AQA A-Level Computer Science: Fundamentals of Computer Systems — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Fundamentals of computer systems study guide | Fundamentals of computer systems revision notes

These questions cover section 4.6 of the AQA A-Level Computer Science (7517) specification: hardware and software (4.6.1), classification of programming languages (4.6.2), program translators (4.6.3), logic gates (4.6.4) and Boolean algebra (4.6.5). This content is assessed on Paper 2.

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Section A: hardware and software

1. Define the terms hardware and software. [2]

2. Classify each as system or application software: (a) a compiler (b) a spreadsheet (c) a disk defragmenter (d) a graphics library. [2]

3. Explain the two roles of an operating system given in the specification. [4]

Section B: programming languages and translators

4. (a) Describe the difference between machine code and assembly language. [2] (b) Give two advantages and one disadvantage of writing a program in assembly language rather than a high-level language. [3] (c) Explain the term imperative high-level language. [2]

5. A games company develops a program and then sells it to customers. (a) Explain why the developers might use an interpreter during development. [2] (b) Explain why the finished program is compiled before it is sold. [3] (c) State the difference between source code and object code. [1]

6. Some compilers produce bytecode. Explain what bytecode is and how it is used, and give one advantage of this approach. [3]

Section C: logic gates and circuits

7. A circuit has inputs A, B and C. A and B are inputs to an AND gate; C is the input to a NOT gate; the outputs of these two gates are the inputs to an OR gate whose output is Q.

(a) Write a Boolean expression for Q. [1] (b) Complete a truth table for the circuit. [4]

8. Draw (or describe) the logic circuit for the expression Q = ¬(A + B) · C. [2]

9. (a) Construct a half-adder, stating the gate used for each output. [3] (b) A full-adder has inputs A = 1, B = 0 and carry in = 1. State the sum and carry out. [2] (c) Explain why full-adders, not half-adders, are chained to add two 8-bit numbers. [2]

10. Describe the use of an edge-triggered D-type flip-flop. [2]

Section D: Boolean algebra

11. Simplify, showing each step and naming the identity used: (a) A · B + A · ¬B [2] (b) A · (¬A + B) [2] (c) ¬(A + ¬B) [2] (d) ¬(A · B) + B [3]

12. Use De Morgan’s laws to show that ¬(¬A · ¬B) = A + B, and state what this means for building an OR gate from other gates. [3]

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Answers

1. Hardware: the physical components of a computer system [1]. Software: the programs and data that run on the hardware [1].

2. (a) System (b) Application (c) System (utility) (d) System (library) [2 – all four correct; 1 for two or three].

3. It hides the complexities of the hardware [1], giving users and programs a simpler interface (a virtual machine) so they do not need to control devices directly [1]. It performs resource management [1], allocating processors, memory and I/O devices among competing processes [1].

4. (a) Machine code is binary instructions executed directly by the processor [1]; assembly language uses mnemonics, each corresponding to one machine code instruction, so it is easier for people to read but must be assembled [1]. (b) Advantages (any two): direct control of hardware and memory; code can be very fast; code can be very compact [2]. Disadvantage (any one): harder to write and debug; processor specific so not portable [1]. (c) A high-level language whose statements are commands that say how to solve a problem, executed in sequence and changing the program’s state [1]; each statement usually translates into several low-level machine code instructions [1].

5. (a) An interpreter runs each statement straight away, so code can be tested without waiting for a full compilation [1], and it stops at the statement that causes an error, making debugging easier [1]. (b) The compiled object code runs faster because it has already been translated [1]; customers do not need the compiler or interpreter to run it [1]; the source code is not distributed, so it cannot easily be copied or changed [1]. (c) Source code is the program as written by the programmer; object code is the translated, executable form [1].

6. Bytecode is an intermediate language produced as the compiler’s final output [1]. It is executed by a virtual machine (a bytecode interpreter) on the target computer [1]. Advantage: the same compiled program runs on any platform that has the virtual machine (portability), without distributing the source [1].

7. (a) Q = A · B + ¬C [1]. (b)

A	B	C	A · B	¬C	Q
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

A · B column [1]; ¬C column [1]; Q column [2] (1 mark for up to two errors).

8. A and B into a NOR gate (or an OR gate followed by a NOT gate) [1]; its output and C into an AND gate, giving Q [1].

9. (a) Inputs A and B both go to an XOR gate, whose output is the Sum [1], and to an AND gate, whose output is the Carry [1]; correct circuit with both gates sharing the inputs [1]. (b) Sum = 1 ⊕ 0 ⊕ 1 = 0 [1]; carry out = 1 · 0 + 1 · (1 ⊕ 0) = 1 [1]. (c) Each bit position after the first must add two bits and the carry from the previous position [1]; a half-adder has only two inputs, while a full-adder has a carry-in so it can be chained [1].

10. It is used as a memory unit that stores one bit [1]. On the triggering edge of the clock signal the output takes the value of the D input, and it holds that value until the next triggering edge [1].

11. (a) = A · (B + ¬B) (distributive) [1] = A · 1 = A (complement, then identity) [1]. (b) = A · ¬A + A · B (distributive) [1] = 0 + A · B = A · B (complement, identity) [1]. (c) = ¬A · ¬¬B (De Morgan) [1] = ¬A · B (double negation) [1]. (d) = ¬A + ¬B + B (De Morgan) [1] = ¬A + 1 (complement) [1] = 1 (null law) [1].

12. ¬(¬A · ¬B) = ¬¬A + ¬¬B (De Morgan) [1] = A + B (double negation) [1]. So an OR gate can be built by inverting both inputs and passing them through a NAND gate – in fact, NAND gates alone can build any circuit [1].

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Where marks are usually lost

• Drawing a truth table without the intermediate columns, then making an error in Q that cannot earn method marks.
• Writing ¬(A + B) as ¬A + ¬B: De Morgan changes the operator as well as splitting the bar.
• Describing an interpreter as producing an executable file.
• Giving “faster” as an advantage of compilation without saying it is the execution that is faster.

Frequently asked questions

Do I need to know how a D-type flip-flop works inside? No. The AQA 7517 specification says knowledge of its internal operation is not required. Know that it stores one bit and changes only on the clock’s triggering edge.

Which gate symbols should I draw? The ANSI/IEEE 91-1984 distinctive shape symbols, which are the ones the specification names. Practise drawing NAND and NOR with the small circle (the inversion bubble) on the output.