® IGCSE is the registered trademark of Cambridge International Examinations. CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge Ordinary Level MARK SCHEME for the May/June 2015 series 2210 COMPUTER SCIENCE 2210/11 Paper 1, maximum raw mark 75 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 2015 series for most Cambridge IGCSE ® , Cambridge International A and AS Level components and some Cambridge O Level components.
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® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge Ordinary Level
MARK SCHEME for the May/June 2015 series
2210 COMPUTER SCIENCE
2210/11 Paper 1, maximum raw mark 75
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 2015 series for most Cambridge IGCSE
®, Cambridge International A and AS Level components and some
1 (a) parallel any one from: – 8 bits/1 byte/multiple bits sent at a time – using many/multiple/8 wires/lines (1 mark) serial any one from: – one bit sent at a time – over a single wire (1 mark) [2] (b) parallel – faster rate of data transmission (1 mark) serial any one from: – more accurate/fewer errors over a longer distance – less expensive wiring – less chance of data being skewed/out of synchronisation/order (1 mark) [2] (c) parallel any one from: – sending data from a computer to a printer – internal data transfer (buses) (1 mark) serial – connect computer to a modem (1 mark) [2]
2 (a) – universal serial bus – description of USB [1]
(b) Any two from:
– devices are automatically detected and configured when initially attached – impossible to connect device incorrectly/connector only fits one way – has become the industry standard – supports multiple data transmission speeds – lots of support base for USB software developers – supported by many operating systems – backward compatible – faster transmission compared to wireless [2]
(A is 1 OR B is 1) (1 mark) AND (1 mark) (B is 1 OR C is NOT 1) (1 mark) accept equivalent ways of writing this: e.g. (A OR B = 1) AND (B OR NOT C = 1) e.g. (A OR B) AND (B OR NOT C) e.g. (A + B) (B + C) [3]
2 web server name accept these three items in any order
3 file name
HTML tags/text
firewall
proxy server
[6]
5 1 mark per device, 1 mark per category
Description of storage device Name of storage device
Category of storage
Primary Secondary Off-line
optical media which uses one spiral track; red lasers are used to read and write data on the media surface; makes use of dual-layering technology to increase the storage capacity
DVD
�
non-volatile memory chip; contents of the chip cannot be altered; it is often used to store the start-up routines in a computer (e.g. the BIOS)
ROM �
optical media which uses concentric tracks to store the data; this allows read and write operations to be carried out at the same time
DVD-RAM � (�)
non-volatile memory device that uses NAND flash memories (which consist of millions of transistors wired in series on single circuit boards)
Solid State Drive/memory
(SSD)
(SD/XD card) (USB storage
device)
�
(�)
optical media that uses blue laser technology to read and write data on the media surface; it uses a single 1.1 mm polycarbonate disc
– program/software that replicates/copies itself – can delete or alter files/data stored on a computer – can make the computer “crash”/run slow
pharming any two from:
– malicious code/software installed on a user’s hard drive/actual web server – this code redirects user to a fake website (without their knowledge) – to obtain personal/financial information/data
phishing any two from:
– legitimate-looking emails sent to a user – as soon as recipient opens/clicks on link in the email/attachment … – … the user is directed to a fake website (without their knowledge) – To obtain personal/financial information/data
[6]
(b) (i) Any two from:
– spyware/key logging software can only pick up key presses – using mouse/touchscreen means no key presses to log – the numbers on the key pad are in random/non-standard format, which makes it more difficult to interpret [2]
(ii) 1 mark for name and 1 mark for description any one from: chip and PIN reader – only the user and the bank know which codes can be generated request user name – additional security together with password/PIN anti-virus – removes/warns of a potential virus threat which can’t be passed on to customers firewall – (helps) to protect bank computers from virus threats and hacking encryption – protects customer data by making any hacked information unreadable security protocol – governs the secure transmission of data Biometric – to recognise user through the use of, e.g. facial/retina/finger print Alerts – users IP/MAC address is registered and user is alerted through, e.g. SMS if account is accessed through an unregistered address [2] 7 (a)
2/3 matches – 2 marks 1 match – 1 mark
[2]
address bus
control bus
data bus
this bus carries signals used to coordinate the computer’s activities
this bi-directional bus is used to exchange data between processor, memory and input/output devices
this uni-directional bus carries signals relating to memory addresses between processor and memory
– reads values in registers “C” and “D” – and checks the values against those stored in registers “A” and “B” (NOTE: the first two statements can be interchanged, i.e. “A” and “B” read first) – If values in corresponding registers are the same – the microprocessor sends a signal to sound alarm/ring [3]
(d) Any three from:
– uses a light sensor – sends signal/data back to microprocessor – signal/data converted to digital (using ADC) – value compared by microprocessor with pre-set/stored value – if value < stored value, signal sent by microprocessor … – … to the voltage supply (unit) – … “value” of signal determines voltage supplied/brightness of LED [3]
(e) Any two from:
– no need to warm up – whiter tint/more vivid colours/brighter image – higher resolution – much thinner monitors possible/lighter weight – more reliable technology/longer lasting – uses much less power/more efficient [2]
generates an error report at the end of translation of the whole program
�
stops the translation process as soon as the first error is encountered
�
slow speed of execution of program loops �
translates the entire program in one go �
[5]
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge Ordinary Level
MARK SCHEME for the May/June 2015 series
2210 COMPUTER SCIENCE
2210/12 Paper 1, maximum raw mark 75
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 2015 series for most Cambridge IGCSE
®, Cambridge International A and AS Level components and some
4/5 matches – 4 marks 3 matches – 3 marks 2 matches – 2 marks 1 match – 1 mark
[4]
malicious code installed on the hard drive of a user’s computer or on the web server; this code will re-direct user to a fake web site without their consent
software that gathers information by monitoring key presses on a user’s computer and relays the information back to the person who sent the software
program or code that replicates itself and is designed to amend/delete/copy data and files on a user’s computer without their consent
the act of gaining illegal access to a computer system without the owner’s consent
creator of code sends out a legitimate-looking email in the hope of gathering personal and financial data from the recipient; it requires the email or attachment to be opened first
(b) 1 mark for correct sensor, 1 mark for its matching application (all THREE applications must be different) sensor application infra-red/motion automatic doors burglar alarm systems temperature chemical process central heating/air con system greenhouse environment oven sound/acoustic burglar alarm systems leak detection system disco lighting moisture/humidity clothes drier environmental control (greenhouse, air con) pressure burglar alarm system traffic light control chemical process carbon dioxide/ pollution monitoring in a river oxygen/gas greenhouse environment (growth control) confined area (e.g. space craft) Fish tank/Aquarium magnetic field mobile phone anti-lock braking CD players
10 (a) 1 mark for two correct lines, 2 marks for four correct lines
[2] (b) 1 mark for each correct binary value 1 mark for each correct hexadecimal value
hexidecimal
L: 1 1 0 1 1 0 0 0 D 8
G: 1 1 0 0 1 1 1 0 C E
[4]
0 1 1 0 1 1 0 0
0 1 1 0 1 0 0 1
0 1 1 0 0 1 1 1
0 1 1 0 1 1 1 0
L (108):
I (105):
G (103):
N (110):
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge Ordinary Level
MARK SCHEME for the May/June 2015 series
2210 COMPUTER SCIENCE
2210/21 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 May/June 2015 series for most Cambridge IGCSE
®, Cambridge International A and AS Level components and some
Section A 1 (a) (i) Many correct answers, they must be meaningful. These are examples only. – MiddayTemperature[1:30]
or MiddayTemperature[0:29]
or MiddayTemperature[30]
or MiddayTemperature[29]
or MiddayTemperature[] (1 mark)
– MidnightTemperature[1:30]
or MidnightTemperature[0:29]
or MidnightTemperature[30]
or MidnightTemperature[29]
or MidnightTemperature[] (1 mark) [2]
(ii) Answers, must match above and the upper bound should have been changed from 30 to 7 or 29 to 6 or no change if not used. These are examples only. – MiddayTemperature[1:7] MidnightTemperature[1:7]
or MiddayTemperature[7] MidnightTemperature[7] [1]
(iii) Any two variables with matching reasons, 1 mark for the variable and 1 mark for the matching reason. The variables and the matching reasons must relate to the tasks in the pre-release. There are many possible correct answers these are examples only. Variable – Counter: (Integer)
Reason – to use as a loop counter when entering the temperature Variable – HighNoon: (Real)
Reason – to store the highest midday temperature [4]
(b) If loop used – initialisation before loop – loop – running total inside loop – calculation of average outside loop – output of average with message outside loop (Max 4 marks) – completion of at least 3 of initialisation, running total, calculation of average and output of average with message for both midday and midnight (1 mark) [5] sample algorithm:
PRINT 'The average midday temperature is ', MiddayAverage
PRINT 'The average midnight temperature is ', MidnightAverage
If loop not used – total of 7 midday temperatures – calculation of midday average (Note could be combined as one calculation, see example below) – total of 7 midnight temperatures – calculation of midnight average (Note could be combined as one calculation, see example below) – output of both averages with suitable messages [5] sample algorithm:
(c) 1 mark for the data set and 1 mark for the matching reason. There are many possible correct answers, these are examples only. Data set – 30, 29, 28, 31.5, 32.3, 33, 29.7 Reason – normal data that should be accepted Data set – twenty, 23.99, seventeen, 501, –273, @#@, seventy seven Reason – abnormal data that should be rejected [2] (d) Maximum 6 marks in total for question part Explanation (max 6) – set variable called HighestMidday to a large minus number – loop (30 or 7) times to check each midday temperature in turn – check midday temperature against HighestMidday / midday temperature > HighestMidday – …replace value in HighestMidday by midday temperature – …store array index in MiddayMonthDay/MiddayWeekday – output HighestMidday outside the loop – output MiddayMonthDay/MiddayWeekday outside the loop Sample algorithm (max 4):
HighestMidday ← -999 FOR Count ← 1 TO 7
IF MiddayTemperature [Count] > HighestMidday
THEN HighestMidday ← MiddayTemperature[Count]
MiddayMonthDay/MiddayWeekday ← Count
ENDIF
NEXT Count PRINT 'The highest midday temperature was ',HighestMidday, ' on
day ', Count
If pseudocode or programming only and no explanation, then maximum 4 marks [6]
4 1 mark for each correct link, up to maximum of 4 marks
[4] 5 Any two points from – a variable is used to store data that can change during the running of a program – a constant is used to store data that will not be changed during the running of a program [2] 6 – FOR (… TO … NEXT)
– REPEAT (… UNTIL)
– WHILE (… DO … ENDWHILE) [3]
7 (a) – 7 [1] (b) – Brochure No – Uniquely identifies each property [2] (c) Garage – Boolean Number of Bedrooms – Number/Integer/Single Price in $ – Number/Single/Real/Currency [3] (d) 399000 H13 450000 H10 [2]
Field: Property Type Garage Price in $ Brochure No
Table: PROPERTY PROPERTY PROPERTY PROPERTY
Sort:
Show: � � � �
Criteria: True < 200000
or:
or
Field: Property Type Garage Price in $ Brochure No
Table: PROPERTY PROPERTY PROPERTY PROPERTY
Sort:
Show: � � � �
Criteria: Yes < 200000
or:
or
Field: Property Type Garage Price in $ Brochure No
Table: PROPERTY PROPERTY PROPERTY PROPERTY
Sort:
Show: � � � �
Criteria: =Yes < 200000
or:
or
Field: Property Type Garage Price in $ Brochure No
Table: PROPERTY PROPERTY PROPERTY PROPERTY
Sort:
Show: � � � �
Criteria: =-1 < 200000
or:
(1 mark) (1 mark) (1 mark) (1 mark)
[4]
® IGCSE is the registered trademark of Cambridge International Examinations.
CAMBRIDGE INTERNATIONAL EXAMINATIONS
Cambridge Ordinary Level
MARK SCHEME for the May/June 2015 series
2210 COMPUTER SCIENCE
2210/22 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 May/June 2015 series for most Cambridge IGCSE
®, Cambridge International A and AS Level components and some
1 (a) (i) Many correct answers, they must be meaningful. This is an example only. – PupilName[1:30]
or PupilName[0:29] or PupilName[30] or PupilName[29] or PupilName[] [1]
(ii) Many correct answers, they must be meaningful. This is an example only.
– StartWeight[1:30]
or StartWeight[0:29] or StartWeight[30] or StartWeight[29] or StartWeight[] [1]
(iii) Answers, must match (i) and (ii) above and the upper bound should have been changed from 30 to 600 or 29 to 599 or no change if not used. – StartWeight[1:600] or StartWeight[600]
– PupilName[1:600] or PupilName[600] [1] (b) any four from – prompt for entry of final weight that includes pupil’s name – input final weight – validation check for final weight – calculation of difference in weight – ……using the initial weight stored in the array – store difference in weight (Max 4 marks) – loop for 600 pupils (1 mark) [5] sample algorithm: FOR Count � 1 TO 600
REPEAT
PRINT 'Please enter weight for ', PupilName[Count] INPUT FinalWeight
(c) (i) any two from – check that the weights are within a given range – check that the weights are numeric – check that the weights are given to one decimal point – character/type check on name – length check on name
[2] (ii) 1 mark for the data and 1 mark for the matching reason. There are many possible correct answers this is an example only. Weight 1 – 35.2 Reason – normal data that should be accepted Weight 2 – twenty Reason – abnormal data that should be rejected [4] (d) Maximum 6 marks in total for question part Explanation (max 6) – loop 30 or 600 times to check each difference in weight – check for a difference in weight – less than -2.5 (final weight – start weight) or greater than 2.5 (start weight – final weight) – …If so output pupil’s name – …if so output difference in weight – …if so output message that it is a fall in weight Sample algorithm (max 4) FOR Count � 1 TO 30 IF WeightDifference [Count] < -2.5 THEN PRINT PupilName[Count], 'The weight loss was ',
WeightDifference [Count] ENDIF
NEXT Count If pseudocode or programming only and no explanation, then maximum 4 marks [6]
Section B 2 1 mark for each error identified + suggested correction Line 1 or Large =9999: this should read Large = 0 Line 3 or WHILE: this should read WHILE Counter < 30 line 6 or IF: this should read IF Num > Large THEN Large = Num line 7 or Counter =…: this should read Counter = Counter + 1 [4]
[4] (b) – (modulo 11) check digit calculation [1] (c) 1 mark for identifying the problem, 2 marks for the solution Problem – doesn’t deal correctly with remainder 10/a check digit of X Solution – check Z for X as a final digit – have a special case where check = 10 – accept where Check = 10 and F = X [3]