computer science for examination from 2023
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Cambridge IGCSE™
COMPUTER SCIENCE 0478/02Paper 2 Algorithms, Programming and Logic For examination from 2023MARK SCHEME
Maximum Mark: 75
Specimen
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Gen
eric
Mar
king
Prin
cipl
es
Thes
e ge
nera
l mar
king
prin
cipl
es m
ust b
e ap
plie
d by
all
exam
iner
s w
hen
mar
king
can
dida
te a
nsw
ers.
The
y sh
ould
be
appl
ied
alon
gsid
e th
e sp
ecifi
c co
nten
t of t
he m
ark
sche
me
or g
ener
ic le
vel d
escr
ipto
rs fo
r a q
uest
ion.
Eac
h qu
estio
n pa
per a
nd m
ark
sche
me
will
als
o co
mpl
y w
ith th
ese
mar
king
pr
inci
ples
.
GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
1:
Mar
ks m
ust b
e aw
arde
d in
line
with
:
•th
e sp
ecifi
c co
nten
t of t
he m
ark
sche
me
or th
e ge
neric
leve
l des
crip
tors
for t
he q
uest
ion
•th
e sp
ecifi
c sk
ills
defin
ed in
the
mar
k sc
hem
e or
in th
e ge
neric
leve
l des
crip
tors
for t
he q
uest
ion
•th
e st
anda
rd o
f res
pons
e re
quire
d by
a c
andi
date
as
exem
plifi
ed b
y th
e st
anda
rdis
atio
n sc
ripts
.
GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
2:
Mar
ks a
war
ded
are
alw
ays
who
le m
arks
(not
hal
f mar
ks, o
r oth
er fr
actio
ns).
GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
3:
Mar
ks m
ust b
e aw
arde
d po
sitiv
ely:
•m
arks
are
aw
arde
d fo
r cor
rect
/val
id a
nsw
ers,
as
defin
ed in
the
mar
k sc
hem
e. H
owev
er, c
redi
t is
give
n fo
r val
id a
nsw
ers
whi
ch g
o be
yond
the
scop
e of
the
sylla
bus
and
mar
k sc
hem
e, re
ferr
ing
to y
our T
eam
Lea
der a
s ap
prop
riate
•m
arks
are
aw
arde
d w
hen
cand
idat
es c
lear
ly d
emon
stra
te w
hat t
hey
know
and
can
do
•m
arks
are
not
ded
ucte
d fo
r err
ors
•m
arks
are
not
ded
ucte
d fo
r om
issi
ons
•an
swer
s sh
ould
onl
y be
judg
ed o
n th
e qu
ality
of s
pelli
ng, p
unct
uatio
n an
d gr
amm
ar w
hen
thes
e fe
atur
es a
re s
peci
fical
ly a
sses
sed
by th
e qu
estio
n as
indi
cate
d by
the
mar
k sc
hem
e. T
he m
eani
ng, h
owev
er, s
houl
d be
una
mbi
guou
s.
GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
4:
Rul
es m
ust b
e ap
plie
d co
nsis
tent
ly, e
.g. i
n si
tuat
ions
whe
re c
andi
date
s ha
ve n
ot fo
llow
ed in
stru
ctio
ns o
r in
the
appl
icat
ion
of g
ener
ic le
vel
desc
ripto
rs.
GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
5:
Mar
ks s
houl
d be
aw
arde
d us
ing
the
full
rang
e of
mar
ks d
efin
ed in
the
mar
k sc
hem
e fo
r the
que
stio
n (h
owev
er; t
he u
se o
f the
full
mar
k ra
nge
may
be
lim
ited
acco
rdin
g to
the
qual
ity o
f the
can
dida
te re
spon
ses
seen
).
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IMEN
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GE
NE
RIC
MA
RK
ING
PR
INC
IPLE
6:
Mar
ks a
war
ded
are
base
d so
lely
on
the
requ
irem
ents
as
defin
ed in
the
mar
k sc
hem
e. M
arks
sho
uld
not b
e aw
arde
d w
ith g
rade
thre
shol
ds o
r gra
de
desc
ripto
rs in
min
d.
Mar
k sc
hem
e ab
brev
iatio
ns
/
se
para
tes
alte
rnat
ive
wor
ds /
phra
ses
with
in a
mar
king
poi
nt//
se
para
tes
alte
rnat
ive
answ
ers
with
in a
mar
king
poi
ntun
derli
ne a
ctua
l wor
d gi
ven
mus
t be
used
by
cand
idat
e (g
ram
mat
ical
var
iant
s ac
cept
ed)
max
indi
cate
s th
e m
axim
um n
umbe
r of m
arks
that
can
be
awar
ded
( )
the
wor
d / p
hras
e in
bra
cket
s is
not
requ
ired,
but
set
s th
e co
ntex
t
Not
e: N
o m
arks
are
aw
arde
d fo
r usi
ng b
rand
nam
es o
f sof
twar
e pa
ckag
es o
r har
dwar
e.
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Que
stio
nA
nsw
erM
arks
1(a)
One
mar
k fo
r eac
h co
rrec
t lin
e.
Pseu
doco
de d
escr
iptio
nPs
eudo
code
sta
tem
ent
a lo
op th
at w
ill a
lway
s ite
rate
at
leas
t onc
e
a co
nditi
onal
sta
tem
ent t
o de
alw
ith m
any
poss
ible
out
com
es
a lo
op th
at w
ill a
lway
s ite
rate
a s
etnu
mbe
r of t
imes
a co
nditi
onal
sta
tem
ent w
ithdi
ffere
nt o
utco
mes
for t
rue
and
fals
e
FOR…TO…NEXT
IF…THEN…ELSE…ENDIF
WHILE…DO…ENDWHILE
CASE…OF…OTHERWISE…ENDCASE
REPEAT…UNTIL
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Que
stio
nA
nsw
erM
arks
1(b)
•ap
prop
riate
loop
con
trols
•re
ad fr
om a
rray
•ou
tput
from
arr
ay (t
he la
st tw
o po
ints
can
be
in o
ne s
tate
men
t, se
e ex
ampl
e)
Not
e: re
adin
g an
d th
e ou
tput
MU
ST b
e w
ithin
the
sam
e lo
op.
For e
xam
ple:
Count ← 0
WHILE Count < 50 DO
OUTPUT Name[Count]
Count ← Count + 1
ENDWHILE
3
Que
stio
nA
nsw
erM
arks
2Va
lidat
ion
chec
k
One
mar
k fo
r des
crip
tion:
•To
test
if th
e da
ta e
nter
ed is
pos
sibl
e / r
easo
nabl
e / s
ensi
ble.
•A
rang
e ch
eck
test
s th
at d
ata
ente
red
fits
with
in s
peci
fied
valu
es.
One
mar
k fo
r exa
mpl
e: •
Allo
w a
ny c
orre
ct v
alid
atio
n ch
eck
as a
n ex
ampl
e (r
ange
, len
gth,
type
, pre
senc
e, fo
rmat
, etc
.).
Verif
icat
ion
chec
k
One
mar
k fo
r des
crip
tion:
•
To te
st if
the
data
inpu
t is
the
sam
e as
the
data
that
was
inte
nded
to b
e in
put.
•A
doub
le e
ntry
che
ck e
xpec
ts e
ach
item
of d
ata
to b
e en
tere
d tw
ice
and
com
pare
s bo
th e
ntrie
s to
che
ck th
ey a
re th
e sa
me.
One
mar
k fo
r exa
mpl
e:A
llow
any
cor
rect
ver
ifica
tion
chec
k as
an
exam
ple
(vis
ual,
doub
le e
ntry
, etc
.).
4
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Que
stio
nA
nsw
erM
arks
3B
1
Que
stio
nA
nsw
erM
arks
4O
ne m
ark
for a
hie
rarc
hica
l stru
ctur
e.O
ne m
ark
for s
uita
ble
nam
es fo
r the
sub
-sys
tem
s.O
ne m
ark
for i
dent
ifiab
le in
puts
.O
ne m
ark
for i
dent
ifiab
le o
utpu
ts.
For e
xam
ple:
Sat
ellit
e na
viga
tion
syst
em
Inpu
t des
tinat
ion
Out
put d
irect
ions
New
des
tinat
ion
Sav
ed d
estin
atio
nM
apLi
st
4
Que
stio
nA
nsw
erM
arks
5(a)
One
mar
k fo
r eac
h er
ror i
dent
ified
and
cor
rect
ion:
•Numbers
sho
uld
be Number
•IF Number > 100
sho
uld
be IF Number >= 100
•
INPUT Number
is m
issi
ng fr
om in
side
the
loop
, ins
ert INPUT Number
afte
r the
ENDIF
sta
tem
ent.
•Th
e fin
al OUTPUT Number
is n
ot n
eede
d, re
mov
e it.
4
5(b)
One
mar
k fo
r bot
h en
ds o
f the
rang
e an
d co
rrec
t ine
qual
ity s
ymbo
ls.
One
mar
k fo
r the
AN
D //
nes
ted
IFs.
The
test
sho
uld
be IF Number >= 100 AND Number <= 200
2
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Que
stio
nA
nsw
erM
arks
6(a)
One
mar
k fo
r eac
h co
rrec
t gat
e, w
ith th
e co
rrec
t inp
ut(s
) as
show
n.
A B C
X
4
6(b)
AB
CX
00
00
00
10
01
01
01
10
10
00
10
10
11
01
11
11
Four
mar
ks fo
r eig
ht c
orre
ct o
utpu
ts.
Thre
e m
arks
for s
ix o
r sev
en c
orre
ct o
utpu
ts.
Two
mar
ks fo
r fou
r or f
ive
corr
ect o
utpu
ts.
One
mar
k fo
r tw
o or
thre
e co
rrec
t out
puts
.
4
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IMEN
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Que
stio
nA
nsw
erM
arks
7(a)
Flag
Count
Name[1]
Name[2]
Name[3]
Name[4]
Temp
Jamal
Amir
Eve
Tara
01
Amir
Jamal
Eve
Tara
Jamal
12
Amir
Eve
Jamal
Tara
Jamal
13
Amir
Eve
Jamal
Tara
Jamal
14
Amir
Eve
Jamal
Tara
Jamal
01
Amir
Eve
Jamal
Tara
Jamal
02
Amir
Eve
Jamal
Tara
Jamal
03
Amir
Eve
Jamal
Tara
Jamal
04
Amir
Eve
Jamal
Tara
Jamal
One
mar
k fo
r Fla
g co
lum
n.O
ne m
ark
for C
ount
col
umn.
One
mar
k fo
r Tem
p co
lum
n.Tw
o m
arks
for a
ll co
rrec
t Nam
e co
lum
ns o
r one
mar
k fo
r tw
o or
thre
e co
rrec
t Nam
e co
lum
ns.
Not
e: R
epea
ted
valu
es d
o no
t nee
d to
be
writ
ten
unle
ss a
val
ue is
rew
ritte
n.
5
7(b)
•bu
bble
sor
t / s
ortin
g th
e na
mes
•as
cend
ing
orde
r / A
to Z
/ lo
wes
t to
high
est /
alp
habe
tical
ord
er2
Que
stio
nA
nsw
erM
arks
8O
ne m
ark
for e
ach.
10.0
0 b
ound
ary
/ abn
orm
al d
ata
// th
e pr
ice
shou
ld b
e re
ject
ed //
val
ue is
out
of r
ange
9.99
b
ound
ary
/ ext
rem
e / n
orm
al d
ata
// th
e pr
ice
shou
ld b
e ac
cept
ed //
val
ue is
with
in n
orm
al ra
nge
ten
a
bnor
mal
dat
a //
inpu
t sho
uld
be re
ject
ed //
val
ue is
wro
ng ty
pe
3
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Que
stio
nA
nsw
erM
arks
9A
ny th
ree
from
: •
data
is n
ot lo
st w
hen
the
com
pute
r is
switc
hed
off /
/ dat
a is
sto
red
perm
anen
tly •
data
can
be
used
by
mor
e th
an o
ne p
rogr
am o
r reu
sed
whe
n a
prog
ram
is ru
n ag
ain
•da
ta c
an b
e ba
cked
up
or a
rchi
ved
•da
ta c
an b
e tra
nspo
rted
from
one
pla
ce /
syst
em to
ano
ther
.
3
Que
stio
nA
nsw
erM
arks
10C
1
Que
stio
nA
nsw
erM
arks
11(a
)20
1
11(b
)(i)
CatNo
1
11(b
)(ii)
it is
a u
niqu
e id
entif
ier
1
11(c
)Tw
o m
arks
for f
our c
orre
ct a
nsw
ers.
One
mar
k fo
r tw
o or
thre
e co
rrec
t ans
wer
s.
Fiel
dD
ata
type
CatNo
Text
Title
Text
Genre1
Text
Streaming
Boo
lean
/ Te
xt
2
11(d
)FROM
"Comedy"
2
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Que
stio
nA
nsw
erM
arks
12(a
)O
ne m
ark
for e
ach
corr
ect l
ine.
DECLARE X : STRING
DECLARE Y : INTEGER
DECLARE Z : INTEGER
3
12(b
)O
ne m
ark
for s
torin
g st
ring
in X
.O
ne m
ark
for c
allin
g th
e fu
nctio
n le
ngth
.O
ne m
ark
for u
sing
the
corr
ect p
aram
eter
X.
One
mar
k fo
r usi
ng th
e su
bstri
ng fu
nctio
n.O
ne m
ark
for c
orre
ct p
aram
eter
s.O
ne m
ark
for o
utpu
tting
leng
th a
nd s
ubst
ring
retu
rn v
alue
s.Fo
r exa
mpl
e:X ← "Programming is fun"
OUTPUT Length(X)
Y ← 16
Z ← 3
OUTPUT SubString(X,Y,Z)
6
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Que
stio
nA
nsw
erM
arks
13R
ead
the
who
le a
nsw
er, a
war
d a
mar
k fro
m b
oth
of th
e fo
llow
ing
tabl
es a
nd a
dd u
p th
e to
tal.
Mar
ks a
re a
vaila
ble
for:
•A
O2
(max
imum
9 m
arks
) •
AO
3 (m
axim
um 6
mar
ks).
The
tech
niqu
es a
nd th
e da
ta s
truct
ures
requ
ired
are
liste
d be
low
. The
requ
irem
ents
may
be
met
usi
ng a
sui
tabl
e bu
ilt-in
fu
nctio
n fro
m th
e pr
ogra
mm
ing
lang
uage
use
d (e
.g. P
ytho
n, V
B.N
ET
or J
ava)
.
Tech
niqu
es re
quire
d:R
1 C
alcu
late
tota
l mar
k fo
r eac
h st
uden
t (ite
ratio
n an
d to
talli
ng).
R2
Cal
cula
te a
vera
ge m
ark
for e
ach
stud
ent r
ound
ed to
the
near
est w
hole
num
ber.
R3
Sel
ectio
n of
gra
de fo
r eac
h st
uden
t (se
lect
ion)
.R
4 O
utpu
t for
eac
h st
uden
t nam
e, to
tal m
ark,
ave
rage
mar
k, g
rade
aw
arde
d (o
utpu
t with
app
ropr
iate
mes
sage
s).
R5
Cal
cula
te, s
tore
and
out
put t
he n
umbe
r of d
istin
ctio
ns, m
erits
, pas
ses
and
fails
for t
he w
hole
cla
ss (i
tera
tion,
co
untin
g an
d ou
tput
with
app
ropr
iate
mes
sage
s).
Dat
a st
ruct
ures
requ
ired:
The
nam
es u
nder
lined
mus
t be
used
as
prov
ided
in th
e sc
enar
io.
Arr
ays
or li
sts StudentName
, StudentMark
,
(TotalMark
and
AverageMark
may
be
seen
but
no
requ
irem
ent t
o st
ore)
Varia
bles
ClassSize
, SubjectNo
, SubjectCounter
, StudentCounter
DistinctionNo
, MeritNo
, PassNo
, FailNo
cou
ld b
e an
arr
ay o
r lis
t
Con
stan
ts
Distinction
, Merit
, Pass
cou
ld b
e va
riabl
es
15
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Que
stio
nA
nsw
erM
arks
13Ex
ampl
e 15
mar
k an
swer
in p
seud
ocod
e.
// meaningful identifier names and appropriate data structures (variables, constants and the
// given arrays) to store all the data required
DECLARE TotalMark : ARRAY[1:50] OF INTEGER
DECLARE AverageMark : ARRAY[1:50] OF INTEGER
DECLARE SubjectCounter : INTEGER
DECLARE StudentCounter : INTEGER
DECLARE DistinctionNo : INTEGER
DECLARE MeritNo : INTEGER
DECLARE PassNo : INTEGER
DECLARE FailNo : INTEGER
CONSTANT Distinction = 70
CONSTANT Merit = 55
CONSTANT Pass = 40
// initialisation processes for this scenario, initialising the running totals used for
// grades and combined totals
DistinctionNo ← 0
MeritNo ← 0
PassNo ← 0
FailNo ← 0
FOR StudentCounter ← 1 to ClassSize
TotalMark[StudentCounter] ← 0
NEXT StudentCounter
// programming techniques of iteration, selection, totalling, counting and output are used
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Que
stio
nA
nsw
erM
arks
13FOR StudentCounter ← 1 to ClassSize
FOR SubjectCounter ← 1 to SubjectNo
TotalMark[StudentCounter] ← TotalMark[StudentCounter] + StudentMark[StudentCounter,
SubjectCounter]
NEXT SubjectCounter
AverageMark[StudentCounter] ← INT((TotalMark[StudentCounter] / SubjectNo) + 0.5)
OUTPUT "Name ", StudentName[StudentCounter]
OUTPUT "Combined total mark ", TotalMark[StudentCounter]
OUTPUT "Average mark ", AverageMark[StudentCounter]
IF AverageMark[StudentCounter] >= Distinction
THEN
DistinctionNo ← DistinctionNo + 1
OUTPUT "Grade Distinction"
ELSE
IF AverageMark[StudentCounter] >= Merit
THEN
MeritNo ← MeritNo + 1
OUTPUT "Grade Merit"
ELSE
IF AverageMark[StudentCounter] >= Pass
THEN
PassNo ← PassNo + 1
OUTPUT "Grade Pass"
ELSE
FailNo ← FailNo + 1
OUTPUT "Grade Fail"
ENDIF
ENDIF
ENDIF
NEXT StudentCounter
OUTPUT "Number of Distinctions ", DistinctionNo
OUTPUT "Number of Merits ", MeritNo
OUTPUT "Number of Passes ", PassNo
OUTPUT "Number of Fails ", FailNo
0478
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SPEC
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LES
202
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AO
2: A
pply
kno
wle
dge
and
unde
rsta
ndin
g of
the
prin
cipl
es a
nd c
once
pts
of c
ompu
ter s
cien
ce to
a g
iven
con
text
, inc
ludi
ng th
e an
alys
is
and
desi
gn o
f com
puta
tiona
l or p
rogr
amm
ing
prob
lem
s
01–
34–
67–
9
No
cred
itabl
e re
spon
se
At l
east
one
pro
gram
min
g te
chni
que
has
been
use
d.
Any
use
of s
elec
tion,
iter
atio
n,
coun
ting,
tota
lling
, inp
ut a
nd o
utpu
t.
Som
e pr
ogra
mm
ing
tech
niqu
es u
sed
are
appr
opria
te to
the
prob
lem
.
Mor
e th
an o
ne te
chni
que
seen
app
lied
to th
e sc
enar
io, r
efer
to th
e lis
t of
tech
niqu
es n
eede
d.
The
rang
e of
pro
gram
min
g te
chni
ques
us
ed is
app
ropr
iate
to th
e pr
oble
m.
All
crite
ria s
tate
d fo
r the
sce
nario
ha
ve b
een
cove
red
by th
e us
e of
ap
prop
riate
pro
gram
min
g te
chni
ques
, re
fer t
o th
e lis
t of t
echn
ique
s ne
eded
.
Som
e da
ta h
as b
een
stor
ed b
ut n
ot
appr
opria
tely.
Any
use
of v
aria
bles
or a
rray
s or
oth
er
lang
uage
-dep
ende
nt d
ata
stru
ctur
es,
e.g.
Pyt
hon
lists
.
Som
e of
the
data
stru
ctur
es c
hose
n ar
e ap
prop
riate
and
sto
re s
ome
of th
e da
ta re
quire
d.
Mor
e th
an o
ne d
ata
stru
ctur
e us
ed
to s
tore
dat
a th
at is
requ
ired
by th
e sc
enar
io.
The
data
stru
ctur
es c
hose
n ar
e ap
prop
riate
and
sto
re a
ll th
e da
ta
requ
ired.
The
data
stru
ctur
es u
sed
stor
e al
l the
da
ta th
at is
requ
ired
by th
e sc
enar
io.
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fro
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Pag
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LES
202
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AO
3: P
rovi
de s
olut
ions
to p
robl
ems
by:
•ev
alua
ting
com
pute
r sys
tem
s •
mak
ing
reas
oned
judg
emen
ts
•pr
esen
ting
conc
lusi
ons
01–
23–
45–
6
No
cred
itabl
e re
spon
se
Pro
gram
see
n w
ithou
t rel
evan
t co
mm
ents
.P
rogr
am s
een
with
som
e re
leva
nt
com
men
t(s).
The
prog
ram
has
bee
n fu
lly
com
men
ted.
Som
e id
entif
ier n
ames
use
d ar
e ap
prop
riate
.
Som
e of
the
data
stru
ctur
es u
sed
have
m
eani
ngfu
l nam
es.
The
maj
ority
of i
dent
ifier
s us
ed a
re
appr
opria
tely
nam
ed.
Mos
t of t
he d
ata
stru
ctur
es u
sed
have
m
eani
ngfu
l nam
es.
Sui
tabl
e id
entif
iers
with
nam
es
mea
ning
ful t
o th
eir p
urpo
se h
ave
been
us
ed th
roug
hout
.
All
the
data
stru
ctur
es u
sed
have
m
eani
ngfu
l nam
es.
The
solu
tion
is il
logi
cal.
The
solu
tion
cont
ains
par
ts th
at m
ay
be il
logi
cal.
The
prog
ram
is in
a lo
gica
l ord
er.
The
solu
tion
is in
accu
rate
in m
any
plac
es.
Sol
utio
n co
ntai
ns fe
w li
nes
of c
ode,
w
ith e
rror
s, th
at a
ttem
pt to
per
form
a
task
giv
en in
the
scen
ario
.
The
solu
tion
cont
ains
par
ts th
at a
re
inac
cura
te.
Sol
utio
n co
ntai
ns li
nes
of c
ode,
with
so
me
erro
rs, t
hat l
ogic
ally
per
form
ta
sks
give
n in
the
scen
ario
. Ign
ore
min
or s
ynta
x er
rors
.
The
solu
tion
is a
ccur
ate.
Sol
utio
n lo
gica
lly p
erfo
rms
all t
he ta
sks
give
n in
the
scen
ario
. Ign
ore
min
or
synt
ax e
rror
s.
The
solu
tion
atte
mpt
s at
leas
t one
of
the
requ
irem
ents
.
Sol
utio
n co
ntai
ns li
nes
of c
ode
that
at
tem
pt a
t lea
st o
ne ta
sk g
iven
in th
e sc
enar
io.
The
solu
tion
mee
ts m
ost o
f the
re
quire
men
ts.
Sol
utio
n co
ntai
ns li
nes
of c
ode
that
pe
rform
mos
t tas
ks g
iven
in th
e sc
enar
io.
The
solu
tion
mee
ts a
ll th
e re
quire
men
ts g
iven
in th
e qu
estio
n.
Sol
utio
n pe
rform
s al
l the
task
s gi
ven
in
the
scen
ario
.
0478
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chem
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Pag
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LES
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0
BLA
NK
PA
GE