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CAMBRIDGE INTERNATIONAL EXAMINATIONS
International General Certificate of Secondary Education
MARK SCHEME for the May/June 2013 series
0607 CAMBRIDGE INTERNATIONAL MATHEMATICS
0607/42 Paper 4 (Extended), maximum raw mark 120
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.
1
2
(a
(b
(a
(b
Pa
a) (
(i
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(iv
b)
a)
b)
age
(i)
ii)
ii)
v)
e 2
2
4
5
7
9
2
7
y
3
2
27.2
49 :
500
748
9 o
(2x
7
2
+y
y
3 ×
5
1
2 (
: 45
0
8.8[
or 8
2( x
1=
+
y
2y
o.
(27
5 f
[0]
8.83
+x
3
1=
= y
.e.
.22
fina
ca
3 (
)1
3
1
y +
2…
al a
ao f
8.8
3=
+ 1
)
answ
fina
829
(3x
o.e
wer
al a
to
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e.
IG
©
r
answ
8.8
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or
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830
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r
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brid
e.
r
rk S
M
dge
Sc
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e In
che
/Ju
nter
eme
une
rnat
e
e 2
tion
201
nal
13
Ex
1
2
3
2
3
M2
B2
M1
B1
B1
xam
2
2
1
1
1
mina
S
o
a
o
M
o
M
÷
S
M
A
o
o
M
o
If
m
B
o
i.
y
atio
SC1
or B
answ
or 2
M2
or M
M1
÷ 12
SC2
M2
Allo
or M
or 6
M2
or M
f M
mis
B1
or fi
.e c
y =
ons
1 fo
B1 f
wer
27.2
for
M1
for
2 d
2 fo
for
ow
M1
650
for
M1
M0,
sin
for
fina
corr
5
1
s 20
or a
for
r)
2 (2
r 72
for
r 72
oes
or a
r lo
1.5
for
mu
r gr
for
, B1
ng b
r x2
al an
rec
013
ans
r 18
27.2
20
r 72
20
s no
ans
og (
54
r 65
ulti
rap
r gr
1 fo
brac
2=
nsw
ct eq
S
3
we
80 s
22.
÷ 1
20
× 1
ot s
we
(10
for
50
ipli
phs
rap
or
cke
x7=
wer
qua
Syll
06
er 4
see
..) :
1.4
= 1
1.0
spo
er 8
000
r 10
× 1
ied
tha
ph o
2x
ets
x
r 0,
atio
lab
607
45:4
n (
: 25
4 o
144
4 o
oil m
8.85
0/65
000
1.0
d by
at i
of y
2( xx
if c
o.e
, 7
on w
bus
7
49 o
ma
5 o
o.e.
4%
o.e.
met
5
50)
0/65
5n =
y 1.
inte
y =
+x
corr
e. o
wit
s
or
ay b
o.e.
se
im
tho
÷ l
50
= 1
.05
erse
= 1.
)1+
rec
or 2
tho
1.0
be i
en
mp
od
log
in m
100
at
ect
.05x
or
ct ex
2(2
ut f
09 (
imp
plie
g(1.
me
00
lea
as x
3r x
xpa
2x +
fra
P
(1.0
plie
d b
.05
etho
ast
sho
(xx
ans
+ 1)
ctio
Pa
4
088
ed b
by 7
)
ods
twi
ow
+x
sion
) =
ons
pe
42
8 to
by
749
.
ice
wn
)3
ns f
3(x
s le
er
o 1.
the
9.
co
Co
foll
x +
eadi
089
eir r
orre
ond
low
+ 3)
ing
9)
rati
ectl
done
w
) o.
g to
io
y
e
e.
3
(c
(a
(b
Pa
) (
(i
a)
b) (
(i
age
(i)
ii)
(i)
ii)
e 3
1
w
2
b
o
w
w
3
y
R
3
w
10
10
+w
2×
bett
or
(ww
2w
3 h
=y
Rul
so
9
0
+
(10
ter
90
+w
9+
20
−=
led
oi
s
(w
9
0
+
9w
mi
−x
lin
soi
9+
)=
3−w
in
1−
ne th
)9 −
2=
36
ww
o.
hro
2×−
2
1
=
ww
.e. f
oug
10×
o.
0
w
fin
gh (
IG
©
0w
.e.
al a
(– 2
GC
Ca
5=
ans
2, 0
M
SE
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Mar
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(ww
er
and
rk S
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))9
, 4)
che
/Ju
nter
o.e
)
eme
une
rnat
e. o
e
e 2
tion
or
201
nal
2
13
Ex
B1
B1
M1
E1
4
2
2
2FT
xam
1
1
1
1
T
mina
If
w
i.
c
a
a
i.
E
li
B
M
q
c
o
a
s
s
T
tr
(u
B
h
B
B
B
o
o
F
S
atio
f no
wx =
.e.
corr
allo
also
.e.
Esta
ine
B3
M2
qua
com
or M
a +
ubs
imp
Tria
rial
unl
B1 F
hou
B1
B1
B1
or
or ti
FT
SC1
ons
ot s
= 1
cor
rect
w
o al
cor
abli
e wi
for
for
dra
mple
M1
b =
stit
plif
al a
ls f
less
FT
urs a
for
for
for
iny
on
1 fo
s 20
see
0 i
rrec
t fo
10(
llow
rrec
ish
ith
r w
r (w
atic
etin
for
= 9
tuti
fyin
and
for
s co
T fo
and
r y
r ru
r
y pa
ly i
or c
013
en o
is n
ct m
orm
(w +
w o
ct c
hed
no
= 3
+w
c sh
ng
r (w
or
ons
ng
im
com
orre
or 1
d m
−=
uled
art n
if p
corr
S
3
only
not
mul
mat
+ 9
over
coll
wi
err
3 or
12+
how
sq
w +
sk
s in
or
mpro
mp
ect
0 ÷
minu
x−
d lin
not
pos
rec
Syll
06
y im
suf
ltip
9) –
r co
lect
ith
ror
r fo
)(2 w
wing
quar
)a+
ketc
n qu
(w
ove
pari
t an
÷ th
ute
c+
ne t
t sh
itiv
ct a
lab
607
mp
ffic
plic
– 10
om
tion
at l
rs o
or w
−w
g z
re r
)(w
ch o
uad
+w
em
son
nsw
heir
s
c o.
thr
had
ve g
rea
bus
7
lied
cien
cati
0w
mmo
n o
leas
or o
w =
)3−
zero
reac
w +
of q
drat
) 22
9
ent
n o
wer
r po
e.
oug
ded
gra
a un
s
d b
nt.
on
= 2
on d
of tw
st o
omi
= 3
or
os o
chi
)b
qua
tic 2−
t –
f di
fou
osi
o
gh
die
nsh
by f
of
2.5
den
wo
one
issi
or
go
or q
ing
wi
adra
for
4
81−
allo
ista
und
tiv
r y
(0,
ent
hade
furt
f an
w(w
nom
ter
e m
ons
– 1
ood
qua
g −
ith
atic
rmu
3=
ow
anc
d w
e ro
=y
, 4)
and
ed
P
ther
eq
w +
min
rms
more
s
12
sk
adra
9 ±
ab
c or
ula
36 o
w M
ces
with
oot
kx
or
d c
Pa
4
r w
quat
+ 9
nato
s
e in
w
ketc
atic
2
±
= –
r co
be
o.e
M2 f
and
h 1
t co
1−
r gr
uts
pe
42
work
tion
) o
or
nter
ww
ch o
c fo
22
– 3
orre
efor
e.
for
d ti
or 2
orre
1 o
adi
s x-
er
kin
n in
.e.
rme
w
of
orm
5
36 o
ect
re
at
ime
2 tr
ectl
o.e.
ient
axi
ng.
n th
edia
mula
or
lea
es
rial
ly i
. se
t of
is
e.g
he
ate
a or
ast 3
ls)
into
een
f 2
g.
r
3
o
n
Page 4 Mark Scheme Syllabus Paper
IGCSE – May/June 2013 0607 42
© Cambridge International Examinations 2013
(c) 73 −= xy o.e. 3 B2 for cxy += 3 o.e. or 7−= kxy o.e.
or M1 for rise/run = 13
42
−
−−
o.e.
and M1 for correct method for finding c
Answer 3x – 7 implies M1 M1
4 (a) 125.7 or 126 2 M1 if at least 2 mid-values seen (60, 125 and
195)
or 2 from 1680, 5625, 5265 or sum of 12570
(b) Columns from 100 to 150 and 150 to 240
Heights 0.9 and 0.3
1
1,1
Accept freehand
5 (a) 534.6… www 3 3 M1 for 9882 + 10602 – 2 × 988 × 1060 cos 30
A1 for 285800 to 285802 or 286000
(b)
99811852
5359981185][cos
222
××
−+=
26.6 (26.62 to 26.65)
M2
A1
Allow use of 534.6… for 535
M1 for correct implicit statement
Strictly dependent on at least M1
SC2 if correct without working
(c) 353 (353.3 to353.4) 1FT FT 380 – their (b) only if answer between 270
and 360.
6 (a) 720 2 M1 for 0.5 × 12 × 6 × 20 o.e.
(b) (i)
700 (700.2 to 700.4)
4 Allow 432 + 120 5 as final answer for full
marks
M1 for [BC2]= 122 + 62
M2 for BC × 20 + 12 × 20 + 6 × 20 + 2 × area
triangle ABC
or M1 if one of the five areas missing or is
incorrect
(ii) 3.5[0] (3.501 to 3.502) 1FT FT their (i) × 0.005
(c) 14.4 (14.42 to 14.43) 3 M1 for 202 + 122 (544) or 202 + 122 + 62 (580)
(Square roots 23.323…, 24.08…)
M1 for tan = 22
1220
6
+their
or
sin = 222
61220
6
++their
or cos = 222
22
61220
1220
++
+
their
their o.e.
7
8
9
(a
(b
(c
(a
(b
(a
(b
(c
Pa
a)
b)
) (
(i
a)
b)
a)
b)
)
age
(i)
ii)
e 5
[
6
7
0
–
0
x
x
x
9
9
9
5
0]6
669
752
0.41
– 1.
0.7
x Y
x >
x Y
9.18
9.77
9.02
6 30
9 (6
24
1
.49
798
Y –1
0
Y=0.
8 (9
7 or
2 to
0 o
669
cao
(–
(0
1.4
798
9.1
r 9
o 9.
o.e.,
9.0
o fi
1.4
0.79
9 o
8 o
77…
.78
.03
, [F
to
inal
496
976
or –
or 0
…)
8 (9
3
Feb
66
l an
6 to
6…
–1.4
0.79
)
9.77
b] 9
9.1
nsw
o –
…)
496
976
73 t
IG
©
9th
1)
wer
1.4
6 to
6…
to 9
GC
Ca
494
o –1
9.7
M
SE
amb
4)
1.4
75
w
Mar
E –
brid
94
….
ww
rk S
M
dge
.)
w 4
Sc
ay/
e In
che
/Ju
nter
eme
une
rnat
e
e 2
tion
201
nal
2
1
1
13
Ex
3
1
1
2FT
B2
B1
B1
1FT
1
1FT
3
2
4
xam
T
2
1
1
T
T
mina
B
B
o
s
F
0
B
S
o
C
F
F
M
o
w
M
A
e
M
o
M
atio
B1
B2
or M
een
FT
0.40
B1
Ske
or a
Con
FT
FT
M2
82
or M
with
M1
Allo
exac
M3
or M
M1
ons
for
for
M1
n
the
089
for
etch
a di
ndo
the
the
for2+
M1
hou
for
ow
ct
for
M1
for
(3
s 20
r [F
r [0
for
eir
9…
r po
h co
ffe
one
eir
eir
r 2
82
+
for
ut s
r 3
7
9
28
r 3
7
for
r 0
9.0
013
Feb]
]6
r 27
(d)
i.e
oor
ould
ren
< f
(a)
(a)
× 8
−
r
A
squa
360
70
9
8π
360
70
r 3
7
.5 ×
09 t
S
3
] 9t
30
72
1
)(i)
no
qu
d b
nt fu
for
if
if
8 ×
8.2
8
2
AB
are
0×π
o.e
0×π
360
70
× 8
to 3
Syll
06
th.
o.e
+
÷
ot 2
uali
be w
func
r Y
on
on
× si
8.8
B
=
e ro
×π
e. a
×π
0×π
8 ×
39.
lab
607
e.
13
184
2 dp
ty s
with
ctio
etc
e n
e p
n35
cos
sin
oot
16×
as f
8×
×π
8 s
10…
bus
7
so
400
p
ske
h d
on e
c
nega
posi
5 o
70s
n 35
6o.e
fina
–
8× a
sin7
…)
s
i or
0 B
etch
diffe
e.g
ativ
itiv
o.e.
0 (
5 o.
e.
al a
0.5
and
70
)
r B
1F
h
fere
g. y
ve r
ve r
. e
( 8
e.
ans
5 ×
d
o.e
B1 f
FT f
ent w
=y
roo
root
.g.
.84
or
we
8 ×
e.
(
P
for
for
win
2
x
ot
t
22.
r ab
er b
× 8
(30
Pa
4
17
0.4
ndo
− x
....
bov
but
sin
.1 o
pe
42
30
409
ow.
3−x
)
ve e
mu
n70
or 3
er
0 o
9 or
.
2−
exp
ust
0 o
30.
or 0
r
2
[1
pres
be
o.e
.07
3 0
14]
ssio
.
….
00
on
..)
10
11
0 (a
(b
(c
(d
(e
(f)
(g
(a
(b
Pa
a) (
(i
b)
)
d)
) (
(i
)
g)
a) (
(i
(ii
b) (
(i
age
(i)
ii)
(i)
ii)
(i)
ii)
ii)
(i)
ii)
e 6
x
(
7
2
0
–
4
C
1
3
6
x =
180
720
2
0 Y
–2 Y
42.9
Cor
6
4
6
2
1 o
6
5,
36
1
18
0, –
0
Y g(
Y g
9 (4
rrec
o.
o.
o.e
6
1
cao
0
– 1
(x)
g(x
42.
ct a
.e.
.e.
.
an
o
)
Y
) Y
94…
area
nd
2
Y 2
…)
a sh
6
5,
), 3
had
6
1
17
ded
an
IG
©
(3
.
nd
GC
Ca
17.
6
5
M
SE
amb
.0 t
Mar
E –
brid
to 3
rk S
M
dge
317
Sc
ay/
e In
7.1)
che
/Ju
nter
)
eme
une
rnat
e
e 2
tion
201
nal
13
Exxam
minaatio
1
1
ons
2
2
1
1,1
1
1
1
1
1,1
1
1
1
1
2
2
s 20013
B1
B1
Co
Co
Co
Co
or
in
Fo
pe
fo
Do
ca
Do
SC
B1
po
M
(0
0.
S
3
1 fo
1 fo
ond
ond
ond
ond
r y f
equ
or a
erc
or 3
o n
anc
o n
C1
1 fo
osit
M1 f
0.02
027
Syll
06
or i
or i
don
don
don
don
for
ual
all
cent
3 sf
not
cell
not
for
or a
tion
for
278
778
lab
607
ina
ina
ne π
ne π
ne 4
ne s
g(x
itie
pa
tag
f.
pe
ing
ac
r 0.
any
n
the
8 or
8)
bus
7
ccu
ccu
π f
π f
4 π
stri
x) a
es in
arts
ges
ena
g or
cep
.33
y on
eir
r 0
s
urat
urat
for
for
for
ct i
and
n p
s ac
wi
lise
r co
pt r
if
ne
6
1×
02.
te s
te s
180
180
r 72
ine
d al
part
cce
ith
e in
onv
rat
no
of t
th×
72
ske
ske
0
0
20
qua
llow
ts (
ept
the
nco
ver
tios
fra
the
heir
or
P
etch
etch
alit
w s
i) a
de
e u
orre
rtin
s or
acti
e 6
5
6
1r
0.0
Pa
4
h of
h of
ties
sepa
and
cim
usua
ect
ng.
r w
ion
in
027
pe
42
f co
f 2
s, al
ara
d (ii
mal
al r
t
word
n se
a c
777
er
osx
2sin
llow
ate
i)
ls o
rul
ds.
en
cor
7 to
x
n( 2
x
w x
or
les
.
rrec
2
x
)
x
ct
Page 7 Mark Scheme Syllabus Paper
IGCSE – May/June 2013 0607 42
© Cambridge International Examinations 2013
(iii)
36
11cao
2 M1 for 1 –6
5
6
5theirtheir ×
(0.306 or 530.0 or 0.3055 to
0.3056)
or 6
5
6
1
6
1×+
or (ii) + 2 × 6
5
6
1theirtheir ×
or 6
1
6
1theirtheir × + 2 ×
6
5
6
1theirtheir ×
(c) 5 2 SC1 for answer of 4 or 65 = 7776
seen or 54 = 625 seen
or M1 for attempted products of
1,6
1
6
5>×
ktheirtheir
k
12 (a) 18.75 (18.7 or 18.8)
18.5
23.5
13
1
1
1
1
(b) (i) r = –4.31t + 120 2 – 4.313…., 120.0…. B1 for
r = –4.31t + c or r = kt + 120
Allow x for t
(ii) Negative 1
(iii) 25 (25.1 to 25.4) 1FT FT their equation only if linear
13 (a) 3
2p + 3
1q o.e.
2 M1 for correct route from O to X or
PQ = q – p o.e. or correct
unsimplified answer
(b) –
3
2p + 3
5q o.e.
3 B2 for kp + 3
5q or –
3
2p + kq,
k≠ 0 or correct unsimplified
expression or
M1 for correct route from X to Y
or –3
2p + mq + nq, m 0,0 ≠≠ n
(c) 4± 1,1 If 0 scored , M1 for 22253 =+ k
o.e.