MARKING SCHEME BANK
2002 – 2011
Compiled & Edited By
Dr. Eltayeb Abdul Rhman
www.drtayeb.tk
First Edition
2011
Page 1 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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 2
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
SC1 if accurate but no arcsSC1 if accurate arcs but no line
Indep. 6
11
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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.
Page 4 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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
If B0, allow SC1 for correct butunsimplified seen
(b) (i) |b| = 5(ii) |a – b| = 5 www
B1B1 (2)
Page 5 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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 3�37½% 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)
Page 6 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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
Page 1 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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 2
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
SC1 if accurate but no arcsSC1 if accurate arcs but no line
Indep. 6
11
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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.
Page 4 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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
If B0, allow SC1 for correct butunsimplified seen
(b) (i) |b| = 5(ii) |a – b| = 5 www
B1B1 (2)
Page 5 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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 3�37½% 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)
Page 6 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2003 0580/0581 4
(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
Page 1 Mark Scheme Syllabus Paper
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
Page 2 Mark Scheme Syllabus Paper
MATHEMATICS – JUNE 2004 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
MATHEMATICS – JUNE 2004 0580/0581 4
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
Page 2 Mark Scheme Syllabus Paper
MATHEMATICS – JUNE 2004 0580/0581 4
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
Page 3 Mark Scheme Syllabus Paper
MATHEMATICS – JUNE 2004 0580/0581 4
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
Page 4 Mark Scheme Syllabus Paper
MATHEMATICS – JUNE 2004 0580/0581 4
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2005 0580/0581 2
* 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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – JUNE 2005 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
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
Page 2 Mark Scheme Syllabus Paper
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)
Page 3 Mark Scheme Syllabus Paper
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
Page 4 Mark Scheme Syllabus Paper
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
Page 5 Mark Scheme Syllabus Paper
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
Page 6 Mark Scheme Syllabus Paper
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
Page 1 Mark Scheme Syllabus Paper
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
Page 2 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 02
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
Page 3 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 02
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
Page 1 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 04
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
Page 2 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 04
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
Page 3 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 04
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
Page 4 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 04
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
Page 5 Mark Scheme Syllabus Paper
IGCSE – May/June 2006 0580 and 0581 04
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
Page 2 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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]
Page 3 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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
Any other method must be complete
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]
Page 4 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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]
Page 5 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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]
Page 6 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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
P1 for 4 or 5 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]
Page 7 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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]
Page 8 Mark Scheme Syllabus Paper
IGCSE – May/June 2007 0580 and 0581 04
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]
Page 2 Mark Scheme Syllabus Paper
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
Page 3 Mark Scheme Syllabus Paper
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
Page 4 Mark Scheme Syllabus Paper
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
Page 2 Mark Scheme Syllabus Paper
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
M1 2 more moves completed correctly
Page 3 Mark Scheme Syllabus Paper
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
Page 4 Mark Scheme Syllabus Paper
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
Page 2 Mark Scheme Syllabus Paper
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]
Page 3 Mark Scheme Syllabus Paper
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
Page 4 Mark Scheme Syllabus Paper
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]
Page 5 Mark Scheme Syllabus Paper
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]
Page 2 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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
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.
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2009 0580, 0581 22
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
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
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.
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
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]
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
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]
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
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]
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
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]
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
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]
Page 7 Mark Scheme: Teachers’ version Syllabus Paper
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 21
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 22
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 22
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 23
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 23
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 41
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 Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 41
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 41
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
P1 for 5 or 6 points ft
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.
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 41
(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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 42
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 Marks
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)
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 42
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 42
(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.
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 42
(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
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 42
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 43
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 Marks
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
Any other method must be complete
(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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 43
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 43
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
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 43
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)
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2010 0580 43
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 21
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 21
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 22
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 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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 22
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 23
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 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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 23
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 23
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 41
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
Qu. Answers Mark Part Marks
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)
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 41
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)
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 41
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)
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 41
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
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 41
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
P2 for 4 or 5 correct plots ft
P1 for 2 or 3 correct plots ft
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
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
Page 7 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 42
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 43
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
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 43
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 43
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)
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 43
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
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – May/June 2011 0580 43
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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)
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
Marks in brackets are totals for questions or part questions.
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)
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 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)
Reasonable curve through 8 points
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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
4/45 o.e. A1 Accept 8/90 etc, 0.0888(or 9), 8.88 or 8.89% www2
(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
Page 4 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
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)
Page 5 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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)
Page 6 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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)
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 2
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
Marks in brackets are totals for questions or part questions.
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)
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 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)
Reasonable curve through 8 points
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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
4/45 o.e. A1 Accept 8/90 etc, 0.0888(or 9), 8.88 or 8.89% www2
(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
Page 4 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
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)
Page 5 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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)
Page 6 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2003 0580/0581 4
(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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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
Page 1 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 4
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)
Page 2 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 4
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
Page 3 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 4
(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
Page 4 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 4
(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]
Page 5 Mark Scheme Syllabus Paper
IGCSE EXAMINATIONS – NOVEMBER 2004 0580/0581 4
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]
Page 1 Mark Scheme Syllabus Paper
IGCSE – NOVEMBER 2005 0580/0581 2
* indicates that it is necessary to look in the working following a wrong answer
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
Page 2 Mark Scheme Syllabus Paper
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
Page 1 Mark Scheme Syllabus Paper
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]
Page 2 Mark Scheme Syllabus Paper
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]
Page 3 Mark Scheme Syllabus Paper
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]
Page 4 Mark Scheme Syllabus Paper
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]
Page 5 Mark Scheme Syllabus Paper
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]
Page 4 Mark Scheme Syllabus Paper
IGCSE - OCT/NOV 2006 0580/0581 2
3
5,
3
5
−
−− xxor
3
5 y−
Page 3 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 02
* indicates that it is necessary to look in the working following a wrong answer
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
Page 4 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 02
* indicates that it is necessary to look in the working following a wrong answer
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
Page 3 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 04
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]
Page 4 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 04
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]
Page 5 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 04
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]
Page 6 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 04
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]
Page 7 Mark Scheme Syllabus Paper
IGCSE – October/November 2007 0580/0581 04
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]
Page 2 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 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) 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
M1 2 more 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)
Page 3 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 21
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
Page 2 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 22
Abbreviations
cao correct answer only ft follow through after an error oe or equivalent SC Special Case www without wrong working
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
M1 2 more 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)
Page 3 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 22
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
Page 3 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 04
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]
Page 4 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 04
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]
Page 5 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 04
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]
Page 6 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 04
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]
Page 7 Mark Scheme Syllabus Paper
IGCSE – October/November 2008 0580/0581 04
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]
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 21
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 21
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 22
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 22
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 22
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
Abbreviations
cao correct answer only
cso correct solution only
dep dependent
ft follow through
isw ignore subsequent working
oe or equivalent
SC Special Case
soi seen or implied
www without wrong working
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
(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]
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
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)
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
(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]
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
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]
Page 7 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2009 0580 04
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]
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 21
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 Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 21
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 22
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 Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 22
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 22
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)
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 23
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 Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 23
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)
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 23
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 41
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
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 41
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 41
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
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 41
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)
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 41
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 42
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
Qu. Answers Mark Part Marks
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)
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 42
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 42
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)
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 42
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
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 42
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
Page 2 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 43
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
Qu. Answers Mark Part Marks
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
Page 3 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 43
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
Page 4 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 43
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
Page 5 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 43
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)
Page 6 Mark Scheme: Teachers’ version Syllabus Paper
IGCSE – October/November 2010 0580 43
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