1.4.3 Boolean Algebra Exam Questions
Paper 1 · Computer Systems · 21 past paper questions · 57 marks
Exam questions for 1.4.3 Boolean algebra
21 OCR past paper questions (57 marks) with mark schemes
Answer each question, reveal the mark scheme, then award yourself marks. Scores and answers are saved in this browser so you can see which topics need more work.
Download Questions & Mark Scheme (PDF)Past paper questions
- Question 1H446/01 · June 20244 marks
Ben installs burglar alarms. The alarm is made up of a door sensor and a motion sensor. When the alarm is set, the siren will sound if either the door sensor or the motion sensor detect movement. The alarm also has a test mode setting. When the test mode setting is enabled, an engineer can check the sensors are working without the siren going off. The inputs to the alarm are as follows. Draw a logic circuit to show the logic that is used in the burglar alarm to determine if the siren goes off.
Input System A Door sensor B Motion sensor C Alarm has been set D Test mode enabled Show mark scheme
1 mark for each correct area including labelled inputs: - NOT gate for test mode - OR gate for door sensor and motion sensor - AND gate combining alarm set with OR output - AND gate combining with NOT test mode Max 3 if any additional gates. NOT gate must have only one input, AND/OR gates must have two inputs.Mark yourself: how many marks did you get?/ 4 - Question 2H446/01 · June 20241 mark
Simplify the Boolean expression ¬A v ¬B using De Morgan's First Law.
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¬(A ∧ B) or NOT (A AND B) (1 mark)Mark yourself: how many marks did you get?/ 1 - Question 3H446/01 · June 20241 mark
Simplify the Boolean expression ¬(¬B) using double negation.
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B (1 mark)Mark yourself: how many marks did you get?/ 1 - Question 4H446/01 · June 20242 marks
Simplify the Boolean expression (A v B) ∧ (A v C) using distribution.
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A v (B ∧ C) 1 mark for A 1 mark for v (B ∧ C) Doesn't need brackets. Allow other correct notation e.g. A+B.CMark yourself: how many marks did you get?/ 2 - Question 5H446/013 marks
Draw the logic gate diagram for this expression: Q = (A v ¬B) ∧ C
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- A going into an OR gate - B going into a NOT gate with the output going into the OR gate - Output of the OR gate going into an AND gate with C (3 marks - 1 per correct component)Mark yourself: how many marks did you get?/ 3 - Question 6H446/01 · June 20232 marks
A computer scientist has created the logic circuit shown in Fig. 6. Give the Boolean expression that represents the logic circuit shown in Fig. 6. Do not attempt to simplify the expression.
Fig. 6 Show mark scheme
¬(A ∨ B) ⊻ C, which can also be written NOT (A OR B) XOR C 1 mark for ¬(A ∨ B). The brackets are required: NOT A OR B is incorrect. 1 mark for XOR C. The expression can be written in a different order, for example C XOR NOT (B OR A), as long as it is logically correct. (2 marks)Mark yourself: how many marks did you get?/ 2 - Question 7H446/01 · June 20234 marks
The following Karnaugh map represents a logic circuit. Use this Karnaugh map to find the simplified expression for this circuit. You should highlight the map as appropriate and write the expression.
CD \ AB 00 01 11 10 00 1 1 1 1 01 1 1 0 0 11 0 0 0 0 10 0 0 1 1 Show mark scheme
(¬A ∧ ¬C) ∨ (A ∧ ¬D) 1 mark for correct highlighting on the Karnaugh map: one group of four where AB is 00 or 01 and CD is 00 or 01, and one group of four where AB is 11 or 10 and CD is 00 or 10 (this group wraps from the top row to the bottom row) 1 mark for NOT A AND NOT C 1 mark for A AND NOT D 1 mark for OR joining the two correct expressions together (4 marks)Mark yourself: how many marks did you get?/ 4 - Question 8H446/014 marks
Elliott has designed a logic circuit. The expression he has created is: Q = (A ∧ ¬B) ∨ (¬A ∧ C ∧ D) ∨ (A ∧ B). Complete the Karnaugh map to simplify this expression.
CD \ AB 00 01 11 10 00 01 11 10 Show mark scheme
1 mark for filling in the table correctly 1 mark for correct first group 1 mark for correct second group 1 mark for the simplified expression: A ∨ (C ∧ D) (4 marks)Mark yourself: how many marks did you get?/ 4 - Question 9H446/013 marks
Draw a Logic diagram for the following expression: Q = ¬(A ∧ B) ∨ (C ∧ ¬D)
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- An AND gate taking A and B as inputs with the output connecting to a NOT gate (or NAND gate) - An AND gate taking C and the NOT of D as the inputs - An OR gate taking the outputs of the NOT and AND gates (3 marks - 1 per bullet)Mark yourself: how many marks did you get?/ 3 - Question 10H446/011 mark
Draw an XOR gate.
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Correct XOR gate symbol drawn (1 mark) Accept diagram of gate only without input/outputMark yourself: how many marks did you get?/ 1 - Question 11H446/012 marks
Explain the difference in the function of OR and XOR gates.
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OR gate outputs true if at least one of its inputs is true XOR gate outputs true if and only if one of its inputs is true (not both) (2 marks - 1 per bullet) Accept appropriate, correctly labelled, truth tables.Mark yourself: how many marks did you get?/ 2 - Question 12H446/012 marks
A half adder has inputs A and B, and outputs S (Sum) and C (Carry). Give the Boolean expressions for the outputs S and C.
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S = A ⊻ B (A XOR B) C = A ∧ B (A AND B) (2 marks - 1 per expression)Mark yourself: how many marks did you get?/ 2 - Question 13H446/01 · June 20214 marks
Describe how a half adder logic circuit can be adapted to add together two 4-bit binary numbers.
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Two half adders can be joined together With an OR gate to form a full adder 4 full adders can be used to add two four bit numbers Carry out on one joined to carry in on next (4 marks - 1 per bullet)Mark yourself: how many marks did you get?/ 4 - Question 14H446/013 marks
Users of a computer system have a unique username and password. The system has two-factor authentication so users must respond to either an email or text message containing a secret code. Let A be valid username, B be matching password, C be email response, D be text response, Q be entry allowed. Complete the Boolean expression for Q.
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Q = A ∧ B ∧ (C ∨ D) 1 mark for (A ∧ B) 1 mark for (C ∨ D) 1 mark for the ∧ joining them Brackets must be included. Accept alternative symbols for AND/OR.Mark yourself: how many marks did you get?/ 3 - Question 15H446/01 · June 20212 marks
Simplify the Boolean expression Q = ¬(¬A ∧ ¬B) so that it does not include any negation. You must explain which Boolean algebra rule(s) you are using at each step.
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Identification of De Morgan's and/or double negation rule Correct final answer: A ∨ B (2 marks)Mark yourself: how many marks did you get?/ 2 - Question 16H446/01 · June 20205 marks
Complete the Karnaugh map below for the Boolean expression (¬A ∧ ¬B) ∨ (A ∧ ¬B). Use the Karnaugh map to find a simplified Boolean expression that is equivalent to (¬A ∧ ¬B) ∨ (A ∧ ¬B).
CD \ AB 00 01 11 10 00 01 11 10 Show mark scheme
Left column filled with 1s for (¬A ∧ ¬B) Right column filled with 1s for (A ∧ ¬B) Middle 2 columns filled with zero or blank Karnaugh map used to show 1s highlighted with overlap Simplified expression: ¬B (NOT B) (5 marks - 3 for map, 2 for expression)Mark yourself: how many marks did you get?/ 5 - Question 17H446/01 · June 20202 marks
State the purpose of a D-type flip-flop circuit.
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Delay/store a value Of 1 bit When a signal is given (2 marks - 1 per bullet, max 2)Mark yourself: how many marks did you get?/ 2 - Question 18H446/01 · June 20204 marks
Describe the inputs and outputs used by a D-type flip-flop circuit, explaining how the inputs are used to control the outputs.
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Data input Clock input Q output When clock input goes high, Q changes to D NOT Q is reverse of Q (4 marks - 1 per bullet, max 4)Mark yourself: how many marks did you get?/ 4 - Question 19H446/013 marks
A cinema offers discounted tickets under these conditions: Customer is under 18 AND has a student card, OR Customer is over 60 AND has ID. Let A=under 18, B=student card, C=over 60, D=has ID, Q=discount issued. Complete the Boolean expression for Q.
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Q = (A ∧ B) ∨ (C ∧ D) 1 mark for (A ∧ B) 1 mark for (C ∧ D) 1 mark for the ∨ joining the two parts Accept alternative notation. Accept answers without brackets.Mark yourself: how many marks did you get?/ 3 - Question 20H446/01 · June 20172 marks
Simplify the expression (E ∧ F) ∨ (E ∧ G).
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E ∧ (F ∨ G) or (F ∨ G) ∧ E 1 mark for (F ∨ G) 1 mark for ∧ E (2 marks)Mark yourself: how many marks did you get?/ 2 - Question 21H446/01 · June 20173 marks
Draw a half adder using logic gates.
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XOR Gate for Sum output AND Gate for Carry output Correct connections and no additional gates (3 marks - 1 per bullet)Mark yourself: how many marks did you get?/ 3
- 1Logic gates: AND (all 1), OR (at least one 1), NOT (invert), XOR (different)
- 2Boolean laws: De Morgan's - (A.B)' = A'+B', (A+B)' = A'.B'
- 3Karnaugh maps: Group 1s in powers of 2; Can wrap around edges
- 4D-type flip-flop: Q follows D at clock edge; Used for registers, memory
- •Practice answering questions under timed conditions
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- •Focus on understanding the marking points, not just memorising answers
- •Review past paper questions from multiple years to spot patterns