.5(0)

200

1

2

M1 0.53 or 23 seen

18 (a)

(b)

27x9

25x4

2

2

B1 kx9 or 27xn

B1 kx4 or 25xn

19 (a)

(b)

32

37.5

2

2

B1 figs 32 or 1 cm to 2.5 km or 8 000 000 seen

B1 (figs 25)2 seen or figs 375 in answer

20 (a)

(b)

(c)

(d)

35

55

55

125

1

1ft

1ft

1ft

90 – (a) but b > 0

= (b)

180 – (c)

21 96 www 5 M1 32 + 42

A1 5

M1 ½ × 6 × “5” (= 15)

M1 4 × their triangle area + 62




22 (a)

(b) (i)

(ii)

159

50

0.208

3

2

2

M1 evidence of using area under graph

M1 stating area correctly

M1 3 × (1000/60) oe

M1 evidence of numerical rise/run or use of

(v – u)/t




Abbreviations

cao correct answer only cso correct solution only dep dependent ft follow through after error isw ignore subsequent working oe or equivalent SC Special Case www without wrong working art anything rounding to soi seen or implied


1 (a) (i) 625362

1380×

+

1 Allow 115 for 62 + 53

(ii) 7.27 (7.271 to 7.272) 1

(iii) 42 2 M1 for

75

3150oe

(b) (i) 235

3 B2 for angle ACS = 55 or angle ACN = 125 B1 for 55 seen

(ii) 12.6 (12.58 to 12.59)

3

M2 for 9.186

4× or 55cos4244 ××++ or

35sin4244 ××++ oe

(M1 for 6

4 soi or 55cos42 ×× or

35sin42 ×× soi oe)

(c) 1500 3 M2 for 08.01

1380

−

oe

(M1 for recognition that 92% = 1380)




2 (a)

Monday

5

3 ,

5

2 1

Tuesday

7

4 ,

7

3 1

7

5 ,

7

2 1

(b) (i) 35

12 oe cao 2 M1

5

3×7

4 ft their tree

(ii)

35

9 oe cao 2 M1

5

3×7

3 ft their tree

(iii)

35

19 oe 2 ft

ft their (b)(ii) + 35

10 ft their tree throughout (iii)

M1 for 5

2×7

5 + their (b)(ii)

or 7

2

5

2

7

4

5

31 ×−×−

(c)

35

34 oe cao 3 ft their tree throughout (iv)

M2 for 1 –

−=××35

11

4

1

7

2

5

2

(M1 for

=××35

1

4

1

7

2

5

2)

or M2 for4

3

7

2

5

2

7

5

5

2

5

3××+×+

(M1 for any two of these)

3 (a) 3 www 3 M1 for ( )1+

=

m

kp oe A1 for k = 36

or M2 for 4 × 9 = p × 12 oe

(b) (i) (x + 5)(x – 5) 1

(ii)

( ))5(

12

−

+

x

x

final answer

3 B2 for factors ( )( )512 ++ xx or SC2 for final

answer 5

2

1

−

+

x

x

(B1 for ( )( )bxax ++2 where ab = 5 or

2b + a = 11 or SC1 for )5)((2

1++ xx )

(c) x < 7 oe final answer 3 M2 for 8x * 56 where * is inequality or = sign

(B1 for 5x – 20 or 36 − 3x)




4 (a) (i) (cos (HFG)) = 1462

12146222

××

−+ M2 M1 for implicit form

58.4 (58.41…) A2 A1 for 0.5238…

(ii) 0.5 × 6 × 14 × sin (their 58.4) oe 35.8 or 35.77 to 35.78

M1 A1ft

ft their (i) Correct or ft their (i)

(b) (sin (RQP)) = 18

12)117sin( ×

36.4 or 36.44...

M2

A1

M1 for implicit form

5 (a) (i) Correct translation (see diagram) 2 SC1 for translation by

−

k

3 or by

− 2

k

(ii) Correct reflection (see diagram) 2 SC1 for reflection in y = −1

(b) (i) Stretch, (factor) 3, y-axis or x = 0 invariant

1 1 1

(ii) Rotation

90° clockwise

(1, − 1)

1 1 1

Accept −90°

(c) (i)

10

03 ft from (b)(i) 2 ft SC1 for

30

01(ft from (b)(i)) or

10

0k

with k algebraic or numeric but ≠ 1 or 0

(ii) Rotation,

180° Origin

1 1 1

Accept O or (0,0)

6 (a) 23.6 (23.60…) 2 M1 for 142 + 192

(b) 2300 or 2303 to 2304 cao

4

M3 for 2 × ½ × 14 × 19 + 14 × 36 + 19 × 36 + their BC × 36 M2 for 4 of these added M1 for ½ × 14 × 19

(c) 4788 or 4790 cao 2 M1 their triangle area × 36

(d) 43(.0) or 43.04 to 43.05 cao 2 M1 for (their (a))2 + 362 or 362 + 192 + 142

(e) 18.9° to 19.02° cao 3 M2 for inv sin

CEtheir

14 or

inv tan

+ 223619

14 or

inv cos

+

CEtheir

361922

or complete longer

methods (M1 for clearly identifying angle CEA)




7 (a) 1(.00) 4(.00) 11.1(1) 1(.00) 0.25 3 B2 for 4 correct, B1 for 3 correct

(b) 10 points plotted Correct shaped curve through 10 points (condone 2 points slightly missed) 2 separate curves not crossing x-axis and not touching or crossing y-axis

P3 ft

C1 ft

B1

B2 for 8 or 9 points correct ft B1 for 6 or 7 points correct ft ft their points if shape correct – ignore anything between – 0.6 and 0.6 Independent

(c) −0.85 to – 0.75 cao 0.75 to 0.85 cao

1 1

(d) Tangent drawn (ruled) at x = 1.5

– 3 to −2

T1 2

Allow slight daylight Dep on T1 M1 evidence rise/run dependent on tangent SC1 for answer in range 2 to 3 Answer implies M but not the T mark

(e) (i) y = x − 2 oe 1

(ii) line ruled to cross curve 2 ft Dependent on (i) in form y = mx + c, m ≠ 0, c ≠ 0 B1 for gradient ft or y intercept ft but again to cross curve at all possible points

(iii) 2.5 to 2.7 cao 1 Dependent on (e)(i) correct

8 14.2 14 13

3

2 1

M1 for Σ fx (10 × 11 + 8 × 12 + 16 × 13 + 11 × 14 + 7 × 15 + 8 × 16 + 6 × 17 + 9 × 18 ) (1065) (allow one error or omission)

M1dep for ÷ Σf (10 + 8 + 16 + 11 + 7 + 8 + 6 + 9) (75) (allow one further error or omission) M1 for 37th, 37.5th or 38th seen

(b) (i) 21, 30, 15 2 B1 for 2 correct

(ii) 20 20 10 (10) 1.05 1.5 1.5 (0.9)

3 1, 1, 1 for each correct vertical pair

(c) )1.3(1210

43125.210=

++

+×+×

n

n

M2 M1 for either numerator or denominator seen

multiplying across and collecting terms (n =) 8 www 4

M1

A1

dep on linear numerator and denominator

their (68.2 − 25 − 36) = their (4 − 3.1) × n




9 (a) x [=3 y [=2 1, 1

(b) x + y Y 9 1

(c) 6x + 14y Y 84 1

(d) x = 3 y = 2

9=+ yx

Line from (0, 6) to (14, 0) Correct quadrilateral unshaded or clearly indicated

1, 1

2

2

1

Accept clear and freehand lines long enough to define the correct quadrilateral SC1 for line through (0, 9) or (9, 0) B1 for through (0, 6) or (14, 0)

(e) $ 70 2 B1 for considering (7, 2)

10 (a) (A 1) 8 27 64 125 (B 4) 8 12 16 20 (C 4) 9 16 25 36

2 1 2

B1 for 3 correct B1 for 3 correct

(b) 512 169

1 1

(c) 25 99

1 1

(d) 145 n3 + 4n oe 16 (n + 1)2 – 4n oe but isw

1, 1 1, 1

Likely oe is (n – 1)2




Abbreviations



dep dependent



oe or equivalent

SC Special Case


art anything rounding to

soi seen or implied


1 (a) (i) 25

(ii) 15.5 (15.46 to 15.47)

(iii) 0.05 oe

1

1

2

B1 for 1/100 or 0.01 seen

(b) 8812.50 final answer www 3 3 Condone 8812.5

M2 for 7500 × 5 × 0.035 + 7500 oe (implied by

final answers 8810, 8812, 8813 or 8812.5(0)

seen)

or B2 for 1312.5 as final answer

or M1 for 7500 × 5 × 0.035 oe (implied by final

answers 1310, 1312, 1313)

(c) (i) 22 × 3 × 5

(ii) 12

(iii) 240

2

2

2

Allow 2 × 2 × 3 × 5

M1 for any correct product of 3 factors = 60 seen

or correct factor ladder or correct tree

(condone 1’s on tree/ladder)

M1 for 22 × 3 or 2 × 2 × 3 oe

M1 for 24 × 3 × 5 or 2 × 2 × 2 × 2 × 3 × 5 oe

SC2 only for both correct answers (ii) (iii)

reversed

2 (a) 3.02 (3.023…) www 4 4 M3 for 2221.71.52 ++ oe may be in two steps

or 9.15...to9.11 (3.018 to 3.026..)

or M2 for 22 + 1.52 + 1.72 oe implied by 9.11 to

9.15….

or M1 for any correct Pythag in 1 of the faces

e.g. 22 + 1.52

(b) 34.1 to 34.3 cao www 3 3 M2 for sin = 1.7/their EC

or cos = their EG/their EC or tan = 1.7/their EG

or complete long method

(M1 for CEG as required angle – accept on

diagram if clear)

(c) (i) 2.95 cao

(ii) Yes and because their (c)(i) < their

(a)

1

1ft

ft their (a) and their (c)(i), must say yes or no oe

and compare the two distances – numerically or

by labels




3 (a) (i) 142 to 150

(ii) (0)59 to (0)63

(iii) 148o to 152o drawn

Distance 6.8 to 7.2 cm drawn

(iv) 328 to 332o

(v) 60 www 2

2

1

1

1

1

2

B1 for 7.1 to 7.5 seen

Both marks available from the position of B as

lines don’t need to be drawn.

M1 for 202 or better seen

(b) 667 (666.6 to 666.7) www 3 3 B1 for 2.25 (h), 135 (mins), 8100 (sec)

and M1 for 1500 ÷ their time in hours (time

must be in range 2.09 to 3.25)

(could be implied by 697 to 698)

(c) (cos =) 79011252

14507901125222

××

−+

96.9 (96.87 to 96.88) www 4

M2

A2

M1 for

14502 = 11252 + 7902 – 2 × 1125 × 790cosQ

A1 for (cos =) –0.1197…(which implies M2)

4 (a) 4

– 5.8 or – 5.75 or – 5.7

– 2

1

1

1

(b) 10 correct plots ft

Correct shape curve through 10 points

(condone 2 points slightly missed)

Two separate branches not crossing y-axis

P3ft

C1ft

B1

ft from their values in (a) generous with

(– 0.25, 12.1)

P2 for 8 or 9 correct plots ft

or P1 for 6 or 7 correct plots ft

ft their points if shape correct – ignore anything

between – 0.25 and 0.25

C1 and B1 are independent

(c) – 2.5 to – 2.3

– 0.5 to – 0.4

2.75 to 2.9

1

1

1

(d) Correct tangent drawn at x = –2

– 4 to – 2.5

T1

2

Allow slight daylight

Dep on T1

M1 Rise/Tread attempt Dep on T1

or SC1 for answer in range 2.5 to 4 after T1




5 (a) 2, 3, 4, 5 3 M2 for 1 < n ≤ 5 seen (M1 for 1 < n or 5≤n )

Allow 2 6<≤ n in M2 or M1 case

If 0, B2 for 3 correct with no extras or 4 correct

with 1 extra.

(b) (i) 2x(x + 5y)

(ii) 3(a – 2b)(a + 2b)

2

3

B1 for x(2x +10y) or 2(x2 + 5xy)

B2 for (3a – 6b)(a + 2b) or (a – 2b)(3a + 6b)

or correct answer seen in working

or B1 for 3(a2 – 4b2)

If B0, SC1 for )2)(2(22bababa +−=−

(c) (i) ½ x(x + 17) = 84 or 842)17( ×=+xx

Correct proof of x2 + 17x – 168 = 0

(ii) (x – 7)(x + 24)

(iii) 7 and –24 ft

M1

E1

2

1ft

Condone ½ x × x + 17 = 84 but only for M mark

No errors or omission of brackets anywhere

SC1 for (x + a)(x + b) where a and b are integers

and a + b = 17 or ab = – 168

Correct or ft from their factors if quadratic

(d) – 3 www 3 3 B2 for 15 – 6 = x – 4x oe or better

M1 for 15 – x = 2(3 – 2x) or better

or 7½ – x/2 = 3 – 2x

(e) 6245)( 2−××−−

p = – –5 and r = 2 × 2

B1

B1

( 73 )

Dependent onr

qp + or

r

qp −

or ( )24

5−x B1

16

253+ B1

3.39, –0.89 final answers

B1B1

SC1 for 3.4 or 3.386… or 3.39 seen and – 0.9 or

– 0.886… or – 0.89 seen




6 (a) (i) 45 < t Y 55

(ii) 52.6 (52.63….. ) www 3

1

3

Allow any indication e.g. 4th interval

M1 for 6 × 10 + 15 × 27.5 + 19 × 40 + 37 × 50

+ 53 × 62.5 + 20 × 75 (= 7895)

Allow 1 error/omission

and M1 dep for ÷ 150

(b) (i) 40, 77, 130, 150

(ii) Correct scales

6 correct plots ft

Curve or ruled lines through the 6

points

2

S1

P3ft

C1ft

B1 for 2 or 3 correct values

ft from (i) if increasing values.

(35, 21) must be inside square 20 – 22

but (55, 77) may be inside or edge of square



ft their points if increasing

condone graph starting at (20, 6)

(c) (i) 54 to 55

(ii) 18.5 – 22.5

(iii) Their reading at 60 – their reading at

50

1

2

1

B1 for UQ = 62.5 to 65 or LQ = 42.5 to 44 seen

(iv)

150

)2( 50atreadingtheir150 ±− oe 2 SC1 for

150

2)50(atreadingtheir ± oe

(v) If their (iv) is

150

k, then ft their

149

1

150

−

×

kk

2ft In (iv) and (v), condone answers as decimals to

3 sf

Penalise first occurence only of 2sf decimals

isw cancelling/conversion

M1 for 149

1

150

−

×

kk




7 (a) 87.5 (87.45 to 87.52) www 4 4 M1 for ½ × 2.5 × 9.5 soi by 11.875 or 71.25

and M2 for ½ × 2.52 × sin60 × 6 oe (16.23 to

16.24)

or M1 for ½ × 2.52 × sin60 (2.706..)

or 1 trapezium (8.1189..)

(b) 107.9 ….. to 108.0…..www3 3

Must see at least 4 figures

M2 for 360

55× π × 152 or M1 for

360

55 seen

(c) (i) 2.29 (2.291 to 2.293) www 2 2 M1 for 108 = 15πr oe allow 107.9 to 108.0…

for their 108

(ii) 14.8 (14.82 to 14.83) cao www 3 3 M2 for 22

2.29their15 −

(M1 for h2

+ their 2.292

= 152

)

(d) 70.9 to 71.5 cao www 3 3 M2 for 3

π

(their 2.292 × their 14.8 – their 1.1452

× their 7.4) (not 15 or 7.5)

or 8

7 ×

3

π

× their 2.292 × their 14.8

or M1 for 1/8 oe e.g. 3

3

15

5.7 or 7/8 or (½ their R

and ½ their h) seen

8 (a) Correct enlargement

2 B1 for any enlargement of 2 in correct

orientation

(b) (i) Stretch only

y- axis oe invariant

(factor) 4

1

1

1

(ii)

10

04 2ft Ft their factor 4

SC1 for

10

0k k ≠ 0, 1≠ or

40

01ft their

factor 4

(c) Shear only

x-axis oe invariant

(factor) 2

1

1

1




9 (a) (i) 3, 8, 15 in correct positions

(ii) 12

2

3

B1 for 2 correct values in correct positions

M2 for 12 × (12 + 2) (= 168) or 12, (12 + 2)

or M1 for n2 + 2n = 168 then

M1 for (n + a)(n + b) where a and b are integers

and ab = – 168 or a + b = 2 oe

(b) (i) 2 + 3n oe

(ii) 2n – 1 oe

2

2

Allow unsimplified e.g. 5 + 3(n – 1)

B1 for 3n oe seen

B1 for 2k seen

(c) a = 2

1, b = 1

2

1 cao 6 B1 for 12 or 30 seen but if 30 clearly only from

Diagram 4 then B0.

M1 for any 1 of a + b + 1 = 3 oe

8a + 4b + 2 = 12 oe

27a + 9b + 3 = 30 oe

M1 for a 2nd of the above equations

M1 (indep) for correctly eliminating a or b from

pair of linear equations

B1 for one correct value




Abbreviations



dep dependent



oe or equivalent

SC Special Case



soi seen or implied


1 (a) (i) 34.65

(ii) 41.58

(iii) 264

1

2

3

M1 for 0.15 × 277.2 implied by 41.6 or 41.58

seen and not spoiled

M2 for 277.2 ÷ (1 + 0.05) o.e.

or M1 for recognition that 105(%) = 277.20

(b) (i) 1000

(ii) 3650

2

2

M1 for 2200 ÷ (2 + 4 + 5) × 5

M1 for 2200 ÷ 44 × 73

2 (a) (i) Image at (4, –4), (6, –4), (6, –6),

(2, –6)

(ii) Image at (–4, –4), (–4, –6), (–6, –6),

(–6, –2)

(iii) Reflection

y = –x

2

2 ft

1 ft

1 ft

SC1 for reflection in y-axis

SC1 ft if rotated 90° anti-clockwise about (0, 0)

ft their Z (name of transformation)

independent (full details)

(b) (i) Image at (2, 2), (3, 2), (3, 3), (1, 3)

2

SC1 for enlargement s.f. 0.5 with correct

orientation, different centre or sf – 0.5,

centre (0, 0)

(ii)

5.00

05.0 cao 2 B1 B1 each column

(c) (i) Image at (0, 4), (2, 4), (0, 6), (–4, 6) 2 SC1 if 3 vertices correct

(ii)

−

10

11 2 SC1 for

10

1 k, 0≠k but can be algebraic or

numeric or for

− 11

01




3 (a) (x + 5)2 – 2x2 = 1 oe

(x + 5)2 = x2 + 10x + 25 or

x2 + 5x + 5x + 25

x2 + 10x + 25 – 2x2 = 1

0 = x2 – 10x – 24

M1

B1

E1

Equiv means equation in the three parts,

allowing (x + 5)2 expanded

For final line reached without any errors or

omissions after any previous line with (x + 5)2

expanded

(b) 12 3 M2 for (x – 12)(x + 2) or full correct expression

from formula.

Allow SC1 for ))(( bxax ++ and ab = – 24 or

a + b = – 10

then SC1 ft (dependent on quadratic factors or

two roots from formula) for correct selection of

+ve root, if only one +ve.

Answer of 12 and –2 scores M2 only

(c) 53.1 to 53.2 www 3 3 M2 for 2 × )(tan2

11− o.e. i.e. any complete

method

or M1 for tan = 2

1 o.e. i.e. any correct method

leading to any angle in diagram (expressions can

be implicit and bod which angle is being worked

out) (Implied by 26.56 to 26.57 or 26.6, 63.43

to 63.44 or 63.4, 126.8 to 126.9)

53 or 127 without working score 0

4 (a) (8.6.2

986))cos(

222−+

=A M2 M1 for correct implicit equation with cosA

78.58… www 4 A2 A1 for 0.1979 to 0.198 (this implies M2)

(b) (i) 78.6 1 Allow 78.58…

(ii)

)6.78sin(

5.4=r oe M2 (M1 for sin(78.6) =

r

5.4)

Allow 78.58… or their angle BOM for M2 or M1

4.590 to 4.591 cao www 3 A1

(c) 35.5 (35.48 to 35.57…) cao www 4 4 M1 Area triangle = 0.5 × 6 × 8 × sin (78.6) oe

Allow 78.58.. (23.52..)

M1 Circle = 259.4×π Allow 4.590 to 4.591

(66.15 to 66.22…)

M1 (dependent) % = triangle / circle × 100

Dependent on first 2 M’s




5 (a) 9.11, 4.25, 2, …, 2, 4.25, 9.11 3 B2 for 4 or 5 correct and B1 for 2 or 3 correct

(b) 12 points plotted

Smooth curve through 12 points

Two branches, neither touching y-axis

5 P3, ft their (a), P2 for 10 or 11 points, P1 for 8

or 9.

C1 correct shape ft their points shape same.

Ignore anything between – 0.5 and 0.5.

B1 independent

(c) (i) x = 0

(ii) tangent at –1.5

–3 to –1.8

(iii) –1.7 to – 1.55, –0.7 to –0.55,

0.55 to 0.7, 1.55 to 1.7

(iv) y = 2x drawn to meet graph twice

1

1.8 to 1.9

1

T1

2

2

B1

B1

B1

Dependent on tangent

M1(also dep on T1) for attempt at rise/run or

SC1 for 1.8 to 3

B1 for 1 or more correct

6 (a) (i) 5.8

(ii) 4.6 to 4.65

(iii) 2.35 to 2.5

(iv) 172 or 171

1

1

1

2

SC1 for 28 or 29

(b) (i) 72 to 76, 38 to 42

(ii) Their correct Σfx ÷ 200

(iii) p ÷ 2, q, where p, q are from (b)(i)

Histogram with two new columns of

correct width

Two correct heights

2

4

2ft

2ft

Must be integers. B1 either.

M1 for 3 or 4 correct mid-values seen 2, 5, 6.5,

8.5

M1 for Σfx, ft their frequencies and x anywhere

in interval, including boundaries

36 × 2 + (72 to 76) × 5 + (38 to 42) × 6.5 + 50 ×

8.5

M1 for ÷ 200 or their 200 (dependent on second

M1)

(74, 40 give 1127 then 5.635 (or 5.64 or 5.63))

Other pairs of frequencies from (b)(i) must have

a sum of 114 to gain the A mark.

B1 either ft (ft their table)

B1

B1 ft (ft their freq. densities)




7 (a) Correct tree diagram. 5 B1 for labels flower and not flower

First pair B1 for 10

7 and 10

3

B1 for next three branches after flowers

B1 for clear labels for colours

B1 for 3

2 , 4

1 and 12

1 in correct places

If three branches at ends of both branches of first

pair, lose final B, unless probabilities of 0

indicated.

(b) 40

33 o.e. (0.825) cao

3 M2 for 1 – 4

1

10

7× (M1 for

4

1

10

7× or

( )4

1

10

7 1−× ) oe

or

M2 for +10

3

10

7 × 3

2 + 10

7 × their 12

1

or 4

3

10

7

10

3×+ oe

(c) 7 cao 2 M1 for 120 × 10

7 × their 12

1

8 (a) Arc centre D, radius 6 cm 1

(b) (i) Perp bisector of AB, with two pairs

of arcs

(ii) Bisector of angle B, with arcs

2

2

At least 3 cm from AB. SC1 accurate without

arcs or accurate arcs (but no choice)

At least 5 cm from B. SC1 accurate without arcs

or accurate arcs (but no choice)

(c) (i) Q at intersection of loci

(ii) 2.7 cm to 2.9 cm cao

1

1

Dependent on at least both SC1’s

Dependent on (c)(i)

(d) Region inside arc, to left of perp bisector

and below angle bisector

1 Dependent on at least both SC1’s in (b)

9 (a) (i) 81

(ii) 8.5

2

2

B1 for (f(2) =) 7

B1 for (f(0.5) =) 2.5

(b) 3

1−xoe 2 M1 for (

3

1)

−

=

yx or

3

1)(f)(

−

=

x

x

or 13 −= xy or 1)(f3 −= xx

or – 1 then ÷3 in flowchart (must be clear)

(c) 3x2 + 12x + 13 final answer 2 M1 for 3(x + 2)2 + 1 or better

(d) (x =) )1(2

)1)(1(433 2−±−

2 B1 for )1)(1(432 − or better Seen anywhere

If in form r

qp + or

r

qp − oe ,

B1 for p = – 3 and r = 2(1)

or ( )22

3+x B1 then 1

4

9− B1

–2.62, – 0.38 final answer 1,1 If 0, SC1 for –2.6 or – 2.62 or –2.618…

and –0.4(0) or – 0.38 or –0.382 to –0.381 seen

Answers only B1 B1




10 (a) (i) (a) p + q 1

(b) 2

1 p – 2

1 q oe 2 M1 for CMLC + o.e. can be written in terms

of p and/or q

(c) 4

3 p +4

3 q oe cao 2 M1 for LNDLAD ++ o.e can be written in

terms of p and/or q ft their (i)(b)

(ii) AN is a multiple of AC o.e 1 Must be vectors (dependent on answers to (a),

(c))

(b) (i) 30

(ii) 135

2

1ft

M1 for 2x + x + 15 + 75 = 180 or better

ft 165 – their x but only if final answer obtuse

11 (a) (i) 10 1

(ii)

2

43× or

2

)13(3 +× (= 6) 1

(iii) 7260

(iv) 12 840

1

2

M1 for S200 – S120 (20100 – 7260) or

)200121(2

80+ o.e.

(v) 160 400 2 M1 for 2(1 + 2 + 3 + ……… + 400) o.e.

(b) (i) 36, 100

(ii) 11025

1, 1

1

Ignore right-hand column

(iii)

2

2

)1(

+nn

oe 1 isw

(iv) 3 348 900

(v) 32

1

2

M1 for square root then × 2 (1056)

or SC1 for answer 33



IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 2


1

0.5 or 2

1 c.a.o.

1

2 (-)4504 1

Allow (-)4500

3 (a) 121 (b) (n + 1)2

1 1

Allow 49, 64, 81, 100, 121 n2 + 2n + 1

4 3/2500, 1/8, 0.00126 2* M1 for all 3 evaluated as decimals (or fractions or percentages or stand. form)

SC1 reversed order

5 (a) -1, 36

(b) 2 , 30

1 1

Allow –1, S6 SC1 (a) –1 and (b) 36 , 2 , 30

6

I = mr/5

2* M1 for 12)(100

mr240

×

××

o.e.

7

66.7

2 M1 for 1006.3

4.2× o.e.

8 (a) -1

(b) 5k

1 1

9 (a) 32000 (b) 25450 25550

1 1, 1

SC1 both correct and reversed

10 11.5(2) 3* M1 F = kv2 M1 k = 18/402 or better

11 (a) 3110 (b) 322

2*

1 √

M1 for 1936 ÷ 0.623 or 1936 x 1.61 Allow 3107.54, 3107.5, 3108 or 3107.3 SC1 3107

1000000 ÷ (a)

12 (a) 45, 225

(b) 157.5

1, 1 1

Allow 158

13 (a) 5.5 or 5½ (b) 21.5

1 2*

M1 172 ÷ 8

14 (a) )1(

3

+

+

xx

x

(b) -3

3*

1 √

M1 3(x + 1) - 2x M1 denominator x(x + 1)




15 (a) angle bisector of angle P (b) radius from T or U

2*

2*

M1 correct construction method A1S1o

SC1 for accurate line but no arcs M1 radius drawn, meets (a) and O labelled. A1S1o

16 (a) A(2,0) B(0,-6) (b) 6.32 (c) (1,-3)

1, 1 2*

1 √

SC1 correct and reversed M1 (AB2) = “(0 –2)”2 + “(-6 -0)”2 from (a)

17 (a) 20 (b) 98 (c) 62 (d) 124 (e) 36

1 1 1 1

1 √

(b) – (c)

18 (a) 5.8 x 108 (b) 98

(c) 10200

1 2*

2*

M1 figs 58 ÷ figs 59 or figs 9830508

M1 figs 59 ÷ figs 58 x 10n or (b)

1 x 10n

n = 3 or 6

19 (a) -6

(b) (i) 0.4 (ii) (0.4, 0.2)

2

2

1

M1 1 – 2(7/2)

M1 2

5x o.e., 2 - 4x = x or better

20 (a) (i) -2/3p + q (ii) -3/4q + p (b) 1/3p – 1/2q

2* 2*

2*

M1 use of AQ = S2/3p S q or AO + OQ

M1 use of BQ = S3/4q S p or BO + OP M1 -1/4q + 1/3BP

21 (a) 60x + 80y @1200 seen (b) x A y (c) line y = x line through (20,0) and (0,15) shading out or R labelled (d) 20 c.a.o.

1 1 1 2* 1 1

Allow 0.6x + 0.8y @ 12 M1 intention A1 accurate Dep. on both lines Allow 20, 0 or 20 + 0

Total 70





1 (a) 144:96

Final answer 3:2 or 1.5:1 or 1:0.667

B1

B1

(2)

After B0, allow SC1 for reversed “correct” final ans. www2

(b) (i) 32 (children) B1

(ii) 54 (adults off) B1

(iii) 110 (adults on) B1

(iv) 26 (=x) w.w.w. B1

(4)

(c)

)456(

4300

++

×

thier

80 children

M1

A1

(2)

www2

(d) (i) Final Ans. 21 13 or (0)9 13 pm B1 Condone hrs but hrs and minutes ⇒ BO

(ii) 7 h 20 min (o.e)

110

10×

×110

100or

M1 Implied by 6 h 40 min or 400 min

40 min A1

(3)

(11)

www2

2 (a) (i) 1.8(02..) B1 Throughout (a)(i)(ii)(iii) NO misreads allowed.

(ii) 1.99

2 =

3600

80h o.e.

(h =) 178(.2 )

M1

A1

Must be h, not h

ww2 (Must be correct – e.g. 178.4

⇒ MO ww)

(iii) A

2 =

3600

hm

3600A2 = hm

m

A2

3600= h

M1

M1

M1

(6)

(First step must be correct from correct formula for first M1.)

Correctly squares at any stage

Correctly multiplies at any stage

Correctly divides at any stage

Only a correct answer in this form can get M3.

(b) (i) (x + 4) (x – 4) B1 i.s.w. solutions in all (b)

(ii) x(x – 16) B1 Condone loss of final bracket in any (b)

(iii) (x – 8)(x – 1) B2

(4)




(c) (i) x(3x – 9) = 2x2 – 8 o.e. M1

2x2 – 8 = 3x

2 – 9x

x2 – 9x + 8 = 0

E1

No error seen and some working to reach final quoted equation. Must have = 0. (E = established)

(ii) x = 1

x = 8

B1

B1

(iii) time = 15 (sec) c.a.o.

distance = 120 (m) c.a.o.

B1

B1

(6)

(16)

3 (a) (i) 172 + 32

2 – 2.17.32 cos40°

√their 479.54

Answer in range 21.89 to 21.91 (m)

M2

M1

A1

Allow M1 for sign error or correct implicit eqn

Dep M2. NOT for o

40cos225 or

2146

www4

(ii)

9.21

40sin

17

sin

their

o

=

T

M1 or 172 = 32

2 + (their 21.9)

2 – 2.32. (their

21.9) cosT

sinT =

9.21

40sin17

their

o

(0.499) M1

cosT = 21.9)r2.32.(thei

their 222 17)9.21(32 −+

29.9° A1

(7)

Accept 29.93° to 29.94°. www3

(b) (i) 125° c.a.o. B1 All bearings must be 0° Y==θ Y=360° to

score

** (ii) 305° B1√ √ (180° + their 125°) correct

** (iii) 335° or 334.9° B1√

(3)

√ (their 305° + their T) correct

(c)

tan( F̂ ) = 32

30 o.e.

M1

or TXF ˆ = tan

–1 30

32 clearly identified.

°

43.2°

A1

(2)

(12)

(43.15239°) www2 NOT 43.1

4 (a) Scale correct S1 0 Y t Y 7 (14 cm) and 0 – 60 ↑ (12 cm)

8 correct plots (0 , 0), (1 , 25),

(2 , 37.5), (3 , 43.8), (4 , 46.9),

(5 , 48.4), (6 , 49.2), (7 , 49.6)


P3

C1

(5)

Allow P2 for 6 or 7 correct

P1 for 4 or 5 correct

Accuracy better than 2mm horizontally.

In correct square ↑

Not for linear or bad quality




(b) (i) f(8) = 49.8 or 49

128

103 o.e.

B1 Do not accept improper fractions

f(9) = 49.9 or 49

256

231 o.e.

B1

(ii) f(t large) ≈ 50 B1

(3)

(c) (i) Tangent drawn at t = 2 B1 Not a chord and not daylight

Uses vert/horiz using scale M1 Can be given after B0 if line not too far out

** Answer correct for their tangent A1 √

(ii) Acceleration or units B1

(4)

Accept ms–2

, m/s2, m/s/s.

(d) (i) Straight line through (0 , 10)

Straight line gradient 6

B1

B1 Must be ruled and full length to earn B2

**

**

(ii) one √ intersection value for t

Second √ t and range

B1√

B1√

(iii) Distance = area (under curve)

First particle (f(t)) goes further

M1

A1

(6)

(18)

Marking final answers throughout this question

5 (a) (i) 0.2 o.e. B1 Accept 2/10, 1/5, 20%

(ii) 0.4 o.e. B1 After first B0, condone “2 in 10” type answers.

(iii) 0.5 o.e. B1 Never condone 2 : 10 type

(iv) 0.1 o.e. B1

(v) 0 B1

(5)

Accept “none”, “nothing”, 0/10, nil, zero

(b) (i) 2/10 x 1/9 M1

1/45 o.e. A1 Accept 2/90, 0.0222 2.22% www2

(ii) 3/10 x 2/9 M1

1/15 o.e. A1 Accept 6/90 etc, 0.0666(or 7), 6.66 or 6.67% www2

(iii) (their) 1/45 + (their) 1/15 M1


(iv) Clearly 1 – (their) 4.45 o.e. M1 Alternative method must be complete

41/45 A1

(8)

(13)

Accept 82/90 etc, 0.911, 91.1% www2




6 (a) π(30)2 (50) M1

141 000 (cm3) A1

(2)

(141 300 to 141 430) www2

(b) (i) 18 (cm) B1

(ii)

∠AOB

2

1cos = (their 18)/30

x2

M1

M1dep

Allow M1 or M2 at similar stages for other methods e.g. sin A = 18/30 then (180° – 2A)

AOB∠ = 106.26° c.a.o A1

(4)

Must have 2 decimal places seen. ww1 (condone = 106.3 afterwards)

(c) (i) (their)

360

3.106 used

M1

π(30)2 used

834 to 835.3 (cm2)

M1

A1

www3

(ii)

2

1.30.30sin (their) 106.3° or

2

1.48.18

M1

431.8 to 432 (cm2) A1 www2

(iii) Ans. Rounds to 403 cm2 A1

(6)

(d) (i) 50 x (their) 403 M1

** 20 100 to 20 200 (cm3) A1√ √ correct for their “403” www2

** (ii) 20.1 to 20.2 (litres) B1√

(3)

√ their previous answer ÷ 1000

(e)

− )their(d)(itheir(a)

2

1k

M1 k = 1 (cm3) k = .001 (litres) k = other ⇒

consistent conversion error.

50.3 to 51 (litres) A1

(2)

(17)

Marking final answer www2

7 (a) (i) F

− 4

2

M1 A1 M marks for letters, A marks for descriptions. If no letter given, allow SC1 for correct description

(ii) D x = 1 M1 A1

(iii) E (2 , –1) M1 A1

(iv) C (s.f.) 3 M1 A1

(v) A Shear M1 A1

(10)




(b) (–1 –2)

75

31 or QP

M1 Penalty –1 for each wrong one thought possible.

(– 11 –17) final ans A2 Allow SC1 for one correct

(1 2 3)

−

3

2

1

or RS M1

(12) A2

(6)

(16)

Brackets essential here.

Allow SC1 for 12 or –1 + 4 + 9

8 (a) (i) 10 < M Y 15 B1 Must clearly mean this and not 32

(ii) Midpoints 5, 12.5, 17.5, 22.5, 32.5

M1 Allow for 3 or 4 correct

∑ fx (60 + 400 + 490 + 540 + 780) M1 (2270) Needs previous M1 or only marginally out

(their) 2270 ÷ 120 M1 dep previous M1

18.9 (2) (kg)

(1)

A1 www4

(iii) 36° B1

(6)

(b) Horizontal scale 2 cm ≡ 5 units

(numbered or used correctly)

S1 0 Y M Y 40. Accuracy < 2 mm.

If S0 (e.g. 1 cm ≡ 5 units) can score B5

If S0 (e.g. 0, 10, 15) can only score on correct width bars. Penalty –1 for polygon superimposed.

Heights 3k, 16k, 14k, 12k, 4k cm B5 If not scored, decide on their “k” and allow SC1 for each “correct” bar.

(Needs [=2 bars to decide on value of

k if k ≠ 1.)

Their k = 1 B1

(7)

(13)

9 (a) (i) (Diagram) 5 only B1

(ii) (Diagram) 4 only B1

(iii) (Diagram) 2 only B1

(3)




(b) Diagram 1 9 (cm2)

Diagrams 2 and 3 have same area

B1

B1

9.00 to 3 s.f.

One of them

2

1 x 3 x 3

M1

42

1 (cm

2)

A1 www2

Diagram 4

4

1 π3

2 s.o.i.

M1 (7.07 cm2)

2

1 x 6 x 6 – their 9π/4

M1 indep. i.e. 18 – kπ where k numerical

10.9 (cm2) A1 www3

Diagram 5 22

2

1° s.o.i

M1

(bc = 72 )

6 tan22

2

1°

M1 (2.485) (This is AD or DE)

2

1 (6 – their 2.485) x 6

dep.M1or 18 –

2

1 x 6 x their 2.485. (o.e.)

10.5 (cm2) A1

(11)

(14)

www4





1

0.5 or 2

1 c.a.o.

1

2 (-)4504 1

Allow (-)4500

3 (a) 121 (b) (n + 1)2

1 1

Allow 49, 64, 81, 100, 121 n2 + 2n + 1

4 3/2500, 1/8, 0.00126 2* M1 for all 3 evaluated as decimals (or fractions or percentages or stand. form)

SC1 reversed order

5 (a) -1, 36

(b) 2 , 30

1 1

Allow –1, S6 SC1 (a) –1 and (b) 36 , 2 , 30

6

I = mr/5

2* M1 for 12)(100

mr240

×

××

o.e.

7

66.7

2 M1 for 1006.3

4.2× o.e.

8 (a) -1

(b) 5k

1 1

9 (a) 32000 (b) 25450 25550

1 1, 1

SC1 both correct and reversed

10 11.5(2) 3* M1 F = kv2 M1 k = 18/402 or better

11 (a) 3110 (b) 322

2*

1 √

M1 for 1936 ÷ 0.623 or 1936 x 1.61 Allow 3107.54, 3107.5, 3108 or 3107.3 SC1 3107

1000000 ÷ (a)

12 (a) 45, 225

(b) 157.5

1, 1 1

Allow 158

13 (a) 5.5 or 5½ (b) 21.5

1 2*

M1 172 ÷ 8

14 (a) )1(

3

+

+

xx

x

(b) -3

3*

1 √

M1 3(x + 1) - 2x M1 denominator x(x + 1)




15 (a) angle bisector of angle P (b) radius from T or U

2*

2*

M1 correct construction method A1S1o

SC1 for accurate line but no arcs M1 radius drawn, meets (a) and O labelled. A1S1o

16 (a) A(2,0) B(0,-6) (b) 6.32 (c) (1,-3)

1, 1 2*

1 √

SC1 correct and reversed M1 (AB2) = “(0 –2)”2 + “(-6 -0)”2 from (a)

17 (a) 20 (b) 98 (c) 62 (d) 124 (e) 36

1 1 1 1

1 √

(b) – (c)

18 (a) 5.8 x 108 (b) 98

(c) 10200

1 2*

2*

M1 figs 58 ÷ figs 59 or figs 9830508

M1 figs 59 ÷ figs 58 x 10n or (b)

1 x 10n

n = 3 or 6

19 (a) -6

(b) (i) 0.4 (ii) (0.4, 0.2)

2

2

1

M1 1 – 2(7/2)

M1 2

5x o.e., 2 - 4x = x or better

20 (a) (i) -2/3p + q (ii) -3/4q + p (b) 1/3p – 1/2q

2* 2*

2*

M1 use of AQ = S2/3p S q or AO + OQ

M1 use of BQ = S3/4q S p or BO + OP M1 -1/4q + 1/3BP

21 (a) 60x + 80y @1200 seen (b) x A y (c) line y = x line through (20,0) and (0,15) shading out or R labelled (d) 20 c.a.o.

1 1 1 2* 1 1

Allow 0.6x + 0.8y @ 12 M1 intention A1 accurate Dep. on both lines Allow 20, 0 or 20 + 0

Total 70





1 (a) 144:96

Final answer 3:2 or 1.5:1 or 1:0.667

B1

B1

(2)

After B0, allow SC1 for reversed “correct” final ans. www2

(b) (i) 32 (children) B1

(ii) 54 (adults off) B1

(iii) 110 (adults on) B1

(iv) 26 (=x) w.w.w. B1

(4)

(c)

)456(

4300

++

×

thier

80 children

M1

A1

(2)

www2

(d) (i) Final Ans. 21 13 or (0)9 13 pm B1 Condone hrs but hrs and minutes ⇒ BO

(ii) 7 h 20 min (o.e)

110

10×

×110

100or

M1 Implied by 6 h 40 min or 400 min

40 min A1

(3)

(11)

www2

2 (a) (i) 1.8(02..) B1 Throughout (a)(i)(ii)(iii) NO misreads allowed.

(ii) 1.99

2 =

3600

80h o.e.

(h =) 178(.2 )

M1

A1

Must be h, not h

ww2 (Must be correct – e.g. 178.4

⇒ MO ww)

(iii) A

2 =

3600

hm

3600A2 = hm

m

A2

3600= h

M1

M1

M1

(6)

(First step must be correct from correct formula for first M1.)

Correctly squares at any stage

Correctly multiplies at any stage

Correctly divides at any stage

Only a correct answer in this form can get M3.

(b) (i) (x + 4) (x – 4) B1 i.s.w. solutions in all (b)

(ii) x(x – 16) B1 Condone loss of final bracket in any (b)

(iii) (x – 8)(x – 1) B2

(4)




(c) (i) x(3x – 9) = 2x2 – 8 o.e. M1

2x2 – 8 = 3x

2 – 9x

x2 – 9x + 8 = 0

E1

No error seen and some working to reach final quoted equation. Must have = 0. (E = established)

(ii) x = 1

x = 8

B1

B1

(iii) time = 15 (sec) c.a.o.

distance = 120 (m) c.a.o.

B1

B1

(6)

(16)

3 (a) (i) 172 + 32

2 – 2.17.32 cos40°

√their 479.54

Answer in range 21.89 to 21.91 (m)

M2

M1

A1

Allow M1 for sign error or correct implicit eqn

Dep M2. NOT for o

40cos225 or

2146

www4

(ii)

9.21

40sin

17

sin

their

o

=

T

M1 or 172 = 32

2 + (their 21.9)

2 – 2.32. (their

21.9) cosT

sinT =

9.21

40sin17

their

o

(0.499) M1

cosT = 21.9)r2.32.(thei

their 222 17)9.21(32 −+

29.9° A1

(7)

Accept 29.93° to 29.94°. www3

(b) (i) 125° c.a.o. B1 All bearings must be 0° Y==θ Y=360° to

score

** (ii) 305° B1√ √ (180° + their 125°) correct

** (iii) 335° or 334.9° B1√

(3)

√ (their 305° + their T) correct

(c)

tan( F̂ ) = 32

30 o.e.

M1

or TXF ˆ = tan

–1 30

32 clearly identified.

°

43.2°

A1

(2)

(12)

(43.15239°) www2 NOT 43.1

4 (a) Scale correct S1 0 Y t Y 7 (14 cm) and 0 – 60 ↑ (12 cm)

8 correct plots (0 , 0), (1 , 25),

(2 , 37.5), (3 , 43.8), (4 , 46.9),

(5 , 48.4), (6 , 49.2), (7 , 49.6)


P3

C1

(5)

Allow P2 for 6 or 7 correct

P1 for 4 or 5 correct

Accuracy better than 2mm horizontally.

In correct square ↑

Not for linear or bad quality




(b) (i) f(8) = 49.8 or 49

128

103 o.e.

B1 Do not accept improper fractions

f(9) = 49.9 or 49

256

231 o.e.

B1

(ii) f(t large) ≈ 50 B1

(3)

(c) (i) Tangent drawn at t = 2 B1 Not a chord and not daylight

Uses vert/horiz using scale M1 Can be given after B0 if line not too far out

** Answer correct for their tangent A1 √

(ii) Acceleration or units B1

(4)

Accept ms–2

, m/s2, m/s/s.

(d) (i) Straight line through (0 , 10)

Straight line gradient 6

B1

B1 Must be ruled and full length to earn B2

**

**

(ii) one √ intersection value for t

Second √ t and range

B1√

B1√

(iii) Distance = area (under curve)

First particle (f(t)) goes further

M1

A1

(6)

(18)

Marking final answers throughout this question

5 (a) (i) 0.2 o.e. B1 Accept 2/10, 1/5, 20%

(ii) 0.4 o.e. B1 After first B0, condone “2 in 10” type answers.

(iii) 0.5 o.e. B1 Never condone 2 : 10 type

(iv) 0.1 o.e. B1

(v) 0 B1

(5)

Accept “none”, “nothing”, 0/10, nil, zero

(b) (i) 2/10 x 1/9 M1

1/45 o.e. A1 Accept 2/90, 0.0222 2.22% www2

(ii) 3/10 x 2/9 M1


(iii) (their) 1/45 + (their) 1/15 M1


(iv) Clearly 1 – (their) 4.45 o.e. M1 Alternative method must be complete

41/45 A1

(8)

(13)

Accept 82/90 etc, 0.911, 91.1% www2




6 (a) π(30)2 (50) M1

141 000 (cm3) A1

(2)

(141 300 to 141 430) www2

(b) (i) 18 (cm) B1

(ii)

∠AOB

2

1cos = (their 18)/30

x2

M1

M1dep

Allow M1 or M2 at similar stages for other methods e.g. sin A = 18/30 then (180° – 2A)

AOB∠ = 106.26° c.a.o A1

(4)

Must have 2 decimal places seen. ww1 (condone = 106.3 afterwards)

(c) (i) (their)

360

3.106 used

M1

π(30)2 used

834 to 835.3 (cm2)

M1

A1

www3

(ii)

2

1.30.30sin (their) 106.3° or

2

1.48.18

M1

431.8 to 432 (cm2) A1 www2

(iii) Ans. Rounds to 403 cm2 A1

(6)

(d) (i) 50 x (their) 403 M1

** 20 100 to 20 200 (cm3) A1√ √ correct for their “403” www2

** (ii) 20.1 to 20.2 (litres) B1√

(3)

√ their previous answer ÷ 1000

(e)

− )their(d)(itheir(a)

2

1k

M1 k = 1 (cm3) k = .001 (litres) k = other ⇒

consistent conversion error.

50.3 to 51 (litres) A1

(2)

(17)

Marking final answer www2

7 (a) (i) F

− 4

2

M1 A1 M marks for letters, A marks for descriptions. If no letter given, allow SC1 for correct description

(ii) D x = 1 M1 A1

(iii) E (2 , –1) M1 A1

(iv) C (s.f.) 3 M1 A1

(v) A Shear M1 A1

(10)




(b) (–1 –2)

75

31 or QP

M1 Penalty –1 for each wrong one thought possible.

(– 11 –17) final ans A2 Allow SC1 for one correct

(1 2 3)

−

3

2

1

or RS M1

(12) A2

(6)

(16)

Brackets essential here.

Allow SC1 for 12 or –1 + 4 + 9

8 (a) (i) 10 < M Y 15 B1 Must clearly mean this and not 32

(ii) Midpoints 5, 12.5, 17.5, 22.5, 32.5

M1 Allow for 3 or 4 correct

∑ fx (60 + 400 + 490 + 540 + 780) M1 (2270) Needs previous M1 or only marginally out

(their) 2270 ÷ 120 M1 dep previous M1

18.9 (2) (kg)

(1)

A1 www4

(iii) 36° B1

(6)

(b) Horizontal scale 2 cm ≡ 5 units

(numbered or used correctly)

S1 0 Y M Y 40. Accuracy < 2 mm.

If S0 (e.g. 1 cm ≡ 5 units) can score B5

If S0 (e.g. 0, 10, 15) can only score on correct width bars. Penalty –1 for polygon superimposed.

Heights 3k, 16k, 14k, 12k, 4k cm B5 If not scored, decide on their “k” and allow SC1 for each “correct” bar.

(Needs [=2 bars to decide on value of

k if k ≠ 1.)

Their k = 1 B1

(7)

(13)

9 (a) (i) (Diagram) 5 only B1

(ii) (Diagram) 4 only B1

(iii) (Diagram) 2 only B1

(3)




(b) Diagram 1 9 (cm2)

Diagrams 2 and 3 have same area

B1

B1

9.00 to 3 s.f.

One of them

2

1 x 3 x 3

M1

42

1 (cm

2)

A1 www2

Diagram 4

4

1 π3

2 s.o.i.

M1 (7.07 cm2)

2

1 x 6 x 6 – their 9π/4

M1 indep. i.e. 18 – kπ where k numerical

10.9 (cm2) A1 www3

Diagram 5 22

2

1° s.o.i

M1

(bc = 72 )

6 tan22

2

1°

M1 (2.485) (This is AD or DE)

2

1 (6 – their 2.485) x 6

dep.M1or 18 –

2

1 x 6 x their 2.485. (o.e.)

10.5 (cm2) A1

(11)

(14)

www4





1

15 1

2 550.6 2 M1 for 550, 551, 550.59, 550.5 seen SC1 87.7 only

3 (a) √16 or 65/13

(b) π or √14

1 1

Allow 4 or 5 Not 22/7

4 x > -4 or -4 < x 2*

M1 -4 seen on answer line or M1 correct movement of 2 terms

5 14 2* M1 correct movement of 2 terms

6 (a) (-)96 (b) 0

2*

1

B1 answers in the range 90 to 100 or 1.5 to 1.7

7 3200 3* M1 R = kv2 M1 k = 2 Note 2400 scores M0

8 3.23 2* M1

124figs

4figs or

124figs

124figs128figs −

Note 3.13 is M0

9 (a) 71 (b) 7n + 1 (c) 37

1 1

1√

Allow 7 x n + 1 Their part (b) = 260 correctly solved

10 766 3* M1 sin 12 = h/864 M1 586 + their “180” cos78 or sine rule

11 (a) B A 41 43 45 48 47 49 46 40 42 44 (b) 2

2*

1√

B1 One region correct The numbers must be completely inside the correct region Count the numbers in the region between A and B Not 45, 49

12 c =

3

5b2+

3* M1 for a correct operation M1 for a second correct operation





13 (a) 115 125 (b) 2400

1

2√

M1 their 1252 – their 1152

14 (a) (-1, 0) (1, -4) (b) -1 < x < 1

1, 1 1

Allow in words provided ± 1 clearly excluded

15 (a) (b) (c)

1*

2*

1

M1 for complete arc radius 5cm ± 1mm M1 for perp. bisector of CD M1 construction arcs B1 shading to the right of (a) and above (b), can be scored if parts (a) and (b) are incomplete but there must be only 4 boundaries

16 P = 54o

q = 51o r = 78o

1

1√

2√

105 – p r = 180 – 2q M1 for use of 180 – 2q

17 3.10 or -7.10

4 M1 for 42 -4 x 1 x -22 or better

M1 for 2

**4 ±−

A1 A1 SCA1 3.09 and -7.09 or 3.1 and -7.1

18 (a)

−

−−

114

87

(b)

220

022

(c) 22

1

− 51

24 o.e

2

2

2

B1 any 2 correct B1 either column correct M1 either adjoint matrix correct or determinant 22 seen

19 (a) 180 (b) 37.7

3*

2*

M1 for 2 x π x 35 oe

M1 dep for 400 - π x 70

M1 for 2 π (41 – 35) or 2 π 41 + (a) – 400





20 (a)

24

11 24

14 24

10

(b) (i) 300

91 o.e (= 0.303)

(ii) 150

77 o.e (= 0.513)

2

2*

2*

B1 any 2 correct and ISW

M1 25

14 x

24

13 only

M1 for adding their R x Y and Y x R probabilities

21 (a) vector lines drawn (b) (5, 1) (c) 5.83

1, 1

1, 1

2*

AB ends at (4,6) BC horizontal 4 units long SC2 for (1, 5) if B is at (6, 4) and C is at (6, 8)

M1 √(32 + 52)

TOTAL 70




1 (a) 15 : 13 or 13 : 15 B1 Allow n : 1, or 1 :n, where n is 15/13, 13/15, 1.15 (3 or 4), 0.866 (6 or 7)

(b) 0.28 × 45 000 o.e. 12 600

M1 A1

(c) 100

00039

00016× o.e.

41.0 or better

M1

A1

Condone 41 41.0 ( 2 or 3)

(d)

25.2

00045o.e.

20 000

M1

A1

SC1 for 36 000

(e) 00084

30

5× o.e

14 000

M1

A1

Their attempt at 45 000 + 39 000 and their ‘30’

[9]

2 (a)(i) p = 12 q = 1.5 r = 1.2

B1 B1 B1

If not labelled, mark in order given

(ii) Scales correct 12 correct points plotted within 1 mm Smooth curve through all points

S1

P3√

C1

To 11 horizontally and 12 √ vertically are possible

P2√ for 10 or 11 correct.

P1√ for 8 or 9 correct. Within ½ small square, none ruled, correct shape.

(iii) Tangent drawn at (3, 3)

Attempts xinincrease

yinincreasefor their

tangent –0.6 to –1.0 www

T1

M1

A1

Allow a parallel line below curve, slight chord, but not an intended chord

dep. on T1. If no working must fit tangent acc (0.1) for 1 cm horizontally

If correct method shown allow answer in range even with slight slip.

(b) Correct straight line ruled and complete for range 0 to 8

B2 B1 for any straight ruled line with y-intercept 8 (except y = 8) or gradient –1

(c)(i) x

x

−=

+

81

12

12 = 8x + 8 – x2 – x o.e. seen x2 – 7x + 4 = 0

M1

E1

Must be seen to expand the brackets correctly

(ii) x = 0.5, 0.6, 0.7 or 0.8

B1

Must be correct for their graph (1 mm)




or 6.2, 6.3, 6.4 or 6.5 B1 B1 maximum for use of formula to get 6.4 and 0.6 unless convinced it is a check. Coordinates get B0

[17]

3 (a) π × 402 × 110 552 600 to 553 000

M1 A1

or 0.553 m3

(b) 1.6 × 14

146.1

)(

×

atheir

6 hours 51 minutes

M1

M1

A2

(22.4) Accept alternate methods Dep. correct answer (24 687.5 secs = 411..mins) A1 for 411..mins or 6.85 to 6.86 hrs

After A0, SC1 for ÷ 3 600 s.o.i. (6 hrs 52 mins)

(c) 70 × 1002

their (a) ÷ (70 × 1002) 8 www

M1 M1 A2

Dep. could be 0.553 ÷ 70 After A0, SC1 for digits 78…, 79 or 8(0)

[10]

4 (a) Correct scales Correct triangle

S1 T1

From –8 to 8 for x and y (Acc is 2 mm)

(b) A1(–7, 5) B1 (–4, 5) C1 (–4, 7) TR2√ SC1 for any translation

(c) A2 (2, –4) B2 (5, –4) C2 (5, –6) R2√ SC1√ for reflection in x = –1 or y = 1

(d) A3 (–2, 4) B3 (4, 4) C3 (4, 8) E2√ SC1 for enlargement SF2 or correct ray method but o.o.r.

(e)(i) A4 (–2, –2) B4 (–2, –5) C4 (–4, –5)

B2√ SC1√ for 2 correct points

(ii) Reflection only in line y = –x o.e.

B1 B1

with no extras

(f)(i) A5 (3, 2) B5 (7.5, 2) C5 (7.5, 4) B2√ SC1√ for 2 correct points Or stretch factor 1.5 with x-axis invariant A5 (2, 3) B5 (5, 3) C5 (5, 6)

(ii)

10

05.1 B2 SC1 for a correct column in correct

position [16]

5 (a)(i) (cosA = )

70402

457040222

××

−+

(0.7991)

37

M2

E1

4 475/ 5 600 M1 for correct implicit form. Accept complete alternate methods.

Accept 36.9–37

(ii) 14 to 14.1

0.5 × 40 × 70 × sin36.9 – 37 o.e. 841.3 to 843 www

B1 M2

A1

Allow complete alternative methods ww3




(b)(i) 70sin51 o.e. ( = 54.4) E2 M1 for

70

p= sin51 o.e.

(ii)

70

q= cos51 o.e.

44.1 or better

M1

A1

Alt. method – Pythagoras’ ww2 (44.0524…)

(c) angle D = 94

(BD = ) 86180sin

54sin45

−

a.r.t. 36.5 c.s.o

B1

M2

A1

M1 for 86180sin

45

54sin −

=

BD

ww4

[15]

6 (a)(i) ( ) ( )125.234

427241030

x

x

+

++++or

61 + 4x = 2.125 (34 + x) o.e. 6

M2

M1

A1

M1 for (0 or 3) + 10 + 24 + 27 + 4x.

Dep. –deals with the fraction correctly www4 or T and I gets 4

(ii) 1 strict f.t. B1√ 1 for x ≤ 18, 2 for 19 ≤ x ≤ 66 If no answer in (i) accept 1

(b)(i) (a) 21 (b) 30

B1 B1

(ii) 1.4 B2 M1 for 42 ÷30 or 1 cm2 = 5 seen

(iii) ( )''

..'.'..'.'.

128

454252721522251530510 ++++

27.57 to 27.6 c.s.o

M2

A1

(3 530 for ∑fx) f.t. values 21 and 30 from (b)(i) Allow 1 slip in figures for M2 M1 for 4 of mid values 5, 15, 22.5, 27.5, 45 or method correct but mid

values up to ± 0.5.

If 0 scored, SC1√ for ‘128’ seen [12]

7 (a)(i) 5 B1

(ii) x2 – 2x – 3 (= 0) x = –1 and 3 (–1, 0) and (3, 0)

M1

A1

A1

Implied by correct factors or use of formula If A0, SC1 for (x – 3) (x + 1) or

( )1.2

31.4222

−−−±

(iii) (1, –4) B2 Or clear 1 and –4 in correct order B1 for either correct value

(b)(i) Reflection in x-axis or turns upside down o.e.

B1

(ii) Correct statement referring to (0, 0) as minimum value

B1

Accept correct sketches in both cases

(c)(i) 0 B1




(ii) 32a + 3b = 0 and 42a + 4b = 8 o.e. Attempts to eliminate a’s or b’s a = 2 b = –6

M1

M1

A1 A1

e.g. accept equates coefficients (2 out of 3 terms) and attempts to subtract their equations www4

[13]

8 (a)(i) 32.2 B1

(ii) 550 B1

(iii) (a) 2 × 9.2 + 1.6 × 8 o.e. 31.2 (b) 8.7 or better

M1 A1

B1√

If 0 scored SC1 for answer 3120

Their 31.2 ÷ 3.6 correctly evaluated 2 s.f. (or better) (8.6 r), accept correct fraction

(b)(i) figs 395 ÷ 25

× 100 indep 15.8

M1 M1 A1

Implied by figs 158 www www3

(ii) figs 128 × 252 80 000 www

M1 A1

Ignore subsequent unit conversions

(iii) figs 250 ÷ 253

× 1000 indep 16

M1 M1 A1

Implied by figs 16 www3

[13]

9 (a)(i) 2 – 3x = 7 – x o.e. –2.5 o.e.

M1 A1

e.g. 5/–2

(ii) Correct first step of rearrangement

3

x2 −

o.e.

M1

A1

e.g. y – 2 = 3x o.e. or division by 3 or (2 – y)/3 SC1 for inverse of 7 – x (from f(x) = 7 – x)

(iii) 26 www B3 B1 for gf(2) = 16 www and B1 for fg(2) = –10 www in correct order.

(iv) 2 – 3x2 B1 Final answer

(b)(i) 4 B1

(ii) 27

1− B1 Accept 1/–27

(iii) 7.57.5

3.65 to 3.66 × 106

M1 A1

Implied by figs 36.. to 37..

or 3.7 × 106

(iv) Square root of a negative number o.e.

B1 Must make reference to square root or square

(v) 5

B1

[14]




10 (a)(i) Reasonable rhombus sketchedRhombus

1 1

(ii) Reasonable kite sketched Kite

1 1

If (i) and (ii) reversed give SC2 if completely correct otherwise

(b) 2x 180 – 2x o.e.

1 1

Ignore repeats but not choice Ignore repeats but not choice

(c) 0.5. × 12 × 20 o.e. 120

M1 A1

(d) Uses Pythagoras’ or considers a correct triangle/rhombus area equation with variables defined 13 www

M1

A2

Equation f.t. from (c) Accept algebraic Pythagoras’ A1 for 10 and 24 as length of diagonals soi e.g. by 5 and 12 as shorter lengths of right-angled triangle. Implies M1 if no working shown ww3

[11]



IGCSE – NOVEMBER 2005 0580/0581 2


1 210 1

2 2

1 or 0.5 1

3 (0) 2 M1 12 + 12 – 24 Ignore brackets around 0

4 (a) –1.8 1

(b) 21 1

5 $10 2* M1

10012

35800

×

××

6 (a) 0.82 1 Allow 0.64

(b) 0.8–1 1 Allow 1.25

7 (a) 3.16(227766) 1

(b) 0 1

8 ca2

12

1− 2* M1 any answer or working simplifying to

2

1a –

2

1c

9 (a) 2380 1

(b) 2381.60 1

(c) 2400 1

10 5.7 x 1026 3* M1 x 95 A1 5.7 B1 1026

11 23 3* M1

360

90x 4 x 2π x 1.2 A1 23.2 – 23.6

B1 round down

12 ( )3 52 −c or 3√(2c – 10) 3* B1, B1, B1 for each completed correct operation

13 7.5 3* M1 F =

2d

k A1 k = 480 B2 30 x 42 = F x 82

14 (a) 7a(c + 2) 1

(b) 6ax(2x2 + 3a2) 2 B1 any 2 factors removed correctly



IGCSE – NOVEMBER 2005 0580/0581 2

15 (a) 54 1

(b) 42 1

(c) 78 1

16

7

4−>x or x > – 0.571 (428571...)

3* M1 any 2 operations correct M1 any 2 more operations correct

17 (a) 72 2* M1 360 ÷ 5 or 180 – 540/5

(b) 36 1 ft 2

1 (a)

18 (a) x18/9 2 B1 B1

(b) 2x 2 B1 kx or B1 2xk where k is number

19 (a) 2

1− or –0.5 2* M1 5/10 or –ve

(b) y = 2

1− x + 5 o.e. 2* ft M1 for y = (a) x + c or y = mx + 5

20 (a) 80.6 2* M1 for area 2

1 x 3 x 12.4 + 5 x 12.4

(b) 7 2* M1 19.4 or 100 – (a) M1 for 12.4 + v

21 (a) 6.93 2* M1 AE2 = 82 – 42 o.e.

(b) 60.6 3* M1 2 x (a) x 8 M1 subtract π x 42

22 (a)

2

2

M1 all construction arcs A1 line accurate ±1o or B1 accurate but no construction seen

M1 all construction arcs A1 line accurate ±1mm or B1 accurate but no construction seen

ignore additional lines

(b) arc radius 7 cm 1 ignore arc continuing outside the triangle

(c) shading 1 below (a)(i) and left of (a)(ii) and right of arc

23 (a) 3.6 2* M1 for 5x = 18 or x – 18 /5 = 0

(b) –0.3, –11.7 www 4* M1 for √132 M1 for

2

12 k±− A1 –0.3 A1 –11.7

www

NOTE SC2 for correct answers and no working

Completing the square scores M1 √33

M1 –6 ±d

TOTAL 70



IGCSE – NOVEMBER 2005 0580/0581 4

1 (a) 1216 B1

(b) 1.47 B1

(c) 100

5.11

75.95.11×

−

M1

15.2 A1 ww2 SC1 for 17.9

(d) 74347 ÷ o.e. M1

621 A1 ww2

(e) 9.04347 ÷ o.e. M1

4830 A1 ww2

(f)(i)

25.3

2350 o.e.

M1 Must deal with the minutes correctly

723 to 723.1 A1 ww2

(ii) 200.9 to 201 A1ft their (i) ÷ 3.6 r.o.t. to 3sf or better

[11]

2 (a) Correct Scales S1 Accuracy 2 mm throughout question. From –8 to 8 for x and y possible.

(b) Correct triangle ABC T1

(c) (i) Correct translation with vertices at (5, –7), (8, –7), (8, –5)

TR2ft SC1ft for any translation

(ii) Correct reflection with vertices at (–4, 2), (–7, 2), (–7, 4)

FR2ft SC1ft for two points correct or reflection in x = 1 or y = –1

(iii) Correct rotation with vertices at (–2, –2), (–5, –2), (–5, –4)

RN2ft SC1ft for 2 points correct

(d) (i) Correct image drawn with vertices at (3, 2), (7.5, 2), (7.5, 4)

B3 B2 for 3 correct points shown in working. B1 for 2 correct vertices s.o.i.

(ii)

5.10

01

15

1 o.e.

B2 SC1 for

5.1

1 or

5.10

01

(iii) Stretch

y-axis invariant o.e.

factor 3

2

B1

B1

B1

[16]



IGCSE – NOVEMBER 2005 0580/0581 4

3 (a) (i) 60 B1

(ii) ( ) 60cos1572157222

××−+=RS M2 M1 – if one error in formula

13 A2 A1 for ( ) 1692

=RS www4

(b) (i) 145 B1

(ii)

14

55sin

15

sin=

Q o.e.

M1

14

55sin15sin =Q

M1 Implies previous method

61.4 A1 www3

(iii) (R =) 63.6 B1

( )

55sin

'6.63sin'14=PQ

M1 their sin(180 – 55 – b(ii)). Could be explicit equivalent cosine rule

15.3 A1 www3

(c) 55sin'.3.15.'15.2

1'60sin'15.72

1 + M2 M1 for one correct triangle area in working (45.466 + 93.998)

139 or 140 www A2 A1 for 139.4 to 139.5 www4

[16]

4 (a) (i) 12 B1

(ii) 3 B1

(iii) 21 B1

(iv) 2 B1

(v) 24

14 o.e B1 Accept probabilities as fractions/decimals/%

(vi) 19

12 o.e. B1

(b) (i) 21

1122

12× M1

462

132 o.e. (0.286) A1 2/7 in simplest form www2

(ii) 21

1222

10× M1

their 221

1222

10×× o.e. M1

462

240 o.e.(0.519) A1 40/77 in simplest form www3

[11]



IGCSE – NOVEMBER 2005 0580/0581 4

5 (a) 0.9 or better B1 (0.8888..)

–10.1 or better B1 –10.1111..)

(b) (i) Correct scales S1 –3 to 3 for x, and –11 to 2 for y possible

(ii) 12 points correctly plotted P3ft P2ft for 10 or 11 correct (acc. is 1 mm)

P1ft for 8 or 9 correct

both branches with correct shape C1ft Acc. 2

1 small square, correct shape, not ruled

Graph does not cross the y-axis B1

(c) Any integer [ 1 B1

(d) Correct ruled line from –3 to +3 B2 SC1 for line with gradient of 2 or passing through (0, –5) but not y = –5.

(e) (i) –0.45 to –0.3 B1

0.4 to 0.49 B1

2.9 to 2.99 B1

(ii) x2 – 1 = 2x3 – 5x2 M1 i.e. correct multiplication to remove fraction

2x3 – 6x2 + 1 = 0 A1 www2

(f) (i) Tangent drawn with gradient 2≈ B1 Parallel by eye to y = 2x – 5

(ii) Linear eqn. in x and y with gradient 2 B1

c = their intercept B1 within 1 mm, dep on linear eqn in x and y

[19]

6 (a) 2 B1

(b) 3563

1××× o.e. M1

30 A1

(c) Isos. triangle or invtan ( )3

3 o.e. M1

45 A1 www2

(d) ( ) 2256 +=BD o.e. M1

BDBF2

1= M1 Dep. (BF = 3.905….)

angle = invtan

BFtheir

3

M1 Dep on previous method

37.5 to 37.54 A1 www4

(e) (l2) = 32 + (their FB)2 o.e. M1 Not for FB = 3

4.92 to 4.93 A1 ww2

[11]



IGCSE – NOVEMBER 2005 0580/0581 4

7 (a) (i) ( ) 351.15.22

1 ×+ o.e. M1

63 A1

(ii) their (a) x 24 M1

1512 A1ft

(iii) 1512000 B1ft their (a)(ii) x 1000

(b) (i) 35.03 x 24 x 2.25 M1

1891.62… A1 www2

(ii) 1900 B1ft their b(i) rounded to nearest 100

(c) (i) 145.122××π M1

6870 or better A1 (6872.2339 or 6873.125 (π = 3.142))

(ii) [their (a)(ii) ÷ their (c)(i)] M1

x 1 000 000 A1 o.e. e.g. using litres

÷ (60 x 60 x 24) M1 Implied by 2.54

2 days 13 hours A1 www4

[14]

8 (a) (i) x

40 B1

(ii) 1

40

2

40−=

+ xx

o.e. M2

SC1 for 2

40

+x

seen

40x = 40(x + 2) – x(x + 2) o.e. M1 Correctly removes the fraction

40x = 40x + 80 – x2 – 2x

x2 + 2x – 80 = 0 E1 Correct conclusion – no errors

(iii) –10 B1

8 B1

(iv) 8 B1ft their positive x dep on one of each sign

(b) (i) m = n + 2.55 o.e. B1

2m = 5n o.e. B1

(ii) 2(n + 2.55) = 5n M1 f.t. their linear equations in n and m any correct method to an equation in one variable

m = 4.25 A1

n = 1.7 A1

[13]



IGCSE – NOVEMBER 2005 0580/0581 4

9 (a) 160 < h Y 170 B1

(b) (i) Mid values 125, 135, 145, 155, 165, 175, 185, 195

M1 Allow 1 slip

(15 x 125 + 24 x 135 + 36 x 145 + 45 x 155 + 50 x 165 + 43 x 175 + 37 x 185 + 20 x 195)

M1 Dep on mid values ± 0.5, allow 1 slip in mid-values (43830)

÷270 M1 Dep on previous method

162 or better A1 (162.333..) www4

(ii) Mid-values are an estimate of each interval o.e.

B1 e.g. exact values not given

(c) p = 15, q = 39, r = 75 B2 B1 for 2 correct. If no labels, take in order given

(d) Correct scales S1

9 points correctly plotted ft P3ft P2ft for 7 or 8 correct acc. 1 mm P1ft for 5 or 6 correct

Curve or line through 9 points C1ft Dep on ‘S’ shape within 2

1 small square of

points

(e) (i) 162 to 164 B1

(ii) 176 to 178 B1

(iii) 28 to 30 B1

(iv) 167.5 to 168.5 B1

(f) Uses 240 or 241 on cumul, freq. axis M1 e.g. annotates graph or shows values in working

186.5 to 188 A1 ww2

[19]



IGCSE – NOVEMBER 2005 0580/0581 4



IGCSE - OCT/NOV 2006 0580/0581 2



IGCSE - OCT/NOV 2006 0580/0581 2

3

5,

3

5

−

−− xxor

3

5 y−



IGCSE - OCT/NOV 2006 0580/0581 04



IGCSE – October/November 2007 0580/0581 02


1 (a) 4.25957(744…) or 4.25958 1 Must have at least 6 figures correct (b) 4.3 1√ Correct answer or ft from (a)

2 5 x 104 or 50000 2* M1 3.6 x 1013 / 7.2 x 108

or M1 8.33…x 10-3 x 6 x 106

3 (a) 4 cao 1 (b) 0 1 Allow zero or none or no symmetry

4 x2 , cos x°, x-1 2* W1 reverse order Numerical values not allowed in answer space 5 2 2* M1 25c/35 or 125c/ 175 or 25c = 50 or 125c = 250 or 875c = 1750 oe 6 (a) 0.003 x 3000 cao 1 No extra zeros allowed. Accept standard form (10 + 20)2

(b) 0.01 or 1/100 1 SC1 for answer 0 if 0 is used for 0.003 in (a) 7 x = 2 y = -6 cao 3* M1 consistent x and + for x

or consistent x and - for y A1A1

8 (a) 0.701 cao 1 Allow 0.70, of course, if 0.701 seen in working (b) (-)190 2* M1 14020 – 20000 x 0.6915 or reversed 9 p = 2 q = -12 3* M1 x2 + 2px + p2 (+q) or (x + 2)2 – 4 – 8 A1 A1 If no marks scored give SC1 for p = 2 in answer

10 170 provided that 22 is not used 3* M1 ½ x π x (12 or 6)2 M1 ½ x π x 122 - ½ x π x 62 7 SC2 54π or SC1 π x 122 - π x 62 seen allow 452… - 113……

11 100 3* M1 M = kr3 A1 k = 0.8 M1 kM = r3 A1 k = 1.25

12 (a) ø 1

(b) ξ 1 No brackets allowed. Not ε or e (c) A 1 No brackets allowed

13 28.2 28.6 exact values only 3* M1 two of 6.05, 6.15, 8.05 or 8.15 seen A1 28.2 or 28.6 in either answer space SC2 both correct reversed





14 (a) 13.5 2* M1 2x = 27 or x = 4.5 or x – 27 = 0 3 2

(b) -1 and 4 cao 2* M1 (x – 4)(x + 1) or 3 ± √(32 – 4 x 1x -4) or 3 ± √(25/4) 2 2

15 C 1 Any clear indication

(b) ½ a + ⅓ b oe 1 Fractions need not be cancelled (c) - 1 a + 2 b oe 1,1 Mark coefficients of a and b independently 2 3

16 (a) 2* M1 connecting volumes A1 cube root of volumes

or M1 cubing A1 connecting volumes (b) 1.12 2* M1 82 or Area sf = 64

17 √ ((6/T)2 -1) or √ (36/T2 -1) oe 4* W1 each of the first 3 completed correct operations ignore ± …

18 (a) - 4 or 4 cao 1 Note that a fraction is required 5 -5

(b) y = - 4 x (+0) oe forms 1√ y = (a)x allow decimal or unsimplified fraction 5 (c) y = - 4 x + 3.4 oe 2 W1√ y = (a) x + c or y = (b)x + c W1 3.4 5 Allow 17/5 oe

19 (a) 3.365 to 3.375 1 Inclusive (b) 0.26 to 0.27 2* M1 3.52 and 3.25 to 3.26 seen (even on diagram) (c) 55, 56 or 57 1 20 (a) 65 All answers cao 1 If answer space is blank check diagram (b) 25 unless √ applied 1√ 90 – (a)

(c) 103 1√ 168 – (a) (d) 206 1√ 2(c) 21 (a) 3x2 1,1 W1 for 3 and ind W1 for x2 must be single term (b) - 6 2* M1 1/64 SC1 2-6 in answer space 22 (a) 0 0 cao 4* W2 for 4 correct or W1 for 2or 3correct of 3 4 in

0 0 W1 2 4 in 2A 2 3 A2 2 2 1 0

(b) I 1 …… allow 0 1

TOTAL 70




1 (a) (i) 385 × 0.9 oe M1 Implied by ans 346 or 347 ($) 346.5(0) cao A1 www2 (ii) 385 ÷ 1.1(0) oe

($) 350 cao M1

A1

www2

(b) (i)

×

+1923

23210 oe

M1

115 cao A1 www2 (ii) their (i) 50.2× + (210 – their (i)) × 1.50

($) 430 cao

M1

A1

(287.5 + 142.5) www2

(iii)

{their (ii) – 410} / 410 ( 100× )oe

4.88

M1

A1

Dep on (ii) being greater than 410 www2 (4.878 …) After M0, SC1 for 104.9 or better or 4.9 ww

(c) 2.6(210 – x) or 1.4(210 – x) seen M1 2.6(210 – x) + 1.4x = 480 M1 Allow 2.6x + 1.4 (210 – x ) = 480 546 – 480 = 2.6x – 1.4x or 2.6x – 1.4x = 480 – 294 M1 Dep on M2 55 cao A1 if trial and error, B4 or B0

if using simultaneous equations

210=+ yx M1

4806.24.1 =+ yx M1

variable eliminated by correct method M1d After 0 scored, SC2 for ans 155 [14]

2 (a) (i) 6 B1

(ii) 4.5 B1

(iii) +×+×+×+×+× 5447342211(

)7268 ×+× (127)

M1 Allow 1 slip

28÷

4.54

M1dep

A1

dep on 1st M1 www 3 4.53571…

(iv)

27

3

28

4×

M1 Accept all probabilities as fracts/dec/%

-1 once for words or 2 sf, do not accept

ratios i.s. cancelling after correct answer.

63

1 o.e.

A1 www2 e.g. (756

12 , 0.0159 etc)

(v)

20

3

21

4×

M1

35

1 o.e.

A1 www2 e.g. (420

12 , 0.0286 etc)

(vi)

26

4

27

23

28

24××

M1

819

92 o.e.

A1 www2 e.g. (19656

2208 , 0.112)

(b) (i) 0.08 o.e. B1

(ii) 0.9 × 0.05 their (b)(i) + 0.9 × 0.05

0.125 o.e.

M1

M1dep

A1

dep on 1st M1 www3

(iii) 7 B1 ft their (ii) × 56 either correct to 3sf or better or r.o.t. [16]




3 (a) (i) (0, 1) B1 Accept w/out brackets/ commas, condone (ii) (4, 0) and (0, 4) B1B1 vectors, or states x = , y = (b) -1 cao B1 (c) (x) < 0 (allow ≤ ) B1 Any other variable < 0 B0 (d)

xx −=+ 412

o.e. B1 must be these 4 terms

(e) p +(-)√q where p = −1 and r = 2×1 M1

r and q = 1² − 4(1)(-3) o.e. M1

Allow second mark if in form p± r

q

-2.30 , 1.30 cao www4

A1A1

If ww ans.correct but wrong acc - SC3 After A0, A0, SC1 for -2.3027756 and 1.3027756 rounded or truncated

(f) (-0.5, 4.5 or 4.49) B1ft f.t (their –2.30 + their 1.30) ÷2 B1 ft ft (4 – their x co-ord dep on attempt at mid

value of x from values in e) [12]

4 (a) (i) 42

5.3π = 153.86 to 153.96 or 154 M1A1 www2

(ii) 3

3

4 5.3π = 179.5 to 179. 62 or 180 M1A1 www2

(iii) their (ii)× 5.6 1005 to 1006 or 1008or 1010 (g)

M1

A1ft

their (ii)× 5.6 correct to 3sf or better (allow in kg)

(b) 88

2×π (1608-1609) M1 Alt d

28π = 2 × their (ii) M1

h2

8π = 2× their (ii) + 882×π M1dep (2× their (a)(ii)) )8( 2

π÷ M1dep

(2× their (ii) + 882×π ) ÷ (

28π ) M1dep add 8 M1dep

9.78 to 9.79 (cm) A1 www4 (c) 1000 (or 1) ÷4.8 π

3

4÷ M1 49.7….. (or 0.0497)

3ans (or 10 × 3 ans ) M1dep Dep on previous M1

3.67 to 3.68 (cm) A1 www3 figs 368 or ans 3.7 gets M2 [13]




5 (a) (i) =−

2247 5.74 (cm)

M1A1 www2 5.74456…

(ii) 6.32 (cm) B1 6.32455….. (b) 2× ×

2

1 8× ’5.74’+2× ×2

1 6× ’6.32’ + 8×6 M1

131.8 to 132(cm2) A1ft www2 ft 48 +8× their (a)(i) + 6× their (a)(ii)

(c) (i) ((PX)² ) = (their (a)(i))² - 3² M1 or their a(ii)² - 4² or 7² - (3² + 4²) 24 soi or 4.898..… seen E1

(ii) Tan(PNX) =

4

))(( ictheir o.e.

M1 Alt correct trig methods involving their (a)(ii) M1 for correct explicit statement

50.7 to 50.84 oe A1 www2 for a trig ratio (iii) (HPN) 180 – 2 × their (ii) M1 78.3 to 79 A1 www2 Alt – cos rule method – M1 at

explicit stage (iv)

tan = 5

))(( ictheir o.e.

M2 M1 for recognition of angle PAX or PAC oe

44.4 to 44.43° A1 Alt trig methods with PA = 7 used www3 44.4153086

(v) PHN or PGM o.e. (letters) B1 B0 if extras [15]

6 (a) (i) AB=13 cm and BD=15 cm (± 2 mm) B1 Angle A = 80° (± 2°) B1 A,B,C,D correct within 4 mm B1 Dep. on B2 (ii) Angle ADB correct (57-61°) (± 2°) B1ft Either in working or written on diagram

Angle DCB correct (101-105°) (± 2°) B1ft

(iii) Acc. bisector of angle A with arcs B2ft B1 for accurate without/wrong arcs (at least 5 cm long) (± 2°)(± 2 mm) (iv) Acc. perp. bisector of AD with at least 1 B2ft B1 for accurate without/wrong arcs pair of arcs (± 2°)(± 2 mm) (at least 5 cm

long) B1 for each if accurate with arcs but short

(v) ‘Correct’ area shaded below their perp. B1 Dep. on at least B1 in (iii) and B1 in (iv) bisector and below their angle bisector (b) (i)

30

80sin

26

sin=

D

M1 No M marks in (b) for measuring + using lengths from diagram e.g. AD = 20 m but allow 13, 15, 9 used for 26, 30, 18 in b

30

80sin26)(sin =D

M1dep dep on 1st M

58.57 to 58.6° A1 www3

(ii) Angle BDC = 41.4 B1 ft Ft 100 – their 58.6 (BC

2 =)182 + 302 – 2 '4.41cos'3018×× M1 Allow 41 or 42 for angle BDC

square root of correct collection M1dep Dep on 1st M (413.88… ) 20.3 to 20.35 (m) cao A1 www4 (iii)

'4.41sin'30185.0

'4.41sin'30265.0

××

+××

oe M2 M1 for correct area of one triangle

(257.9 or 178.6). Must see calc for trapezium height if used (30sin ‘41.4’) Allow 41 or 42 for angle BDC

436 to 437 (m2) cao A1 www3 [20]




7 (a) Correct axes S1 must fit on paper 2mm acc throughout Ignore labels on triangles throughout

(b) Correct triangle drawn (T) T1 vertices at (8, 6), (6, 10) and (10, 12) (c) (i) Correct reflection in y = x drawn (P) P2ft ft their T, P1 for two correct vertices drawn

(6, 8), (10, 6), (12, 10) or line y = x correctly drawn (within 2mm of (12,12) if extended)

(ii)

01

10

B2 B1 for a correct column

(d) (i) Correct enlargement, scale factor 0.5,

centre (0,0) drawn (Q) Q2ft (4, 3), (3, 5), (5, 6)

Q1 for any enlargement s.f. ½ or 2 correct vertices drawn SC1 for 3 points within 5 mm if rays method used or for correct enlargement but of P

(ii) Enlargement only (scale factor) 0.5

(centre) (0, 0) o.e.

B1

B1

B1

indep indep

(e) Correct stretch drawn (R) R2ft R1 for two correct vertices ft

(4, 6), (3, 10), (5, 12) [13]

8 (a) 2 B1 (b)

112

3+

−x

M1

12

123

−

−+

x

x

M1 Dep on 1st M1

12x

2x2

−

+o.e. final ans

A1 www3

(c)

13+=

xy

1

3+=

yx

xy

31 =− or xy = 3 + x

M1 Alt

yx

31 =−

3)1( =−yx M1dep Dep on 1st M1 3)1( =−xy

1x

3

−

o.e. final answer A1

www3 1

3

−x

o.e

If answer is x =

1

3

−x

allow M2

(d) 256 B2 B1 for 23 = 8 or

82 seen

(e)

13

2

724

+

−

=x

M1 M for r.h.s. followed by attempt at

recognising 2x = …………………

-3 A1 After M0, SC1 for 1/8 o.e seen www2 [11]




9 (a) -7, 512, 9

8 , 81, 2187, -2106 B6 B1 each. Allow in any order ignore letters

(b) (i) (P) 9 – 2n B1 Accept correct expressions in any form

e.g. 7 – 2(n – 1) (ii) (Q) n3

B1 If ‘n =’ withhold the first mark earned (iii)

(R) 1n

n

+

B1

(iv) (S) (n + 1)2 B1

(v) (T) 1n

3−

B1

(vi) (U) (n + 1)2 -

1n3

−

B1ft their (iv)-their (v) dep on both algebraic expressions

(c) their(b)(i) = – 777 M1 393 cao A1 www2

(d) 12 B2 SC1 for 11 or n - 1 = 11 or

11123,3 seen [16]




Abbreviations


1 (a) 2

(b) 0

1

1

Allow none oe

2 a = 3

b = 4

2

W1 one correct

If no marks scored M1 (3 × 2)(2 × 4) oe

3 1.59(459…) or 59/37 or 37

221 2 M1

37

22 or 0.5945… seen

4 (a) 2.67 × 10–2

(b) 0.0267(00…)

1

1ft

cao – must be correct notation

correct or ft

5 Correct locus 2 M1 arc through D radius BD

A1 some indication that the arc is from D to D′

6 60

120

2

W1 one correct Allow 60.00.. or 120.00..

or if W0, SC1 the angles add up to 180°

7 50.1225 cao 2 M1 6.15 and 8.15 seen

8 x2(a + b)

(±) ((p2 + d2)/(a + b))

1

2

M1 2 moves completed correctly

9 (a) y = 2x – 4

(b) (2, 0)

2

1ft

W1 2x + c or W1 mx – 4

For y = 2x + k only, allow (–k/2, 0)

10 x = 8 y = 5 3 M1 ×2 and add or ×3 and subtract

A1

11

)3)(32(

18

−+

−

xx

oe 3 W1 denominator correct in answer space (including

any brackets)

M1 4(x – 3) –2(2x + 3) A1 –18

12 x > –0.16 or –0.16 < x

or x > 25

4−

3 M1 2 moves completed correctly


Final mark must be given for answer line

M1 p = k/(q + 2)2

or p(q + 2)2 = k

A1 k = 125

M1 p = (k/(q + 2))2

A1 k2 = 125 or

k = 125

13 1.25 3

If no marks awarded

SC1 5 : k/25 in this form

p : k/100 (colon optional)

or SC1 for either

5 = k/(3 + 2)2 or 5 = k/52

Allow 5/4

14 (a) 45498 or 4.5498 × 104 cao

(b) 7240

2

2

M1 2.656 × 109 ÷ 58376

M1 )(2

r=

π

(a)

maigna





15 (a) 0.5 or 2

1

(b) –1 or –1.(00) cao www

(c) 2

4cos −x

oe

1

2

2

M1 cos180

M1 subtracting 4 and then dividing by 2 seen

e.g. 2

4−x or

2

4−y or

2

4)(f −x

16 (a) 1000 1400 1960 2744 3842

(2740) (3840)

(b)

(c) 3.2 or 3.3

2

2

1ft

W1 three correct 3 sf answers or better

P1ft 4 or 5 plots correct or ft from their table

C1 smooth curve cao

To half a small square

If a curve and a line are drawn mark the curve

cao or ft from their (b)

17 (a) (i) –3p – 2q

(ii) –3p + 4q

(iii) –4p

(b) 8

1

1

2

1

allow –(3p + 2q)

allow –(3p – 4q)

M1 (ii) – (p + 4q) or BC – AC = BA

or (ii) – p – 4q

18 (a) 1.05

(b) 3360

(c) 18.7

2

3

1ft

M1 clear attempt at y–step/x–step

M1 attempting the area under the graph

W1 2

21)180140( ×+

May be done by triangles and rectangles

(b) / 180 evaluated correctly

(a) 53.4

3

M1 50/360 × π ×122 or 30/360 × π ×62

M1 50/360 × π ×122 – 30/360 × π ×62

19

(b) 49.6 3 M1 50/360 × 2 × π ×12 or 30/360 × 2 × π × 6

M1 12 + 6 + 12 + 6 + both their arcs

20 (a) 600x + 1200y ≥ 720000

(b) x + y ≤ 900

(c)

(d) 300

1

1

4

1ft

seen

W1 drawing x + y = 900

W1 drawing x + 2y = 1200

W1 R is below x + y = 900

W1 R is above x + 2y = 1200

The lines must be in the right place

Accurate to one small square

Correct or ft from their labelled R,

accuracy ± 10 on the lowest y value in R

70

maigna





Abbreviations


1 (a) 2

(b) 0

1

1

Allow none oe

2 a = 4

b = 3

2

W1 one correct

If no marks scored M1 (4 × 2)(2 × 3) oe

3 1.59(459…) or 59/37 or 37

221 2 M1

37

22 or 0.5945… seen

4 (a) 3.85 × 10–2

(b) 0.0385(00…)

1

1ft

cao – must be correct notation

correct or ft

5 Correct locus 2 M1 arc through D radius BD

A1 some indication that the arc is from D to D′

6 45

135

2

W1 one correct Allow 45.. or 135.00..

or if W0, SC1 the angles add up to 180°

7 15.8025 cao 2 M1 2.45 and 6.45 seen

8 x2(a + b)

(±) ((p2 + d2)/(a + b))

1

2

M1 2 moves completed correctly

9 (a) y = 2x – 6

(b) (3, 0)

2

1ft

W1 2x + c or W1 mx – 6

For y = 2x + k only, allow (–k/2, 0)

10 x = 5 y = 2 3 M1 ×4, ×3 and add or ×3 and subtract

A1

11

)32)(15(

17

−+

−

xx

oe 3 W1 denominator correct in answer space (including

any brackets)

M1 5(2x – 3) –2(5x + 1) A1 –17

12 x > –0.16 or –0.16 < x

or x > 25

4−

3 M1 2 moves completed correctly


Final mark must be given for answer line

M1 p = k/(q + 2)2

or p(q + 2)2 = k

A1 k = 64

M1 p = (k/(q + 2))2

A1 k2 = 64 or

k = 8

13 0.64

25

16

3

If no marks awarded

SC1 4 : k/16 in this form

p : k/100 (colon optional)

or SC1 for either

4 = k/(2 + 2)2 or 4 = k/42

14 (a) 45498 or 4.5498 × 104 cao

(b) 7240

2

2

M1 2.656 × 109 ÷ 58376

M1 )(2

r=

π

(a)

maigna





15 (a) 1

(b) 0

(c) 2

6tan −x

oe

1

2

2

M1 tan180

M1 subtracting 6 and then dividing by 2 seen

e.g. 2

6−x

or 2

6−y or

2

6)(f −x

16 (a) 1000 1400 1960 2744 3842

(2740) (3840)

(b)

(c) 3.2 or 3.3

2

2

1ft

W1 three correct 3 sf answers or better

P1ft 4 or 5 plots correct or ft from their table

C1 smooth curve cao

To half a small square

If a curve and a line are drawn mark the curve

cao or ft from their (b)

17 (a) (i) –3p – q

(ii) –4p + 2q

(iii) –5p

(b) 10

1

1

2

1

allow –(3p + q)

allow –(4p – 2q) or –2(2p – q) or 2(q – 2p)

M1 (ii) – (p + 2q) or BC – AC = BA

or (ii) – p – 2q

18 (a) 1.05

(b) 3360

(c) 18.7

2

3

1ft

M1 clear attempt at y–step/x–step

M1 attempting the area under the graph

W1 2

21)180140( ×+

May be done by triangles and rectangles

(b) / 180 evaluated correctly

(a) 37.1

3

M1 50/360 × π × 102 or 30/360 × π × 52

M1 50/360 × π × 102 – 30/360 × π × 52

19

(b) 41.3 3 M1 50/360 × 2 × π × 10 or 30/360 × 2 × π × 5

M1 10 + 5 + 10 + 5 + both their arcs

20 (a) 600x + 1200y ≥ 720000

(b) x + y ≤ 900

(c)

(d) 300

1

1

4

1ft

seen

W1 drawing x + y = 900

W1 drawing x + 2y = 1200

W1 R is below x + y = 900

W1 R is above x + 2y = 1200

The lines must be in the right place

Accurate to one small square

Correct or ft from their labelled R,

accuracy ± 10 on the lowest y value in R

70

maigna





1 (a) (i) ($) 6 000 cao B2

M1 for 0.1 × 10 000 + 0.25 × 20 000 oe

(ii) 15 (%) cao B2

M1 for 40000

(a)(i)their× 100

(b) ($) 11 200 ft B1 ft ft 17200 – their (a)(i)

(c) (i) ($) 7500 cao B2 M1 for

35

12000

+

× 5 oe

After M0, SC1 for 4500

(ii) 9/80 cao B1

Ignore decimals or %’s seen Mark final fraction

(d) ($) 8640 cao B2 M1 for 10 800 ÷ 1.25 oe

[10]

2 (a) (i) x(x + 4) / 2 = 48 oe

x2 + 4x – 96 = 0

M1

E1

Eqn must include 48 Dep on M1 + shows one intermediate algebraic step with no errors seen

(ii) – 12 or 8 B1B1 Allow deletion of negative root

(iii) 12 (cm) correct or ft B1ft Accept 12 or ft their positive root in part (ii) (if only one) + 4.

(b) 54 oe B2 M1 for

4+x

x

= 6

1 oe

(c) (i) (x + 4)2 + x2 = 92 oe or x2 + 8x + 16 + x2 = 81 2x2 + 8x – 65 = 0

M1

E1

Accept 2nd line for M1 or 2x2 + 8x +16 = 81 Dep on M1 with no errors, expanded brackets step needed

(ii) p +(-)√q where p = −8 and r = 2 × 2

r and q = 8² − 4(2)(–65) oe (584) – 8.04, 4.04 cao www

M1

M1

A1A1

Allow second mark if in form p± r

q

SC2 if correct solutions but no working shown or SC1 for –8.041522987 and 4.041522987 rounded or truncated

(iii) 21.08 or 21.1 (cm) strict ft B1ft

dep

ft 4.04 in part (ii) or 2 × a positive root + 13 [14]




3 (a) 5.(04), 0(.0), 8.7 or 8.66(6…) or better seen

B3 1 each

(b) Correct axes for domain and range 10 correct points, on correct grid line or within correct 2mm square vertically Reasonable curve through 10 points condone curvature around x = –0.2 and 0.2 Two separate branches

S1

P3ft

C1ft

B1ft

P2ft for 8 or 9 correct P1ft for 6 or 7 correct Correct shape, not ruled, within 1 mm of points (curves could be joined) Independent but needs two ‘curves’ on either side of y-axis

(c) (i) y = –3x ruled correctly

–2.95 to – 2.6, – 0.75 to – 0.6, 0.5 to 0.6

L1

B2

Check at (–1, 3) to (1, –3) within1 mm (can be shorter) B1 for 2 correct. isw y – values No penalty for each extra value if curve is cut more than 3 times

(ii) (a =) 3 (b =) –1 B1B1 After 0,0 SC1 for 01323

=−+ xx

(d) Tangent to their curve ruled at x = –2 rise/run using correct scales

–4.5 to –3

T1

M1

A1

Must be a reasonable tangent allow slight daylight <1mm Dep on T1 (implied by answer 3 to 4.5) Must show working if answer out of range

[17]

4 (a) 72 B1

(b) (i) 0.5 × 15 × 15 sin (their 72) oe 106.9 to 107 (cm2) cso

M1

A1

not for 90° www2

(ii) 534.5 to 535 (cm2) ft B1 ft ft their (i) × 5

(iii) π × 152 × 50

their (ii) × 50 Vol of cylinder – prism 8590 – 8625 (cm3) cao

M1

M1

M1

A1

(707 or 35350) or π × 152

(26750) or π × 152 – their (b) (ii)

Dep on M2 then × 50 www4

(c) (AB =) 15sin(their36) × 2 oe (17.63) (not 30° or 45°)

Area of one rectangle = their AB × 50 5 (50 × a length) + 2 × their (b)(ii) 5470 – 5480 (cm2) cao

M1

M1

M1

A1

or )72cos(151521515 22their×××−+

Not for 90° or 60° or sine rule dep on 1st

M (881.5) not 15 × 50 Indep (4407.5 + 1070) www4 [12]




5 (a) (60 + 40)/35 Correct method to convert a decimal time to minutes 14 46 or 2 46 pm cao

M1

M1

A1

(2.857…) could be in parts ft a decimal either full answer or decimal part × 60 (e.g. 51.(428), 171.(4.. )or 2hrs 51 or 51 m) www3

(b) (i) 260 B1

(ii) 145 B1ft ft their (b) (i) – 115

(c) (AC2 = ) 402 + 602 – 2 × 40 × 60 × cos115

(AC=) of a correct combination

85(.0 km) cao

M2

M1

A1

M1 for correct implicit version

dependent (7229)

www4

(d)

(c)their

A 115sin

60

sin= oe

(sinA =) (c)their

115sin × 60

39.76 to 39.8 cao

M1

M1

A1

Implicit equation Could use cosine rule M1 for implicit and M1 for explicit form Dep on M1 Explicit equation www3

(e) 40sin80 + 60sin35 oe (39.4) (34.4)

73.76 – 73.81 (km) cao

M2

A1

their (c) × sin(100 – their (d)) or their (c) × cos (their (d) – 10) M1 for either 40sin80 or 60sin35 or implicit trig version using their (c) www3 [15]

6 (a) (i) 30 B1

(ii) 30, 30.5, 31 B1 B1

B1

Penalty 1 for each extra value Ignore repeated values

(iii)

x

x

++

×+×+×

710

323173010= 30.65

correct clearance of fraction 3 cao

M1

M1

A1

Dep on M1 e.g. 517 + 32x = 521.05 + 30.65x oe www3

(b) (i)

200

27242326211151535 ×+×+×+×

20.93 or 20.9 cao

M3

A1

(4186/200) M1 for use of 15, 21, 23, 27 (allow one error)

and M1 for use of ∑ fx with value of x in

correct range used (allow one further error)

and M1 dep on 2nd M for dividing by ∑ f or

200 www4 Accept 21 after M3 earned

(ii) 2.6 cao

0.7 and 0.8

B1

B4

B3 for one correct or B2 for 3.5 and 4 seen or B1 for 4 seen [16]




7 (a) (i)

Translation only

−11

0oe

B1

B1

Throughout parts (i) to (v) if more than one transformation is given then no marks at all for that part Accept T

(ii) Reflection only x = 1 oe only

B1

B1

Accept M

(iii) Reflection only y = -x oe only

B1

B1

Accept M

(iv) Enlargement only (centre)(2, 0), only (scale factor) 0.5 oe only

B1

B1

B1

Accept E

(v) Stretch only (factor) 2, only x-axis oe invariant cao only

B1

B1

B1

Accept S Ignore parallel to y-axis

(b) (i)

−

−

01

10

B2

B1 each column

(ii)

20

01

B2

B1 for right hand column [16]

8 (a) x = 78 alternate angles either y = 144 or z = 102 (opposite angles of) cyclic quad (= 180) and z = 102 or y = 144 Angles (in (a)) quadrilateral (= 360) or (opp angles of) cyclic quad (= 180)

B1

R1

B1

R1

B1

R1

Dep on B1 Accept Z angle, extras can spoil Accept longer reasons using correct language and clarity with angles used. e.g. allied angles gives 102° and angles on a straight line = 180° Dep on B1, extras can spoil Dep on B1 extras can spoil

(b) Their z + 36 ≠ 180 oe R1 Could also use their angles x and y provided x + y ≠ 180. Could be a longer reason involving angles must be clearly explained.

(c) 72 or 288 B1

(d) 51 cao B1 [9]




9 (a) (p =) 5 cao, (q =) 12 cao (r =) 1 ft

B1

B1

B1ft

Accept in correct order if no labels ft for r = 18 – their p – their q provided r not negative

(b) (i) 17 cao B1

(ii) 12 cao B1

(c) (i) 26 cao B1

(ii) 57 ft B1ft ft 45 + their q

(d) (i)

100

8 oe isw

B1

(ii)

100

45 oe isw

B1

(e) Any fraction with denominator 74 seen

73

36

74

37×

73

18 oe isw cao

B1

M1

A1

ft their fraction i.e. one taken off each part

1

1

−

−

×

l

k

l

k N.B

73

36

2

1× gets B1M1

5402

1332 www3 (if decimal then 0.247 or better)

Do not accept ratio or in words [12]

10 (a) (i)

2

)18(8 +×

= 36

1 + 2 + 3 + …. + 8 = 36

E1

E1

(ii) 80 200 B1

(b) (i) 2 (1 + 2 + 3 + ….. + n) =

2 × 2

)1( +nn

= n (n + 1)

E1

both steps must be shown

(ii) 40 200 B1

(iii) 40 000 B1ft ft their (a)(ii) – their(b)(ii) or their (b)(ii) – 200 ft Not for zero or negative answer

(c) (i)

2

)12(2 +nn

oe final answer

B1 e.g. 2n² + n

(ii) n² cao B2 M1 for their (c)(i) – n(n + 1) or n(n + 1) – n

or n/2(2+2(n-1)) [9]



IGCSE – October/November 2009 0580 21

Qu Answers Mark Part Marks

1 (a) 6 (b) 0

1 1

2 47, 53 2 B1, B1 independent

3 –0.577 or

3

3−

or 3

1−

2 B1 numerator 0.5

or B1 denominator –0.866……or 2

3−

4 1.25 x4 (or 14

1 x4) 2 B1 1.25 B1 x4

5 161 2 M1 1.322 × 109 / 8.2 × 108 (× 100)

6 5

2 M1 |A| = 0 × –4 – 1 × –8 or better or |B| = 7 × –5 – 0 × 1 or better det symbol can be implied by the working

7

2 B1, B1

8 5 www 2 M1 (–4 – –1)2 + (8 – 4)2 or better

9 x = 0.5 y = 3 www 3 M1 consistent × and – for y or consistent × and + for x

A1 one correct provided M1 scored

10 245 3 M1 d = kv2

A1 k = 1/20 or M1 v2 = kd A1 k = 20

11 258 cao 3 M1 18.5 or 24.5 seen M1 6 × sum of their two upper bounds

12 –36x2 + 48x or 12x(4 – 3x) oe or other partly factorised versions

3 M1 squaring to “9x2 –12x + 4” algebraic M1 multiplying by –4 terms M1 adding 16 only

13 x [ 0.8 or x [ 5

4 cao 3 B1 12 – 18x B1 –4 + 8x these terms may be reversed if moved to the other side of the inequality allow >=

14 $11.50 3 M1 198 × r3 r can be anything

dep M1 r = 1.019 and subtracting 198 SC2 209.50 on answer line




15 (a) (i) OQ

(ii) RM or MP

(b)

×S

1 1 2

Allow ½RP B1, B1 correct position wrt each direction of the vector ± 1 mm

16 (a) (0)810 or 8:10 etc. (b) 4 (c) 265

1 2 1

M1 (3 + 3)/(1 + 0.5)

17 (a) 261.48 cao (b) (±)3.86(48…) or 3.865

2 2

M1 4000 / 15.2978 M1 (15.9128 – 15.2978)/15.9128 (× 100) or (“261.48 – 4000/15.9128) / “261.48”

18 m = 2 c = –8 4 B1 B(4, 0) or A(–2, 0) seen or used B1 m = 2

M1 substituting (4, 0) into y = 2x + c or 04

0

−

− c

= 2

19 (a) 44 (b) 158

2 2

M1 OCB = 68

20 (a) 38 (b) 45 to 46 (c) 15 to 16 (d) 10 or 11

1 1 1 2

SC1 70 on answer line

M1 speed/time

21 (a) 0.8 or 4/5 cao

(b) 960 www

2

3 M1 30 × (12 + 36)/2

M1 10 × (12 + 36)/2

M1 12 × 40 M1 ½ × 40 × 24




22 (a) 2 (b) 4x3 + 5

(c) 2

)13( −x

2 2

2

M1 f(0) = 1 M1 4(x3 + 1) + 1

M1 rearranging y = (2x + 1)/3 to make x the subject and interchanging x and y. Allow any one error in the working

70




Qu Answers Mark Part Marks

1 (a) 6 (b) 0

1

1

2 37, 41 2 B1, B1 independent

3 –0.577 or

3

3−

or 3

1−

2 B1 numerator 0.5 oe

or B1 denominator –0.866……or 2

3−

4 1.25 x4 (or 14

1 x4) 2 B1 1.25 B1 x4

5 139 2 M1 1.322 × 109 / 9.5 × 108 (× 100)

6 8 2 M1 |A| = 0 × 12 – 1 × –4 or better or |B| = 3 × –4 – 0 × 4 or better det symbol can be implied by the working

7

2 B1, B1

8 10 www 2 M1 (–2 – –8)2 + (10 – 2)2 or better

9 x = 0.5 y = 3 www 3 M1 consistent × and – for y or consistent × and + for x

A1 one correct provided M1 scored

10 128 3 M1 d = kv2

A1 k = 2/25 (= 0.08) or M1 v2 = kd A1 k = 12.5

11 198 cao 3 M1 12.5 and 20.5 seen M1 6 × sum of their two upper bounds

12 –36x2 + 48x or 12x(4 – 3x) oe or other partly factorised versions

3 M1 squaring to “9x2 –12x + 4” algebraic M1 multiplying by –4 terms M1 adding 16 only

13 x [ 0.8 or x [ 4/5 cao 3 B1 12 – 18x B1 –4 + 8x these terms may be reversed if moved to the other side of the inequality allow >=

14 $12.92 3 M1 249 × r3 r can be anything

dep M1 r = 1.017 and subtracting 249 SC2 261.92 on answer line




15 (a) (i) OQ

(ii) RM or MP

(b)

×S

1

1

2

Allow ½RP B1, B1 correct position wrt each direction of the vector ± 1 mm

16 (a) (0)810 or 8:10 etc. (b) 4 (c) 265

1

2

1

M1 (3 + 3)/(1 + 0.5)

17 (a) 261.48 cao (b) (±)3.86(48…) or 3.865

2

2

M1 4000 / 15.2978 M1 (15.9128 – 15.2978)/15.9128 (×100) oe or (“261.48” – 4000/15.9128) / “261.48”

18 m = 2 c = –10 4 B1 B(5, 0) or A(–4, 0) seen or used B1 m = 2

M1 substituting (5,0) into y = 2x + c or 05

0

−

− c

= 2

19 (a) 44 (b) 158

2

2

M1 OCB = 68

20 (a) 38 (b) 45 to 46 (c) 15 to 16 (d) 10 or 11

1

1

1

2

SC1 70 on answer line

M1 speed/time

21 (a) 0.8 or 4/5 cao

(b) 960 www

2

3 M1 30 × (12 + 36)/2

M1 10 × (12 + 36)/2

M1 12 × 40 M1 ½ × 40 × 24




22 (a) 2 (b) 4x3 + 5

(c) 2

)13( −x

2

2

2

M1 f(0) = 1 M1 4(x3 + 1) + 1

M1 rearranging y = (2x + 1)/3 to make x the subject and interchanging x and y. Allow any one error in the working

70




Abbreviations



dep dependent

ft follow through


oe or equivalent

SC Special Case

soi seen or implied


1 (a) (i) 8.4(0) B2 B1 for 1.2 or 3.6 seen

or SC1 for figs 84 in answer

(ii) 100

20×

(i)their oe

42 ft www2

M1

A1ft

ft their 8.4 × 5

After 0 scored SC1 ft for 58% or

10020

20×

− (i)their correctly given

(b) 6 B2 M1 for 9 or 8 ÷ (1 + 8 + 3) soi

(c) 3

2

4.2× oe (= 3.6 seen)

or their (a) (i) ÷ 7 × 3

912

3× oe (= 2.25 seen)

1.6(0) cao www3

M1

M1

A1

(d)

25.1

40.2 oe

1.92 www2

M1

A1

Implied by figs 192

[11]

2 (a) (i) Reflection (M), x = 1 B1,B1 If extra transformations given in part (a) then

zero scored

(ii) Rotation (R)

180

(centre) (1, 0)

B1

B1

B1

Must be “rotation”.

Allow half turn for 180.

Allow other clear forms of (1, 0)

(iii) Enlargement (E)

(centre) (6, 4)

(scale factor) 3

B1

B1

B1

Must be “enlargement”

Allow other clear forms of (6, 4) e.g. vector

Accept 3 : 1 or 1 : 3

(iv) Shear (H)

y-axis invariant oe

(factor) –1

B1

B1

B1

Must be “shear”

Allow other explanation for invariant but not

“parallel to”

isw after y-axis invariant seen




(b) (i)

−10

01

B2 B1 for correct right–hand column in 2 by 2

matrix

(ii)

− 11

01

ft

B2ft Ft only their factor in (a) (iv) provided not zero

B1ft for left-hand column in 2 by 2 matrix

provided shear factor is not zero

or SC1 for

11

01 if not ft

[15]

3 (a) (i) 1 B1 Penalty of –1 in question if any answers given

as decimals or percentages (to 3sf) alone, but

isw cancelling/conversion after correct answer

(ii) 6

3 oe B1

(b) (i) 30

2 oe www2 B2 M1 for 5

1

6

2×

(ii) 6–12 and 12–6 and 7–11 and 11–7 soi

k × 6

1 × 5

1 for k = integer

30

4 oe www3

M1

M1

A1

Evidence of all pairs adding up to 18 but no

extras e.g. 4/6 × 1/6

Without seeing the first M, 6

4 × 5

1 oe scores

M2, 6

2 × 5

1 oe scores M1

(iii) 5

2

6

4×

30

8 oe www2

M1

A1

(c) 5

2

6

4

6

2×+ oe

30

18 oe cao www2

M1

A1

6

2 + their (b) (iii)

(d) 4 B2 M1 for (1 + 1 + 6 + 7 + 11 + 12 + x) ÷ 7 = 6 or

better

[13]




4 (a) (i) Accurate triangle with 2 arcs seen,

2 mm accuracy for lines AC and BC

B2 SC1 if accurate but no arcs or one arc or if AC

and BC are wrong way round with arcs

(ii) Accurate bisector of angle ACB, 2°

accuracy and both pairs of arcs shown

(accept equidistant marks on edges for 1st

set of arcs) + must meet AB

B2ft Ft their triangle

SC1ft if accurate but no/one pair of arcs or

short with arcs

In both (ii) and (iii) isw

(iii) Accurate perpendicular bisector of AD

2 mm accuracy at mid-point and 2° for

right angle and shows both sets of arcs

+ must meet AC

B2ft ft their D, which must be on AB

SC1ft if accurate but no/one pair of arcs or

short with arcs

(iv) Correct region shaded cao B1 Dependent on correct triangle, accurate

bisectors of angle ACB and side AD with correct

D

(b) (i) (cos C) =

1801402

240180140222

××

−+ oe

– 0.111(1)…or better or 96.37 to 96.38

M2

E1

(–5600/50400 or –14/126)

Allow use of 7, 9 and 12

M1 for correct implicit statement

Verification using 96.4 scores M2 max

Accept 9

1− but not a non-reduced fraction

(ii) 0.5 × 140 × 180 sin (their 96.4) oe

12521 to 12523 or 12 500 or 12520 cao

www2

M1

A1

(s = 280), allow use of 7, 9 (31.3…)

(iii) (Sin B =)

240

)4.96sin(140 their oe

35.4 or 35.42 to 35.44 cao www3

M2

A1

Allow use of 7, 12

M1 for correct implicit statement

SC2 for correct answer by other method

[15]

5 (a) (i) (x + 3)(2x + 5) – x(x + 4) = 59 oe

2x2 + 6x + 5x + 15 –x2 – 4x = 59 oe

x2 + 7x – 44 = 0

M1

A1

E1

Implies M1 (allow 11x for 6x + 5x)

Correct conclusion – no errors or omissions

(ii) (x + 11)(x – 4) B2 SC1 any other (x + a)(x + b) where

a × b = – 44 or a + b = 7

(iii) –11, 4 www ft B1ft Strict ft dep on at least SC1 in (ii)

allow recovery if new working seen

(iv) tan =

5) (2

3) (

++

++

rootvetheir

rootvetheir oe

28.3 (00…) ft www2

M1

A1ft

Could be alt trig method

oe M1 where trig function is explicit

ft one of their positive roots

(27.4° (27.40 – 27.41) from x = 11)




(b) (i)

4

52

+

+

x

x

= x

x 3+ oe

x2 + 4x + 3x + 12 = 2x2 + 5x

x2 – 2x – 12 = 0

M1

A1

E1

Must be seen. Allow ratio or correct products

Correct expansion of brackets seen (allow 7x for

4x + 3x)

Correct conclusion – no errors or omissions M1

must be seen

(ii)

)1(2

)12)(1(4)2()2( 2−−−±−−

or (x – 1)² –12 – 1 (B1)

and x – 1 = ± 13 (B1)

– 2.61, 4.61 final answers www4

B1,B1

B1,B1

In square root B1 for (–2)2 – 4(1)(–12) or better

If in form r

qp + or

r

qp −,

B1 for – (–2) and 2(1) or better

If B0, SC1 for –2.6 and 4.6 or both answers

correct to 2 or more dps rot

– 2.6055…, 4.6055….

(iii) 26.4 (26.42…. to 26.44….) ft B1ft ft 4 × a positive root + 8

[16]

6 (a) (i) –16 B1

(ii) 18 to 19 B1

(b) (i) –4.3 to –4.2, 1.5 to 1.6 B1,B1

(ii) –4.5 to –4.4 , 1.3 to 1.4 B1,B1

(iii) –4.5 to –4.4 < x < 1.3 to 1.4 ft B1ft Ft their (ii). Allow clear worded explanations

and condone Y signs

(c) 7

30− oe isw conversion B2 Accept

7

24− , 30/–7

M1 for 30/7 oe fracts, isw conversion or for

–30/7 oe soi

(d) Ruled line passing within 2 mm of

(–5, 30) and (2, 0)

B2 B1 for ruled line parallel to g(x). By eye (21°

to 25° to horizontal if in doubt) allow broken

line

(e) (i) Ruled horizontal line through (–3, –27) B1 No daylight, not chord (allow broken)

(ii) y = –27 B1

(f) Ruled lines x = –3, x = –2, y = 40

Region enclosed by lines x = –3,

x = –2, y = 40 and y = g(x)

B1

B1

Long enough to be boundary of region – allow

broken or solid ruled lines

Allow any clear indication

[15]




7 (a) (i)

360

60 × π × 2 × 24 oe

25.1 (25.12 to 25.14) www2

M1

A1

Accept 8 π

(ii)

360

60 × π × 242 oe

301 or 302 or 301.4 to 301.7 www2

M1

A1

Accept 96 π

(b) (i) πd = their (a) (i) oe

4 (3.99 – 4.01) cao www2

M1

A1

(ii) 242 – (their radius)2

23.7 (23.66 to 23.67) cao www2

M1

A1

Alt trig method for h explicit

Accept 354,1402,560

(iii) 3

1 × π × (their r)2 × (their h)

394 – 398 cao www2

M1

A1

Not for h = 24

(c) (i) 27W B1

(ii) 4W B1 If B0, B0 in (c), SC1 for 27 and 4 alone

[12]

8 (a) 5.5 < t Y 6 B1 Condone poor notation

(b) 4.25, 4.75, 5.25, 5.75, 6.25, 6.75

(2 × 4.25 + 7 × 4.75 + 8 × 5.25 + 18 × 5.75

+ 10 × 6.25 + 5 × 6.75) (= 283.5)

÷ 50 or their ∑ f

5.67 www4

M1

M1

M1

A1

At least 5 correct mid-values seen

∑ fxwhere x is in the correct interval allow

one further slip

Depend on second method

After M3 allow 5.7

isw conversion to mins/secs and reference to

classes

(c) (i) 17, 15 B1

(ii) Rectangular bars of heights 11.3

and 15

Correct widths of 1.5 and 1 – no gaps

B1ft

B1ft

B1

ft their 17 divided by 1.5

ft their 15

11.3 plot between 11 and 12 include lines and

15 to be touching the 15 line

(iii) 2.5 cao B1 [10]




9 (a) 3(m – 3) + 4(m + 4) = –7 × 12

3m – 9 + 4m + 16 = –84

–13 www4

M2

A1

A1

Allow all over 12 at this stage

M1 for 3(m – 3) + 4(m + 4) seen

Allow all over 12 at this stage

May be seen in stages

(b) (i) 0.5 oe B1

(ii)

)3)(1(

)1(2)3(3

+−

−−+

xx

xx

)3)(1(

11

+−

+

xx

x

final answer

M1

A1

If brackets not seen allow

3x + 9 – 2x ± 2 as numerator with a correct

denominator

isw incorrect expansion of denominator if

correct brackets seen

(iii) 1

)3)(1(

)11(=

+−

+

xx

xx

ft or

x + 11 = x

1 (x – 1)(x + 3) or better ft

x2 + 11x = x2 + 3x – x – 3

3

1− oe cso www3

M1

M1

A1

Must clear one denominator correctly

Ft their (b)(ii) dep on fraction in (ii) with

(x –1)(x +3) oe as denominator

Depend on previous M1

– 0.33(33…)

(c) p(q – 1) = t oe

pq = t + p

p

pt + oe final answer www3

M1

M1

M1

Multiplying by (q – 1)

Ft their first step

e.g. pq only term on one side

Ft their 2nd step

e.g. dividing by p

Note: q – 1 = p

tis M2 and then q =

p

t + 1 is

M1

[13]

10 (a) 21 + 23 + 25 + 27 + 29 = 125

31 + 33 + 35 + 37 + 39 + 41 = 216

B1

B1

(b) Cubes B1

(c) (i) n oe B1

(ii) n3 oe B1

(d) 42 – 4 + 1 = 13 www E1 Allow 16 for 42, otherwise all must be seen

(e) 7 × 43 + 2 + 4 + 6 + 8 + 10 + 12 B1 All must be seen

(f) n(n – 1) final answer oe B1

(g) n(n2 – n + 1) + their (f)

n3 – n2 + n + n2 – n = n3

M1

E1

All must be seen, no errors or omissions

[10]




Abbreviations

cao correct answer only cso correct solution only dep dependent ft follow through after error isw ignore subsequent working oe or equivalent SC Special Case www without wrong working


1 20 (but 3, 4 and 8 must be seen www) 2 M1 3, 4 and 8 seen www

2 1.2496 cao 2 Allow 1625

156

M1 1 + 0.2 + 0.04 + 0.008 + 0.0016

3 2 2 M1 3x – 1 – 3x + 3

4 0.93 0.92 9.0 3 9.0 2 M1 0.94(8683…) 0.96(5489….) 0.8(1) 0.7(29)

5 (a) 5 (b) 2

1 1

6 1.15(2) × 10–2 2 M1 figs 115(2)

7 x

x

2

5 + 2 M1 4 + 1 + x seen

or M1 x

x

4

210 + oe

8 40.5 2 M1 6.75 seen or 6 × their LB

9 $674.92, 674.9(0) or 675 3 M2 600 × (1 + (4/100))3 or better oe or M1 600 × 1.042 oe

10 x = 4 y = –3 3 M1 consistent mult and sub/add A1 one correct value but M must be scored

11 D P C

12 3

A O B

0

2

211

3 Marks allocated for R in one of the regions shown

12 x = +/- √(5y) – 3 or x = +/- y5 – 3

3 M1 correct move of the 5 completed M1 correct move of the square completed M1 correct move of the 3 completed




13 x < –3 3 M1 correct move M1 correct move M1 correct move

14 (a) 10(.0) 1

(b) 2

2

1, 2.5(0) 2 M1 2n – 3 = 2

15 31.4 cao 3 M1 2

1 × 2 × π × 3 oe

M1 6 + 8 + 6 + 1 + 1 + k π

16 2

3

+

−

x

x 4 B2 (x – 3)(x – 2) or B1 (x + a)(x + b)

where ab = 6 or a + b = –5 B1 (x – 2)(x + 2)

17 (a)

80

08 oe 2 B1 for one column (or row) correct

(b)

−

4

1

4

1

4

1

4

1

oe 2 B1 for –1/8

db

ca or B1 for

−

−−22

22 seen

18 (a) (i) Tangent (ii) 4.4 to 6 (b) 780

1 2 2

Correct tangent drawn

dep M1 attempting to find gradient of their tangent

M1 evidence of finding the area under the graph ONLY from t = 12 to t = 25

19 (a) 20200 (b) 1260

2 2

M1 65 × 300 + 700 M1 71190 / 56.5

20 x = 0.84 or 7.16 4 B1 2

8 k± B1 √(82 – 4 × 1 × 6) or better

A1 A1

21 (a) Bisector (b) (4, 2) (c) y = –2x + 10 oe

2 1 3

B1 accurate line B1 two sets of correct arcs

B1 correct m B1 correct c

M1 correct use of y = mx + c oe on answer line

22 (a)

D140

3 2 12L

E

4

B1 0 and 14 in correct place B1 2 in correct place

B1 3 in correct place B1 12 in correct place

(b) 11 (c) 23

1ft

1ft

B1ft 8 + their 3 B1ft 21 + their 2




Abbreviations



dep dependent



oe or equivalent

SC Special Case



1 (a) 5

(b) 0

1

1

2 10 2 M1 33 – 25

or 38 – 30

M1 30 – 15 – 5 oe

with no further working

3 uv

Jm

−

= 2 M1 m(v – u) seen

4 (a) 40

(b) 65

1

1

5 23.6 2 M1 sin R = 20/50 or 90sin

50

sin

20=

R

6 (a) 6.58 × 10–3

(b) 0.0066 cao

1

1

× and 10 essential

Allow 6.6 × 10–3

7 t = 22

1 2 M1 (b)t = (b)(3t – 5)

8 Answer given so only working scores

marks

2 M1 7/27 + 48/27 or 7/27 + (1)21/27

M1 completely correct finish

9 2390

2410

2

M1 119.5 and 120.5

or B1 for one correct answer

10 60 3 B1 540 used

M1 [their 540 – 3 × 140]/2

11 128 3

M1 R = kv2

A1 k = 2

1

12 )2)(1(

7

+−

−

xx

x

3 M1 3(x – 1) – 2(x + 2) seen

B1 denominator correct seen

A1 all correct




13 245 or 246 3 M1 π × 52

M1 182 – their kπ

14

3 M1 2 lines correct length

M1 2 compass arcs correct length

A1 complete accurate drawing with all lines and

arcs solid

15 36 cao 3 M1 1900/2.448 (= 776.14)

A1 “776.(14…)” – 740 (= 36.14…)

16 (a) 9

4x8

2 B1 9

4 B1 x8

(b) 2y–1 2 B1 2 B1 y–1

17 (a)

Boys Girls Total

Asia 62 28 90

Europe 35 45 80

Africa 68 17 85

Total 165 90 255

3 B1 two or three correct

or B2 four or five correct

(b)

17

3 or 0.176(47…) 1 Allow

255

45, 85

15, 51

9

18 (a)

−−

140

014 2 B1 two or three correct answers

(b) –14 1

(c)

−−

45

45 2 B1 two or three terms correct

19 (a) 14.1

(b) 3.74 or 3.78

2

3

M1 (BD2) = 102 + 102 or sin45 = 10/CD

M1 (a)/2 M1 (their (a)/2)2 + PM2 = 82

20 (a)

R

(b)

4

1

B1 y = 2

single line thro B1 (6, 0) and B1 (0,6)

B1 y = 2x

Correct R cao




21 (a) 2

(b) 6.7 to 7.3

(c) 203

1

1

3

M1 intention to find area under the graph

M1 2

1 × 7 × 14 + 9 × 14 +

2

1 × 4 × 14 oe

22 (a) (0, 7)

(b) (i) y = 2x + 3

1

2

B1 y = 2x + c, c ≠ 7 or B1 y = kx + 3, k ≠ 0

(ii) (1, 4) 3 B1 y = 5

M1

++

2

"5"3,

2

20 A1 (1, ft4)




Abbreviations



dep dependent



oe or equivalent

SC Special Case



1 –8.3 1 Allow –810

3

2 21 55 1 Allow 9.55 pm

3 1.6305 cao 2 B1 4.33(44…) seen or answer 1.63, 1.630,

1.6304….

4

1, 1

5 Correct working 2 M1 4

15 +

3

4 =

12

16

12

45+

M1 12

15

12

61=

6 4.93% < 41

20 < 0.492 <

161

80 2 Allow decimal equivalents in answer space

M1 decimals 0.48(78..), 0.496(8..), 0.0493

7 1.14 2 M1 3.38 ÷ 1.04 (= 3.25)

or M1 4.39 × 1.04

8 1200 2 M1 figs 8 ÷ 40 × figs 9 ÷ 15

or M1 (figs 8 × figs 9) ÷ (40 × 15)

9 9.6 cao 2 M1 10

12

8=

x

oe

10 216.32 cao 2 M1 200 × (1 + (4/100))2 oe

11 13 2 M1 21 + 15 – 23

or M1 15 – x + x + 21 – x + 1 = 24 oe

12 (a) 25

(b) 0.4

1

1

If zero scored SC1 for 250 and 4 or

6.25 and 6.35

13 10a + b or a × 101 + b (× 100) 2 M1 [a × 107 + b × 106] ÷ 106




14 10.8 or 1083

70 3 M1 figs 10 ÷ time

M1 10 ÷ 0.92r, 0.922 or 83/90

15 y = –2x + 8 cao oe 3 M1 (m =) 30

28

−

−

oe B1 c = 8 or y = mx + 8

or subst. correct point in y = “m” x + c

16 2

4

g

h or

2

2

gh 3 M1 squaring correctly

M1 clearing denominator correctly

M1 dividing by coefficient of i

or SC2 for correct unsimplified expression

17 x = –1, y = 5 3 M1 consistent multiplication and either add or

subtract

A1 for one correct after M1

18 315 3 M1 360

x

× 2 × π × 8 oe

M1 360

x

× 2 × π × 8 (+ 16) = (16 +) 14π

19 2.88 3 M1 403 oe seen A1 2 880 000

B1ft their 2 880 000 ÷ 1003

or B1 0.000045 M1 403 A1 cao

or M1 0.43 M1 45 × 0.43 A1

20 (a) 63.4 2 M1 tan(M) = 2

4 oe

(b) Vertices at (4, 1), (8, 1) and (10, 3) 2 B1 two vertices correct

21 (a) 2.4 oe

(b) 680

1

3

M1 an area found

M1 40 × 20 – 2

1 × 20 × 12 oe

22 y [ 1, x Y 3, y Y x + 5 oe 5

B1 y R 1

B1 x R 3

B2 y R x + 5 or B1 y R – x + 5

where R is any inequality

B1 all 3 inequalities correct

23 (a) (Angles in) same segment

(b) (i) 100

(ii) 43

1

1

1

Allow (angles on) the same arc

(iii) 3 2 B1 OBC or OCB =

2

1(180 – 86) (= 47)




24 (a) xy

yx 2−

2 B1 correct numerator

B1 correct denominator

(b)

3

x

www 3 M1 x(x + 1) M1 3(x + 1)

25 (a) –3 2 B1 g(2

1) = 2 or fg(x) =

x

2 – 7 oe

(b)

72

1

−x

1

(c)

2

7+x 2 M1 for y + 7 = 2x or x = 2y – 7




Abbreviations



1 (a) (i) 1088 (ii) Their 1088 × 2 and (3136 – their 1088) × 4.5 2176 + 9216

2

M1 E1

M1 for 3136 ÷ (17 + 32) soi by 64 or 2048 2048 may be 32 × 64

(b) 11.9 to 11.9031 www 3 M2 for

( )11392

1001139212748 ×−

oe

or M1 for 11392

1139212748 − soi by 0.1119

or ( )10011392

12748× soi by 111.9 or 112 or 1.119

(c) 8900 3 M2 for 11392 ÷ 1.28 oe or M1 for 11392 = 128(%) oe

2 (a) (i) Correct reflection (1, –1) (4, –1) (4, –3) (ii) Correct rotation (–1, 1) (–1, 4) (–3, 4) (iii) Reflection only y = x oe or y = – x oe

2

2

1dep

1

SC1 for reflection in y-axis or vertices only of correct triangle SC1 for rotation 90 clockwise about O or vertices only of correct triangle Two transformations scores 0 Dependent on at least SC1 scored in both (i) and (ii) Only from 2 and 2 or SC1 and SC1 scored Only from 2 and SC1 or SC1 and 2 scored

(b) (i)

− 01

10 oe 2

B1 for either column correct or determinant = 1

(ii) Rotation, 90° clockwise, origin oe 2 B1 for rotation and origin B1 for 90° clockwise oe




3 (a) 72 – 2x oe seen x (72 – 2x) = 72x – 2x2

M1 E1

No errors or omissions

(b) 2x(36 – x) or –2x(x – 36) 2 isw solutions B1 for answers 2(36x – x2) or x(72 – 2x) or correct answer spoiled by incorrect simplification

(c) 630, 640, 70 3 B1 for each correct value

(d) 8 correct plots P3ft

C1

ft for their values ft P2 for 6 or 7 correct plots ft P1 for 4 or 5 correct plots Curve of correct shape through minimum of 7 of their points No ruled sections

(e) (i) 7.5 to 8.5 27.5 to 28.5 (ii) 641 to 660

2 1

B1 for either value correct

(f) 41 2 M1 for 500 ÷ 12 soi by 41.6… to 42

4 (a) 1.52 + 22 (l =) 2.5 π × 1.5 × their 2.5 2 × π × 1.5 × 4 Addition of their areas for cone and cylinder 49.45 to 49.5

M1 A1 M1 M1 M1

A1

soi by 6.25 May be on diagram Their 2.5 ≠ 2 soi by 11.77 to 11.8 or 3.75π soi by 37.68 to 37.715 or 12π soi by 15.75π This M mark is lost if any circles are added www 6

(b) (i) π × 1.52 × 4 M1 soi by 28.26 to 28.3 or 9π

3

1π × 1.52 × 2 M1 soi by 4.71 to 4.72 or 1.5π

Addition of their volumes 32.9(7) to 32.99… (ii) 84(.0) to 84.1 www

M1 E1 3

10.5π implies M3

M1 for ½ π × 0.52 soi by 0.392 to 0.393 or π/8 and M1 for their 33 ÷ (½ π × 0.52) soi by 264/π or SC1 for 42 to 42.1 as answer

(c) (i) 33000 (ii) 18min 20s cao

1 2

M1 for their 33000 ÷ 1800 soi by 18.3(3…) or correct in mins and secs for their 33000




5 (a) 8 correct plots Joined by curve or ruled lines

P3

C1ft

P2 for 6 or 7 correct plots P1 for 4 or 5 correct plots ft their points Must join minimum of 7 points

(b) (i) 161 to 162 (ii) 171 to 172 (iii) Their (b)(ii) – 150

1 1

1ft

Strict ft provided > 0

(c) (i)

200

55 oe

40

11 1 isw incorrect cancelling for both parts of (c)

(ii)

39800

1100 oe

398

11 3 M2 for 2 × their

200

55 ×

199

10 oe soi by 0.0276…

or M1 for their200

55 ×

199

10 oe

796

11 soi by

0.0138…

(d) (i) 30, 35, 20 (ii) Blocks in correct position w = 1cm, fd = 4 w = 1cm, fd = 6 w = 2cm, fd = 3.5

2

1 1ft

1ft

B1 for 1 correct value Strict ft from their 30 unless 0 Strict ft from their 35 unless 0

6 (a) (i) 13 cao www 2 M1 for 5.19

PQ =

5.16

11 oe or sf = 2/3 or 1.5 seen

or correct trig (ii) 10.39 to 10.4 www 3 M2 for 22

5.165.19 − or explicit trig or M1 for x2 + 16.52 = 19.52 or implicit trig

(iii) 57.76 to 57.81 www 2 M1 for sin =

5.19

5.16 oe

(iv) 655 to 655.4 2 M1 for 0.02 × (32)3

(b) (i) 163.5 to 164 www (ii) 100.8 to 100.9 or 101 www

4

4

M2 for 672 + 1052 – 2 × 67 × 105cos143 or M1 for implicit form A1 for 26732 to 26896 B1 for (DEF =) 78o May be on diagram

and M2 for 78their sin

70sin105× provided their 78 ≠ 32

or 70

or M1 for 70sin

EF =

78their sin

105 oe their 78 ≠ 32

or 70




7 (a) w = 59 (angle in) isosceles (triangle) x = 31 (angle in) semicircle (= 90) oe

y = 62 (angles in) same segment or (on) same arc (are =) z = 28 (angles in) triangle (= 180)

1 1

1ft

1

1

1

1ft

1

The marks for the reasons are dependent on the correct angle or correct ft angle Any incorrect statement in reason loses that mark ft 90 – their w Allow diameter

ft 180 – their(w + x + y) or 90 – their y

(b) (i)

3

2 1

(ii)

−

4

2 2ft ft

7

0 – their (i)

B1 ft for one correct element

(c) (i)

3

1t final answer 1

(ii)

3

1(– t + r) final answer 2 M1 for correct unsimplified answer

or TR = –t + r oe

or TP = 3

1TR oe

(iii)

3

1r final answer 2 M1 for correct unsimplified answer

or QT + TP oe for any correct path

or 3

1t + their (ii)

(iv) QP =

3

1OR oe 1dep Dependent on correct answer in (iii)

QP is parallel to OR or r 1dep Dependent on multiple of r as answer in (iii)




8 (a) (i) 3 (ii) 4 (iii) 4x – 3 final answer

1 1 2

M1 for 2(2x – 1) – 1

(iv) 2

1+x oe final answer 2 M1 for x = 2y – 1 or

2

1+y oe or

2

1)( +xfoe

(v) –

2

1 and 1

2

1 4 B1 for (2x – 1)2 soi

M2 for 2x – 1 = ± 2 M1 for 4x2 – 2x – 2x + 1 or M1 for 2x – 1 = 2 and M1 for (2x + 1)(2x – 3) or correct substitution in formula soi by (4 ± √64)/8

(b) (i) y =

x

16 oe 2 Condone y = k/x and k = 16 stated

M1 for y = x

k oe

(ii) 32 1

9 (a) (i) 21 (ii) P6 = ½ × 6 × 7 or better (= 21) (iii) 1275 (iv) 3825 (v) 11325 (vi) 7500

1 1 1

1ft

1 1ft

Allow 3(6 + 1) ft for 3 × their (iii) ft their (v) – their (iv) provided > 0

(b) (i) 56 2 M1 for 1 × 6 + 2 × 5 + 3 × 4 + 4 × 3 + 5 × 2 + 6 × 1

(ii) S6 = 6

1 × 6 × 7 × 8 or better (= 56) 1

(iii) 1540 1

(c) 56 – 35 = 21 1

(d) Correct algebraic proof with no errors 3 M1 for

6

1n(n + 1)(n + 2) –

6

1 (n – 1)(n)(n + 1) oe

and M1 for 6

1n(n + 1)(3) oe




Abbreviations



1 (a) 432 2 M1 for 756 ÷ 7 × 4 oe

(b) (i) 8970 2 M1 for 7800 × 1.15 oe After 0 scored, SC1 for 1170 as answer

(ii) )100(

7800

)7800(9867their ×

−

or 1.15 × 1.10

M2

Their 9867 is their (b)(i) × 1.1 Implied by 1.265 or 0.265 or 126.5 or M1 for their (b)(i) × 1.10 (9867 seen or 2067 seen)

26.5 % cao A1 www3

(c) 8100 3 M2 for 9720 ÷ 1.2 oe or M1 for 120% = 9720 oe

(d) 562.43 or 562 or 562.4(0) or 562.432 3 M2 for 500 × 1.04³ or alt complete method or M1 for 1.04² or 1.04³ oe soi e.g. $540.80 or 562.(43..) seen in working

2 (a) (i) 11 (ii) 22

1 1

(b)

4

1+x oe final answer 2 M1 for x + 1 = 4y or

4

1)(g +x

or 4

1+y

(c) 16x² – 8x + 7 final answer 3 M1 for 6 + (4x – 1)² and B1 for 16x² – 4x – 4x + 1 or better seen

(d) 0.5 or ½ www 3 M2 for 16x – 4 – 1 = 3 or better or M1 for 4(4x – 1) – 1 (= 3) Alt method

M2 allow g–1g–1(3) complete method or M1 for g(x) = g–1(3)




3 (a) (i) 63 to 63.5

(ii) 50 to 50.5 (iii) 21.5 to 22.5

1 1 1

(b) 46 2 B1 for 34 seen (could be on graph)

(c) (i) 12, 14 (ii) {35 × 8 + 45 × their 12 + 55 × 14 +

65 × 22 + 75 × their 14 + 85 × 10} ÷ their 80 (or 80)

1, 1

M3

M1 for mid-values soi (allow 1 error/omit)

and M1 for use of ∑ fx with x in correct

boundary including both ends (at least 4 products) (4920 seen implies M2) and M1 depend on 2nd M for dividing by their 80 (or 80) (not 54 or less)

61.5 cao A1 www4

4 (a) (i) 218 (217.7 to 218) (ii) 501 (500.7 to 501.4) (iii) 99

2 1ft 2ft

M1 for 1/3π × 42 × 13 ft their (a) × 2.3 ft 50 000 ÷ their (a)(ii) and truncated to whole number M1 for 50 000 ÷ their (a)(ii) oe or answers 99.8 or 100

(b) their (a)(i) ×

3

13

5.32

oe M2 or 1/3π × 102 × 32.5

or M1 for (32.5 ÷ 13)³ (=15.625) seen or (13 ÷ 32.5)³ (= 0.064) seen

3400 or 3410 (3401 to 3407) A1 www3

(c) (r² =) 550 ÷ 12π 3.82 (3.818 to 3.821)

M2

A1

(14.58 to 14.6)

or M1 for 12π r² = 550 or better www3




5 (a) (i) x² + (x + 7)² = 17² oe x² + x² + 7x + 7x + 49 = 17² or better 2x² + 14x – 240 = 0 x² + 7x – 120 = 0 (ii) (x + 15)(x – 8) (iii) –15 and 8 (iv) 15

B1 B1

E1 2

1ft 1ft

Must be seen Must be shown – correct 3 terms With no errors seen M1 for (x + a)(x + b) where a and b are integers and a × b = –120 or a + b = 7 Ignore solutions after factors given Correct or ft dep on at least M1 in (ii) Correct or ft their positive root from (ii) + 7 dep on a positive and negative root given

(b) (i) 3x(2x – 1) = (2x + 3)² oe 4x² + 6x + 6x + 9 or better seen 6x² – 3x = 4x² + 12x + 9 oe

2x² – 15x – 9 = 0

M1

B1

E1

e.g. 6x² – 3x = 4x² + 12x + 9 must see equation before simplification

Indep With no errors seen and both sets of brackets expanded

(ii) )2(2

)9)(2(4)15)((15)(2

−−−±−−oe

1 1

In square root B1 for ((–)15)2 – 4(2)(–9) or better (297)

If in form r

qp +or

r

qp −,

B1 for –(–15) and 2(2) or better

8.06 and -0.56 cao (iii) 76.5 (76.46 to 76.48)

1, 1 1ft

SC1 for –0.6 or –0.558… and 8.1 or 8.058… ft 8 times a positive root to (b)(ii) add 12

6 (a) (i) 54802 + 33002 – 2 × 5480 × 3300 × cos165

8709.5..

M2

E2

(75 856 005) M1 for implicit version If E0, A1 for 75800000 to 75900000

(ii) (sinL =)

8710

165sin × 3300

(0.09806…)

M2

M1 for 8710

165sin

3300

sin=

L oe (allow 8709.5.)

Could use cosine rule using 8710 or better – M2 for explicit form or M1 for implicit form (allow 5.6 to 5.63 for A mark)

5.6 (5.62 to 5.63) A1 www3

(b) 22 35 or 10 35 pm 2 Accept 22 35 pm B1 for 15 35 or 3 35 pm seen or answers 22h 35 mins or (0)8 35(am) or 10 35(am)

(c) 8710 ÷ 800 10.88 to 10.9 with no conversion to

h/min or 10 (hrs) 52 (mins) to 10 (hrs) 54

(mins) oe 13 hrs 45 mins – their time in hrs and

mins oe or 13.75 – their decimal time and a

correct conversion to hrs and mins or minutes

2 hr 52 mins cao

M1 A1

M1

A1

Implied by correct final ans 2hrs 52 mins if not shown Dep on first M1 e.g. 13 hrs 45mins – 11 hrs 29 mins or 13.75 – 10.9 then 2hrs 51 mins www4 (2 hrs 51.75 mins)




7 (a) –3, –4.25, –3 1, 1, 1 Allow – 4.2 or – 4.3 for – 4.25

(b) 10 correct points plotted Smooth curve through their 10 points

and correct shape Two separate branches

P3ft

C1

B1ft

P2ft for 8 or 9 correct P1ft for 6 or 7 correct Correct shape not ruled, (curves could be joined) Indep but needs two ‘curves’ on either side of y-axis

(c) (i) 0.7 to 0.85

(ii) Any value of k such that k Y –3 and must be consistent with their graph

1 1ft

–1 each extra ft consistent with their graph (If curves are joined then k = –3 only)

(d) y = 5x drawn – 0.6 to –0.75, 0.55 to 0.65

L1 1, 1

Ruled and long enough to meet curves Indep –1 each extra

(e) Tangent drawn at x = –2 y change / x change attempt 2.7 to 4.3

T1

M1

A1

Must be a reasonable tangent, not chord, no clear daylight Depend on T and uses scales correctly. Mark intention – allow one slight slip e.g. sign error from coords but not scale misread If no working shown and answer is out of range – check their tangent for method Answer in range gets 2 marks after T1 earned

8 (a) (i) Correct translation to (3, –5), (5, –6) and (4, –4)

2 SC1 for translation of

k

3 or

− 7k

or vertices

only

(ii) Correct reflection to (4, 1), (5, 3) and (6, 2)

2

SC1 for reflection in y = 3 or vertices only

(iii) Correct rotation to (–2, 0), (–1, 2) and (–3, 1)

2 SC1 for rotation 90 clockwise around (0, 0) or vertices only

(iv) Correct enlargement to (0, –3), (–8, 1) and (–4, –7)

2 SC1 for two correct points or vertices only

(b) 16 cao 1

(c) (i) Correct transformation to (–4, 0), (5, 3) and (–2, 0)

(ii) Shear only

x-axis oe invariant (factor) 3

3

1

1 1

B2 for 3 correct points shown in working but not plotted

or B1 for incorrect shear drawn with x-axis invariant or two correct points shown If more than one transformation given – no marks available Accept fixed, constant oe for invariant

(iii)

−

10

31oe 2 B1 for determinant = 1 or k

−

10

31 oe




9 (a) 11

4 and

10

4, 1

10

7

10

3 1, 1

Accept fraction, %, dec equivalents (3sf or better) throughout but not ratio or words i.s.w. incorrect cancelling/conversion to other forms Pen –1 once for 2 sf answers

(b) (i)

11

7 ×

10

6 M1

110

42 oe

55

21 A1 www2 0.382 (0.3818…)

(ii)

11

7 ×

10

4 +

11

4 ×

10

7 M2 ft their tree

M1 for either pair seen

110

56 oe

55

28 A1 www3 0.509(0..)

(c) (i)

9

5

10

6

11

7×× or their (b)(i) ×

9

5 M1

990

210 oe

33

7 A1 www2 0.212(1..)

(ii) 1 –

××9

2

10

3

11

4 oe M2 Longer methods must be complete

M1 for 4/11, 3/10 and 2/9 seen

990

966 oe

165

161 A1 www3 0.976 (0.9757…)

10 (a) 21 and 34 1

(b) –5 8 1 + 1

(c) (i) 4, 6

(ii) x = 28 y = –5 z = 23

3

5

M1 for 2 + d = e oe or d + e = 10 oe seen and either M1 for a correct eqn in d or e seen e.g. 2e = 12 oe or 2d = 8 oe or B1 for either correct B4 for any two correct or M3 for any of 18 = 3x – 66 oe or 3y + 33 = 18 oe or 33 – 3z = -36 oe

or M1 for 2 of y = x – 33 oe or y + z = 18 oe or x + y = z oe

and M1 for combining two of the previous equations correctly isw (does not have to be simplified) after 0 scored SC1 for –33 + their x = their y or their x + their y = their z or their y + their z = 18




Abbreviations



dep dependent



oe or equivalent

SC Special Case



soi seen or implied


1 (a) 200 ÷ 10 × 3 oe

200 ÷ 10 × 2 oe

M1

M1

(b) 65 2 M1 for 100

60

39× oe 35 is M0

(c) 46 3 M2 for 36.80 ÷ 0.8 oe

or M1 for 80% = 36.80 oe

(d) 0.6(0) 3 M2 for 5(x + 12) + 2x = 64.2 oe

or (64.2 – 5 × 12) ÷ 7

or 5x + 2(x – 12) = 64.2 oe or (64.2 + 2 × 12) ÷ 7

or M1 for y = x + 12 and 5y + 2x = 64.2

or y = x – 12 and 5x + 2y = 64.2

After M0, SC1 for k(x ± 12) seen

2 (a) (cosQ =)5.442

75.44222

××

−+o.e.

110.74….

M2

E2

M1 for 72 = 42 + 4.52 – 2 × 4 × 4.5 × cos(Q)

If E0 then A1 for – 0.354(1….)

(b)

85sin

40sin7)( =RS M2 M1 for

85sin

7

40sin=

RS o.e.

4.516 … E1 Can be implied by second M

(c) Angle R = 55°

0.5 × 7 × 4.52 × sin(their 55) o.e.

0.5 × 4 × 4.5 × sin110.7 o.e.

Triangle PRS + Triangle PQR

21.4 (21.36 – 21.42)

B1

M1

M1

M1

A1

(May be seen on diagram)

(12.95 – 13.0) their 55 is (180 – 40 – 85)

(8.418 – 8.42) (s = 7.75)

Dependent on M1, M1

www 5




3 (a) 5x2 – x or x(5x – 1) 2 M1 for x2 + 3x or 4x2 – 4x correct

(b) 27x9 2 B1 for 27 or for x9

(c) (i) 7x7(1 + 2x7)

(ii) (y + w)(x + 2a)

(iii) (2x + 7)(2x – 7)

2

2

1

M1 for any correct partially factorised

expression

or 7x7(1 + ...)

M1 for x(y + w) + 2a(y + w) or

y(x + 2a) + w(x + 2a)

(d) ( )

)2(2

)1(24552

−±− oe 2

In square root B1 for 5² – 4(2)(1) or better (17)

If in form r

qp +or

r

qp −

B1 for p = – 5 and r = 2(2)

–2.28

–0.22

1

1

SC1 for –2.3 or –2.281 to –2.280 and

–0.2 or –0.220 to –0.219

4 (a) (i)

43

25

1

1

If 0, 0 then SC1 for 25 and 43 seen

(ii) (16) 2 B1 for 16 without brackets

(iii)

−

−

− 24

35

2

1 isw

or

−

−

12

2

3

2

5

2 B1 for determinant = –2

or B1 for

−

−24

35k

(b) Reflection only

x-axis oe

1

1

If more than one transformation given – no

marks available

independent

(c)

−

01

10 2 B1 for one correct column




5 (a) (i) Accurate perpendicular bisector,

with 2 pairs of arcs, of CD.

(ii) Accurate angle bisector, with two

pairs of arcs, of angle A.

2

2

SC1 if accurate without arcs.

SC1 if accurate without arcs.

(b) SHOP written in correct region S1 Dependent on at least SC1 in (i) and (ii) and

intersection

(c) (i) Arc, centre B, radius 5cm,

reaching across ABCD.

(ii) Area outside their arc centre B

and outside SHOP shaded

1

1ft

Allow good freehand

dep on S1

6 Accept fraction, %, dec equivalents (3sf or

better) throughout but not ratio or words

i.s.w. incorrect cancelling/conversion to other

forms

Pen –1 once for 2 sf answers

(a) (i) 33 1

(ii)

3125

243 (0.07776) 2 Accept 0.0778. M1 for

5

5

3

oe

(b) (i)

5

2, 4

3, 8

1, 8

7 3 B1 for

5

2 and

4

3 B1 for

8

1 B1 for

8

7

(ii)

20

1 (0.05) cao 2 M1 for their

5

2 × their

8

1

(iii)

5

1 (0.2) ft 2ft ft

20

3 + their (b)(ii) or M1 for

4

1

5

3×

7 (a) – 5.4

3.7

1

1

(b) 8 points correctly plotted ft

Smooth cubic curve through all 8

points

P3

C1

P3ft their table.

P2ft for 6 or 7 points. P1ft for 4 or 5 points

Only ft points if shape not affected.

(c) –2, –4, 4 2 B1 for 2 correct

(d) 7 points correctly plotted ft

Two separate smooth branches of

rectangular hyperbola

P2

C1

P2ft P1ft for 5 or 6 points

Must pass through all 7 points, only ft if shape

not affected and no contact with either axis.

(e) (i) –2.9 Y x Y– 2.8

2.05 Y x Y 2.15

(ii) a = 10

b = –40

1

1

1

1

Not with y coordinates




8 (a) (i) 396 (395.6 – 396) 4 M1 for 3

2 × π × 33 and M1 (independent) for

π × 32 × 12,

M1 (dependent on M2) for adding

126 π implies M3

(ii) 3.13 (3.125 – 3.128….) ft

(iii) 144 (144 – 144.4) ft

2ft

2ft

ft their (i) × 7.9 ÷ 1000 .

M1 for × 7.9 soi by figs 313 or 3125 – 3128…

ft 15 × 6 × 6 – their (a)(i)

M1 for 6 × 6 × 15 oe

(b) (i) 311 (310.8 – 311.1)

(ii) 3.50 (3.496 to 3.50) ft

5

2ft

M1 for 2 × π × 32 and M1 (independent) for

π × 6 × 12 and M1 for π × 32,

M1 (dependent on M3) for adding.

(99π implies M4)

ft their (b)(i) × 0.01125

M1 for their (b)(i) ÷ 8 and × figs 9

implied by figs 3496 to 350

9 (a) (i)

5

9 1

(ii)

7

4 1

1 If 0, SC1 for

−

=2

5CB seen

(iii) BA or – AB 1 BA not indicated as a vector is not enough.

(iv) 10.3 (10.29 – 10.30) 2 M1 for (their 9)2 + (their 5)2

(b) (i) 2u 1

(ii)

2

1(t – u ) oe 2 M1 for

2

1 (their DCADBA ++ ) or equivalent

correct route for BM , along obtainable vectors

in terms of t and u

or M1 for correct unsimplified answer

(iii)

2

3u +

2

1t oe ft 2ft ft their (i) + their (ii) simplified

or t + u – their (b)(ii) simplified

M1 for correct (or ft) unsimplified (i) + (ii)

or t + u – their (b)(ii)




10 (a) 7, 8, 8, 10, 11, 16

and 8, 8, 8, 10, 10, 16

5 Mark answer spaces only or clearly indicated

lists. Allow numbers in any order but must be

lists of 6 integers

B4 for either correct list

If not B4 then

B1 for a series with mode 8

and B1 for a series with median 9

and B1 for a series with sum 60

(b) (i) (30 × 65 + 35 × 85 + 40 × 95 +

40 × 110 + 15 × 135) ÷ 160

4

M1 for mid-values soi (allow 1 error/omission)

and M1 for use of ∑ fx with x in correct

interval including both boundaries allow one

further error/omission

and M1 (dependent on second M) for ÷ 160

94.7 (94.68 – 94.69)

(ii) Heights of 4, 2, 0.5 with correct

interval widths

4

www 4

B3 for 2 correct

or B2 for 1 correct

or B1 for all three freq. densities correct but

no/incorrect graph

11 (a) 30 42

42 56

71 97

4 B3 for 2 correct rows

or B2 for 1 correct row

or B1 for any term in column 5 correct

(b) (i) 2550

(ii) 30

1

1

(c) (n + 1)(n + 2) oe final ans 1

(d) (i) 2n2 + pn + 1 = t

Uses a value of n up to 6 and a

matching t from the table

e.g. puts n = 3 and t = 31

2 × 3² + 3p + 1 = 31 M1

OR

Use p = 4 to get 2n² + 4n + 1 = 31

and simplifies to 3 term eqn M1

OR both

2 × 9 + 4 × 3 + 1 (= 31) M1

with one part evaluated

OR

n(n + 1) + (n + 1)(n + 2) – 1

or better M1

(ii) 241

(iii) 12

2

1

3

Correct solution shown with 1 intermediate step

to p = 4 E1

Solve correctly to get n = 3 E1

Conclusion e.g. 31 = 31 E1

Correct simplification to 2n2 + 4n + 1 E1

M1 for 2n2 + 4n + 1 = 337

and M1 for (n – 12)(n +14) or correct expression

for n using formula

(e) 1−+= DAL oe 1


MARKING SCHEME BANK

2002 – 2011

Compiled & Edited By

Dr. Eltayeb Abdul Rhman

www.drtayeb.tk

First Edition

2011

MS BANK 580 - WELCOME IGCSE | Dr. Tayeb's Website · 2011-12-27MS BANK 580 - WELCOME IGCSE | Dr....

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