plumiLogical operations: Eduqas A‑Level Computer Science questions
Every Eduqas A-Level Computer Science exam question on 1.2 Logical operations, 2016–2025, with its mark scheme and examiner’s report.
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Calculations
Q4(a)
The whole paper ↗Q4(b)
The whole paper ↗Q5(a)
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Q5(b)
The whole paper ↗Q1(a)
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Q1(b)
The whole paper ↗Q5(a)
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Q5(b)
The whole paper ↗Q2(a)
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Q2(b)
The whole paper ↗Q4(a)
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Q4(b)
The whole paper ↗Q2(a)
The whole paper ↗
Q2(b)
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All 56 questions on Logical operations, as a list
- June 2025, Component 1, Q4(a) (5 marks): Draw a truth table to demonstrate the value of Q: Q = (NOT A OR B) XOR (A AND C) [5]
- June 2025, Component 1, Q4(b) (4 marks): Draw a truth table to prove the following: NOT (A AND B) = NOT A OR NOT B [4]
- June 2025, Component 1, Q5(a) (5 marks): B.(A + C).A.(B + C).A.B [5]
- June 2025, Component 1, Q5(b) (4 marks): 5. Simplify the following Boolean expressions using Boolean algebra identities and De Morgan’s law where appropriate. Clearly show each step. (b) A.(B + C) [4]
- June 2025, AS Component 1, Q1(a) (1 mark): Input Output P Q R 0 0 0 1 0 0 0 1 1 1 1 0 R = [1]
- June 2025, AS Component 1, Q1(b) (1 mark): Input Output X Y Z 0 0 1 1 0 0 0 1 0 1 1 0 Z = [1]
- June 2025, AS Component 1, Q6 (8 marks): Clearly showing each step, simplify the following Boolean expression: P. (P + Q) + R. (P + Q) + P. (Q + R) [8]
- June 2024, Component 1, Q2 (6 marks): Draw a truth table to prove the following Boolean expression: [6] (A AND NOT B) OR (NOT A AND B) = A XOR B
- June 2024, Component 1, Q5(a) (5 marks): A.B + A.(1.C + C.1) [5]
- June 2024, Component 1, Q5(b) (5 marks): X.Y + Z.(Y.Z + 0.X +0) [5]
- June 2024, AS Component 1, Q2 (7 marks): Clearly showing each step, simplify the following expression using only Boolean identities and rules. [7] Do not use truth tables in your answer. (A + B).(A + B
- June 2023, Component 1, Q2(a) (6 marks): ( X + Z ).(Z + X) [6]
- June 2023, Component 1, Q2(b) (4 marks): B. A + C ( ).A.(A + C) [4]
- June 2023, Component 1, Q4(a) (4 marks): 4. Draw a truth table to prove or disprove the following Boolean rules: (a) A + B = A.B [4]
- June 2023, Component 1, Q4(b) (4 marks): A + 0 = A.1= A + A = A.A [4]
- June 2023, AS Component 1, Q4 (3 marks): The following data is stored in a 4-bit register: 1 1 0 0 Demonstrate how the state of the second least significant bit can be discovered using a logical operat
- June 2023, AS Component 1, Q6 (7 marks): Simplify the following Boolean expression using Boolean identities and rules. [7] X.(X + Y + 0) + X.(1+ Y) + Y.(Y.Y)
- June 2022, Component 1, Q2(a) (5 marks): B.(C + A.B) + B.(A + C) [5]
- June 2022, Component 1, Q2(b) (5 marks): 2. Clearly showing each step, simplify the following Boolean expressions using Boolean algebra, identities and De Morgan's Law. (b) A.B.B + C [5]
- June 2022, AS Component 1, Q2 (5 marks): Draw a truth table for the Boolean expression. [5] _ A ⊕ B + (B + _ C) P =
- June 2022, AS Component 1, Q5 (8 marks): Clearly showing each step, simplify the following expression using only Boolean identities and rules. [8] Do not use truth tables in your simplification. _ _ (
- Autumn 2021, Component 1, Q4(a) (5 marks): (A + C) . (A.B) + (C.B) [5]
- Autumn 2021, Component 1, Q4(b) (5 marks): B.A . (B + A) .A + C.1 [5]
- Autumn 2021, Component 1, Q9 (3 marks): This is a signed eight-bit integer: 110011002 Include this integer in a worked example to demonstrate how masking can be used to determine the sign of the integ
- Autumn 2021, Component 1, Q11(a) (4 marks): NOT (X NAND Y) = X AND Y [4]
- Autumn 2021, Component 1, Q11(b) (4 marks): A OR (A NOR B) = A OR NOT B [4]
- Autumn 2020, Component 1, Q2(a) (5 marks): Clearly showing each step, simplify the following Boolean expressions using Boolean algebra, identities and De Morgan’s Law. ( ) + X + Y + Z.Y (a) Z + Z + X [5]
- Autumn 2020, Component 1, Q2(b) (5 marks): A.(A + C).B.(B + C) + A [5]
- Autumn 2020, Component 1, Q4(a) (2 marks): A OR (A AND B) = A [2]
- Autumn 2020, Component 1, Q4(b) (6 marks): ( A + B ).(A + C) = A + (B.C) [6]
- Autumn 2020, AS Component 1, Q1(a)(i) (1 mark): Input Output A B C 0 0 1 1 0 0 0 1 0 1 1 0 [1]
- Autumn 2020, AS Component 1, Q1(a)(ii) (1 mark): Input Output X Y Z 0 0 1 1 0 1 0 1 0 1 1 1 [1]
- Autumn 2020, AS Component 1, Q1(b) (3 marks): Draw a truth table for the following Boolean expression: P = Q + (R.S) ¯ ¯ [3]
- Autumn 2020, AS Component 1, Q6 (7 marks): Clearly showing each step, simplify the following Boolean expression using Boolean algebra and identities: [7] X.(Y + X) + X.(Y + Y) + Z.(Y + Z) ¯ ¯ ¯
- June 2019, Component 1, Q3(a) (5 marks): 3. Clearly showing each step, simplify the following Boolean expressions using Boolean algebra, identities and De Morgan’s Law. (a) A.A + A.B + A.B [5]
- June 2019, Component 1, Q3(b) (5 marks): A. (B + B) + A. (A + B) [5]
- June 2019, Component 1, Q6 (6 marks): Draw a truth table to prove the distributive law: A. (B + C) = (A.B + A.C) [6]
- June 2019, Component 1, Q7 (3 marks): Two eight-bit numbers are: X 01011011 Key 11001110 Include these two numbers in a worked example to explain how the XOR operation can be used for encrypting dat
- June 2019, AS Component 1, Q6 (6 marks): Clearly showing each step, simplify the following Boolean expression using Boolean algebra and identities: [6] R.(S + R) + Q.(Q + P) + Q.(1 + P)
- June 2018, Component 1, Q1 (4 marks): Complete the following truth table. [4] A B A + B A.B (A+B).(A+B) (A.B) + (A.B) 0 0 0 1 1 0 1 1
- June 2018, Component 1, Q4(a)(i) (1 mark): 4. (a) Complete the following Boolean identities: (i) A + 1 = [1]
- June 2018, Component 1, Q4(a)(ii) (1 mark): (ii) A.A = [1]
- June 2018, Component 1, Q4(a)(iii) (1 mark): (iii) A + 0 = [1]
- June 2018, Component 1, Q4(b) (5 marks): Simplify the following Boolean expression: B.C.(C + D) + C.D + C + A [5]
- June 2018, AS Component 1, Q3(a) (4 marks): Complete the following truth table. [4] (A or B) Not ((A or B) A B A and B A or B XOR XOR (A and B)) (A and B) 0 0 0 1 1 0 1 1
- June 2018, AS Component 1, Q3(b)(i) (3 marks): Demonstrate a process that can be used to discover the state of the most significant bit in the register. [3]
- June 2018, AS Component 1, Q3(b)(ii) (3 marks): Demonstrate how this register can be cleared using a logical operator. [3]
- June 2018, AS Component 1, Q10 (8 marks): Clearly showing each step, simplify the following Boolean expression: [8] P.(Q + R) + Q.(P + Q) + R.(P + R) + S.S
- June 2017, Component 1, Q4(a) (4 marks): Using the laws of Boolean algebra and De Morgan’s theorem simplify the following Boolean expression: A .(A . B) [4]
- June 2017, Component 1, Q4(b) (4 marks): Simplify the following Boolean expression: A.B.C + A.B.C + A.B.C + A.B.C [4]
- June 2017, Component 1, Q7(a) (4 marks): Draw a truth table to show the value of P for all possible values of A, B and C for the following Boolean expression: P = (A NAND B) NOR C [4]
- June 2017, Component 1, Q7(b) (3 marks): Using the data in the 8 bit register below, design a mask and use it to demonstrate how a logical operation can be used to extract the value of bit 4 and bit 5
- June 2017, AS Component 1, Q6 (6 marks): Clearly showing each step, simplify the following Boolean expression: A.(B + C) + B.(A + B) + C.(A + C) [6]
- June 2016, AS Component 1, Q1(a) (4 marks): Complete the truth table below. [4] A B A OR B A AND B A XOR B A OR (NOT B) 0 0 0 1 1 0 1 1
- June 2016, AS Component 1, Q1(b) (3 marks): Using the following number: 101011112 Show how a logical operation can be used to discover the state of the most significant (leftmost) bit. [3]
- June 2016, AS Component 1, Q10 (6 marks): Clearly showing each step, simplify the following Boolean expression: [6] A.(A + C) + A.(C + B) + C.(C + B)
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