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CAMBRIDGE INTERNATIONAL EXAMINATIONS
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.
om
.c
MARK SCHEME for the May/June 2013 series
s
er
International General Certificate of Secondary Education
Pa
age
e2
1 (a
a) (i)
(
M
Marrk Sc
Sche
eme
e
IG
GCSE
E – May//Ju
une
e 201
2 13
S
Sylllab
bus
s
06
607
7
Pape
P
er
4
42
227.2
2 (27
( .22
2…)
1
(iii)
449 : 45
5 fina
f al aansw
werr
2
S
SC1
1 fo
or ans
a weer 45:4
4 49 or
o 1.009 (1.0
( 088
8 to
o 1.089
9)
oor B1
B for
f r 18
80 sseen (maay be
b iimp
plieed by
b theeir rati
r io
aansw
werr)
oor 27.2
2 2 (2
27.2
22...) : 25
5 o.e.
o
(iiii)
5500
0
3
M
M2 forr 72
20 ÷ 11.44 o.e.
o
oor M1
M forr 72
20 = 144
1 4% seen
(iv
v)
7748
8.8[[0] caao ffinaal answ
a werr
2
im
M
M1 forr 72
20 × 11.04 o.e.
o
mpplied by
b 749
7 9.
÷ 12
2 doess no
ot sspo
oil met
m tho
od
9 or
o 8.83
8 3 (8.8829 to 8.8
830
0)
3
S
SC2
2 fo
or ans
a weer 8.85
8 5
M
M2 forr lo
og (10
( 000
0/65
50) ÷ log
l g(1..05)
A
Allo
ow 1.5
54 forr 10
000
0/65
50 in m
meetho
ods.
oor M1
M forr 65
50 × 1.0
1 5n = 100
1 00
oor 650
6 mu
ultiipliied
d by
y 1..05 at leaast twiice coorreectly
(b
b)
M
M2 forr grrap
phs thaat inte
i erseect as sho
ow
wn
oor M1
M forr grrapph of
o y = 1..05x
2 (a
a)
(b
b)
22x( 2 x + 1) = 33x( x + 3) o.ee.
M2
2
Iff M0,
M , B1
1 fo
or 2xx ( 2 x + 1) orr 3x (xx + 3) Coond
donee
m
missin
ng brac
b ckeets if corr
c recct exxpaanssion
ns foll
f low
w
7
B2
2
B
B1 forr x 2 = 7 x o.ee. or
o 2(2
2 2x + 1)) = 3(xx + 3)) o.e.
oor fina
fi al an
nsw
werr 0,, 7
2y
1
=
y +1 3
M1
1
3 × 2y = y + 1 o.ee. or bettterr
1
5 o..e.
B1
1
i..e corr
c recct eq
quaatio
on wit
w thout ffractio
onss leeadiing
g to
1
y=
5
B1
1
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3
Pa
age
e3
(c) (i)
(
M
Marrk Sc
Sche
eme
e
IG
GCSE
E – May//Ju
une
e 201
2 13
110
so
oi
w
10
0
s
soi
w+9
B1
1
B1
1
3 (a
a)
(b
b) (i)
(
(iii)
Pape
P
er
4
42
Iff no
ot see
s en only
o y im
mplied
d by
b ffurttherr work
w kinng. e.g
g.
w
wx = 10 is
i not
n suffficcien
nt.
2 × 10(w
( + 9) − 2 × 10
0w = 5w
5 w(w + 9) ) o.ee. oor
bbettter
oor
M1
1
i..e. corrrecct mul
m ltip
pliccation off an eq
quattion
n inn th
he
ccorrrectt fo
orm
mat
aallow 10((w + 99) – 10
0w = 2.5
2 w(w
w + 9) o.e.
aalso
o alllow
w over
o r co
om
mmo
on den
d nom
min
nato
or
i..e. corrrecct coll
c lecttion
n of
o tw
wo terrmss
w 2 + 9w
9 w − 36
3 =0
E1
1
E
Estaabliish
hed wiith at leas
l st one
o e more
m e in
nterrmeediaate
liinee wiith no errrorrs or
o omi
o issionss
3 h 20 miin ww
ww
w
4
B
B3 forr w = 3 orr fo
or w = 3 or – 12
1 www
w
M
M2 forr (w
w + 122)(w − 3) or goood sk
ketcch of
o
qquadraaticc sh
how
wing
g zero
z os or
o qquaadraaticc fo
orm
mulaa orr
− 9 ± 225
ccom
mpleetin
ng sqquarre reac
r chiingg
2
oor M1
M forr (w
3 oor
w + a)(w
) w + b) wiith ab = – 36
a + b = 9 or skketcch of
o qua
q adraaticc orr co
orreect
subsstittutionss inn qu
uad
drattic forrmu
ula beeforre
pliffyin
ng or (w
o e.
simp
w + 92 ) 2 − 814 = 36
3 o.e
T
Triaal and
a im
mprooveementt – alloow
w M2
M for
f at leaast 3
trrialls for
f com
mpparison
n of diistaancces and
d tiimees
(uunllesss co
orreectt an
nsw
wer fou
undd with
w h 1 or 2 trrialls)
B
B1 FT
FT fo
or 10 ÷ th
heirr po
ositive ro
oott co
orreectlly into
i o
hhou
urs and
a d minu
m utes
f al ans
a sweer
y = − x − 1 o..e. fin
2
B
B1 forr y = − x + c o.e.
R
Rulled lin
ne th
hrooug
gh (–
( 2,
2 0)
0 aand
d (0, 4))
2
B
B1 forr ru
uled
d linne thr
t oug
gh (0,, 4) orr gradiientt off 2
90
0
= 2 12 o..e.
w(w
w + 9)
(iii)
S
Sylllab
bus
s
06
607
7
2 T
2FT
o . seeen
n
or y = kx − 1 o.e.
B
B1 forr
oor
oor tiiny
y paart not
n t sh
had
ded
F
FT only if
i pos
p itiv
ve gra
g dieent and
d cutss x-axiis
S
SC1
1 fo
or corr
c recct areaa un
nsh
hadeed
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Page 4
Mark Scheme
IGCSE – May/June 2013
y = 3 x − 7 o.e.
(c)
3
Syllabus
0607
Paper
42
B2 for y = 3 x + c o.e. or y = kx − 7 o.e.
2− − 4
or M1 for rise/run = 3−1 o.e.
and M1 for correct method for finding c
Answer 3x – 7 implies M1 M1
125.7 or 126
4 (a)
2
Columns from 100 to 150 and 150 to 240
Heights 0.9 and 0.3
(b)
534.6…
5 (a)
www 3
1
1,1
3
M1 if at least 2 mid-values seen (60, 125 and
195)
or 2 from 1680, 5625, 5265 or sum of 12570
Accept freehand
M1 for 9882 + 10602 – 2 × 988 × 1060 cos 30
A1 for 285800 to 285802 or 286000
(b)
(c)
6 (a)
(b) (i)
(ii)
(c)
1185 2 + 998 2 − 535 2
2 × 1185 × 998
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
353 (353.3 to353.4)
1FT
FT 380 – their (b) only if answer between 270
and 360.
[cos =]
720
2
M1 for 0.5 × 12 × 6 × 20 o.e.
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
3.5[0] (3.501 to 3.502)
1FT
14.4 (14.42 to 14.43)
3
FT their (i) × 0.005
M1 for 202 + 122 (544) or 202 + 122 + 62 (580)
(Square roots 23.323…, 24.08…)
6
M1 for tan =
sin =
or
20 2 + 12 2 + 6 2
their
or cos =
their 20 2 + 12 2
6
their
their
© Cambridge International Examinations 2013
20 2 + 12 2
20 2 + 12 2 + 6 2
o.e.
Pa
age
e5
7 (a
a)
M
Marrk Sc
Sche
eme
e
IG
GCSE
E – May//Ju
une
e 201
2 13
[0]6
6 30
0 o.e.,
o , [F
Feb
b] 9th
9
3
(b
b)
6669
9 (6
669
9.0 to 669.1
1)
1
(c) (i)
(
7752
24 cao
o fiinall an
nsw
wer
1
(iii)
00.41
1
2 T
2FT
a)
8 (a
B2
2
– 1..49 (– 1.4
4966 to
o – 1.4
494
4)
00.7
798 (0
0.79
9766…
…)
(b
b)
9 (a
a)
x Y –1
1.49 or
o ––1.4
496
6 to
o –1
1.494
x>0
x Y=0.
Y 798
8 or
o 00.79
976
6…
99.18
8 (9
9.177…
…))
S
Sylllab
bus
s
06
607
7
Pape
P
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4
42
B
B1 forr [F
Feb]] 9tth.
B
B2 forr [0]6 30 o.ee.
oor M1
M forr 27
7 12 + 13 soi orr B
B1 for
f 17 30
0 oor 03 00
0
seen
n
F
FT theeir (d))(i) ÷ 184
400
0 B1F
FT for
f 0.4
409
9 orr
00.40
089
9…i.e noot 2 dp
p
B
B1 forr po
oor quuality ske
s etch
h
S
Skeetch
h co
ould
d bbe with
w h diffe
d fereent win
w ndo
ow..
oor a differen
nt fu
funcctio
on e.g
e g. y =
2 3
−x −2
x
B1
1
B1
1
1FT
1
T
1
1 T
1FT
C
Con
ndo
one < for
f r Y etcc
F
FT theeir (a) if one nega
n ativ
ve rroo
ot
3
M
M2 forr 2 × 8 × sin35
5 o.e.
o . e.g.
F
FT theeir (a) if one posi
p itiv
ve rroott
[1
14]
84.22.... )
8 2 + 8 2 − 2.8.8 coss 70
0 ( 8
AB
AB
oor M1
M forr 2 = sin
n 35
5 o.e. orr ab
bov
ve exp
e presssio
on
8
w
with
hou
ut squa
s aree ro
oot
(b
b)
99.77
7 orr 9.78
8 (99.77
73 to
t 9.7
9 75…..)
(c)
99.02
2 to
o 9..03
3
www
w4
70
7
× π × 16
6 o.ee.
3 0
360
28
8π
A
Allo
ow
o.ee. as
a fina
f al ans
a weer but
b mu
ust be
9
eexacct
2
M
M1 forr
4
M
M3 forr
70
7
× π × 8 – 0.55 × 8 × 8 sinn700 oo.e.
3 0
360
7
70
× π × 8 and
oor M1
M forr
a d
3 0
360
M
M1 forr 0.5 × 8 × 8 sin7
s 70 o.ee.
(39.0
09 to
t 339.10…
…))
(30
( .1 or
o 30.
3 .07…...)
© Ca
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mina
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ons
s 20
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3
Pa
age
e6
M
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e
IG
GCSE
E – May//Ju
une
e 201
2 13
S
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bus
s
06
607
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Pape
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4
42
10
0 (a
a) ((i)
(iii)
2
B1
1 foor ina
i ccu
uratte sskeetch
h off co
osxx
2
B1
1 foor ina
i ccu
uratte sskeetch
h off 2sin
2 n( 2 )
1
Co
onddon
ne π for
f 1800
1,1
1
Co
onddon
ne π for
f 1800
x
(b
b)
x = 180
(c)
(180
0, – 1)
d)
(d
7720
0
2
1
1
Co
onddon
ne 4 π forr 7220
(e) (i)
(
0 Y g((x) Y 2
1
Co
onddon
ne stri
s ct ine
i quaalittiess, alllow
wx
orr y ffor g(xx) and
a d alllow
w sepa
s araate
inequualitiees in
n part
p ts (i) and
a d (iii)
––2 Y g(x
g )Y2
1
(iii)
(f))
442.9
9 (4
42.94…
…)), 317 (317..0 tto 317
3 7.1))
(g
g)
C
Corrrecct area
a a shhad
ded.
a) (i)
(
11 (a
4
6
o..e.
(iii)
2
6
o..e.
(iiii)
b) (i)
(
(b
(iii)
1
1 o.e
o .
5
6
1
1
or aall pa
artss accceept decim
malls oor
Fo
peerccenttag
ges wiith thee usua
u al rul
r les
fo
or 3 sff.
Do
on
not peenalisee in
nco
orreectt
ca
anccelling
g orr co
onvverrtin
ng.
Do
on
not accep
pt rat
r tioss orr word
w ds..
1
SC
C1 forr 0..33 if no fraactiion
n seen
1
5
1
5
, 6 an
nd 6 , 6 an
nd 6
1
cao
o
336
1,1
1
2
1 foor any
a y on
ne of tthee
B1
po
osittion
n
2
M ffor theeir
M1
5
6
in a cor
c rrecct
1
1
× thheirr
6
6
(0
0.02278
8 orr 0.02
2 7 or 0.00277777 to
0.027778
8)
© Ca
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Page 7
(iii)
Mark Scheme
IGCSE – May/June 2013
11
cao
36
Syllabus
0607
2
Paper
42
5
5
× their
6
6
(0.306 or 0.30 5 or 0.3055 to
0.3056)
M1 for 1 – their
or
1 1 5
+ ×
6 6 6
or
(ii) + 2 × their
1
5
× their
6
6
1
1
× their + 2 ×
6
6
1
5
their × their
6
6
or their
(c)
5
2
18.75 (18.7 or 18.8)
18.5
23.5
13
1
1
1
1
r = –4.31t + 120
2
(ii)
Negative
1
(iii)
25 (25.1 to 25.4)
12 (a)
(b) (i)
13 (a)
2
3
p+
1
3
q o.e.
1FT
2
SC1 for answer of 4 or 65 = 7776
seen or 54 = 625 seen
or M1 for attempted products of
k
5
1

 their  × their , k > 1
6
6


– 4.313…., 120.0…. B1 for
r = –4.31t + c or r = kt + 120
Allow x for t
FT their equation only if linear
M1 for correct route from O to X or
PQ = q – p o.e. or correct
unsimplified answer
(b)
–
2
3
p+
5
3
q
o.e.
3
5
or –
(c)
±4
1,1
2
B2 for kp + 3 q or – 3 p + kq,
k ≠ 0 or correct unsimplified
expression or
M1 for correct route from X to Y
2
3
p + mq + nq, m ≠ 0, n ≠ 0
If 0 scored , M1 for 32 + k 2 = 52
o.e.
© Cambridge International Examinations 2013