Level:
Correct: 0
Incorrect: 0
Marks: 0/0
Exam Practice
Subject:
Topics:
Cambridge IGCSE Computer Science (Paper 1) Smart practice A mixed round aimed at your weak spots
Unit 1
Data representation
Number systems
Binary, denary and hexadecimal conversions
0%
Uses of hexadecimal
0%
Binary addition and overflow
0%
Logical binary shifts
0%
Two's complement
0%
Text, sound and images
Character sets: ASCII and Unicode
0%
Representing sound
0%
Representing images
0%
Data storage and compression
Data storage units and file size calculations
0%
Data compression
0%
Unit 2
Data transmission
Types and methods of data transmission
Packets and packet switching
0%
Methods of data transmission
0%
The USB interface
0%
Methods of error detection
Parity, checksum and echo check
0%
Check digits and ARQ
0%
Encryption
Encryption
0%
Unit 3
Hardware
Computer architecture
The CPU, Von Neumann architecture and the FDE cycle
0%
Cores, cache and clock speed
0%
Instruction sets and embedded systems
0%
Input and output devices
Input devices
0%
Output devices
0%
Sensors
0%
Data storage
Primary storage: RAM and ROM
0%
Secondary storage: magnetic, optical and solid-state
0%
Virtual memory and cloud storage
0%
Network hardware
Network hardware: NIC, MAC and IP addresses, routers
0%
Unit 4
Software
Types of software and interrupts
System software, application software and operating systems
0%
Firmware and interrupts
0%
Programming languages, translators and IDEs
High-level languages, low-level languages and assembly language
0%
Compilers and interpreters
0%
Integrated development environments (IDEs)
0%
Unit 5
The internet and its uses
The internet and the world wide web
The internet, the world wide web, URLs and HTTP(S)
0%
Web browsers, locating web pages and cookies
0%
Digital currency
Digital currency and blockchain
0%
Cyber security
Cyber security threats
0%
Keeping data safe: security solutions
0%
Unit 6
Automated and emerging technologies
Automated systems
Automated systems
0%
Robotics
Robotics
0%
Artificial intelligence
Artificial intelligence
0%
Avg
score:
–
Topics
done:
0/39

Primary storage: RAM and ROM

Practise 24 exam-style questions on Primary storage: RAM and ROM for Cambridge IGCSE Computer Science (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

  • 24 questions
  • 1–6 marks each
  • Instant marking
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

State the name given to the set of start-up instructions stored in ROM that the CPU runs when a computer is switched on.

Model answer

The bootstrap.

Explanation

The bootstrap (part of the BIOS firmware) is a small program stored in ROM. It runs first when the computer is switched on and loads the operating system from secondary storage into RAM.

Mark scheme
  1. 1 mark

    Bootstrap (loader)

    accept 'BIOS' or 'boot program'; do not accept 'operating system'

Question 2

2 marks

Describe the role of ROM in a computer.

Model answer

ROM stores the start-up instructions (the bootstrap) that the CPU needs as soon as the computer is switched on, so that it can check the hardware and load the operating system. ROM is non-volatile and read-only, so these instructions are kept permanently and cannot be changed.

Explanation

ROM holds the small set of instructions needed to get the computer started. They must be there the moment the power comes on, and must not be changed, so non-volatile, read-only memory is ideal. Do not say ROM stores the operating system: the OS is on secondary storage and is loaded into RAM by the start-up instructions.

Mark scheme
  1. 1 mark

    Stores the start-up instructions / bootstrap / BIOS / firmware

    max 2

  2. 1 mark

    That the CPU needs as soon as the computer is switched on (to start the computer and load the operating system)

  3. 1 mark

    Stores them permanently: non-volatile, so not lost when the power is off / read-only, so they cannot be changed

Question 3

3 marks

Amir is typing an essay on his computer when there is a power cut. When the power returns, he switches the computer back on. It starts up normally, but the changes he made to his essay since he last saved it have been lost.

Explain why.

Model answer

While Amir was editing, his essay and the unsaved changes were held in RAM. RAM is volatile, so everything in it was lost when the power was cut. The computer still started because the start-up instructions are stored in ROM, which is non-volatile, so they were not affected by the power cut.

Explanation

The scenario contrasts the two kinds of primary storage: volatile RAM lost the unsaved work, non-volatile ROM kept the start-up instructions. The last saved version survived because saving writes it to secondary storage, which is also non-volatile.

Mark scheme
  1. 1 mark

    The unsaved changes/the essay being edited were held only in RAM

  2. 1 mark

    RAM is volatile, so its contents were lost when the power was cut

  3. 1 mark

    The start-up instructions are stored in ROM, which is non-volatile, so they were not lost and the computer could start

Question 4

4 marks

The table shows the specification of a laptop.

Calculate how many times larger the RAM is than the ROM, and explain why the RAM needs to be so much larger than the ROM. Show your working.

Laptop specification. Table with columns "Component", "Capacity". Component: RAM; Capacity: 16 GiB | Component: ROM; Capacity: 4 MiB | Component: Solid-state drive; Capacity: 1 TiB.

Model answer

16 GiB = 16 × 1024 MiB = 16 384 MiB
16 384 ÷ 4 = 4096
The RAM is 4096 times larger than the ROM.

The RAM has to hold the operating system, all the applications that are open and the data being used, which together take up a lot of space. The ROM only has to hold the small set of start-up instructions (the bootstrap), so it needs very little capacity.

Explanation

Convert both values to the same unit before dividing: 1 GiB = 1024 MiB, never 1000. The size difference reflects their roles: RAM is the working memory for everything in use, while ROM holds just enough instructions to start the computer.

Mark scheme
  1. 1 mark

    16 GiB = 16 × 1024 = 16 384 MiB

    working: converting to the same unit

  2. 1 mark

    16 384 ÷ 4 = 4096 (times larger)

    do not accept an answer using 1000 (e.g. 4000)

  3. 1 mark

    RAM must hold the operating system, applications and data currently in use, which are large / many at once

  4. 1 mark

    ROM only needs to hold the small start-up program/bootstrap

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