A-level Computer Science Example answers Paper 2 - 76 marks

A-level Computer Science Example answers Paper 2 - 76 marks
Student responses with examiner
commentary
V1.0 17/10/2014
Strictly confidential
Student responses with examiner
commentary
A-level Computer Science 7517
Paper 2 7517/2]
For teaching from September 2015
For assessment from summer 2017
Specimen Assessment Paper 2 7517/2
Introduction
These resources should be used in conjunction with the Specimen Assessment material (7517/2)
from the AQA website. This document illustrates how examiners intend to apply the mark scheme
in live papers. While every attempt has been made to show a range of student responses
examiners have used responses, and subsequent comments, which will provide teachers with the
best opportunity to understand the application of the mark scheme.
Strictly confidential
2 of 27
A-level
COMPUTER SCIENCE
Total Mark 76
Paper 2
TBC
Student 2
am/pm
2 hours 30 minutes
Materials

There are no additional materials required for this paper.
Instructions
 Use black ink or black ball-point pen.
 Fill in the boxes at the bottom of this page.
 Answer all questions.
 Do all rough work in this book. Cross through any work that you do not want to be marked.
 You may use a calculator.
Information
 The marks for questions are shown in brackets.
 The maximum mark for this paper is 100.
Advice
 In some questions you may be required to indicate your answer by shading a lozenge. If you
wish to change your answer make sure that the incorrect answer is clearly crossed through with
an x.
Please write clearly, in block capitals, to allow character computer recognition.
Centre number
Candidate number
Surname
Forename(s)
Candidate signature
Paper 2 V1.1 (FINAL DRAFT)
4
Answer all questions in the spaces provided.
0 1
Figure 1 shows how some of the components of a computer system can be connected
together.
Figure 1
B
E
Controller
C
D
Controller
A
Table 1 lists the names of six components in the column headings and the five letters
(A-E) from Figure 1 in the row headings.
For each row in Table 1, shade one lozenge, in the appropriate column, to indicate
As an example, the first row has been completed for you, to indicate that component A
in Figure 1 is the Address bus.
[4 marks]
Table 1
Processor
Address
bus
Data bus
Main
memory
Keyboard
Visual
display
unit
A

B
C

4

D

E
4
4
The student has correctly identified all four components. Initially, a mistake was
made on row D, but the student has corrected this by crossing through the
incorrectly shaded lozenge.
Paper 2 V1.1 (FINAL DRAFT)
5
0 2
The internal buses in a computer use parallel communication while most peripherals
communicate with a computer using serial communication.
0 2 . 1
Explain the differences between the ways in which parallel and serial communication
is carried out.
[2 marks]
Parallel communication is faster than serial as parallel sends multiple data bits
simultaneously whilst serial only sends one bit at a time. 
1
The student has correctly identified the distinction between the number of bits sent
simultaneously by each method, so has achieved one mark. The initial statement
that parallel communication is faster was not relevant to the question which is about
how the communication is carried out. It is also no longer clear that parallel
communication is faster than serial as a result of higher clock speeds, lack of crosstalk in serial communications etc.
Most peripherals, such as printers and keyboards, communicate with a computer
using a serial connection.
0
2
. 2
Apart from the widespread availability of USB (Universal Serial Bus) ports, explain
why peripherals usually use a serial communication method such as USB instead of
parallel communication.
[1 mark]
1

If the peripheral was a long distance away then parallel would not work.
This is enough for a mark, as it is just the point that parallel will only work over short
distances that is on the mark scheme, but made in reverse.
In communications systems, a distinction is made between the bit rate and the baud
rate.
0 2 . 3
Define the term baud rate.
[1 mark]
0
This is the rate at which data can be sent.
This is not enough. To get a mark, the student needs to relate the response to the
rate at which the signal on the transmission medium can change.
0 2 . 4
Explain how it is possible for the bit rate to be higher than the baud rate.
[1 mark]
This could be achieved by having four different voltage levels in the system , with
1
each voltage level representing two bits of data e.g. 0v = 00, 1v = 01, 2v = 10,
3v =11
This is a very detailed explanation; the student has given more detail than is
required for the mark.
Paper 2 V1.1 (FINAL DRAFT)
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3
5
6
0 3
A burglar alarm system is to be implemented that has the following sensors:
 a door sensor D that outputs TRUE when the door is open and FALSE when the
door is shut
 a pressure mat sensor M that outputs TRUE while a weight is detected on it and
FALSE when no weight is detected on it.
The alarm also has a key K that turns the alarm on and off. K outputs a TRUE signal
when the alarm is switched on and FALSE when the alarm is off.
The alarm output A sounds a bell. It should be TRUE if:
 the alarm is on AND
 either of the sensors D or M are set to the value TRUE.
0 3 . 1
In the space below, draw a logic circuit that will behave as described above for the
inputs D, M and K and the output A.
[2 marks]
K

D
A
2

M
0 3 . 2
Write a Boolean expression to represent the logic of this alarm system.
[2 marks]
A=

2
The student has not given the expected response which is slightly simpler (D+M).K
but this is a logically equivalent expression so both marks have been awarded. This
does not represent the logic of the circuit (which would need 1 OR gate and 2 AND
gates), but does represent the logic of the alarm system.
Paper 2 V1.1 (FINAL DRAFT)
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7
0 3 . 3
In this alarm system, the alarm bell will sound only while the door is open or a weight
is placed on the pressure mat. If someone who has stepped on to the mat moves off
it, or an open door is closed, the alarm bell will stop ringing.
A D-type flip-flop could be incorporated into the logic circuit so that the alarm bell
would continue to sound after a person closed the door or moved off the pressure
mat.
Explain how this could be achieved. In your answer refer to:



why a D-type flip-flop would be suitable for this task
where the D-type flip-flop would need to be inserted into the circuit
what additional input the D-type flip-flop would need.
[3 marks]
The flip-flop would need to be inserted at the very end of the circuit and would
be used as it could remember that the alarm had been triggered  and keep its
2
output to the bell active.
“At the very end of the circuit” is just enough for a mark, as from the diagram given it
can be seen that this means after the AND gate. Although the term “memory” has
not been used, the second part of the description explains that the flip-flop would be
memory so a second mark has also been awarded. The student needed to identify
that a trigger input would be required for the third mark.
Turn over for the next question
Paper 2 V1.1 (FINAL DRAFT)
Turn over 
6
7
8
0 4
The phrase “Internet of Things” is used to describe the connection of many everyday
devices such as home heating controls, utility meters, cars and environmental sensors
to the Internet. It is believed that tens of billions of devices will be connected to the
Internet of Things by the end of the decade.
One anticipated use of the Internet of Things is to monitor the food that consumers
have inside their fridges. This data could be gathered automatically from consumers’
devices by retailers who sell food. Retailers could use the data to analyse consumer
consumption habits or automatically prepare deliveries for customers.
In the context of an Internet connected fridge, discuss the technologies that will be
required to make the Internet of Things work.
You may wish to consider how the data might be captured, how networking
technologies are changing to provide the necessary infrastructure, and how the data
gathered by retailers could be stored and processed, from a hardware and software
viewpoint.
[12 marks]
The fridge would need to be able to identify which products were inside it and
when they would expire or run out. The food could be identified by
incorporating a bar code scanner into the fridge [Capturing Point], or alternatively
if RFID tags were added onto the products these could be scanned automatically as
as items were taken in and out because it is wireless. [Capturing Point] Some
products like fruit might not have a barcode so the person who the fridge belongs
to would have to input their details, maybe using a keyboard or a smart phone that
connects to the fridge by Bluetooth [Capturing Point]. The bar code could be
used to look up the expiry dates from a database on the Internet.
If very many devices were connected to the Internet then more IP addresses would
be needed than are currently available, which is being addressed by IPv6
IPv6 [Networking Point]. Also, it is likely that Internet connections would need to
be faster due to the vast amount of data being collected so cable connections
might need to be replaced by fibre optic ones which are faster [Networking Point].
The vast amount of data captured would be “big data” [Storage Point] which is
very hard to deal with. Hardware like parallel processing [Processing Point] could be
used so that more data could be processed at the same time. Functional
programming might be applied [Processing Point] to this as it is good at scaling
Paper 2 V1.1 (FINAL DRAFT)
Turn over 
9
to work on parallel systems [Processing Point].
There would also be a lot of data to store. Probably the company would have a
big data storage warehouse that would include lots of hard disks. As the amount
of data would grow every day, the company will probably need to keep the data only
for a certain period and then delete it to avoid running out of storage capacity
9
[Storage Point].
This is a Level 3 response. The student has made 3 capturing points, 2 networking
points, 2 storage points and 3 processing points so does not meet the required criteria
of having made 3 points per topic area to achieve Level 4. However, as the response
covers all four areas and two or three points have been made per area, this happily
meets the requirements for Level 3. Level 3 only requires that three areas are covered
and in fact four have been, so a mark towards the top end of Level 3 is appropriate.
9
12
Paper 2 V1.1 (FINAL DRAFT)
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10
0 5
The icon in Figure 2 is represented in a computer's memory as a bitmap image.
Figure 2
Row
1
2
3
4
5
6
7
8
9
10
Four different colours have been used in the icon.
Row 1 of the icon is represented in the computer's memory as the bit pattern:
1
0 5 . 1
0 5 . 2
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
What are the bit patterns that have been used to represent a grey pixel and a white
pixel?
[1 mark]
Grey pixel: 00
White pixel: 11

1
State one possible 20-bit representation for Row 4 of the icon in Figure 2.
[1 mark]
1
1
1
1
0
0
1
0
0
1
0
1
1
0
1
0
0
0
1
1

This is a fully correct response. The pattern 10 has been used for brown pixels and
01 for blue pixels.
Paper 2 V1.1 (FINAL DRAFT)
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1
11
0 5 . 3
Calculate the number of bytes required to represent all the pixel data in the icon as a
bitmap.
Show your working.
[2 marks]
1
Answer: 25 
The student has given the correct answer but has only achieved one mark because
their working is not shown.
0 5 . 4 When the bitmap is saved as a file, the file size is bigger than the answer to
. This is because metadata is saved in the file with the pixel data
0 5 . 3
State one item of metadata that would be stored in a bitmap file.
[1 mark]
Width of the image 
1
Run-length encoding (RLE) is an example of a compression method that could be
used to reduce the amount of memory required to store the icon in Figure 2.
0 5 . 5
Describe the principle used by RLE to compress a file and explain why RLE is an
appropriate compression method for compressing images such as icons.
[3 marks]
Run length encoding is a method of data compression that identifies sequences of
the same data item  and shortens these by recording the data item and a count
of how many times it occurs in the sequence.  It is suitable for images because
2
images are often large files.
The student had given a good description of RLE, worth two marks. However the
reason that has been given is about why images are often compressed, not why
RLE is a suitable method for this. A better response would be that images often
contain areas that are the same colour and so long sequences of identical data
values.
Paper 2 V1.1 (FINAL DRAFT)
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6
8
12
Figure 3 shows the structure of an example machine code instruction, taken from the
instruction set of a particular processor.
0 6
Figure 3
Opcode
Basic Machine
Addressing
Operation
Mode
0 1 1 0 1 0
1
0
0 6 .
1
Operand(s)
0
1
0
1
0
1
1
How many different basic machine operations could be supported by the instruction
set of the processor used in the example in Figure 3?
[1 mark]
64 
1
Figure 4 shows an assembly language program together with the contents of a
section of the main memory of the computer that the program will be executed on.
The assembly language instruction set that has been used to write the program is
listed in Table 2.The lines of the assembly language program have been numbered to
help you answer question parts 0 6 . 2 to 0 6 . 4 .
Figure 4
Line
1
2
3
4
5
0 6 .
2
3
4
Main Memory
Contents
(in decimal)
23
10
62
18
contents of address 100 instead of the number 100.
0
What value will be stored in register R2 immediately after the program has executed
the commands from line 1 through to line 3?
Student understood ADD instruction but has used the wrong [1 mark]
33 
0 6 .
Memory
Address
(in decimal)
100
101
102
103
What value will be stored in register R2 immediately after the command in line 1 has
been executed?
[1 mark]
Student has used the wrong addressing mode, loading
23
0 6 .
Command
MOV R2, #100
LDR R3, 101
ADD R2, R2, R3
LSL R3, R2, #1
HALT
addressing mode in the previous part. However, the
functionality of adding 10 is correct so the mark is awarded.
What value will be stored in register R3 after the complete program has finished
executing?
[1 mark]
The student has added one instead of doing a logical shift
left.
34
Paper 2 V1.1 (FINAL DRAFT)
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1
0
13
Table 2
LDR Rd, <memory ref>
STR Rd, <memory ref>
ADD Rd, Rn, <operand2>
SUB Rd, Rn, <operand2>
MOV Rd, <operand2>
CMP Rn, <operand2>
B <label>
B<condition> <label>
Load the value stored in the memory location specified by
<memory ref> into register d.
Store the value that is in register d into the memory location
specified by <memory ref>.
Add the value specified in <operand2> to the value in register n
and store the result in register d.
Subtract the value specified by <operand2> from the value in
register n and store the result in register d.
Copy the value specified by <operand2> into register d.
Compare the value stored in register n with the value specified by
<operand2>.
Always branch to the instruction at position <label> in the
program.
Conditionally branch to the instruction at position <label> in the
program if the last comparison met the criteria specified by the
<condition>. Possible values for <condition> and their
meaning are:



AND Rd, Rn, <operand2>
ORR Rd, Rn, <operand2>
EOR Rd, Rn, <operand2>
MVN Rd, <operand2>
LSL Rd, Rn, <operand2>
LSR Rd, Rn, <operand2>
HALT
EQ: Equal to.
NE: Not equal to.
GT: Greater than.
 LT: Less than.
Perform a bitwise logical AND operation between the value in
register n and the value specified by <operand2> and store the
result in register d.
Perform a bitwise logical OR operation between the value in
register n and the value specified by <operand2> and store the
result in register d.
Perform a bitwise logical exclusive or (XOR) operation between
the value in register n and the value specified by <operand2>
and store the result in register d.
Perform a bitwise logical NOT operation on the value specified by
<operand2> and store the result in register d.
Logically shift left the value stored in register n by the number of
bits specified by <operand2> and store the result in register d.
Logically shift right the value stored in register n by the number of
bits specified by <operand2> and store the result in register d.
Stops the execution of the program.
Interpretation of <operand2>
<operand2> can be interpreted in two different ways, depending upon whether the first symbol is a
# or an R:
 # - use the decimal value specified after the #, eg #25 means use the decimal value 25.
 Rm - use the value stored in register m, eg R6 means use the value stored in register 6.
The available general purpose registers that the programmer can use are numbered 0 to 12.
Question 6 continues on the next page
Paper 2 V1.1 (FINAL DRAFT)
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14
Programs written in a high-level language can be compiled or interpreted.
Companies that develop computer programs to sell usually compile the final version of
a program before distributing it to customers.
0 6 . 5
Explain why the final version of a computer program is usually translated using a
compiler.
[2 marks]
This is done so that the final user does not need to buy translation software
 and
also to stop users seeing the source code which may the company will want to
keep secret.
2

The JavaScript programming language can be used to write programs that are
executed in a web browser on any Internet user’s computer.
0 6 . 6
Explain why programs written in the JavaScript language, to be executed in a web
browser, are interpreted rather than compiled.
[2 marks]
This is because, if there are any errors in a program, it will still execute until the
point where the error is.
0
The student has missed the point of the question, giving a general response about
why an interpreter might be chosen instead of relating it to the context of the
question. A better response would explain the programs would be executed on a
range of computers which might have different types of processor so compiled
machine code could not be used.
4
Paper 2 V1.1 (FINAL DRAFT)
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8
15
0 7
Figure 5 shows the physical topology of a local area network (LAN) and its connection
to the Internet. The LAN uses the IPv4 protocol.
Figure 5
Subnet
192.168.0.0
Wireless
Access
Point
Subnet
192.168.1.0
Switch
Router 1
B
A
Router 2
Router and
Firewall
0 7 . 1
Subnet
192.168.2.0
Switch
157.44.12.32
Internet
State suitable IP addresses for:
[2 marks]
The ‘Router 2’ port labelled A : 192.168.2.1 
The computer network interface card labelled B :192.168.2.2 
0 7 . 2
2
State one advantage of the star topology over the bus topology, and explain how this
is achieved.
[2 marks]
One advantage is that it will be more secure.

1
A mark is awarded for the advantage, but the reason has not been given so the
second mark cannot be awarded.
Question 7 continues on the next page
Paper 2 V1.1 (FINAL DRAFT)
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16
Laptop computers connect to the network using WiFi. They use carrier sense multiple
access with collision avoidance (CSMA/CA) to determine when to transmit data.
0 7 . 3
Describe how the CSMA/CA method is used.
[6 marks]
A computer wanting to transmit data will check to see if any data signals are
already being sent.

If a signal is present then the computer will wait
there is no signal before starting to transmit.

 until
If two computers send at the same
time a collision might occur but wireless networks cannot detect this. Therefore,
the receiver has to send an acknowledgement back to the transmitter when the
data has been correctly received.  If no acknowledgement arrives, the receiver
will know that it must retransmit the data.

5
This is a good response that covers all parts of the process and shows a good level
of understanding so that it falls into Level 3. There is no explanation of how long the
sender will wait before attempting a retransmission so the mark awarded is 5 instead
of 6.
Each packet of data transmitted around the LAN includes a checksum, which is used
for error detection.
0 7 . 4
Describe how the checksum is used for error detection.
[3 marks]
The checksum is calculated when the data is transmitted and again when it is
2
received.  If the two results differ than an error has been detected so the data
will need to be corrected or transmitted again. 
The student has not explained that the checksum is calculated from the data so the
third mark cannot be awarded.
Paper 2 V1.1 (FINAL DRAFT)
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10
13
17
0 8
A particular computer uses a normalised floating point representation with an 8-bit
mantissa and a 4-bit exponent, both stored using two’s complement.
Four bit patterns that are stored in this computer’s memory are listed in Figure 6 and
are labelled A, B, C, D. Three of the bit patterns are valid floating point numbers and
one is not.
Figure 6
A
0
1
0
1
1
1
0
0
1
0
0
0
1
1
0
1
0
0
1
0
0
0
0
0
0
0
0
1
0
1
1
1
1
1
1
1
0
1
1
1
1
0
0
0
Exponent
Mantissa
0 8 . 1
1
Exponent
Mantissa
D
0
Exponent
Mantissa
C
1
Exponent
Mantissa
B
0
Complete Table 3. In the Correct letter (A-D) column shade the appropriate lozenge
A, B, C or D to indicate which bit pattern in Figure 6 is an example of the type of value
described in the Value description column.
Do not use the same letter more than once.
[3 marks]
Table 3
Value description
A positive normalised value
Correct letter (A-D)

A
B
C
D

A
B
C
D

A
B
C
D
The most negative value that can be
represented
A value that is not valid in the
representation because it is not
normalised
Question 8 continues on the next page
Paper 2 V1.1 (FINAL DRAFT)
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3
18
0 8 . 2
The following is a floating point representation of a number:
0
1
0
1
1
0
0
0
0
1

Mantissa
0
1
Exponent
Calculate the decimal equivalent of the number. Show how you have arrived at your
answer.
[2 marks]
Answer: 22 
2
Two marks have been awarded. Indicating that the binary point is moved five places
to the right is sufficient for the working mark. As with all such questions showing all
the working is more secure.
0 8 . 3
Write the normalised floating point representation of the negative decimal value -6.75
in the boxes below. Show how you have arrived at your answer.
[3 marks]
1001.01 
Answer:
1
0
0
1
0
1
0
0
Mantissa
0
0
1
1

Exponent
There is just enough in the working for all three marks to be awarded. The student
has managed to write down -6.75 in fixed point directly and has indicated that the
binary point is moved three places left. This latter point could have been made
clearer in the working so that both working marks were more secure.
Paper 2 V1.1 (FINAL DRAFT)
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3
19
An alternative two's complement format representation is proposed. In the alternative
representation 6 bits will be used to store the mantissa and 6 bits will be used to store
the exponent.
Existing Representation (8-bit mantissa, 4-bit exponent):
Mantissa
Exponent
Proposed Alternative Representation (6-bit mantissa, 6-bit exponent):
Mantissa
0 8 . 4
Exponent
Explain the effects of using the proposed alternative representation instead of the
existing representation.
[2 marks]
The range would be decreased and the representation would be more accurate.
0
8
This student has made the comparison the wrong way around so no marks can be
awarded. In fact, the range would be increased and the precision reduced.
Turn over for the next question
0 9
A school stores information about its sports day in a relational database.
Paper 2 V1.1 (FINAL DRAFT)
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10
20
The details of the track events are stored using the three relations in Figure 7.
Figure 7
Athlete (AthleteNumber, Forename, Surname, Class, Gender, DateOfBirth)
Race (RaceNumber, Gender, Distance, Type, StartTime)
RaceEntryAndResult (RaceNumber, AthleteNumber, TimeSet)
Each athlete who takes part in a race is given a unique AthleteNumber. Athletes can
run in more than one race. If they do, they keep the same AthleteNumber for the entire
day.
Many races are run throughout the day. An example race would be the boys 80m
hurdles, the third race of the day, which starts at 13:30. The entry in the Race table for
this race is shown in Table 4:
Table 4
RaceNumber Gender Distance
Type
StartTime
3
Boys
80
Hurdles
13:30
When an athlete is entered into a race, a record of the entry is created in the
RaceEntryAndResult table. Initially, the TimeSet is recorded as 00:00.00 (meaning
0 minutes, 0 seconds, 0 hundredths of a second) to indicate that the race has not yet
been run. After the race has been run, if the athlete successfully completes it, then
their TimeSet value is updated to record the time that they achieved in minutes,
seconds and hundredths of a second. The TimeSet value remains at 00:00.00 for
athletes who fail to complete the race.
The primary keys in the Athlete and Race relations have been identified in Figure 7 by
underlining them. The correct primary key for the RaceEntryAndResult relation has
not been identified.
0 9 . 1
In Figure 8 below, underline the appropriate attribute name(s) to identify the correct
primary key for this relation.
[1 mark]
Figure 8
RaceEntryAndResult(RaceNumber, AthleteNumber, TimeSet)
Paper 2 V1.1 (FINAL DRAFT)

1
Turn over 
21
0 9 . 2
Relations in a database should usually be fully normalised.
Define what it means for a database to be fully normalised.
[2 marks]
A database is fully normalised when all of the fields in the primary key depend
on the whole primary key  and no other fields. 
2
The response does not cover First Normal Form, ie that data must be atomic, but
two points have been made so full marks are awarded.
0 9 . 3
On the incomplete Entity-Relationship diagram below show the degree of the three
relationships that exist between the entities.
[2 marks]
Athlete
RaceEntryAndResult

1
Race
The student has only drawn two relationships so only one mark can be awarded. It is
possible that the student read the number of marks from the question instead of the
instruction and so believed only two relationships were required (one per mark).
Athlete number 27 is to be entered into race number 6.
0 9 .
4
Write the SQL commands that are required to make this entry.
INSERT INTO RaceEntryAndResult 
[2 marks]
2
VALUES (6, 27, 0)

The response is acceptable for two marks. The 0 is not strictly in time format, but
enough understanding has been shown for this mark to be awarded.
Question 9 continues on the next page
Paper 2 V1.1 (FINAL DRAFT)
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22
Figure 7 is repeated below.
Figure 7 (repeated)
Athlete(AthleteNumber, Forename, Surname, Class, Gender, DateOfBirth)
Race(RaceNumber, Gender, Distance, Type, StartTime)
RaceEntryAndResult(RaceNumber, AthleteNumber, TimeSet)
Athlete number 27 sets a time of 0:18.76 (0 minutes, 18 seconds, 76 hundredths of a
second) for race number 6.
0 9 . 5
Write the SQL commands that are required to update the athlete’s entry for this race,
to store this time in the TimeSet field.
[3 marks]
UPDATE RaceEntryAndResult
SET TimeSet = 0:18:76
WHERE RaceNumber = 6 
2
The two AO3 (programming) marks are awarded for correct SQL syntax in all three
clauses. The omission of a condition in the WHERE clause for the athlete number is
not significant enough to prevent the award of these marks. However, without the
second condition the AO2 (analysis) mark is not awarded.
Paper 2 V1.1 (FINAL DRAFT)
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23
The competition organisers want to produce a list of all of the athletes who took part in
race number 6 with the athlete who won (set the lowest time) at the top and the other
athletes below the winner in the order in which they finished.
Only athletes who finished the race should be included in the list.
The following information should appear for each athlete: AthleteNumber, Forename,
Surname and TimeSet.
0 9 . 6
Write an SQL query to produce the list.
[5 marks]
SELECT AthleteNumber, Forename, Surname, TimeSet
FROM Athlete, RaceEntry 
WHERE RaceNumber = 6 AND TimeSet<>0 
4
ORDERBY TimeSet
Both AO3 (programming) marks can be awarded as there is correct SQL syntax in
all four clauses and the query is very close to being fully correct. Writing “ORDER
BY” incorrectly as one word is acceptable. Two of the AO2 (analysis) marks can be
awarded. The third cannot as the linking condition between the two tables is
missing.
The database system is to be extended for use in an inter-school athletics league.
Users at any school in the county will be able to access the system to input the results
of races.
It is possible that two users might try to access or update the system at the same time.
0 9 . 7
Explain the conditions under which simultaneous access to a database could cause a
problem, and how this could be dealt with.
[3 marks]
The problem would occur if the two people tried to open the same record at the
same time. One way to deal with this would be to have a queue of all accesses
to the database  and to deal with each one in turn. 
2
The first point made is not sufficient for a mark – simultaneous access only causes a
problem if two records are updated simultaneously, not if they are merely read
simultaneously. Then, two good points are made about the use of a transaction
system and how transactions in a queue are dealt with in a FIFO manner.
Paper 2 V1.1 (FINAL DRAFT)
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18
24
1 0
Two computers, A and B, are involved in a secure communication that uses
asymmetric encryption. A is sending a message to B.
Each computer has a public key and a private key.
1 0 .
1 Complete the missing words in the following paragraph.
[2 marks]
A will encrypt the message using _A’s public_ key. The message will be
decrypted by B using _ B’s private __ key.
1
The student has failed to understand that the keys must be used in pairs ie a
message encrypted with one of B’s keys must be decrypted with the other of B’s
keys. However, the second response is correct so one mark is awarded.
The security of the communication could be improved by the addition of a digital
signature.
1 0 . 2
State two benefits of including a digital signature.
[2 marks]
A digital signature is a hash value calculated from the data to be transmitted that is
the encrypted. The hash is recalculated when the data is received and if the
1
recalculated hash does not match the transmitted one then it can be detected that
the message has been tampered with.

The student has explained how a digital signature works rather than addressing the
question. However, there is a clear implication that a benefit would be that tampering
with a message could be detected so one mark can be awarded.
2
4
Paper 2 V1.1 (FINAL DRAFT)
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1 1 . 1
Table 5 lists six Boolean equations. Three of them are correct, the others are not.
Shade the lozenges next to the three equations are correct.
[3 marks]
Table 5
Equation
Correct? (Shade three)
AA 1
A  B  A B
A  1 1
A  ( A  B)  A



A  ( A  B)  B
A 1  1
1 1 .
3
2 Use Boolean algebra to simplify the following expression:
A B B A
Show your working.
[3 marks]


̅
̅
2
Two marks are awarded for a correct application of DeMorgan’s laws and then
recognising that B is a common term in both parts of the expression. However, the
student has then gone wrong by equating A+NOT A with 0 instead of 1, so has not
arrived at a correct answer.
Turn over for the next question
Paper 2 V1.1 (FINAL DRAFT)
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6
26
1 2
In a functional programming language, a recursively defined function named map and
a function named double are defined as follows:
map f []
= []
map f (x:xs) = f x : map f xs
double x
= 2 * x
The function map has two parameters, a function f, and a list that is either empty
(indicated as []), or non-empty, in which case it is expressed as (x:xs) in which x is
the head and xs is the tail, which is itself a list.
1 2 . 1
In Table 6, write the value(s) that are the head and tail of the list
[ 1, 2, 3, 4 ].
[1 mark]
Table 6
Head
1
Tail
2, 3, 4 
Brackets missed
around tail but this
is still acceptable.
1
The result of making the function call double 3 is 6.
1 2 . 2
Calculate the result of making the function call listed in Table 7.
[1 mark]
Table 7
Function Call
Result
map double [ 1, 2, 3, 4 ]
2, 4, 6, 8 
1
The mark has been awarded despite the fact that the result has not been written in
list notation, ie the square brackets are missing.
1 2 . 3
Explain how you arrived at your answer to question 1 2 .
steps that you followed.
2 and the recursive
[3 marks]
The double function is executed on each of the items in the list.  It was
3
first applied to the head  and then a recursive call was made to apply it to the
rest of the list. 
5
5
END OF QUESTIONS
Acknowledgement of copyright holders and publishers
Permission to reproduce all copyright material has been applied for. In some
cases, efforts to contact copyright holders have been unsuccessful and AQA
will be happy to rectify any omissions of acknowledgements in future papers if
notified.
Copyright © 2014 AQA and its licensors. All rights reserved.
Paper 2 V1.1 (FINAL DRAFT)
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27
Version 1.0
First published (07/10/2014)
Last updated (17/10/2014)
Paper 2 V1.1 (FINAL DRAFT)
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