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Question 1 Report
Character sets are used to represent text characters in binary.
(a) Describe what a character set is. [2]
(b) State one difference between ASCII and Unicode. [1]
(a) A character set is a collection of characters where each character is assigned a unique binary code. [1] It defines the mapping between text characters (letters, digits, punctuation and symbols) and their binary representations, allowing computers to store, transmit and display text consistently. [1]
(b) The key difference is capacity: ASCII uses 7 bits per character, allowing a maximum of 128 different characters (covering the English alphabet, digits and basic symbols). Unicode can use up to 32 bits per character, supporting over a million unique characters. [1] This means Unicode can represent virtually every written language in the world, including Chinese, Arabic, Hindi and emoji, while ASCII is limited to basic Latin characters.
(a) A character set is a collection of characters where each character is assigned a unique binary code. [1] It defines the mapping between text characters (letters, digits, punctuation and symbols) and their binary representations, allowing computers to store, transmit and display text consistently. [1]
(b) The key difference is capacity: ASCII uses 7 bits per character, allowing a maximum of 128 different characters (covering the English alphabet, digits and basic symbols). Unicode can use up to 32 bits per character, supporting over a million unique characters. [1] This means Unicode can represent virtually every written language in the world, including Chinese, Arabic, Hindi and emoji, while ASCII is limited to basic Latin characters.
Question 2 Report
(a) The logic circuit below has two inputs (A and B) and one output (X).
Complete the truth table for this circuit.
| A | B | NOT A | X |
|---|---|---|---|
| 0 | 0 | ||
| 0 | 1 | ||
| 1 | 0 | ||
| 1 | 1 |
[4]
(b) Write the Boolean expression for output X.
[2]
(c) Describe in words a real-world situation that this logic circuit could model.
[2]
(d) Draw a truth table for the Boolean expression X = A OR (NOT B).
[3]
(e) State two uses of logic gates in everyday electronic devices.
[2]
(a) The circuit passes input A through a NOT gate, then feeds the result (NOT A) and input B into an AND gate to produce output X. The truth table is completed by evaluating each row:
| A | B | NOT A | X = (NOT A) AND B |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 |
Row 1: NOT 0 = 1; 1 AND 0 = 0. [1]
Row 2: NOT 0 = 1; 1 AND 1 = 1. [1]
Row 3: NOT 1 = 0; 0 AND 0 = 0. [1]
Row 4: NOT 1 = 0; 0 AND 1 = 0. [1]
(b) X = (NOT A) AND B [1+1]
This reads: the output X is 1 only when A is 0 (NOT A = 1) and B is 1.
(c) A real-world example: a security light that turns on (X = 1) when it is dark outside (A = 0, meaning NOT A = 1) AND motion is detected (B = 1). During daytime (A = 1) or when there is no motion (B = 0), the light stays off. [2]
(d) Truth table for X = A OR (NOT B):
| A | B | NOT B | X = A OR (NOT B) |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
Row 1: NOT 0 = 1; 0 OR 1 = 1. [1]
Row 2: NOT 1 = 0; 0 OR 0 = 0. [1]
Row 3: NOT 0 = 1; 1 OR 1 = 1.
Row 4: NOT 1 = 0; 1 OR 0 = 1. [1]
(e) Two uses of logic gates in everyday electronic devices:
(a) The circuit passes input A through a NOT gate, then feeds the result (NOT A) and input B into an AND gate to produce output X. The truth table is completed by evaluating each row:
| A | B | NOT A | X = (NOT A) AND B |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 1 | 0 | 0 |
Row 1: NOT 0 = 1; 1 AND 0 = 0. [1]
Row 2: NOT 0 = 1; 1 AND 1 = 1. [1]
Row 3: NOT 1 = 0; 0 AND 0 = 0. [1]
Row 4: NOT 1 = 0; 0 AND 1 = 0. [1]
(b) X = (NOT A) AND B [1+1]
This reads: the output X is 1 only when A is 0 (NOT A = 1) and B is 1.
(c) A real-world example: a security light that turns on (X = 1) when it is dark outside (A = 0, meaning NOT A = 1) AND motion is detected (B = 1). During daytime (A = 1) or when there is no motion (B = 0), the light stays off. [2]
(d) Truth table for X = A OR (NOT B):
| A | B | NOT B | X = A OR (NOT B) |
|---|---|---|---|
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 1 |
Row 1: NOT 0 = 1; 0 OR 1 = 1. [1]
Row 2: NOT 1 = 0; 0 OR 0 = 0. [1]
Row 3: NOT 0 = 1; 1 OR 1 = 1.
Row 4: NOT 1 = 0; 1 OR 0 = 1. [1]
(e) Two uses of logic gates in everyday electronic devices:
Question 3 Report
The analogue-to-digital conversion (ADC) process is used to convert analogue signals to digital data. The diagram shows the stages of this process.
(a) Describe the process of analogue-to-digital conversion for a sound recording. [4]
(b) Explain what quantisation is and why it can introduce errors. [2]
(c) Describe the reverse process (digital-to-analogue conversion) that occurs when a digital sound file is played through a speaker. [2]
(a) The process of analogue-to-digital conversion for a sound recording:
(b) Quantisation is the process of rounding each measured amplitude to the nearest available level, because there are only a finite number of levels determined by the bit depth (for example, 256 levels with 8-bit depth). [1] The difference between the actual amplitude and the nearest available level is called quantisation error. With a low bit depth, fewer levels are available, so the rounding error is larger, producing audible noise or distortion in the recording. [1]
(c) The reverse process (digital-to-analogue conversion):
(a) The process of analogue-to-digital conversion for a sound recording:
(b) Quantisation is the process of rounding each measured amplitude to the nearest available level, because there are only a finite number of levels determined by the bit depth (for example, 256 levels with 8-bit depth). [1] The difference between the actual amplitude and the nearest available level is called quantisation error. With a low bit depth, fewer levels are available, so the rounding error is larger, producing audible noise or distortion in the recording. [1]
(c) The reverse process (digital-to-analogue conversion):
Question 4 Report
(a) Explain the difference between firmware and software.
[2]
(b) State one example of firmware.
[1]
(c) Explain why firmware is stored in ROM rather than RAM.
[2]
(d) A computer's BIOS/UEFI is an example of firmware. Describe the role of the BIOS/UEFI during the boot process.
[3]
(a) Firmware is software that is permanently (or semi-permanently) stored in hardware, typically in ROM or flash memory. It controls the fundamental low-level operations of a device and is not routinely changed by the user. [1] Regular software (application or system software) is stored on secondary storage, loaded into RAM when needed, and can be easily installed, updated, or removed by the user at any time. [1]
(b) The BIOS/UEFI in a computer is a common example of firmware. It runs every time the computer is switched on, before the operating system loads. [1]
(c) Firmware is stored in ROM (Read-Only Memory) because ROM is non-volatile, meaning it retains its contents when the power is turned off. [1] This is essential because the firmware must be available immediately when the device is switched on, before the operating system or any other software has loaded from secondary storage. If it were in RAM (which is volatile), it would be lost every time the power was cut. [1]
(d) During the boot process, the BIOS/UEFI performs the following role:
(a) Firmware is software that is permanently (or semi-permanently) stored in hardware, typically in ROM or flash memory. It controls the fundamental low-level operations of a device and is not routinely changed by the user. [1] Regular software (application or system software) is stored on secondary storage, loaded into RAM when needed, and can be easily installed, updated, or removed by the user at any time. [1]
(b) The BIOS/UEFI in a computer is a common example of firmware. It runs every time the computer is switched on, before the operating system loads. [1]
(c) Firmware is stored in ROM (Read-Only Memory) because ROM is non-volatile, meaning it retains its contents when the power is turned off. [1] This is essential because the firmware must be available immediately when the device is switched on, before the operating system or any other software has loaded from secondary storage. If it were in RAM (which is volatile), it would be lost every time the power was cut. [1]
(d) During the boot process, the BIOS/UEFI performs the following role:
Question 5 Report
(a) Convert the denary number 347 to hexadecimal. Show your working.
[3]
(b) Verify your answer by converting it back to denary. Show your working.
[2]
(c) Explain one situation where a programmer would need to convert between denary and hexadecimal.
[1]
(d) State how many bytes are needed to store the denary value 347.
[2]
(a) To convert denary 347 to hexadecimal, use repeated division by 16:
Read the remainders from bottom to top: 347 in hexadecimal = 15B. [3]
(b) Convert 15B hex back to denary to verify:
256 + 80 + 11 = 347. Verified. [2]
(c) A programmer would need this conversion when defining colours for web design (e.g. converting RGB values from denary to hex colour codes) or when reading hexadecimal memory addresses during debugging and needing to understand the denary value stored at that location. [1]
(d) A single byte can store values from 0 to 255. Since 347 > 255, it cannot fit in 1 byte. [1] Two bytes (16 bits) can store values from 0 to 65535, which is more than enough for 347. Therefore 2 bytes are needed. [1]
(a) To convert denary 347 to hexadecimal, use repeated division by 16:
Read the remainders from bottom to top: 347 in hexadecimal = 15B. [3]
(b) Convert 15B hex back to denary to verify:
256 + 80 + 11 = 347. Verified. [2]
(c) A programmer would need this conversion when defining colours for web design (e.g. converting RGB values from denary to hex colour codes) or when reading hexadecimal memory addresses during debugging and needing to understand the denary value stored at that location. [1]
(d) A single byte can store values from 0 to 255. Since 347 > 255, it cannot fit in 1 byte. [1] Two bytes (16 bits) can store values from 0 to 65535, which is more than enough for 347. Therefore 2 bytes are needed. [1]
Question 6 Report
The diagram shows the components of a simple automated security system.
(a) Using the diagram, describe the sequence of events when the motion sensor detects movement. [4]
(a) When the motion sensor detects movement, the following sequence occurs:
(a) When the motion sensor detects movement, the following sequence occurs:
Question 7 Report
A company is choosing between a wired and wireless network for its new office.
| Factor | Wired network | Wireless network |
|---|---|---|
| Speed | ||
| Security | ||
| Installation cost | ||
| Mobility |
(a) Complete the table by comparing wired and wireless networks. [4]
(b) Recommend which type of network the company should use. Give two reasons for your recommendation. [1]
(a)
| Factor | Wired network | Wireless network |
|---|---|---|
| Speed | Faster and more consistent data transfer speeds, with modern Ethernet supporting 1 Gbps or more. [1] | Generally slower and speeds can vary depending on the distance from the access point, physical obstructions, and interference from other devices. |
| Security | More secure, as an attacker would need physical access to the cable or network port to intercept data. [1] | Less secure, as radio signals can be intercepted by anyone within range, requiring strong encryption (e.g. WPA3) to protect the data. |
| Installation cost | Higher installation cost due to the need to run cables through walls, floors and ceilings. [1] | Lower installation cost, as only wireless access points need to be positioned and configured, with no physical cabling required to each device. |
| Mobility | Devices are fixed to the location of their cable connection, limiting movement. [1] | Devices can be used anywhere within the signal range, allowing staff to move freely with laptops and mobile devices. |
(b) The best approach for most offices is a combination of both: wired connections for desktop computers and servers that need consistent speed and maximum security [1], and wireless access points for laptops, tablets and mobile devices so that staff can work flexibly in meeting rooms and open-plan areas. [1]
(a)
| Factor | Wired network | Wireless network |
|---|---|---|
| Speed | Faster and more consistent data transfer speeds, with modern Ethernet supporting 1 Gbps or more. [1] | Generally slower and speeds can vary depending on the distance from the access point, physical obstructions, and interference from other devices. |
| Security | More secure, as an attacker would need physical access to the cable or network port to intercept data. [1] | Less secure, as radio signals can be intercepted by anyone within range, requiring strong encryption (e.g. WPA3) to protect the data. |
| Installation cost | Higher installation cost due to the need to run cables through walls, floors and ceilings. [1] | Lower installation cost, as only wireless access points need to be positioned and configured, with no physical cabling required to each device. |
| Mobility | Devices are fixed to the location of their cable connection, limiting movement. [1] | Devices can be used anywhere within the signal range, allowing staff to move freely with laptops and mobile devices. |
(b) The best approach for most offices is a combination of both: wired connections for desktop computers and servers that need consistent speed and maximum security [1], and wireless access points for laptops, tablets and mobile devices so that staff can work flexibly in meeting rooms and open-plan areas. [1]
Question 8 Report
(a) A school stores student records in a database. Define the following database terms.
(i) Record
[1]
(ii) Field
[1]
(iii) Primary key
[1]
(b) The student database has the following fields: StudentID, FirstName, LastName, DateOfBirth, YearGroup.
State which field would be the most suitable primary key. Give a reason.
[2]
(c) State appropriate data types for each field in the table.
| Field | Data type |
|---|---|
| StudentID | |
| FirstName | |
| DateOfBirth | |
| YearGroup |
[4]
(a) Database term definitions: [3]
(b) The most suitable primary key is StudentID. [1] This is because it is guaranteed to be unique for every student. Other fields such as FirstName or DateOfBirth could be shared by multiple students (two students could have the same name or birthday), but each student is assigned a distinct StudentID. [1]
(c) Appropriate data types for each field: [4]
| Field | Data type | Reason |
|---|---|---|
| StudentID | Integer (or String if it contains letters) | A numeric identifier for each student. [1] |
| FirstName | String / Text | Names consist of alphabetic characters of varying length. [1] |
| DateOfBirth | Date | A date data type stores day, month, and year and allows date-based sorting and calculations such as calculating age. [1] |
| YearGroup | Integer | Year groups are whole numbers (e.g. 7, 8, 9, 10, 11) with no decimal component. [1] |
(a) Database term definitions: [3]
(b) The most suitable primary key is StudentID. [1] This is because it is guaranteed to be unique for every student. Other fields such as FirstName or DateOfBirth could be shared by multiple students (two students could have the same name or birthday), but each student is assigned a distinct StudentID. [1]
(c) Appropriate data types for each field: [4]
| Field | Data type | Reason |
|---|---|---|
| StudentID | Integer (or String if it contains letters) | A numeric identifier for each student. [1] |
| FirstName | String / Text | Names consist of alphabetic characters of varying length. [1] |
| DateOfBirth | Date | A date data type stores day, month, and year and allows date-based sorting and calculations such as calculating age. [1] |
| YearGroup | Integer | Year groups are whole numbers (e.g. 7, 8, 9, 10, 11) with no decimal component. [1] |
Question 9 Report
The diagram shows a simplified view of a central processing unit (CPU).
(a) State the purpose of the ALU. [2]
(b) State the purpose of the Control Unit. [2]
(c) State what each of these registers is used for:
(i) Program Counter (PC) [1]
(ii) Memory Address Register (MAR) [1]
(iii) Memory Data Register (MDR) [1]
(iv) Accumulator (ACC) [1]
(a) Purpose of the ALU [2]
The ALU (Arithmetic Logic Unit) performs arithmetic operations such as addition, subtraction, multiplication, and division on numerical data. [1]
It also performs logical operations such as comparisons (AND, OR, NOT) and determining whether values are equal, greater than, or less than each other. These logical comparisons are essential for decision-making in programs (e.g. IF statements). [1]
(b) Purpose of the Control Unit [2]
The Control Unit (CU) controls and coordinates the activities of all other components within the CPU and the wider computer system. [1]
It fetches instructions from memory, decodes them to determine what operation is required, and sends the appropriate control signals to other components (ALU, registers, memory, I/O devices) to execute each instruction in the correct sequence. [1]
(c) Register purposes [4]
(i) Program Counter (PC): Holds the memory address of the next instruction to be fetched and executed. After each fetch, it is incremented (usually by 1) so it points to the following instruction. [1]
(ii) Memory Address Register (MAR): Holds the address of the memory location that is about to be read from or written to. This address is placed on the address bus. [1]
(iii) Memory Data Register (MDR): Holds the data that has been fetched from memory (during a read operation) or the data that is about to be written to memory (during a write operation). It acts as a buffer between the CPU and main memory. [1]
(iv) Accumulator (ACC): Holds the results of calculations and logical operations performed by the ALU. It is the primary working register for arithmetic and serves as a temporary store for intermediate results. [1]
(a) Purpose of the ALU [2]
The ALU (Arithmetic Logic Unit) performs arithmetic operations such as addition, subtraction, multiplication, and division on numerical data. [1]
It also performs logical operations such as comparisons (AND, OR, NOT) and determining whether values are equal, greater than, or less than each other. These logical comparisons are essential for decision-making in programs (e.g. IF statements). [1]
(b) Purpose of the Control Unit [2]
The Control Unit (CU) controls and coordinates the activities of all other components within the CPU and the wider computer system. [1]
It fetches instructions from memory, decodes them to determine what operation is required, and sends the appropriate control signals to other components (ALU, registers, memory, I/O devices) to execute each instruction in the correct sequence. [1]
(c) Register purposes [4]
(i) Program Counter (PC): Holds the memory address of the next instruction to be fetched and executed. After each fetch, it is incremented (usually by 1) so it points to the following instruction. [1]
(ii) Memory Address Register (MAR): Holds the address of the memory location that is about to be read from or written to. This address is placed on the address bus. [1]
(iii) Memory Data Register (MDR): Holds the data that has been fetched from memory (during a read operation) or the data that is about to be written to memory (during a write operation). It acts as a buffer between the CPU and main memory. [1]
(iv) Accumulator (ACC): Holds the results of calculations and logical operations performed by the ALU. It is the primary working register for arithmetic and serves as a temporary store for intermediate results. [1]
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