CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge International General Certificate of Secondary Education
MARK SCHEME for the October/November 2015 series
0478 COMPUTER SCIENCE
0478/23 Paper 2, maximum raw mark 50
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 October/November 2015 series for most
Cambridge IGCSE®, Cambridge International A and AS Level components and some
Cambridge O Level components.
® IGCSE is the registered trademark of Cambridge International Examinations.
Page 2 Mark Scheme Syllabus Paper
Cambridge IGCSE – October/November 2015 0478 23
Section A
1 (a) (i) Any two variables with matching uses, one mark for the variable and one mark for the
matching use. The variables and the matching uses must relate to the tasks on the exam
paper. There are many possible correct answers these are examples only.
Variable 1 – Counter(: INTEGER)
Use – to use as a loop counter when entering the temperatures
Variable 2 – HighestTemperature(: REAL)
Use – to store the highest temperature recorded [4]
(ii) Any two constant with matching uses, one mark for the constant and one mark for the
matching use. The constants and the matching uses must relate to the tasks on the
exam paper. There are several possible correct answers these are examples only.
Constant 1 – MinAppartmentTemperature = 21.5/22
Use – to keep the temperature when the air-conditioning should be
switched off
Constant 2 – MaxAppartmentTemperature = 24.5/24
Use – to keep the temperature when the air-conditioning should be
switched on [4]
(b) Any four from:
– initialisation, set highest apartment temperature to a low value, set lowest apartment
temperature to a high value outside loop
– input temperature
– store in array
– test for temperature > highest apartment temperature reset highest apartment
temperature if this is the case
– test for temperature < lowest apartment temperature reset lowest apartment temperature
if this is the case
– calculate range
– output highest temperature, lowest temperature and the range outside loop
(Max four marks)
– loop 60 times must have both tests within the loop, initialisation before the loop and
output after the loop (One mark) [5]
sample algorithm:
HighestTemp 0; LowestTemp 100
FOR Count 1 to 60
INPUT Temperature
ApartmentTemp[Count] Temperature
IF ApartmentTemp[Count] > HighestTemp
THEN HighestTemp ApartmentTemp[Count]
ENDIF
IF ApartmentTemp[Count] < LowestTemp
THEN LowestTemp ApartmentTemp[Count]
ENDIF
NEXT Count
Range HighestTemp – LowestTemp
PRINT 'Highest Temperature recorded ', HighestTemp
PRINT 'Lowest Temperature recorded ', LowestTemp
PRINT 'Range ', Range
© Cambridge International Examinations 2015
Page 3 Mark Scheme Syllabus Paper
Cambridge IGCSE – October/November 2015 0478 23
(c) (i) Explanation six marks from:
1 – check if highest temperature <= 24 and lowest temperature >= 22…
– … message temperature always within acceptable range then exit
2 – check if highest out of range
– so count number of times temperature goes above range
– message recorded temperature too high on counted number of occasions
3 – check if lowest out of range
– so count number of times temperature goes below range
– message recorded temperature too low on counted number of occasions
General
– check all recorded temperatures (loop)
[6]
(ii) Any one from:
– only checks necessary conditions
– uses results from task 2 [1]
© Cambridge International Examinations 2015
Page 4 Mark Scheme Syllabus Paper
Cambridge IGCSE – October/November 2015 0478 23
Section B
2 One mark for each error identified + suggested correction
line 5 or IF Num < 0: this should read IF Num > 0 (THEN Total = Total + Num)
line 6 or (IF Num > 0 ) THEN Counter = Counter + 1:
this should read (IF Num > 0 THEN)Poscount = Poscount + 1
line 7 Average = Total/Poscount: this should come after the end of the repeat loop
line 9 or PRINT Num: this should read PRINT Average [4]
3 (a) Number 1 Trace Table
X T1 T2 Output
37 2 5 5
2 2
(1 mark) (1 mark)
Number 2 Trace Table
X T1 T2 Output
191 11 15 F
11 B
(1 mark) (1 mark) [4]
(b) – convert a denary number to hexadecimal
– and output it in reverse order [2]
4 (a) (i) Normal
(ii) Acceptable data to test that the results are as expected. [2]
(b) One mark for the data set, one mark for the type and one mark for the matching reason
There are many possible correct answers this is an example only.
Set 1 – Age 4, height 0.9
Type – Boundary/Extreme
Reason – Data to test the validation that is just within the limits of acceptability
Set 2 – Age 10, height 1.4
Type – Abnormal
Reason – Data that should be rejected and produce an error message [6]
© Cambridge International Examinations 2015
Page 5 Mark Scheme Syllabus Paper
Cambridge IGCSE – October/November 2015 0478 23
5 One mark for every two correct types
Boat Name – text
Model – text
Engine Power – number
Number of Seats – number
Life Raft – “yes/no”/text/Boolean
Day Price – currency/number
0, 1 no marks
2, 3 one mark
4, 5 two marks
6 three marks [3]
(b) One mark for each correct different check
Boat Name Presence Check/Type Check/Character Check
Model Format check/Type check/Presence Check/Length check/
Use of Drop-down box to select
Number of Seats Type check/Presence Check/Range Check/
Use of Drop-down box to select
Day Price Type check/Presence Check/Range Check [4]
(c)
Field: Boat Name Model Day Price Number of Engine
Seats Power
Table: BOAT BOAT BOAT BOAT BOAT
Sort:
Show:
Criteria: =4 > 100
or:
(1 mark) (1mark) (1 mark) (1 mark) (1 mark) [5]
© Cambridge International Examinations 2015
