0580 Mathematics: MARK SCHEME For The May/June 2013 Series
0580 Mathematics: MARK SCHEME For The May/June 2013 Series
0580 Mathematics: MARK SCHEME For The May/June 2013 Series
0580 MATHEMATICS
0580/32 Paper 3 (Core), maximum raw mark 104
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 should be read in conjunction with the question paper and the Principal Examiner
Report for Teachers.
Cambridge will not enter into discussions about these mark schemes.
Cambridge is publishing the mark schemes for the May/June 2013 series for most IGCSE, GCE
Advanced Level and Advanced Subsidiary Level components and some Ordinary Level components.
Page 2 Mark Scheme Syllabus Paper
IGCSE – May/June 2013 0580 32
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
(ii) 10 1
(iii) 8 1
(iv) 7 1
(v) Mode 1
8
(b) (i) oe 1 Must be a fraction
24
2 (a) (i) 3m 1
(ii) m+4 1
(c) Correct length and bearing 2 B1 for correct length 7.8 to 8.2
B1 for correct bearing 302º to 306º
5 (a) (i) Perpendicular bisector with 2 sets of 2 B1 correct line with some or no arcs
correct arcs
(iii) Angle bisector with 2 sets of correct arcs 2 B1 correct line with some or no arcs
(iv) Trapezium 1
(b) 7 correctly plotted points 3ft P2ft for 5 or 6 correctly plotted points
P1ft for 3 or 4 correctly plotted points
Correct smooth curve going below 1
y = –4 at lowest point
(iii) 0.5 to 0.9, 4.1 to 4.5 1ft,1ft ft is the x coordinates of the intersection
of their line and their curve
(d) (– 5, 2) 1
B1 for y = kx + 3 k≠0
4
8 (a) 6 2 M1 for [× 60] oe
40
(b) (i) Line from (1450,4) to (1510,4) 1
Line from (1510,4) to (1530,0) 1ft Ft is (their 1510,4) to (their 1510 + 20,0)
(ii) Positive 1
360
(c) 11.6 3 M1 for × 29 or better, implied by
9
1160
and M1 indep for ‘their 1160’ / 100 soi
or 0.29 seen
(d) [x] = 2.5, [y] = 0.5 3 M1 for correct method to eliminate one
variable.
A1 for x or y correct.