Computer Science
Class X
Answer of Mock Exam 6 (Paper 1)
3
• Used in HTML colour codes / Notations for colour in HTML
e.g., red - # FF0000, green - # 00FF00, blue - # 0000FF (Must be in proper format i.e., RGB)
• Used in memory dump / error codes (shows the memory location of the error codes)
e.g., 5F 3A 09 F1
• Used in MAC (Media Access Control) addresses
e.g., 01-23-45-67-89-AB (in 48 bits) or 01-23-45-67-89-AB-CD-12 (in 64 bits) (Must be in
proper format)
• Used to represent IP addresses
e.g., B1.2A.1F.A2 (in 32 bits for IPv4 version) (Must be in proper format)
e.g., B213:ABC2:C123: AABC: 2323:1322:9392:11AB (in 128 bits for IPv6 version) (Must be in
proper format)
• To generate error messages on the internet
e.g., error #404 page not found, error #503 Service Unavailable, error #502 Bad Gateway server
(Must be with valid value)
• Used in Assembly language
e.g., 5F 3A 09
• Used in Machine Language
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e.g., AB 2C F2
• Used as locations (address) in memory
e.g., AA 09 F1
(a) Solid state technology
(b) How SSD operates:
• Flash storage
• Uses transistors/controls gates/floating gates
• Can be NAND/NOR technology // Can use flip-flops
• Stores data by flashing it onto the chips/device
• Controlling/using the flow of electrons through/using transistors/chips/gates
• The electric current reaches the control gate and flows through to the floating gate to be
stored
• When data is stored, the transistor is converted from 1 to 0 / 0 to 1
5
Parity check / parity byte check:
- An extra bit (parity bit) added to a string of binary code to ensure the number of 1-bits is either
even or odd, depending upon the parity check system used.
- If the number of 1s is even then it is regarded as ‘even’ parity and if the number of 1s is odd then
it is regarded as ‘odd’ parity.
Method:
• The sending and receiving computers agree the protocol to be used (even or odd)
• The sending computer adds the correct parity bit to the binary data (either an extra 1 or 0)
• The sending computer sends the binary data, including the parity bit
• The receiving computer checks to make sure the overall parity of the data received is as agreed
(an even or odd number of 1 bits)
• If the parity of the data is incorrect, the receiving computer will request that the data is transmitted
again
Check digit:
• It is a validation method
• Used to check data entry
• The final digit is calculated from data
• The digit is appended / added to data
• The final digit is recalculated when data has been input
• The two final digits are compared
• If digits are different, error is detected // If digits match, no error is detected
Checksum:
• A value is calculated from the original data following a particular algorithm. This value is
known as the checksum value
• Checksum value is transmitted with the original data
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• Checksum value is recalculated after transmission by the receiver using the same algorithm on
the received data
• If the calculated checksum value matches with the received checksum value, then the data is
accurate
• If not, then a request is sent to resend the data
Automatic Repeat Request / Query (ARQ):
• uses acknowledgement (a return message) and timeout (set time agreed between the parties)
• parity check / checksum error checking may be included with the original data
• when no error is found then a positive acknowledgement is sent by the receiver to the
sender
• when the receiver detects an error in data transmission it asks for the packet to be resent
(i.e., a negative acknowledgement is sent to the sender)
• if the sender does not receive anything back from the receiver before timeout occurs, the data is
automatically resent
• This will keep happening until the sender receives a positive acknowledgement/until the ARQ
limit is reached
(a)
“The robot is a mechanical device”:
• The chassis and robotic arm are a mechanical device
• a robot is essentially a machine made of physical parts, like gears, motors, and actuators
“Movable”
The paint sprayer arm must be able to position correctly to spray all parts of the car
“Can sense its surroundings”
Sensors will sense when a car is in position // determine when an obstacle is encountered / edge of the car
is reached
“It is a controlled by a computer program”
The computer program sets the parameters/type of car/paint to be used
(b)
Robotic arm: To position the spray nozzle to the correct position
Sensor: To capture data
Actuator // Motor: To drive various motors to perform the robot’s movement
Microprocessor: To process the various inputs and execute the control program
Memory: To temporarily store input data // store program
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Speaker // beeper: To provide audio output
(a)
Explanation of advantages / Benefits of USB:
• It is a universal standard - so it is likely to be compatible with the computer
• It can only be inserted one way - so there is less chance of connecting a device incorrectly
• It is a high-speed connection - so data will be transmitted quicker
• It uses serial transmission - so it is cheaper to manufacture/buy and also less chance of skewing /
errors
• It doesn’t require a network - therefore, can be used if a network is down
• It is backwards compatible - so no additional technology is needed
• It can power the device - therefore, no separate source of power is needed
• Drivers are automatically downloaded // device is automatically identified - so no need to find
them online / install them manually
N.B. Explanation is not required for getting mark
Disadvantages / Drawbacks of USB:
❖ Even though USB is backward compatible, very early USB standards may not always be
supported by the latest computers
❖ Data transmission speed is still relatively slow
❖ After a certain cable length transmission becomes problematic
(b) (i)
• Paper jam
• Out of paper
• Out of toner/ink
• Buffer full
• Awaiting input
• Print complete
• Printer ready
(b) (ii)
Any one from:
− Operating system
− Interrupt handler
− Interrupt service routine
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8a
Four from:
Similarities
• They both involve the use of sensors
• They both do not require human input
• They both require Analogue to Digital Converters to convert input
Differences
• Control systems act in real time
• Control systems use output devices such as motors/actuators
• In a monitoring system the output does not affect the input
Must have one of each to gain full marks
8b
Advantages:
• Blu-ray discs can store higher quality videos than DVDs
• Blu-ray stores up to 128 GB against 8.7GB max
• A single Blu-ray disc stores more data than a single DVD
• Blu-ray discs cost less to buy per unit memory
• Blu-ray drives can read DVD but DVD drives cannot read Blu-ray discs
• Blu-ray disks automatically come with a secure encryption system that helps to prevent piracy
and copyright infringement
• the data transfer rate for a DVD is 10 Mbps and for a Blu-ray disc it is 36 Mbps (this equates to
1.5 hours to transfer 25 GiB of data).
Disadvantages:
• Blu ray disc drives are more expensive to buy than DVD drives
• A single Blu-ray disc costs more to buy than a single DVD
• Bly-ray consumes more power than DVD
9 (a)
Sends the URL of the website
• … to a DNS to find the IP address
• Connects to the webserver (using the IP address) …
• … using HTTP / HTTPS
• Renders/Translates the HTML
• Manages SSL/TLS certificate process
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• Stores/retrieves cookies
9b
10
Hacking
Spyware / Key logging software
Pharming
Phishing
Cookies
Virus / Worms
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11
12
Compiler - three max from:
• Translates the whole program as one complete unit/at once
• Translates the program into machine code
• Creates an executable file
• Produces a report of errors in the code after compilation
• Can optimise source code to run as fast or as efficiently as possible
Linker – two max from:
A linker takes one or more object files and combines them…into a single executable file
Both needed – two max from:
• Many programming languages allow the writing of different pieces of code/modules separately
• Programming tasks are simplified as large programs can be broken into smaller, more manageable
pieces……the linker is used to put all the modules together
• Without the compiler the linker would have no object files to combine
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At least one point from each section must be made to gain full marks
13 (a)
+ 78 in 10 bits = 0 0 0 1 0 0 1 1 1 0
One’s complement = 1 1 1 0 1 1 0 0 0 1
+1 +1
_____________________________________
Two’s complement = 1 1 1 0 1 1 0 0 1 0
13 b
• The result of the calculation is greater than 1023 // The value generated is larger than can be
stored in the register
• The result of the calculation would require more than 10 bits to be represented / A register has a
predetermined number of bits and there are too many bits for it
END OF PAPER
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