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UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS
for the guidance of teachers
0444 MATHEMATICS (US)
0444/11
Paper 1 (Core), maximum raw mark 56
This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of
the examination. It shows the basis on which Examiners were instructed to award marks. It does not
indicate the details of the discussions that took place at an Examiners’ meeting before marking began,
which would have considered the acceptability of alternative answers.
Mark schemes must be read in conjunction with the question papers and the report on the
examination.
• Cambridge will not enter into discussions or correspondence in connection with these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2012 question papers for most IGCSE,
GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level
syllabuses.
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MARK SCHEME for the May/June 2012 question paper
s
er
International General Certificate of Secondary Education
Page 2
Mark Scheme: Teachers’ version
IGCSE – May/June 2012
Syllabus
0444
Paper
11
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
soi
seen or implied
Qu
Answers
1
30
1
(a)
Equilateral
1
(b)
3
1
3
532
2
M1 for 5(h)33(min) + 3(h)19(min) seen
4
12 400
2
M1 for
2
Mark
Part marks
5
× 248 000 oe or better
100
or SC1 for answer 260400 www
21
5
2
M1 for 2 × 3 – 5 × (–3) or better
or B1 for 6 and –15 i.e. both terms evaluated
0.85b + 7.5n
2
B1 for 0.85b OR 7.5n seen
Perpendicular bisector drawn
with 2 pairs of arcs and ruled,
within ±2mm and ±2°
2
SC1 for a ruled perpendicular
without arcs or only one pair
or 2 pairs of correct arcs with no line drawn
(a)
Rhombus
1
(b)
131°
1
9
2.25 oe
2
M1 4x = 7 + 2 OR x –
10
1000
2
M1 for
11
2a 2
oe final answer
3b 2
2
SC1 for answer of k
6
OR
7
8
85n + 750n
final answer
100
2
7
=
or better.
4
4
1
× 600 oe
0.6
a2
2a n
OR
for
3b 2
b2
where n is an integer and k ≠ 0
© University of Cambridge International Examinations 2012
Page 3
Mark Scheme: Teachers’ version
IGCSE – May/June 2012
Syllabus
0444
If zero scored, then M1 for
Paper
11
x+4
y +1
= 0 and
= 0 seen
2
2
−4
9
1
1
(a)
1, 3, 5, 15
1
(b)
3 p (5 p + 8t ) final answer
2
B1 for answer of 3(5 p 2 + 8 pt ) or p (15 p + 24t )
or SC1 for correct answer seen
14
Triangle drawn correctly with ruler
and arcs
3
M1 for one side drawn to correct length
and M1 for clear method of crossing arcs
even if wrong scale or inaccurate
15
2
12
13
16
52
82
oe or (1)
oe
30
30
55
27
25 27
or M1 for
oe or (1) +
oe
+
30
30
30 30
or SC1 for common denominator of 30k seen
M2 for
51°
1
(b)
90°
1
(c)
66°
1
x = −2
y=3
3
(a)
(–1, 2)
1
(b)
 4
 
 − 5
1
(c)
(1, 5)
1
330
1
1000 or 1 × 103
2
B1 for 1000000 or 1 × 106 or 106 seen
(a)
9p – 4q final answer
2
SC1 for answer of 9p ± jq
OR ±kp – 4q j, k are integers
or for continued work after correct answer
(b)
x=
2
M1 for correct first step
19
20
3
(a)
17
18
11
cao
15
g−y
oe
2
M1 for consistent multiplication and subtraction.
Allow computational errors
A1 for x = –2 or y = 3
i.e. either g − y = 2 x oe OR
or SC1 for answer x =
y−g
2
© University of Cambridge International Examinations 2012
g
y
=x+
2
2
Page 4
21
18
1
(ii) 17
2
21
1
36 000 π
4
(a) (i)
(b)
22
Mark Scheme: Teachers’ version
IGCSE – May/June 2012
Syllabus
0444
Paper
11
M1 for clear attempt to find the middle number
B1 for radius = 30 soi
4
and M1 for π × their30 3 soi,
3
4
and M1 ft dep for π × 27000
3
If zero scored, then
SC3 for answer of 36 000 provided π seen in method
or for answer of 288 000 π
or for art 113000 calculated from 36000 × π
4
or SC2 for π × 216000
3
4
or SC1 for π × 60 3 soi
3
© University of Cambridge International Examinations 2012